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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?

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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 $\mathbb{Q}$.

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  • $\begingroup$ 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. $\endgroup$ – CKA Jun 28 '14 at 16:36
  • $\begingroup$ Is it possible that the typo is cubic $\longrightarrow$ quadratic? $\endgroup$ – Andreas Caranti Jun 28 '14 at 16:40
  • $\begingroup$ 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. $\endgroup$ – CKA Jun 28 '14 at 16:46

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