Commutative Algebra 28

Posted on April 15, 2020 by limsup

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 a, b 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.

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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.

warningEvery 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}

warningEven 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.

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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 M, N 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:
    April 25, 2020 at 5:08 am

    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” ?

    Reply

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