Question

Let R and S be rings and f : R→ S be a homomorphism. If x is an idempotent in R,

show that f(x) is an idempotent in S. Hence, or otherwise, determine all ring

homorphisms from Z×Z to Z .

Answer 1

Since x is idempotent in R, we have that $x^2 = x$

Consider,

$f(x) = f(x^2) = f(xx) = f(x)f(x) = [f(x)]^2$

Hence f(x) is idempotent in S.

The idempotent of $\mathbb{Z×Z}$ are (0,0), (0,1), (1,0) and (1,1)

Aside from the zero morphism and identity morphism, we seek to see if there are others.

To do this consider,

(1,1) = (0,1) +(1,0)

Also, it is expected that f(1,1) =1 since f is a ring homomorphism.

So, we have 1 = f(1,1) = f ((0,1)+ (1,0)) = f(0,1) + f(1,0)

Case 1

If f(0,1) = 0 then f(1,0) = 1

And we must have

f(a,b) = f(a(0,1) + b(1,0)) = a f(0,1) + b f(1,0) = b

i.e f(a,b) = b

Case 2

If f(0,1) = 1 then f(1,0) = 0

And we must have

f(a,b) = f(a(0,1) + b(1,0)) = a f(0,1) + b f(1,0) = a

i.e f(a,b) = a