conference_presentation (1)

Integral Closures and Nonsingular Plane Curves
Mentor: Sander Mack-Crane
Brandon Van Over
Directed Reading Program, 2016
Brandon Van Over Integral Closures and Nonsingular Plane Curves
Directed Reading Program, 2016 1 /
27

Outline
1 The Algebra
The Ordered Triple
The Integral Closure
2 The Geometry
Plane Curves
Singularities
3 Algebraic Geometry
Rings of Functions
4 How does this all connect?
Integral Closures of Rings of Functions
27

Outline
1 The Algebra
The Ordered Triple
2 The Geometry
Plane Curves
Singularities
Rings of Functions
27

(A, K, L)
The Ordered Triple
A is a commutative integral domain with unit
K is the field of fractions of A
L is a finite field extension of K
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Example 1
A = Z
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Example 1
A = Z
K = Q
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Example 1
A = Z
K = Q
L = Q(
√
d) for d ∈ Z squarefree
27

Example 2
A = Z[x]
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Example 2
A = Z[x]
K = Q(x)
27

Example 2
A = Z[x]
K = Q(x)
L = Q(x)[
√
f ] = {m + n
√
f |m, n ∈ Q(x)}
Here f is a squarefree nonconstant polynomial.
27

Outline
1 The Algebra
The Ordered Triple
2 The Geometry
Plane Curves
Singularities
Rings of Functions
27

First Motivating Question
We want to know:
”What are all the elements of L which are zeroes of monic polynomials in
A[x]?”
27

Integral Elements
Deﬁnition 1
We say that an element α ∈ L is integral over A if it is the root of a monic
polynomial in A[x].
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Integral Elements
Deﬁnition 1
We say that an element α ∈ L is integral over A if it is the root of a monic
polynomial in A[x].
Example 1
Every integer a is an integral over Z as it a root of x − a = 0
27

Integral Closure
Deﬁnition 2
The set of all elements of L integral over A is the integral closure of A in L.
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Integral Closure
Deﬁnition 2
But how do we ﬁnd such elements?
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Integral Closure
Definition 2
But how do we find such elements?
Theorem
Let A be a UFD, then α ∈ L is integral over A iff the minimal polynomial
of α has coefficients in A.
27

Integral Closure Examples
Example 1
As a result of Gauss’ lemma, the elements integral over Z in Q are just Z.
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Example 1
Note that this is an example of a ring being integrally closed in its ﬁeld of
fractions.
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Example 1
Note that this is an example of a ring being integrally closed in its ﬁeld of
fractions.
Example 2
The ring of integers Z[
√
d] for d squarefree is the integral closure of Z in
Q[
√
d] if d ≡4 2, 3. In the case that d ≡4 1 the integral closure is
Z[1+
√
d
2 ].
27

Function Fields Example
Polynomials
The integral closure of Z[x] in Q(x)[
√
f ] is Z[x][
√
f ] for f a squarefree
nonconstant polynomial
27

Outline
1 The Algebra
The Ordered Triple
2 The Geometry
Plane Curves
Singularities
Rings of Functions
27

Plane Curves
The geometry of our story begins with the zeroes of polynomials of two
variables, namely f (x, y) ∈ k[x, y], where k is the algebraic closure of a
ﬁeld k.
27

Plane Curves
The geometry of our story begins with the zeroes of polynomials of two
variables, namely f (x, y) ∈ k[x, y], where k is the algebraic closure of a
ﬁeld k.
Deﬁnition
Zf (k) = {(a, b) ∈ k × k : f (a, b) = 0}.
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y2
− x3
Figure: y2
− x3
27

Outline
1 The Algebra
The Ordered Triple
2 The Geometry
Plane Curves
Singularities
Rings of Functions
27

Singular Plane Curves
Deﬁnition
A point (a, b) in Zf (k) is singular if
∂f
∂x
(a, b) =
∂f
∂y
(a, b) = 0.
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Singular Plane Curves
Definition
A point (a, b) in Zf (k) is singular if
∂f
∂x
(a, b) =
∂f
∂y
(a, b) = 0.
We can see how this definition fits with the curve in the previous slide as
in the case where f (x, y) = y2 − x3,
∂f
∂x
= −3x2 and
∂f
∂y
= 2y. These are
simultaneously zero at the cusp.
27

Outline
1 The Algebra
The Ordered Triple
2 The Geometry
Plane Curves
Singularities
Rings of Functions
27

Let f (x, y) ∈ k[x, y] be an irreducible polynomial.
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We study the ring k[x, y]/(f ) to gain information about our curve Zf (k).
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How do we do this?
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How do we do this?
Main Idea
In order to gain information about our curve described by Zf (k), we give
Zf (k) and k a topology so that we can identify the equivalence classes of
polynomials in the ring k[x, y]/(f ) with continuous functions C(Zf (k), k)
deﬁned on the curve Zf (k).
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The Zariski Topology
Closed sets in k are ﬁnite sets or all of k
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The Zariski Topology
Closed sets in k are finite sets or all of k
Closed sets in Zf (k) are either finite unions of points or all of Zf (k)
We look at g−1(a) for a ∈ k, which is the same as looking at the zeroes of
g(x, y) − a. The following theorem establishes continuity:
Theorem
Let f ∈ k[x, y] be an irreducible polynomial. Let g(x, y) be a polynomial
not divisible by f (x, y). Then Zf (k) ∩ Zg (k) is finite.
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Rings of Functions
We can now identify our ring with the continuous functions C(Zf (k), k) in
the following way:
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Rings of Functions
the following way:
If we deﬁne g ∈ C(Zf (k), k) by g(a, b) := g(a, b), then we have an
injection if : k[x, y]/(f ) → C(Zf (k), k) given by sending g → g.
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Rings of Functions
the following way:
If we define g ∈ C(Zf (k), k) by g(a, b) := g(a, b), then we have an
injection if : k[x, y]/(f ) → C(Zf (k), k) given by sending g → g.
If if (g) = if (h) then their difference is 0 as continuous functions on our
curve Zf (k), and so f must divide their difference in k[x, y]/(f ), otherwise
Zg−h(k) ∩ Zf (k) is finite.
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Outline
1 The Algebra
The Ordered Triple
2 The Geometry
Plane Curves
Singularities
Rings of Functions
27

The Ordered Triple
In terms of our ordered triple, we let
A = k[x]
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The Ordered Triple
A = k[x]
K = k(x)
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The Ordered Triple
A = k[x]
K = k(x)
L = Quot(k[x, y]/(f ))
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The Ordered Triple
A = k[x]
K = k(x)
L = Quot(k[x, y]/(f ))
Theorem
Let f ∈ k[x, y] be an irreducible polynomial. The ring k[x, y]/(f ) is
integrally closed iﬀ the curve Zf (k) is nonsingular.
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Example
Recall from the previous example that y2 − x3 has a singularity at
(x, y) = (0, 0).
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Example
(x, y) = (0, 0).
We know that y2 − x3 = 0 in k[x, y]/(y2 − x3), hence
y2 − x3 = x2((
y
x
)2 − x) = 0.
Since x = 0 in Quot(k[x, y]/(f )), we must have that (
y
x
)2 − x = 0 and
hence
y
x
is integral over k[x].
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Example
(x, y) = (0, 0).
We know that y2 − x3 = 0 in k[x, y]/(y2 − x3), hence
y2 − x3 = x2((
y
x
)2 − x) = 0.
Since x = 0 in Quot(k[x, y]/(f )), we must have that (
y
x
)2 − x = 0 and
hence
y
x
is integral over k[x].
It can be shown that
y
x
is not an element of k[x, y]/(f )). So it is not
integrally closed.
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Smoothing out y2
− x3
Note that
Since y2 = x3 in our ring of functions, we have that (
y
x
)2 = x
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Smoothing out y2
− x3
Note that
y
x
)2 = x
we also see that (
y
x
)3 = (
y
x
)(
y
x
)2 = (
y
x
)x = y
27

Smoothing out y2
− x3
Note that
y
x
)2 = x
we also see that (
y
x
)3 = (
y
x
)(
y
x
)2 = (
y
x
)x = y
It turns out that is the integral closure of k[x] in Quot(k[x, y]/(f )) is
(k[x, y]/(f ))[
y
x
].
27

Smoothing out y2
− x3
Note that
y
x
)2 = x
we also see that (
y
x
)3 = (
y
x
)(
y
x
)2 = (
y
x
)x = y
(k[x, y]/(f ))[
y
x
].
As a subring of the ﬁeld Quot(k[x, y]/(f )), (k[x, y]/(f )) can be written as
sums and products of k, x, y.
27

Smoothing out y2
− x3
Note that
y
x
)2 = x
we also see that (
y
x
)3 = (
y
x
)(
y
x
)2 = (
y
x
)x = y
(k[x, y]/(f ))[
y
x
].
As a subring of the ﬁeld Quot(k[x, y]/(f )), (k[x, y]/(f )) can be written as
sums and products of k, x, y.
As we saw above, these can be obtained with the element
y
x
, so this just
the subring of Quot(k[x, y]/(f )) given by k and
y
x
. Namely k[y
x ].
27

Smoothing out y2
− x3
If we let t =
y
x
we get the following isomorphism:
k[
y
x
] ∼= k[x][t]/(t2
− x)
27

Smoothing out y2
− x3
If we let t =
y
x
we get the following isomorphism:
k[
y
x
] ∼= k[x][t]/(t2
− x)
This isomorphism gives a ring of functions on the following smooth curve
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Figure: t2
− x
27

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