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Let $\phi:R\to S$ be a finite ring homomorphism, in the sense that $S$ is a finitely generated module over $R$. Then $\phi^*:\text{Spec }S\to\text{Spec }R$ has finite fibers.

I have not found a proof anywhere of this relatively simple fact, and to me, it's not completely obvious.

Let $\phi$ be as above. Consider any prime $\mathfrak p\in\text{Spec }R$ and primes $\mathfrak q_i\in\text{Spec }S$ mapped to $\mathfrak p$ under $\phi^*$, ie the points in the fiber, or rather $\mathfrak p=R\cap\mathfrak q_i$. If $\mathfrak q_i\subset \mathfrak q_j$, we automatically have $\mathfrak q_i=\mathfrak q_j$, since this is the case for integral ring homomorphisms, and finite homormorphisms are integral. Hence, we can assume the $\mathfrak q_i$ are not subsets of each other. Passing to localisation by $\mathfrak p$ in each ring, we get an induced ring homomorphism $\phi_i:R _\mathfrak p\rightarrow S_{\mathfrak q_i}$, which is finite since $\phi$ is finite. Further, taking the quotient by the maximal ideal, we get an induced morphism of fields $k(\mathfrak p)\hookrightarrow k(\mathfrak q_i)$, which is a finite field extension again.

I'm stuck here. If it were the case that there was a slightly larger finite field extension of $k(\mathfrak p)$ containing all the $k(\mathfrak q_i)$, then I could imagine that this directly proves the theorem since there are only finitely many intermediate fields.

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    $\begingroup$ An easier proof: compute the fiber via a tensor product. Since this preserves finite generation as a module, you get that the ring whose spec is your fiber is a finite module over the residue field which then must have finitely many prime ideals. $\endgroup$ – KReiser May 31 at 18:28
  • $\begingroup$ @KReiser I'm not very sure what you mean (or rather, how you get) by the fiber via a tensor product. $\endgroup$ – George Jun 2 at 23:20
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An easier solution to this is to compute the fiber of this morphism and observe that it's finite. The fiber of the morphism $\operatorname{Spec} S\to\operatorname{Spec} R$ over a point $x\in \operatorname{Spec} R$ is given by the fiber product $\operatorname{Spec} S \times_{\operatorname{Spec} R} \{x\}$. In order to calculate what this fiber product is, first we note that this fiber product is affine - it's a closed subscheme of an affine scheme, which is automatically affine. Next, we use the fact that the category of affine schemes is dual to the category of commutative rings to see that we're looking to compute the pushout of the diagram $k\leftarrow R\to S$, where the left arrow is the map given by modding out by the maximal ideal determining $x$.

Pushouts in commutative rings are computed via tensor products, so we know that the fiber over our point $x$ is going to be $\operatorname{Spec}$ of the tensor product $S\otimes_{R} k$. Since $S$ is finitely generated over $R$ as a module, we see that this tensor product is a $k$-algebra which is finitely generated $k$-module. As a $k$-algebra which is finite-dimensional $k$-vector space can only have finitely many maximal ideals, we see that the fiber has finitely many points.

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  • $\begingroup$ Although I tagged it as AG and I appreciate the answer, it uses much more machinery than what is (supposedly) needed to answer this question. For reference, the question is asked in a strictly commutative-algebraic section of a book. In fact, tensor products are not defined yet. Perhaps it might be helpful to add that this is asked in the same section that the Cohen-Seidenberg going up/down theorems are proved. $\endgroup$ – George Jun 3 at 0:04

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