# Ring Homomorphism Preserves Subrings

## 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 $R_2$.

### Corollary

The image of a ring homomorphism $\phi: R_1 \to R_2$ is a subring of $R_2$.

## Proof 1

Since $S \ne \O$, $\phi \sqbrk S \ne \O$.

From Group Homomorphism Preserves Subgroups, $\struct {\phi \sqbrk S, +_2}$ is a subgroup of $\struct {R_2, +_2}$.

From Homomorphism Preserves Subsemigroups, $\struct {\phi \sqbrk S, \circ_2}$ is a subsemigroup of $\struct {R_2, \circ_2}$.

Thus, as $\struct {R_2, +_2}$ is a group and $\struct {R_2, \circ_2}$ is a semigroup, the result follows.

$\blacksquare$

## Proof 2

From Morphism Property Preserves Closure, $\phi \sqbrk {R_1}$ is a closed algebraic structure.

From Epimorphism Preserves Rings, $\phi \sqbrk S$ is a ring.

Hence the result, from the definition of subring.

$\blacksquare$

## Proof 3

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.

$\blacksquare$