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?

Almost. But, you should just snip out ", $R$ is also a UFD and hence". The fact that every non-zero prime is maximal is a fact true about PIDs but NOT about UFDs. Indeed, consider that $\mathbb{Z}[x]/(x)$ is an integral domain but not a field. 
