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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?

Xander Henderson
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  • There are two other equivalent conditions: every short exact sequence of $R$-modules is split, and $R$ is semisimple Artinian. – Geoffrey Trang Apr 17 '20 at 21:44

1 Answers1

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It is correct. Another way to prove is through short exact sequences:

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.

  • for future visitors, I think this answer uses the following fact: https://math.stackexchange.com/questions/1877465/q-is-an-injective-module-iff-injections-from-q-always-split – D.R. Mar 06 '22 at 23:30