# Every $R$-module is projective if and only if every $R$-module is injective

I am solving an exercise from Dummit, Foote. Let $$R$$ be a ring with $$1$$. Then the following are equivalent:

1. Every $$R$$-module is projective.

2. Every $$R$$-module is injective.

Proof:

(1) implies that for any R-mod $$A$$ we have $$Ext^i(-,A)$$ is a zero functor. But this means that for any R-mod $$Ext^i(L,A) = 0$$ for all $$A$$ so $$L$$ is injective. I show $$(2) \Rightarrow (1)$$ exactly the same way.

It is quite easy so I am afraid I missed something. Is my proof correct?

• There are two other equivalent conditions: every short exact sequence of $R$-modules is split, and $R$ is semisimple Artinian. – Geoffrey Trang Apr 17 at 21:44

A $$R$$-module $$P$$ is projective if and only if every short exact sequence of $$R$$-modules $$0 \longrightarrow M \longrightarrow N \longrightarrow P \longrightarrow 0$$ splits and a $$R$$-module $$I$$ is injective if and only if every short exact sequence of $$R$$-modules $$0 \longrightarrow I \longrightarrow M \longrightarrow N \longrightarrow 0$$ splits.
Therefore, using these equivalences we have: every $$R$$-module is projective $$\Leftrightarrow$$ every short exact sequence of $$R$$-modules splits $$\Leftrightarrow$$ every $$R$$-module is injective.