For $k$ a field, $k[x]$ is Noetherian but not Artinian?

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?

• The ring $k[x]$ has a certain trivial prime ideal that is not maximal. Actually all PIDs have that trivial ideal. – Jyrki Lahtonen Oct 9 '16 at 18:47
• 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 – Chill2Macht Oct 9 '16 at 18:48

$\cdots (x^3)\subseteq (x^2)\subseteq (x)$ is an infinite decreasing sequence of ideals.