Preimage of Image of Subring under Ring Homomorphism
Jump to navigation
Jump to search
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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 2^\circ$