# A problem about an $R$-module that is both injective and projective.

Let $R$ be a domain that is not a field, and let $M$ be an $R$-module that is both injective and projective. Prove that $M= \left \{ 0 \right \}$.

This is exercise 7.52 of Rotman's Advanced Modern Algebra. Using theorems before exercises, because $M$ is injective and $R$ is a domain, I conclude that $$\forall m\in M ,\forall r\in R\ (r\neq 0) ,\exists {m}'\in M \Rightarrow m=r{m}'$$ and also because $M$ is projective there is a surjective $\psi$ from free $R$-module $F$ with basis $\left \{ e_{i} \right \}_{i\in I}$ to $M$ and thus we can conclude that for every $m\in M$ we have $$m=\sum r_{i}\Psi (e_{i})$$ now I don't know how should I use these together.

The idea of what is happening or a suggestion or a hint will be great.

• I came by this question because this is an exercise of Robert Ash, Abstract Algebra (the very last exercise of the book). The main idea is to prove that for an injective $R$-module $M$, the ring $Hom_R(M, R)=0$. For some reason in the book there is an assumption that $M$ must also be projective, but it seemed to me that for this part it is redundant (and user89712's solution confirms my hypothesis). Commented Jul 30, 2022 at 9:23

As you remarked, injective modules are divisible, that is, $rM=M$ for all $r\in R$, $r\ne 0$.

The key step is to show that every non-zero homomorphism $f:M\to R$ is surjective. Let $x\in M$ such that $f(x)\ne 0$. Set $r=f(x)$. Since $rM=M$ there exists $y\in M$ such that $ry=x$. Then $rf(y)=r$, so $f(y)=1$, and this is enough.

Now use that $M$ is projective: there exists $F$ a free module of base $(e_i)_{i\in I}$ such that $0\to M\to F$. Consider $p_i:F\to Re_i\cong R$ the canonical projection. If $M\ne 0$ there is an $i\in I$ such that $p_i(M)\ne 0$, so the homomorphism $M\to F\stackrel{p_i}\to R\stackrel{r\cdot}\to R$ is non-zero for each $r\in R$, $r\ne 0$. This homomorphism must be surjective, so $1=rp_i(z)$ for some $z\in M$, and this shows that $R$ is a field, a contradiction.

• I have a question,can you tell me how this solution came to your mind?I Mean what is your technic to solve?I know that maybe you dont want answer but consider it as a big help to someone!
– kpax
Commented Nov 29, 2013 at 17:52
• @kpax I wanted to prove that $R$ is a field and this happens, for example, if one can prove that $rR=R$ for all $r\ne 0$. Then I've thought that I have to use somehow a free module and the natural way to arrive from a free module (over $R$) to $R$ is to use the canonical projections. That's all!
– user89712
Commented Nov 29, 2013 at 19:13
• thanks a lot,it was instructive for me.
– kpax
Commented Nov 29, 2013 at 19:29

The result is equivalent to a seemingly stronger result:

Proposition. Let $$R$$ be a domain, $$F$$ a free $$R$$-module. If $$F$$ has a nontrivial injective submodule, then $$R$$ is a field.

To see that this is equivalent to the result in the question, note that for an injective submodule $$M\subset F$$ we have a SES $$0 \rightarrow M\xrightarrow i F\rightarrow F/M\rightarrow 0,$$ which must split since $$M$$ is injective. Hence $$F\cong M\oplus F/M$$, so $$M$$ is projective.

Here is an element-oriented proof of the proposition:

Proof. Let $$\mathcal X$$ be a basis for $$F$$. For a nonzero element $$m\in M$$, write $$m = a_1x_1+\dots +a_nx_n$$ where $$a_i \in R$$ and $$x_i\in\mathcal X$$ ($$i=1,\dots ,n$$). We may assume that $$a_1\neq 0$$. We can find $$m'\in M$$ with $$a_1m' = m$$. Writing $$m' = b_1 x_1+\dots +b_nx_n$$ where $$b_i\in R$$ ($$i=1,\dots ,n$$), we see that $$a_1m'=m$$ implies $$b_1 = 1$$.

For arbitrary $$a\in R\setminus 0$$, we must have $$m''\in M$$ with $$am'' = m'$$. Writing $$m'' = c_1x_1 + \dots + c_nx_n,$$ we see that $$ac_1 = 1$$. Hence $$a$$ is a unit. QED.

• ... And as a supplement, we see immediately the fact that "the only divisible subgroup of $\mathbb{Z}$ is trivial". This fact is frequently used without mention in several more advanced literature (as it is kind of trivial). An interesting application is to show that $\mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z}$ is a monomorphism in the category of divisible abelian groups, hence the category is not abelian. This is quite disappointing anyway. Sorry to be irrelevant. Commented Aug 25, 2022 at 11:22