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)$.

*Proof.*Apply the functor $ \mathbb{Z}/n\otimes_\mathbb{Z}-$ to the exact sequence $ 0\to \mathbb{Z}\xrightarrow{m}\mathbb{Z}\to\mathbb{Z}/m\to 0$ to get that $ \mathbb{Z}/n\otimes_\mathbb{Z}\mathbb{Z}/m$ is isomorphic to the quotient $ (\mathbb{Z}/n)/m(\mathbb{Z}/n)$, which is a quotient of $ \mathbb{Z}$ having $ \mathrm{gcd}(m,n)$ elements.

Here's another application of the tensor functor being a left-adjoint:

**Proposition.**A if $ V,W$ are vector spaces of dimension $ m$ and $ n$ respectively then $ \mathrm{dim}_k(V\otimes_k W) = nm$.

*Proof.*Since the tensor functor is a left-adjoint it commutes with direct sums, so distribute: $ (\oplus^m k)\otimes_k(\oplus^n k)$.

What about fields? Suppose $ k$ is a field and $ E,F\supseteq k$ are two extensions. What can we say about $ E\otimes_k F$? Even after imposing reasonable assumptions on the type of extensions, there is more to this question than meets the eye.

**Proposition.**If $ E,F\supseteq k$ are fields and $ E/k$ is a finite separable extension, then $ E\otimes_k F$ is isomorphic as rings to a finite product of fields.

*Proof. See p. 132 of Bump's book "Algebraic Geometry".*

In fact this gives rise to the interesting example in scheme theory where the pullback of $ \mathrm{Spec}(E)\rightarrow\mathrm{Spec}(k)\leftarrow\mathrm{Spec}(F)$ where all these schemes have one element in their underyling topological space, can have any finite number of elements.

For a good overview of the properties of the tensor product of two extensions over a base field, see Jacobson's book "Basic Algebra II".

## Flatness

A fundamental property of modules is *flatness*. We call a right $ R$-module flat if $ M\otimes_R-$ is an exact functor. The same definition with the suitable modifications goes for left modules.

**Proposition.**As an abelian group, $ \mathbb{Z}/n$ is not flat for any integer $ n > 1$.

*Proof.*Apply $ \mathbb{Z}/n\otimes_\mathbb{Z}-$ to the exact sequence $ 0\to \mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}\to 0$.

Actually:

**Proposition.**An abelian group is flat if and only if it is torsion-free.

*Proof.*Let $ A$ be the abelian group. Recall that $ A$ is the direct limit of its finitely generated subgroups. If $ A$ is torsion free, then its finitely generated subgroups are flat, and the result follows because $ \mathrm{Tor}$ commutes with direct limits. If $ A$ is flat, then $ \mathrm{Tor}_1^\mathbb{Z}(\mathbb{Q}/\mathbb{Z},A)$ is its torsion subgroup, which is zero because $ A$ is flat.

As a corollary, we get that $ \mathbb{Q}$ is flat as an abelian group. So $ \mathbb{Q}\otimes_\mathbb{Z}-$ preserves exact sequences.

## Flat and Other Properties

How are flat modules related to projective modules?

**Proposition.**A module is flat if and only if it is isomorphic to a direct limit of projective modules. This is Lazard's theorem. The "only if" condition is the easy part.

*Proof.*See Lazard, "Autour de la platitude", Bulletin de la S. M. F., 97 (1969), p. 81-128.

A easier characterisation of flatness comes from the character module: if $ M$ is a left $ R$-module then its character module, sometimes called the Pontryagin dual module, is $ M' = \mathrm{Hom}_\mathbb{Z}(M,\mathbb{Q}/\mathbb{Z})$ with the right $ R$-action given by $ (fr)(x) = f(rx)$ (exercise: verify that $ M'$ is a right $ R$-module)

**Proposition.**A left $ R$-module $ M$ is flat if and only if $ M'$ is injective as a right $ R$-module.

*Proof.*For a short proof see Weibel's book "An Introduction to Homological Algebra".

Finally let's consider the Harada-Auslander theorem:

**Proposition.**Let $ R$ be a ring. Every left (right) $ R$-module is flat if and only if $ R$ is von Neumann regular.

*Proof.*See Harada, "A Note on the Dimension of Modules and Algebras" or Auslander "On Regular Group Rings".

A ring is von Neumann regular if for each $ a\in R$ there exists an $ x\in R$ such that $ axa = a$. Fields, for instance, have this property (if $ a$ is nonzero $ x= a^{-1}$) so every vector space is flat, but we already *knew* that because every vector space is free. However, arbitrary products of fields are also easily seen to be von Neumann regular, and the endomorphism ring of a vector space is von Neumann regular, too. In fact, the endomorphism ring of any semisimple module is von Neumann regular (see Rowen's "Ring Theory").