Coordinate Rings as k-algebras
Let k be an algebraically closed field. Recall that a closed subset is identified by its coordinate ring k[V], which is a finitely generated k-algebra since
An affine k-variety is a finitely generated k-algebra A which is a reduced ring. Formally, we write V for the variety and for the algebra instead.
We say V is irreducible if A is an integral domain. A morphism of affine k-varieties is a homomorphism of the corresponding k-algebras
Thus, each closed set V gives an affine variety; isomorphic closed sets give isomorphic affine varieties. Conversely, we have:
Any affine k-variety A is the coordinate ring of some closed .
Indeed since A is finitely generated as a k-algebra, there is a surjective homomorphism (of k-algebras) ; its kernel is a radical ideal because A is a reduced ring. Hence . ♦
Note that taking the set:
recovers the set of points for B = k[V]. This gives another application of the “Hom” construction.
Now, the reader might question the utility of this point of view, since we are just back to dealing with closed subsets . The advantage here is that now we can expand the class of objects of interest.
Recall that to study multiplicity of intersection, we should really be looking at general ideals of k[V] and not just radical ones. Hence we define the following.
An affine k-scheme is a finitely generated k-algebra A. Again we write V for the scheme and .
An affine k-scheme V is a k-variety if and only if is reduced.
From our understanding of closed sets, we may now define the following for any affine k–schemes V and W. Note that now we have the capacity to define more general constructions (see the last few rows).
|What we say||What we actually mean|
|P is a point on V.||is a maximal ideal of k[V]; equivalently we have a k-algebra homomorphism .|
|is a morphism of affine k-schemes.||is a homomorphism of k-algebras.|
|W is a closed subvariety of V.||for some radical ideal of .|
|W is an irreducible closed subvariety of V.||for some prime ideal of .|
|is a set-theoretic intersection in V.||For , we have .|
|is a union in V.||For , we have .|
|W is a closed subscheme of V.||for some ideal of .|
|is a scheme intersection in V.||For , we have .|
Let A and B be any rings. Prove the following.
- An ideal of must be of the form , where (resp. ) is an ideal of A (resp. B).
- A prime ideal of must be of the form or , where (resp. ) is a prime ideal of A (resp. B).
- A maximal ideal of must be of the form or , where (resp. ) is a maximal ideal of A (resp. B).
Define the corresponding construction for disjoint union in the above table.
The affine k-variety for a singleton point is just . A point on an affine k-scheme V is thus a morphism .
Now we take the affine k-scheme # such that . We will write this as with . Geometrically, corresponds to a “small perturbation” such that . Note that # has exactly one point, so there is a unique map ; the corresponding homomorphism takes .
We imagine # as a point with an arrow sticking out. Thus takes the point to a point P on V and the arrow to a tangent vector at P.
A tangent vector on the k-scheme V is a morphism of k-schemes .
The base point of the tangent vector is given by composing . The tangent space of a point is the set of all tangent vectors with base point P, denoted by .
In terms of k-algebras, a tangent vector is a k-algebra homomorphism
If has base point P, then in the composition . This means and so . Thus factors through the following:
So it suffices to consider all k-algebra homomorphisms which satisfies . This gives a k-linear map , i.e. the dual space of as a k-vector space.
Conversely, a k-linear map also gives a k-algebra homomorphism by mapping k to k. Hence we have shown:
is parametrized by the set of all k-linear maps . In particular, it forms a finite-dimensional vector space over k.
Explain why is finite-dimensional over k.
Consider a simple example: with . A point corresponds to . A tangent vector at P then corresponds to a k-linear map
But is generated by so g is uniquely determined by and . Clearly .
To compute the resulting we take, for each ,
ignoring the terms of degree 2 or more in . Hence
Suppose is cut out by , so . Take the point P = (0, 0), which corresponds to . Let us compute the space of tangent vectors at P.
[Edited from GeoGebra plot.]
As before we have . Instead of looking at and , we look at their preimages as ideals of . (Remember the correspondence of ideals between a ring and its quotient!) Thus
since Y is a basis for this space. Hence .
Now take cut out by so . Take P = (0, 0) again.
[Edited from GeoGebra plot.]
As before and . Again, we take their preimages and as ideals of . This time we get and so
Geometrically, the tangent space at P is larger than the dimension when there is a singularity at P, as the reader can see from the graphs above. Thus this gives an algebraic definition of singularity at various points. We will of course need an algebraic definition of dimension for this.
In each of the following varieties, compute the dimension of the tangent space at the origin. Intuitively, which varieties are singular at the origin?