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 $f^{-1}(\mathfrak p)$ is a prime ideal of $A$.

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

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

By definitin of preimage, $f(a)f(b) = f(ab) \in \mathfrak p$.

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

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

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

Proof 2
By Preimage of Ideal under Ring Homomorphism is Ideal, $f^{-1}(\mathfrak p)$ 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 $f^{-1}(\mathfrak p)$.

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

.

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

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

Also see

 * Definition:Induced Mapping on Spectra of Rings