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For a slightly more down-to-earth explanation, notice over a commutative ring $R$, any simple $R$-module has the form $R/\mathfrak m$ for some maximal ideal $\mathfrak m$. If $R/\mathfrak m$ is projective, then it is a summand of some free $R$-module. Since free $R$-modules are torsion-free, $\mathfrak m = 0$, hence $R$ is field. In particular, $R/\mathfrak m$ = $R$ is injective!