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Hi everyone,

let $D$ be a skew field, which is finite dimensional over its center $k$. Assume that $k$ is a number field, and let $\mathcal{O}_D$ be the set of elements $z\in D$ which are roots of a monic polynomial with coefficients in $\mathcal{O}_k$ . Is $\mathcal{O}_D$ a subring of $D$ ?

Thanks!

G.

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    $\begingroup$ I don't know. If it is a subring, I would expect that you could show it for the case of the quaternions, and then generalize that argument. Gerhard "Not Yet Ready For Counterexamples" Paseman, 2012.10.12 $\endgroup$ Commented Oct 12, 2012 at 16:06

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No, the "integral" elements are not a subring. There are at least two ways to understand this.

In the case of $D$ being the integral Hamiltonian quaternions, or even the Hurwitz integers therein (adjoining $(1+i+j+k)/2$ to give a maximal subring), we can easily conjugate ourselves to another subring: using the model of $D$ inside two-by-two complex matrices spanned over $\mathbb R$ by $\pmatrix{1 & 0 \cr 0 & 1}$, $\pmatrix{i & 0 \cr 0 & -i}$, $\pmatrix{0 & 1 \cr -1 & 0}$, and $\pmatrix{0 & i \cr i & 0}$, conjugating by diagonal elements $\pmatrix{a+bi & 0 \cr 0 & a-bi}$ where $a\pm bi$ have some odd Gaussian prime factors not in common will move us out of the Hurwitz integers.

This is a direct manifestation of the point that subrings of $D$'s finitely-generated as $\mathbb Z$-modules are obtained by taking products of maximal compact subrings of all the $D\otimes_{\mathbb Q} \mathbb Q_p$ and intersecting with $D$. At primes $p$ where $D$ becomes the matrix algebra, there is by-far not a unique such maximal subring. Approximating this globally gives the kind of counterexample in the previous paragraph.

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    $\begingroup$ Nice. Maybe I should repost my comment saying "If it isn't a subring..." . Gerhard "I Call A Do Over" Paseman, 2012.10,12 $\endgroup$ Commented Oct 12, 2012 at 19:07
  • $\begingroup$ thanks a lot! that was a silly question, especially because I already had this kind of counterexamples in the split case before posting my question ^^... $\endgroup$
    – GreginGre
    Commented Oct 16, 2012 at 9:38
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    $\begingroup$ Perhaps, but it did not look silly to me, and now it has the potential to be used as a teaching example for many who will follow. Gerhard "Handing Down Silliness To Generations" Paseman, 2012.10.16 $\endgroup$ Commented Oct 16, 2012 at 20:41
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    $\begingroup$ Indeed, this question is reasonable and natural. Who could have guessed that it was already so definitely dispatched? Not me, except that I'd by-chance experienced the dispatch-ing thereof. Presumably this is why this web-site exists. :) $\endgroup$ Commented Oct 17, 2012 at 1:27

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