# Preimage of Prime Ideal under Ring Homomorphism is Prime Ideal

## Theorem

Let $A$ and $B$ be commutative rings with unity.

Let $f : A \to B$ be a ring homomorphism.

Let $\mathfrak p \subseteq B$ be a prime ideal.

Then its preimage $\map {f^{-1} } {\mathfrak p}$ is a prime ideal of $A$.

## Proof 1

By Preimage of Ideal under Ring Homomorphism is Ideal, $\map {f^{-1} } {\mathfrak p}$ is a ideal of $A$.

Let $a, b \in A$ with $a b \in \map {f^{-1} } {\mathfrak p}$.

By definitin of preimage, $\map f a \, \map f b = \map f {a b} \in \mathfrak p$.

Because $\mathfrak p$ is prime, $\map f a \in \mathfrak p$ or $\map f b \in \mathfrak p$.

Thus $a \in \map {f^{-1} } {\mathfrak p}$ or $b \in \map {f^{-1} } {\mathfrak p}$.

Thus $\map {f^{-1} } {\mathfrak p}$ is a prime ideal.

$\blacksquare$

## Proof 2

By Preimage of Ideal under Ring Homomorphism is Ideal, $\map {f^{-1} } {\mathfrak p}$ is a ideal of $A$.

Let $B / \mathfrak p$ be the quotient ring.

By Prime Ideal iff Quotient Ring is Integral Domain, $B / \mathfrak p$ is an integral domain.

By Kernel of Quotient Ring Epimorhpism, the quotient ring epimorphism $B \to B / \mathfrak p$ has kernel $\mathfrak p$.

By Kernel of Composition of Ring Homomorphisms, the composition $A \overset f \to B \to B / \mathfrak p$ has kernel $\map {f^{-1} } {\mathfrak p}$.

By Universal Property of Quotient Ring, there exists a ring homomorphism $A / \map {f^{-1} } {\mathfrak p} \to B / \mathfrak p$.

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By Subring of Integral Domain is Integral Domain, $A / \map {f^{-1} } {\mathfrak p}$ is an integral domain.

By Prime Ideal iff Quotient Ring is Integral Domain, $\map {f^{-1} } {\mathfrak p}$ is a prime ideal of $A$.

$\blacksquare$