Advanced Algebra II
Tensor product
In this section, we are going to define an important notion, called tensor product.
Let’s start by recalling that, in linear algebra, there are various sit- uation we need to consider bilinear maps. More precisely, let V, W be vector spaces over a field k. We know that all linear maps from V to k again form a vector space which is the dual vector space of V . Moreover, we can consider maps f : V × W → k which is linear with respect to V and W . Such maps are called bilinear maps. (One can even consider multilinear maps). We can characterize them in terms of basis.
Now in general, we would like to work on modules over rings. Let M = MRbe a right R-module and N =RN be a left R-module. And let C be an abelian group. We would like to consider maps f : M ×N → C satisfying
(1)
f (a1+ a2, b) = f (a1, b) + f (a2, b) f (a, b1+ b2) = f (a, b1) + f (a, b2) f (ar, b) = f (a, rb)
for all a, a1, a2 ∈ M, b, b1, b2 ∈ N, r ∈ R. A map satisfying equation (1) is called middle linear.
Theorem 0.1 ( Existence and universal property of tensor product).
Let R be a ring and M = MR be a right R-module and N = RN be a left R-module. There is an abelian group, denoted M ⊗RN together with a middle linear map ı : M × N → M ⊗RN. Moreover, for any middle linear map f : M × N → C to an abelian group C, there is a unique homomorphism ¯f : M ⊗RN → C such that ¯f ı = f .
As an easy exercise of universal property, one can easily see that:
Corollary 0.2. The tensor product is unique up to isomorphism.
Proof. We first construct M ⊗RN. Let F be the free abelian group on the set M ×N. Let K be the subgroup generated by following elements (for all a, a1, a2 ∈ M, b, b1, b2 ∈ N, r ∈ R):
(a1+ a2, b) − (a1, b) − (a2, b);
(a, b1+ b2) − (a, b1) − (a, b2);
(ar, b) − (a, rb).
Let M⊗RN := F/K. Clearly, elements in M⊗RN is generated by cosets (a, b) + K. We denote the coset (a, b) + K by a⊗b.
Let ı : M × N → M⊗RN by ı(a, b) = a⊗b. It’s clear that ı is middle linear.
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It remains to show that M⊗RN has the required universal property.
To see this, let f : M × N → C be a middle linear map. Naturally, we consider ¯f : M⊗RN → C by ¯f (a⊗b) := f (a, b). ( Or one can say that f induces a homomorphism f0 : F → C. K is in the kernel, hence it induces ¯f : F/K → C.) The uniqueness actually follows from the universal property of free abelian group F . ¤ Example 0.3.
Z2⊗ZZ3 = 0.
Z2⊗ZZ2 ∼= Z2.
The notion of tensor product is quite general. The point is that it makes sense in various setting.
Example 0.4. Let R, S be rings. Let M = MR be a right R-module and N = RN be a left R-module. If M = SM is a left S-module, then M⊗RN is a left S-module naturally. Similarly, if N = NS, then M⊗RN is a right S-module naturally.
Proposition 0.5. Let MR, MR0 be right R-modules and RN,RN0 be left R-modules. And let f : M → M0, g : N → N0 be module homomor- phisms. Then there is a unique group homomorphism f ⊗g : M⊗RN → M0⊗RN0.
Proof. Consider a middle linear map (f, g) : M × N → M0⊗RN by (a, b) 7→ f (a)⊗g(b). By the universal property, we are done. ¤ Proposition 0.6. Let R be a ring with identity, then there are R- module isomorphism M⊗RR ∼= M and R⊗RN ∼= N.
Proof. There is a natural map : N → R⊗RN by (x) = x⊗1. It’s clear that this is an RM homomorphism. We then construct f : R × N → N by f (r, x) = rx. It’s clear that this is middle linear and thus induces a group homomorphism ¯f : R⊗RN → N by ¯f (r⊗x) = rx. It’s also easy to see that this is a module homomorphism.
Therefore, it suffices to check that ¯f = 1N (which is clear) and
¯f = 1R⊗RN. This mainly due to Xri⊗xi =X
(1⊗rixi) = 1⊗X rixi.
¤ Theorem 0.7 (Tensor product is right exact). Let N1 → Nf 2 → Ng 3 → 0 be an exact sequence of left R-module. Then for any right R-module M, we have
M⊗RN1 −→ M⊗1⊗f RN2 −→ M⊗1⊗g RN3 → 0 is exact.
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Proof. For y ∈ N3, y = g(z) for some z ∈ N2, thus for x ∈ M, x⊗y = (1⊗g)(x⊗z). Hence Im(1⊗g) generate M⊗RN3. It follows that 1⊗g is surjective.
(1⊗g)(1⊗f )(x⊗w) = x⊗gf (w) = x⊗0 = 0.
Therefore, Im(1⊗f ) ⊂ Ker(1⊗g). There is thus an induced map α : M⊗RN2/Im(1⊗f ) → M⊗RN3. It suffices to show that α is an isomorphism. To this end, we intend to construct the inverse map.
Consider x⊗y ∈ M⊗RN3, there is z ∈ N2 such that g(z) = y. We define β0 : M × N3 → M⊗RN2/Im(1⊗f ) by β0(x, y) = x⊗z. We first check that this is well-defined. If z, z0 ∈ N2 such that g(z) = g(z0) = y, then z−z0 ∈ Kerg = Imf . Thus there is w ∈ N1such that z−z0 = f (w).
One verifies that
x⊗z = x⊗(z0+ f (w)) = x⊗z0+ x⊗f (w)
= x⊗z0+ (1⊗f )(x⊗w) = x⊗z0.
It’s routine to check that β0 is middle linear, hence it induces β : M⊗RN3 → M⊗RN2/Im(1⊗f ). One can check that
αβ(x⊗y) = αx⊗z = x⊗g(z) = x⊗y., βα(x⊗z) = β(x⊗g(z)) = x⊗z.
¤ We now concentrate on the case that R is commutative. Let M, N, C be R-modules. By a bilinear map we mean f : M × N → C such that
(2)
f (a1+ a2, b) = f (a1, b) + f (a2, b) f (a, b1+ b2) = f (a, b1) + f (a, b2) f (ar, b) = f (a, rb) = rf (a, b)
for all a, a1, a2 ∈ M, b, b1, b2 ∈ N, r ∈ R. One can have M⊗RN as in the previous construction. Note that M⊗RN is naturally an R- module. And the unique homomorphism M⊗RN → C is an R-module homomorphism.
Example 0.8. Let V, W be a vector space over k with basis {v1, ..., vn}, {w1, ..., wm} respectively. Then V ⊗W is a vector space over k with basis {vi⊗wj}i=1,...,n,j=1,...,m.
Exercise 0.9 (base change). Let ϕ : R → S be a homomorphism of commutative ring with identity. Then S is an R-module. We have R[x]⊗RS ∼= S[x].
Theorem 0.10. Let R, S be rings and MR,RNS,SL be modules. Then there is an isomorphism of abelian groups
HomS(M⊗RN, L) ∼= HomR(M, HomS(N, L)).
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We are not going to prove this. The moral for this is that the ”func- tor” tensor product and the ”functor” Hom are adjoint. We did see this before in linear algebra:
Example 0.11. Let V be a vector space over k. Then Homk(V ⊗V, k) represents the bilinear maps to k. It’s adjoint, Homk(V, Homk(V, k)) = Homk(V, V∗), represents the linear maps from V to V∗, where V∗ de- note the dual vector space of V .
