The question is:

Let f : R ! S be a homomorphism. Let

T = { f(r) : r 2 R}.

T is called the image of f. Show that T is a subring of S.

Hint: In this problem, you can use the following facts:

f(0R) = 0S,f(−x) = −f(x) for x 2 R, and f(x − y) = f(x) − f(y) for x, y 2 R. These facts were stated in class for isomorphisms only but the same properties (with the same proofs) hold

for homomorphisms.

Thanks very much. :confused: