# Splitting Field of Cubic Polynomial Over the Rationals

I'm having a hard time wrapping my head around some of concepts Pinter's Abstract Algebra introduces about splitting fields (or root fields, as it calls them). Hopefully if I can be pointed in the right direction with a short question in the exercises I might be able to make more headway with the rest.

What I'm asked to prove is "If c is a complex root of a cubic $a(x) \in \mathbb{Q}[x]$, then $\mathbb{Q}(c)$ is the root field of $a(x)$ over $\mathbb{Q}$".

First I tried taking $\{1,c,c^2\}$ as a basis of $\mathbb{Q}(c)$ as a vector space over $\mathbb{Q}$ and proving that the other roots (the conjugate $\bar{c}$ and some real root $r$) could be expressed in the form $k_0 + k_1c + k_2 c^2 \ (k_0,k_1,k_2 \in \mathbb{Q})$, but this lead nowhere.

In the chapter this exercise comes from the book introduces a few theorems about extending field isomorphisms to isomorphisms of the extensions. I played around with extending the identity automorphism of $\mathbb{Q}$ to an automorphism of the splitting field of $a(x)$ that fixes $\mathbb{Q}$, then showing that to be precisely $\mathbb{Q}(c)$, but again got nowhere; either I'm going down the wrong route or don't understand the material well enough to use it.

Could someone point me in the right direction with this?

Maybe I'm missing something, but the statement does not appear to hold as it stands.

Consider $a(x) = x^{3} - 2$.

Let $\omega$ be a primitive third root of unity. Let $c = \omega \sqrt[3]{2}$.

Then $\mathbb{Q}(c)$, of degree $3$ over over $\mathbb{Q}$, is definitely not the splitting (= root) field of $a(x)$ over $\mathbb{Q}$.

The splitting field $\mathbb{Q}(\sqrt[3]{2}, \omega \sqrt[3]{2}, \omega^{2}\sqrt[3]{2})= \mathbb{Q}(\omega, \sqrt[3]{2})$ has degree $6$ over $\mathbb{Q}$.

• I just double checked and that's the statement verbatim in my copy of the text; no prior or later questions refer to it directly either. If I've been trying to prove something that's in the text through a typo or other error, no wonder I've been having a hard time. – CKA Jun 28 '14 at 16:36
• Is it possible that the typo is cubic $\longrightarrow$ quadratic? – Andreas Caranti Jun 28 '14 at 16:40
• That's possible; I've proven already that a quadratic extension of the rationals using a minimum polynomial p(x) contains both roots of p(x), so it would make sense. I'll carry on assuming that's the typo, then: thanks for the spot. – CKA Jun 28 '14 at 16:46