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How would I show that if $a+bi\in K$ where $a$,$b$ are real and $\mathbb{Q}\subset K$ is some field, then $a-bi\in K$. This is the last step of a proof I'm trying to do.

Thanks in advance!

EDIT1: How about if $a$,$b$ are algebraic?

EDIT2: I'm suppose to find the smallest normal extension of $Q(\theta)$ where $\theta$ is the only real root of an irreducible degree 3 real polynomial. I know the smallest normal will be the splitting field. So it will be $\mathbb{Q}(\theta,\alpha,\bar{\alpha})$ which is the same as $\mathbb{Q}(\theta,\alpha)$. Therefore, I'm trying to prove that $\bar{\alpha}\in\mathbb{Q}(\theta,\alpha)$.

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I suppose that you mean $K\subseteq\Bbb C$. – Andrea Mori Jan 25 '13 at 20:48
Consider $ \mathbb{Q}(e + i \pi) \subseteq \mathbb{C} $. I find it highly unlikely that $ e - i \pi $ is contained in this field... – Haskell Curry Jan 25 '13 at 20:49
Smallest normal extension of $\mathbb Q(\theta)$ or smallest normal extension of $\mathbb Q$ containing $\mathbb Q(\theta)$? Certainly the first question does not have a unique answer. – JSchlather Jan 25 '13 at 21:25
up vote 2 down vote accepted

There's a much more direct approach to the problem you're trying to solve: just factor $f$ over in $\mathbb{Q}(\theta)$ using knowledge that $\theta$ is a root in that field. Or over $\mathbb{Q}(\theta, \alpha)$ using knowledge of two of its roots in that field.

Incidentally, this same example gives a negative answer to your first question: $\bar{\alpha}$ can't be in $\mathbb{Q}(\alpha)$, because that would imply $\theta \in \mathbb{Q}(\alpha)$ as well, but $\mathbb{Q}(\alpha)$ can't be the splitting field of your polynomial. (it is a degree $3$ extension, but the splitting field has an order-2 automorphism)

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Once you have this idea, there are a number of shortcuts. e.g. if $f$ is monic, $\theta \alpha \bar{\alpha}$ is the negation of the constant term of $f$, and $\theta + \alpha + \bar{\alpha}$ is the negation of the coefficient on $x^2$. – Hurkyl Jan 25 '13 at 21:38
But doesn't $\mathbb{Q}(\theta,\alpha,\bar\alpha)=\mathbb{Q}(\theta,\alpha)$? Or else it would contradict the tower theorem. – user44322 Jan 25 '13 at 21:42
$\mathbb{Q}(\theta, \alpha) = \mathbb{Q}(\alpha, \bar{\alpha}) = \mathbb{Q}(\theta, \bar{\alpha}) = \mathbb{Q}(\theta, \alpha, \bar{\alpha})$. However, $\mathbb{Q}(\theta)$, $\mathbb{Q}(\alpha)$, and $\mathbb{Q}(\bar{\alpha})$ are three different fields. (and none of them are equal to $\mathbb{Q}(\theta, \alpha, \bar{\alpha})$) – Hurkyl Jan 25 '13 at 22:11
But the degree of $\mathbb{Q}(\theta,\alpha,\bar{\alpha})$ over $\mathbb{Q}(\theta,\alpha)$ is 1 because $\mathbb{Q}(\theta,\alpha,\bar{\alpha})$ over $\mathbb{Q}$ is 6. – user44322 Jan 25 '13 at 23:46

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