0
$\begingroup$

I want to show that the field containing $\mathbb{Q}$, and a pair of complex conjugates is closed under complex conjugation.

I'm not sure if this is even true--just a step that I think would work well for a proof I'm working on. Is this true? If so, how can I approach this?

$\endgroup$
  • $\begingroup$ You could use the field axioms $\endgroup$ – Shailesh Mar 18 '16 at 5:15
  • $\begingroup$ @EthanAlwaise I figured that $|z|^2$ may not be in $\mathbb{Q}$, is it necessarily in $K$? $\endgroup$ – user323886 Mar 18 '16 at 5:18
  • $\begingroup$ @Shailesh I did try that, but for any $a+bi$ I'm not sure how to show $a^2+b^2$ or $2a$ is in $K$. If they're rational, of course, but they may not be. $\endgroup$ – user323886 Mar 18 '16 at 5:38
0
$\begingroup$

I'm not sure I understand your question, but assuming we take what you've said literally:

Consider $\mathbb{Q}(i, \pi + \sqrt{2}i)$. This contains a pair of (non-real) complex conjugates, namely $\pm i$, but it does not contain the complex conjugate $\pi - \sqrt{2} i$ of $\pi + \sqrt{2}i$.

$\endgroup$
  • $\begingroup$ How about if the extension is known to be algebraic? Sorry as I didn't realize that it could be relevant. $\endgroup$ – user323886 Mar 18 '16 at 6:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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