Prove that if $F$ is a field, every proper prime ideal of $F[X]$ is maximal.
Should I be using the theorem that says an ideal $M$ of a commutative ring $R$ is maximal iff $R/M$ is a field? Any suggestions on this would be appreciated.
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Hint: Every prime ideal $P$ of $F[x]$ is of the form $P=(f(x))$ for some polynomial $f$. Use this to show that $F[X]/P$ is a domain which is a finite dimensional vector space, and so a field (why?).
Hint $\ $ Polynomial rings over fields enjoy a (Euclidean) division algorithm, hence every ideal is principal, generated by an element of minimal degree (= gcd of all elements). But for principal ideals: contains $\!\iff\!$ 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), $ $ and $ $ irreducible $\!\iff\!$ prime, again by the Euclidean algorithm (or Euclid's Lemma, or $\,F[x]\,$ a UFD).
If $F$ is a field, $F[x]$ is a Euclidean Domain (just the normal division of polynomials you learn in high school). Thus $F[x]$ is a PID and a UFD. In a PID or UFD, proper prime ideals are maximal.