# Field homomorphism $\varphi:\mathbb{Q}(\sqrt[3]{2})\to\mathbb{K}$ where $\mathbb{K}$ is the splitting field of $x^3-2$

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.

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If two homomorphisms agree on $\mathbb{Q}$ and agree on $\sqrt[3]{2}$, then they are the same homomorphism; and as you note, there are exactly three possible choices for $\varphi(\sqrt[3]{2})$. To show every choice yields a homomorphism, recall that if $\alpha$ is a root of the monic irreducible polynomial $x^3-2$, then $$\mathbb{Q}[x]/(x^3-2)\cong\mathbb{Q}(\alpha)$$ via an isomorphism that restricts to the identity on (the image of) $\mathbb{Q}$ (in the quotient) and maps (the class of) $x$ to $\alpha$, since the map $\mathbb{Q}[x]\to\mathbb{K}$ given by $x\mapsto \alpha$ has kernel the multiples of $x^3-2$. The fact that the assignment $x\mapsto \alpha$ induces a homomorphism that is the identity on $\mathbb{Q}$ is the "universal property of the polynomial ring" (perhaps that is what you are remembering? $\mathbb{Q}[x]$ is the free $\mathbb{Q}$-algebra on one generator).

In particular, $$\mathbb{Q}(\sqrt[3]{2}) \cong\frac{\mathbb{Q}[x]}{(x^3-2)} \cong\mathbb{Q}(\alpha)$$ via the corresponding composition of maps, which shows for each $\alpha$ there is a homomorphism such that $\sqrt[3]{2}\mapsto\alpha$.

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