I was wondering what the prime ideals of the ring $K[x]$ are, where $K$ is a ring. My guess is that it's any ideal generated by a set of irreducible polynomials over the ring $K$. Have I covered all prime Ideals in $K[x]$ or are there any specific prime ideals I may have missed

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    $\begingroup$ Not all the prime ideals are covered. What about the extensions of the prime ideals of $K$ in $K[x]$? These are also prime ideals. There may be more. $\endgroup$ – Krish Jan 28 '15 at 18:04
  • $\begingroup$ @GeorgesElencwajg are you saying that the prime ideals must have atleast one irreducible polynomial, or that it must have exactly one irreducible polynomial, and no other generators? $\endgroup$ – Andrew Brick Jan 28 '15 at 18:13
  • $\begingroup$ @Andrew: sorry, I misread your question and I deleted my comment. Also: think about Krish's very judicious comment. $\endgroup$ – Georges Elencwajg Jan 28 '15 at 19:32
  • $\begingroup$ If $K$ is a noetherian integral domain, then you are right: every prime ideal is generated by a finite set of irreducible polynomials. $\endgroup$ – user26857 Jan 29 '15 at 21:53
  • $\begingroup$ If $K$ is a pid you still have a nice answer. The prime ideals are $(0)$, $(f)$ for an irreducible polynomial and the maximal ones $m=(p,f)$ where $p \in K$ is a prime and $f \in K[x]$ is a polynomial whose reduction in $K/p[x]$ is irreducible. $\endgroup$ – Bogdan Feb 2 '15 at 8:49

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