I'm doing an exercise in Stewart and Tall's Algebraic Number Theory which has me confused.

Suppose $\theta$ satisfies $p(\theta) = \theta^3 - 7\theta + 6 = 0$, but that $\theta \notin \Bbb R$. Find the degree of the field extension $\Bbb R(\theta)/\Bbb R$.

What is meant by $\theta \notin \Bbb R?$ The polynomial $p(x)$ splits entirely over $\Bbb R$ as $(x + 3)(x-2)(x-1)$ so that, considered over $\Bbb R$, the extension is trivial.. how should I think about $\theta$ when all of the roots of $p$ lie in $\Bbb R$?

  • $\begingroup$ Well, it won't be a field, but you can just look at the ring of rational functions in the variable $\theta$ with the property that $\theta^3$ may be exchanged for $7\theta-6$. At least I think that's what they mean. $\endgroup$ – Arthur Aug 22 '17 at 15:57
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    $\begingroup$ @Arthur The exercise suggests that the result be a field, perhaps it's just a typo? $\endgroup$ – Edward Evans Aug 22 '17 at 15:58
  • $\begingroup$ @Cauchy Where the parentheses refer to the ideal generated by $x-1$? $\endgroup$ – Edward Evans Aug 22 '17 at 16:02
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    $\begingroup$ @Cauchy I'm fairly certain that $\Bbb R[x]/(x-1) \cong \Bbb R$ anyway $\endgroup$ – Edward Evans Aug 22 '17 at 16:03
  • $\begingroup$ @Cauchy It seems a little stupid anyway. There are no intermediate extensions between $\Bbb R$ and $\Bbb C$ so if $\theta \notin \Bbb R$ then $\theta \in \Bbb C$, so the degree should just be $2$? $\endgroup$ – Edward Evans Aug 22 '17 at 16:09

This surely must be some typo: as you say, no such $\theta$ exists, since every root of $p(x)$ is real. I'm guessing the problem statement meant to have a slightly different cubic which has non-real roots, in which case the conclusion would be that $\mathbb{R}(\theta)$ is $\mathbb{C}$.


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