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Suppose that both $a$ and $a^2$ are roots of an irreducible polynomial $f$ over $\mathbb{Q}$. Show that $a^4,a^8\ldots$ are also roots of $f$.

My attempt: I think if I can say there exist automorphism of splitting field of this polynomial which sends $a$ to $a^2$ then I have done but I can't see it.

Thanks in advance.

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    @Peter: $x$, $x-1$. For less trivial cases, think of roots of unity. – 2017-02-23
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    If $|a|\neq1$, this would imply an infinite number of roots. Thus, necessarily $a$ is a certain $n$th root of unity that the splitting field must contain... – 2017-02-23
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    You have an automorphism $\varphi$ of the splitting field sending $a$ to $a^{2}$, because they are both roots. What is $\varphi(a^{2})$? What can you assume about $\varphi(a^{2})$ in relation to your polynomial? – 2017-02-23

3 Answers 3

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Let $L/\mathbb{Q}$ be a splitting field for $f$. Since $f$ is irreducible, $G=\mathrm{Aut}(L)$ acts transitively on the roots of $f$. In particular, this means

  1. if $\psi\in G$ and $\alpha$ is a root of $f$, then so is $\psi(\alpha)$ and

  2. if $\alpha$ and $\beta$ are two roots of $f$, then there exists $\psi\in G$ with $\psi(\alpha)=\beta$.

That means there exists $\phi\in G$ with $\phi(a)=a^2$ (by 2).

Now $\phi(a^2)=\phi(a)^2=(a^2)^2=a^4$ is a root of $f$ (by 1), $\phi(a^4)=\phi(a)^4=a^8$ is a root of $f$ (by 1), etc.

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We have $a,a^2$ roots same polynomial, then $\mathbb Q(\alpha)\cong \mathbb Q(\alpha^2)$.

In other hand if $a$ is an algebraic element over $\mathbb Q$ with odd degree, then $\mathbb Q(a)= \mathbb Q(a^2)$.

Therefore $[\mathbb Q(a):\mathbb Q]=[\mathbb Q(a^2):\mathbb Q]=[\mathbb Q(a^2):\mathbb Q(a^4)][\mathbb Q(a^4):\mathbb Q] $ and $[\mathbb Q(a^2):\mathbb Q(a^4)]$ is odd degree, then $\mathbb Q(a^2)= \mathbb Q(a^4)=\mathbb Q(a)$, then $a^4$ is root of same polynomial.

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If $f$ is irreducible in $\mathbb{Q}[x]$, and $\alpha$ is any root of $f$, then the map $\phi_{\alpha}$ which sends $h(x) \mapsto h(\alpha)$ is an onto homomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}(\alpha)$. The kernel is $(f)$ and so $\phi_{\alpha}$ is an isomorphism from $\mathbb{Q}[x]/(f)$ to $\mathbb{Q}(\alpha)$.

Apply this to $\alpha = a$ and to $\alpha = a^2$, compose the two isomorphisms, and what you find is an isomorphism from $\mathbb{Q}(a)$ to $\mathbb{Q}(a^2)$ that sends $a$ to $a^2$. But since $\mathbb{Q}(a^2) \subseteq \mathbb{Q}(a)$, and since they are isomorphic, you find $\mathbb{Q}(a^2) = \mathbb{Q}(a)$. Therefore, the isomorphism from $\mathbb{Q}(a)$ to $\mathbb{Q}(a^2)$ that sends $a$ to $a^2$ is actually an automorphism of $\mathbb{Q}(a)$.