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## Affine varieties and regular maps

I have to admit I’ve been using somewhat unconventional definitions. The usual definition of an affine variety is as an irreducible Zariski-closed subset of $\text{MaxSpec } k[x_1, ... x_n] \simeq \mathbb{A}^n(k)$, affine $n$-space over an algebraically closed field $k$. A generic Zariski-closed subset is usually referred to instead as an algebraic set (although some authors also call these varieties), and the terminology does not apply to non-algebraically closed fields. The additional difficulty that arises in the non-algebraically-closed case is that it’s harder to think about points. For example, $\text{MaxSpec } \mathbb{R}[x]$ has two types of points corresponding to the two types of irreducible polynomials: the usual points $(x - a), a \in \mathbb{R}$ on the real line and additional points $(x^2 - 2ax + (a^2 + b^2)), a, b \in \mathbb{R}$. These points can be thought of as orbits of the action of $\text{Gal}(\mathbb{C}/\mathbb{R})$ on $\mathbb{C}$, hence $\text{MaxSpec } \mathbb{R}[x]$ can be thought of as the quotient of $\text{MaxSpec } \mathbb{C}[x]$ by this group action. This picture generalizes.

Anyway, for convenience let’s stick to $k = \mathbb{C}$. In this case, and more generally in the algebraically closed case, there is a reasonably simple description of what the category of affine varieties looks like, but first we have to describe what the morphisms look like and then we have to take the strong Nullstellensatz on faith, since we haven’t proven it yet.

Polynomial maps

As in the case of the algebras $C(X)$, we want any notion of a morphism between varieties to come from a homorphism between the corresponding algebras. How does one define the algebra of a variety? Well, to really do this we first need the strong Nullstellensatz.

Theorem (strong Nullstellensatz): $\mathbb{C}[x_1, ... x_n]$ is Jacobson.

Corollary: For polynomial rings, $I(V(J)) = \text{rad}(J)$.

I’d like to postpone the proof of the strong Nullstellensatz until after we introduce localization. For now, if $J$ is an ideal in $\mathbb{C}[x_1, ... x_n]$ of polynomial equations which describe a variety $V(J)$, then $I(V(J)) = \text{rad}(J)$, so a reasonable description of the affine coordinate ring of $V$ is as the quotient $\mathbb{C}[V] \simeq \mathbb{C}[x_1, ... x_n]/(I(V)) \simeq \mathbb{C}[x_1, ... x_n]/(\text{rad}(J))$. The definition of this quotient formalizes what it means for $\mathbb{C}[V]$ to be the ring of functions on $V$. Now recall that $V$ is irreducible if and only if $I(V)$ is prime and that $\text{rad}(J)$ contains all nilpotents, so it follows that $V$ is an affine variety if and only if $\mathbb{C}[V]$ is a reduced integral domain.

Now what we want to do is to define the category of affine varieties in such a way that it becomes equivalent to the opposite of the category of finitely-generated reduced integral domains over $\mathbb{C}$. The morphisms in this category are $\mathbb{C}$-algebra homomorphisms $\mathbb{C}[x_1, ... x_n]/I \to \mathbb{C}[y_1, ... y_m]/J$, and any such homomorphism is determined by the images of the generators $x_1, ... x_n$. These images must be elements of $\mathbb{C}[y_1, ... y_m]/J$, hence polynomials $f_1, ... f_n$ of $y_1, ... y_m$, and they must satisfy the conditions imposed by $I$. This is perhaps clearest with an example.

Example. There is a homomorphism $\phi : \mathbb{C}[x, y]/(x^2 + y^2 - 1) \to \mathbb{C}[t][t^{-1}] \simeq \mathbb{C}[t, s]/(ts - 1)$ given by $x \mapsto \frac{t + s}{2}, y \mapsto \frac{t - s}{2i}$. One can verify that these polynomials satisfy the constraint $x^2 + y^2 = 1$, hence this is a legitimate homomorphism, and in fact it has an inverse which is also given by polynomial maps and which therefore defines an isomorphism between the two algebras. This is the algebraic side of the isomorphism between the “complex circle” and the punctured complex plane.

Polynomial maps define maps between varieties in the opposite direction just by plugging in. For example, in the above example given any point $(t, s)$ on the variety $\mathbb{C}[t, s]/(ts - 1)$, we get a point $(x, y)$ on $\mathbb{C}[x, y]/(x^2 + y^2 - 1)$ just by using the formulas $x = \frac{t + s}{2}, y = \frac{t - s}{2i}$. A function between affine varieties given by polynomial equations in this way is called a regular map, and regular maps form the morphisms in the category of affine varieties. (Note that by essentially the same argument as in the previous post, regular maps are always continuous in the Zariski topology.) From an abstract perspective, the reason this works is that the weak Nullstellensatz guarantees that the quotient of a finitely-generated $\mathbb{C}$-algebra by a maximal ideal is $\mathbb{C}$, so given a maximal ideal and the corresponding quotient $\mathbb{C}[V] \to \mathbb{C}$, we can compose with a homomorphism $\phi : \mathbb{C}[W] \to \mathbb{C}[V]$ to get a homomorphism $\mathbb{C}[W] \to \mathbb{C}$ which must be surjective, hence which must define another maximal ideal.

In this way we have, almost by fiat, declared the following to be true.

Proposition: The category of complex affine varieties and regular maps is contravariantly equivalent to the category of finitely-generated reduced integral $\mathbb{C}$-algebras.

Note that, now that we have defined regular maps, we can define $\mathbb{C}[V]$ as the ring of regular functions $V \to \mathbb{C}$, so our treatment of affine varieties is now almost identical to our treatment of compact Hausdorff spaces, with regular functions substituting for continuous functions.

Subvarieties of the affine plane

A subvariety of a variety $V$ is just a variety contained in $V$, and the subvarieties of $V$ are in one-to-one correspondence with the prime ideals of $\mathbb{C}[V]$. The subvarieties of $\mathbb{A}^1(\mathbb{C})$ are straightforward to describe, since the prime ideals of $\mathbb{C}[x]$ are precisely $(0)$ and the maximal ideals $(x - a), a \in \mathbb{C}$: in other words, only the entire affine line and points are subvarieties. Similarly, the subvarieties of $\mathbb{A}^2(\mathbb{C})$ are precisely the prime ideals of $\mathbb{C}[x, y]$. It is straightforward to describe these, but the nicest way to give the proof is to replace $\mathbb{C}[x]$ by an abstract principal ideal domain $R$ and consider the prime ideals of $R[y]$. This approach has the distinct advantage of also working in the case $R = \mathbb{Z}$, which will become important.

Proposition: The prime ideals of $R[y]$ are precisely the zero ideal $(0)$, the ideals $(f(y))$ where $f(y)$ is irreducible, and the maximal ideals $(p, f(y))$ where $p \in R$ is prime and $f(y)$ is irreducible in $(R/p)[y]$.

Proof. Recall that in a principal ideal domain, every nonzero prime ideal is maximal. It follows that $R/p$ is a field for every prime $p$, hence $(R/p)[y]$ is an integral domain, so $(p)$ is prime in $R[y]$. Now, $(R/p)[y]$ is a polynomial ring over a field, hence its prime ideals are precisely the principal ideals generated by irreducible polynomials. The quotient of $(R/p)[y]$ by any such polynomial is a finite extension of $R/p$, so these give us the maximal ideals $(p, f(y))$. This exhausts the set of prime ideals of $R[y]$ containing some prime ideal of $R$.

Now recall that if $R$ is a UFD, then so is $R[y]$. It follows that a prime ideal is principal if and only if it is generated by an irreducible element. Now suppose that $P$ is a prime ideal which is not principal and which contains no prime ideal of $R$. By Gauss’s lemma any two polynomials $f, g \in R[y]$ with no common factor in $R[y]$ also have no common factor in $K[y]$, where $K$ is the fraction field of $R$. But then we know that there exist $p, q \in K[y]$ such that $pf + qg = 1$, hence clearing denominators we obtain a contradiction.

Corollary: The prime ideals of $\mathbb{C}[x, y]$ are $(0)$, the principal ideals $(f(x, y))$ where $f$ is irreducible, and the maximal ideals $(x - a, y - b)$.

So the subvarieties of the affine plane $\mathbb{A}^2(\mathbb{C})$ are the affine plane itself, its points, and the hypersurfaces $f(x, y) = 0$ for irreducible polynomials $f$. In the plane, hypersurfaces have (complex) dimension 1 in a sense that will be made precise later, so we will refer to them as algebraic curves. Dimension 1 is the simplest case after dimension 0 (points), so it is natural to begin there, and it is also the case that is relevant to understanding number rings since it turns out that those also have dimension 1 in the appropriate sense. In a sense weaker than isomorphism, it also turns out that every algebraic curve is equivalent to one in the affine plane.

### 6 Responses

1. Can you help clarify the proof of the second proposition above for me? If R is a PID then when you say p is a prime and R/p is a field do you mean p is a nonzero prime element, hence irreducible since R is a PID, and the ideal (p) is a prime ideal (hence maximal) so R/(p) is a field. So then R/(p)[y] is an integral domain (it’s a PID correct?). After that I’m kinda lost, still trying to work through this. I’m obviously not seeing something here because I’m confused about why you say (p) is prime in R[y], but it’s not one of the prime ideals in the statement of the proposition. Is it? Ugh……

• Wait… (p) prime in R[y] so p irreducible in R[y] so it’s one of the ideals (f(y)) for f(y) irreducible in R[y]?

2. What does it mean if pf + qg = 1? (the last proof)
Does that mean f, g dont have common factors and therefore R is UFD.

3. Hi, in your proof on the prime ideals of R[y], I don’t understand a line in the first paragraph: “The quotient of (R/p)[y] by any such polynomial is a finite extension of R/p, so these give us the maximal ideals (p, f(y)).” Could you explain why is the quotient a finite extension and how does it give the maximal ideals you state? Thanks!

• Oh, finite field extension is due to Nullstellensatz. It now remains to understand the maximal ideals.

• The Nullstellensatz is way overkill. $R/p$ is a field, so $(R/p)[y]$ is polynomials in one variable over a field. Quotienting by a polynomial of degree $n$ produces an $n$-dimensional vector space over $R/p$. This quotient is a field iff the polynomial is irreducible, and quotients which are fields give maximal ideals.