# Is being finitely generated a local property?

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!

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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

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 !]

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@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
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.

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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!

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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