# The Preimage of Prime ideals are Prime Ideals

## Problem 412

Let $f: R\to R’$ be a ring homomorphism. Let $P$ be a prime ideal of the ring $R’$.

Prove that the preimage $f^{-1}(P)$ is a prime ideal of $R$.

## Proof.

The preimage of an ideal by a ring homomorphism is an ideal.
(See the post “The inverse image of an ideal by a ring homomorphism is an ideal” for a proof.)

Thus, $f^{-1}(P)$ is an ideal of $R$.

We prove that the ideal $f^{-1}(P)$ is prime.
Suppose that we have $ab\in f^{-1}(P)$ for $a, b\in R$. Then we have $f(ab) \in P$.
Since $f$ is a ring homomorphism, we obtain
\begin{align*}
f(a)f(b)=f(ab)\in P.
\end{align*}

Since $P$ is a prime ideal, it follows that either $f(a)\in P$ or $f(b)\in P$.
Hence we have either $a\in f^{-1}(P)$ or $b\in f^{-1}(P)$.
This proves that the ideal $f^{-1}(P)$ is prime.

Let $f:R\to R'$ be a ring homomorphism. Let $I'$ be an ideal of $R'$ and let $I=f^{-1}(I)$ be the preimage...