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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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    `\emph` does not work in this site; for italics, use either `*` or `_`, like so: `*italics*` gives *italics*, as does `_italics_`, which yields _italics_. For boldface, use `**`: typing `**bold**` will give **bold.**2011-08-24
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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 $k[V]$?2011-08-24
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    Good point -- I should have read more carefully. I'll think about it.2011-08-24
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    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).2011-08-25
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    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.2011-08-25

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