TITLE: Element of $\text{Aut}_{\mathbb Q}(\mathbb Q(\sqrt[3]2),\mathbb Q^{alg})$.
QUESTION [2 upvotes]: I have to show that the extension $\mathbb Q(\sqrt[3]2)/\mathbb Q$ is not a Galois extension by showing that $$\text{Aut}(\mathbb Q(\sqrt[3]2))\neq \text{Hom}_{\mathbb Q}(\mathbb Q(\sqrt[3]2),\mathbb Q^{alg})$$
I think we always have that 
$$\text{Aut}(\mathbb Q(\sqrt[3]2))\subset \text{Hom}_{\mathbb Q}(\mathbb Q(\sqrt[3]2),\mathbb Q^{alg}),$$
so I have to show that there is an element in $\text{Hom}_{\mathbb Q}(\mathbb Q(\sqrt[3]2),\mathbb Q^{alg})$ that is not in $\text{Aut}(\mathbb Q(\sqrt[3]2))$.
To me $$\text{Aut}(\mathbb Q(\sqrt[3]2))=\{id,\sigma\}$$
where 
\begin{align*}
\sigma_1 :1&\longmapsto \sqrt[3]2\\
\sqrt[3]2&\longmapsto  \sqrt[3]{4}
\end{align*}
Question 1 : how can I show it's the only one ?
Since the extension $\mathbb Q(\sqrt[3]2)/\mathbb Q$ is separable (since $\mathbb Q$ is of characteristic $0$), we have that  $$|\text{Hom}_{\mathbb Q}(\mathbb Q(\sqrt[3]2),\mathbb Q^{alg})|=[\mathbb Q(\sqrt[3]{2}):\mathbb Q]=3$$
and thus, we have our result. 
Question 2 : I don't see what could be the other homomorphism of $\text{Hom}_{\mathbb Q}(\mathbb Q(\sqrt[3]2),\mathbb Q^{alg})$, would it be $\sqrt[3]{2^k}\longmapsto i\sqrt[3]{2^{k+1}}$ with $k=0,1,2$ ?

REPLY [1 votes]: A ring homomorphism $\mathbb Q(\sqrt[3]2) \to \mathbb Q^{alg}$ must send $\sqrt[3]2$ to another root of $x^3-2$.
There are 3 roots of $x^3-2$ in $\mathbb Q^{alg}$, but only one of them is real.
In particular, an automorphism of $\mathbb Q(\sqrt[3]2)$ can only send $\sqrt[3]2$ to itself and so is the identity.
Therefore, $\text{Aut}(\mathbb Q(\sqrt[3]2))$ is the trivial group and $\text{Hom}_{\mathbb Q}(\mathbb Q(\sqrt[3]2),\mathbb Q^{alg})$ has 3 elements.
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