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I feel like one of the hypotheses in the problems I am working on is not necessary and I will explain why.

Let $R$ be a commutative ring with no divisors of zero. Show every nonzero element of a projective $R$ module is free.

Is the conclusion still true is we do not assume the module is projective?

My reasoning is as follows. Let $M$ be a projective $R$-module and let $x\in M$ be a nonzero element. To show $x$ is free it to show the set $\{x\}$ is free as a subset so it suffices to check the cyclic module $\langle x\rangle = Rx$ is free of rank 1. But this amounts to show there is an isomorphism $Rx \cong R$. This is easy because the map is trivially surjective and injective by the fact that there are no zero divisors... Where did we need projective here?

The only thing I see projectivity giving is the fact that $M$ is a direct factor of the direct sum $\oplus_{t\in T}R$ so we can get a representation for each $x = (r_t)_{t \in T}$ but I don't see how this representation is needed.

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I think we only need that $P$ is a submodule of a free module. If you not require anything, let $P=R/rR$, then $1$ is not free, right? –  wxu Sep 13 '11 at 8:56
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$\LaTeX$ tips: (i) to get the curly brackets to appear, you need to use the escape character: \{ and \} inside math mode. For the angle brackets, it is better to use \langle and \rangle and not < and >; the former are delimiters, the latter are operators. The spacing is completely different. –  Arturo Magidin Sep 13 '11 at 21:59

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You're right: if $R$ is an integral domain, then for all $x \in R$, the principal ideal $(x)$ is a free $R$-module. More specifically it is the zero module (which is free!) if $x = 0$. If $x$ is nonzero, then the map $a \mapsto ax$ gives an $R$-module isomorphism $R \stackrel{\sim}{\rightarrow} (x)$, where we have used the fact that since $R$ is a domain, $x$ is not a zero-divisor. (This is, of course, almost exactly what you said...)

A slightly more sophisticated take on the above breaks it down into the following three basic facts.

1) A ring $R$ is a domain iff every ideal $I$ of $R$ is torsionfree: for $0 \neq a \in R$ and $0 \neq x \in I$, $ax \neq 0$.
2) If an $R$-module $M$ is torsionfree then it has zero annihilator: $\operatorname{ann} M = 0$. (In general $\operatorname{ann} M = \{a \in R \ | \ aM = 0 \}$.)
3) If $M = \langle x \rangle_R$ is a monogenic $R$-module, then the map $a \mapsto ax$ induces an isomorphism $R/\operatorname{ann} M \stackrel{\sim}{\rightarrow} M$.

Note also that Georges Elencwajg also said essentially the same thing in an answer which he recently deleted because he found it "stylistically vapid and garrulous". That seems kind of harsh, Georges...

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Thank you , Pete: it's always a pleasure to hear from you, and now even more... –  Georges Elencwajg Sep 13 '11 at 23:18

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