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HW 11, Problem 1, Part (2): For each z ∈ P , we construct a linear map
φz : M ⊗R N → M ⊗R N ⊗R P, φz (x ⊗ y) = x ⊗ y ⊗ z.
Indeed, it is clear that such a map exists since the map
M × N → M ⊗R N ⊗R P, (x, y) 7 → x ⊗ y ⊗ z,
is R-bilinear. Since φz+z′ = φz + φz′ and φrz = rφz for r ∈ R, the map
(M ⊗R N ) × P → M ⊗R N ⊗R P, (α, z) 7 → φz (α),
is R-bilinear. Hence this induces an R-linear map
(1) (M ⊗RN )⊗RP → M ⊗RN ⊗RP, (x⊗y)⊗z 7 → φz (x⊗y) = x⊗y⊗z.
Conversely, the map
M × N × P → M ⊗R N ⊗R P, (x, y, z) 7 → x ⊗ y ⊗ z,
is trilinear, hence induces an R-linear map
(2) M ⊗R N ⊗R P → (M ⊗R N ) ⊗R P, x ⊗ y ⊗ z 7 → (x ⊗ y) ⊗ z.
We see that the maps given by (1) and (2) are inverses of each other, hence there is an isomorphism
(M ⊗R) ⊗R P ' M ⊗R N ⊗R P, (x ⊗ y) ⊗ z 7 → x ⊗ y ⊗ z. §
HW 11, Problem 7. Note that we can consider B, C as A-modules via the maps f, g, respectively. Consider the multilinear map
B × C × B × C → B ⊗A C
given by (b, c, b′, c′) 7 → bb′^ ⊗ cc′.
This induces an A-linear map
B ⊗A C ⊗A B ⊗A C → B ⊗A C,
sending b ⊗ c ⊗ b′^ ⊗ c′^ to bb′^ ⊗ cc′. Now there is a natural isomorphism
(B ⊗A C) ⊗A (B ⊗A C) ' B ⊗A C ⊗A B ⊗A C,
of A-modules sending (b ⊗ c) ⊗ (b′^ ⊗ c′) to b ⊗ c ⊗ b′^ ⊗ c′. Therefore we get a natural A-linear map
(B ⊗A C) ⊗A (B ⊗A C) → B ⊗A C,
sending (b ⊗ c) ⊗ (b′^ ⊗ c′) to bb′^ ⊗ cc′, and hence an A-bilinear map
(B ⊗A C) × (B ⊗A C) → B ⊗A C, 1
sending (b ⊗ c, b′^ ⊗ c′) to bb′^ ⊗ cc′. This shows that B ⊗A C has a natural ring structure, the multiplication being determined by
(b ⊗ c) · (b′^ ⊗ c′) = bb′^ ⊗ cc′.
The map
A → B ⊗A C, a 7 → f (a) ⊗ 1 = 1 ⊗ g(a),
is clearly a ring homomorphism and makes B ⊗A C into an A-algebra. §
HW 12, Problem 1. Here is a useful lemma:
Lemma 0.1. Suppose A, B are two rings, M is an A-module, P is a B- module and N is an (A, B)-bimodule i.e. N is both an A-module and a B- module in such a way that the actions of A and B commute. Then M ⊗A N (respectively N ⊗B P ) is naturally a B-module (resp. is naturally an A- module) and there is a natural isomorphism of (A, B)-bimodules
(M ⊗A N ) ⊗B P ' M ⊗A (N ⊗B P ).
Proof. We show first that M ⊗A N is naturally a B-module. For each b ∈ B, consider the map
M × N → M ⊗A N, (m, n) 7 → m ⊗ b · n.
This map is A-bilinear since for instance (m, a · n) 7 → m ⊗ b · (a · n) = m ⊗ a · (b · n) = a · (m ⊗ b · n). (We are using at this point that the actions of A and B on N commute.) Hence it induces an A-linear map
φb : M ⊗A N → M ⊗A N, φb(m ⊗ n) = m ⊗ b · n.
Since φb+b′ = φb + φb′ and φbb′ (·) = φb(φb′ (·)), b 7 → φb defines an action of B on M ⊗A N making M ⊗A N into a B-module. Likewise, N ⊗B P is naturally an A-module. Now, for each z ∈ P , the map
M × N → M ⊗A (N ⊗B P ), (x, y) 7 → x ⊗ (y ⊗ z),
is A-bilinear, hence induces an A-linear map
ψz : M ⊗A N → M ⊗A (N ⊗B P ), x ⊗ y 7 → x ⊗ (y ⊗ z).
Thus we get a map
(M ⊗A N ) × P → M ⊗A (N ⊗B P ), (x ⊗ y, z) 7 → x ⊗ (y ⊗ z).
which is easily checked to be B-bilinear. This gives a B-linear map
(M ⊗A N ) ⊗B P → M ⊗A (N ⊗B P ), (x ⊗ y, z) 7 → x ⊗ (y ⊗ z).
which is also easily seen to be A-linear. Likewise there is an (A, B)-linear map in the opposite direction sending x ⊗ (y ⊗ z) to (x ⊗ y) ⊗ z. This shows the required isomorphism. §
HW 13, Problem 2. Let M be a free module over the PID A and N a submodule. Let S be a basis for M. Look at the collection Σ of pairs (T, U ) consisting of subsets T ⊆ S such that N ∩ 〈T 〉 is a free A-module with basis U , ordered by inclusion. i.e. (T, U ) ≤ (T ′, U ′) if T ⊆ T ′^ and if U ⊆ U ′. Clearly Σ is nonempty. Suppose (T 1 , U 1 ) ⊂ (T 2 , U 2 ) ⊂... is a chain in Σ. Let T = ∪iTi and U = ∪iUi. It is easy to see that N ∩ 〈T 〉 is free with basis U. Thus every chain in Σ has an upper bound. By Zorn’s lemma, Σ has a maximal element (T, U ). We claim that T = S. If not pick an element s ∈ S such that s 6 ∈ T , and look at the subset T ′^ = T ∪ {s} of S. If N ∩ 〈T ′〉 = N ∩ 〈T 〉, then (T ′, U ) ∈ Σ and (T, U )〈(T ′, U ) so we have contradicted the choice of (T, U ). If N ∩ 〈T 〉 ( N ∩ 〈T ′〉, the proof we gave in class in the finitely generated case shows that we can pick an element w ∈ 〈T ′〉 such that U ∪ {w} is a basis for 〈T ′〉 ∩ N , so again (T, U ) < (T ′, U ′). Thus we must have T = S, which proves what is required. §
HW 14, Problem 1.
(a) Since M is isomorphic to a sum of modules of the from R/qn^ with q prime, it is enough to check this when M = R/qn. If q 6 = p, and M = R/qn, then M [p] and M/pM are both zero. If M = R/pnR, M [p] = pn−^1 R/pnR ' R/pR and M/pM ' (R/pn)/(pR/pn) ' R/pR also.
(b) Let x/s ∈ M (q)P , where x ∈ M (q) and s ∈ A \ P. Then qnx = 0 for some n. But qn^ ∈ A \ P , since q is not an associate of p. Thus x/s = 0 in M (q)P , hence M (q)P = 0. Now consider M (p). For any s ∈ A \ P , we claim the map
M (p) −→s M (p)
is an isomorphism. This is clear if s is a unit, so it suffices to check this for s = q where q is irreducible and prime to p. Clearly M (p)[q] = 0. Also if x ∈ M (p), then pnx = 0 for some n. Since pn, q are coprime, we can write 1 = apn^ + bq for some a, b ∈ R. Then x = q(bx), so that multiplication by q is surjective on M (p). This shows the claim that multiplication by q (and hence by any s 6 ∈ P ) is an isomorphism on M (p). So it makes sense to talk of s−^1 x for any x ∈ M (p). Now make M (p) into an RP -module by defining
a s
· x = a · s−^1 x.
Consider the canonical map
M (p) → M (p)P.
If x/s ∈ M (p)P , with x ∈ M (p) and s ∈ A \ P , we can write
x s
s
· x =
s−^1 x 1
so that the map above is surjective. If x ∈ M (p) and x/1 = 0 in MP , then there exists s ∈ A \ P such that sx = 0. Since x ∈ M (p) we also
have pmx = 0 for some m, hence x = 0. This shows that the map above is injective also, hence an isomorphism.
(c) Again, it suffices to check this when M = M (q) for some q. If q 6 = p, both sides are zero. If q = p, i.e. M = M (p), this follows from part (b).
(d) Existence was proved in class. For uniqueness, suppose M ' R/(a 1 ) ⊕ · · ·⊕R/(an) and also M ' R/(b 1 )⊕· · ·⊕R/(bm). Pick a prime p dividing a 1. Then p | ai for all i, hence M/pM ' (R/p)n. But also, M/pM ' (R/pR)N where N is the number of bj s divisible by p. Hence n = N ≤ m. Likewise m ≤ n, so we must have m = n; this argument also shows that the primes dividing a 1 are the same as the primes dividing b 1. Let p be any such prime and P the ideal generated by p. Let α 1 ,... , αn and β 1 ,... βn be the powers of p dividing a 1 ,... , an and b 1 ,... bn respectively. Then we see that
MP ' R/pα^1 ⊕... ⊕ R/pαn^ ' R/pβ^1 ⊕ · · · ⊕ R/pβn^ ,
so that αi = βi by what we have proven in class. §
