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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 !

abc
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2 Answers2

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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 $A^4-1=0$.

orangeskid
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    As a small note - you didn't have to factor $x^{4}-1$ , we just ended up multiplying again its factors. If you wanted to demonstrate that $gcd(x^{4}-1,x^{4}+1)=1$ then it is easy to use $\frac{1}{2}\cdot(x^{4}+1)-\frac{1}{2}\cdot(x^{4}-1)=1$ – Belgi Oct 11 '14 at 09:04
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    @Belgi: You are right! You only need that its complementary factor $x^4+1$ is irreducible in $\mathbb{Q}[x]$. – orangeskid Oct 11 '14 at 09:14
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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 $\Bbb{Q}$?

Jyrki Lahtonen
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  • How do you deal with the two cases of the minimal polynomial of $A$ being of degree $3$ and the two cases of it being of degree $2$ ? (the other cases are trivial as $A$ is scalar) – Belgi Oct 11 '14 at 08:41
  • The roots of x^4+1 are complex roots.A 3x3 matrix must have a real root in it's characteristic equation. So at most only two conjugate roots can be factors of the minimal polynomial.... ? – abc Oct 11 '14 at 08:49
  • Ah, got it. thanks! – Belgi Oct 11 '14 at 08:50
  • @abc: What do you know about irreducibility of polynomials over $\Bbb{Q}$? Ever heard of Eisenstein test? – Jyrki Lahtonen Oct 11 '14 at 08:50
  • @Jyrki Lahtonen.No i haven't. – abc Oct 11 '14 at 08:51
  • Ok. What about cyclotomic polynomials (also works here)? – Jyrki Lahtonen Oct 11 '14 at 08:51
  • I am a beginner Mathematics student :| – abc Oct 11 '14 at 08:53
  • Ok. Then write down the factorization of $x^4+1$ over the reals that you get from conjugate pairs of roots. Are the coefficients of those factors rational? Helper: $$x^4+1=(x^4+2x^2+1)-2x^2=(x^2+1)^2-(\sqrt2 x)^2.$$ – Jyrki Lahtonen Oct 11 '14 at 08:53
  • Sorry to bother again, but how did you intend to use Eisenstein or cyclotomic polynomials ? Eisenstein doesn't apply (directly, at least) and the cyclotomic polynomial is $\frac{x^{p}-1}{x-1} = 1+x+...+x^p$ for a prime $p$ which also doesn't seem to help.. – Belgi Oct 11 '14 at 08:56
  • @Belgi: If $f(x)=x^4+1$, then $f(x+1)=x^4+4x^3+6x^2+4x+2$. Also $f(x)=\Phi_8(x)$, the eighth cyclotomic polynomial, and $\Phi_n(x)$ is irreducible over $\Bbb{Q}$ for all $n$. – Jyrki Lahtonen Oct 11 '14 at 08:57
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    @abc: If you have not been given any tools to verify the irreducibility of $x^4+1$ over $\Bbb{Q}$, then IMO whoever gave you this exercise was a bit cruel. As you see from the other answer (expanding what I was hinting at), this irreducibility is absolutely essential for the argument to work out. – Jyrki Lahtonen Oct 11 '14 at 09:12
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    @ Jyrki Lahtonen: I was not given the tools and as a matter of fact i have just learnt what irreducibility is.But thank you for being very patient and persistent in your teaching. – abc Oct 11 '14 at 09:18