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I have a question regarding some Commutative Algebra facts. I am using the book A Course in Commutative Algebra by Gregor Kemper. On page 24 he states that if $k$ is a field, the polynomial ring $k[x]$ is Noetherian, and not Artinian. And in theorem 2.8 he states:

Let $R$ be a ring. Then the following statements are equivalent:

$R$ is Artinian $\Leftrightarrow$ $R$ is Noetherian and every prime ideal of $R$ is maximal.

Now I learned in my basic algebra courses that if $k$ is a field, then $k[x]$ is a principal ideal domain. And in a PID, every prime ideal is maximal. But this would be in contradiction with theorem 2.8, since $k[x]$ is Noetherian and PID and therefore $k[x]$ must be Artinian. What goes wrong here?

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    $\begingroup$ The ring $k[x]$ has a certain trivial prime ideal that is not maximal. Actually all PIDs have that trivial ideal. $\endgroup$ – Jyrki Lahtonen Oct 9 '16 at 18:47
  • $\begingroup$ You are correct that if $k$ is a field, that $k[x]$ is a PID, because $k[x]$ is a Euclidean domain, with the degree of the polynomial the Euclidean valuation, and every Euclidean domain is a principal ideal domain $\endgroup$ – Chill2Macht Oct 9 '16 at 18:48
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$\cdots (x^3)\subseteq (x^2)\subseteq (x)$ is an infinite decreasing sequence of ideals.

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In a PID, every nonzero prime ideal is maximal. But the condition for an artinian ring is that every prime ideal must be maximal, even the zero ideal (if it is prime). Therefore, the only PIDs that are artinian are fields.

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