Determine all ring homomorphisms from Z(+)Z to Z.

Answer is (a,b)-->a, (a,b)-->b, (a,b)-->0. Any help would be great!

Let F: Z x Z --> Z be a ring homomorphism.

Since Z + Z is generated by (1, 0) and (0, 1), it suffices to find to find F(1, 0) and F(0, 1).
Why: F(a,b) = F(a(1,0) + b(0,1)) = a F(1, 0) + b F(0, 1), since a,b in Z.
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Note that F(0,0) = 0.
Also, F(1,1) = F((1,1) * (1,1)) = F(1,1) * F(1, 1) = F(1,1)^2.
==> F(1,1)^2 - F(1, 1) = 0
==> F(1,1) * [F(1,1) - 1] = 0
==> F(1,1) = 0 or F(1,1) = 1.

Moreover, F(1, 0) = F(1, 0) * F(1, 0) = [F(1, 0)]^2
==> F(1, 0) = 0 or 1.

Similarly, F(0, 1) = 0 or 1.

We have four possibilities:
(i) F(1, 0) = 0 and F(0, 1) = 0 ==> F(a,b) = 0 for all (a,b) in Z x Z [the zero map].

(ii) F(1, 0) = 0 and F(0, 1) = 1 ==> F(a,b) = b for all (a,b) in Z x Z

(iii) F(1, 0) = 1 and F(0, 1) = 0 ==> F(a,b) = a for all (a,b) in Z x Z

(iv) F(1, 0) = 1 and F(0, 1) = 1 ==> F(a,b) = a+b for all (a,b) in Z x Z.

It's easy to check that the first three maps yield ring homomorphisms.

The fourth possibility is not a ring homomorphism, because it implies that F(1,1) = 2, which is impossible by the remark that F(1,1) = 0 or 1.

I hope this helps!

What about (a,b) -> 2a-3b . That should give you a hint.

Looks like Homomorphism(Z x Z, Z) is isomorphic to Z x Z.