You forgot to mention it, but R and S must be rings.Originally Posted by suedenation
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:
3. An identity (0) for addition exists.
4. An additive inverse exists for all elements of T.
6. An identity (1) for multiplication exists.
7. * distributes over +.
You can use the homomorphism to assist you. For example, the first property:
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:
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.