# Introduction to Algebraic Geometry (3 in a series)

Published: Tue 04 March 2008

I'm going to move on with the identification of geometric objects with functions from these objects down to a field soon enough, but I'd like to spend a little time nailing down the categorical language of this association. Basically, we have two functors I and V going back and forth between two categories. And the essential statement of the last post is that these two functors form an equivalence of categories.

Now, first off in this categorical language, I want to nail down exactly what the objects are. In the category $$\mathcal{AV}ar_k$$ the objects are solution sets of systems of polynomial equations. And in the category $$\mathcal{RA}lg_k$$, the objects are finitely presented Noetherian reduced k-algebras.

The functor $$V:\mathcal{RA}lg_k\to \mathcal{AV}ar_k$$ acts on objects by sending an algebra R to the solution set of the polynomial equations generating the ideal in a presentation of the algebra.

And the functor $$I:\mathcal{AV}ar_k\to \mathcal{RA}lg_k$$ takes a variety to the algebra of polynomial functions on the variety. This is a slight modification to the way I've introduced I in the previous posts - but the good news is that I(V) is the quotient of the right polynomial ring with the previously defined I(V).

## Morphisms on varieties

In order to define a category, it's not enough with the objects - we want morphisms as well. Since everything else is defined by polynomials, we're going to define a morphism of varieties $$V\to W$$ to be a map $$\mathbb A^n\to\mathbb A^m$$, polynomial in each coordinate, such that the image of the restriction to the variety $$V\subseteq\mathbb A^n$$ is contained in $$W\subseteq\mathbb A^m$$.

In other words, it is a map $$(x_1,\dots,x_n)\mapsto(f_1(x_1,\dots,x_n),\dots,f_m(x_1,\dots,x_n))$$ defined by polynomials $$f_1,\dots,f_m$$.

An isomorphism of varieties is exactly what we expect it to be - it is a morphism that has an inverse.

One specific kind of highly interesting isomorphisms are the linear automorphisms of $$\mathbb A^n$$. These are given, essentially, by invertible change-of-coordinate matrices in the way we are used to from linear algebra.

### Examples

Recall from the earlier posts the twisted cubic curve $$V(x-y^2,x-z^3)$$. Points on it have the form $$(s,s^2,s^3)$$ - and this, incidentially, gives us precisely a map $$\mathbb A^1\to\mathbb A^3$$ displaying an isomorphism between the twisted cubic curve and the affine line. The inverse is given by $$(x,y,z)\mapsto x$$.

Consider the parabola $$V(x-y^2)$$. This is also isomorphic to the affine line, over the maps $$t\mapsto(t,t^2)$$ and $$(x,y)\mapsto x$$.

On the other hand, the affine line is not isomorphic to the nodal curve $$V(y^2-x^2-x^3)$$. The easiest way to show this is to go over smoothness of curves and singular points - which I hope to deal with later at some point. Essentially, smoothness is an invariant of varieties under isomorphisms, and since the point (0,0) is singular on the nodal curve, and the affine line has no singular points, the two varieties can not possibly be isomorphic.

Note that images of varieties need not be affine algebraic varieties - they will, however, always be quasi-projective varieties. We'll see if I get into this later on.

## Morphisms of algebras and functoriality of V and I

We really do already know what morphisms look like in $$\mathcal{RA}lg_k$$. This category is the full subcategory of the category of k-algebras - by which we mean that it picks out objects among k-algebras, and have all k-algebra maps between objects as morphisms.

The really awesome bit happens when we start considering the morphisms we've defined. Given a morphism $$F:V\to W$$, we define the pullback $$F^\#:k[W]\to k[V]$$ by $$F^\#(f)=f\circ F$$. This takes a map $$f:W\to k$$ and makes a map $$F^\#(f):V\to k$$. Since this is a composition of polynomials, it is also a polynomial function. If $$f\in I(V)$$, then $$F^\#(f)(p)=f(F(p))$$, and since $$F(p)\in V$$, it follows that $$f(F(p))=0$$, and thus $$F^\#(f)\in I(W)$$.

In the other direction, suppose that R and S are reduced finitely generated k-algebras. Then $$R=k[x_1,\dots,x_r]/I$$, and $$S=k[y_1,\dots,y_s]/J$$. We fix a homomorphism $$\sigma:R\to S$$, and we wish to construct a variety morphism $$F$$ such that $$F^\#=\sigma$$.

Let $$F_i\in k[y_1,\dots,y_s]$$ be a representative of $$\sigma(x_i)$$, and define $$F:\mathbb A^s\to \mathbb A^r$$ by $$a\mapsto(F_1(a),\dots,F_r(a))$$. We need to verify that F maps V to W. This follows if we can only show that for every $$a\in V$$, all polynomials in I vanish on F(a). Let $$g\in I$$. Then
$$g(F(a))=g(F_1(a),\dots,F_r(a))=g(\sigma(x_1)(a),\dots,\sigma(x_r)(a))=\sigma(g)(a)$$
and since $$g\in I$$, it represents the zero class of $$k[x_1,\dots,x_r]/I$$, so $$\sigma(g)=0$$ and hence $$\sigma(g)\in J$$. But J is the ideal of all functions vanishing on V. Hence $$F(a)\in W$$, and the proof is complete.

In essence, what this proves to us is that the operations V and I form a contravariant equivalence of categories between $$V:\mathcal{RA}lg_k$$ and $$\mathcal{AV}ar_k$$.