## Homomorphism and Subrings

Let f: R -> S be a homomorphism of Rings and T a subring of S.
Let P = { r belongs to R | f(r) belongs to T}
Prove P is a subring of R.

Theorems used:
If S and nonempty subset of R such that S is closed under multiplication and addition, then S is a subring of R.

If f : R -> S a homomorphism of rings, then f(OR) = 0S
(0R is the 0th element of R and similar for 0S)

First i showed P nonempty. R is a ring So O(R) belongs to R. Then f(0R) = 0S because f is a homomorphism and f maps the zero element to the zero element ( prevevious result)

But T is a Subring of S so 0S belongs to T thus P is nonempty.

(There is a theorem that says if I show P is nonempty I just now show closure under subtraction and multiplication to show P is a subring)

So let x and y belong to P
Now f(x-y) = f(x) - f(y). Doesn't this have to belong to T?
Both f(x) and f(y) are in T since each x and y belong to P
But because T is a subring of S isn't it closed under subtraction already so f(x) - f(y) belongs to T?

Then f(xy) = f(x)f(y) and a similar argument holds?