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

A backup of a deleted PSQ :

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.


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.

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