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.