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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