Answer to Question #146091 in Abstract Algebra for Sourav Mondal

Question

Let R and S be commutative rings and

f : R -->S be a ring homomorphism.

Show that the inverse image of a prime ideal in S is a grime ideal in R

Answer 1

A proper ideal $P$  of a commutative ring $R$  is prime if it has the following property: if $a$  and $b$  are two elements of $R$  such that their product $ab$  is an element of $P$, then $a$ is in $P$  or  $b$ is in $P$.

Let $R$ and $S$ be commutative rings and $f : R \to S$ be a ring homomorphism. Let us show that the inverse image of a prime ideal $P$ in $S$ is a prime ideal in $R$. Let $Q=f^{-1}(P)\subset S$ and $ab\in Q$. Then $f(ab)\in P$. Since $f$ is a homomorphism, $f(ab)=f(a)f(b)$, and thus $f(a)f(b)\in P$. Since $P$ is prime, $f(a)\in P$ or $f(b)\in P$. Therefore, $a\in f^{-1}(P)=Q$ or $b\in f^{-1}(P)=Q$. It follows that $Q$

is a prime ideal.