Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

up vote 3 down vote accepted

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 $\sqrt[3]{2}\mapsto\alpha$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.