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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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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?2012-05-19
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    @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.2012-05-19
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    OK this is good to know, thanks for all the clarification.2012-05-19
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    @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.2012-05-20
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    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 $B = 0$.2012-05-20
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    Also, we proved that if $A$ is a ring and if $S$ is a multiplicatively closed subset (not containing 0) then primes in $S^{-1} A$ correspond to primes in $A$ which do not intersect $S$. Does this still hold if $S^{-1} A$ is the zero ring? - I don't think so because $S$ contains 0 means that any prime must intersect it. Is there still a correspondence?2012-05-20
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    @Paul Hi again. The correspondence seems to remain true, since the zero ring does not have any prime ideals. I'll have to think about your other question. I'm worried that I'll say something that contradicts lying over, but the key is probably that $f$ need not be injective.2012-05-20
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    @Paul I think $0 \in f(A - \mathfrak p)$ can happen in natural examples. I believe $f\colon k[x, y] \to k[y]$ sending $p(x, y)$ to $p(0, y^2)$ is finite, but there are no primes of $k[y]$ that pull back to $(y)$.2012-05-20
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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.2012-05-19
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    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?2012-05-19
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    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.2012-05-19
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    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?2012-05-19
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    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 $\mathfrak{p}$.2012-05-20
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    Thanks I understand that part now. The problem was that in my course we proved the correspondence abstractly (using a universal property) so I wasn't sure where prime ideals got send under the map. Got it now :)2012-05-20