Prove that the only automorphism of $\mathbb{Q}(\sqrt[3]{2})$ is the identity automorphism. I am having difficulty with this proof. I have previously proved that for an automorphism on the extension field $K$ of $F$ that when restricted to $F$ is the identity automorphism must take roots of $f(x)\in F[x]$ in $E$ to roots of $f(x)$ in $E$. I was thinking of using the minimal polynomial $x^{3}-2\in\mathbb{Q}[x]$ and the result of previous problem I solved, but I suspect there are automorphisms that don't fix the subfield, so the result can't be used or I need to prove that any automorphism of an extension field must fix the subfield that is being extended or always sends roots in $E$ to roots in $E$.
I am not asking for a solution to this, but maybe a hint or pointer in the right direction. This problem is introduced in the text prior to Galois theory, so please keep that in consideration.
 A: The other answers are correct.  If you have a field extension $K/F$, then each $F$-automorphism is determined by its action on the roots of the minimal polynomial, which is $x^3 -2$.  To add to what's already been said, I'll throw in some Galois theory that hopefully you can appreciate very soon:
If $F$ is a field containing $\mathbb{Q}$, then $K/F$ is a Galois extension $\iff$ $K$ is the splitting field for some $f \in F[x]$.  If $K/F$ is a Galois extension, then $|\text{Aut}(K/F)| = [K:F]$.  Otherwise, $|\text{Aut}(K/F)| < [K:F]$.  In our case, $\mathbb{Q}[\sqrt[3]{2}]$ is not a Galois extension because the minimal polynomial doesn't split completely in it; hence, the size of its automorphism group must be strictly less than $3$.
Furthermore, it is a theorem that $|\text{Aut}(K/F)|$ divides $[K:F]$, even if $K/F$ is not Galois.  We have $[K:F] = 3$, so $|\text{Aut}(K/F)|$ must be a proper divisor of $3$.  We conclude $|\text{Aut}(K/F)| = 1$, and if this group has only $1$ element, it must be the identity.
A: Hint: Think the roots of $x^3-2$, it has only one real root which is $\sqrt[3]{2}$ and automorphism of $K=\mathbb{Q}(\sqrt[3]{2})$ must permute the roots of $x^3-2$. ($K$ clearly does not contain any complex number)
A: An automorphism $\phi$ of $\mathbb{Q}(\sqrt[3]{2})$ must send $\sqrt[3]{2}$ to a root of its minimal polynomial, i.e. $x^3 -2$.
But the only root of $x^3 -2$ contained in $\mathbb{Q}(\sqrt[3]{2})$ is $\sqrt[3]{2}$ itself, so $$\phi(\sqrt[3]{2}) = \sqrt[3]{2}$$ But this implies that $\phi$ is the identity of $\mathbb{Q}(\sqrt[3]{2})$ .
