I have been reading through Dummit and Foote, and they note the following result on page 477:

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> **Corollary.** Let $K/F$ be an extension of the field $F$. Let $A \in M\_{n \times n}(F)$. Then the rational canonical form for $A$, the minimal polynomial for $A$ and the characteristic polynomial for $A$ are the same whether these objects are computed over $F$ or over $K$.
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[Note: I am not asking for a proof of this result.]

Later in this section (p. 487), they present an example where they try to find all similarity classes of matrices $A$ with entries from $\mathbb{Q}$ satisfying $A^6 = I$. It is noted that the minimal polynomial $m(x)$ must divide $x^6 - 1$, and that the factorization of $x^6 - 1$ over $\mathbb{Q}[x]$ is:
\begin{equation\*}
x^6 - 1 = (x - 1)(x + 1) (x^2 + x + 1)(x^2 - x + 1)
\end{equation\*}
They note that $m(x)$ must be of degree less than or equal to 3 (by Cayley-Hamilton), and then proceed to make a list of possible minimal polynomials under these constraints.

The list they produce is:

(1) $x-1$

(2) $x + 1$

(3) $x^2 + x +1$

(4) $x^2 - x + 1$

(5) $(x - 1)(x + 1)$

(6) $(x - 1)(x^2 - x +1)$

(7) $(x - 1)(x^2 + x + 1)$

(8) $(x + 1)(x^2 - x +1)$

(9) $(x + 1)(x^2 + x + 1)$

However, I was thinking about the corollary when I was reading this example. If I am interpreting this corollary correctly, there should also only be 9 options for the minimal polynomial if we think of $A$ being in $M\_{n \times n}(\mathbb{C})$. But in $\mathbb{C}[x]$, the factorization of $x^6 - 1$ is
\begin{equation\*}
x^6 - 1 = (x-1)(x+1)(x- \zeta\_3)(x - \zeta\_3^2)(x - \zeta\_6)(x - \zeta\_6^5)
\end{equation\*}
where $\zeta\_3 = e^{2\pi i(\frac{1}{3})}$ and $\zeta\_6 = e^{2 \pi i(\frac{1}{6})}$. From this factorization, it would seem that the polynomial
\begin{equation\*}
(x - 1)(x - \zeta\_3)
\end{equation\*}
could be one option for the minimal polynomial. 

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> But this polynomial does not appear in (1) - (9). Doesn't the above corollary state that it should be there?
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