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I'm having difficulties proving the following:

Let $R$ be an integral domain and $P$ a projective and injective $R$-module. Show that $P=0$ or $R=Q(R)$, where $Q(R)$ denotes the field of fractions of $R$.

(I haven't found a similar question here on MSE, but maybe I'm searching for the wrong terms).

This is how far I have come so far:

  • I could prove that if $M$ is a free, divisible $R$-module, then $M=0$ or $R=Q(R)$
  • I could prove that an injective $R$-module is divisible.

(my definition of 'divisible is': for each $r \in R$, where $r$ is not a zero divisor, the map $M \to M, m \mapsto rm$ is surjective)

Then I wanted to use that if $P$ is projective, there exists a free module $F$ and a module $Q$, so that $F = P \oplus Q$, and then wanted to show that $F$ is also divisible. But I don't think that needs to be.

Any help/hints/links in the right direction are much appreciated!

Thanks in advance!

johnnycrab
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    I've just found this link which was suggested after posting: http://math.stackexchange.com/questions/585495/a-problem-about-an-r-module-that-is-both-injective-and-projective?rq=1 – johnnycrab Jun 26 '16 at 10:19
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    Usually before posting such links will be suggested on the right-hand side also, which would change as you type more in your question. – user21820 Jun 26 '16 at 10:50

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