Count the similarity classes of 3×3 matrices over an extension of QQ

Prove that the number of similarity classes of 3 \times 3 matrices over \mathbb{Q} which have a given characteristic polynomial c(x) in \mathbb{Q}[x] is the same as the number of similarity classes having characteristic polynomial c(x) in F[x], where F is any extension of \mathbb{Q}. Demonstrate that the analogous result does not hold for 4 \times 4 matrices.


Let c(x) \in \mathbb{Q}[x] be a degree 3 polynomial, and suppose c(x) is the characteristic polynomial of a 3 \times 3 matrix over \mathbb{Q}. There are five distinct possibilities for the irreducible factorization of c(x).

  1. c(x) = (x-\alpha)^3 for some \alpha \in \mathbb{Q}. Now (x-\alpha)^3 is the irreducible factorization of c(x) over any extension F of \mathbb{Q}. Over both F and \mathbb{Q}, the possible lists of invariant factors whose product is c(x) are x-\alpha, x-\alpha, x-\alpha; x-\alpha, (x-\alpha)^2, and (x-\alpha)^3. Thus over both F and \mathbb{Q} there are 3 similarity classes of matrices having characteristic polynomial c(x).
  2. c(x) = (x-\alpha)^2(x-\beta) for some distinct \alpha,\beta \in \mathbb{Q}. Similarly, this is the irreducible factorization of c(x) over any extension F of \mathbb{Q}. Over both fields, the possible lists of invariant factors whose product is c(x) are x-\alpha, (x-\alpha)(x-\beta) and (x-\alpha)^2(x-\beta). Thus over both F and \mathbb{Q} there are 2 similarity classes of matrices having characteristic polynomial c(x).
  3. c(x) = (x-\alpha)(x-\beta)(x-\gamma) for some distinct \alpha,\beta,\gamma \in \mathbb{Q}. Again this is the irreducible factorization of c(x) over any extension F of \mathbb{Q}. Over both fields, there is only one possible list of invariant factors whose product is c(x), and that is (x-\alpha)(x-\beta)(x-\gamma). Thus over both F and \mathbb{Q} there is only one similarity class of matrices having characteristic polynomial c(x).
  4. c(x) = (x-\alpha)(x^2-\beta x - \gamma) for some \alpha, \beta, \gamma \in \mathbb{Q}. Now over any extension F of \mathbb{Q}, the polynomial r(x) = x^2 - \beta x - \gamma is either irreducible or factors as a product of linear polynomials. We claim that r(x) is not a perfect square. To see this, suppose we have r(x) = (x-\eta)^2 for some \eta \in F. Now r(x) = x^2 -2 \eta x + \eta^2, but comparing coefficients, we see that \eta = \beta/2 is rational- a contradiction since r(x) is irreducible over \mathbb{Q}. So in both cases, over both F and \mathbb{Q} any list of invariant factors whose product is c(x) must include, as divisors of its largest term, all factors of r(x) as well as x-\alpha. Thus over both F and \mathbb{Q} there is only one similarity class of matrices having characteristic polynomial c(x).
  5. c(x) = x^3 - \alpha x^2 - \beta x - \gamma is irreducible in \mathbb{Q}[x]. There is then only one similarity class of matrices over \mathbb{Q} having characteristic polynomial c(x). Now let F be an extension of \mathbb{Q}. There are five possibilities for the irreducible factorization of c(x) over F.
    1. c(x) is irreducible in F[x]. In this case, there is only one similarity class of matrices over F having characteristic polynomial c(x), as desired.
    2. c(x) factors into one linear and one irreducible quadratic. In any list of irreducible factors whose product is c(x), both of these factors must appear in the largest term. Thus there is only one similarity class of matrices over F having characteristic polynomial c(x), as desired.
    3. c(x) factors into three distinct linear terms. Again, in any list of irreducible factors whose product is c(x), all three factors must appear in the largest term. So there is only one similarity class of matrices over F having characteristic polynomial c(x), as desired.
    4. c(x) = (x-a)(x-b)^2 for some distinct a,b \in F. We claim that in fact a,b \in \mathbb{Q}. To see this, note that by comparing coefficients, we have (1) a+2b = \alpha, (2) 2ab + b^2 = -\beta, and (3) ab^2 = \gamma. Substituting (1) into (2) for a, we have (I) b^2 - \frac{2}{3} \alpha b - \frac{1}{3}\beta = 0. Substituting (1) into (3) for a, we have (II) b^3 - \frac{1}{2} \alpha b^2 + \frac{1}{2} \gamma = 0. Multiplying (2) by b and then substituting (3), we have (III) b^3 + \beta b + 2 \gamma = 0. Now (III) - (II) - \frac{1}{2} \alpha (I) = 0, and thus b = -\frac{9 \gamma + \alpha \beta}{18 \beta + 2\alpha^2}. So b \in \mathbb{Q}, and thus a \in \mathbb{Q}. But then c(x) is reducible over \mathbb{Q}, a contradiction.
    5. c(x) = (x-a)^3 for some a \in F. Comparing coefficients, we see that a = \alpha/3 is rational, a contradiction since c(x) is irreducible over \mathbb{Q}.

Now consider c(x) = x^4 + 2x^2 + 1. Over \mathbb{Q}, c(x) factors into irreducibles as c(x) = (x^2+1)^2, but over \mathbb{C}, it factors into irreducibles as c(x) = (x+i)^2(x-i)^2. Over \mathbb{Q} there are two similarity classes of matrices having characteristic polynomial c(x), but over \mathbb{C} there are 4 such classes.

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