Section 18.3-19.1.
Today we will discuss finite-dimensional associative algebras
and their representations.
Definition 1. Let A be a finite-dimensional associative algebra
over a field F . An element
a ∈ A is nilpotent if an = 0 for some positive integer n. An
algebra is said to be nilpotent if
all of its elements are.
Exercise 2. Every subalgebra and factor algebra of a nilpotent
algebra are nilpotent. Con-
versely, if I ⊂ A is a nilpotent ideal, and the quotient algebra
A/I is nilpotent, then so is
A.
Example. The algebra of all strictly lower- (or upper-)
triangular n×n matrices is nilpotent.
Proposition 3. If the algebra A is nilpotent, then An = 0 for
some n ∈ Z+, that is the
product of any n elements if the algebra A equals 0.
Proof. Let B ⊂ A be the maximal subspace for which there
exists n ∈ Z+ such that Bn = 0.
Note that B is closed under multiplication, i.e. B ⊃ Bk for any k
∈ Z+. Assume B 6= A
and choose an element a ∈ AB. Since aBn = 0, there exists k ∈
Z+ such that aBk 6⊂ B.
Replacing a with a non-zero element in aBk we obtain aB ⊂ B.
Recall that there exists

m ∈ Z+ so that am = 0. Now, let us set C = B ⊕〈a〉. Then we
have
Cmn = 0
which contradicts the definition of the subspace B. �
Unless A is commutative, the set of all nilpotent elements of A
does not have to be an
ideal (in general, not even a subspace). On the other hand, if I,J
are nilpotent ideals in A,
then so is
I + J = {x + y |x ∈ I,y ∈ J} .
Therefore, there exists the maximal nilpotent ideal which
contains every other nilpotent ideal
of A.
Definition 4. The radical of A is the maximal nilpotent ideal of
A it is denoted rad(A).
The algebra A is semisimple if rad(A) = 0.
If char(F) = 0, there exists an alternative description of
semisimple algebras. Consider
the regular representation of the algebra A
ρ: A → L(A), ρ(a)(b) = ab,
and define a “scalar product” on A via
(a,b) = tr(ρ(ab)) = tr(ρ(a)ρ(b)).
One can see that (·, ·) is a symmetric bilinear function on A
satisfying (ab,c) = (a,bc).
Definition 5. For any ideal I ⊂ A, its orthogonal complement I⊥
is defined as

I⊥ = {a ∈ A |(a,i) = 0 for all i ∈ I} .
Proposition 6. For any ideal I ⊂ A, its orthogonal complement
I⊥ ⊂ A is also an ideal.
Proof. For any a ∈ A, x ∈ I⊥ , and y ∈ I we have
(xa,y) = (x,ay) = 0 and (ax,y) = (y,ax) = (ya,x) = 0.
�
1
2
Proposition 7. If A is an algebra over a field F with char(F) = 0,
then every element a ∈ A
orthogonal to all of its powers is nilpotent.
Proof. Let a ∈ A be such that
(a,an) = tr ρ(a)n+1 = 0 for all n ∈ Z+.
Let L ⊃ F be the splitting field of the characteristic polynomial
f(x) of the operator ρ(a).
Then, over L we have
f(x) = tk0
s∏
i=1
(t−λi)ki,
where λi are distinct for i = 1, . . . ,s, and
tr ρ(a)n+1 =

s∑
i=1
kiλ
n+1
i = 0.
Letting n run through the set 1, . . . ,s, the above equation yields
a system of s homogeneous
linear equations in variables k1, . . . ,ks. The determinant of this
system equals
λ21 . . .λ
2
nV (λ1, . . . ,λn),
where V (λ1, . . . ,λn) is the Vandermonde determinant.
Therefore, the system is non-degenerate
and we derive that kj = 0 in F for all j = 1, . . . ,s. If char(F) = 0
the latter is equivalent to
f(x) = tk0,
which implies that the operator ρ(a) is nilpotent. Therefore, for
some m ∈ Z+ we have
am+1 = T(a)m(a) = 0,
and we conclude that a ∈ A is nilpotent. �
Definition 8. The scalar product (·, ·) is non-degenerate if for
any a ∈ A there exists a′ ∈ A
such that (a,a′) 6= 0. Equivalently, it is nilpotent if A⊥ = 0.
Theorem 9.

(1) If the scalar product (a,b) = tr(ρ(ab)) is non-degenerate, then
the algebra A is
semisimple.
(2) If the algebra A is semisimple and char(F) = 0, then the
scalar product is non-
degenerate.
Proof. (1) Let I ⊂ A be a nilpotent ideal. Then for any x ∈ I and
any a ∈ A, their
product xa is nilpotent and hence
(a,x) = tr ρ(ax) = 0.
Therefore, I ⊂ A⊥ = 0.
(2) Conversely, if char(F) = 0 the previous two propositions
show that A⊥ is a nilpotent
ideal, therefore A⊥ = 0.
�
Definition 10. An algebra is simple if it has no proper ideals.
Note that a simple algebra is semi-simple unless it is nilpotent.
However, in that case
A2 ⊂ A is a proper ideal of A, hence A2 = 0. Therefore, a
simple algebra is semi-simple,
unless A is 1-dimensional with 0 multiplication.
3
Theorem 11. Any semisimple associative algebra A can be
decomposed into a direct sum of
simple algebras

A = A1 ⊕···⊕As,
moreover, any ideal of A is a sum of some direct summands in
the above decomposition.
Proof. We will prove this theorem in the assumption char(F) =
0. If A is simple, we have
s = 1 and the statement is trivial. Otherwise, let A1 ⊂ A be a
minimal ideal of A. Then
either
A = A1 ⊕A⊥ 1 ,
or A1 ⊂ A⊥ 1 . In the second case, A1 has to be nilpotent,
which would contradict the semisim-
plicity of A. Then, indeed A is a direct sum of A1 and A
⊥
1 , in which case every ideal of A1 or
A⊥ 1 is also an ideal of A. This allows us to conclude that A1 is
simple and A
⊥
1 is semisimple.
We know apply the same reasoning to the algebra A⊥ 1 .
Let now I ⊂ A be any ideal. Denote by πk : A → Ak the
canonical projections, and set
Ik = πk(I). Then Ik is an ideal of the algebra Ak. If Ik 6= 0 we
have Ik = Ak which yields
Ak = A
2
k = AkIk = AkI ⊂ I,
which means that I is the sum of several summands Aj. �

Theorem 12. Any non-trivial simple associative algebra A over
an algebraically closed field
F is isomorphic to L(V ) for some vector space V over F .
Moreover, any non-trivial irre-
ducible representation of A is isomorphic to the tautological
representation of L(V ).
Proof. Let ρ: A → L(A) be the regular representation, and V be
a minimal invariant
subspace, equivalently a minimal left ideal in A. Denote φ = ρ|V
: A → L(V ). Using
the fact that ker(φ) ⊂ A is an ideal, and one of the results from
the previous lecture, we get
either
A ' φ(A) = L(V )
or dim(V ) = 1 and φ(A) = 0. In the latter case, we see that AV
= 0 and therefore the
subspace
A0 = {x ∈ A |Ax = 0}
is nonzero. Since A0 ⊂ A is an ideal, we must have A = A0
which would imply that A is
trivial.
Therefore, φ(A) = L(V ) for some vector space V , and φ is a
tautological representation of
A in V . In that case we have previously shown that the regular
representation ρ: A → L(A)
is isomorphic to the sum of φ with itself n times:
ρ = nφ.
Now, let τ : A → L(U) be any irreducible representation of A,
and u0 ∈ U be any nonzero

vector. Then the map
T : A → U, T(a) = τ(a)(u0)
is a morphism between the regular representation (A,ρ) and the
representation (U,τ), indeed
for any x ∈ A we have
T(ρ(a)(x)) = T(ax) = τ(ax)(u0) = τ(a)τ(x)(u0) = τ(a)T(x).
If the representation (U,τ) is non-trivial, we have im(T) = U,
therefore U is a factor repre-
sentation of the regular representation ρ. Since U is irreducible,
we get (U,τ) ' (V,φ). �
4
Definition 13. Subalgebra Z(A) ⊂ A of an associative algebra A
defined as
Z(A) = {z ∈ A |za = az ∀ a ∈ A}
is called the center of A.
Corollary 14. Every semisimple associative algebra over an
algebraically closed field F is
isomorphic to the algebra
A = L(V1) ⊕···⊕L(Vs) where s = dim Z(A)
for some vector spaces V1, . . . ,Vs over F . If dim(Vi) = ni for i
= 1, . . . ,s, we get
dim(A) = n21 + · · · + n
2

s.
Proof. The only statement that have not been yet proven is that
the number of direct
summands is equal to dim Z(A). It follows from the fact that the
center of each summand is
1-dimensional and consists of scalar matrices. The latter
statement is left as an exercise. �
Corollary 15. Let ρj for j = 1, . . . ,s be the irreducible
representation of the algebra
A ' L(V1) ⊕···⊕L(Vs)
in the space Vj obtained via projection πj : A → L(Vj). Then
any non-trivial irreducible
representation of A is isomorphic to one of the representations
ρj.
Remark 16. Note that representations ρj are pairwise non-
isomorphic since their kernels
are distinct.
Let us now get back to the representation theory of finite
groups.
Theorem 17. Let G be a finite group of order n. Then the
algebra FG is semisimple provided
char(F) does not divide n.
Proof. Let ρ: FG → FG be the regular representation. Then it is
easy to see that
tr ρ(g) =
{

n, g = e,
0, g 6= e.
Therefore, for any g,h ∈ G we have
(g,h) =
{
n, gh = e,
0, gh 6= e.
If char(F) does not divide n, this scalar product is
non0degenerate, therefore FG is semisim-
ple. �
In what follows we will assume that F is an algebraically closed
field whose characteristic
does not divide |G|. By the results of the previous lecture, FG is
isomorphic to a direct sum
of matrix algebras L(Vi).
Theorem 18. For any finite group G with |G| = n, there are only
finitely many non-
isomorphic irreducible representations of G over the field F .
Dimensions n1, . . . ,ns of these
representations satisfy
n21 + · · · + n
2
s = n,
and s is the number of conjugacy classes in G.

5
Proof. The first statement and the equality
n21 + · · · + n
2
s = n
follow from the fact that FG is a semisimple associative
algebra, and a correspondence
between representations of FG and G. We also know that s =
dim Z(FG). An element
a =
∑
g∈ G agg belongs to the center Z(FG) if and only if
hah−1 =
∑
g∈ G
aghgh
−1 =
∑
g∈ G
ah−1ghg = a
for all h ∈ G. The latter is equivalent to the statement that the
coefficients ag are constant
on conjugacy classes of G:
ag = ahgh−1

for all g ∈ G. Therefore, Z(FG) is a linear span of elements of
the form
∑
g∈ C g, where
C ⊂ G is a conjugacy class, and dim Z(FG) equals the number
of conjugacy classes of G. �
Example. (1) Any irreducible representation of an abelian group
is 1-dimensional, there-
fore a finite abelian group of order n has precisely n non-
isomorphic irreducible rep-
resentations.
(2) The permutation group S3 has a trivial representation, a 1-
dimensional sign repre-
sentation, where σ 7→ sgn(σ), and a 2-dimensional irreducible
representation, where
S3 acts by permuting vertices of a regular triangle. Up to
isomorphism, these are all
irreducible representations of S3 because
6 = 12 + 12 + 22.
(3) In a similar fashion, one can find the following non-
isomorphic irreducible represen-
tations of S4:
• the trivial one;
• sign representation;
• a composition of the projection S4/K4 ' S3 with the irreducible
2-dimensional
representation of S3 considered above;
• an isomorphism with the group of rotations of a cube
(composed with its natural

3-dimensional representation);
• an isomorphism with the permutation group of the vertices of
a regular tetrahe-
dron composed (composed with its natural 3-dimensional
representation).
Checking dimensions we get
24 = 12 + 12 + 22 + 32 + 32,
and conclude that these are all irreducible representations of S4
up to isomorphism.
Let us summarize, what we have proved so far. Let G be a finite
group, and ρi : V → Vi,
i = 1, . . . ,s be all distinct (up to isomorphism) irreducible
representations of G over an
algebraically closed field F , whose characteristic does not
divide |G|. Then the group algebra
FG can be written as
FG ' L(V1) ⊕···⊕L(Vs),
and ρi is simply a projection onto the i-th summand. Then the
subspaces L(Vi) are the
isotypic components of the regular representation ρ of the
algebra FG, and the restriction of
ρ onto L(Vi) is isomorphic to niρi where ni = dim(Vi).
Therefore, for any a,b ∈ FG, we have
(a,b) =
s∑
i=1
ni tr ρi(a)ρi(b).

6
Now, let us consider the (finite-dimensional) space F [G] of F-
valued functions on G. Every
function f ∈ F[G] can be uniquely continued to a linear function
on FG by
f(
∑
g∈ G
agg) =
∑
g∈ G
agf(g),
which allows us to identify F[G] with the space of linear
functionals on FG, dual to FG. On
the other hand, the non-degenerate scalar product
(·, ·) : FG×FG −→ F
provides a bijection between FG and its dual via∑
g∈ G
agg 7→
∑
g∈ G
agfg where fg(h) = (g,h).
Since

(g,h) =
{
n, gh = e,
0, gh 6= e,
we see that
fg = nδg−1,
where δg−1 is the Kronecker delta.
Above isomorphisms allow us to drag the scalar product (·, ·)
from FG onto F[G]. In terms
of δ-functions we obtain
(δg,δh) =
1
n2
(g−1,h−1) =
{
1
n
, gh = e,
0, gh 6= e,
and for any functions f1,f2 ∈ F[G] we have
(f1,f2) =
1
n

∑
g∈ G
f1(g)f2(g
−1).
Now, recall that
FG = L(V1) ⊕···⊕L(Vs).
Let us now choose a basis 〈ei〉, i = 1, . . . ,ni in the
representation Vi, and denote by fijk ∈ F[G]
the (j,k) matrix coefficient in the representation Vi, that is
fijk(g) = 〈ej,ρi(g)ek〉 .
Then one can see that
(fijk,f
i′
j′k′) =
1
ni
δii′δjk′δj′k.
Definition 19. A character χφ ∈ F [G] of the representation φ: G
→ GL(V ) is defined by
χ(g) = tr φ(g)
for any g ∈ G.
It is clear that he character of a sum of two representations
equals the sum of the characters:
χφ+ψ = χφ + χψ,

and that a character is a class function:
χ(g) = χ(hgh−1)
for any g,h ∈ G. The following theorem is a direct corollary of
above calculations:
7
Theorem 20. Let χi, i = 1, . . . ,s be the characters of irreducible
representations ρi of the
group G. Then
(χi,χj) = δij,
equivalently, characters χi form an orthonormal basis in the
subspace of class functions in
F[G].
Corollary 21. Let χ be the character of a representation φ: G →
GL(V ). Then,
V '
s∑
i=1
(χ,χi)Vi.
In other words, (χ,χi) is the multiplicity of the irreducible
representation Vi in the decompo-
sition of the representation V .
Proof. Representation V can be written as

V =
s∑
i=1
kiVi
for some ki ∈ Z>0. Therefore
χ =
s∑
i=1
kiχi and (χ,χi) = ki.
�
Corollary 22. A representation φ: G → GL(V ) with character χ
is irreducible if and only
if (χ,χ) = 1.
Proof. By above,
(χ,χ) =
s∑
i=1
k2i
which is equal to 1 if and only if one of the ki is 1, and all the
other are 0. �
Remark 23. Above corollaries show that a character, which is
clearly less data than a
representation, since it only knows about traces of operators

rather than operators themselves,
is enough to recover the whole representation.
Finally, if F = C it is more convenient for computations to
replace the bilinear form (·, ·)
with the following Hermitian scalar product:
(f1 |f2) =
1
n
∑
g∈ G
f1(g)f2(g),
where x denotes the complex conjugate to x. If now in each of
the spaces Vi, one chooses a
basis orthonormal with respect to the Hermitian inner product,
they would get
fikj(g
−1) = fijk(g),
which yields
(fijk |fi
′
j′k′) = (f
i
jk,f
i′
k′j′) =

1
ni
δii′δjj′δkk′.
Example.
8
(1) A character of a 1-dimensional representation coincides with
the representation. If
G = 〈g |gn = e〉 is a cyclic group of order n, its irreducible
representations are defined
by
ρk(g) = ω
k−1,
where ω = e2πi/n. Therefore, the table of characters of G takes
the following form
χ0 χ1 . . . χn−1
e 1 1 . . . 1
g 1 ω . . . ωn−1
g2 1 ω2 . . . ω2(n−1)
...
...
...
. . .

...
gn−1 1 ωn−1 . . . ω(n−1)
2
(2) Using the description of irreducible representations of the
group S4, we can write out
its table of characters as well
χ1 χ
′
1 χ2 χ3 χ
′
3
e 1 1 2 3 3 1
(12) 1 -1 0 -1 1 6
(12)(34) 1 1 2 -1 -1 3
(123) 1 1 -1 0 0 8
(1234) 1 -1 0 1 -1 6
where the first column shows representatives of conjugacy
classes in S4, and the last
column shows number of elements in the corresponding
conjugacy class. For example
one gets
(χ2,χ3) = (χ2 |χ3) =
1
24
(2 · 3 · 1 + 0 · (−1) · 6 + 2 · (−1) · 3 + (−1) · 0 · 8 + 0 · 1 · 6) =
0.

(3) Let V be a 6-dimensional space of functions on the set of
faces of a cube. An iso-
morphism between S4 and the rotation group of a cube defines a
representation of
φ: S4 → GL(V ), let χ be its character. Every element g ∈ S4
permutes the faces
of a cube, and therefore permutes δ-functions of these faces
accordingly. Therefore,
χ(g) = tr φ(g) equals the number of faces fixed by g. This allows
us to compute the
character χ:
e (12) (12)(34) (123) (1234)
χ 6 0 2 0 2
Computing scalar products of χ with those of irreducible
representations of S4 we
find
(χ |χ1) = 1, (χ |χ′1) = 0, (χ |χ2) = 1, (χ |χ3) = 1, (χ |χ
′
3) = 0.
Therefore, φ ' ρ1 ⊕ρ2 ⊕ρ3.
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Title: Phaedra
Author: Jean Baptiste Racine
Translator: Robert Bruce Boswell
Release Date: October 30, 2008 [EBook #1977]
Last Updated: February 7, 2013
Language: English
Character set encoding: ASCII
*** START OF THIS PROJECT GUTENBERG EBOOK
PHAEDRA ***
Produced by Dagny, John Bickers, and David Widger
PHAEDRA
By Jean Baptiste Racine
Translated by Robert Bruce Boswell
Contents
INTRODUCTORY
NOTE

PHAEDRA
ACT I
ACT II
Phaedra, by Jean Baptiste Racine
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h.htm[10/4/2019 1:44:51 PM]
ACT III
ACT IV
ACT V
INTRODUCTORY NOTE
JEAN BAPTISTE RACINE, the younger contemporary of
Corneille, and his rival for supremacy in French
classical tragedy, was born at Ferte-Milon, December 21, 1639.
He was educated at the College of Beauvais,
at the great Jansenist school at Port Royal, and at the College
d'Harcourt. He attracted notice by an ode written
for the marriage of Louis XIV in 1660, and made his first really
great dramatic success with his
"Andromaque." His tragic masterpieces include "Britannicus,"
"Berenice," "Bajazet," "Mithridate,"
"Iphigenie," and "Phaedre," all written between 1669 and 1677.
Then for some years he gave up dramatic
composition, disgusted by the intrigues of enemies who sought
to injure his career by exalting above him an

unworthy rival. In 1689 he resumed his work under the
persuasion of Mme. de Maintenon, and produced
"Esther" and "Athalie," the latter ranking among his finest
productions, although it did not receive public
recognition until some time after his death in 1699. Besides his
tragedies, Racine wrote one comedy, "Les
Plaideurs," four hymns of great beauty, and a history of Port
Royal.
The external conventions of classical tragedy which had been
established by Corneille, Racine did not
attempt to modify. His study of the Greek tragedians and his
own taste led him to submit willingly to the rigor
and simplicity of form which were the fundamental marks of the
classical ideal. It was in his treatment of
character that he differed most from his predecessor; for
whereas, as we have seen, Corneille represented his
leading figures as heroically subduing passion by force of will,
Racine represents his as driven by almost
uncontrollable passion. Thus his creations appeal to the modern
reader as more warmly human; their speech, if
less exalted, is simpler and more natural; and he succeeds more
brilliantly with his portraits of women than
with those of men.
All these characteristics are exemplified in "Phaedre," the
tragedy of Racine which has made an appeal to
the widest audience. To the legend as treated by Euripides,
Racine added the love of Hippolytus for Aricia,
and thus supplied a motive for Phaedra's jealousy, and at the
same time he made the nurse instead of Phaedra
the calumniator of his son to Theseus.
PHAEDRA
CHARACTERS

THESEUS, son of Aegeus and King of Athens.
PHAEDRA, wife of Theseus and Daughter of Minos and
Pasiphae.
HIPPOLYTUS, son of Theseus and Antiope, Queen of the
Amazons.
ARICIA, Princess of the Blood Royal of Athens.
OENONE, nurse of Phaedra.
THERAMENES, tutor of Hippolytus.
ISMENE, bosom friend of Aricia.
PANOPE, waiting-woman of Phaedra.
GUARDS.
The scene is laid at Troezen, a town of the Peloponnesus.
Phaedra, by Jean Baptiste Racine
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ACT I
SCENE I
HIPPOLYTUS, THERAMENES
HIPPOLYTUS
My mind is settled, dear Theramenes,
And I can stay no more in lovely Troezen.
In doubt that racks my soul with mortal anguish,
I grow ashamed of such long idleness.
Six months and more my father has been gone,
And what may have befallen one so dear
I know not, nor what corner of the earth
Hides him.
THERAMENES

And where, prince, will you look for him?
Already, to content your just alarm,
Have I not cross'd the seas on either side
Of Corinth, ask'd if aught were known of Theseus
Where Acheron is lost among the Shades,
Visited Elis, doubled Toenarus,
And sail'd into the sea that saw the fall
Of Icarus? Inspired with what new hope,
Under what favour'd skies think you to trace
His footsteps? Who knows if the King, your father,
Wishes the secret of his absence known?
Perchance, while we are trembling for his life,
The hero calmly plots some fresh intrigue,
And only waits till the deluded fair—
HIPPOLYTUS
Cease, dear Theramenes, respect the name
Of Theseus. Youthful errors have been left
Behind, and no unworthy obstacle
Detains him. Phaedra long has fix'd a heart
Inconstant once, nor need she fear a rival.
In seeking him I shall but do my duty,
And leave a place I dare no longer see.
THERAMENES
Indeed! When, prince, did you begin to dread
These peaceful haunts, so dear to happy childhood,
Where I have seen you oft prefer to stay,
Rather than meet the tumult and the pomp
Of Athens and the court? What danger shun you,
Or shall I say what grief?
HIPPOLYTUS
That happy time
Is gone, and all is changed, since to these shores
The gods sent Phaedra.

THERAMENES
I perceive the cause
Of your distress. It is the queen whose sight
Offends you. With a step-dame's spite she schemed
Your exile soon as she set eyes on you.
But if her hatred is not wholly vanish'd,
It has at least taken a milder aspect.
Besides, what danger can a dying woman,
One too who longs for death, bring on your head?
Can Phaedra, sick'ning of a dire disease
Of which she will not speak, weary of life
And of herself, form any plots against you?
HIPPOLYTUS
It is not her vain enmity I fear,
Another foe alarms Hippolytus.
I fly, it must be own'd, from young Aricia,
The sole survivor of an impious race.
THERAMENES
What! You become her persecutor too!
The gentle sister of the cruel sons
Of Pallas shared not in their perfidy;
Why should you hate such charming innocence?
HIPPOLYTUS
I should not need to fly, if it were hatred.
THERAMENES
May I, then, learn the meaning of your flight?
Is this the proud Hippolytus I see,
Than whom there breathed no fiercer foe to love
And to that yoke which Theseus has so oft
Endured? And can it be that Venus, scorn'd
So long, will justify your sire at last?

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Has she, then, setting you with other mortals,
Forced e'en Hippolytus to offer incense
Before her? Can you love?
HIPPOLYTUS
Friend, ask me not.
You, who have known my heart from infancy
And all its feelings of disdainful pride,
Spare me the shame of disavowing all
That I profess'd. Born of an Amazon,
The wildness that you wonder at I suck'd
With mother's milk. When come to riper age,
Reason approved what Nature had implanted.
Sincerely bound to me by zealous service,
You told me then the story of my sire,
And know how oft, attentive to your voice,
I kindled when I heard his noble acts,
As you described him bringing consolation
To mortals for the absence of Alcides,
The highways clear'd of monsters and of robbers,
Procrustes, Cercyon, Sciro, Sinnis slain,
The Epidaurian giant's bones dispersed,
Crete reeking with the blood of Minotaur.
But when you told me of less glorious deeds,
Troth plighted here and there and everywhere,
Young Helen stolen from her home at Sparta,
And Periboea's tears in Salamis,
With many another trusting heart deceived

Whose very names have 'scaped his memory,
Forsaken Ariadne to the rocks
Complaining, last this Phaedra, bound to him
By better ties,—you know with what regret
I heard and urged you to cut short the tale,
Happy had I been able to erase
From my remembrance that unworthy part
Of such a splendid record. I, in turn,
Am I too made the slave of love, and brought
To stoop so low? The more contemptible
That no renown is mine such as exalts
The name of Theseus, that no monsters quell'd
Have given me a right to share his weakness.
And if my pride of heart must needs be humbled,
Aricia should have been the last to tame it.
Was I beside myself to have forgotten
Eternal barriers of separation
Between us? By my father's stern command
Her brethren's blood must ne'er be reinforced
By sons of hers; he dreads a single shoot
From stock so guilty, and would fain with her
Bury their name, that, even to the tomb
Content to be his ward, for her no torch
Of Hymen may be lit. Shall I espouse
Her rights against my sire, rashly provoke
His wrath, and launch upon a mad career—
THERAMENES
The gods, dear prince, if once your hour is come,
Care little for the reasons that should guide us.
Wishing to shut your eyes, Theseus unseals them;
His hatred, stirring a rebellious flame
Within you, lends his enemy new charms.
And, after all, why should a guiltless passion
Alarm you? Dare you not essay its sweetness,
But follow rather a fastidious scruple?

Fear you to stray where Hercules has wander'd?
What heart so stout that Venus has not vanquish'd?
Where would you be yourself, so long her foe,
Had your own mother, constant in her scorn
Of love, ne'er glowed with tenderness for Theseus?
What boots it to affect a pride you feel not?
Confess it, all is changed; for some time past
You have been seldom seen with wild delight
Urging the rapid car along the strand,
Or, skilful in the art that Neptune taught,
Making th' unbroken steed obey the bit;
Less often have the woods return'd our shouts;
A secret burden on your spirits cast
Has dimm'd your eye. How can I doubt you love?
Vainly would you conceal the fatal wound.
Has not the fair Aricia touch'd your heart?
HIPPOLYTUS
Theramenes, I go to find my father.
THERAMENES
Will you not see the queen before you start,
My prince?
HIPPOLYTUS
That is my purpose: you can tell her.
Yes, I will see her; duty bids me do it.
But what new ill vexes her dear Oenone?
SCENE II
HIPPOLYTUS, OENONE, THERAMENES
OENONE

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Alas, my lord, what grief was e'er like mine?
The queen has almost touch'd the gates of death.
Vainly close watch I keep by day and night,
E'en in my arms a secret malady
Slays her, and all her senses are disorder'd.
Weary yet restless from her couch she rises,
Pants for the outer air, but bids me see
That no one on her misery intrudes.
She comes.
HIPPOLYTUS
Enough. She shall not be disturb'd,
Nor be confronted with a face she hates.
SCENE III
PHAEDRA, OENONE
PHAEDRA
We have gone far enough. Stay, dear Oenone;
Strength fails me, and I needs must rest awhile.
My eyes are dazzled with this glaring light
So long unseen, my trembling knees refuse
Support. Ah me!
OENONE
Would Heaven that our tears
Might bring relief!
PHAEDRA
Ah, how these cumbrous gauds,
These veils oppress me! What officious hand

Has tied these knots, and gather'd o'er my brow
These clustering coils? How all conspires to add
To my distress!
OENONE
What is one moment wish'd,
The next, is irksome. Did you not just now,
Sick of inaction, bid us deck you out,
And, with your former energy recall'd,
Desire to go abroad, and see the light
Of day once more? You see it, and would fain
Be hidden from the sunshine that you sought.
PHAEDRA
Thou glorious author of a hapless race,
Whose daughter 'twas my mother's boast to be,
Who well may'st blush to see me in such plight,
For the last time I come to look on thee,
O Sun!
OENONE
What! Still are you in love with death?
Shall I ne'er see you, reconciled to life,
Forego these cruel accents of despair?
PHAEDRA
Would I were seated in the forest's shade!
When may I follow with delighted eye,
Thro' glorious dust flying in full career,
A chariot—
OENONE
Madam?
PHAEDRA
Have I lost my senses?

What said I? and where am I? Whither stray
Vain wishes? Ah! The gods have made me mad.
I blush, Oenone, and confusion covers
My face, for I have let you see too clearly
The shame of grief that, in my own despite,
O'erflows these eyes of mine.
OENONE
If you must blush,
Blush at a silence that inflames your woes.
Resisting all my care, deaf to my voice,
Will you have no compassion on yourself,
But let your life be ended in mid course?
What evil spell has drain'd its fountain dry?
Thrice have the shades of night obscured the heav'ns
Since sleep has enter'd thro' your eyes, and thrice
The dawn has chased the darkness thence, since food
Pass'd your wan lips, and you are faint and languid.
To what dread purpose is your heart inclined?
How dare you make attempts upon your life,
And so offend the gods who gave it you,
Prove false to Theseus and your marriage vows,
Ay, and betray your most unhappy children,
Bending their necks yourself beneath the yoke?
That day, be sure, which robs them of their mother,
Will give high hopes back to the stranger's son,
To that proud enemy of you and yours,
To whom an Amazon gave birth, I mean
Hippolytus—
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PHAEDRA
Ye gods!
OENONE
Ah, this reproach
Moves you!
PHAEDRA
Unhappy woman, to what name
Gave your mouth utterance?
OENONE
Your wrath is just.
'Tis well that that ill-omen'd name can rouse
Such rage. Then live. Let love and duty urge
Their claims. Live, suffer not this son of Scythia,
Crushing your children 'neath his odious sway,
To rule the noble offspring of the gods,
The purest blood of Greece. Make no delay;
Each moment threatens death; quickly restore
Your shatter'd strength, while yet the torch of life
Holds out, and can be fann'd into a flame.
PHAEDRA
Too long have I endured its guilt and shame!
OENONE
Why? What remorse gnaws at your heart? What crime
Can have disturb'd you thus? Your hands are not
Polluted with the blood of innocence?
PHAEDRA
Thanks be to Heav'n, my hands are free from stain.
Would that my soul were innocent as they!

OENONE
What awful project have you then conceived,
Whereat your conscience should be still alarm'd?
PHAEDRA
Have I not said enough? Spare me the rest.
I die to save myself a full confession.
OENONE
Die then, and keep a silence so inhuman;
But seek some other hand to close your eyes.
Tho' but a spark of life remains within you,
My soul shall go before you to the Shades.
A thousand roads are always open thither;
Pain'd at your want of confidence, I'll choose
The shortest. Cruel one, when has my faith
Deceived you! Think how in my arms you lay
New born. For you, my country and my children
I have forsaken. Do you thus repay
My faithful service?
PHAEDRA
What do you expect
From words so bitter? Were I to break silence
Horror would freeze your blood.
OENONE
What can you say
To horrify me more than to behold
You die before my eyes?
PHAEDRA
When you shall know
My crime, my death will follow none the less,
But with the added stain of guilt.

OENONE
Dear Madam,
By all the tears that I have shed for you,
By these weak knees I clasp, relieve my mind
From torturing doubt.
PHAEDRA
It is your wish. Then rise.
OENONE
I hear you. Speak.
PHAEDRA
Heav'ns! How shall I begin?
OENONE
Dismiss vain fears, you wound me with distrust.
PHAEDRA
O fatal animosity of Venus!
Into what wild distractions did she cast
My mother!
OENONE
Be they blotted from remembrance,
And for all time to come buried in silence.
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PHAEDRA
My sister Ariadne, by what love

Were you betray'd to death, on lonely shores
Forsaken!
OENONE
Madam, what deep-seated pain
Prompts these reproaches against all your kin?
PHAEDRA
It is the will of Venus, and I perish,
Last, most unhappy of a family
Where all were wretched.
OENONE
Do you love?
PHAEDRA
I feel
All its mad fever.
OENONE
Ah! For whom?
PHAEDRA
Hear now
The crowning horror. Yes, I love—my lips
Tremble to say his name.
OENONE
Whom?
PHAEDRA
Know you him,
Son of the Amazon, whom I've oppress'd
So long?
OENONE

Hippolytus? Great gods!
PHAEDRA
'Tis you
Have named him.
OENONE
All my blood within my veins
Seems frozen. O despair! O cursed race!
Ill-omen'd journey! Land of misery!
Why did we ever reach thy dangerous shores?
PHAEDRA
My wound is not so recent. Scarcely had I
Been bound to Theseus by the marriage yoke,
And happiness and peace seem'd well secured,
When Athens show'd me my proud enemy.
I look'd, alternately turn'd pale and blush'd
To see him, and my soul grew all distraught;
A mist obscured my vision, and my voice
Falter'd, my blood ran cold, then burn'd like fire;
Venus I felt in all my fever'd frame,
Whose fury had so many of my race
Pursued. With fervent vows I sought to shun
Her torments, built and deck'd for her a shrine,
And there, 'mid countless victims did I seek
The reason I had lost; but all for naught,
No remedy could cure the wounds of love!
In vain I offer'd incense on her altars;
When I invoked her name my heart adored
Hippolytus, before me constantly;
And when I made her altars smoke with victims,
'Twas for a god whose name I dared not utter.
I fled his presence everywhere, but found him—
O crowning horror!—in his father's features.
Against myself, at last, I raised revolt,

And stirr'd my courage up to persecute
The enemy I loved. To banish him
I wore a step—dame's harsh and jealous carriage,
With ceaseless cries I clamour'd for his exile,
Till I had torn him from his father's arms.
I breathed once more, Oenone; in his absence
My days flow'd on less troubled than before,
And innocent. Submissive to my husband,
I hid my grief, and of our fatal marriage
Cherish'd the fruits. Vain caution! Cruel Fate!
Brought hither by my spouse himself, I saw
Again the enemy whom I had banish'd,
And the old wound too quickly bled afresh.
No longer is it love hid in my heart,
But Venus in her might seizing her prey.
I have conceived just terror for my crime;
I hate my life, and hold my love in horror.
Dying I wish'd to keep my fame unsullied,
And bury in the grave a guilty passion;
But I have been unable to withstand
Tears and entreaties, I have told you all;
Content, if only, as my end draws near,
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You do not vex me with unjust reproaches,
Nor with vain efforts seek to snatch from death
The last faint lingering sparks of vital breath.
SCENE IV
PHAEDRA, OENONE, PANOPE

PANOPE
Fain would I hide from you tidings so sad,
But 'tis my duty, Madam, to reveal them.
The hand of death has seized your peerless husband,
And you are last to hear of this disaster.
OENONE
What say you, Panope?
PANOPE
The queen, deceived
By a vain trust in Heav'n, begs safe return
For Theseus, while Hippolytus his son
Learns of his death from vessels that are now
In port.
PHAEDRA
Ye gods!
PANOPE
Divided counsels sway
The choice of Athens; some would have the prince,
Your child, for master; others, disregarding
The laws, dare to support the stranger's son.
'Tis even said that a presumptuous faction
Would crown Aricia and the house of Pallas.
I deem'd it right to warn you of this danger.
Hippolytus already is prepared
To start, and should he show himself at Athens,
'Tis to be fear'd the fickle crowd will all
Follow his lead.
OENONE
Enough. The queen, who hears you,
By no means will neglect this timely warning.

SCENE V
PHAEDRA, OENONE
OENONE
Dear lady, I had almost ceased to urge
The wish that you should live, thinking to follow
My mistress to the tomb, from which my voice
Had fail'd to turn you; but this new misfortune
Alters the aspect of affairs, and prompts
Fresh measures. Madam, Theseus is no more,
You must supply his place. He leaves a son,
A slave, if you should die, but, if you live,
A King. On whom has he to lean but you?
No hand but yours will dry his tears. Then live
For him, or else the tears of innocence
Will move the gods, his ancestors, to wrath
Against his mother. Live, your guilt is gone,
No blame attaches to your passion now.
The King's decease has freed you from the bonds
That made the crime and horror of your love.
Hippolytus no longer need be dreaded,
Him you may see henceforth without reproach.
It may be, that, convinced of your aversion,
He means to head the rebels. Undeceive him,
Soften his callous heart, and bend his pride.
King of this fertile land, in Troezen here
His portion lies; but as he knows, the laws
Give to your son the ramparts that Minerva
Built and protects. A common enemy
Threatens you both, unite them to oppose
Aricia.
PHAEDRA
To your counsel I consent.
Yes, I will live, if life can be restored,

If my affection for a son has pow'r
To rouse my sinking heart at such a dangerous hour.
ACT II
SCENE I
ARICIA, ISMENE
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ARICIA
Hippolytus request to see me here!
Hippolytus desire to bid farewell!
Is't true, Ismene? Are you not deceived?
ISMENE
This is the first result of Theseus' death.
Prepare yourself to see from every side.
Hearts turn towards you that were kept away
By Theseus. Mistress of her lot at last,
Aricia soon shall find all Greece fall low,
To do her homage.
ARICIA
'Tis not then, Ismene,
An idle tale? Am I no more a slave?
Have I no enemies?
ISMENE
The gods oppose
Your peace no longer, and the soul of Theseus
Is with your brothers.

ARICIA
Does the voice of fame
Tell how he died?
ISMENE
Rumours incredible
Are spread. Some say that, seizing a new bride,
The faithless husband by the waves was swallow'd.
Others affirm, and this report prevails,
That with Pirithous to the world below
He went, and saw the shores of dark Cocytus,
Showing himself alive to the pale ghosts;
But that he could not leave those gloomy realms,
Which whoso enters there abides for ever.
ARICIA
Shall I believe that ere his destined hour
A mortal may descend into the gulf
Of Hades? What attraction could o'ercome
Its terrors?
ISMENE
He is dead, and you alone
Doubt it. The men of Athens mourn his loss.
Troezen already hails Hippolytus
As King. And Phaedra, fearing for her son,
Asks counsel of the friends who share her trouble,
Here in this palace.
ARICIA
Will Hippolytus,
Think you, prove kinder than his sire, make light
My chains, and pity my misfortunes?
ISMENE

Yes,
I think so, Madam.
ARICIA
Ah, you know him not
Or you would never deem so hard a heart
Can pity feel, or me alone except
From the contempt in which he holds our sex.
Has he not long avoided every spot
Where we resort?
ISMENE
I know what tales are told
Of proud Hippolytus, but I have seen
Him near you, and have watch'd with curious eye
How one esteem'd so cold would bear himself.
Little did his behavior correspond
With what I look'd for; in his face confusion
Appear'd at your first glance, he could not turn
His languid eyes away, but gazed on you.
Love is a word that may offend his pride,
But what the tongue disowns, looks can betray.
ARICIA
How eagerly my heart hears what you say,
Tho' it may be delusion, dear Ismene!
Did it seem possible to you, who know me,
That I, sad sport of a relentless Fate,
Fed upon bitter tears by night and day,
Could ever taste the maddening draught of love?
The last frail offspring of a royal race,
Children of Earth, I only have survived
War's fury. Cut off in the flow'r of youth,
Mown by the sword, six brothers have I …

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Title: Phaedra
Author: Jean Baptiste Racine
Translator: Robert Bruce Boswell
Release Date: October 30, 2008 [EBook #1977]
Last Updated: February 7, 2013
Language: English
Character set encoding: ASCII
*** START OF THIS PROJECT GUTENBERG EBOOK
PHAEDRA ***
Produced by Dagny, John Bickers, and David Widger
PHAEDRA

By Jean Baptiste Racine
Translated by Robert Bruce Boswell
Contents
INTRODUCTORY
NOTE
PHAEDRA
ACT I
ACT II
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ACT III
ACT IV
ACT V
INTRODUCTORY NOTE
JEAN BAPTISTE RACINE, the younger contemporary of
Corneille, and his rival for supremacy in French
classical tragedy, was born at Ferte-Milon, December 21, 1639.
He was educated at the College of Beauvais,
at the great Jansenist school at Port Royal, and at the College

d'Harcourt. He attracted notice by an ode written
for the marriage of Louis XIV in 1660, and made his first really
great dramatic success with his
"Andromaque." His tragic masterpieces include "Britannicus,"
"Berenice," "Bajazet," "Mithridate,"
"Iphigenie," and "Phaedre," all written between 1669 and 1677.
Then for some years he gave up dramatic
composition, disgusted by the intrigues of enemies who sought
to injure his career by exalting above him an
unworthy rival. In 1689 he resumed his work under the
persuasion of Mme. de Maintenon, and produced
"Esther" and "Athalie," the latter ranking among his finest
productions, although it did not receive public
recognition until some time after his death in 1699. Besides his
tragedies, Racine wrote one comedy, "Les
Plaideurs," four hymns of great beauty, and a history of Port
Royal.
The external conventions of classical tragedy which had been
established by Corneille, Racine did not
attempt to modify. His study of the Greek tragedians and his
own taste led him to submit willingly to the rigor
and simplicity of form which were the fundamental marks of the
classical ideal. It was in his treatment of
character that he differed most from his predecessor; for
whereas, as we have seen, Corneille represented his
leading figures as heroically subduing passion by force of will,
Racine represents his as driven by almost
uncontrollable passion. Thus his creations appeal to the modern
reader as more warmly human; their speech, if
less exalted, is simpler and more natural; and he succeeds more
brilliantly with his portraits of women than
with those of men.
All these characteristics are exemplified in "Phaedre," the
tragedy of Racine which has made an appeal to

the widest audience. To the legend as treated by Euripides,
Racine added the love of Hippolytus for Aricia,
and thus supplied a motive for Phaedra's jealousy, and at the
same time he made the nurse instead of Phaedra
the calumniator of his son to Theseus.
PHAEDRA
CHARACTERS
THESEUS, son of Aegeus and King of Athens.
PHAEDRA, wife of Theseus and Daughter of Minos and
Pasiphae.
HIPPOLYTUS, son of Theseus and Antiope, Queen of the
Amazons.
ARICIA, Princess of the Blood Royal of Athens.
OENONE, nurse of Phaedra.
THERAMENES, tutor of Hippolytus.
ISMENE, bosom friend of Aricia.
PANOPE, waiting-woman of Phaedra.
GUARDS.
The scene is laid at Troezen, a town of the Peloponnesus.
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ACT I
SCENE I
HIPPOLYTUS, THERAMENES
HIPPOLYTUS
My mind is settled, dear Theramenes,

And I can stay no more in lovely Troezen.
In doubt that racks my soul with mortal anguish,
I grow ashamed of such long idleness.
Six months and more my father has been gone,
And what may have befallen one so dear
I know not, nor what corner of the earth
Hides him.
THERAMENES
And where, prince, will you look for him?
Already, to content your just alarm,
Have I not cross'd the seas on either side
Of Corinth, ask'd if aught were known of Theseus
Where Acheron is lost among the Shades,
Visited Elis, doubled Toenarus,
And sail'd into the sea that saw the fall
Of Icarus? Inspired with what new hope,
Under what favour'd skies think you to trace
His footsteps? Who knows if the King, your father,
Wishes the secret of his absence known?
Perchance, while we are trembling for his life,
The hero calmly plots some fresh intrigue,
And only waits till the deluded fair—
HIPPOLYTUS
Cease, dear Theramenes, respect the name
Of Theseus. Youthful errors have been left
Behind, and no unworthy obstacle
Detains him. Phaedra long has fix'd a heart
Inconstant once, nor need she fear a rival.
In seeking him I shall but do my duty,
And leave a place I dare no longer see.
THERAMENES
Indeed! When, prince, did you begin to dread
These peaceful haunts, so dear to happy childhood,

Where I have seen you oft prefer to stay,
Rather than meet the tumult and the pomp
Of Athens and the court? What danger shun you,
Or shall I say what grief?
HIPPOLYTUS
That happy time
Is gone, and all is changed, since to these shores
The gods sent Phaedra.
THERAMENES
I perceive the cause
Of your distress. It is the queen whose sight
Offends you. With a step-dame's spite she schemed
Your exile soon as she set eyes on you.
But if her hatred is not wholly vanish'd,
It has at least taken a milder aspect.
Besides, what danger can a dying woman,
One too who longs for death, bring on your head?
Can Phaedra, sick'ning of a dire disease
Of which she will not speak, weary of life
And of herself, form any plots against you?
HIPPOLYTUS
It is not her vain enmity I fear,
Another foe alarms Hippolytus.
I fly, it must be own'd, from young Aricia,
The sole survivor of an impious race.
THERAMENES
What! You become her persecutor too!
The gentle sister of the cruel sons
Of Pallas shared not in their perfidy;
Why should you hate such charming innocence?
HIPPOLYTUS

I should not need to fly, if it were hatred.
THERAMENES
May I, then, learn the meaning of your flight?
Is this the proud Hippolytus I see,
Than whom there breathed no fiercer foe to love
And to that yoke which Theseus has so oft
Endured? And can it be that Venus, scorn'd
So long, will justify your sire at last?
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Has she, then, setting you with other mortals,
Forced e'en Hippolytus to offer incense
Before her? Can you love?
HIPPOLYTUS
Friend, ask me not.
You, who have known my heart from infancy
And all its feelings of disdainful pride,
Spare me the shame of disavowing all
That I profess'd. Born of an Amazon,
The wildness that you wonder at I suck'd
With mother's milk. When come to riper age,
Reason approved what Nature had implanted.
Sincerely bound to me by zealous service,
You told me then the story of my sire,
And know how oft, attentive to your voice,
I kindled when I heard his noble acts,
As you described him bringing consolation
To mortals for the absence of Alcides,

The highways clear'd of monsters and of robbers,
Procrustes, Cercyon, Sciro, Sinnis slain,
The Epidaurian giant's bones dispersed,
Crete reeking with the blood of Minotaur.
But when you told me of less glorious deeds,
Troth plighted here and there and everywhere,
Young Helen stolen from her home at Sparta,
And Periboea's tears in Salamis,
With many another trusting heart deceived
Whose very names have 'scaped his memory,
Forsaken Ariadne to the rocks
Complaining, last this Phaedra, bound to him
By better ties,—you know with what regret
I heard and urged you to cut short the tale,
Happy had I been able to erase
From my remembrance that unworthy part
Of such a splendid record. I, in turn,
Am I too made the slave of love, and brought
To stoop so low? The more contemptible
That no renown is mine such as exalts
The name of Theseus, that no monsters quell'd
Have given me a right to share his weakness.
And if my pride of heart must needs be humbled,
Aricia should have been the last to tame it.
Was I beside myself to have forgotten
Eternal barriers of separation
Between us? By my father's stern command
Her brethren's blood must ne'er be reinforced
By sons of hers; he dreads a single shoot
From stock so guilty, and would fain with her
Bury their name, that, even to the tomb
Content to be his ward, for her no torch
Of Hymen may be lit. Shall I espouse
Her rights against my sire, rashly provoke
His wrath, and launch upon a mad career—

THERAMENES
The gods, dear prince, if once your hour is come,
Care little for the reasons that should guide us.
Wishing to shut your eyes, Theseus unseals them;
His hatred, stirring a rebellious flame
Within you, lends his enemy new charms.
And, after all, why should a guiltless passion
Alarm you? Dare you not essay its sweetness,
But follow rather a fastidious scruple?
Fear you to stray where Hercules has wander'd?
What heart so stout that Venus has not vanquish'd?
Where would you be yourself, so long her foe,
Had your own mother, constant in her scorn
Of love, ne'er glowed with tenderness for Theseus?
What boots it to affect a pride you feel not?
Confess it, all is changed; for some time past
You have been seldom seen with wild delight
Urging the rapid car along the strand,
Or, skilful in the art that Neptune taught,
Making th' unbroken steed obey the bit;
Less often have the woods return'd our shouts;
A secret burden on your spirits cast
Has dimm'd your eye. How can I doubt you love?
Vainly would you conceal the fatal wound.
Has not the fair Aricia touch'd your heart?
HIPPOLYTUS
Theramenes, I go to find my father.
THERAMENES
Will you not see the queen before you start,
My prince?
HIPPOLYTUS
That is my purpose: you can tell her.
Yes, I will see her; duty bids me do it.

But what new ill vexes her dear Oenone?
SCENE II
HIPPOLYTUS, OENONE, THERAMENES
OENONE
Phaedra, by Jean Baptiste Racine
https://www.gutenberg.org/files/1977/1977-h/1977-
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Alas, my lord, what grief was e'er like mine?
The queen has almost touch'd the gates of death.
Vainly close watch I keep by day and night,
E'en in my arms a secret malady
Slays her, and all her senses are disorder'd.
Weary yet restless from her couch she rises,
Pants for the outer air, but bids me see
That no one on her misery intrudes.
She comes.
HIPPOLYTUS
Enough. She shall not be disturb'd,
Nor be confronted with a face she hates.
SCENE III
PHAEDRA, OENONE
PHAEDRA
We have gone far enough. Stay, dear Oenone;
Strength fails me, and I needs must rest awhile.
My eyes are dazzled with this glaring light
So long unseen, my trembling knees refuse

Support. Ah me!
OENONE
Would Heaven that our tears
Might bring relief!
PHAEDRA
Ah, how these cumbrous gauds,
These veils oppress me! What officious hand
Has tied these knots, and gather'd o'er my brow
These clustering coils? How all conspires to add
To my distress!
OENONE
What is one moment wish'd,
The next, is irksome. Did you not just now,
Sick of inaction, bid us deck you out,
And, with your former energy recall'd,
Desire to go abroad, and see the light
Of day once more? You see it, and would fain
Be hidden from the sunshine that you sought.
PHAEDRA
Thou glorious author of a hapless race,
Whose daughter 'twas my mother's boast to be,
Who well may'st blush to see me in such plight,
For the last time I come to look on thee,
O Sun!
OENONE
What! Still are you in love with death?
Shall I ne'er see you, reconciled to life,
Forego these cruel accents of despair?
PHAEDRA
Would I were seated in the forest's shade!

When may I follow with delighted eye,
Thro' glorious dust flying in full career,
A chariot—
OENONE
Madam?
PHAEDRA
Have I lost my senses?
What said I? and where am I? Whither stray
Vain wishes? Ah! The gods have made me mad.
I blush, Oenone, and confusion covers
My face, for I have let you see too clearly
The shame of grief that, in my own despite,
O'erflows these eyes of mine.
OENONE
If you must blush,
Blush at a silence that inflames your woes.
Resisting all my care, deaf to my voice,
Will you have no compassion on yourself,
But let your life be ended in mid course?
What evil spell has drain'd its fountain dry?
Thrice have the shades of night obscured the heav'ns
Since sleep has enter'd thro' your eyes, and thrice
The dawn has chased the darkness thence, since food
Pass'd your wan lips, and you are faint and languid.
To what dread purpose is your heart inclined?
How dare you make attempts upon your life,
And so offend the gods who gave it you,
Prove false to Theseus and your marriage vows,
Ay, and betray your most unhappy children,
Bending their necks yourself beneath the yoke?
That day, be sure, which robs them of their mother,
Will give high hopes back to the stranger's son,
To that proud enemy of you and yours,

To whom an Amazon gave birth, I mean
Hippolytus—
Phaedra, by Jean Baptiste Racine
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PHAEDRA
Ye gods!
OENONE
Ah, this reproach
Moves you!
PHAEDRA
Unhappy woman, to what name
Gave your mouth utterance?
OENONE
Your wrath is just.
'Tis well that that ill-omen'd name can rouse
Such rage. Then live. Let love and duty urge
Their claims. Live, suffer not this son of Scythia,
Crushing your children 'neath his odious sway,
To rule the noble offspring of the gods,
The purest blood of Greece. Make no delay;
Each moment threatens death; quickly restore
Your shatter'd strength, while yet the torch of life
Holds out, and can be fann'd into a flame.
PHAEDRA
Too long have I endured its guilt and shame!

OENONE
Why? What remorse gnaws at your heart? What crime
Can have disturb'd you thus? Your hands are not
Polluted with the blood of innocence?
PHAEDRA
Thanks be to Heav'n, my hands are free from stain.
Would that my soul were innocent as they!
OENONE
What awful project have you then conceived,
Whereat your conscience should be still alarm'd?
PHAEDRA
Have I not said enough? Spare me the rest.
I die to save myself a full confession.
OENONE
Die then, and keep a silence so inhuman;
But seek some other hand to close your eyes.
Tho' but a spark of life remains within you,
My soul shall go before you to the Shades.
A thousand roads are always open thither;
Pain'd at your want of confidence, I'll choose
The shortest. Cruel one, when has my faith
Deceived you! Think how in my arms you lay
New born. For you, my country and my children
I have forsaken. Do you thus repay
My faithful service?
PHAEDRA
What do you expect
From words so bitter? Were I to break silence
Horror would freeze your blood.
OENONE

What can you say
To horrify me more than to behold
You die before my eyes?
PHAEDRA
When you shall know
My crime, my death will follow none the less,
But with the added stain of guilt.
OENONE
Dear Madam,
By all the tears that I have shed for you,
By these weak knees I clasp, relieve my mind
From torturing doubt.
PHAEDRA
It is your wish. Then rise.
OENONE
I hear you. Speak.
PHAEDRA
Heav'ns! How shall I begin?
OENONE
Dismiss vain fears, you wound me with distrust.
PHAEDRA
O fatal animosity of Venus!
Into what wild distractions did she cast
My mother!
OENONE
Be they blotted from remembrance,
And for all time to come buried in silence.

Phaedra, by Jean Baptiste Racine
https://www.gutenberg.org/files/1977/1977-h/1977-
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PHAEDRA
My sister Ariadne, by what love
Were you betray'd to death, on lonely shores
Forsaken!
OENONE
Madam, what deep-seated pain
Prompts these reproaches against all your kin?
PHAEDRA
It is the will of Venus, and I perish,
Last, most unhappy of a family
Where all were wretched.
OENONE
Do you love?
PHAEDRA
I feel
All its mad fever.
OENONE
Ah! For whom?
PHAEDRA
Hear now
The crowning horror. Yes, I love—my lips
Tremble to say his name.

OENONE
Whom?
PHAEDRA
Know you him,
Son of the Amazon, whom I've oppress'd
So long?
OENONE
Hippolytus? Great gods!
PHAEDRA
'Tis you
Have named him.
OENONE
All my blood within my veins
Seems frozen. O despair! O cursed race!
Ill-omen'd journey! Land of misery!
Why did we ever reach thy dangerous shores?
PHAEDRA
My wound is not so recent. Scarcely had I
Been bound to Theseus by the marriage yoke,
And happiness and peace seem'd well secured,
When Athens show'd me my proud enemy.
I look'd, alternately turn'd pale and blush'd
To see him, and my soul grew all distraught;
A mist obscured my vision, and my voice
Falter'd, my blood ran cold, then burn'd like fire;
Venus I felt in all my fever'd frame,
Whose fury had so many of my race
Pursued. With fervent vows I sought to shun
Her torments, built and deck'd for her a shrine,
And there, 'mid countless victims did I seek
The reason I had lost; but all for naught,

No remedy could cure the wounds of love!
In vain I offer'd incense on her altars;
When I invoked her name my heart adored
Hippolytus, before me constantly;
And when I made her altars smoke with victims,
'Twas for a god whose name I dared not utter.
I fled his presence everywhere, but found him—
O crowning horror!—in his father's features.
Against myself, at last, I raised revolt,
And stirr'd my courage up to persecute
The enemy I loved. To banish him
I wore a step—dame's harsh and jealous carriage,
With ceaseless cries I clamour'd for his exile,
Till I had torn him from his father's arms.
I breathed once more, Oenone; in his absence
My days flow'd on less troubled than before,
And innocent. Submissive to my husband,
I hid my grief, and of our fatal marriage
Cherish'd the fruits. Vain caution! Cruel Fate!
Brought hither by my spouse himself, I saw
Again the enemy whom I had banish'd,
And the old wound too quickly bled afresh.
No longer is it love hid in my heart,
But Venus in her might seizing her prey.
I have conceived just terror for my crime;
I hate my life, and hold my love in horror.
Dying I wish'd to keep my fame unsullied,
And bury in the grave a guilty passion;
But I have been unable to withstand
Tears and entreaties, I have told you all;
Content, if only, as my end draws near,
Phaedra, by Jean Baptiste Racine

https://www.gutenberg.org/files/1977/1977-h/1977-
h.htm[10/4/2019 1:44:51 PM]
You do not vex me with unjust reproaches,
Nor with vain efforts seek to snatch from death
The last faint lingering sparks of vital breath.
SCENE IV
PHAEDRA, OENONE, PANOPE
PANOPE
Fain would I hide from you tidings so sad,
But 'tis my duty, Madam, to reveal them.
The hand of death has seized your peerless husband,
And you are last to hear of this disaster.
OENONE
What say you, Panope?
PANOPE
The queen, deceived
By a vain trust in Heav'n, begs safe return
For Theseus, while Hippolytus his son
Learns of his death from vessels that are now
In port.
PHAEDRA
Ye gods!
PANOPE
Divided counsels sway
The choice of Athens; some would have the prince,
Your child, for master; others, disregarding
The laws, dare to support the stranger's son.
'Tis even said that a presumptuous faction
Would crown Aricia and the house of Pallas.

I deem'd it right to warn you of this danger.
Hippolytus already is prepared
To start, and should he show himself at Athens,
'Tis to be fear'd the fickle crowd will all
Follow his lead.
OENONE
Enough. The queen, who hears you,
By no means will neglect this timely warning.
SCENE V
PHAEDRA, OENONE
OENONE
Dear lady, I had almost ceased to urge
The wish that you should live, thinking to follow
My mistress to the tomb, from which my voice
Had fail'd to turn you; but this new misfortune
Alters the aspect of affairs, and prompts
Fresh measures. Madam, Theseus is no more,
You must supply his place. He leaves a son,
A slave, if you should die, but, if you live,
A King. On whom has he to lean but you?
No hand but yours will dry his tears. Then live
For him, or else the tears of innocence
Will move the gods, his ancestors, to wrath
Against his mother. Live, your guilt is gone,
No blame attaches to your passion now.
The King's decease has freed you from the bonds
That made the crime and horror of your love.
Hippolytus no longer need be dreaded,
Him you may see henceforth without reproach.
It may be, that, convinced of your aversion,
He means to head the rebels. Undeceive him,
Soften his callous heart, and bend his pride.
King of this fertile land, in Troezen here

His portion lies; but as he knows, the laws
Give to your son the ramparts that Minerva
Built and protects. A common enemy
Threatens you both, unite them to oppose
Aricia.
PHAEDRA
To your counsel I consent.
Yes, I will live, if life can be restored,
If my affection for a son has pow'r
To rouse my sinking heart at such a dangerous hour.
ACT II
SCENE I
ARICIA, ISMENE
Phaedra, by Jean Baptiste Racine
https://www.gutenberg.org/files/1977/1977-h/1977-
h.htm[10/4/2019 1:44:51 PM]
ARICIA
Hippolytus request to see me here!
Hippolytus desire to bid farewell!
Is't true, Ismene? Are you not deceived?
ISMENE
This is the first result of Theseus' death.
Prepare yourself to see from every side.
Hearts turn towards you that were kept away
By Theseus. Mistress of her lot at last,
Aricia soon shall find all Greece fall low,
To do her homage.

ARICIA
'Tis not then, Ismene,
An idle tale? Am I no more a slave?
Have I no enemies?
ISMENE
The gods oppose
Your peace no longer, and the soul of Theseus
Is with your brothers.
ARICIA
Does the voice of fame
Tell how he died?
ISMENE
Rumours incredible
Are spread. Some say that, seizing a new bride,
The faithless husband by the waves was swallow'd.
Others affirm, and this report prevails,
That with Pirithous to the world below
He went, and saw the shores of dark Cocytus,
Showing himself alive to the pale ghosts;
But that he could not leave those gloomy realms,
Which whoso enters there abides for ever.
ARICIA
Shall I believe that ere his destined hour
A mortal may descend into the gulf
Of Hades? What attraction could o'ercome
Its terrors?
ISMENE
He is dead, and you alone
Doubt it. The men of Athens mourn his loss.
Troezen already hails Hippolytus
As King. And Phaedra, fearing for her son,

Asks counsel of the friends who share her trouble,
Here in this palace.
ARICIA
Will Hippolytus,
Think you, prove kinder than his sire, make light
My chains, and pity my misfortunes?
ISMENE
Yes,
I think so, Madam.
ARICIA
Ah, you know him not
Or you would never deem so hard a heart
Can pity feel, or me alone except
From the contempt in which he holds our sex.
Has he not long avoided every spot
Where we resort?
ISMENE
I know what tales are told
Of proud Hippolytus, but I have seen
Him near you, and have watch'd with curious eye
How one esteem'd so cold would bear himself.
Little did his behavior correspond
With what I look'd for; in his face confusion
Appear'd at your first glance, he could not turn
His languid eyes away, but gazed on you.
Love is a word that may offend his pride,
But what the tongue disowns, looks can betray.
ARICIA
How eagerly my heart hears what you say,
Tho' it may be delusion, dear Ismene!
Did it seem possible to you, who know me,

That I, sad sport of a relentless Fate,
Fed upon bitter tears by night and day,
Could ever taste the maddening draught of love?
The last frail offspring of a royal race,
Children of Earth, I only have survived
War's fury. Cut off in the flow'r of youth,
Mown by the sword, six brothers have I …
Section 18.1-2.
In the next 2-3 lectures we will have a lightning introduction to
representations of finite
groups.
For any vector space V over a field F, denote by L(V ) the
algebra of all linear operators
on V , and by GL(V ) the group of invertible linear operators.
Note that if dim(V ) = n, then
L(V ) = Matn(F) is isomorphic to the algebra of n×n matrices
over F , and GL(V ) = GLn(F)
to its multiplicative group.
Definition 1. A linear representation of a set X in a vector space
V is a map φ: X → L(V ),
where L(V ) is the set of linear operators on V , the space V is
called the space of the rep-
resentation. We will often denote the representation of X by
(V,φ) or simply by V if the
homomorphism φ is clear from the context. If X has any
additional structures, we require
that the map φ is a homomorphism. For example, a linear
representation of a group G is a
homomorphism φ: G → GL(V ).

Definition 2. A morphism of representations φ: X → L(V ) and
ψ : X → L(U) is a linear
map T : V → U, such that
ψ(x) ◦T = T ◦φ(x)
for all x ∈ X. In other words, T makes the following diagram
commutative
V
φ(x)
//
T
��
V
T
��
U
ψ(x)
// U
An invertible morphism of two representation is called an
isomorphism, and two representa-
tions are called isomorphic (or equivalent ) if there exists an
isomorphism between them.
Example.
(1) A representation of a one-element set in a vector space V is
simply a linear operator

on V . Two such representations are isomorphic if two operators
have the same Jordan
Normal Form.
(2) For any vector space, identity maps id : GL(V ) → GL(V )
and id : L(V ) → L(V ) give
tautological representations of the group GL(V ) and the algebra
L(V ) respectively.
(3) For each representation φ: X → L(V ) one can define its
dual representation
φ∗ : X → L(V ∗ ), φ∗ (x)(α)(v) = α(φ(x)(v))
for all x ∈ X, v ∈ V , and α ∈ V ∗ .
(4) The trivial representation of a group G is a homomorphism
φ: G → GL(V ), where
V is a 1-dimensional vector space over the field F, and φ(g) = 1
∈ GL(V ) = F× for
all g ∈ G.
(5) Recall the dihedral group Dn =
⟨
r,s |rn = s2 = 1,rs = sr−1
⟩
. The following map
r 7→
(
cos (2π/n) −sin (2π/n)
sin (2π/n) cos (2π/n)
)

, s 7→
(
0 1
1 0
)
defines a 2-dimensional representation φ: Dn → GL(2,R).
1
2
(6) The following map gives a representation of a symmetric
group S4:
(12) 7→
0 1 0 0
1 0 0 0
0 0 1 0
0 0 0 1
1 0 0 0
0 0 1 0

0 1 0 0
0 0 0 1
1 0 0 0
0 1 0 0
0 0 0 1
0 0 1 0
(7) There is a natural representations of Sn on the space of
polynomials in n variables:
(σf)(x1, . . . ,xn) = f(xσ(1), . . . ,xσ(n)).
Note, that symmetric polynomials stay invariant under this
action.
(8) Any Galois group Gal(K/F) comes with its representation in
K over the field F.
Definition 3. An invariant subspace of a representation φ: X →
L(V ) is a subspace W ⊂ V
invariant under φ(x) for all x ∈ X. An invariant subspace gives
rise to a subrepresentation
φW : X → L(W), φW (x) = φ(x)|W ,
and a factor representation
φV/W : X → L(V/W), φV/W (x)(v + W) = φ(x)(v).

Remark 4. If one chooses a basis of the space V of a
representation φ: X → L(V ) in such
a way that the first k vectors span an invariant subspace W ⊂ V
, then any operator φ(x) is
written in matrix form as
φ(x) =
(
φW (x) ∗
0 φV/W (x)
)
Example.
(1) Let Sn act on F
n by permuting basis vectors, as in example (4) above. Then the
subspace {a,a, . . . ,a) |a ∈ F} ⊂ Fn is a 1-dimensional invariant
subspace. There is
a complementary invariant subspace of dimension n− 1 defined
by
{(a1, . . . ,an) |a1 + · · · + an = 0} .
Note that Fn is isomorphic to the direct sum of these two
subspaces.
(2) For any positive integer d, the subspace Poldn of
polynomials in n variables of degree
d in the space Poln of all polynomials of n variables gives a
subrepresentation of Sn.
Definition 5.

(1) A representation is irreducible (or simple) if it (or rather the
underlying vector space)
has no invariant subspaces other than 0 and itself, otherwise it
is reducible.
(2) A representation is indecomposable if it can not be written
as a direct sum of two
invariant subrepresentations.
(3) A representation is completely reducible if it can be written
as a direct sum of irre-
ducible representations.
Example. The representation of a 1-element set given by a 2-
dimensional space V = F2 and
an operator
A =
(
a b
0 c
)
is reducible, since the subspace 〈e1〉 ⊂ V is invariant,
however it is indecomposable unless
b = 0. If b = 0, the representation V is completly reducible: V =
〈e1〉⊕ 〈e2〉.
Exercise 6. If T : V → U is a morphism of representations of X,
then ker(T) ⊂ V and
im(T) ⊂ U are invariant subspaces.
3

Corollary 7. If V and U are irreducible representations of X, any
morphism T : V → U is
either 0 or an isomorphism.
Proof. Since V is irreducible, the subrepresentation ker(T) is
either 0, in which case T is
injective, or coincides with the whole V , in which case T = 0.
Since U is irreducible, we must
now have im(T) = U unless T = 0, which forces T to be an
isomorphism. �
Corollary 8 (Schur’s lemma). Any endomorphism T : V → V of
an irreducible represen-
tation V of X over an algebraically closed field is a scalar, that
is T = λ · Id for some
λ ∈ F .
Proof. Let T ∈ EndX(V ) = HomX(V,V ) be an endomorphism
of V . Then for any λ ∈ F, so
is T−λ·Id. Choosing λ to be an eigenvalue of T (which we can
always do over an algebraically
closed field) we see that ker(T −λ · Id) is a subrepresentation of
X in V . Therefore, since V
is irreducible, we must have T = λ · Id. �
Corollary 9. Let φ: X → L(V ) and ψ : X → L(U) be a pair of
irreducible representations
over an algebraically closed field. Then any morphisms T,S : V
→ U differ by a scalar factor.
Proof. If one of the two morphisms is zero, the statement is
obvious. Otherwise, both mor-
phisms are isomorphisms, hence ST−1 is an endomorphism of
an irreducible representation
U which implies ST−1 = λ Id. �
Corollary 10. Any irreducible representation of an abelian group
over an algebraically closed
field is 1-dimensional.

Proof. If G is abelian and φ: G → GL(V ) is a representation of
G, operators φ(g) commute
for all g ∈ G, hence any of them can be viewed as an
endomorphism of V . If V is irre-
ducible, By Schur’s lemma, all of them must be scalar, therefore
any subspace is invariant
and representation V can be irreducible if and only if dim(V ) =
1. �
Example. The 2-dimensional plane R2 is an irreducible
representation of the abelian group
SO(2,R) of plane rotations
G =
{(
cos θ −sin θ
sin θ cos θ
)∣ ∣ ∣ ∣ θ ∈ R
}
.
However, over the field of complex numbers, we see that(
cos θ −sin θ
sin θ cos θ
)
∼
(
eiθ 0
0 e−iθ
)
.

Proposition 11. Every subrepresentation and factor
representation of a completely reducible
representation is completely reducible.
Proof. Let V be a completely reducible representation, and U be
its subrepresentation.
Then for any U1 ⊂ U, there exists an invarinat subspace V ′ ⊂
V such that V ' U1 ⊕ V ′.
Setting U2 = V
′ ∩U we see that U2 is invariant and U ' U1 ⊕U2.
Now, let π : V → V/W be a canonical projection to a factor
representation, and U1 ⊂ V/W
be the subrepresentation. Then V1 = π
−1(U1) is an invariant subspace of V , which con-
tains W . Now, we have V = V1 ⊕ V2 for some invariant
subspace V2, which implies that
V/W ' U1 ⊕U2 where U2 = π(V2) is an invariant subspace in
V/W . �
Proposition 12. If a representation φ: X → L(V ) is completely
reducible, then V can be
written as a direct sum of minimal (nonzero) irreducible
subspaces. Conversely, if V can be
written as a sum of minimal invariant subspaces V = V1 + · · ·
+ Vn, then V is completely
reducible.
4
Proof. The first statement is immediate: choose any minimal
invariant subspace V1 ⊂ V

and write V = V1⊕V2, continue the same process for V2. Now,
let V = V1 +· · ·+Vn for some
minimal invariant subspaces Vi ⊂ V , and U ⊂ V be any
invariant subspace. Then U ∩Vj is
either 0 or coincides with Vj for any j = 1, . . . ,n. Setting
U ′ =
∑
j:U∩Vj=∅
Vj
we get V = U ⊕U ′. �
Definition 13. A sum of representations φ: X → L(V ) and ψ : X
→ L(U) is a representa-
tion
φ + ψ : X → V ⊕U, (φ + ψ)(x) = (φ(x),ψ(x)).
Remark 14. In matrix form we have
(φ + ψ)(x) =
(
φ(x) 0
0 ψ(x)
)
Corollary 15. (1) A representation is completely reducible if
and only if it is isomorphic
to a sum of irreducible subrepresentations:
V = V1 ⊕···⊕Vn, φ = (φ1, . . . ,φn).

(2) In that case, any subrepresentation and factor representation
of V is isomorphic to a
sum of some of the representations Vj.
Proof. Left as an exercise. �
Definition 16. Note that in the above corollary, some of the
irreducible summands Vj might
be isomorphic.
(1) Let S be an irreducible subrepresentation of a completely
reducible representation V ,
then the S-isotypic component of V is the subspace VS ⊂ V
defined as
VS =
⊕
j |Vj'S
Vj.
(2) A representation V is isotypic if it has single isotypic
component.
(3) We will say that the representation V has simple spectrum if
it can be written as a
direct sum of pairwise non-isomorphic irreducible
representations, equivalently if all
of its isotypic components are irreducible.
Corollary 17. Let φ: X → L(V ) be a completely reducible
representation with simple spec-
trum, and V1, . . . ,Vn be its irreducible subrepresentations.
Then the following holds.

(1) The decomposition of V into a sum of irreducible
subrepresentations Vj is unique.
(2) If the base field F is algebraically closed, then every
endomorphism of V takes the
form
T(x) = λjx, for any x ∈ Vj.
Proof. Left as an exercise. �
It is convenient to describe isotypic representations in the
following way. Let φ: X → L(V )
be a representation, and Z be any vector space (over the same
field). Define a representation
φ̃: L(V ⊗Z), φ̃(x)(v ⊗z) = φ(x)(v) ⊗z.
5
If {z1, . . .zn} is a basis of Z we get
V ⊗Z ' (V ⊗z1) ⊕···⊕ (V ⊗zn),
which provides a decomposition of an isotypic representation V
⊗Z into irreducible subrep-
resentations.
Proposition 18. Let φ: X → L(U) be an irreducible
representation of X over an alge-
braically closed field F , Z be a vector space over F , and ψ : X
→ L(U ⊗ Z) be the isotypic
representation of X discussed above. Then every X-invariant

subspace of U ⊗ Z is of the
form U ⊗Z0, for some subspace Z0 ⊂ Z.
Proof. It is enough to prove the statement for a minimal
invariant subspace W ⊂ U ⊗ Z.
Any element w ∈ W can be written as
w = S1(w) ⊗z1 + · · · + Sn(w) ⊗zn,
where Sj : W → U are certain morphisms between
representations (W,ψ|W ) and (U,φ). We
know that any such isomorphisms have to differ by scalar
multiples, hence Sj = λjS for some
morphism S : W → U and λ1, . . . ,λn ∈ F . This way we obtain
w = S(w) ⊗ (λ1z1 + · · · + λnzn),
and therefore
W = U ⊗ (λ1z1 + · · · + λnzn).
�
Theorem 19. Let φ: X → L(V ) be an irreducible representation
of X over an algebraically
closed field F . Then the subalgebra of L(V ) generated by
operators φ(x), x ∈ X coincides
with L(V ) unless dim(V ) = 1 and φ = 0.
Proof. Recall the following isomorphism of vector spaces:
V ⊗V ∗ ' L(V ), v ⊗α 7−→ lα,v
where `α,v(u) = α(u)v for all α ∈ V ∗ and u,v ∈ V . Note that
under this isomorphism, for
any operator T ∈ L(V ) we have
T`α,v = `α,T(v) = `T∗ (α),v, (∗ )

where T∗ ∈ L(V ∗ ) defined by
T∗ (α)(v) = α(Tv).
Due to a canonical bijection between subspaces of V and V ∗ ,
which sends each subspace
U ⊂ V to its annihilator Ann(U) ⊂ V ∗ , we see that the dual
representation φ∗ : X → L(V ∗ )
is also irreducible. Let us now define representations Tl and Tr
of X on the space L(V ),
precisely
Tl(x)(A) = φ(x)A and Tr(x)A = Aφ(x).
Thanks to the formula (∗ ), representations Tl and Tr are
isotypic (we can think that they act
on only one tensor factor in V ⊗V ∗ ).
Note now that the subspace φ(X) ⊂ L(V ) is invariant under
both Tl and Tr. Using the
previous Proposition, we conclude that φ(X) can be written
simultaneously as V ⊗W0 and
as V0 ⊗ V ∗ , where V0 and W0 are subspaces of V and V ∗
respectively. The latter can only
happen if φ(X) = L(V ) or φ(X) = 0 in which case one has to
assume dim(V ) = 1 since
otherwise V would not be irreducible. �
6
Theorem 20. Any representation of a finite group G over a field
F such that char(F) does
not divide |G| is completely reducible.

Proof. Let G be a finite group, and φ: G → GL(V ) be its
representation over a field F with
char(F) not dividing |G|. It is enough to show that for any
invariant subspace U ⊂ V there
exists a complementary invariant subspace W ⊂ V such that V '
U ⊕ W . The latter is
equivalent to finding a projector π from V onto U, which is a
morphism of representations
of G. Indeed, recall that a projector is a linear operator P : V →
U such that P(V ) ⊂ U and
P(u) = u for any u ∈ U. Then setting W = ker(P) we obtain an
invariant subspace W ⊂ V
satisfying V ' U ⊕W.
Note that the space of all projectors P : V → U forms a linear
subspace S in the space
L(V ) of all linear operators on V . Consider an action of G on
L(V ) by conjugation, that is
g ◦A = φ(g)Aφ(g)−1 for all g ∈ G and A ∈ L(V ).
Since, U is a subrepresentation of G we see that the subspace S
⊂ L(V ) is preserved under
this action. Let us choose any projector P0 ∈ S and define P to
be the “centre of mass” of
the orbit of P0 under the action of G:
P =
1
|G|
∑
g∈ G
(g ◦P0).

Then the projector P ∈ S is invariant under the adjoint action of
G, or equivalently, commutes
with all operators φ(g) ∈ L(V ). �
Finally, let us discuss representations of the group algebra FG
of a finite group G, where
F is a field. Recall that an algebra A over a ring R is an R-
module, which is a ring itself, in
particular if R = F is a field, then an algebra A over F is a
vector space over F, with a ring
structure. As a vector space, algebra FG consists of linear
combinations∑
g∈ G
agg where ag ∈ F,
addition on FG is defined component-wise, and multiplication
has the form
agg ·ahh = agah(gh).
In other words, FG is an algebra over the field F with basis
vectors labelled by elements of
the finite group G, and with multiplication defined from that of
G.
As we discussed in the beginning of this lecture, a
representation of an algebra A (over
the ring R) is a homomorphism φ: A → L(V ) which respects the
operations on A. In
other words, a representation of A is simply an A-module V
(which then forced to be an
R-module itself). Now, it is easy to see that there is a bijection
between representations of
the group G and its group algebra FG, moreover this bijection
respects subrepresentations,

factor representations, etc. Indeed, given a representation φ: G
→ GL(V ) of the group G,
we define (and denote by the same symbol) a representation
φ: FG → L(V ), φ(
∑
g∈ G
agg) =
∑
g∈ G
agφ(g).
Conversely, every representation of FG yields a representation
of G when restricted to the
basis vectors of FG.
7
Example. (1) The trivial representation of G corresponds to a 1-
dimensional represen-
tation of FG where
(agg)(v) = agv for any ag ∈ F, g ∈ G, v ∈ V.
(2) Considering FG as a (left) module over itself, we obtain a
(left) regular representation
of G, which is a representation of dimension |G| defined by
g(
∑
h∈ G

ahh) =
∑
h∈ G
ah(gh) for all ah ∈ F, g,h ∈ G.
The bijection described above allows us to study representations
of finite groups by study-
ing those of finite-dimensional associative algebras.
Section 14.7.
Today we discuss the insolvability of a general quantic in
radicals.
Definition 1. An extension K/F is cyclic if it is Galois with
cyclic Galois group.
Proposition 2. Let F be a field of characteristic not dividing n,
which contains a primitive
n-th root of unity. Then for any a ∈ F , the extension F( n
√
a)/F is a cyclic extension of
degree dividing n, here n
√
a denotes any root of the polynomial xn −a.
Proof. Since all roots of xn−a can be obtained by multiplying n
√
a by n-th roots of unity all

of which lie in F , we see that F( n
√
a)/F is the splitting field for xn −a and hence is Galois.
Since the Galois group G = Gal(F( n
√
a)/F) permutes the roots of xn −a, for any σ ∈ G we
have σ( n
√
a) = ζσ
n
√
a for certain n-th root of unity ζσ. Therefore, we get a map
ψ : G −→ Z/nZ, σ 7−→ ζσ,
and we leave it as an exercise to show that ψ is a
homomorphism. The kernel of ψ consists
of those elements which fix n
√
a and is therefore trivial. This shows that ψ is an embedding,
and hence the order of G divides n. �
Definition 3. A character χ of a group G with values in a field L
is a homomorphism
χ: G −→ L×.
Definition 4. Character χ1, . . . ,χn of a group G with values in
L are linearly dependent if
there exist elements a1, . . . ,an ∈ L such that
a1χ1(g) + · · · + anχn(g) = 0 for all g ∈ G,

and linearly independent otherwise.
Proposition 5. Distinct characters of a group G with values in a
field L are linearly inde-
pendent.
Proof. Suppose that characters χ1, . . . ,χm are linearly
dependent. Choose the depen-
dence relation with the minimal number n of coefficients ai. Let
g0 ∈ G be such that
χ1(g0) 6= χn(g0), it exists since χ1 6= χn. Then we have
n∑
j=1
ajχj(g0g) =
n∑
j=1
ajχj(g0)χj(g) = 0.
Subtracting expression
χn(g0)
n∑
j=1
ajχj(g) =
n∑
j=1
ajχn(g0)χj(g) = 0
from the left hand side of the previous equality, we get
n−1∑

j=1
aj (χj(g0) −χn(g0)) χj(g) = 0 for all g ∈ G,
which contradicts our assumption that n was the minimal
number of non-zero coefficients in
a dependence relation for characters χ1, . . . ,χm. �
Proposition 6. Let K/F be a cyclic extension of degree n, where
the field F contains a
primitive n-th root of unity and char(F) does not divide n. Then
K = F( n
√
a) for some
a ∈ F .
1
2
Proof. Let σ ∈ G = Gal(K/F) be the generator of the cyclic
Galois group. For any element
α ∈ K and any n-th root of unity ζ, consider the Lagrange
resolvent
(α,ζ) =
n−1∑
k=0
ζkσk(α) = α + ζσ(α) + · · · + ζn−1σn−1(α).
Since ζ ∈ F and therefore σ(ζ) = ζ, we have

σ(α,ζ) =
n−1∑
k=0
ζkσk+1(α) = ζ−1
n−1∑
k=0
ζk+1σk+1(α) = ζ−1
n∑
k=1
ζkσk(α) = ζ−1(α,ζ),
where the last equality follows from the fact that ζn = 1 ∈ F and
σn = 1 ∈ G. Now, we see
that
σ ((α,ζ)n) =
(
ζ−1
)n
(α,ζ)n = (α,ζ)n,
which implies (α,ζ)n ∈ F for any α ∈ K.
Let us now choose such an α ∈ K that (α,ζ) 6= 0, which exists
because the characters
1,σ, . . . ,σn−1 are independent. Let us also pick ζ to be a
primitive root of unity. Then
σk(α,ζ) = ζ−k(α,ζ),
which means that (α,ζ) ∈ K is not fixed by any nontrivial

element of G, and therefore
does not lie in any proper subfield of K. We now conclude that
K = F ((α,ζ)), and since
(α,ζ)n ∈ F , we can write K = F( n
√
a) for a = (α,ζ)n. �
In what follows, for simplicity we will assume char(F) = 0.
Nonetheless, all the results
will be valid for the fields whose characteristic does not divide
orders of the roots taken.
Definition 7.
(1) Let K = F( n
√
a) for some element a ∈ F. Then K/F is a simple radical
extension.
(2) A field K is a root extension of a field F if there exists a
chain of subfields
F = K0 ⊂ K1 ⊂ ···⊂ Ks = K,
such that Ki/Ki−1 is s simple radical extension for all 1 = 1, . . .
,s.
(3) An element α, which is algebraic over F , can be expressed
by radicals if α ∈ K for
some root extension K/F.
(4) A polynomial f(x) ∈ F[x] can be solved by radicals if all of
its roots can be expressed
by radicals.

Remark 8. Simply speaking, a polynomial can be solved by
radicals over a field F, if
all of its roots can be obtained by applying successive
operations of addition, subtraction,
multiplication, division, and extraction of n-th roots to elements
of F .
Lemma 9. For every root extension K/F , there exists a field E
⊃ K, such that E/F is a
Galois root extension, and each simple radical extension
Ei/Ei−1 is cyclic.
Proof. Let L be the Galois closure of K. Then for any σ ∈
Gal(L/F), there exists a chain
of subfields
F = σ(K0) ⊂ σ(K1) ⊂ ···⊂ σ(Ks) = σ(K),
where σ(Ki/Ki−1) is again a simple radical extension:
Ki = Ki−1 (
ni
√
ai) implies σ(Ki) = σ(Ki−1)
(
ni
√
σ(ai)
)
.
3

Note also, that a composite of two root extensions is also a root
extension. Indeed, if K/F
and K′/F are root extensions with intermediate extensions Ki
and K
′
j respectively, one can
consider a chain
F = K0 ⊂ ···⊂ Ks = K = KK′0 ⊂ KK
′
1 ⊂ KK
′
r = KK
′.
It follows that the composite of all fields of the form σ(K), σ ∈
Gal(L/F), is again a root
extension, but the latter composite is precisely L.
Now we have a Galois root extension L/F with a chain of
subfields
F = L0 ⊂ L1 ⊂ ···⊂ Lk = L,
where Li = Li−1( ni
√
ai) for all i = 1, . . . ,k. Choose a number n such that ni |n for all
1 6 i 6 k, and set F ′ = F(ζn), where ζn is a primitive n-th root of
unity. Then the extension
F ′/F is a simple radical extension. Finally, consider the
composition E = LF ′, which is

Galois over F since both L and F ′ are. Moreover, we have the
chain of subfields
F ⊂ F ′ = F ′K0 ⊂ F ′K1 ⊂ ···⊂ F ′Ks = E,
and each extension F ′Ki/F
′Ki−1 is cyclic by the first proposition in this lecture. �
Recall that a group G is solvable if there exists a chain of
normal subgroups
1 = Gs CGs−1 C · · ·CG1 CG0 = G,
with cyclic quotient groups Gi−1/Gi for all i = 1, . . . ,s. Recall
also that subgroups and
quotient groups of a solvable group are themselves solvable,
and that if both H ⊂ G and
G/H are solvable, then so is G.
Theorem 10. A polynomial can be solved in radicals if and only
if its Galois group is solvable.
Proof. Suppose that f(x) ∈ F [x] can be solved in radicals. Then
each root of f(x) is con-
tained in an extension as in the lemma above. Consider the
composite L of these extensions
for all roots of f(x). Then L/F is again a Galois root extension
with intermediate cyclic
extensions Li/Li−1. Now, let G = Gal(L/F) and Gi be the
subgroups of G corresponding to
the subfields Li. Since
Gal(Li/Li−1) = Gi−1/Gi,
we see that G is solvable. Moreover, since L contains the
splitting field of the polynomial
f(x), the Galois group H of f(x) is a subgroup of G and therefore

solvable.
Suppose now that the Galois group H of f(x) is solvable, and let
K be the splitting field
of f(x). Consider the chain of subgroups
1 = Hs CHs−1 C · · ·CH1 CH0 = H,
with cyclic quotient groups Hi−1/Hi, and let Ki = K
Hi be the fixed fields of these subgroups.
Then we get a chain of cyclic extensions
F = K0 ⊂ K1 ⊂ ···⊂ Ks = K.
Considering an extension F ′/F as in the preceding lemma, we
have a chain of cyclic extensions
F ⊂ F ′ = F ′K0 ⊂ F ′K1 ⊂ ···⊂ F ′Ks = F ′K.
Now however, each extension F ′Ki/F
′Ki−1 is a simple radical extension by one of the Propo-
sitions we proved today, since F ′Ki−1 contains the appropriate
roots of unity. Finally, since
all roots of f(x) are contained in F ′K, we see that f(x) can be
solved in radicals. �
Corollary 11. The general equation of degree > 5 is not solvable
in radicals.
4
Proof. The Galois group of a general polynomial of degree n is
isomorphic to Sn, and Sn
is not solvable for n > 5. To show that Sn is not solvable, it is
enough to notice that the

subgroup An ⊂ Sn is simple for n > 5. The latter follows from
the following easy to shoe
facts: An is generated by 3-cycles for n > 5, any normal
subgroup of An contains a 3-cycle,
hence contains all 3-cycles. �
Section 14.6.
Today we discuss Galois groups of polynomials. Let us briefly
recall what we know already.
If L/F is a Galois extension, then L is the splitting field of a
separable polynomial f(x) ∈ F[x].
If f(x) factors into a product of irreducible polynomials
f(x) = f1(x) . . .fk(x)
with degrees deg(fj) = nj, the Galois group Gal(L/F) embeds
into the following product of
symmetric groups:
Gal(L/F) ⊂ Sn1 ×···×Snk ⊂ Sn,
where n = n1 + · · · + nk. Moreover, Gal(L/F) acts transitively
on the set of roots of each
irreducible factor. The following two examples are also well-
known to us by now. The Galois
group of (x2 − 2)(x2 − 3) is isomorphic to the Klein 4-group K4
⊂ S4 and the Galois group
of x3 − 2 is isomorphic to S6.
In what follows we shall describe Galois groups of polynomials
of degrees less than or equal
to 4, and prove that the Galois group of a “general” polynomial
of degree n is isomorphic to

Sn.
Definition 1. Let x1, . . . ,xn be indeterminates. Define k-th
elementary symmetric function
by
ek = ek(x1, . . . ,xn) =
∑
16i1<···<ik6n
xi1 . . .xin.
For example, we have
e0(x1,x2,x3) = 1,
e1(x1,x2,x3) = x1 + x2 + xn,
e2(x1,x2,x3) = x1x2 + x1x3 + x2x3,
e3(x1,x2,x3) = x1x2x3.
Definition 2. Again, let x1, . . . ,xn be indeterminates. Then the
general polynomial of
degree n is defined as the product
(x−x1) . . . (x−xn).
It is easy to see that
(x−x1) . . . (x−xn) =
n∑
k=0
(−1)kek(x1, . . . ,xn)xn−k
With the above definition, we see that for any field F, the

extension F(x1, . . . ,xn) is
the splitting field for the general polynomial of degree n. We
now make the permutation
group Sn act on F(x1, . . . ,xn) by permuting the variables. It is
clear that Sn preserves
F(e1, . . . ,en), which shows that Sn is a subgroup of the
automorphism group of F(x1, . . . ,xn)
over F(e1, . . . ,en). On the other hand, the named automorphism
group is itself a subgroup
of Sn, therefore we have
Aut(F(x1, . . . ,xn)/F(e1, . . . ,en)) ' Sn.
Now, if the general polynomial was non-separable, it would
force the order of the auto-
morphism group Aut(F(x1, . . . ,xn)/F(e1, . . . ,en)) to be strictly
less than n!. This allows
us to conclude that the F(x1, . . . ,xn) is the splitting field of a
separable polynomial with
coefficients in F(e1, . . . ,en), and hence the extension F(x1, . . .
,xn)/F(e1, . . . ,en) is Galois.
1
2
Corollary 3. Any symmetric rational function f(x1, . . . ,xn) is a
rational function in vari-
ables e1, . . . ,en.
Proof. By the Fundamental Theorem of Galois Theory, F(e1, . . .
,en) is precisely the Sn-fixed
subfield of F(x1, . . . ,xn). �
Remark 4. We have just proved that F(x1, . . . ,xn)

Sn = F(e1, . . . ,en). However, a stronger
statement is true: F[x1, . . . ,xn]
Sn = F [e1, . . . ,en]. Exercises in this chapter outline a boring
proof of this statement. A more conceptual proof uses the notion
of an integral extension.
Check it out if you are curious!
Now, let f(x) ∈ F [x] be a polynomial of degree d with roots x1,
. . . ,xd. We will say that
f(x) is generic if its roots are algebraically independent, that is
there is no such polynomial
g(t1, . . . , td) that g(x1, . . . ,xd) = 0. We can see that the Galois
group of a generic polynomial
f(x) is Sn, because there are no algebraic obstructions for
realizing a permutation σ ∈ Sn
as an element of the Galois group. Let us now make the
following two remarks. First, the
Galois group of a polynomial over a finite field is cyclic (since
every finite extension is cyclic),
therefore, there are no generic polynomials over finite groups.
Second, the condition that
roots of f(x) are algebraically dependent is equivalent to the
condition that its coefficients
are algebraically dependent. We will prove one direction and
leave the other as an exercise.
Assume that the roots x1, . . . ,xd are dependent, that is there
exists a polynomial g(t1, . . . , td)
such that g(x1, . . . ,xd) = 0. Define the polynomial
g̃(t1, . . . , td) =
∏
σ∈ Sd
g(tσ(1), . . . , tσ(d)).

Now one can see that the condition g̃(x1, . . . ,xd) = 0 provides
an algebraic dependence of
the coefficients of f(x) which are nothing else but the
elementary symmetric functions of the
roots x1, . . . ,xd. Therefore, we can conclude that the general
polynomial is separable over
F(e1, . . . ,en).
Recall that [Sn : An] = 2, and for n > 5 An is the unique normal
subgroup of Sn. Therefore,
for n > 5 there is a unique field K such that
F(e1, . . .en) ⊂ K ⊂ F(x1, . . .xn)
and K/F(e1, . . .en) is Galois. Moreover, K is an extension of
degree 2.
Definition 5. Define the discriminant of x1, . . . ,xn by
D =
∏
i<j
(xi −xj)2.
The discriminant of a polynomial is defined to be the
discriminant of its roots.
Note the following properties of the discriminant which will be
handy in what follows:
(1) D ∈ F [e1, . . . ,en], and if D is viewed as a discriminant of
the polynomial f(x) ∈ F(e1, . . .en)[x],
then
√

D =
∏
i<j(xi −xj) is contained in the splitting field of f.
(2) A permutation σ fixes
√
D if and only if σ ∈ An. Therefore, given a field F with
char(F) 6= 2, the element
√
D generates the fixed field of An, which is a quadratic
extension of F(e1, . . . ,en).
(3) Let D be the discriminant of a polynomial f(x) ∈ F(e1, . .
.en)[x]. Then f is separable
if and only if D 6= 0. Therefore, over a perfect field f(x) is
reducible if and only if
D = 0.
3
(4) Assume that f(x) ∈ F[x] is separable. Reformulating
property (2) above we see that
the Galois group of f(x) is a subgroup of An if and only if
√
D ∈ F .
Now, let us finally discuss Galois groups of (the splitting fields
of) polynomials of degrees
2, 3, and 4.

Polynomials of degree 2. Consider f(x) = x2 + ax + b with roots
α and β. Then
D = (α−β)2 = (α + β)2 − 4αβ = e21 − 4e2 = (−a)
2 − 4b = a2 − 4b.
Clearly, f(x) is separable if and only if D 6= 0. Moreover, the
Galois group is isomorphic to
Z/2Z if D 6= 0 and is trivial otherwise.
Polynomials of degree 3. Consider f(x) = x3 + ax2 + bx + c.
Making a substitution
x = y −a/3, the polynomial becomes
g(y) = y3 + py + q
where
p =
3b−a2
3
and q =
2a3 − 9ab + 27c
27
.
The above substitution does not change the splitting field or the
discriminant, so we shall
only work with g(y) from now on. Let α,β,γ be its roots, then
D =
(
(α−β)(α−γ)(β −γ)

)2
.
As before, we can write out D in terms of the roots in a
straightforward manner, and express
the result in terms of coefficients of g(y). However, we are
going to use the following trick,
which will simplify the work a little: observe that
D = −g′(α)g′(β)g′(γ).
Using the fact that
g′(y) = 3y2 + p,
we can derive
D = −4p3 − 27q2.
Now, if g(y) is reducible over F , it either splits into three linear
factors, in which case the
Galois group is trivial, or into a linear and an irreducible
quadratic factor. In the latter case,
the Galois group is isomorphic to Z/2Z.
If g(y) is irreducible then a root of g(y) generates an extension
of degree 3, and therefore
the degree of the splitting field is divisible by 3. Since the
Galois group G is a subgroup of
S3 we only have two possibilities: G ' A3 ' Z/3Z or G ' S3. The
former is realized if and
only if
√
D ∈ F .
Polynomials of degree 4. As before, we consider a polynomial

f(x) = x4+ax3+bx2+cx+d,
and make a substitution x = y −a/4. Then f(x) turns into
g(y) = y4 + py2 + qy + r
where
p =
−3a2 + 8b
8
, q =
a3 − 4ab + 8c
8
, and r =
−3a4 + 16a2b− 64ac + 256d
256
.
Let the roots of g(y) be α1, . . . ,α4, and let G be the Galois
group of the splitting field of
g(y).
If g(y) is reducible it either splits into a linear and a cubic, in
which case G coincides
with the Galois group of the cubic which we already considered,
or into a pair of quadratics.
4
In the latter case the splitting field is the extension F(

√
D1,
√
D2), where D1 and D2 are
the discriminants of the corresponding quadratics. Then G is
either isomorphic to the Klein
4-group if
√
D1/D2 6∈ F , or to Z/2Z otherwise.
Now, we only need to consider the case when g(y) is
irreducible. In that case, the Galois
group G ⊂ S4 acts transitively on the roots. Inspecting the
subgroups of S4 we conclude that
the only subgroups with transitive action on the set of 4
elements are
(1) S4
(2) A4
(3) Any of the 3 Sylow 2-subgroups, each isomorphic to D8
(4) K4, the Klein 4-group
(5) any of the cyclic groups Z/4Z generated by the cycle of
length 4
In order to express the discriminant in terms of the coefficients
of g(y) we once again use
a trick. Namely, we consider elements
θ1 = (α1 + α2)(α3 + α4), θ1 = (α1 + α3)(α2 + α4), θ1 = (α1 +
α4)(α2 + α3).
Each of them is fixed by a separate Sylow 2-subgroup, and all 3
of them are fixed by the
Klein 4-group, which is precisely the intersection of all Sylow

2-subgroups in S4. It is also
clear that S4 permutes elements θj, therefore
h(x) = (x−θ1)(x−θ2)(x−θ3) ∈ F [x].
Moreover, it is rather easy to compute that
h(x) = x3 − 2px2 + (p2 − 4r)x + q2.
This polynomial is called the resolvent cubic for the quartic
g(y). It is also easy to see that
the discriminants of g(y) and h(x) coincide, which allows us to
compute
D = 16p4r − 4p3q2 − 128p2r2 + 144pq2r − 27q4 + 256r3.
Let us also notice that the splitting field of the resolvent cubic
is a subfield in that of g(y),
so the Galois group of the former is a quotient of G.
We are finally ready to explore the Galois group G. Suppose
first that h(x) is irreducible. If√
D 6∈ F, then G 6⊂ A4, and the Galois group of h(x) is S3,
which implies that |G| is divisible
by 6. This leaves us with the only option G ' S4. If the resolvent
cubic is irreducible but√
D ∈ F, then G ⊂ A4 and |G| is divisible by 3, the order of the
Galois group of h(x).
Therefore G = A4.
Now, it is only left to consider cases when the cubic is
reducible. If it splits completely,
then G fixes all of θj which implies G ⊂ K4 and therefore G =
K4. Otherwise, if the cubic

splits into a linear and a quadratic, then precisely one of θj lies
in F , hence G ⊂ D8 and
G 6⊂ K4. This leaves us with 2 possibilities: G ' D8 or G '
Z/4Z. To distinguish between
the 2, note that D8∩A4 ' K4 while Z/4Z∩A4 ' Z/2Z, the former
group is transitive on the
roots of g(y) and the latter is not. Recall also that the fixed field
of A4 is F(
√
D). Therefore,
we can determine G completely by factoring g(y) over F(
√
D).
Fundamental theorem of algebra. We finish this lecture by
proving the Fundamental
Theorem of Algebra.
Theorem 6. Every polynomial f(x) ∈ C[x] of degree n has
precisely n roots in C (counted
with multiplicities). Equivalently, C is algebraically closed.
Proof. We shall use the following two statements:
5
(1) Any polynomial g(x) ∈ R[x] of odd degree has a root in R.
Equivalently, there are
no non-trivial extensions of R of odd degree.
(2) Any polynomial h(x) ∈ C[x] of degree 2 has roots in C.
Equivalently, there are no