# Morphisms, Splitting Fields, and Primitive Cube Roots

Let $\alpha=\sqrt[3]2$ and $\omega=e^{2\pi i/3}$, and let $K=\mathbb{Q}[\alpha,\omega]$ be the splitting field of $f(x)=x^3-2$ over $\mathbb{Q}.$ Determine all morphisms $K\to K$.

Ok, so I have an exam tomorrow and I am a little confused on this kind of stuff. I know ω is a primitive cube root of unity. I know $f(x)$ has 1 real root, $\alpha$, and 2 complex roots. I just am a little confused on how you determine morphisms from $K\to K$. Can someone please help explain this to me? It would really help me out a lot, thanks.

• Welcome to Math.SE! For future reference we use Mathjax here (e.g. $\theta$ for $\theta$): meta.math.stackexchange.com/questions/5020/…. Since this is revision, please say so with a tag :) – Shaun Nov 19 '13 at 19:46
• By "morphisms $\;K\to K\;$" do you mean $\;\Bbb Q$-homomorphisms of the field $\;K\;$ ? – DonAntonio Nov 19 '13 at 20:10
• @Shaun: please don't. See meta.math.stackexchange.com/questions/11694/… for recent discussion about the revision tag, which I am about to remove. – Willie Wong Nov 22 '13 at 11:17

A morphism $K \to K$ is fully determined by the images of $\alpha$ and $\omega$, as the elements of $\mathbb Q$ must map to themselves. Because $\alpha$ is a root of $f$, so must the image of $\alpha$ be; therefore, $\alpha$ must map to one of $\alpha$, $\omega\alpha$, $\omega^2\alpha$. Similarly, $\omega$ is a root of $x^2 + x + 1$ (i.e., of $(x^3 - 1)/(x - 1)$), and therefore must map to either $\omega$ or $\omega^2$. This gives 6 options for a morphism. They are all 6 really possible; how you see this depends on what you already know, but the key is that $K$ has degree 6 over $\mathbb Q$, i.e., has dimension 6 as a $\mathbb Q$-vectorspace.