Show that there do not exist 3 $\times$ 3 matrices $A$ over $\mathbb{Q}$ such that $A^8 = I $and $A^4 \neq I.$.

Question

Show that there do not exist 3 $\times$ 3 matrices $A$ over $\mathbb{Q}$ such that $A^8 = I $and $A^4 \neq I.$.

I am aware that the minimal polynomial of $A$ divides $(x^8−1)=(x^4−1)(x^4+1)$.If the minimal polynomial divides $x^4+1$ then it will have roots outside $\mathbb{Q}$.The roots of the minimal polynomial are also roots of Characteristic polynomial of A , thus Characteristic polynomial of $A$ has roots outside $\mathbb{Q}$. I am unable to progress from this point onwards.

I would really appreciate some help.

Thanks !

Answer 1

Hint: Prove that $x^4+1$ is irreducible in $\Bbb{Q}[x]$. What factors of $x^4+1$ can thus occur as factors of the minimal polynomial of a $3\times3$ matrices with entries from $\Bbb{Q}$?

Answer 2

We'll prove this:

Let $A\in M_{3\times 3}(\mathbb{Q})$ so that $A^8-1=0$. Then $A^4 -1=0$.

Indeed, let $P(x)$ the characteristic polynomial of $A$. $P$ is a polynomial monic of degree $3$ with coefficients in $\mathbb{Q}$. We know (Cayley-Hamilton)that $P(A)=0$.

Denote by $Q(x) = x^8-1$. By hypothesis we have $Q(A)=0$.

Let $D = GCD(P,Q)$. By the Euclid's algorithm we know that $D$ is a combination of $P$ and $Q$. Therefore

$$D(A)=0$$

Now $Q$ decomposes in $\mathbb{Q}[x]$ into irreducible factors as

$$x^8-1= (x-1)(x+1)(x^2+1)(x^4+1)$$

Now $D$ is a factor of $P$ and so $\deg D \le 3$. But $D$ is also a factor of $x^8-1$ and being of degree $\le 3$ it must be relatively prime to $x^4+1$. It follows that $D$ divides $(x-1)(x+1)(x^2+1)= x^4-1$. Since $D(A)=0$ we conclude $A^4-1=0$.