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What is an example of an algebraic field not closed under complex conjugation?

In all subfields of $\mathbb C$ I think of, complex conjugation is a transposition.

I think I understand that it is often preferable to use "smallest" possible fields. So if we need a field in which a polynomial of degree n has n roots (is algebraically closed), we take $\mathbb Q$ and adjoin the new roots, so every element of the new, extended field (splitting field) has form $a + b\alpha + c \beta + ...$, where $a,b,...\in \mathbb Q$ and $\alpha, \beta, ...$ are the roots (and conversely, for given roots, we can find a (unique?) minimal polynomial). Now, it is claimed here, that for $\alpha = exp(\frac{2i\pi}{3})\cdot 2^{\frac{1}{3}}$, $\overline \alpha \notin Q(\alpha)$. Why is that?

Also, it is stated here that, more generally, for every $\gamma$ being one of the complex solutions to an irreducible (that is "infactorable" in $\mathbb Q$) polynomial of degree 3, $\overline \gamma \notin \mathbb Q(\gamma)$. Could you please explain that in simple, non-Galois terms?

Also, I was advised to take a look at transcendental numbers to answer the question, but am struggling to find a reasonable connection.

It is apparent I don't have but basic knowledge of linear algebra (vector spaces, transformations, affine geometry, matrices, basic polynomial theorems...) and in effort to answer the question, I soon found I need to understand the concepts of field extensions, Galois theory, or even constructible numbers, so I immersed myself in those topics, but feel rather overwhelmed right now, as I don't know which parts of the theory do I need to be able answer the question. So if you don't feel like explaining in detail, please at least point me in the right direction.

THANK YOU SO MUCH FOR ANSWERING!

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1 Answer 1

up vote 5 down vote accepted

Let’s look at three fields, of which the first is $K_0=\mathbb Q(\lambda)$, where $\lambda$ is the real cube root of $2$. I hope you know that $K_0$ consists of all numbers of the form $a+b\lambda+c\lambda^2$, where $a,b,c$ are rational numbers. And of course you see that $K_0$ is a subfield of $\mathbb R$.

Now let’s call $\omega=e^{2i\pi/3}=-1/2+i\sqrt3/2$, the cube root of unity in the second quadrant. And let’s call $K_i=\mathbb Q(\omega^i\lambda)$ for $i=1,2$. You should be able to see that the fields $K_i$, all three of them, are isomorphic to each other, since $\omega\lambda$ is just as good a cube root of $2$ as $\lambda$ is. And they are all different, for instance if $K_1$ were equal to $K_2$, then this supposed field would contain $\omega^2\lambda/(\omega\lambda)=\omega$. But if $\omega$ were in $K_1$, the isomorphic field $K_0$ would also contain a cube root of unity different from $1$, which it doesn’t, since it’s a real field.

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1) So $\omega \notin K_1$ and $\omega ^2 = \overline \omega \in K_1$, therefore $K_1$ is not closed under comp. conj.? In other words, each $K_i$ contains exactly one cube root of unity, because they are isomorfic? 2) Why is $K_0$ of that form, not just $a+b \lambda$? How precisely are these extensions constructed? This prevents me from understanding why $\omega ^2 \in K_1$ and $\omega \in K_2$. –  mirgee Feb 20 at 5:53
    
3) How does the fact that they are isomorfic and different imply that thing about containing\not containing $\omega$ and $\overline \omega$? That isomorphism could be defined arbitrarily, couldn't it? –  mirgee Feb 20 at 6:00
    
4) What is so special about cube roots? This seems to apply to all odd roots of unity. –  mirgee Feb 20 at 6:07
    
5) Also, I think if a field contains $\omega$, then it must contain $\overline \omega$, since $\overline \omega = \omega ^{-1}$. I am totally lost now. –  mirgee Feb 20 at 6:56
    
OK, I know answer to 2)... Also, please feel free to answer only the least stupid questions :) –  mirgee Feb 20 at 7:20

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