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تسجيل

Modules-Tensor Products of Abelian Groups

المؤلف:  David R. Wilkins

المصدر:  Algebraic Topology

الجزء والصفحة:  94-96

2-7-2017

2012

Proposition 1.1  Z_m ⊗_Z Z_n≅ Z_gcd(m,n)  for all positive integers m and n,  where Z_n = Z/nZ and gcd(m, n) is the greatest common divisor of m and n.

Proof The cyclic groups Z_m and Z_n are generated by a and b respectively,  where a = 1 + Z_m and b = 1 + Z_n. Moreover Z_m = {j.a : j ∈ IZ}, Z_n = {k.b : k ∈ Z}, j.a = 0 if and only if m divides the integer j, and k.b = 0 if and only if n divides the integer k.

Now Z_m ⊗_Z Z_n is generated by elements of the form x ⊗ y, where x ∈ Z_m and y ∈ Z_n. Moreover (j.a) ⊗ (k.b) = jk(a ⊗ b) for all integers j and k. It follows that Z_m ⊗_Z Z_n = {ja⊗b : j ∈ Z}. Thus the tensor product Zm ⊗_Z Z_n is a cyclic group generated by a ⊗ b. We must show that the order of this generator is the greatest common divisor of m and n.

Let r = gcd(m, n). It follows from a basic result of elementary number theory that there exist integers s and t such that r = sm + tn. Then

                r(a ⊗ b) = sm(a ⊗ b) + tn(a ⊗ b) = s((ma) ⊗ b) + t(a ⊗ (nb))

                                  = s(0 ⊗ b) + t(a ⊗ 0) = 0.

It follows that the generator a ⊗ b of Z_m ⊗_Z Z_n is an element of finite order,  and the order of this element divides r.

It remains to show that a ⊗ b is of order r. Now if j, j`, k and k` are integers, and if j.a = j`.a and k.b = k`.b then m divides j – j` and n divides k – k` . But then the greatest common divisor r of m and n divides jk – j`k`, since jk – j`k` = (j – j`)k + j` (k – k`). Let c be the generator 1 + rZ of Z_r. Then there is a well-defined bilinear function f: Z_m × Z → Zr, where f(j.a, k.b) = jk.c for all integers j and k. This function induces a unique group homomorphism ϕ: Z_m ⊗_Z Z → Z_r, where ϕ(x ⊗ y) = f(x, y) for all x ∈ Z_m and y ∈ Z_n. Then ϕ(ja ⊗ b) = jc for all integers j. Now the generator c of Z_r is of order r, and thus jc = 0 only when r divides j. It follows that ja ⊗ b = 0 only when r divides j. Thus the generator a ⊗ j of Z_m ⊗_Z Z_n is of order r, and therefore Z_m ⊗_Z Z_n≅ Z_r, where r = gcd(m, n),  as required.

  There is a fundamental theorem concerning the structure of finitelygenerated Abelian groups, which asserts that any finitely-generated Abelian group is isomorphic to the direct sum of a finite number of cyclic groups.

Thus, given any Abelian group A, there exist positive integers n₁, n₂, . . . , n_k and r such that

                  A≅Z_n1 ⊕ Z_n2 ⊕ · · · ⊕ Z_nk ⊕ Z^r.

Now Corollary 1.4in(Direct Sums and Tensor Products) ensures that Z ⊗Z B ≅ B for any Abelian group B. It follows from Lemma 1.6(Direct Sums and Tensor Products)  that

                          A ⊗_Z B ≅(Z_n1 ⊗_Z B) ⊕ (Z_n2 ⊗_Z B) ⊕ · · · ⊕ (Z_nk ⊗_Z B) ⊕ B^r.

On applying Proposition 1.1, we find in particular that

                      A ⊗_Z Z_m≅ Z_gcd(n₁,m) ⊕ Z_gcd(n₂,m) ⊕ · · · ⊕ Z_gcd(n_k,m) ⊕ Z^r_m

for any positive integer r. Also A ⊗_Z Z ≅ A, by Corollary 1.4in(Direct Sums and Tensor Products).

Note that that Z₁ is the zero group 0, and therefore 0 ⊕ B ≅B for any Abelian group. (Indeed 0 × B = {(0, b) : b ∈ B}, and this group of ordered pairs of the form (0, b) with b ∈ B is obviously isomorphic to B.) We are thus in a position to evaluate the tensor product of any two finitely-generated Abelian groups Note also that if integers m and n are coprime, then Z_mn≅ Z_m ⊕ Z_n.

Indeed let a ∈ Z_m be an element of order m (which therefore generates Z_m),  and let b ∈ Z_n be an element of order n. Then the order of the element (a, b)  of Z_m ⊕ Z_n is divisible by both m and n, and is therefore divisible by mn.  It then follows that (a, b) generates the group Z_m ⊕ Z_n, and this group istherefore isomorphic to Z_mn.

Example Let