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

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?).

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  • $\begingroup$ so every prime Ideal is just a single polynomial? I don't know this notation $P=(f(X))$ $\endgroup$ – tmpys Feb 14 '14 at 9:46
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    $\begingroup$ @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? $\endgroup$ – Alex Youcis Feb 14 '14 at 9:48
  • $\begingroup$ sorry I'm super confused. I have been told what Ideals are and what maximals are, and I know what f[x] is but that's about it. This is not something I am graded on.... $\endgroup$ – tmpys Feb 14 '14 at 10:08
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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).

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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.

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