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I am trying to prove the following:

Let $R$ be a commutative ring and $M$ be a free $R$-module having a finite basis of $n$ elements. Let $T:M\to M$ be a surjective $R$-module homomorphism. Then $T$ is injective.

Let $I$ be a maximal ideal of $R$. Note that $I$ annihilates $\widehat M:=M/IM$ and thus $\widehat M$ is an $\bar (R/I)$-module. Write $\bar R=R/I$.

For each $r\in R$, write $\bar r$ to denote $r+I$ and for each $m\in M$ write $\widehat m$ to denote $m+IM$.

Define $\widetilde T:\widehat M\to \widehat M$ as $$\widetilde T(\widehat m)=\widehat{Tm},\quad \forall \widehat m\in \widehat M$$ It is easy to see that $\widetilde T$ is well defined. Note that $\widetilde T$ is a surjective linear operator on a vector space and hence is also injective.

Note that if $\mathcal B=\{e_1,\ldots,e_n\}$ is a basis of $M$, then $\widehat B=\{\widehat e_1,\ldots,\widehat e_n\}$ is a basis for $\widehat M$.

Now suppose $T(r_1e_1+\cdots+r_ne_n)=0$.

Then we get $\widetilde T(\bar r_1\widehat e_1+\cdots+\bar r_n\widehat e_n)=\widehat 0$. Since $\widetilde T$ is injective, we get $\bar r_i=\bar 0$ for all $i$. But this doesn't lead to $r_i=0$.

Can anybody see how to complete the proof from here?

caffeinemachine
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    Have a look at this post including the comments. – Ben Nov 01 '14 at 20:16
  • I do not know anything about tensors so I haven't read it. – caffeinemachine Nov 01 '14 at 20:18
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    Well, the $M\otimes k$ in the other question is just your $\widehat{M}$. – Ben Nov 01 '14 at 20:23
  • Thanks. But it is still too advanced for me. I have just started reading module theory. Read the concept of a free module and some obvious questions came to mind. – caffeinemachine Nov 01 '14 at 20:30
  • Consensus of the comments to the other question: you might be loosing too much information by looking at the residue fields. If you don't know the Nakayama lemma, then I think you have to make use of freeness. Suppose it were vector spaces, how would you construct the inverse map? – Ben Nov 01 '14 at 20:32
  • In case of vector spaces I had the rank-nullity theorem. Let me see if an inverse map can be directly constructed. Never thought of it this way. – caffeinemachine Nov 01 '14 at 20:35

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