Homomorphism and Subrings: Proving P is a Subring of R

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SUMMARY

The discussion focuses on proving that the set P, defined as P = { r ∈ R | f(r) ∈ T } for a homomorphism f: R -> S and a subring T of S, is a subring of R. The proof establishes that P is nonempty by demonstrating that the zero element of R, 0R, maps to the zero element of S, 0S, confirming that 0S is in T. Furthermore, closure under subtraction and multiplication is verified by leveraging the properties of homomorphisms and the closure of T as a subring, confirming that both f(x - y) and f(xy) belong to T for any x, y ∈ P.

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  • Understanding of ring theory and the definition of a subring.
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Homework Statement



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.


Homework Equations


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)


The Attempt at a Solution



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?
 
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This is entirely correct. And the multiplication is indeed analogous.
 

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