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Let $A$ be a commutative ring and let $B$ be a finite $A$-algebra. Let $f:A \to B$ be a ring homomorphism. I want to show that whenever $\mathfrak{p} \subseteq A$ is a prime ideal, then there are finitely many prime ideals $\mathfrak{q} \subseteq B$ such that $f^{-1}(\mathfrak q) = \mathfrak p$. Now I am a bit confused because I have been told that this is a local property so I can assume that $A$ is a local ring. However I am not sure why it is a local property or even what the corresponding statement is for each localisation of A at a prime ideal.

would really appreciate any advice on this, thank you.

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up vote 3 down vote accepted

This $f$ had better be the homomorphism giving $B$ its $A$-algebra structure. If it is allowed to be arbitrary then I don't think the statement is true: let $k$ be an infinite field, and define $f\colon k[x] \to k[x]$ by $p(x) \mapsto p(0)$. Then $f^{-1}((x - a)) = (x)$ for all $a \in k$. [Geometrically, this is the map from the affine line to itself sending all closed points to the origin.]

To attack the problem, localize at $\mathfrak p$ and get a finite homomorphism $f_\mathfrak p\colon A_\mathfrak p \to B_\mathfrak p$. This fits into a commutative diagram of rings \begin{array}{ccc} A_\mathfrak p & \stackrel{f_\mathfrak p}\rightarrow & B_\mathfrak p \\ \uparrow & & \uparrow \\ A & \stackrel f\to & B \end{array} and there is a corresponding commutative diagram of spectra. I think the key is to draw that diagram and figure out what you know about the four maps involved.

The ring $B_\mathfrak p$ is canonically isomorphic to $T^{-1}B$, where $T$ is the multiplicative set $f(A - \mathfrak p)$. So the primes of $B_\mathfrak p$ correspond to the primes $\mathfrak q$ of $B$ such that $f^{-1}(\mathfrak q) \subset \mathfrak p$. [See Proposition 6.1 of Milne] And after staring at your diagram for a while you'll discover that $f_\mathfrak p^{-1}(\mathfrak qB_\mathfrak p) = \mathfrak pA_\mathfrak p$ if and only if $f^{-1}(\mathfrak q) = \mathfrak p$.

Now you can reduce to the case where $(A, \mathfrak p)$ is local, in which you never need worry that $f^{-1}(\mathfrak q) \not\subset \mathfrak p$. Can you think of a ring whose spectrum describes the primes of $B$ containing $\mathfrak pB$? Then you can apply the answers given by Andrea and I here.

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OK, so it's like $B_{\mathfrak p} \cong B \otimes_A A_{\mathfrak p }$ as an $A_{\mathfrak p}$-module. Thanks. I get confused by all the multiple structures the same objects have. So how does the multiplication work in that ring, and how do I know that $B_{\mathfrak p}$ is finite as an $A_{\mathfrak p}$-module? – Paul Slevin May 19 '12 at 18:42
@PaulSlevin That's even better: it's a tensor product of $A$-algebras, so the ring structure becomes clear. For the last bit: I claim that if $b_1, \ldots, b_n$ generate $B$ as an $A$-module, then their images in $B_\mathfrak p$ generate it as an $A_\mathfrak p$-module. – Dylan Moreland May 19 '12 at 19:03
OK this is good to know, thanks for all the clarification. – Paul Slevin May 19 '12 at 23:21
@PaulSlevin I think the example works, but I mixed up the ideal $(0)$ and the point $0 = (x)$ because I'm a moron. I tried to get stuff out of the comments [Hence making it look like you're having a conversation with yourself; sorry about that.] and into the answer; let me know whether it makes sense. – Dylan Moreland May 20 '12 at 0:33
I deleted my comments too to clean things up a bit. I just had a question about this multiplicative set business - when we take $f(A\setminus \mathfrak p)$, is it possible for $0$ to be in there? In my course we defined a multiplicatively closed set could not contain 0. I can see that if $0$ is in $f(A\setminus \mathfrak p)$ then $B_{\mathfrak p} =0$, but not why $B = 0$. – Paul Slevin May 20 '12 at 14:37

You can suppose that $A$ is not only a local ring, but actually a field. Let $f \colon A \to B$ be a finite ring homomorphism, let $\mathfrak{p}$ be a prime ideal of $A$, let $f^* \colon \mathrm{Spec} B \to \mathrm{Spec} A$ the associated map, i.e. $f^*(\mathfrak{q}) = f^{-1}(\mathfrak{q})$ for every prime ideal $\mathfrak{q}$ of $B$.

(1) You can suppose that $A$ is local and $\mathfrak{p}$ is the unique maximal ideal of $A$. Consider $S = A \setminus \mathfrak{p}$. $S^{-1}f \colon A_\mathfrak{p} \to S^{-1}B$ is a finite homomorphism of rings. You should check that $(f^*)^{-1}(\mathfrak{p})$ is in bijection with $((S^{-1}f)^*)^{-1}(\mathfrak{p}A_{\mathfrak{p}})$. Hence you can replace $A$ and $\mathfrak p$ with $A_\mathfrak{p}$ and $\mathfrak{p} A_\mathfrak{p}$.

(2) You can suppose that $A$ is a field and $\mathfrak p = 0$. $f$ induces a finite homomorphism $\bar{f} \colon A/\mathfrak{p} \to B/\mathfrak{p}B$. Check that the fiber of $f^*$ over $\mathfrak{p}$ is in bijection with the fiber of $\bar{f}^*$ over $\overline{\mathfrak p} = 0$. You can replace $A$ with the field $A/\mathfrak{p}$ and $B$ with $B/\mathfrak{p}B$.

Steps (1) and (2) can be unified in a unique step: replace $A$ with $k(\mathfrak{p})$, and $B$ with $B \otimes_A k(\mathfrak{p})$, where $k(\mathfrak{p})$ is the residue field of $\mathfrak{p}$. This is described in Proposition 3.1.16 of Liu's Algebraic geometry and arithmetic curves.

The proof continues in my answer here.

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Very similar to my answer, but better :) I love that book of Liu's. – Dylan Moreland May 19 '12 at 17:22
So $S^{-1} f : A_{\mathfrak p}\to S^{-1} B$ finite means that it is a ring homomorphism, and $S^{-1} B$ is a finite $A_{\mathfrak p}$-algebra right? How do I know that this is the case? – Paul Slevin May 19 '12 at 18:06
Yes, it means that. You can see that this is the case as follows: if $M$ is a finite $A$-module, then $S^{-1}M$ is a finite $S^{-1}A$-module. – Andrea May 19 '12 at 18:45
OK Thanks. Also, I am having trouble seeing that $(f^*)^{-1}(\mathfrak p)$ is in bijection with $( ( S^{-1}f)^* )^{-1} ( \mathfrak p A_{\mathfrak p})$, how do I see this? – Paul Slevin May 19 '12 at 20:04
You should think of that on your own! Hint: use the correspondence between prime ideals of $\mathfrak{p}A_{\mathfrak{p}}$ and prime ideals of $A$ which are contained in $\mathfrak{p}$. – Andrea May 20 '12 at 7:39

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