Let $a$ be a prime element in PID. Show that $R/(a)$ is a field. Let $a$ be a prime element in PID. Show that $R/(a)$ is a field.
My attempt: Since $a$ is prime, $(a)$ is a prime ideal of $R$. Since $R$ is a PID,  every nonzero prime ideal of $R$ is maximal. This implies that $(a)$ is maximal and hence $R/(a)$ is a field. 
Is my proof correct?

Hint $\ $ For principal ideals: contains $\equiv$ divides, i.e. $\rm\: (a)\supseteq (b)\!\iff\! a\mid b.\:$ Thus, having no proper containing ideal (maximal) is equivalent to having no proper divisor (irreducible). Further, since PIDs are UFDs, $ $ irreducible $\equiv$ prime in any PID. Summing up, we have deduced that, in a PID, $\rm\ (f)\:$ is maximal $\rm\!\iff\! f\:$ is irreducible $\rm\iff f\:$ is prime. Conclude the proof by invoking the result that $\rm\:R/I\:$ is a field $\rm\iff I\:$ is maximal.
