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By the way, we're assuming R is an integral domain. I'm guessing we're going to want to show that R has no nontrivial proper ideals. So, let I be an ideal in R.

$0\rightarrow I \rightarrow R\rightarrow R/I\rightarrow 0$ splits, since R/I is an R-module, thus projective. so R is isomorphic to $I \oplus R/I$, but I'm not too sure what to do from there.

Alternatively, every R-module is projective iff every R-module is injective, so Baer's criterion might be useful, but again I'm not sure where to go from that.

Any hints?

WLOG
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mathelp
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2 Answers2

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Every module being projective characterizes the class of semisimple rings. These are classified by the Wedderburn-Artin theorem as direct products of matrix rings over skew field. If you include skew field in your concept of `field', then your assertion is true. In any case, you do need that $R$ is an integral domain, for otherwise you could take $M_2(\mathbb{C})$ for example.

Karsten
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Suppose $I$ is a nonzero, proper ideal of $R$. Then there is some $x \neq 0 \in I$. This nonzero element $x$, of the ring, kills the nonzero element $(0,\bar 1) \in I \oplus R/I$. If this module was isomorphic to $R$, then $R$ would have a nonzero torsion element and this is not the case.

Tim
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