# Showing an extension field is closed under complex conjugation

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?

• You could use the field axioms – Shailesh Mar 18 '16 at 5:15
• @EthanAlwaise I figured that $|z|^2$ may not be in $\mathbb{Q}$, is it necessarily in $K$? – user323886 Mar 18 '16 at 5:18
• @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. – user323886 Mar 18 '16 at 5:38

## 1 Answer

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 $\pi + \sqrt{2}i$.

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