With $F$ a field and $I$ an ideal of $F[x]$, then $F[x]/I$ is Artinian I want to show that if $F$ is a field and $I$ an ideal of $F[x]$, then $F[x]/I$ is Artinian.
We know immediately that $F[x]$ is a PID. So $I=(a)$ for some $a\in F[x]$. Now, if we can show that $(a)$ is prime (hence maximal in PID), we'll be done, as any field is Artinian.
But I can't see why this must be true, as it would imply that any ideal of $F[x]$ is prime, right? How would I show this? Alternatively, is there a better (maybe more intuitive) way to do this?
 A: As you say, it is certainly not true that every ideal of $F[x]$ is prime (just take a reducible polynomial as the generator). 
Notice that $F[x]$ has a basis $\{1,x,x^2,\dots\}$ as a vector space. When we mod out by a polynomial of degree $d,$ we create a linear dependence among $1,x,\dots,x^d.$ So the quotient is always a finite-dimensional $F$-algebra, which means it is Artinian (since a descending chain of ideals is a descending chain of subspaces, and the dimension can only drop finitely many times).
Edit: Of course, CPM is quite right that it doesn't work if $I$ is the zero ideal. My proof assumes that $I$ contains a non-zero polynomial!
A: This doesn't seem true as stated.  If $I=(0)$, then $F[x]/I \cong F[x]$, but an integral domain is artinian if and only if it is a field.  Since artinian is the same as dimension 0 and noetherian.  So $0$ would be prime and therefore maximal since the dimension is 0.
If $I$ is a non-zero ideal, you know any PID is a Dedekind domain, and If $D$ is a Dedekind domain, and $I$ is a non-zero ideal, then $D/I$ is a principal Artinian ring.  See Theorem 20.11 http://alpha.math.uga.edu/~pete/integral.pdf
