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Let $f(x) \in \mathbb{Z}[x]$ be (edit: monic) irreducible such that if $\theta$ is a root of $f$ then $\mathbb{Q}(\theta)/\mathbb{Q}$ is a Galois extension. In other words, all Galois conjugates of $\theta$ lie in $\mathbb{Q}(\theta)$. Is it necessarily the case that all Galois conjugates of $\theta$ lie in $\mathbb{Z}[\theta]$?

Unfortunately such polynomials $f$ are fairly hard to come by in the wild, so 'fair' examples seem to be scarce. Quadratics are silly, I've seen a few contrived cubics, but none yield a counterexample, cyclotomics are silly etc.

Of course, we need an example so that $\mathbb{Z}[\theta]$ is not the full ring of integers...

...or the result may be true...

I wouldn't be surprised if the proof/counterexample is trivial by the way...

EDIT

The original question has been answered below. Here are one or two follow up questions that I am still very interested in:

  1. (Suggested by JyrkiLahtonen): Suppose we insist that $\mathbb{Z}[\theta]$ is maximal among the subrings of the form $\mathbb{Z}[a]$ with $a$ an algebraic integer of $\mathbb{Q}(\theta)$; can the result be proven, or is there a counterexample?
  2. What can we say about the denominators that appear? I wouldn't be surprised if the only primes that appear in denominators are those that divide the conductor of $\mathbb{Z}[\theta]$ (or $\mathbb{Z}[a]$) in the ring of integers...
abcdefg
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  • My initial reaction was What a cool question! Wonder what's going on? But, if my counterexample checks out, then it sorta ruins the fun. I am all for finding and formulating an improved version of this question that rules out my trick. – Jyrki Lahtonen Jan 10 '17 at 08:18
  • May be we want to insist that $\Bbb{Z}[\theta]$ is maximal among the subrings of the form $\Bbb{Z}[a]$ with $a$ an algebraic integer of $\Bbb{Q}(\theta)$? – Jyrki Lahtonen Jan 10 '17 at 08:34
  • Even with quadratics, I think things will go wrong if $f$ is not monic. What happens with $\theta=\frac{1+\sqrt5}4$? – Mathmo123 Jan 10 '17 at 09:01
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    @Mathmo123: True. But I thought it was "implied" that $\theta$ is an algebraic integer. Also, with your $\theta$ we get $$1-2\theta^2=\frac{1-\sqrt{5}}4.$$ If you take a non-integer $\theta$, then the ring $\Bbb{Z}[\theta]$ becomes larger, making it easier to include the conjugates in some sense. I found it more natural to go to a non-maximal order within the ring of algebraic integers. – Jyrki Lahtonen Jan 10 '17 at 09:30
  • @Mathmo123 , @ jyrki , Yes, sorry, I should have specified that $f$ here is monic, so that $\theta$ is an algebraic integer and, like @ jyrki says, $\mathbb{Z}[\theta]$ is a (potentially non-maximal) order of the ring of algebraic integers. Of course, the question is only ever interesting when $\mathbb{Z}[\theta]$ is a non-maximal order. – abcdefg Jan 10 '17 at 14:35
  • Side question though: given $\theta$, a root of a not-necessarily-monic quadratic (exactly like what @Mathmo123 provided), is there an easy way to see whether $\sigma(\theta)$ lies in $\mathbb{Z}[\theta]$? – abcdefg Jan 10 '17 at 15:18
  • @JyrkiLahtonen I think your counterexample checks out - thanks :). I've edited the question to reflect follow up questions that I'm interested in (like the one you included in your comment). By the way, what is the protocol on marking the answer below as the accepted answer in light of having follow up questions that aren't yet answered? – abcdefg Jan 11 '17 at 09:02
  • I think the way you are doing it is fine. Some users (most often those who answered an accidentally easy question) insist that the follow-up should always be a separate question. I suppose there is some merit to doing it that way, but I fully endorse the way you did it, with the history of the question in plain view (those complaints are often about the original version of the question disappearing completely). – Jyrki Lahtonen Jan 11 '17 at 09:09
  • Anyway, don't accept my answer. If you do that some users will view it as a signal not to look at your question. – Jyrki Lahtonen Jan 11 '17 at 09:11

1 Answers1

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I think there is the following rather trivial way of producing cubic counterexamples, namely using integer multiples of non-counterexamples.

Let $u=2\cos(2\pi/9)$. It is well known that the minimal polynomial of $u$ is $g(x)=x^3-3x-1$, and that $\sigma(u)=u^2-2=2\cos(4\pi/9)$ is a conjugate (the third conjugate is then $\sigma^2(u)=2\cos(8\pi/9)=2-u-u^2$ as the sum of all three conjugates is obviously zero). Therefore $L=\Bbb{Q}(u)$ is the splitting field of $g(x)$, and hence cyclic Galois of degree three.

Consider the number $\theta=3u$. It immediately follows that $L=\Bbb{Q}(\theta)$. One of its conjugates is $$\theta':=\sigma(\theta)=3(u^2-2)=3u^2-6=\frac{\theta^2}3-6.$$ Because $\theta$ is an algebraic integer its (monic) minimal polynomial $m(x)$ has integer coefficients. So if $p(x)\in\Bbb{Z}[x]$ is any polynomial, then $p(x)=q(x)m(x)+r(x)$ for some at most quadratic $r(x)\in\Bbb{Z}[x]$. So if $p(\theta)=\theta'$ then also $r(\theta)=\theta'$. This is impossible because the representation of any element of $L$ as a quadratic polynomial in $\theta$ with rational coefficients is unique.

Jyrki Lahtonen
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  • Seems related to http://math.stackexchange.com/questions/1767252/expressing-the-roots-of-a-cubic-as-polynomials-in-one-root. – lhf Jan 10 '17 at 09:18