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Math 614: Lecture notes
Sep 24, 2007
More on Hom,
, and Exactness, cont’d
Proposition 1. If i : M ′^ ↪→ M and j : N ′^ ↪→ N are R-module inclusions, then M M ′^
⊗R
N
N ′
∼= M^ ⊗R^ N
Im(i ⊗R 1 N ) + Im(1M ⊗R j)
Proof. We have short exact sequences
S : 0 → M ′^ i −→ M π −→ M ′′^ → 0
and T : 0 → N ′^ i −→ N π −→ N ′′^ → 0
and the right-exactness of tensor product means we have the following com- mutative diagram, where the rows and column are exact:
M ⊗R N ′^ b (^) //
d
M ′′^ ⊗R N ′^ //
e
M ′^ ⊗R N
f (^) // M ⊗R N
g (^) // M ′′^ ⊗R N h
M ′′^ ⊗R N ′′^ //
By the exactness of the column, M ′′⊗RN ′′^ ∼= M^ ′′⊗RN Im e. Since^ g^ is surjective, M ′′^ ⊗R N = Im g, so that M ′′^ ⊗R N ′′^ ∼= Im Im^ ge. (In particular, Im e ⊆ Im g, but
this already follows from the surjectivity of b and g). The exactness of the middle row means that g induces an isomorphism ˜g : M Im^ ⊗R fN ∼ −→ Im g. Moreover, g−^1 (Im e) = Im d + Im f. To see this, first take some z ∈ g−^1 (Im e). Then by surjectivity of b, there is some x with g(z) = e(b(x)) = g(d(x)), so z −d(x) ∈ Ker g = Im f , whence z ∈ Im d+Im f. Conversely, take some element d(x) + f (y) ∈ Im d + Im f. Then g(d(x) + f (y)) = g(d(x)) = e(b(x)) ∈ Im e, so that d(x) + f (y) ∈ g−^1 (Im e). Thus, under the isomorphism ˜g, we have ˜g−^1 (Im e) = (Im d + Im f )/ Im f. Hence, we have
M ′′^ ⊗R N ′′^ ∼=
M ′′^ ⊗R N
Im e
∼= Im^ g Im e
∼= (M^ ⊗R^ N^ )/^ Im^ f (Im d + Im f )/ Im f
∼= M^ ⊗R^ N
Im d + Im f
(The third isomorphism above is induced by ˜g−^1 .) But d = 1M ⊗R j and f = i ⊗R 1 N , so we are done.
The right-exactness of tensor product will allow us to see what extension of scalars does to an R-module, at least in terms of its presentation. We detour here to discuss this important topic:
Presentations
Let M be any R-module. Then there is an R-linear surjection from a free module onto R. Say X is a set and R⊕X^ is a free module that admits an R- linear surjection π : R⊕X^ M. This map has some kernel, say K, which also admits an R-linear surjection from a free module onto it; say π′^ : R⊕Y^ K. Let j : K ↪→ R⊕X^ be the inclusion map, and g = j ◦ π′^ : R⊕Y^ → R⊕X^ the composition. Then it follows easily that the following sequence is (right)- exact: R⊕Y^
g −→ R⊕X^ π −→ M → 0. (1) A right-exact sequence where the first two modules are free (i.e. a se- quence in the form of (1)) is called a presentation of M. We have shown that every R-module has a presentation. If X can be chosen to be a finite set, we already know this is equivalent to M being finitely generated. If both X and Y can be chosen to be finite sets, we say that M is finitely presented. Note that any map of finitely generated free modules with fixed bases (say g : Rn^ → Rm) can be represented uniquely by an m × n matrix of elements of R, in the following sense: Let e 1 ,... , en be the fixed basis of Rn^ and e′ 1 ,... , e′ m
Extension of scalars, applied to presentations
Let ϕ : R → S be a ring homomorphism, let M be an R-module. We have a presentation as in (1) above, which is, of course, right exact. Moreover, we know that tensor product commutes with direct sum, and that S⊗R R ∼= S by a natural isomorphism. Hence, we have the following commutative diagram, with exact rows because extension of scalars is right-exact:
S ⊗R R⊕Y^
1 ⊗g (^) //
(∼=)^ u
S ⊗R R⊕X^
1 ⊗π (^) //
(∼=)^ v
S ⊗R M // 0
S⊕Y^
g′^ // S⊕X^
π′^ //S ⊗ R M^ //^0
So to represent S ⊗R M as a cokernel, it suffices to understand what the action of g′^ is. Fix bases {ey} for R⊕Y^ and {e′ x} for R⊕X^ , and let’s say that the R-matrix of representing g in terms of these bases is (rxy)x∈X,y∈Y. We have v(1 ⊗ re′ x) = ϕ(r)e′ x and u(1 ⊗ rey) = ϕ(r)ey for all x ∈ X, y ∈ Y , and r ∈ R. Thus, for any y ∈ Y , we have
g′(ey) = v((1 ⊗ g)(u−^1 (ey))) = v((1 ⊗ g)(1 ⊗ ey)) = v(1 ⊗ g(ey))
= v
x∈X
rxye′ x
x∈X
v(1 ⊗ rxye′ x) =
x∈X
ϕ(rxy)e′ x
Hence, the matrix representing g′^ is exactly (ϕ(rxy))x∈X,y∈Y , which means that S ⊗R M is the cokernel of this matrix. Thus, the representation of M as the cokernel of an R-matrix leads di- rectly to the representation of S ⊗R M as the cokernel of an S-matrix, simply by applying ϕ to each of the entries!
Flat and surjective base change on tensor products
Let ϕ : R → S be a ring homomorphism and let M , N be S-modules. Then in general, M ⊗R N has two different S-module structures! Let α(M, N ) = M ⊗R N with the S-module structure inherited from M , and β(M, N ) = M ⊗R N with the S-module structure inherited from N. In symbols, the S-action on α(M, N ) is given by s(m ⊗ n) := (sm) ⊗ n, and the S-action on β(M, N ) is given by s(m ⊗ n) := m ⊗ (sn). To see that this gives two different structures in general, let R be any ring, letS = R[x], where x is an indeterminate over R, and let ϕ : R → R[x] be the
obvious ring inclusion. Let M = N = S. Then it is clear that x ⊗ 1 6 = 1 ⊗ x in S ⊗R S. However, in α(S, S), we have x · (1 ⊗ 1) = x ⊗ 1, and in β(S, S), we have x · (1 ⊗ 1) = 1 ⊗ x, so the S-module structures are distinct. From another point of view, for any two S-modules M and N , one can form the tensor products M ⊗S N and M ⊗RN. In general there is a surjection σM,N : M ⊗R N M ⊗S N , since the defining relations on the tensor product over R form a subset of the defining relations on the tensor product over S. [In particular (rm, n) = (ϕ(r)m, n) and (m, rn) = (m, ϕ(r)n).] It happens that σM,N is an isomorphism iff α(M, N ) = β(M, N ). However, for some ring homomorphisms ϕ : R → S, we have α(M, N ) = β(M, N ) (i.e. σM,N is injective) for all pairs M, N of S-modules. One such example is where S = R/I and ϕ is the canonical surjection. Then for any S-module M and r ∈ R, we have for any z ∈ M that rz = ¯rz. So for any s = ¯r ∈ S (r ∈ R), m ∈ M , and n ∈ N , we have in M ⊗R N that
sm ⊗ n = ¯rm ⊗ n = rm ⊗ n = m ⊗ rn = m ⊗ rn¯ = m ⊗ sn,
which shows that α(M, N ) = β(M, N ). Another example is when S = W −^1 R, and ϕ = lW is the localization map. For any r ∈ R, w ∈ W , m ∈ M , and n ∈ N , we have
m
r w
⊗ n =
m w
⊗ rn =
m w
wrn w
m w
⊗ w
rn w
m w
w ⊗
rn w
= m ⊗
r w
n.
Proposition 2. Let ϕ : R → S be a ring homomorphism with the property that σP,Q is an isomorphism for all pairs P, Q of S-modules. Then for any pair M, N of R-modules, we have
S ⊗R (M ⊗R N ) ∼= (S ⊗R M ) ⊗S (S ⊗R N ).
In other words, tensor product commutes with such a base change.
Proof.
S ⊗R (M ⊗R N ) ∼= (S ⊗S S) ⊗R (M ⊗R N ) ∼= (S ⊗R S) ⊗R (M ⊗R N ) (via σS,S ) ∼= (S ⊗R M ) ⊗R (S ⊗R N ) (by associativity of tensor) ∼= (S ⊗R M ) ⊗S (S ⊗R N ) (via σS⊗RM,S⊗RN )
