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.
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