Matrix Similarity Classes Over Extension Fields of Q (12.2.13)

Dummit and Foote Abstract Algebra, section 12.2, exercise 13:

Show that there are the same number of similarity classes of $3 \times 3$ matrices over $\mathbb{Q}$ for a given characteristic polynomial over $\mathbb{Q}[x]$ as there are for when the entries are over any extension field of $\mathbb{Q}$. Give an example to show this is not true in general for $4 \times 4$ matrices.

Proof: Consider all the cases of the decomposition of $c_A(x)∈\mathbb{Q}[x]$. Let $a \neq b \neq c$.

$(x+a)(x+b)(x+c)$: The only choice for minimal polynomial is $c_A(x)$, and there is one similarity class over $\mathbb{Q}$ and $F$.

$(x+a)^2(x+b)$: Whether viewed over $\mathbb{Q}$ or $F$, there are two similarity classes.

$(x+a)^3$: As before, there are invariably three similarity classes. These are all a result of $x+a$ being irreducible in both $\mathbb{Q}[x]$ and $F[x]$.

$(x^2+ax+b)(x+c)$: There is only one similarity class over $\mathbb{Q}$. If $x^2+ax+b=(x+v_1)(x+v_2)$ for $v_1 \neq v_2$, then there is again only one similarity class. If $x^2+ax+b=(x+v_1)^2=x+2v_1x+v_1^2$, then $2v_1 ∈ \mathbb{Q}$ so $v_1 ∈ \mathbb{Q}$, even though $x^2+ax+b$ doesn't factor and thus doesn't have zeros in $\mathbb{Q}$.

$x^3+ax^2+bx+c$: If this polynomial decomposes in $F[x]$ to three distinct linear factors, or an irreducible quadratic and a linear factor, or doesn't decompose further, then the minimal polynomial remains the same. If it decomposes into $(x+v)^3$, then comparing coefficients of $x^2$ we obtain $3v∈\mathbb{Q}$ so $v∈\mathbb{Q}$. Therefore $x^3+ax^2+bx+c = (x+v_1)^2(x+v_2)$ and we obtain the relations$$2v_1+v_2 = a$$$$v_1^2+2v_1v_2 = b$$$$v_1^2v_2 = c$$We first observe $v_2 = -2v_1 + a$ to manipulate the second equation$$v_1^2 = \dfrac{2}{3}av_1-\dfrac{1}{3}b$$Substituting these both into the third equation yields a rational quadratic expression over $v_1$, so employing the first observation again and rearranging yields$$(6b-2a^2)v_1 = 9c-ab$$Since the original polynomial can't have zeros in $\mathbb{Q}$, $-v_1 \not ∈ \mathbb{Q}$ so $v_1 \not ∈ \mathbb{Q}$ implying $6b-2a^2=0$ implying $b = \dfrac{1}{3}a^2$ and $c = \dfrac{1}{27}a^3$. Now $x^3+ax^2+bx+c=(x+\dfrac{1}{3}a)^3$ decomposes in $\mathbb{Q}[x]$, a contradiction.

Observe the polynomial $x^4+2x^2+1=(x^2+1)^2$ in $\mathbb{Q}[x]$. There are two lists of invariant factors for matrices with this characteristic polynomial, and thus there are two similarity classes over $\mathbb{Q}$. In $\mathbb{C}$ this polynomial decomposes to $(x-i)^2(x+i)^2$, and there are seen to be four lists of invariant factors and thus four similarity classes over $\mathbb{C}$.$~\square$