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Let $R=F[x]$ where $F$ is a field, the $n\times n$ matrix $A=[v_1,\ldots,v_n]\in \operatorname{Mat}_n(R)$. Let $M=R^n/(Rv_1+\cdots+Rv_n)$ be the $R$-module. Show that $M$ is finite dimensional over $F$ iff $f=\det(A)\ne 0$, and in this case, $\dim_FM=\deg f$.

Since $N:=Rv_1+\cdots+Rv_n$ is a submodule of $R^n$, there exists a basis $\{x_1,\ldots,x_n\}$ s.t. $\{a_1x_1,\ldots,a_mx_m\}$ is a basis of $N$ where $m\leq n$ and $a_1\mid a_2\mid\cdots\mid a_n$ and $a_i\in R$. Then $M\cong Rx_1/(a_1x_1)\oplus \cdots \oplus Rx_m/(a_mx_m)\oplus R^{n-m}$. Since $R$ is infinite dimensional over $F$, $M$ is finite dimensional over $F$ if and only if $m=n$. On the other hand, $m=n \Leftrightarrow \{v_1,\ldots, v_n\} $ is a basis of $N$ $\Leftrightarrow v_1,\ldots, v_n$ are linearly independent over $F \Leftrightarrow f\ne0$.

Is this proof right and how to prove $\dim_FM=\deg f$?

Edelweiss Ntu
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  • In the very last line, you mean that the $v_i$ are linear independent over $R=F[x]$. The rest is fine. For the equality, you should take a look at the proof of the structure theorem for finitely generated modules over a PID. You obtain those $a_i$ by multiplying $A$ with matrices in $SL_n(R)$, hence the determinant does not change up to constants. Thus $\deg f = \sum \deg a_i$ and you are done. – MooS Dec 06 '16 at 07:48
  • @MooS Thank you. I'm going to have a look. – Edelweiss Ntu Dec 06 '16 at 07:59
  • @EdelweissNtu : Please look at the recent edits to your question. Proper ways of using \deg, \det, \dim, and \operatorname{} exist. – Michael Hardy Dec 06 '16 at 08:03

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