Preimage of Image of Subring under Ring Homomorphism

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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 $K = \map \ker \phi$ be the kernel of $\phi$.

Let $J$ be a subring of $R_1$.


Then:

$\phi^{-1} \sqbrk {\phi \sqbrk J} = J + K$


Proof

Let $x \in \phi^{-1} \sqbrk {\phi \sqbrk J}$.

Then:

\(\ds x\) \(\in\) \(\ds \phi^{-1} \sqbrk {\phi \sqbrk J}\)
\(\ds \leadsto \ \ \) \(\ds \map \phi x\) \(\in\) \(\ds \phi \sqbrk J\) Definition of Preimage of Element under Mapping
\(\ds \leadsto \ \ \) \(\ds \exists b \in J: \, \) \(\ds \map \phi x\) \(=\) \(\ds \map \phi b\) Definition of Image of Element under Mapping
\(\ds \leadsto \ \ \) \(\ds \map \phi x + \paren {-\map \phi b}\) \(=\) \(\ds 0_{R_2}\) Definition of Ring Negative on $R_2$
\(\ds \leadsto \ \ \) \(\ds \map \phi {x + \paren {-b} }\) \(=\) \(\ds 0_{R_2}\) as $\phi$ is a homomorphism
\(\ds \leadsto \ \ \) \(\ds x + \paren {-b}\) \(\in\) \(\ds K\) Definition of Kernel of Ring Homomorphism
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds J + K\) as $x = b + \paren {x + \paren {-b} }$


So we have shown that:

$\phi^{-1} \sqbrk {\phi \sqbrk J} \subseteq J + K$


Now suppose that $x \in J + K$.

Then:

\(\ds x\) \(\in\) \(\ds J + K\)
\(\ds \leadsto \ \ \) \(\ds \exists b \in J, a \in K: \, \) \(\ds x\) \(=\) \(\ds b + a\)
\(\ds \leadsto \ \ \) \(\ds \map \phi x\) \(=\) \(\ds \map \phi b + \map \phi a\)
\(\ds \) \(=\) \(\ds \map \phi b + 0_{R_2}\) as $a \in K$
\(\ds \) \(=\) \(\ds \map \phi b\)
\(\ds \leadsto \ \ \) \(\ds \map \phi x\) \(\in\) \(\ds \phi \sqbrk J\) as $b \in J$
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \phi^{-1} \sqbrk {\phi \sqbrk J}\) Definition of Preimage of Subset under Mapping


So we have shown that:

$J + K \subseteq \phi^{-1} \sqbrk {\phi \sqbrk J}$


Hence the result by definition of set equality.

$\blacksquare$


Sources

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