Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi to everyone. I am trying ot figure out a proof of the following fact, that I believe is true, but it seems to me that something is lacking. Suppose we have commutative, unitary rings $A,B$ and a (unit preserving) homomorphism $\varphi \colon A to B$ which makes $B$ an $A$-algebra. Suppose also we have elements $f_1,\dots,f_n \in B$ which generate the unit ideal and such that $B_{f_i}$, namely the localisation of $B$ with respect to $f_i$, is a finitely generated $A$-algebra. Show that then $B$ is a finitely generated $A$-algebra. Could someone give me a rigorous proof of this fact (or a counterexample, if this is false)? Thank you!

share|improve this question
5  
This is a special case of Ex.II.3.3(c) in Hartshorne. Maybe a homework problem? –  Sándor Kovács Jan 14 '11 at 0:16

3 Answers 3

Dear Andrea, the fact you mention is indeed true. Consider the associated morphism of affine schemes $Spec (\phi) = f : X=Spec B \to Y=Spec A$. Your hypothesis on unit ideal generation translates into the fact that the open subsets $U_i= Spec B_{f_i}$ cover $X$. The finite generation of $B_{f_i} $ as an $A$-algebra shows that the scheme $U_i$ is of finite type over $Y$. Hence $X$ is of finite type over $Y$ according to Definition (6.3.1) on page 144 in EGA I. The result you request, that $B$ is finitely generated over $A$, follows from Proposition (6.3.3) on page 145 [ heartfelt thanks to roman who spotted that my initial references were mixed up]

Of course one could unpack all this and give a purely algebraic proof. But I think it is nice to see things geometrically and, in case you don't know elementary scheme theory yet, this might motivate you to learn it . At the level used here it is little more than a language in which to speak of algebra. [ By the way, don't be afraid of the hypothesis that the morphisms appearing in this context be quasi-compact: all morphisms between affine schemes are quasi-compact !]

share|improve this answer
    
@Georges: After all, the proof (after some reductions) is an algebraic one. –  Martin Brandenburg Jan 14 '11 at 7:50
    
Martin: absolutely, what else could it be? I view practically all of EGA as algebra formulated in a geometric way , in order to help the visual intuition of human beings. But this is certainly controversial and I wouldn't want to bore users with my fuzzy philosophy, especially not here... –  Georges Elencwajg Jan 14 '11 at 8:45
1  
I thought EGA is geometry formulated in an algebraic way. –  Martin Brandenburg Jan 14 '11 at 12:59
    
:-) . –  Georges Elencwajg Jan 14 '11 at 18:05
    
@Martin & Georges: algebra=geometry –  Sándor Kovács Jan 18 '11 at 5:42

Suppose we have an element $x\in B$. Then it's image in $B_{f_i}$ is equal to some $F^i( \frac{b^i_1}{f_i^{k^i_1}} ,\ldots ,\frac{b^i_{j_i}}{f_i^{k^i_{j_i}}})$, where $\frac{b^i_j}{f_i^{k^i_j}}$ are the finite set of generators of $B_{f_i}$ over $A$, with $b^i_j\in A$, and $F^i$ some polynomials with coefficients in $A$. After multiplying by a large power of $f_i$ this gives us $n$ equalities in $B$ looking like $f_i^{N}x=F'^i(b^i_1,\ldots,b^i_{j_i},f_i)$, again with coefficients in $A$. But as $f_i$ generate unit ideal in $B$, there is an expression of 1 in terms of $f_i$: $a_1f_1+\cdots+a_nf_n=1$, with $a_i\in B$. Exponentiate it to the nN-th power and multiply by $x$, and you'll get $x=G(f_1,\ldots,f_n,a_1,\ldots,a_n,\ldots b^i_j \ldots)$, with $G$ polynomial with coefficients in $A$ (because after exponentiation each monomial of $a_i,f_i$ include at least one $f_j$ in power greater or equal to $N$, so after multiplying by x we could substitute $F'^j$ for $xf_j^N$). So any $x\in B$ can be expressed as a polynomial of $a_i,f_i,b^i_j$ with coefficients in $A$, which means that $B$ is finitely-generated $A$-algebra.

Looking it up in EGA as Georges suggests is also a good idea, I just thought you might not be ready for that yet.

share|improve this answer
    
roman, you are absolutely right. I have edited my text according to your rectification: thanks a lot for your careful reading. By the way, it is great that you have given users a purely algebraic proof: they can now choose, according to their preferences, between an algebraic and a geometric formulation. –  Georges Elencwajg Jan 14 '11 at 8:49
    
Well, in this case I don't think there's a lot to choose from: from what I understood in EGA I (6.3.3), Grothendieck's proof is almost literally the same. I just wanted to provide an option of not looking in EGA: I myself don't read french, so I really don't like it when made to look there. –  14555 Jan 15 '11 at 1:30

Thank you very much for both answers! I already know a bit of scheme theory, so the reference in EGA is nice indeed. Anyway, I was really looking for a completely algebraic proof, and this one convinced me, so thanks again!

share|improve this answer
4  
If you like one of the answers, then you should vote it up and accept it, so that people know this is a "answered" question. –  Yemon Choi Jan 14 '11 at 5:37

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.