Find all field homomorphisms $\sigma:L\to \mathbb{C}$, where $L:=\mathbb{Q}(5^{1/3})$.
Here is what I've thought about:
First of all, since $\sigma(1)=1$, then necessarily $\sigma(a)=a \,\,\,\forall a\in \mathbb{Z}$ and $\sigma(b^{-1})=b^{-1} \,\,\,\forall b\in \mathbb{Z}-\{0\}$, so $\sigma(a/b)=a/b \Rightarrow \sigma$ fixes $\mathbb{Q}$, i.e., $\sigma$ is a $\mathbb{Q}$-homomorphism.
Let $m$ the number of homomorphisms we are looking for. It is easy to see that $L$ is generated by $\{1, 5^{1/3}, 5^{2/3}\}$ as a $\mathbb{Q}$-vector space, so $[L:\mathbb{Q}]=3$. The extension is separable, since $5^{1/3}$ is clearly separable in $\mathbb{Q}$, so that means $m=3$.
Since the extension is not normal ($\mathbb{Q}(5^{1/3})$ is not a splitting field of $\mathbb{Q}$), we have $|\text{Aut}(L|\mathbb{Q})|<m$.
Since $\text{Aut}(L|\mathbb{Q})$ has only two elements (the identity and the permutation of $5^{1/3}$ and $5^{2/3}$), which are homomorphisms, this means I'm supposed to look for one last homomorphism, which is not an automorphism.
Now I'm stuck because I don't know where to look for that one.
Any ideas? Thank you!
Notice that $\Bbb Q (\sqrt[3] 5) = \Bbb Q [\sqrt[3] 5]$ because $\sqrt[3] 5$ is obviously algebraic over $\Bbb Q$, so you have to study where to send an abstract element $a + b \sqrt[3] 5$. Since you've shown that such a morphism fixes $\Bbb Q$, it remains to study where to send $\sqrt[3] 5$.
Notice, on the other hand, that if $\sigma$ is such a morphism, then $\sigma(\sqrt[3] 5) ^3 = \sigma(\sqrt[3] 5 ^3) = \sigma(5) = 5$, so $\sigma(\sqrt[3] 5)$ may be any of the three roots of $x^3 - 5 = 0$.
This means that there are three possibilities for $\sigma$, indexed by the three roots of $x^3 - 5 = 0$: if $r$ is any such root, then $\sigma(a + b \sqrt[3] 5) = a + br$.