Exercise A0.V.1.14

Proposition

Let $R$ be a ring, and let $a, b \in R.$ Then the class of $a$ in $R/(b)$ is prime if and only if the class of $b$ in $R/(a)$ is prime.

Proof

Suppose the class of $a$ in $R/(b)$ is prime.

Then $(R/(b))/(((a)+(b))/(b))$ is an integral domain. But

$$(R/(b))/(((a)+(b))/(b)) \cong R/((a)+(b)).$$

And since $(R/(a))/(((a)+(b))/(a))$ is also isomorphic to $R/((a)+(b))$ we know that $(R/(a))/(((a)+(b))/(a))$ is an integral domain.

Thus, $((a)+(b))/(a)$ is a prime ideal in $R/(a).$

Since $((a)+(b))/(a)$ is the image of $(b)$ in $R/(a),$ it is generated by the equivalence class of $b.$ Being a generator of a prime ideal, the class of $b$ is prime in $R/(a).$

The converse follows by symmetry.

$\blacksquare$