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

Question

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.

Answer 1

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.

Answer 2

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.