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Let $M$ be a finite module over a local integral domain $(A,m)$. Let $k$ be its residue field and $Q$ its fraction field. Consider the $k$-vector space $M \otimes_A k$ and the $Q$-vector space $M \otimes_A Q$.

Show that $\dim_Q (M \otimes_A Q) \leq \dim_k(M \otimes_A k)$ and the equality holds if and only if $M$ is free.

My old and wrong idea (1):

We have two natural maps $\pi : (A,m) \to k$ and $\eta : (A,m) \to Q$, so using the universal property of the localization ($\eta$ satisfies all the required conditions) we obtain a module morphism $\phi : Q \to k$. Notice that because the diagram commutes, i.e. $\phi \circ \eta = \pi$ and $\pi$ is surjective, the same must be for $\phi$. We have the the short exact sequence: $ 0 \to \ker{\phi} \to Q \to k \to 0$ (surjectivity of $\phi$). Using now the fact that the tensor functor is rigth-exact, we should obtain the exact sequence $ M \otimes_A \ker{\phi} \to M \otimes_A Q \to M \otimes_A k \to 0$ and in particular the last nontrivial map must be surjective. It leads to $\dim(M \otimes_a Q) \geq \dim(M \otimes_a k)$ (notice that the sign of the inequality is wrong). Please, could help me in finding the mistake? (done, thanks!)

My old and incomplete idea (2):

I am trying to solve completely the exercise. Here is my new point. Observe that since $(A,m)$ is locally $m$ is the only principal idea and so it coincides with the Jacobson radical (and, in particular, it is contained in it). Let's now recall that $M \otimes_A k \simeq M/mM$, and focus the attention on the second member. Since $M$ is finitely generated, the same will be for its quotient, so choose here a family of generators ($\bar{m_1}$,...,$\bar{m_t}$). Does it suffice to claim that $\dim_k(M \otimes_A k)$ is less (or equal) than $t$? I think not, but I am a bit confused because we are working with generators in an $A$-module sense, while now I am looking for a basis for a vector space (but the set is the same).Then, I continue:using Nakayama's lemma on the previous generators, we obtain that $m_1$,..,$m_t$ is a set of generator for $M$. Let's now analyze the vector space $(M \otimes_A Q)$. Is order to obtain a set of generator for it, I should take all possible combinations of generators for both the modules...but being a field, $Q$ has just one generator. So I should obtain something like "$\dim_K(M \otimes_A Q)$ less or equal than $t+1$...

Edit: if you are interested, you can find here a related post (An exercise on tensor product over an integral domain.).

Thank you in advance. Cheers

Community
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  • I can't see how you get a homomorphism $Q\to k$. – user26857 Apr 28 '15 at 18:30
  • I used the property explained on Atiyah-MacDonald, page 37, proposition 3.1 (I'll write it entirely in case you don't have the book). –  Apr 28 '15 at 18:40
  • I don't need any book to realize that you used in a wrong way the universal property of rings (modules) of fractions. (Your mistake is earlier than the one described by the answer below.) – user26857 Apr 28 '15 at 18:44
  • Ooops, I am really sorry for the misunderstanding in your previous comment . Please, may you explain why I was wrong? –  Apr 28 '15 at 18:49
  • You didn't check that all elements in $A-{0}$ are invertible in $k=A/m$ (and in fact they aren't!). – user26857 Apr 28 '15 at 18:50
  • Why not? Isn't $k$ a field? –  Apr 28 '15 at 18:52
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    Yes, it is! But what about the non-zero elements from $m$? – user26857 Apr 28 '15 at 18:53
  • Of course you are right! Thanks :-) –  Apr 28 '15 at 18:57
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    In my opinion you are far from a solution, that's why I give you few hints: $\dim_k M\otimes k$ equals the minimal number of generators of $M$ (use Nakayama), while $\dim_Q M\otimes Q$ is the maximal number of linearly independent elements in $M$. This could be a good start. – user26857 Apr 28 '15 at 20:22
  • Thanks for the hint - I'll think on it! Please, if possible, could you say me in any case if I made some particular mistake in the previous attempt? –  Apr 28 '15 at 20:28
  • Yes, I almost see it!!! To be more precise: "..equals the minimal number of generators of M (use Nakayama)" I did it and works! "...is the maximal number of linearly independent elements in M" I see that it is less or equal that number - that is however a good point to start too. Now my problem is to show that the second number is always less or equal than the first :-) –  Apr 29 '15 at 19:49

1 Answers1

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The right-exact sequence is one of $A$-modules. Note that in the claim, we have dimensions over two different fields. Also, as a hint, you might want to look at Nakayama's lemma.

Thomas Poguntke
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  • You're right!!! In other words, the fact that there is an A-Module surjection between $M \otimes_A Q$ and $M \otimes_A k$ doesn't allow use to compare their dimension as vector space (because they are vector spaces of different fields). Is it right? (I have repeated the idea just to be sure to have properly understood). Furthermore, thanks for the hint: I'll take a look at Nakayama's lemma...and I'll rewrite in case of problems. Cheers –  Apr 28 '15 at 18:45
  • Yes. Of course the surjection doesn't even exist, as explained in the comments - but even if it did... – Thomas Poguntke Apr 28 '15 at 18:54