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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... –  Jon Beardsley Oct 17 '11 at 21:53
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Your second paragraph is exactly the right approach :) It works for the ring of integers in any number field, too. –  Zev Chonoles Oct 17 '11 at 21:57
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A related fact is that Galois will (transitively) permute the primes above a given rational prime. –  Dylan Moreland Oct 17 '11 at 22:42
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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. –  Gerry Myerson Oct 18 '11 at 0:35
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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). –  Alex B. Oct 18 '11 at 14:38

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