Is the number field closed for conjugate operations? If $a \in \mathbb{F}$,where $\mathbb{F}$ is a number field, then does $\bar{a}\in \mathbb{F}$?
For rational number field,real number field and complex number field, the answer is true. I wonder whether it is always ture even for such field as $\mathbb{Q}(\pi＋i)$.
Any hint will be appreciated.
 A: Consider the field $\Bbb{Q}(\alpha)$ where $\alpha$ is one of the non-real zeros of $p(x)=x^3-2$. The other two zeros are $\overline{\alpha}$ and the real cube root of two.


*

*You get the splitting field of $p(x)$ when you adjoin all but one of the zeros (Vieta relations give you the last zero free of charge).

*That splitting field is a degree six extension so you cannot get it by adjoining a single zero of $p(x)$.

*Therefore $\overline{\alpha}\notin\Bbb{Q}(\alpha)$.



Sketching an argument for the case $a=\pi +i$. This needs more theory.


*

*If we have $\pi-i\in K=\Bbb{Q}(\pi+i)$, then we also have $\pi=\dfrac12((\pi+i)+(\pi-i))\in K$. Consequently we also have $i\in K$.

*But, by transcendence of $\pi$ we know that $\pi +i$ is also transcendental. Therefore $K\simeq L=\Bbb{Q}(x)$ = the field of rational functions in the indeterminate $x$.

*It is not too hard to show that the field $L$ has no non-trivial elements that are algebraic over $\Bbb{Q}$.

*This is a contradiction because our assumption lead to the conclusion that $K$ contains non-rational elements that are algebraic over $\Bbb{Q}$. Most notably, $i\in K$ is algebraic over $\Bbb{Q}$.

