Preimage of Prime Ideal under Ring Homomorphism is Prime Ideal/Proof 1

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

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.

$\blacksquare$