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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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    @JBeardz Rush or not... Any chance o$f$ 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.