Preimage of Subring under Ring Homomorphism is Subring

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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 $S_2$ 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