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Prove that if an integral domain $R$ is an injective $R$-module then $R$ is a field.

"Proof"

Choose a non-zero element $x$ of $R$ and consider the map $f: R \rightarrow R$ given by $f(t)=tx$.

This yields an exact sequence $0 \rightarrow R \rightarrow R \rightarrow R/\langle x \rangle \rightarrow 0$ where the first map is $f$.

By assumption $R$ is injective and thus $R \cong R \oplus R/\langle x \rangle$.

Now the claim is that $1 \in \langle x \rangle$. Otherwise $(1,0 + \langle x \rangle)$, $(0,1+ \langle x \rangle)$ are zero divisors of $R \oplus R/\langle x \rangle$ which is impossible since $R$ is an integral domain.

Thus $1 \in \langle x \rangle$ so $x$ is invertible and we're done. Is this OK?

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  • $\begingroup$ In principle, the direct decomposition you found is of R as an R-module, and not as a ring. You should probably elaborate that point a bit. $\endgroup$ Commented Oct 13, 2012 at 22:40
  • $\begingroup$ Curious that we have two answers that are more interested in showing off their own proof than answering the question you actually asked. $\endgroup$ Commented Oct 14, 2012 at 0:26
  • $\begingroup$ Note that you need to prove every non-zero $x$ is invertible, not just a chosen non-zero $x$. While I'm sure the former was your intent, the latter is what you actually wrote. You should probably explicitly write down what the product operation is on $R \oplus R / \langle x \rangle$, so that you can verify the claimed zero divisors really are zero divisors (or find some other means of showing this). One last thing that gives me pause is the claim $R \cong R \oplus R / \langle x \rangle$ -- but that's probably just because I don't use injective modules much. $\endgroup$ Commented Oct 14, 2012 at 0:35
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    $\begingroup$ @Hurkyl: Maybe it's curious, but I think it will be helpful to the OP to see how the various definitions of injectivity can be used to prove the statement in question. In my opinion, the direct sum decomposition property used by the OP isn't very elegant here. But, of course, that's a matter of taste ... $\endgroup$ Commented Oct 14, 2012 at 0:47

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Define $f$ as in your answer. By injectivity of $R$, there is an $R$-linear map $g: R \to R$ such that $g \circ f = id_R$ (draw a commutative triangle). Hence $1=g(f(1))=g(x)=g(x \cdot 1)=x\cdot g(1)$, i.e. $x$ has a multiplicative inverse.

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