Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|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 their own proof than answering the question you actually asked. – Hurkyl Oct 14 '12 at 0:26
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. – Hurkyl Oct 14 '12 at 0:35
@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 ... – Ralph Oct 14 '12 at 0:47
up vote 2 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.