# Basic question about the definition of a finite map

I've been looking through Milne's notes, and I've gotten a bit tripped up by the section on finite maps. This is just a basic question about the definition, so I hope it's not too stupid.

A regular map of algebraic varieties $\varphi: W \rightarrow V$ is finite if for all open affines $U \subset V$, $\varphi^{-1}(U)$ is an affine variety and $k[\varphi^{-1}(U)]$ is a finite $k[U]$-algebra.

I'm trying to understand the basic case when $W,V$ are affine algebraic varieties and $k[W]$ is a finite $k[V]$-algebra. This is a finite map, but I'm having a bit of trouble digesting how this fits into the definition.

The explanation is that $k[\varphi^{-1}(U)] \simeq k[W] \otimes_{k[V]} k[U]$; I know this holds if $k[W] \otimes_{k[V]} k[U]$ is reduced, but how do we know that?

Milne says this implies that $\varphi^{-1}(U) \rightarrow \text{Spm}(\Gamma(\varphi^{-1}(U), \mathcal{O}_W)$ is an isomorphism (showing that $\varphi^{-1}(U)$ is affine). As I understand it, this follows from the general fact that

$$\text{Spm}(A) \times_{\text{Spm}(R)} \text{Spm}(B) \simeq \text{Spm}(A \otimes_R B / \mathfrak{N})$$

applied to the situation above, where we have $W \simeq \text{Spm}(k[W])$, etc. due to affineness.

So to summarize, I'm wondering why the isomorphism $k[\varphi^{-1}(U)] \simeq k[W] \otimes_{k[V]} k[U]$ holds (why is the RHS reduced?) and if my reasoning for how this implies that $\varphi$ is a finite map (in particular, that $\varphi^{-1}(U)$ is affine) is correct.

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@Dylan -- as stated, Corollary 18.4 applies only to algebras over a field; is it still true when the tensor product is over $k[V]$? – Tony Aug 24 '11 at 21:04
Good point -- I should have read more carefully. I'll think about it. – Dylan Moreland Aug 24 '11 at 21:06
Dear Tony: The algebra you describe will be reduced (think of the case when $V$ is a basic open subset of $U$; in this case it is a localization, and $V$ can be covered by basic open subsets). In general, though, the tensor product of two reduced rings over another ring (even when all are f.g. algebras over an alg. closed field) is not reduced (think of the fiber of the projection from the parabola $x = y^2$ to the $x$-axis over zero). – Akhil Mathew Aug 25 '11 at 2:13
Incidentally, the general claim about finite maps you write is true for schemes; it is a special case of the so-called "affine communication lemma" (see the notes by Ravi Vakil) where you want to show that if an affine cover has a certain property, then every open affine has this property. – Akhil Mathew Aug 25 '11 at 2:22

The trick here is to realize that, given a morphism of varieties $\phi:W\to V$ and an open subset $U\subset V$, affine or not,the fiber product $U\times_V W$ is $\phi^{-1}(U)$, with its obvious morphisms to $U$ and $W$.
As for finite morphisms, it is very easy to check finiteness of a morphism $\phi:W\to V \:$ if $V$ and $W$ are affine varieties . Namely $\phi$ is finite iff the corresponding ring morphism $\phi:k[V] \to k[W]$ is finite.
In other words when the varieties are affine, you don't have to check that for all open affines $U \subset V$, $\varphi^{-1}(U)$ is an affine variety and $k[\varphi^{-1}(U)]$ is a finite $k[U]$-algebra: this is automatic (but not at all trivial), as mentioned in Proposition 8.2.
Finally, since you are learning the subject, you might enjoy the little exercise of proving that the projection $\phi:(x,y)\mapsto x$ of the hyperbola $W=\{(x,y)\in \mathbb A^2_k|xy=1\}$ into the affine line $V=\mathbb A^1_k$ is not finite even though the fibers of this morphism are certainly finite sets (singletons or $\emptyset$)