## Tensor Product over Noncommutative Rings

Following the earlier article on tensor products of vector spaces, we will now look at tensor products of modules over a ring R, not necessarily commutative. It turns out we have to distinguish between left and right modules now. Indeed recall the isomorphism from earlier: $\text{Hom}_K(V, W)\cong V^* \otimes_K W$ for vector spaces. If we replace V and W by left R-modules, then V* becomes a right module, so this hints that we should consider a tensor product between a right module and a left one.

Definition. Let M be a right R-module, N a left R-module and X an abelian group. A bilinear map of modules is a map $B:M \times N \to X$ such that

• when we fix n∈N, B(-, N): M→X is additive;
• when we fix m∈M, B(m, -): N→X is additive;
• B(mr, n) = B(m, rn) for any r∈R, m∈M, n∈N.

The tensor product of M and N, denoted $M\otimes_R N$ is an abelian group together with a bilinear map $\psi: M\times N \to M\otimes_R N$ such that the following universal property holds:

• for any bilinear map $B:M \times N\to X$ there is a unique additive map $f:M\otimes N \to X$ such that $f\circ \psi = B.$

As before, the element $m\otimes n:= \psi(m,n)$ for any $m\in M, n\in N$ is called a pure tensor.

The universal property again guarantees that the tensor product is unique if it exists.

Proof of Existence

The proof is identical to earlier. Let T be the free abelian group with basis $\{e_{m,n} : m\in M, n\in N\}.$ Take the subgroup U generated by elements of the form:

$e_{m+m', n} - e_{m,n} - e_{m', n}, \quad e_{m,n+n'} - e_{m,n} - e_{m, n'}, \quad e_{mr,n} - e_{m,rn},$

for $r\in R, m,m'\in M, n,n'\in N$. Now our desired group is T/U, and $M\times N \to T/U$ is given by $(m,n)\mapsto e_{m,n} \pmod U$ so mn is the image of $e_{m,n}$ in T/U. ♦

Note

A bilinear map M × N → X corresponds to a homomorphism of right modules $M\to \text{Hom}_{\mathbf{Z}}(N, X).$ [ Recall that since N is a left R-module, Hom(NX) becomes a right R-module. ]  Hence, the above universal property can also be written as:

$\text{Hom}_\mathbf{Z}(M\otimes_R N, X) = \text{Hom}_{\text{right}-R}(M, \text{Hom}_{\mathbf{Z}}(N, X)).$

Often, we would like the tensor product to be a module instead of merely an abelian group:

Proposition. Let R, S be rings. If M is an (R, S)-bimodule and N is a left S-module, then $M\otimes_S N$ is a left R-module satisfying $r(m\otimes n) := (rm)\otimes n.$

The bilinear map $\psi : M\times N \to M\otimes_S N$ is also R-linear in M, i.e. $r\cdot \psi(m, n) = \psi(rm, n).$

Proof

By definition $M \otimes_S N$ is an abelian group. For each rR, consider the map

$f_r: M \times N \to M\otimes_S N, \quad (m,n) \mapsto (rm)\otimes n.$

The map is R-bilinear (e.g. fr(msn) = (r(ms))⊗= (rm)snrm⊗sn = fr(msn)). So it induces $\phi_r : M\otimes_S N \to M\otimes_S N$ taking $m\otimes n \mapsto (rm)\otimes n.$ One now checks that this turns $M\otimes_S N$ into a left R-module such that $M \times N \to M\otimes_S N$ is R-linear in M. ♦

Properties.

• For any left R-module M, $R\otimes_R M \cong M$ as left R-modules.
• For any right R-module M, (R, S)-bimodule N, and left S-module P, we have $M\otimes_R (N\otimes_S P)\cong (M \otimes_R N)\otimes_S P.$
• For any right R-module M, and left R-modules $\{N_i\}$, we have $M\otimes_R (\oplus_i N_i) \cong \oplus_i (M\otimes_R N_i).$

Sketch of Proof.

We’ll briefly sketch the proof of the second claim. Fixing mM, the map $f_r: N\times P \to (M\otimes_R N)\otimes_S P$ taking $(n, p) \mapsto (m\otimes n)\otimes p$ is bilinear over S, so it induces a group homomorphism $\psi_r : N\otimes_S P \to (M\otimes_R N)\otimes_S P$ which is easily checked to be left R-linear. Check that this is right R-linear in r, so we obtain a group homomorphism $M\otimes_R (N\otimes_S P) \to (M\otimes_R N)\otimes_S P$ taking (mn)⊗p to m⊗(np). ♦

## Base Extension

Suppose S is an R-algebra and M a left R-module. Since S is an (SR)-bimodule, the tensor product $M_S := S \otimes_R M$ is now a left S-module. This is called the base extension of M. Since $(T\otimes_S S)\otimes_R M \cong T\otimes_R M,$ we have:

Proposition. Base extension commutes, i.e. if T is an S-algebra, then $T\otimes_S (S\otimes_R M) \cong T\otimes_R M$ as T-modules.

Base extension satisfies the following universal property.

Universal Property of Base Extension. The canonical map $\psi: M \to M_S$ taking $m\mapsto 1\otimes m$ is R-linear.

For any left S-module N and R-module homomorphism $\phi: M\to N$ there is a unique S-module homomorphism $f: M_S \to N$ such that $f\circ\psi = \phi.$

Thus for any S-module N:

$\text{Hom}_R(M, N) = \text{Hom}_S(M_S, N)$

Proof

The first statement is obvious. For the second, let us prove existence of f. The map $S\times M \to N$ mapping $(s,m)\mapsto s\cdot \phi(m)$ is R-bilinear so it induces $f: M_S = S\otimes_R M\to N$ satisfying $f(s\otimes m) = s\cdot \phi(m).$ This is S-linear since

$f(s'\cdot (s\otimes m)) = f(s's\otimes m) = s's\cdot \phi(m) = s'\cdot f(s\otimes m)$

and it satisfies $f\circ\psi(m) = f(1\otimes m) = \phi(m).$ Uniqueness follows from the fact that M generates MS as an S-module.

Example

Suppose RR, the real field and V is the space of polynomials with real coefficients and degree ≤ 2. Then $\mathbf{C}\otimes_{\mathbf{R}} V$ is naturally identified with the space of polynomials with complex coefficients and degree ≤ 2.

## For Group Representations

Tensor products are useful in group representations in two different ways. First, suppose V and W are K[G]-modules for a field K and finite group G. Then $V\otimes_K W$ (note: the tensor product over K, not the group ring!) becomes a K[G]-module as well. Indeed, each g induces isomorphisms VV and WW and thus a bilinear map V×WVW taking (vw) to gvgw. This in turn gives us a linear map VW → VW  taking vw to gvgw. [ It is an isomorphism since g-1 induces the inverse. ]

Another use of tensor product is via the induced representation. If H ⊆ G is a subgroup and V is a K[H]-module, then one can define: $V^G := K[G] \otimes_{K[H]} V$ to obtain a representation of G. The universal property of base extension gives us: for any K[G]-module W,

$\text{Hom}_{K[H]}(V, W) = \text{Hom}_{K[G]}(V^G, W).$

This is the Frobenius reciprocity theorem in representation theory and is often written as:

$\left_H = \left<\text{ind}^G_H V, W\right>_G.$

## Tensor Product is Right Exact

We have the following

Theorem. If $N' \to N\to N'' \to 0$ is an exact sequence of left R-modules, then for any right R-module M, we get an exact sequence of abelian groups:

$M\otimes_R N' \to M\otimes_R N\to M\otimes_R N'' \to 0.$

Proof

We use the property from above:

$\text{Hom}_\mathbf{Z}(M\otimes_R N, X) = \text{Hom}_{\text{right}-R}(M, \text{Hom}_{\mathbf{Z}}(N, X)).$

Now, for any abelian group X, since $N' \to N \to N'' \to 0$ is exact by condition, the following is exact since Hom is left-exact:

$0\to \text{Hom}_{\mathbf{Z}}(N'', X)\to\text{Hom}_{\mathbf{Z}}(N, X) \to\text{Hom}_{\mathbf{Z}}(N', X).$

Again since Hom is left-exact, we have:

$0\to \text{Hom}_R(M, \text{Hom}_{\mathbf{Z}}(N'',X)) \to \text{Hom}_R(M, \text{Hom}_{\mathbf{Z}}(N, X)) \to \text{Hom}_R(M, \text{Hom}_{\mathbf{Z}}(N', X))$

which is precisely the same as saying:

$0\to \text{Hom}_\mathbf{Z}(M\otimes_R N'', X) \to\text{Hom}_\mathbf{Z}(M\otimes_R N, X)\to \text{Hom}_\mathbf{Z}(M \otimes_R N', X)$

is exact for all X. Thus, $M\otimes_R N' \to M\otimes_R N \to M\otimes_R N''\to 0$ is exact. ♦

Example

Let I be a two-sided ideal of R, so R/I becomes an R-algebra via the canonical map RR/I. We claim that for any left R-module M, base extension gives

$R/I \otimes_R M \cong M/IM.$

Indeed, consider the map $R/I\times M\to M/IM$ taking (r+Im) to rm+IM. This map is well-defined on R/I and R-bilinear so it induces a map $R/I \otimes_R M \to M/IM.$ Conversely, take the map $M\to R/I \otimes_R M$ where $m\mapsto (1+I)\otimes m.$ Everything in IM maps to 0, so this factors through $M/IM \to R/I\otimes_R M.$ It’s easy to check that both maps are mutually inverse.

Note

From our example above, it is easy to find examples where the tensor product is not left-exact. For example, consider 0 → 2Z → Z. Tensoring with Z/2 is the same as taking M to M/2M; so we obtain 0 → 2Z/4Z → Z/2Z which is not exact since the second map takes everything to 0.

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