# Tensor Products

In this article (and the next few), we will discuss tensor products of modules over a ring. Here is a motivating example of tensor products.

Example

If $V = \{a + bx + cx^2 : a,b,c \in \mathbb R\}$ and $W = \{a' + b'y + c'y^2 : a', b', c' \in \mathbb R\}$ are real vector spaces, then $V\otimes_{\mathbb R} W$ is the vector space with basis $x^i y^j$ over $0 \le i \le 2, 0 \le j \le 2$. Hence it follows that $\dim (V\otimes W) = \dim V \times \dim W$.

Also $\mathbb C \otimes_{\mathbb R} V$ is the complex vector space of all $a+bx+cx^2$ where $a,b,c\in \mathbb C$. Thus $\dim_{\mathbb C} (\mathbb C\otimes_{\mathbb R} V) = \dim_{\mathbb R} V$.

These properties will be proven as we cover the theory.

Throughout this article, A is a fixed ring and all modules and linear maps are over A.

Definition.

Given modules M, N, P, an A-bilinear map is a function

$B : M\times N \longrightarrow P$

such that:

• for each $m\in M$, the function $B(m, -) : N\to P$ is A-linear;
• for each $n\in N$, the function $B(-, n) : M\to P$ is A-linear.

For example over the base ring $\mathbb Z$, the Euclidean inner product

$\mathbb Z^n \times \mathbb Z^n \longrightarrow \mathbb Z, \quad ((x_1, \ldots, x_n), (y_1, \ldots, y_n)) \mapsto \sum_i x_i y_i$

is bilinear. It is important to distinguish between a bilinear $M\times N\to P$ and a linear $M\times N\to P$. For example, in a bilinear map, we must have $B(m, 0) = 0$ for all $m\in M$ while this is hardly ever true for a linear map.

Philosophically, we want the tensor product of M and N to be the module that classifies all bilinear maps from $M\times N$.

Definition.

The tensor product of modules M, N comprises of:

$(P, \phi : M\times N \to P)$

where $P = M\otimes_A N$ is a module, $\phi$ is a bilinear map, such that for any pair

$(Q, \psi : M\times N \to Q)$

where Q is a module and $\psi$ is bilinear, there is a unique linear map $f : P \to Q$ such that $f\circ\phi = \psi$.

Exercise A

1. Prove that the tensor product is unique up to unique isomorphism.

2. For our initial example of $V\otimes_{\mathbb R} W$, let X be the space of all $a+bx$ with ab real. Now take the bilinear

$V \times W \longrightarrow X, \quad (f(x), g(y)) \mapsto f(1)g(-1) + \frac{df}{dx} g(1).$

Compute the resulting $V\otimes_{\mathbb R} W \to X$.

# Construction

To prove that the tensor product exists for any A-modules M and N, let $F$ be the free A-module generated by $M\times N$ as a set. The standard basis of F is denoted by $e_{m, n}$ for $(m,n) \in M\times N$. Now let P be the quotient of F by the submodule generated by elements of the form:

\begin{aligned} e_{m+m', n} - e_{m,n} - e_{m', n}, &\quad e_{am, n} - a\cdot e_{m, n},\\ e_{m, n+n'} - e_{m,n} - e_{m,n'}, &\quad e_{m,an} - a\cdot e_{m,n},\end{aligned}

over all $m,m'\in M, n, n'\in N, a\in A$. The map $\phi : M\times N \to P$ is defined by $(m, n) \mapsto \overline e_{m,n}$; by definition of P this is A-bilinear.

Lemma.

The pair $(P, \phi:M\times N\to P)$ forms the tensor product of M and N.

Proof

Let Q be any module and $\psi : M\times N\to Q$ be any bilinear map. We first define a linear $f': F \to Q$ by taking $e_{m, n}\mapsto \psi(m, n)$. This map factors through P because:

• by bilinearity of $\psi$ we have $\psi(m + m', n) - \psi(m, n) - \psi(m', n) = 0$ and thus $f'$ takes $e_{m+m', n} - e_{m,n} - e_{m',n}$ to 0; Similarly it also takes all elements of the form $e_{am, n} - a\cdot e_{m, n}$, $e_{m, n+n'} - e_{m,n} - e_{m,n'}$ and $e_{m,an} - a\cdot e_{m,n}$ to zero.

Hence we obtain a linear $f:P \to Q$ which takes $\overline e_{m,n} \mapsto \psi(m, n)$. We have

$f\circ\phi(m,n) = f(\overline e_{m,n}) = \psi(m,n)$

and so $f\circ\phi = \psi$.

To prove f is unique, since $f\circ \phi = \psi$ we must have $f(\overline e_{m,n}) = f(\phi(m,n)) = \psi(m,n)$. But the set of $\overline e_{m,n}$ generates F so this proves uniqueness. ♦

Note

Let us set some notation. For the given $\phi : M\times N\to M\otimes N$, any $m\in M$ and $n\in N$ gives us $\phi(m, n)$, denoted by $m\otimes_A n$. The universal property of tensor product thus says the following.

• Any A-bilinear $\psi : M\times N\to Q$ induces a unique linear $M\otimes_A N\to Q$ which sends $m\otimes_A n\mapsto \psi(m, n)$.

From the explicit construction of the tensor product, we also have:

Corollary 1.

The module $M\otimes_A N$ is generated by $m\otimes_A n$ over all $m\in M$ and $n\in N$.

Every element of $M\otimes_A N$ is thus a sum of various $m\otimes n$; but in general not every element is of the form $m\otimes n$. Counter-examples should be easy to find if you take $V\otimes_{\mathbb R} W$ in the beginning of this article.

Exercise B

Prove functoriality: if $f:M\to M'$ and $g:N\to N'$ are linear maps, they induce a linear $h: M\otimes_A N \to M'\otimes_A N'$ satisfying $h(m\otimes n) = f(m) \otimes g(n)$. We denote this map by $f\otimes g$.

In other words, we get a covariant functor

$\otimes : A\text{-}\mathbf{Mod} \times A\text{-}\mathbf{Mod} \longrightarrow A\text{-}\mathbf{Mod}$

Even if f is injective, $f\otimes 1_N : M\otimes_A N \to M'\otimes_A N$ is not injective in general. Thus the functor $- \otimes_A N$ does not take injective maps to injective maps. An example will be given at the end of this article.

# Basic Properties

Now we prove some basic properties of the tensor product.

Proposition 1.

For any module M, $A\otimes_A M \cong M$, where $a\otimes m\mapsto am$.

Note

The isomorphism is natural. To be exact, both sides are functorial in M and we get a natural isomorphism between the functors.

Proof

Let us construct A-linear maps

$f:A\otimes_A M \to M, \quad g : M\to A\otimes_A M.$

To define f we take the map $A\times M\to M$, $(a, m) \mapsto am$. This is easily checked to be A-bilinear so it induces an $f: A\otimes_A M \to M$ satisfying $f(a\otimes m) = am$.

Defining g is easy: set $g(m) = 1 \otimes m$, which is A-linear. It remains to show the maps are mutually inverse. Let $a\in A$, $m\in M$.

• We have $g(f(a \otimes m)) = g(am) = 1 \otimes am = a\otimes m$ due to bilinearity. Since the pure tensors $a\otimes m$ generate $A\otimes_A M$ we have $g\circ f = 1$.
• Also $f(g(m)) = f(1\otimes m) = m$ so $f\circ g = 1$ too. ♦

Proposition 2.

For any modules M, N, P, we have

$(M\otimes_A N)\otimes_A P \cong M\otimes_A (N\otimes_A P), \quad (m\otimes n) \otimes p \mapsto m\otimes (n\otimes p).$

Note

Again, this isomorphism is functorial in MN and P.

Proof

Fix $m\in M$ and consider the map $\phi_m : N\times P \to (M\otimes_A N)\otimes_A P$ which takes $(n, p) \mapsto (m\otimes n)\otimes p$. This is an A-bilinear map so it induces a linear

$f_m : N\otimes_A P \longrightarrow (M\otimes_A N)\otimes_A P, \quad (n\otimes p) \mapsto (m\otimes n) \otimes p.$

Now the map $\phi : M \times (N\otimes P) \to (M\otimes_A N)\otimes_A P,$ which takes $(m, x) \mapsto f_m(x)$ is A-bilinear so it induces a linear

$f : M\otimes_A (N \otimes_A P) \longrightarrow (M\otimes_A N)\otimes_A P, \quad m\otimes(n \otimes p) \mapsto (m\otimes n)\otimes p.$

Similarly, we can define a linear $g : (M\otimes_A N)\otimes_A P \to M\otimes_A (N \otimes_A P)$ which takes $(m\otimes n)\otimes p \mapsto m\otimes (n \otimes p)$. Thus $g\circ f$ takes elements of the form $m\otimes (n\otimes p)$ back to themselves; since these generate the whole module we have $g\circ f = 1$. Similarly $f\circ g = 1$. ♦

Proposition 3.

For any modules M, N, we have

$M\otimes_A N \cong N\otimes_A M, \quad (m\otimes n) \mapsto (n\otimes m).$

Proof

Exercise. ♦

Example

Consider $\mathbb Z \stackrel 2\to \mathbb Z$, the multiplication-by-2 map. Taking $- \otimes_{\mathbb Z} (\mathbb Z/2\mathbb Z)$ gives us $\mathbb Z/2\mathbb Z \to \mathbb Z/2\mathbb Z$, still multiplication-by-2 but now it is the zero map. Hence tensor products do not take injective maps to injective maps in general.

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### 2 Responses to Commutative Algebra 28

1. Vanya says:

In the statement ” Thus the functor – \otimes_A N does not take submodules to submodules. ” did you mean ” does not take injective maps to injective maps” ?

• limsup says:

Agreed: that would be more accurate.