Ring Homomorphism Preserves Subrings/Proof 3
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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.
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $5$: Rings: $\S 24$. Homomorphisms: Theorem $46 \ \text{(i)}$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 57.3$ Ring homomorphisms: $\text{(i)}$