Preimage of Ideal under Ring Homomorphism is Ideal

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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 an ideal of $R_2$.


Then $S_1 = \phi^{-1} \sqbrk {S_2}$ is an ideal of $R_1$ such that $\map \ker \phi \subseteq S_1$.


Proof

From Preimage of Subring under Ring Homomorphism is Subring‎ we have that $S_1 = \phi^{-1} \sqbrk {S_2}$ is a subring of $R_1$ such that $\map \ker \phi \subseteq S_1$.

We now need to show that $S_1$ is an ideal of $R_1$.


Let $s_1 \in S_1, r_1 \in R_1$.

Then:

\(\ds \map \phi {r_1 \circ_1 s_1}\) \(=\) \(\ds \map \phi {r_1} \circ_2 \map \phi {s_1}\) as $\phi$ is a homomorphism
\(\ds \) \(=\) \(\ds S_2\) as $S_2$ is an ideal of $R_2$

Thus:

$r_1 \circ_1 s_1 \in \phi^{-1} \sqbrk {S_2} = S_1$

Similarly for $s_1 \circ_1 r_1$.


So $S_1$ is an ideal of $R_1$.

$\blacksquare$


Sources

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