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Suppose we have a quadratic (Galois) extension of $\mathbb{Q}$, call it $k$ with Galois group $G$. If we look at the ring of integers inside of $k$, call it $\mathcal{O}_k$, is it true that $\mathcal{O}_k$ is stable under $G$? That is, do we have that $\sigma x\in\mathcal{O}_k$ for every $x\in \mathcal{O}_k$ and $\sigma\in G$?

I don't even really know what the routes are that someone would take to look at such a question. I guess this is probably true since elements of the ring of integers are solutions to certain (monic) polynomials with integer coefficients, so the galois group would just change to a different solution of the polynomial, hence remaining in $\mathcal{O}_k$?

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    On second thought, this shouldn't be hard to prove...2011-10-17
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    Your second paragraph is exactly the right approach :) It works for the ring of integers in any number field, too.2011-10-17
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    Okay yeah, thanks. Just didn't want to get my hands dirty, but this all works out rather nicely.2011-10-17
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    For the general case, there's the [normal basis theorem](http://en.wikipedia.org/wiki/Normal_basis) and the important question of *normal integral basis*. See http://en.wikipedia.org/wiki/Normal_integral_basis#Galois_module_structure_of_algebraic_integers.2011-10-17
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    A related fact is that Galois will (transitively) permute the primes above a given rational prime.2011-10-17
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    May I recommend that you write up (and accept) the answer to your own question? This may seem like a strange idea, but it is entirely acceptable behavior.2011-10-18
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    Some people, including yours truly make a living off the fact that the Galois group acts on the ring of integers (and accordingly also on the units therein). Understanding the structure of these $G$-modules is an old and exciting problem, and there is plenty of work left (in a sense, we have barely started).2011-10-18
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    @JBeardz: To prevent this from going off unanswered, and since you figured out the answer on your own, why not post the answer and proof as an answer to your own question?2011-10-18
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    Sorry, I will. Is there any rush?2011-10-19
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    @JBeardz Rush or not... Any chance of your commitment ("I will") ever being coined?2013-06-07

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Let's prove a more general theorem:

Fix an algebraic closure $\overline{\Bbb Q}$ of $\Bbb Q$, and let $\mathcal{O}_{\overline{\Bbb Q}}$ be the ring of integers of $\overline{\Bbb Q}$, i.e. the algebraic integers, i.e. roots of a monic polynomial with integer coefficients. Then, $\mathcal{O}_{\overline{\Bbb Q}}$ is invariant under the action of $\operatorname{Gal}(\overline{\Bbb Q}/\Bbb Q)$.

Let $x$ be an element of $\mathcal{O}_{\overline{\Bbb Q}}$. Then, it satisfies a monic polynomial $f \in \Bbb Z[X]$. Let $\sigma \in \operatorname{Gal}(\overline{\Bbb Q}/\Bbb Q)$. Since $\sigma$ is a field automorphism, in particular a ring homomorphism, we have $f(\sigma(x)) = \sigma(f(x))$. Since $f(x) = 0$, we have $f(\sigma(x)) = 0$, i.e. $\sigma(x)$ is also a root of $f$. Since $f$ is monic, it follows that $\sigma(x) \in \mathcal{O}_{\overline{\Bbb Q}}$, so the theorem is proved.