If we have a ring homomorphism f:R->S, I know that
f(x+y) = f(x) + f(y)
f(xy) = f(x)f(y)
But is it also true that f(x) - f(y) = f(x-y)?
let's see if we can prove it:
f(x-y) = f(x+(-y)) = f(x)+f(-y).
does f(-y) = -f(y)? well, let's see if we can prove it:
f(y) + f(-y) = f(y+(-y)) = f(0) = 0, so f(-y) must be the additive inverse of f(y), -f(y).
therefore (going back to our first proof, now) f(x-y) = f(x) + f(-y) = f(x) + -f(y) = f(x) - f(y). yep, it must be true.
Going back to what was said earlier, I be the OP has already seen a fair amount of group theory, from where he should already know this since every ring homomorphism is a group homomorphism of the rings underlying abelian group.