Ring Homomorphism Preserves Subrings/Proof 3

Theorem

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.

Let $S$ be a subring of $R_1$.

Then $\phi \sqbrk S$ is a subring of $R_2$.

Proof

Let $S$ be a subring of $R_1$.

Since $S \ne \O$ it follows that $\phi \sqbrk S \ne \O$.

Let $x, y \in \phi \sqbrk S$.

Then:

$\exists s, t \in S: x = \map \phi s, y = \map \phi t$

So:

 $\ds x +_2 \paren {-y}$ $=$ $\ds \map \phi s +_2 \paren {-\map \phi t}$ $\ds$ $=$ $\ds \map \phi {s +_1 \paren {-t} }$ as $\phi$ is a homomorphism

As $S$ is a subring of $R_1$, it is closed under $+_1$ and the taking of negatives.

Thus $s +_1 \paren {-t} \in S$ and so $x +_2 \paren {-y} \in \phi \sqbrk S$.

Similarly:

 $\ds x \circ_2 y$ $=$ $\ds \map \phi s \circ_2 \map \phi t$ $\ds$ $=$ $\ds \map \phi {s \circ_1 t}$ as $\phi$ is a homomorphism

Because $S$ is a subring of $R_1$, it is closed under $\circ_1$.

Thus $s \circ_1 t \in S$ and so $x \circ_2 y \in \phi \sqbrk S$.

The result follows from Subring Test.

$\blacksquare$