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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?

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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.

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  • $\begingroup$ 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. $\endgroup$ – Daniel Schepler Mar 27 at 19:56

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