Why does $A^8=I_3$ implies $A^4=I_3$?

Question

Let $A\in M_3(\mathbb{Q})$ such that $A^8=I_3$. Prove that $A^4=I_3$

(We denote by $m_X$ the minimal polynomial of matrix X and by $p_X$ its characteristic polynomial and by $gr \space X$ its degree)

The solution is the following:

$1)$ $m_a$ divides $P\in\mathbb{Q}[X]$, $$P(x)=x^8-1=(x-1)(x+1)(x^2+1)(x^4+1)$$

$2)$ Since $A^4\neq I_3$, then $p_A$ would have at least one common root with $x^4+1$

$3)$ Since $x^4+1$ is irreducible over $\mathbb{Q}[X]$ we have that $x^4+1|m_A$

$4)$ $A\in M_3(\mathbb{Q})$ so $gr\space m_A\leq3$ and $gr\space x^4+1=4$ , contradiction

I understood only points $1)$ and $4)$. Can you please help me to understand $2)$ and $3)$? Are they even right?

Answer 1

2) Since $m_A\mid(x^4-1)(x^4+1)$, every root of $p_A$ is a root of $(x^4-1)(x^4+1)$. If $A^4\neq\operatorname{Id}_3$, then no root of $x^4-1$ is a root of $p_A$. Therefore, $p_A$ must have at least one root of $x^4+1$.

3) The polynomial $x^4+1$ is irreducible in $\mathbb{Q}[x]$ and it has at least a common root with $m_A$. Since $x^4+1$ and $m_A$ have a common factor, $\gcd(x^4+1,m_A)\neq1$. Since $\gcd(x^4+1,m_A)\mid x^4+1$ and this polynomial is irreducible, $x^4+1\mid\gcd(x^4+1,m_A)$. In particular, $x^4+1\mid m_A$.

Answer 2

I think this version of the same argument is clearer and does involve contradiction:

• $m_A$ divides $x^8-1=(x-1)(x+1)(x^2+1)(x^4+1)$.

• the degree of $m_A$ is at most $3$ because $A\in M_3(\mathbb{Q})$.

• Therefore, $m_A$ divides $(x-1)(x+1)(x^2+1)=x^4-1$.

• Write $x^4-1=m_A(x) q(X)$.

• Then, $A^4-I=m_A(A) q(A)=0$.