AG2. Dimensions, Tangent Spaces and Smooth Affine Varieties - Ryan Lok-Wing Pang

Dimensions, Tangent Spaces and Smooth Affine Varieties

Ryan, Lok-Wing Pang

Department of MathematicsHong Kong University of Science and Technology

Introduction to Algebraic Geometry, 2015

Ryan, Lok-Wing Pang (HKUST) Dimensions, Tangent Spaces and Smooth Affine VarietiesIntroduction to Algebraic Geometry, 2015 1

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Outline

1 Dimension of an Affine Variety

2 Tangent Spaces

3 Smooth Varieties: Motivations


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Dimension of an Affine Variety

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

The idea to define dimension in algebraic topology using Zariskitopology is the following:

If X is an irreducible topological space, then any proper closed subsetof X must have dimension (at least) one smaller.

Surprisingly, it turns out that the right definition is purely topological:it depends only on the topological space.

Defintion (Dimension)

Let X be a topological space. Then the dimension dimX of X is definedto be the supremum of all integers n such that there exists a chain ofirreducible closed subsets of X :

X0 ( X1 ( · · · ( Xn

contained in X . In particular, if X is an affine variety, then we define thedimension of the variety to be its dimension as a topological space.


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X0 ( X1 ( · · · ( Xn



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X0 ( X1 ( · · · ( Xn



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X0 ( X1 ( · · · ( Xn



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X0 ( X1 ( · · · ( Xn



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Example

Recall from last time we proved that the only proper subvarieties of A1k are

the whole space and single points, hence dimA1k = 1.

This definition has the advantage of being short and intuitive, but ithas the disadvantage that it is very hard to apply in actualcomputations. To this end, we have an alternative and equivalentdefinition:

Theorem

Let X be an affine variety. Then dimX = tr .deg (k(X )/k), thetranscendence degree of k(X ) over k , where k(X ) is the field of rationalfunctions of X .

Proving the equivalence requires a lot of commutative algebra.

Example

The function field k(Ank) = k(x1, · · · , xn). Hence dimAn

k = n.


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Example




Theorem



Example


k = n.


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Example




Theorem



Example


k = n.


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Example




Theorem



Example


k = n.Ryan, Lok-Wing Pang (HKUST) Dimensions, Tangent Spaces and Smooth Affine Varieties

Introduction to Algebraic Geometry, 2015 4/ 17

Maximal Ideal - Point Correspondence

Recall that we have a 1-1 correspondence between affine varieties Vand prime ideals in k[x1, · · · , xn]:{V ⊆ An

k} ←→ Spec k[x1, · · · , xn].

We also have:

Theorem (Maximal Ideal - Point Correspondence)

Let X be an affine variety with coordinate ring R = k[x1, · · · , xn]/I (X ).Then there is a bijective correspondence between the points p ∈ X andthe maximal ideals m E R.

Proof.

A maximal (hence prime) ideal m of k[x1, · · · , xn] corresponds to aminimal affine variety of An

k (because the correspondence is inclusionreversing), which must be a point p. The result follows from the 4thisomorphism theorem.

It implies that every maximal ideal of k[x1, · · · , xn] is of the formm = (x1 − a1, · · · , xn − an) for some ai ∈ k .


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k} ←→ Spec k[x1, · · · , xn]. We also have:



Proof.





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Proof.





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Proof.





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Tangent Space revisited

In calculus, we define the tangent plane to a surface by taking thespan of tengent vectors to arcs lying on it. We want to translate thisinto algebraic geometry.

Defintion (Derivations)

Let A→ B be a homomorphism of commutative algebras, and M aB-module. We define the derivations

DerA(B,M) = {D : B → M|D satisfying (1) and (2)}

(1) D(b1b2) = D(b1)b2 + b1D(b2) ∀b1, b2 ∈ B.(2) D(a) = 0 ∀a ∈ A.


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Tangent Space revisited

In calculus, we define the tangent plane to a surface by taking thespan of tengent vectors to arcs lying on it. We want to translate thisinto algebraic geometry.

Defintion (Derivations)

Let A→ B be a homomorphism of commutative algebras, and M aB-module. We define the derivations

DerA(B,M) = {D : B → M|D satisfying (1) and (2)}

(1) D(b1b2) = D(b1)b2 + b1D(b2) ∀b1, b2 ∈ B.(2) D(a) = 0 ∀a ∈ A.


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Motivation

Example

If A = k ,B = k[x1, · · · , xn], then M = k is a B-module. Define the map∂/∂xi : B → k by xj 7→ δij . We have

Derk(B, k) =

{D =

∑i

λi∂

∂xi: λi ∈ k

}.

Note that we have k ∼= OX ,p/mp. This is because the map k[X ]→ kgiven by evaluation at p has, by definition, kernel mp, and so does themap OX ,p

∼= k[X ]mp → k .

Defintion (Tangent Space)

Let X be an affine variety. The Zariski Tangent Space of X at p is definedas TpX = Derk(OX ,p, k).


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Motivation

Example


Derk(B, k) =

{D =

∑i

λi∂

∂xi: λi ∈ k

}.


∼= k[X ]mp → k .




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Motivation

Example


Derk(B, k) =

{D =

∑i

λi∂

∂xi: λi ∈ k

}.


∼= k[X ]mp → k .




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Tangent Spaces

Theorem

Let X be an affine variety and p ∈ X , then TpX ∼= (mp/m2p)∗, where V ∗ is

the dual vector space of V .

Proof.

Consider the map TpX → Hom (mp/m2p, k) defined by D → D = D|mp . D

vanishes on m2p because D(fg) = fD(g) + D(f )g ≡ 0 (mod mp) for

f , g ∈ mp. As a k-vector space, OX ,p∼= k ⊕mp, we see that D is

determined by mp. Hence D induces a linear map D : mp/m2p → k and

TpX → Hom (mp/m2p, k) is injective.On the other hand, let C = mp \m2

p

be the complement of m2p so that OX ,p

∼= k ⊕ C ⊕m2p. If T : C → k is

linear, then it is clear that the extension of T to a linear map D on OX ,p

by putting T |k⊕m2p

= 0 is a derviation in p.


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Tangent Spaces

Theorem



Proof.

Consider the map TpX → Hom (mp/m2p, k) defined by D → D = D|mp .

Dvanishes on m2

p because D(fg) = fD(g) + D(f )g ≡ 0 (mod mp) forf , g ∈ mp. As a k-vector space, OX ,p

∼= k ⊕mp, we see that D isdetermined by mp. Hence D induces a linear map D : mp/m

2p → k and


p


∼= k ⊕ C ⊕m2p. If T : C → k is





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Tangent Spaces

Theorem



Proof.




determined by mp.

Hence D induces a linear map D : mp/m2p → k and


p


∼= k ⊕ C ⊕m2p. If T : C → k is





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Tangent Spaces

Theorem



Proof.





TpX → Hom (mp/m2p, k) is injective.

On the other hand, let C = mp \m2p


∼= k ⊕ C ⊕m2p. If T : C → k is





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Tangent Spaces

Theorem



Proof.






p


∼= k ⊕ C ⊕m2p. If T : C → k is





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Smooth Varieties

Defintion (Hypersurfaces)

A hyersurface in Ank is a subvariety defined by a single equation

f (x1, · · · , xn) = 0.

Defintion (Singularities of a Hypersurface)

Let X be a hypersurface f (x1, · · · , xn) = 0 in Ank . A point p ∈ X is a

singularity of X if ∂f /∂xi (P) = 0 for all i = 1, · · · , n.

The set of singularities forms a subvariety of X , defined by f = 0together with the equations ∂f /∂xi = 0 for all i = 1, · · · , n.

Defintion (Smoothness)

A hypersurface X in Ank is called smooth (or nonsingular) if there are no

singularities in X .


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Smooth Varieties



f (x1, · · · , xn) = 0.









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Smooth Varieties



f (x1, · · · , xn) = 0.









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Smooth Varieties



f (x1, · · · , xn) = 0.









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Motivating Example

Example (Elliptic Curve)

Let E be the elliptic curve y2 = x3 + 1 in A2Q, then the singular locus is

defined by the equations y2 − x3 − 1 = −3x2 = 2y = 0, which have nocommon solution in Q, so the curve is smooth.

Example (Nodes in a Singular Curve)

Let X be the ”nodal cubic” y2 = x3 + x2, then the singular locus isdefined by the equations y2 − x3 − x2 = −3x2 − 2x = 2y = 0,which have a unique solution (0, 0). Hence X is singular. It has two”branches” crossing at (0, 0). Such a singularity is called a node.

Example (Cusps in a Singular Curve)

Let Y be the curve y2 = x3, then Y is singular with a unique singularity(0, 0). It has no tangent at (0, 0). Such a singularity is called a cusp.


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Motivating Example









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Motivating Example









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Smooth Varieties

More generally:


An affine variety X = V (f1, · · · , fm) is smooth (or nonsingular) at p if them × n Jacobian matrix has rank n − dimX , i.e.

rk

(∂fi∂xj

(p)

)1≤i≤m1≤j≤n

= n − dimX .

X is nonsingular (or smooth) if X is nonsingular at every point.

When X is generated by a single non-constant polynomial, then thedefiniton reduces to the case of hypersurface.

If X = V (f ) is a hypersurface, then p ∈ X is a singularity iff∂f /∂xi (p) = 0 for all i = 1, · · · , n and f (P) = 0, this gives n + 1equations in n variables. Thus for a randomly chosen hypersurface X ,we would expect X to be nonsingular.


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Smooth Varieties

More generally:



rk

(∂fi∂xj

(p)

)1≤i≤m1≤j≤n

= n − dimX .





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Smooth Varieties

More generally:



rk

(∂fi∂xj

(p)

)1≤i≤m1≤j≤n

= n − dimX .





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Equivalent Definitions of Smoothness

Our definiton of smoothness has a drawback in the sense that itdepends on the embedding of X in affine space.

Zariski proved that nonsingularity is intrinsic:

Theorem (Zariski, 1947)

Let X ⊆ Ank be an affine variety and p ∈ X . Then X is nonsingular at P iff

dimk[X ]/mp(mp/m

2p) = dimX .

This theorem is important because it implies that nonsingularity isintrinsic, i.e. it could be described in terms of the local rings, ratherthan a set of generators.


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dimk[X ]/mp(mp/m

2p) = dimX .



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dimk[X ]/mp(mp/m

2p) = dimX .



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dimk[X ]/mp(mp/m

2p) = dimX .



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Proof.

First we can identify k[X ]/mp with k (why?).

Let p = (a1, · · · , an) ∈ Ank

and let ap = (x1 − a1, · · · , xn − an). We define a map byϕ : k[x1, · · · , xn]→ kn by ϕ(f ) = (∂f /∂x1(p), · · · , ∂f /∂xn(p)). Then it isclear that ϕ(xi − ai ), i = 1, · · · , n form a basis of kn and ϕ(a2

p) = 0.Hence ϕ induces an isomorphism ϕ : ap/a

2p → kn. Now write X = V (b)

and let f1, · · · , ft be a set of generators of b E k[x1, · · · , xn] (this can bedone because k[x1, · · · , xn] is Noetherian). Thenrk(∂fi/∂xj(p)) = dimϕ(b).Using the isomorphism ϕ, this is the same asdim b/(b ∩ a2

p) = dim(b + a2p)/a2

p. Now, OX ,p∼= k[X ]ap . Hence if m is the

maximal ideal of OX ,p, we have m/m2 ∼= ap/(b + a2p). Counting

dimensions of vector spaces, we have dimm/m2 + rk(∂fi/∂xj(p)) = n. Theresult follows immediately.


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Proof.

First we can identify k[X ]/mp with k (why?). Let p = (a1, · · · , an) ∈ Ank

and let ap = (x1 − a1, · · · , xn − an). We define a map byϕ : k[x1, · · · , xn]→ kn by ϕ(f ) = (∂f /∂x1(p), · · · , ∂f /∂xn(p)).

Then it isclear that ϕ(xi − ai ), i = 1, · · · , n form a basis of kn and ϕ(a2









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Proof.




2p → kn.

Now write X = V (b)and let f1, · · · , ft be a set of generators of b E k[x1, · · · , xn] (this can bedone because k[x1, · · · , xn] is Noetherian). Thenrk(∂fi/∂xj(p)) = dimϕ(b).Using the isomorphism ϕ, this is the same asdim b/(b ∩ a2






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Proof.





and let f1, · · · , ft be a set of generators of b E k[x1, · · · , xn] (this can bedone because k[x1, · · · , xn] is Noetherian).

Thenrk(∂fi/∂xj(p)) = dimϕ(b).Using the isomorphism ϕ, this is the same asdim b/(b ∩ a2






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Proof.





and let f1, · · · , ft be a set of generators of b E k[x1, · · · , xn] (this can bedone because k[x1, · · · , xn] is Noetherian). Thenrk(∂fi/∂xj(p)) = dimϕ(b).

Using the isomorphism ϕ, this is the same asdim b/(b ∩ a2






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Proof.







p.

Now, OX ,p∼= k[X ]ap . Hence if m is the




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Proof.







p. Now, OX ,p∼= k[X ]ap .

Hence if m is themaximal ideal of OX ,p, we have m/m2 ∼= ap/(b + a2

p). Countingdimensions of vector spaces, we have dimm/m2 + rk(∂fi/∂xj(p)) = n. Theresult follows immediately.


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Proof.








maximal ideal of OX ,p, we have m/m2 ∼= ap/(b + a2p).

Countingdimensions of vector spaces, we have dimm/m2 + rk(∂fi/∂xj(p)) = n. Theresult follows immediately.


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Proof.









dimensions of vector spaces, we have dimm/m2 + rk(∂fi/∂xj(p)) = n.

Theresult follows immediately.


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Proof.











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In fact, an even stronger result is true:

Theorem

Let X ⊆ Ank be an affine variety with corrdinate ring k[X ] ∼= O(X ), then

X is nonsingular at p iff the localization k[X ]mp∼= OX ,p of k[X ] at mp is a

UFD.



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In fact, an even stronger result is true:

Theorem

Let X ⊆ Ank be an affine variety with corrdinate ring k[X ] ∼= O(X ), then

X is nonsingular at p iff the localization k[X ]mp∼= OX ,p of k[X ] at mp is a

UFD.



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To summarize:

Theorem

Let X = V (f1, · · · , fm) ⊆ Ank be an affine variety and let p ∈ X . Then the

following are equivalent:(1) X is smooth (or nonsingular) at p,(2) the m × n Jacobian matrix has rank n − dimX , i.e.

rk

(∂fi∂xj

(p)

)1≤i≤m1≤j≤n

= n − dimX ,

(3) dimk[X ]/mp(mp/m

2p) = dimX ,

(4) The localization k[X ]mp∼= OX ,p of k[X ] at mp is a UFD.

Food for thought: Verify that the localization of k[x , y ]/(y2 − x3)and k[x , y ]/(y2 − x3 − x2) at p = (0, 0) is not a UFD.


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To summarize:

Theorem



rk

(∂fi∂xj

(p)

)1≤i≤m1≤j≤n

= n − dimX ,


2p) = dimX ,




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To summarize:

Theorem



rk

(∂fi∂xj

(p)

)1≤i≤m1≤j≤n

= n − dimX ,


2p) = dimX ,




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References

[1] M. Atiyah, Introduction to Commutative Algebra, Westview Press,1994.

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

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

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

[5] K. Ueno - Algebraic Geometry 1: From Algebraic Varieties to Schemes,American Mathematical Society, 1999.


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Thanks

Happy Easter everyone!


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