1.
Denitions you should know:
• Ring
• homomorphism (and isomorphism, automorphism)
• R[X]
• integral domain, eld
• unit, zero divisor, nilpotent
• characteristic
Theorems you should know:
• all the nice polynomial stu (divisibility, gcds, roots, degree, etc.)
• subring criterion
• basic ring facts (0R · a = 0R for all a ∈ R, etc.)
How to show a map f : R → S is an isomorphism:
• Show it's a homomorphism:
f(x + y) = f(x) + f(y) for all x, y ∈ R
f(xy) = f(x)f(y) for all x, y ∈ R
• Show it's injective:
If f(x) = f(y), then x = y.
• Show it's surjective:
For all s ∈ S, there is x ∈ R such that f(x) = s.
Example. Show that complex conjugation f : C → C, a + bi → a − bi is an
automorphism.
This is a homomorphism:
For all a1 + b1i, a2 + b2i ∈ C:
f((a1 + b1i) + (a2 + b2i)) = f((a1 + a2) + (b1 + b2)i)
= (a1 + a2) − (b1 + b2)i
= (a1 − b1i) + (a1 − b2i)
= f(a1 + b1i) + f(a2 + b2i)
f((a1 + b1i)(a2 + b2i) = f((a1a2 − b1b2) + (a1b2 + b1a2)i)
= (a1a2 − b1b2) − (a1b2 + b1a2)i
= (a1 − b1i)(a2 − b2i)
= f(a1 + b1i)f(a2 + b2i)
This is injective:
If f(a1 + b1i) = f(a2 + b2i), a1 − b1i = a2 − b2i, so a1 = a2 and (−b1) = (−b2), so
b1 = b2, and a1 + b1i = a2 + b2i. So f is injective.
This is surjective:
If a + bi ∈ C, f(a + (−b)i) = a − (−b)i = a + bi, so f is surjective.
Thus f is an isomorphism
How to show two rings are not isomorphic: assume they are, and get a contradiction
somehow (look for some way the ring structures are dierent.
1
2.
2
Example. Show R × R × R × R and M2(R) are not isomorphic.
Idea: R × R × R × R is commutative, but M2(R) is not.
Suppose there is an isomorphism f : M2(R) → R × R × R × R.
Consider a =
1 0
0 0
and b =
0 1
0 0
.
f(ab − ba) = f(a)f(b) − f(b)f(a) = f(a)f(b) − f(a)f(b) = 0 (since R × R × R × R
is commutative).
But ab − ba =
0 1
0 0
, so then f
0 1
0 0
=
0 0
0 0
= f
0 0
0 0
, so f is
not injective.
Example. Show Z × Z2 not isomorphic to Z.
Several possible ways:
(0, 1) is a nontrivial idempotent in Z × Z2, but Z has no nontrivial idempotent.
(1, 0) · (0, 1) = (0, 0), but there are no zero divisors in Z.
(0, 1) + (0, 1) = (0, 1), but there is no such nonzero element in Z.
(Assume there is an isomorphism, and use one of these (or something else), to get
a contradiction)
To show a polynomial is irreducible:
• If degree is 2 or 3: show it does not have a root (rational roots theorem is
one way in Q[x])
• If in Q[x] or Z[x]: Eisenstein.
To show a polynomial is reducible:
• Factor it.
• Show it has a root.
(long division of polynomials in Zp[x])
Characteristic of a ring.
If R is a commutative ring with identity 1R, and z ∈ Z, even though z is not
necessarily in R, we can kind of see z as being in R by taking 1R + · · · + 1R
z times
, which
we write z·1R for short. If there is a positive integer n such that 1R + · · · + 1R
n times
= 0,
we can actually view the elements of Zn as being contained in R in the same way:
identify [z] with 1R + · · · + 1R
z times
. We call the smallest such n the characteristic of R.
If there is no such n, we say that R has characteristic 0.
On the rst exam, you showed that Z has characteristic 0 and Zn has characteristic
n.
3.
3
Example. In this way, you can think of Z being in M2(R) by thinking of
1 0
0 1
+ · · · +
1 0
0 1
z times
=
z 0
0 z
as z ∈ Z.
You can think of Z2 being in M2(Z2) by thinking of
[1] [0]
[0] [1]
+ · · · +
[1] [0]
[0] [1]
z times
=
[z] [0]
[0] [z]
as [z] ∈ Z.
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