MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $M$ is a finitely generated module over a local ring $(R, \mathfrak{m})$, we can detect whether $M$ has a nonzero free direct summand as follows: Consider the natural map $$\phi_M\colon \mathrm{Hom}_R(M,\mathfrak{m}) \longrightarrow \mathrm{Hom}_R(M,R)$$ induced by the inclusion of $\mathfrak m$ into $R$. Then $M$ has a nonzero free direct summand if and only if $\\phi_m$ is not surjective. (If $M$ has no free direct summand, then the image of every homomorphism $M \longrightarrow R$ must be inside $\mathfrak m$, so $\phi_M$ is surjective. The converse is easy as well.)

I'd like a more precise statement to be true. The cokernel of $\phi_M$ is a submodule of $\mathrm{Hom}_R(M,R/\mathfrak{m})$, so is a finite-dimensional vector space.

Is the maximal rank of a free direct summand of $M$ equal to the dimension of $\mathrm{coker} \;\phi_M$?

One inequality is obvious -- a surjection $M \longrightarrow R^r$ will give a subspace of dimension $r$.

share|cite|improve this question
Isn't the other direction just Nakayama's lemma? – Eric Wofsey Mar 2 '13 at 17:15
Entirely possible I'm missing something silly. – Graham Leuschke Mar 2 '13 at 19:04
up vote 3 down vote accepted

If $F$ is a free direct summand of $M$ of maximal rank, then $M=F\oplus N$, where $N$ has no free direct summand. So $\mathrm{coker} \;\phi_M\cong\mathrm{coker} \;\phi_F\oplus\mathrm{coker} \;\phi_N$, and, as you've pointed out, the dimension of $\mathrm{coker} \;\phi_F$ is the rank of $F$ and $\mathrm{coker} \;\phi_N=0$.

share|cite|improve this answer

Your Answer


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.