Element of $\text{Aut}_{\mathbb Q}(\mathbb Q(\sqrt[3]2),\mathbb Q^{alg})$.

Question

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 $k=0,1,2$ ?

Answer 1

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.