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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?

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    $\begingroup$ Simply that if $P\simeq F[x]^n$, it is not the canonical injection which is this isomotphism. Counter-example: the ideal $n\mathbf Z$ is isomorphic to $\mathbf Z$, yet $\mathbf Z/n\mathbf Z\ne 0$. $\endgroup$ – Bernard Aug 1 '18 at 23:12
  • $\begingroup$ It make sense now, thanks for clarification @Bernard. $\endgroup$ – Jo' Aug 1 '18 at 23:24

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