Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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.

share|improve this answer
    
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
1  
@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
1  
@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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.