You forgot to mention it, but R and S must be rings.Originally Posted bysuedenation

T is, of course, a subset of S.

By definition of the homomorphism, f(a+b)=f(a)+f(b) and f(a*b)=f(a)*f(b).

If a set T is a ring, it must have the following properties:

1. a+b=b+a

2. (a+b)+c=a+(b+c)

3. An identity (0) for addition exists.

4. An additive inverse exists for all elements of T.

5. (a*b)*c=a*(b*c)

6. An identity (1) for multiplication exists.

7. * distributes over +.

You can use the homomorphism to assist you. For example, the first property:

f(a+b)=f(a)+f(b)

f(b+a)=f(b)+f(a)

Since f(a+b)=f(b+a) we have that f(a)+f(b)=f(b)+f(a). So addition in T is commutative.

The second property follows the same pattern.

The third property: f(a+0)=f(a)+f(0)=f(a), so f(0)=0T exists.

The fourth property: f(a+-a)=f(a)+f(-a)=f(0)=0T, so f(-a)=-f(a) exists.

The fifth property:

f((a*b)*c)=(f(a)*f(b))*f(c)

f(a*(b*c))=f(a)*(f(b)*f(c))

and the two of these are equal.

The sixth property f(a*1)=f(a)*f(1)=f(a) so f(1)=1T exists.

The seventh property: f(a*(b+c))=f(a)*(f(b)+f(c)) etc.

And, of course, you should put in a statement that 0T and 1T correspond to 0S and 1S respectively, which is easy to show.

-Dan