For cyclic Z-modules Zm and Zn with generators a and b, respectively, show that Zm ⊗Z Zn is isomorphic to Z(m,n) with generator a ⊗ b, where(m,n) is the greatest common divisor of m and n.Everywhere below the sign (x) is used to denote the tensor product. Proof. It is well-known that for any abelian group A, and any natural m there is an isomorphism
h:A/mA Z_m (x)A defined as follows : h(c+mA)=u(x)c, where u is any generator of Z_m, for example u=1+mZ.
Applying this fact for A=Z_n, we obtain that there is an isomorphism
h:Z_n/mZ_nZ_m(x)Z_n (1)
It remains to show that Z_n/mZ_n is isomorphic to Z_((m,n)). To this end let us consider the mapping
〖f: Z〗_nZ_((m,n)), f([c]+mZ_n)=c+(m,n)Z, where c is any integer. F is defined correctly. Indeed, if [c]=[c], then (c-d) is divided by n, and hence by (n,m). Obviously f is a homomorphism, and is surjective. Moreover, the kernel of f is mZ_n. Indeed , if [c] belongs to the kernel, then c is divided by (n,m), and hence c=d(n,m), for some integer d. As is well-known, there are integers l and k such that
(n,m)=nl+mk.
Therefore
c=d(n,m)=n(dl)+m(dk).
Hence (c-m(dk)) belongs to mZ. Thus [c] belongs to mZ_n. For the converse, if [c] belongs to mZ_n, then c=nt+mr, for some integers t and r. Then c is divided by (n,m). Thus we have proved that kerf=mZ_n. This implies that Z_n/mZ_n is isomorphic to Z_((m,n)).
Under the isomorphism (1), the inverse image of 1(x)1 is [1]+mZ_n. Under the second isomorphism [1]+mZ_n is mapped to 1+(m,n)Z. Thus the generator of Z_m(x)Z_n is 1(x)1.




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