Preimage of Subring under Ring Homomorphism is Subring
Theorem
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $S_2$ be a subring of $R_2$.
Then $S_1 = \phi^{-1} \sqbrk {S_2}$ is a subring of $R_1$ such that $\map \ker \phi \subseteq S_1$.
Proof
Let $K = \map \ker \phi$ be the kernel of $R_1$.
We have that $0_{R_2} \in S_2$ and so $\set {0_{R_2} } \subseteq S_2$.
From Subset Maps to Subset:
- $\phi^{-1} \sqbrk {\set {0_{R_2} } } \subseteq \phi^{-1} \sqbrk {S_2} = S_1$
But by definition, $K = \phi^{-1} \sqbrk {\set {0_{R_2} } }$
and so $S_1$ is a subset of $R_1$ containing $K$, that is:
- $K \subseteq S_1 \subseteq R_1$
Now we need to show that $S_1$ is a subring of $R_1$.
Let $r, r' \in S_1$.
Then $\map \phi r, \map \phi {r'} \in S_2$.
Hence:
\(\ds \map \phi {r +_1 r'}\) | \(=\) | \(\ds \map \phi r +_2 \map \phi {r'}\) | as $\phi$ is a ring homomorphism | |||||||||||
\(\ds \) | \(\in\) | \(\ds S_2\) | because $S_2$ is a subring |
So:
- $r + r' \in S_1$
Then:
\(\ds \map \phi {-r}\) | \(=\) | \(\ds -\map \phi r\) | Group Homomorphism Preserves Inverses | |||||||||||
\(\ds \) | \(\in\) | \(\ds S_2\) | because $S_2$ is a subring |
So:
- $-r \in \phi^{-1} S_1$
Finally:
\(\ds \map \phi {r \circ_1 r'}\) | \(=\) | \(\ds \map \phi r \circ_2 \map \phi {r'}\) | as $\phi$ is a ring homomorphism | |||||||||||
\(\ds \) | \(\in\) | \(\ds S_2\) | because $S_2$ is a subring |
So:
- $r \circ_1 r' \in S_1$
So from the Subring Test we have that $\phi^{-1} \sqbrk {S_2}$ is a subring of $R$ containing $K$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 3^\circ$