Show: For every $\alpha \in \mathbb{Q}$, the polynomial $X^2+\alpha$ has endless different roots in $\mathbb{Q}^{2\times2}$.

Question

Show that for every $\alpha \in \mathbb{Q}$, the polynomial $X^2+\alpha$ has endless different roots in $\mathbb{Q}^{2\times2}$.

What I did:

Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathbb{Q}^{2\times2}, f=X^2+\alpha \in \mathbb{Q}[X].$

Consider $$f(A) = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix}+\begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} = \\ \begin{pmatrix} a^2+bc & ab+bd \\ ac+cd & bc+d^2 \end{pmatrix} + \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} = \\\begin{pmatrix} a^2+bc + \alpha & ab+bd \\ ac+cd & bc+d^2 + \alpha \end{pmatrix}.$$

How do I get to the claim from here on? Thanks in advance!

Answer 1

By definition,

$\alpha \in \Bbb Q \Longrightarrow \exists p, q \in \Bbb Z, \; \alpha = \dfrac{p}{q} = pq^{-1}; \tag 1$

set

$Y = \begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix} ; \tag 2$

then

$Y^2 = \begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix} \begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix}$ $= \begin{bmatrix} -pq^{-1} & 0 \\ 0 & -pq^{-1} \end{bmatrix} = \begin{bmatrix} -\alpha & 0 \\ 0 & -\alpha \end{bmatrix} = -\alpha I, \tag 3$

that is,

$Y^2 + \alpha I = 0; \tag 4$

thus $Y$ satisfies

$X^2 + \alpha \in \Bbb Q[X]; \tag 5$

note further for

$S \in \Bbb Q^{2 \times 2}, \tag 6$

with $S$ invertible, that

$(S^{-1}YS)(S^{-1}YS) = S^{-1}Y(SS^{-1})YS = S^{-1}YIYS$ $= S^{-1}Y^2S = S^{-1}(-\alpha I)S = -\alpha S^{-1}IS = -\alpha I, \tag 7$

or

$(S^{-1}YS)^2 + \alpha I = 0; \tag 8$

thus $S^{-1}YS$ also satisfies (5).

To see that there are an infinite number of such matrices, let

$S = \begin{bmatrix} s & 0 \\ 0 & 1 \end{bmatrix}, \;0 \ne s \in \Bbb Q; \tag 9$

we have

$S^{-1}YS = \begin{bmatrix} s^{-1} & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix} \begin{bmatrix} s & 0 \\ 0 & 1 \end{bmatrix}$ $= \begin{bmatrix} s^{-1} & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & -p \\ q^{-1}s & 0 \end{bmatrix} = \begin{bmatrix} 0 & -ps^{-1} \\ q^{-1}s & 0 \end{bmatrix}, \tag{10}$

which are distinct for distinct values of $s$.

By further adjusting the vaues of $p$ and $q$ even more such matrices may be obtained; for example, every

$\forall p, q \in \Bbb Z, \; \alpha = pq^{-1} \tag{11}$

yields such a matrix $S^{-1}YS$.

Answer 2

HINT: What happens if $a=-d$ and $bc=-a^2-\alpha$?

Answer 3

Pick some $X$ that maps some vector $v$ to another (independent) vector $w$ and that $w$ maps to $-\alpha v$.

• What does that imply for $X^2$?
• How many different $X$ can you obtain this way? (Say, let $v=e_1$, then how many choices for $w$ are possible?)