# Weakening the hypotheses in showing that every nonzero element of a projective $R$-module is free

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
$\LaTeX$ 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

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...)
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 $R/\operatorname{ann} M \stackrel{\sim}{\rightarrow} M$.