Let $F$ be a field, $P$ a $F[x]$-submodule of $F[x]^n$.
Suppose $M=F[x]^n/P$ is Artinian.
Claim M is finite dimensional over $F$.
Attempt Since $F[x]$ is PID, and $M$ finitely generated, \begin{equation} M \cong F[x]^r\oplus F[x]/(p_1)\oplus \cdots\oplus F[x]/(p_k) \end{equation} by Classification Theorem. If $r>0$, $M$ cannot be Artinian. Hence $M$ is f.d. over $F$.
Question If I am not mistaken, submodule of a free module is free (over PID).
Isn't that imply $P\cong F[x]^n$ and $M\equiv0$? What I am missing here?