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 $k$ be an algebraically closed field and $A$ a finitely generated $k$-algebra which is an integral domain. Let $m$ be a maximal ideal of $A$. Does the proof that $A_m/mA_m$ (i.e. the residue field of the localization of $A$ at $m$) is isomorphic to $A/m$ require anything deep like the Nullstelensatz? I.e. is there a basic ring theoretic argument that doesn't bring in any algebraic geometry or deeper results in commutative algebra? My difficulty is in proving the surjectivity of the map $A \rightarrow A_m/mA_m$. Thank you for your time.

share|improve this question
    
It is not hard to show that $A_\mathfrak{m} / \mathfrak{m}_\mathfrak{m} \cong A / \mathfrak{m}$, but to determine that $A / \mathfrak{m} \cong k$ (as $k$-algebras!) requires a form of the Nullstellensatz. –  Zhen Lin Oct 21 '12 at 7:53
    
Thanks Zhen. I've altered the question because I suppose I am willing to accept that A/m is isomorphic to k. –  Michael haneke Oct 21 '12 at 8:21
5  
This is an easy exercise in commutative algebra. Try a more general form (that could be easier to prove!): let $A$ be a commutative ring, $I\subset A$ an ideal and $S\subset A$ a multiplicatively closed set. Then $S^{-1}A/S^{-1}I$ is isomorphic to $T^{-1}(A/I)$, where $T$ is the image of $S$ in $A/I$ via the canonical projection. In particular, if $p$ is a prime ideal of $A$, then $A_p/pA_p$ is isomorphic to the fraction field of $A/p$. –  user26857 Oct 21 '12 at 8:35
    
@Michaelhaneke Do you know that localisation is an exact functor? You can try taking the localisation of the exact sequence $0 \rightarrow m \rightarrow A \rightarrow A/m \rightarrow 0$. –  user38268 Oct 21 '12 at 11:14
1  
@BenjaLim Localization is an exact functor in the category of modules. Here is needed a ring isomorphism. –  user26857 Oct 21 '12 at 17:31

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.