Denote $\mathbb{K}$ as the splitting field of $x^3-2$ . I wish to find $\Gamma:=\left\{ \varphi:\mathbb{Q}(\sqrt[3]{2})\to\mathbb{K}|\forall q\in\mathbb{Q}:\varphi(q)=q\right\} $.
Sind $\varphi$ is a homomorphism : $\varphi(2)=2\varphi(1)=2$ , $\varphi(2)=\varphi(\sqrt[3]{2}\times\sqrt[3]{2}\times\sqrt[3]{2})=\varphi(\sqrt[3]{2})\varphi(\sqrt[3]{2})\varphi(\sqrt[3]{2})=(\varphi(\sqrt[3]{2}))^{3}$.
Hence $\varphi(\sqrt[3]{2})$ is a root of $x^3-2$ thus $|\Gamma|\leq3$.
How can I argue that $|\Gamma|=3$ without actually checking if the maps $\sqrt[3]{2}\to\alpha$ are all field homomorphism where $\alpha$ is any root of $x^3-2$ ?
When I took a course in module theory I remember there was some way (or at least the terminology) to say when we can map in this way and result with a homomorphism, maybe this is related (although the course I'm taking now on field theory assumes no knowledge in modules so I would prefer to also (in addition to an argument from module theory, if there is one) to have an argument that explains this without using module theory.