Let Y:R->S be a homomorphism from a ring R to a ring S. Prove if T is a subring of R, then the set Y(T) = {f in S : f = Y(T) , for an element t in T} is a subring of S.

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- March 27th 2007, 07:00 PMtttcomraderHomorphism Problem
Let Y:R->S be a homomorphism from a ring R to a ring S. Prove if T is a subring of R, then the set Y(T) = {f in S : f = Y(T) , for an element t in T} is a subring of S.

- March 27th 2007, 08:05 PMThePerfectHacker
Y[T]={f(x)|x in T}

If x in T and y in T then xy in T (why?)

Thus, phi(x+y)=phi(x)+phi(y) thus, phi(x)+phi(x) in Y[T].

And phi(0) is identity element.

And phi(x^-1) is inverse.

And it is associative because R is associative.

Finally, phi(x)+phi(y)=phi(x+y)=phi(y+x)=phi(y)+phi(x)

So it is commutative.

This show that <Y[T],+> forms abelian group.

I leave it to you to show everything else.

The same idea appiles.