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.

Let $R$ be a ring and $M$ be an $R$-module. The flat dimension of $M$ is the infimum over all lengths of flat resolutions of $M$. Usually, the flat dimension of $M$ is denoted by $\mathrm{fd}_R(M)$. For example, $\mathrm{fd}_{\mathbb{Z}}(\mathbb{Q}) = 0$. Since $\mathbb{Q}$ has projective dimension $1$, the flat dimension and projective dimension of a module can be different. Sometimes they can be the same: $\mathbb{Z}/n$ for $n$ a positive integer has the same flat and projective dimension as $\mathbb{Z}$-modules.

The weak dimension of a ring $R$ is defined to be $\mathrm{w.dim}(R) = \sup_{M} \{ \mathrm{fd}_R(M) \}$ where $M$ runs over all left $R$-modules. Due to the symmetric nature of the tensor product, we can also take the supremum over all right $R$-modules, in contrast to the asymmetric nature of global dimension.
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The tensor product is one of the most important constructions in mathematics, and here we shall see my favourite examples of the tensor product in action, hopefully to illuminate its properties for beginners. Proofs or references are provided, but since the emphasis is on examples, the proofs that are given are terse and details are left to the interested reader.

Let $ R$ be a ring.

Proposition. If $ M$ is a left $ R$-module and we consider $ R$ as a right $ R$-module then $ R\otimes_R M \cong M$.
Proof. Multiplication $ R\times M\to M$ is bilinear, so it extends to a map $ R\otimes_R M\to M$. The map $ M\to R\otimes_R M$ given by $ m\mapsto 1\otimes m$ is its inverse.
Proposition. If $ M$ is a left (resp. right) $ R$-module then the functor $ -\otimes_R M$ (resp. $ M\otimes_R-$) is right-exact.
Proof. The tensor functor is a left-adjoint so it is right-exact.

Here is an application of the above result.

Proposition. Let $ m,n\geq 1$ be integers. In the category of abelian groups $ \mathbb{Z}/n\otimes_\mathbb{Z}\mathbb{Z}/m\cong \mathbb{Z}/\mathrm{gcd}(m,n)$.

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