# Constructing the Tensor Product of Modules

## The Basic Idea

Today we talk tensor products. Specifically this post covers the construction of the tensor product between two modules over a ring. But before jumping in, I think now's a good time to ask, "What are tensor products good for?" Here's a simple example where such a question might arise:

Suppose you have a vector space $V$ over a field $F$. For concreteness, let's consider the case when $V$ is the set of all $2\times 2$ matrices with entries in $\mathbb{R}$ and let $F=\mathbb{R}$. In this case we know what "$F$-scalar multiplication" means: if $M\in V$ is a matrix and $c\in \mathbb{R}$, then the new matrix $cM$ makes perfect sense. But what if we want to multiply $M$ by *complex* scalars too? How can we make sense of something like $(3+4i)M$? That's precisely what the tensor product is for! We need to create a set of elements of the form $$\text{(complex number) "times" (matrix)}$$ so that the mathematics still makes sense. With a little massaging, this set will turn out to be $\mathbb{C}\otimes_{\mathbb{R}}V$.

So in general, if $F$ is an arbitrary field and $V$ an $F$-vector space, the tensor product answers the question "How can I define scalar multiplication by some *larger* field which contains $F$?" And of course this holds if we replace the word "field" by "ring" and consider the same scenario with modules.

Now this isn't the only thing tensor products are good for (*far* *from it!*), but I think it's the most intuitive one since it is readily seen from the definition (which is given below). You can read about another motivation for the tensor product here. And for all my physics friends, this blurb by K. Conrad sheds light on the relationship between the physicist's tensor and the mathematician's tensor (see p. 46).

So with this motivation in mind, *let's go!*

## From English to Math

Let $R$ be a ring with 1 and let $M$ be a right $R$-module and $N$ a left $R$-module and suppose $A$ is any abelian group. Our goal is to create an abelian group $M\otimes_R N$, called the **tensor product** of $M$ and $N$, such that if there is an $R$-balanced map $i\colon M\times N\to M\otimes_R N$ and any $R$-balanced map $\varphi\colon M\times N\to A$, then there is a unique abelian group homomorphism $\Phi\colon M \otimes_R N\to N$ such that $\varphi=\Phi\circ i$, i.e. so the diagram below commutes.

Notice that the statement above has the same flavor as the universal mapping property of free groups!

**Definition: **Let $X$ be a set. A group $F$ is said to be a **free group** on $X$ if there is a function $i\colon X\to F$ such that for any group $G$ and any set map $\varphi\colon X\to G$, there exists a unique group homomorphism $\Phi\colon F\to G$ such that the following diagram commutes: (i.e. $\varphi=\Phi\circ i$)

So of course we'd like $X$ to be the set $M\times N$ and $G$ to be the abelian group $A$. To obtain a map $i\colon X\to F$ we want to let $F$ be the *free abelian group generated by $M\times N$*. That is, let $F$ be the set of all finite commuting sums of elements of the form $(m_i,n_i)$ where $m_i\in M, n_i\in N$ - this is an abelian group where there are no relations between any distinct pairs $(m,n)$ and $(m',n')$ (see *Dummit and Foote*, p. 360). Also, $\varphi\colon X\to G$ can be *any* set map, so in particular we just want our's to be $R$-balanced:

**Definition**: Let $R$ be a ring with 1. Let $M$ be a right $R$-module, $N$ a left $R$-module, and $A$ an abelian group. A map $\varphi:M\times N\to R$ is called $\mathbf{R}$-balanced if for all $m,m_1,m_2\in M$, all $n,n_1,n_2\in N$ and all $r\in R$,

$\varphi(m_1+m_2,n)=\varphi(m_1,n)+\varphi(m_2,n)$$\varphi(m,n_1+n_2)=\varphi(m,n_1)+\varphi(m,n_2)$$\varphi(mr,n)=\varphi(m,rn)$

Lastly, we observe that in the free group definition, $i\colon X\to F$ is just *some* map, whereas in the tensor product definition, $i\colon M\times N\to M\otimes_R N$ must be an $R$-balanced map! *How can we reconcile this?* By "replacing" F by a certain quotient group $F/H$! (We'll define $H$ precisely below.)

These observations give us a road map to construct the tensor product. And so we begin:

### Step 1

Let $F$ be a free abelian group generated by $M\times N$ and let $A$ be an abelian group. Then by definition (of free groups), if $\varphi:M\times N\to A$ is any set map, and $M\times N \hookrightarrow F$ by inclusion, then there is a unique abelian group homomorphism $\Phi:F\to A$ so that the following diagram commutes.

*In English, this just means, "set maps from $M\times N \to A$ are the same as (i.e. are isomorphic to) homomorphisms from $F$ to $A$."*

### Step 2

Observe that the commuting diagram above is (almost) exactly what we want, *except* that the inclusion map $M\times N\hookrightarrow F$ is not $R$-balanced! To fix this, we must "modify" the target space $F$ by replacing it with the quotient $F/H$ where $H\leq F$ is the subgroup of $F$ generated by elements of the form

- $(m_1+m_2,n)-(m_1,n)-(m_2,n)$
- $(m,n_1+n_2)-(m,n_1)-(m,n_2)$
- $(mr,n)-(m,rn)$

where $m_1,m_2,m\in M$, $n_1,n_2,n\in N$ and $r\in R$. Why elements of this form? Because if we define the map $i:M\times N\to F/H$ by $$i(m,n)=(m,n)+H,$$ we'll see that $i$ is indeed $R$-balanced! Let's check:

So, are we done now? Can we really just replace $F$ with $F/H$ and replace the inclusion map with the map $i$, and *still* retain the existence of a unique homomorphism $\Phi:F/H\to A$? No! Of course not. $F/H$ is *not* a free group generated by $M\times N$, so the diagram below is bogus, right?

Not *totally*. We haven't actually disturbed any structure!

How can we relate the pink and blue lines? We'd really like them to be the same. But we're in luck because they basically are!

### Step 3

By the Fundamental Homomorphism Theorem, homomorphisms from $F/H$ to $A$ are the same as (i.e. isomorphic to) homomorphisms $f$ (or $\Phi$ in the case of the picture above) from $F$ to $A$, *as long as* $H\subseteq \ker(f)$, that is as long as $f(h)=0$ for all $h\in H$. And notice that this condition, $f(H)=0$, forces $f$ to be $R$-balanced!

Let's check:

*Sooooo...* homomorphisms $f:F\to A$ such that $H\subseteq \ker(f)$ are the same as $R$-balanced maps from $M\times N$ to $A$! (Technically, I should say *homomorphisms $f$ restricted to $M\times N$*.) In other words, we have

In conclusion, to say "abelian group homomorphisms from $F/H$ to $A$ are the same as (isomorphic to) $R$-balanced maps from $M\times N$ to $A$" is the simply the *hand-wavy* way of saying

Whenever $i:M\times N\to F$ is an $R$-balanced map and $\varphi:M\times N\to A$ is an $R$-balanced map where $A$ is an abelian group, there exists a unique abelian group homomorphism $\Phi:F/H\to A$ such that the following diagram commutes:

And this is just want we want! The last step is merely the final touch:

### Step 4

At last, we *define* the abelian quotient group $F/H$ to be the tensor product of $M$ and $N$,

whose elements are cosets,

where $m\otimes n$ for $m\in M$ and $n\in N$ is referred to as a *simple tensor*. And there you have it! The tensor product, constructed.

*Reference: *

D. Dummit and R. Foote, *Abstract Algebra,* 3rd ed., Wiley, 2004. (ch. 10.4)