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.
$\blacksquare$