Take the 2-minute tour ×  Prove that if an integral domain $R$
is an injective $R$-module then $R$ is a field.   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?  share|improve
this question 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. –  Mariano Suárez-Alvarez Oct 13 '12
at 22:40 Curious that we have two answers that are more
interested in showing off thei