0

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

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 $K$.

[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.

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

Sam Y.
  • 614
  • The characteristic polynomial is the same because it is $\det (A-xI)$, which only depends on the entries of $A$ and never goes outside $F$. – lhf May 11 '17 at 11:04

2 Answers2

2

Suppose that $M$ is defined over $F$ and you have an extension $K/F$. Let $\sigma\in Gal(\overline K/F)$ be an automorphism (where $\overline{K}$ is the Galois closure of $K/F$). If $p(M)=0$, then $(\sigma p)(M)=0$, where $\sigma$ acts on polynomials by acting on the coefficients. (Verify this!) In particular, if $M$ satisfies a polynomial, $p$, then it satisfies $\gcd(p,\sigma p)$ too. If $p$ is not fixed by $\sigma$, then in particular, $p$ cannot be the minimal polynomial, because the gcd is of strictly lower degree.

Going to your example, if $M$ satisfied $(x-1)(x-\zeta_3)$, then $M$, being defined over $\mathbb Q$, would also satisfy $(x-1)(x-\zeta_3^2)$, and hence would satisfy $(x-1)$.

Aaron
  • 24,207
  • Thank you for this answer. I got a lot out of what you wrote. But two things to mention/ask: (1) In the result I stated, K is any extension of F...it need not be Galois (2) In the case where K is Galois, how do you get the part about M satisfying gcd($\sigma p, p$)? – Sam Y. May 10 '17 at 20:05
  • @SamY. Ok, so there are problems here, and some of what I said can be repaired easily, some cannot. You don't need $K$ to be Galois, just to be contained in some Galois extension (because if the minimal polynomial is no longer minimal over one extension, it is still no longer minimal in an even larger one). This isn't a problem in characteristic zero, but is in positive characteristic. For (2), if $M$ satisfies two polynomials, then it satisfies every linear combination of them. Now, use the fact that the GCD is a linear combination. – Aaron May 10 '17 at 20:16
  • @SamY. Also, for (2), I am making the assumption that $p$ is monic, so that if it isn't invariant, then (using the Euclidean algorithm), the gcd will be of strictly smaller degree. – Aaron May 10 '17 at 21:30
0

I think this might be helpful...

The example wants us to find all similarity classes of matrices with entries from $\mathbb{Q}$. If the matrices were allowed to have complex coefficients, then $(x - 1)(x - \zeta_3)$ could be included in the list of possible minimal polynomials.

But the matrices must have rational coefficients, and it is impossible for a matrix with rational coefficients to be annihilated by $(x - 1)(x - \zeta_3)$. (This last sentence seems plausible to me, though I do not have a rigorous proof for it. I imagine the proof of this could be quite involved...)

Sam Y.
  • 614