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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?

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  • $\begingroup$ In a PID a prime ideal is maximal because if $(p)\subset (q)$ then $q$ is a divisor of $p$. As the answer below notes, this is actually not true for UFDs in general, but it is pretty obvious for PIDs. $\endgroup$ – Thomas Andrews Apr 9 '13 at 14:32
  • $\begingroup$ Related to math.stackexchange.com/questions/360940/factor-rings-of-pids $\endgroup$ – user26857 Apr 14 '13 at 22:46
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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.

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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.

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