Is every left ideal of $M_n (F)[x]$ principal?

Question

Let $F$ be a field and $M_n (F)$ the ring of $n\times n$ matrices with entries in $F$ and $n>1$. I know that that $F[x]$ is a PID. Trivially, $M_n (F)$ is not commutative and not a field. My question is that is every left ideal of $M_n (F)[x]$ principal?

Answer 1

We might have better luck with the isomorphic ring $M_n(F[x])$.

In doing a little quick research I've read that Warfield proved that if $R$ is Bezout, then $M_n(R)$ is Bezout. (I don't have a copy of Warfield's paper to confirm this but regardless $M_n(F)$ is left and right Bezout.)

Since $F[x]$ is Noetherian, so is $M_n(F[x])$ (on both sides.) That would mean $M_n(F[x])$ is Noetherian and Bezout, hence a principal right and left ideal ring.

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Just noticed the slides I linked to also suggest such an argument. One can prove that if $R$ is a right principal ideal ring, then so is $M_n(R)$.