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This is the problem I'm trying to solve:

Let $K$ be a sub-field of $\mathbb{C}$ such that $K \nsubseteq \mathbb{R}$. Show that $|K:K\cap\mathbb{R}|=2$ if and only if $\overline{k} \in K$ whenever $k \in K$.

Some examples:

  1. If $K=\mathbb{Q}(\omega)$, where $\omega = e^{2\pi i/3}$, then $\overline{\omega} = \omega^2$ and $|K:\mathbb{Q}|=2$, and this works fine because $K\cap\mathbb{R}=\mathbb{Q}$.

  2. If $K=\mathbb{Q}(\xi)$, where $\xi = e^{2\pi i/5}$, then $\overline{\xi} = \xi^4$ and $\overline{\xi}^2 = \xi^3$ and $|K:\mathbb{Q}|=4$, but this works fine still because $K\cap\mathbb{R}=\mathbb{Q}(\xi+\xi^4) = \mathbb{Q}(\xi^2+\xi^3)$ has degree $2$ over $\mathbb{Q}$. It is easy to see that $\mathbb{Q}(\xi+\xi^4) = \mathbb{Q}(\xi^2+\xi^3)$, because the minimal polinomial of $\xi$ over $\mathbb{Q}$ is $t^4+t^3+t^2+t+1 \in \mathbb{Q}[t]$, so that $\xi^4+\xi^3+\xi^2+\xi+1=0$, yielding $\xi+\xi^4=-1-(\xi^2+\xi^3)$.

  3. If $K = \mathbb{Q}(\omega\sqrt[3]{2})$, with $\omega$ as above, then $K$ is not closed under taking conjugates. I know that $|K:\mathbb{Q}|=3$, which is prime, so I suppose that it has to be $K\cap \mathbb{R}=\mathbb{Q}$ and so another confirmation of the result.

I've tried many other examples but failed to see an argument for the general case. This is an exercise in a book on Galois Theory, in the chapter in which the Galois correspondence was introduced, so I suppose I'm expected to use that correspondence, the automorphisms...

Thanks in advance.

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  • $\begingroup$ Is your field galois over $\mathbb{Q}$? $\endgroup$ – Matt B Apr 1 '16 at 20:43
  • $\begingroup$ If so, then consider the subfield of $K$ fixed by complex conjugation and show this equals $K \cap \mathbb{R}$. $\endgroup$ – Matt B Apr 1 '16 at 20:44
  • $\begingroup$ You don't need $K$ to be Galois over $\Bbb{Q}$. A subfield of $\Bbb{C}$ that is stable under complex conjugation will be Galois over $K\cap \Bbb{R}$ regardless. $\endgroup$ – Jyrki Lahtonen Apr 1 '16 at 21:26
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If $K$ is stable under complex conjugation, then complex conjugation generates a group $G$ of two automorphisms of $K$. Obviously the fixed field of $G$ is then $K^G=K\cap \Bbb{R}$. It follows from Artin's lemma that $[K:K^G]\le2$. See for example Noam Elkies' write-up for that piece of theory. Because $K^G$ is a proper subfield of $K$ it then follows that $[K:K^G]=2$. If you don't fancy using Artin's lemma here, you can avoid it in this special case as follows.

Let $w\in K\setminus K^G$ be a fixed non-real element. Then $u=w-\overline{w}\in K$ is purely imaginary. The claim follows, if we can show that $K=K^G\oplus u K^G$ as a vector space over $K^G$. To that end let $z\in K$ be arbitrary. Then $z_1=(z+\overline{z})/2\in K^G$, and $z_2=z-z_1$ is pure imaginary. Therefore $z_2/u$ is a real number, and thus an element of $K^G$. Hence we can write $$ z=z_1+\frac{z_2}uu $$ as a sum of elements from $K^G$ and $uK^G$ respectively. The sum $K^G+uK^G$ is obviously direct.


Switching to extended hints for the other direction. Let $z\in K$ be such that $\overline{z}\notin K$. If we denote $F:=K\cap\Bbb{R}$, then clearly $z\notin F$. What can you say about the minimal polynomial $m(x)$ of $z$ over $F$?

  • Show that $z$ and $\overline{z}$ are both zeros of $m(x)$.
  • Show that if $m(x)$ is of degree two only, then the other zero of $m(x)$ is also in $K$.
  • Why does this imply that $[K:F]>2$?
  • What happens if $z$ has no minimal polynomial over $F$?
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  • $\begingroup$ In some sense in the first part I am replacing Galois theory with representation theory of $G$. Complex conjugation has the obvious two eigenvalues, and the field is thus a direct sum of the isotypical subrepresentations (=the eigenspaces). Furthermore, the presence of field operations implies that the two eigenspaces have the same dimension. $\endgroup$ – Jyrki Lahtonen Apr 1 '16 at 21:58
  • $\begingroup$ Thanks a lot. It really helped me solving the problem, although in the direction you presented the extended hints, I could only prove the first and the second point under the assumption that $[K:F]=2$. I know that in general, the sum of all roots, the sum of all products of two roots,..., the product of all roots must belong to $F$, but does it follow, foe any degree of the polynomial, that if $z$ is a root then so is $\overline{z}$, or you were making the same assumption? $\endgroup$ – amrsa Apr 2 '16 at 9:31

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