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


*

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

*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.
A: 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.
A: 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 $\mathbb{Q}(a)$. 
