Prove or Disprove Zxmn Zxm ZxnSolution If m is any nonzero

(Prove or Disprove) Z^x_mn = Z^x_m × Z^x_n

Solution

If m is any non-zero integer, there is a ring homomorphism g : Z -> Zm given by g(x) = x (mod m). g(x+y) = (x+y) mod m = x mod m + y mod m = g(x) + g(y) g(xy) = (xy) mod m = (x mod m) (y mod m) = g(x)g(y) In the same way, if n is another non-zero integer, we have a homomorphism h : Z -> Zn given by h(x) = x mod n. Now, we can combine these homomorphisms into a single homomorphism f : Z -> Zm X Zn, by defining: f(x) = (g(x), h(x)) = (x mod m, x mod n) The ring operations on Zm X Zn are component-wise addition and multiplication: (a,b) + (c,d) = (a+b,c+d) (a,b)(c,d) = (ac,bd) with a, c in Zm and b, d in Zn. f is also a homomorphism, since: f(x+y) = (g(x+y), h(x+y)) = (g(x)+g(y), h(x)+h(y)) = (g(x), h(x)) + (g(y), h(y)) = f(x) + f(y) with a similar result for multiplication. Note that: f(x + mn) = ((x + mn) mod m, (x + mn) mod n) = (x mod m, x mod n) = f(x) and this shows that f(x) only depends on (x mod mn); in other words, we can consider f as a homomorphism from Zmn to Zm X Zn. It remains to show that f is a bijection from Zmn to Zm X Zn. We first show that f is injective (one-to-one). A ring homomorphism f is injective if and only if f(x) = 0 implies x = 0. Let us assume that: f(x) = 0 = (0 mod m, 0 mod n) = (x mod m, x mod n) This means that: x = 0 (mod m) x = 0 (mod n) The first congruence shows that x is a multiple of m, and the second one shows that x is a multiple of n. As m and n are relatively prime, we conclude that x is a multiple of mn, which means that x = 0 in Zmn. This shows that f is injective. To show that f is surjective (onto), we must show that, given any two integers a and b, there exists an integer x such that: f(x) = (a mod m, b mod n) which means: x = a (mod m) [1] x = b (mod n) Because m and n are relatively prime, there exist integers u and v such that: um + nv = 1 [2] x = anv + bmu [3] is a solution of the equations [1]. To see this, multiply [2] by a, this gives: aum + anv = a This shows that anv = a (mod m), and, as bmu is obviously a multiple of m, we have indeed x = a (mod m), and a similar argument shows that x = b (mod m). In summary, f is a ring homomorphism between Zmn and Zm X Zn, and that homomorphism is a bijection; it is therefore an isomorphism.