# Find the rank of the $\mathbb{C}[x]$-module $\mathbb{C}^3$ given by a matrix

As the title says, I'm trying to solve a problem which asks me to find the rank of the $$\mathbb{C}[x]$$-module $$N=\mathbb{C}^3$$ given by

$$A= \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}.$$

I'm struggling to see how to do this, because I can't seem to think of a basis. The vector $$e_1$$ clearly spans $$N$$ since $$A(e_1)=e_2,A(e_2)=e_3$$ and hence $$(ax^2 +bx + c)e_1 = (a, b, c)^t$$, but is obviously not linearly independent since $$(x^3 - 1)e_1 = (0, 0, 0)^t$$ where clearly $$x^3-1 \neq 0$$ in $$N$$.

Can anyone suggest an alternative basis? Or is there another way to find the rank of $$N$$ as a $$\mathbb{C}[x]$$-module?

The problem you have has nothing to do with $$e_1$$. For any $$v\in\mathbb C^3$$, you will have $$(x^3-1)\cdot v=0$$. So $$N$$ cannot have a basis in the setup you are looking at.
What one usually does is to consider, instead of $$\mathbb C[x]$$, the quotient $$\mathbb C[x]/(x^3-1)$$. With this new ring of coefficients, the set $$\{e_1\}$$ will be a basis. In fact, any one-element set will be a basis, and so your module is free of rank one.
• Or, if by rank the question means the dimension of $N \otimes_{\mathbb{C}[x]} \mathbb{C}(x)$ as a $\mathbb{C}(x)$-vector space - then the rank would be 0. – Daniel Schepler Mar 27 at 19:56