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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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    @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 ...2012-10-14

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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.