Non-zero prime ideals of $F[x]$ are maximal

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.

• all posters have questions about something. something like "ideals of $F[X]$" would be a more meaningful title – miracle173 Feb 14 '14 at 9:44
• Got it. Thank you for the edit. – tmpys Feb 14 '14 at 10:59

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?).
• so every prime Ideal is just a single polynomial? I don't know this notation $P=(f(X))$ – tmpys Feb 14 '14 at 9:46
• @tmpys Is generated by a single polynomial. Polynomial rings over fields are called $PIDs$. You could alternatively just use this to prove that every prime is maximal. If $(f(X))$ is prime, then $f(X)$ is irreducible. Show then that if $(f(X))$ were not prime, then there would be some proper ideal $(g(X))\supseteq (f(X))$. Why is that bad? – Alex Youcis Feb 14 '14 at 9:48
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.