An $R$-module $M$ is called injective if the functor $\Hom_R(-,M)$ is exact. The well-known Baer criterion states that an $R$-module $M$ is injective if and only if for every ideal $I$ of $R$, every map $I\to M$ can actually be extended to a map $R\to M$.

For example, $\Q$ is an injective $\Z$-module.

If every $R$-module is injective, then it turns out that every $R$-module is also projective. That's just because every $R$-module being projective or injective is just another way of saying that every short exact sequence of $R$-modules splits.

What about if we weaken the Baer criterion? Let's say that an $R$-module $M$ is called p-injective if the Baer criterion holds for principal ideals. That is, $M$ is p-injective if for every principal left ideal $I$ of $R$, every map $I\to M$ extends to a map $R\to M$.

Every injective $R$-module is also p-injective, but the converse is not true. Indeed: it was R. Yue Chi Ming that noticed that a ring $R$ has every module p-injective if and only if $R$ is von Neumann regular. I've talked about this rings before because they're cool.

Briefly, a ring $R$ is called von Neumann regular if for each $a\in R$ there exists an $x\in R$ such that $axa = a$. Let's suppose we have one of these von Neumann regular rings $R$. Let $M$ be an $R$-module. We claim it's p-injective. Indeed, suppose $g:Rb\to M$ is a map from a principal left ideal $Rb$ to $M$. Choose an $x\in R$ such that $bxb = b$. Such an $x$ exists by definition. Then $G:R\to M$ defined by $G(r) = rg(xb)$ is an extension of $g$. Therefore, $M$ is p-injective.

Therefore, every $R$-module is p-injective if $R$ is von Neumann regular. The converse is similar, and I'll leave it as an exercise.

Let's take an example: an infinite product of fields. Such a ring is easily verified to be von Neumann regular. Therefore, every module is p-injective. But such a ring is not semisimple, and therefore some of its modules are not injective.

Posted by Jason Polak on 05. October 2016 · 2 comments · Categories: commutative-algebra, homological-algebra · Tags:

For finite commutative rings, integral domains are the same as fields. This isn't too surprising, because an integral domain $R$ is a ring such that for every nonzero $a\in R$ the $R$-module homomorphism $R\to R$ given by $r\mapsto ra$ is injective. Fields are those rings for which all these maps are surjective. But injective and surjective coincide for endofunctions of finite sets. Therefore, domains are the same thing as fields for finite rings.

But did you know that there is another class of commutative rings for which fields are the same as integral domains? Indeed, for self-injective rings, fields are the same as domains. By definition, a commutative ring $R$ is self-injective if $R$ is injective as an $R$-module. Note: for noncommutative rings, which we don't consider here, there is a difference between left and right self-injective; that is, an arbitrary ring may be injective as a left module over itself, but not right self-injective, and vice-versa.

In other words, self-injective integral domains are fields. And, the proof is sort of along the lines of the one for finite rings:

Proof. Let $a\in R$ be nonzero. Then the multiplication map $R\xrightarrow{a} R$ is injective, and fits into a diagram


Where the dotted arrow exists because $R$ is injective as an $R$-module; since it is a map $R\to R$ it is given by multiplication by some $b\in R$. Therefore $1 = ab$. QED.

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