# Deleted Question on Semi-Simple and Projective but not Injective Module

A backup of a deleted PSQ : https://math.stackexchange.com/q/3955443/290189

OP : irfanmat

It has a detailed answer by Atticus Stonestrom. It’s pity that his post got deleted. As there’s no reason for undeletion, I’m posting it here so as to preserve the contents.

### Question body

Is there a semi-simple and projective but not injective module? I will be glad if you help.

### Response(s)

In the non-commutative case, the answer is yes. Consider $R$ the ring of upper triangular $2\times 2$ matrices over a field $F$, and denote by $e_{ij}$ the element of $R$ with the $ij$-th entry equal to $1$ and all other entries equal to $0$. We can decompose $R$ as a direct sum of left ideals $$Re_{11}\oplus (Re_{12}+Re_{22})=Re_{11}\oplus Re_{22},$$ so let $M=Re_{11}$. Then $M$ is clearly simple, and – as a direct summand of the free module $R$ – is also projective. However, $M$ is not injective; consider the map $f:Re_{11}\oplus Re_{12}\to M$ taking $e_{11}$ and $e_{12}$ to $e_{11}$. $Re_{11}\oplus Re_{12}$ is a left ideal of $R$, but there is no way to extend $f$ to a map $R\rightarrow M$, so this gives the desired example.

In the case of commutative Noetherian rings, the answer is no.

Lemma: If $R$ is a commutative ring, and $M$ is a simple and projective $R$-module, then $M$ is injective.

Proof: Let $m\in M\setminus{0}$, and consider the map $f:R\to M$ taking $r\mapsto r\cdot m$. By simplicity, $f$ is surjective, and thus by projectivity there is a map $g:M\to R$ such that $f\circ g=\operatorname{id}_M$. In particular, $g$ is injective, so $I:=g(M)$ is an ideal of $R$ isomorphic to $M$.

It suffices therefore to show that $I$ is injective. Letting $J=\ker f$, we have $R=I\oplus J$, and in particular $1=i+j$ for some $i\in I$ and $j\in J$. We have $$i=i\cdot 1=i(i+j)=i^2+ij,$$ but $ij\in I\cap J$ and hence equals $0$, whence $i=i^2$ is idempotent. Now, since $I$ is non-zero, $i$ is non-zero, and thus by simplicity $I=Ri$; also, since $IJ\leqslant I\cap J={0}$, $iJ={0}$. Thus $i$ acts as the identity on $I$ but sends every element of $J$ to $0$, and so any map $h:K\rightarrow I$ from an ideal $K\leqslant R$ must send every element of $K\cap J$ to $0$. Furthermore, by simplicity of $I$, $K\cap I$ equals either ${0}$ or all of $I$. In the first case, $h$ is identically zero, so we may extend it to the zero map from $R$ to $I$. In the second case, we can extend $h$ to the map $\overline{h}:R\to I$ given by $\overline{h}\rvert_I = h\rvert_I$ and $\overline{h}\rvert_J=0$. Thus we can extend any map $K\to I$ to a map $R\to I$, and so by Baer’s criterion $I$ is injective. $\square$

Theorem: If $R$ is a commutative Noetherian ring, and $M$ is a semi-simple and projective $R$-module, then $M$ is injective.

Proof: By semisimplicity, write $M=\bigoplus_{i\in I}M_i$, where each $M_i\leqslant M$ is simple. Since projectivity is preserved under direct summands, each $M_i$ is projective, and hence – by our lemma – is injective. Thus $M$ is a direct sum of injective modules, and so, because $R$ is Noetherian, the Bass-Papp theorem gives that $M$ is injective. $\square$

I’m not sure about the commutative, non-Noetherian case. You’d need an infinite direct sum as an example, since the argument above extends to the non-Noetherian case for finite direct sums.