How to find all automorphisms of $\mathbb{Q}(\sqrt[3]{5})$? 
Find all automorphisms of $\mathbb{Q}(\sqrt[3]{5})$.

How can I solve the above problem ? Please help someone.
 A: An automorphism of $\mathbb{Q}(\sqrt[3]{5})$ is (by definition) a isomorphism $f:\mathbb{Q}(\sqrt[3]{5})\to \mathbb{Q}(\sqrt[3]{5})$.
Note that
$$\mathbb{Q}(\sqrt[3]{5})=\{a+b\sqrt[3]{5}+c(\sqrt[3]{5})^2\mid a,b,c\in\mathbb{Q}\}.$$
Thus, if $f$ is an automorphism of $\mathbb{Q}(\sqrt[3]{5})$ and I tell you where $f$ sends $\sqrt[3]{5}$, i.e. if I tell you that $f(\sqrt[3]{5})=\alpha$ for some $\alpha\in\mathbb{Q}(\sqrt[3]{5})$, then you will know what $f$ does to any element of $\mathbb{Q}(\sqrt[3]{5})$ is because $f(t)=t$ for all $t\in\mathbb{Q}$ and $f$ is a ring homomorphism, hence
$$f(a+b\sqrt[3]{5}+c(\sqrt[3]{5})^2)=a+bf(\sqrt[3]{5})+cf(\sqrt[3]{5})^2=a+b\alpha+c\alpha^2.$$
Thus, to classify the automorphisms of $\mathbb{Q}(\sqrt[3]{5})$, it suffices to figure out which elements $\alpha\in\mathbb{Q}(\sqrt[3]{5})$ an automorphism $f$ is allowed to send $\sqrt[3]{5}$ to.
Now note that, if $\alpha=f(\sqrt[3]{5})$ for an automorphism $f$, then because $(\sqrt[3]{5})^3-5=0$, we must also have that
$$0=f(0)=f((\sqrt[3]{5})^3-5)=f(\sqrt[3]{5})^3-5=\alpha^3-5.$$
Thus, if $f$ is an automorphism of $\mathbb{Q}(\sqrt[3]{5})$ and $f(\sqrt[3]{5})=\alpha\in\mathbb{Q}(\sqrt[3]{5})$, then $\alpha^3=5$.
Clearly, one option is $\alpha=\sqrt[3]{5}$; this corresponds to the identity automorphism (see the computation above). Can you figure out if there are any other $\alpha\in\mathbb{Q}(\sqrt[3]{5})$ that work? 

If I have put real effort into figuring it out from the above, but you are still stuck, you can mouse over the gray area below for a further hint.

 Hint: what are the $\alpha\in\mathbb{C}$ with the property that $\alpha^3=5$? Are they elements of $\mathbb{Q}(\sqrt[3]{5})$?

