Resolution of Singularities - WordPress.com · 2015-05-10 · Ryan, Lok-Wing Pang (HKUST) Resolution of Singularities Introduction to Algebraic Geometry, 2015 5 / 18. Dedekind Domain

Resolution of Singularities

Ryan, Lok-Wing Pang

Department of MathematicsHong Kong University of Science and Technology

Introduction to Algebraic Geometry, 2015

Ryan, Lok-Wing Pang (HKUST) Resolution of SingularitiesIntroduction to Algebraic Geometry, 2015 1

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Outline

1 Dedekind Domains

2 Normality of Varieties

3 Normalization of Singular Affine Varieties

4 Resolution of Singularities


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Some Algebraic Number Theory

Throughout the presentation, we assume the underlying field k to bealgebraically closed.

Recall that a number field K is a subfield of C such that [K : Q] isfinite.Then we can define the ring of integres of K to be OK = K ∩A.

It is easy to see that rings of integers are not always UFD (egZ[√−5]). However, we can prove that the non-zero ideals in a ring of

integers always factor uniquely into prime ideals.

This can be regarded as a generalization of unique factorization in Z,where the ideals are the principal ideals (n) and the prime ideals arethe ideals (p), where p is a rational prime.

We will show that rings of integers have three special properties, andthat any integral domain with these properties also have the uniquefactorization property for ideals.

According, we make the following definition:


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Recall that a number field K is a subfield of C such that [K : Q] isfinite.

Then we can define the ring of integres of K to be OK = K ∩A.







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Dedekind Domain

Defintion (Dedekind Domain)

A Dedekind domain is an integral domain R such that(1) Every ideal is finitely generated;(2) Every non-zero prime ideal is a maximal ideal;(3) R is integrally closed in its field of fractions K .

Condition (1) is equivalent to each of the conditions

(1’) Every increasing sequence of ideals eventually stabilized:I1 ⊂ I2 ⊂ I3 ⊂ · · · implies that In = In+1 = · · · for sufficiently large n;

(1”) Every non-empty set S of ideals has a maximal member: thereexists M ∈ S such that M ⊆ I ∈ S ⇒ M = I .

It is not hard to prove the equivalence of these three conditions, Aring satisfying them is called a Noetherian ring.


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Dedekind Domain








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Dedekind Domain








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Dedekind Domain

Theorem

Every ring of integers OK is a Dedekind Domain.

Proof.

It is known that every ring of intgers is a free Z-module of finite rank n.An ideal I is an additive subgroup, hence it too is a free Z-module of finiterank. Hence I is finitely generated as an ideal. That proves (1).

To show that every nonzero prime ideal P is maximal, it suffices to showthat OK/P is a field.For any α 6= 0 ∈ P, N(α) = m ∈ Z. Note thatm ∈ P since m = αβ, where β is the product of the conjugates of α.β = m/α ∈ K and β ∈ A implies β ∈ OK and hence m ∈ P. Hence(m) = (N(α)) ⊆ P. Therefore |OK/P| ≤ |OK/(m)| ≤ m[K :Q]. So OK/Pis a finite integral domain, which is a field.

Suppose α is a root of a monic polynomialxn + an−1x

n−1 + · · ·+ a0 ∈ OK [x ], then consider the ringR = Z[a0, a1, · · · , an−1, α].


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Dedekind Domain

Theorem


Proof.






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Dedekind Domain

Theorem


Proof.


To show that every nonzero prime ideal P is maximal, it suffices to showthat OK/P is a field.

For any α 6= 0 ∈ P, N(α) = m ∈ Z. Note thatm ∈ P since m = αβ, where β is the product of the conjugates of α.β = m/α ∈ K and β ∈ A implies β ∈ OK and hence m ∈ P. Hence(m) = (N(α)) ⊆ P. Therefore |OK/P| ≤ |OK/(m)| ≤ m[K :Q]. So OK/Pis a finite integral domain, which is a field.




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Dedekind Domain

Theorem


Proof.


To show that every nonzero prime ideal P is maximal, it suffices to showthat OK/P is a field.For any α 6= 0 ∈ P, N(α) = m ∈ Z. Note thatm ∈ P since m = αβ, where β is the product of the conjugates of α.β = m/α ∈ K and β ∈ A implies β ∈ OK and hence m ∈ P.

Hence(m) = (N(α)) ⊆ P. Therefore |OK/P| ≤ |OK/(m)| ≤ m[K :Q]. So OK/Pis a finite integral domain, which is a field.




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Dedekind Domain

Theorem


Proof.






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Dedekind Domain

Theorem


Proof.




n−1 + · · ·+ a0 ∈ OK [x ], then consider the ringR = Z[a0, a1, · · · , an−1, α].Ryan, Lok-Wing Pang (HKUST) Resolution of Singularities

Introduction to Algebraic Geometry, 2015 5/ 18

Dedekind Domain

Proof.


To show that every nonzero prime ideal P is maximal, it suffices to showthat OK/P is a field. For any α 6= 0 ∈ P, N(α) = m ∈ Z. Note thatm ∈ P since m = αβ, where β is the product of the conjugates of α.β = m/α ∈ K and β ∈ A implies β ∈ OK and hence m ∈ P. Hence(m) = (N(α)) ⊆ P. Therefore |OK/P| ≤ |OK/(m)| ≤ m[K :Q]. So OK/Pis a finite integral domain, which is a field.


n−1 + · · ·+ a0 ∈ OK [x ], then consider the ringR = Z[a0, a1, · · · , an−1, α]. Then am0

0 am11 · · · a

mn−1

n−1 αmn for 0 ≤ mi ≤ di

and 0 ≤ mn ≤ n, where di is the degree of the minimal polynomial havingai as a root, certainly generates R as a Z-module. Hence Z[α] ⊂ R is alsoa finitely generated Z-module.Ryan, Lok-Wing Pang (HKUST) Resolution of Singularities

Introduction to Algebraic Geometry, 2015 6/ 18

Dedekind Domain

Some properties of Dedekind Domains:

Theorem

Every ideal in a Dedekind domain factor uniquely into prime ideals.

Theorem

A Dedekind domain is a PID iff it is a UFD.

Theorem

Every ideal in a Dedekind domain is generated as an ideal by at most twoelements. One of them can be chosen arbitrarily.


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Dedekind Domain


Theorem


Theorem


Theorem



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Dedekind Domain


Theorem


Theorem


Theorem



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Normality of Varieties

Defintion (Normal Varieties)

Let X ⊆ Ank be an affine variety. Then X is nomral at a point p ∈ X if

OX ,p is integrally closed in k(X ). We call X a normal variety if X isnormal at every p ∈ X .

Lemma

If X is affine, then X is normal iff O(X ) is integrally closed in k(X ).


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Normality of Varieties

Defintion (Normal Varieties)

Let X ⊆ Ank be an affine variety. Then X is nomral at a point p ∈ X if

OX ,p is integrally closed in k(X ). We call X a normal variety if X isnormal at every p ∈ X .

Lemma

If X is affine, then X is normal iff O(X ) is integrally closed in k(X ).


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Proof of Lemma

Proof.

Suppose O(X ) is integrally closed, we must show that each local ring isintergrally closed in k(X ). Let u ∈ k(X ) whereun + an−1u

n−1 + · · ·+ a0 = 0, for ai ∈ OX ,p.

Write ai = bi/ci wherebi , ci ∈ O(X ) and ci 6∈ mp. Multiplying out denominators yieldsdnu

n + dn−1un−1 + · · ·+ d0 = 0, for di ∈ O(X ). Let v = dnu, then

vn + en−1vn−1 + · · ·+ e0 = 0 for ei ∈ O(X ) and hence u = v/dn ∈ OX ,p.

Conversely, if all the local rings OX ,p are integrally closed in k(X ), andu ∈ k(X ) is integral over O(X ), i.e.un + an−1u

n−1 + · · ·+ a0 = 0, ai ∈ O(X ).But then ai ∈ OX ,p for allp ∈ X , and since OX ,p is integrally closed by assumption, we haveu ∈ OX ,p for all p ∈ X and u ∈

⋂p∈X OX ,p.The result follows from the

fact that O(X ) =⋂

p∈X OX ,p.


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Proof of Lemma

Proof.


n−1 + · · ·+ a0 = 0, for ai ∈ OX ,p.Write ai = bi/ci wherebi , ci ∈ O(X ) and ci 6∈ mp. Multiplying out denominators yieldsdnu







p∈X OX ,p.


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Proof of Lemma

Proof.






n−1 + · · ·+ a0 = 0, ai ∈ O(X ).

But then ai ∈ OX ,p for allp ∈ X , and since OX ,p is integrally closed by assumption, we haveu ∈ OX ,p for all p ∈ X and u ∈



p∈X OX ,p.


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Proof of Lemma

Proof.







⋂p∈X OX ,p.

The result follows from thefact that O(X ) =

⋂p∈X OX ,p.


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Proof of Lemma

Proof.









p∈X OX ,p.


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Smoothness Implies Normality

Theorem

A smooth affine variety X is normal.

Proof.

By the lemma, it suffices to show that if p is a smooth point, then OX ,p isintegrally closed in k(X ). Since OX ,p is a UFD, we can write α ∈ k(X ) inthe form α = u/v , u, v ∈ OX ,p and have no common factors.Suppose α isintegral over OX ,p, then αn + an−1α

n−1 + · · ·+ a0 = 0 for someai ∈ OX ,p.Hence un + an−1vu

n−1 + · · ·+ a0vn = 0, and we have v |un. But

(u, v) = 1 and OX ,p is a UFD, therefore α ∈ OX ,p and we are done.


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Theorem


Proof.

By the lemma, it suffices to show that if p is a smooth point, then OX ,p isintegrally closed in k(X ). Since OX ,p is a UFD, we can write α ∈ k(X ) inthe form α = u/v , u, v ∈ OX ,p and have no common factors.

Suppose α isintegral over OX ,p, then αn + an−1α





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Theorem


Proof.


n−1 + · · ·+ a0 = 0 for someai ∈ OX ,p.

Hence un + an−1vun−1 + · · ·+ a0v

n = 0, and we have v |un. But(u, v) = 1 and OX ,p is a UFD, therefore α ∈ OX ,p and we are done.


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Theorem


Proof.






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It would be nice if the converse was true.

For example, the variety X : xy = z2 is normal but not smooth.(Exercise: k[x , y , z ]/(z2 − xy) is integrally closed.)

So being normal is a weaker condition than being smooth for generalvarieties.

But the beauty is, for algebraic curves, normality and smoothness areequivalent conditions.

Theorem

A normal affine algebraic curve is smooth.

The standard proof is to show that the locus of singular points of anormal variety has codimension at least 2.


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Theorem




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Theorem




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Theorem




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Theorem




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Theorem




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Proof

Theorem


Proof.

Let X be a normal affine algebraic curve over k . It suffices to prove thatthe localization k[X ]mp of k[X ] at mp is a UFD for any p ∈ X .

First I claimthat the coordinate ring k[X ] is a Dedekind domain. k[X ] is a finitelygenerated k-algebra, hence Noetherian. Next, dimX = 1 implies everynon-zero ideal is maximal.Since X is normal, k[X ] is integrally closed.Hence k[X ] is a Dedekind Domain.The result follows from the fact that aDedekind domain D is equivalent to that D being Noetherian integraldomain but not a field and the localization at each maximal ideal is aDiscrete Valuation Ring, in particular a UFD.


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Proof

Theorem


Proof.

Let X be a normal affine algebraic curve over k . It suffices to prove thatthe localization k[X ]mp of k[X ] at mp is a UFD for any p ∈ X .First I claimthat the coordinate ring k[X ] is a Dedekind domain.

k[X ] is a finitelygenerated k-algebra, hence Noetherian. Next, dimX = 1 implies everynon-zero ideal is maximal.Since X is normal, k[X ] is integrally closed.Hence k[X ] is a Dedekind Domain.The result follows from the fact that aDedekind domain D is equivalent to that D being Noetherian integraldomain but not a field and the localization at each maximal ideal is aDiscrete Valuation Ring, in particular a UFD.


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Proof

Theorem


Proof.

Let X be a normal affine algebraic curve over k . It suffices to prove thatthe localization k[X ]mp of k[X ] at mp is a UFD for any p ∈ X .First I claimthat the coordinate ring k[X ] is a Dedekind domain. k[X ] is a finitelygenerated k-algebra, hence Noetherian.

Next, dimX = 1 implies everynon-zero ideal is maximal.Since X is normal, k[X ] is integrally closed.Hence k[X ] is a Dedekind Domain.The result follows from the fact that aDedekind domain D is equivalent to that D being Noetherian integraldomain but not a field and the localization at each maximal ideal is aDiscrete Valuation Ring, in particular a UFD.


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Proof

Theorem


Proof.

Let X be a normal affine algebraic curve over k . It suffices to prove thatthe localization k[X ]mp of k[X ] at mp is a UFD for any p ∈ X .First I claimthat the coordinate ring k[X ] is a Dedekind domain. k[X ] is a finitelygenerated k-algebra, hence Noetherian. Next, dimX = 1 implies everynon-zero ideal is maximal.

Since X is normal, k[X ] is integrally closed.Hence k[X ] is a Dedekind Domain.The result follows from the fact that aDedekind domain D is equivalent to that D being Noetherian integraldomain but not a field and the localization at each maximal ideal is aDiscrete Valuation Ring, in particular a UFD.


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Proof

Theorem


Proof.

Let X be a normal affine algebraic curve over k . It suffices to prove thatthe localization k[X ]mp of k[X ] at mp is a UFD for any p ∈ X .First I claimthat the coordinate ring k[X ] is a Dedekind domain. k[X ] is a finitelygenerated k-algebra, hence Noetherian. Next, dimX = 1 implies everynon-zero ideal is maximal.Since X is normal, k[X ] is integrally closed.Hence k[X ] is a Dedekind Domain.

The result follows from the fact that aDedekind domain D is equivalent to that D being Noetherian integraldomain but not a field and the localization at each maximal ideal is aDiscrete Valuation Ring, in particular a UFD.


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Proof

Theorem


Proof.

Let X be a normal affine algebraic curve over k . It suffices to prove thatthe localization k[X ]mp of k[X ] at mp is a UFD for any p ∈ X .First I claimthat the coordinate ring k[X ] is a Dedekind domain. k[X ] is a finitelygenerated k-algebra, hence Noetherian. Next, dimX = 1 implies everynon-zero ideal is maximal.Since X is normal, k[X ] is integrally closed.Hence k[X ] is a Dedekind Domain.The result follows from the fact that aDedekind domain D is equivalent to that D being Noetherian integraldomain but not a field and the localization at each maximal ideal is aDiscrete Valuation Ring, in particular a UFD.


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Normalization - Motivation

Example

Consider the nonnormal variety V : y2 = x2 + x3. Then theparametrization t = y/x defines a map f : A1

k → V , or equivalently aninclusion k[V ] ↪→ k[t].

Since f is birational, we havek[V ] ⊂ k[t] ⊂ k(t) ∼= k(V ).

A1 is normal since k[t] is integrally closed.

We can identify the ring k[t] with the set Ok(V ),k[V ] of u ∈ k(V ) that areintegral over k[V ]: t2 = 1 + x , hence k[t] ⊆ Ok(V ),k[V ].Conversely, ifu ∈ k(V ) is integral over k[V ], then it is also integral over k[t], henceu ∈ k[t] since k[t] is integrally closed.

Finally, in geometric terminology, k[t] integral over k[V ] says that f is afinite map.

In light of the example, we make the following definition:


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Example


k → V , or equivalently aninclusion k[V ] ↪→ k[t].Since f is birational, we havek[V ] ⊂ k[t] ⊂ k(t) ∼= k(V ).






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Example








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Example




We can identify the ring k[t] with the set Ok(V ),k[V ] of u ∈ k(V ) that areintegral over k[V ]: t2 = 1 + x , hence k[t] ⊆ Ok(V ),k[V ].

Conversely, ifu ∈ k(V ) is integral over k[V ], then it is also integral over k[t], henceu ∈ k[t] since k[t] is integrally closed.




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Example








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Example








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Example








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Normalization

Defintion (Normalization)

A normalization of an affine variety X is an irreducible normal variety X̃ ,together with a regular birational finite map f : X̃ → X .

Theorem

Normalization of a singular affine variety exists.

Lemma

An algebra A over a field k is isomorphic to a coordinate ring k[X ] ofsome closed subset X iff A has no nilpotent elements and is finitelygenerated k-algebra.

Proof.

Necessity is clear. Sufficiency is left as an exercise.


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Normalization



Theorem


Lemma


Proof.



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Normalization



Theorem


Lemma


Proof.



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Normalization

Proof.

Let A be the integral closure of k[X ] in k(X ). Suppose we can find anaffine variety X̃ such that A = k[X̃ ]. Then X̃ is normal and the inclusionk[X ] ↪→ k[X̃ ] defines a regular birational map f : X̃ → X . Clearly X̃ is anormalization of X .

By the above lemma, it suffices to prove that A is afinitely generated k[X ]-module.Now, by Noether normalization theorem,there exists a subring B ⊆ k[X ] such that B is isomorphic to a polynomialring B ∼= k[T1, · · · ,Tr ] and k[X ] is integral over B. It is not hard to seethat A is equal to the integral closure of B in k(X ). Also, k(X ) is a finiteextension of k(T1, · · · ,Tr ) since Ti ’s form a transcendence basis ofk(X ).Finally, B is integrally closed, since Ar

k is smooth, hence normal.Theresult follows from the fact that if B = k[T1, · · · ,Tr ], L = k(T1, · · · ,Tr )and E = k(X ) is a finite extension of L, then the integral closure of B inE is a finite B-module.


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Normalization

Proof.

Let A be the integral closure of k[X ] in k(X ). Suppose we can find anaffine variety X̃ such that A = k[X̃ ]. Then X̃ is normal and the inclusionk[X ] ↪→ k[X̃ ] defines a regular birational map f : X̃ → X . Clearly X̃ is anormalization of X .By the above lemma, it suffices to prove that A is afinitely generated k[X ]-module.

Now, by Noether normalization theorem,there exists a subring B ⊆ k[X ] such that B is isomorphic to a polynomialring B ∼= k[T1, · · · ,Tr ] and k[X ] is integral over B. It is not hard to seethat A is equal to the integral closure of B in k(X ). Also, k(X ) is a finiteextension of k(T1, · · · ,Tr ) since Ti ’s form a transcendence basis ofk(X ).Finally, B is integrally closed, since Ar



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Normalization

Proof.

Let A be the integral closure of k[X ] in k(X ). Suppose we can find anaffine variety X̃ such that A = k[X̃ ]. Then X̃ is normal and the inclusionk[X ] ↪→ k[X̃ ] defines a regular birational map f : X̃ → X . Clearly X̃ is anormalization of X .By the above lemma, it suffices to prove that A is afinitely generated k[X ]-module.Now, by Noether normalization theorem,there exists a subring B ⊆ k[X ] such that B is isomorphic to a polynomialring B ∼= k[T1, · · · ,Tr ] and k[X ] is integral over B. It is not hard to seethat A is equal to the integral closure of B in k(X ).

Also, k(X ) is a finiteextension of k(T1, · · · ,Tr ) since Ti ’s form a transcendence basis ofk(X ).Finally, B is integrally closed, since Ar



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Normalization

Proof.

Let A be the integral closure of k[X ] in k(X ). Suppose we can find anaffine variety X̃ such that A = k[X̃ ]. Then X̃ is normal and the inclusionk[X ] ↪→ k[X̃ ] defines a regular birational map f : X̃ → X . Clearly X̃ is anormalization of X .By the above lemma, it suffices to prove that A is afinitely generated k[X ]-module.Now, by Noether normalization theorem,there exists a subring B ⊆ k[X ] such that B is isomorphic to a polynomialring B ∼= k[T1, · · · ,Tr ] and k[X ] is integral over B. It is not hard to seethat A is equal to the integral closure of B in k(X ). Also, k(X ) is a finiteextension of k(T1, · · · ,Tr ) since Ti ’s form a transcendence basis ofk(X ).

Finally, B is integrally closed, since Ark is smooth, hence normal.The

result follows from the fact that if B = k[T1, · · · ,Tr ], L = k(T1, · · · ,Tr )and E = k(X ) is a finite extension of L, then the integral closure of B inE is a finite B-module.


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Normalization

Proof.

Let A be the integral closure of k[X ] in k(X ). Suppose we can find anaffine variety X̃ such that A = k[X̃ ]. Then X̃ is normal and the inclusionk[X ] ↪→ k[X̃ ] defines a regular birational map f : X̃ → X . Clearly X̃ is anormalization of X .By the above lemma, it suffices to prove that A is afinitely generated k[X ]-module.Now, by Noether normalization theorem,there exists a subring B ⊆ k[X ] such that B is isomorphic to a polynomialring B ∼= k[T1, · · · ,Tr ] and k[X ] is integral over B. It is not hard to seethat A is equal to the integral closure of B in k(X ). Also, k(X ) is a finiteextension of k(T1, · · · ,Tr ) since Ti ’s form a transcendence basis ofk(X ).Finally, B is integrally closed, since Ar

k is smooth, hence normal.

Theresult follows from the fact that if B = k[T1, · · · ,Tr ], L = k(T1, · · · ,Tr )and E = k(X ) is a finite extension of L, then the integral closure of B inE is a finite B-module.


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Normalization

Proof.

Let A be the integral closure of k[X ] in k(X ). Suppose we can find anaffine variety X̃ such that A = k[X̃ ]. Then X̃ is normal and the inclusionk[X ] ↪→ k[X̃ ] defines a regular birational map f : X̃ → X . Clearly X̃ is anormalization of X .By the above lemma, it suffices to prove that A is afinitely generated k[X ]-module.Now, by Noether normalization theorem,there exists a subring B ⊆ k[X ] such that B is isomorphic to a polynomialring B ∼= k[T1, · · · ,Tr ] and k[X ] is integral over B. It is not hard to seethat A is equal to the integral closure of B in k(X ). Also, k(X ) is a finiteextension of k(T1, · · · ,Tr ) since Ti ’s form a transcendence basis ofk(X ).Finally, B is integrally closed, since Ar



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We have shown that normalization always exist. More specifically, wecan ask whether every variety V has a resolution, this is also knownas the problem of Resolution of Singularities.

For curves, normalization is the resolution.

In 1964, Heisuke Hironaka electrified the mathematical world bysolving the problem of resolution of singularities completely forvarieties over fields of characteristic zero.

He was awarded the Fields Medal in 1970.

The problem is still open for varieties over fields of characteristic p indimensions at least 4.


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References

[1] H. Matsumura, Commutative Ring Theory (Cambridge Studies inAdvanced Mathematics), Cambridge University Press 1989.

[2] D. Eisenbud, Commutative Algebra: with a View Toward AlgebraicGeometry. Springer-Verlag, 1990.

[3] R. Hartshorne, Algebraic Geometry. Springer-Verlag, 1977.

[4] D. Marcus - Number Fields, Springer-Verlag, 1977.

[5] I. Shafarevich - Basic Algebraic Geometry 1, Springer-Verlag, 2007.


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Thanks

Thank you!


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Documents

Resolution of Singularities - WordPress.com · 2015-05-10 · Ryan, Lok-Wing Pang (HKUST) Resolution of Singularities Introduction to Algebraic Geometry, 2015 5 / 18. Dedekind Domain