# Morphisms of affine (embedded) varieties = morphisms as quasi-projective varieties?

The following question occurred to me while reading An invitation to algebraic geometry by Smith (2000). The definitions are as in the book (and not up to debate). Everything is over $$\mathbb{C}$$ here.

An affine variety $$V \subset \mathbb{A}^n$$ is the common zero set of polynomials. A morphism $$V \to W$$ of affine varieties $$V \subset \mathbb{A}^n$$ and $$W \subset \mathbb{A}^m$$ is a map $$F \colon V \to W$$ of the form $$x \mapsto (F_1(x),\ldots,F_m(x))$$ for a collection $$F_1,\ldots,F_m$$ of polynomials in $$n+1$$ variables, i.e. a (global) polynomial map.

A quasi-projective variety is a locally closed subset of $$\mathbb{P}^n$$. A morphism $$V \to W$$ of quasi-projective varieties $$V \subset \mathbb{P}^n$$ and $$W \subset \mathbb{P}^m$$ is a map $$F \colon V \to W$$ such that for every $$p \in V$$ there is an open neighborhood $$U$$ of $$p$$ and homogeneous polynomials $$F_0,\ldots,F_m$$ such that $$F|_U$$ is the map $$q \mapsto [F_0(q),\ldots,F_m(q)]$$, i.e. a local homogeneous polynomial map.

Any affine variety is quasi-projective (consider $$\mathbb{A}^n \subset \mathbb{P}^n$$).

Question. Consider affine varieties $$V \subset \mathbb{A}^n$$ and $$W \subset \mathbb{A}^m$$. Are morphisms $$V \to W$$ of affine varieties (first definition) the same thing as when considering them as quasi-projective (second definition)?

I think I can prove one direction: if $$F \colon V \to W$$ is a morphism of affine varieties, then by "homogenization" with the highest degree of the $$F_i$$ one gets a morphism $$\tilde{F} \colon V \to W$$ as quasi-projective varieties which is equal to $$F$$.

But I cannot prove the converse, which I strongly believe is true as well. I would need a global collection of polynomials from the local collections. How can one show this (on the level of the book!)?

I find this to be an obvious question by there is no comment in the book.

Remark: The question is not about the fact that an affine variety (in our sense) could be isomorphic to a non-affine (in our sense) quasi-projective variety, e.g. $$\mathbb{A}^1 \setminus \{ 0 \}$$ is isomorphic to the cone as quasi-projective varieties. The first one is not affine in our definition since it is an open subset and zero sets of polynomials have empty interior.

• This is true. Note that Hartshorne gives a totally local definition of morphisms of affine varieties (probably equivalent to your second definition) and then proves that this is equivalent to your global definition. The point is that a morphism of affine varieties is equivalent to a morphism of their coordinate rings (going in the opposite direction). Check the relevant section of his book (somewhere in chapter I). Nov 29, 2023 at 17:09

## 1 Answer

Let $$f: V \to W$$ be a morphism of affine varieties (considered as quasiprojective varieties) according to the second definition. This means we have embedded $$V$$ and $$W$$ in the chart $$\mathbb{A}_n \cong U_0 = \{x_0 = 1\} \subset \mathbb{P}^n$$. Since $$x_0 = 1$$ on $$V$$ and $$W$$, you don't gain anything by assuming that the morphism is locally given by homogeneous polynomials, as opposed to just locally given by any polynomials at all. So there is an open cover $$\{V_i\}_{i \in I}$$ of $$V$$ with the morphism given by polynomials $$\hat{F}_i = (F_{i1}(x_1,...,x_n),...,F_{in}(x_1,...,x_n))$$ on $$V_i$$. For all $$i, j \in I$$, and $$1 \le k \le n$$, we have $$F_{ik} = F_{jk}$$ on $$V_i \cap V_j$$. $$V$$ is irreducible, so any nonempty open subset is dense, and this implies that $$F_{ik} = F_{jk}$$ on all of $$V$$. Therefore the morphism is given on all of $$V$$ by $$\hat{F}_i$$, for any fixed choice of $$i$$.

• An affine variety is not assumed to be irreducible. Does this work without this assumption?
– Guest439
Nov 30, 2023 at 6:51
• There's also something that puzzles me: wouldn't the exact same argument prove that morphisms between (irreducible) quasi-projective varieties are globally given by (homogeneus) polynomials? This is certainly not true.
– Guest439
Nov 30, 2023 at 6:55
• Ok, I think all this comes from the fact that "locally polynomial" maps of affine varieties are already globally polynomial, and this comes from the fact that regular functions on an affine variety are global polynomial functions. This fact holds for reducible varieties as well. This is unfortunately not well-explained in the book I find...
– Guest439
Nov 30, 2023 at 9:31
• @Guest439 The usual definition of an affine variety assumes it to be irreducible. This doesn’t prove that morphisms between any quasi-projective varieties are globally given by polynomials, because the proof uses the fact that any polynomial function on a subset of an affine variety extends to a well-defined function on the entire variety. What the proof shows is that the extension is unique Nov 30, 2023 at 16:21
• The statement holds in much more generality for affine schemes (Hartshorne chapter 2, prop 2.2). Not sure where to find a reference for just the case of reducible affine algebraic sets over $\mathbb{C}$, but Hartshorne’s proof is straightforward commutative algebra once you know the definition of the structure sheaf. I imagine any proof of the special case you want will be essentially the same as his general proof Nov 30, 2023 at 16:50