Algebraic integers in skew fields - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T21:07:16Z http://mathoverflow.net/feeds/question/109464 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109464/algebraic-integers-in-skew-fields Algebraic integers in skew fields GreginGre 2012-10-12T15:35:54Z 2012-10-12T16:23:04Z <p>Hi everyone,</p> <p>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 $D$ ?</p> <p>Thanks!</p> <p>G.</p> http://mathoverflow.net/questions/109464/algebraic-integers-in-skew-fields/109467#109467 Answer by paul garrett for Algebraic integers in skew fields paul garrett 2012-10-12T16:23:04Z 2012-10-12T16:23:04Z <p>No, the "integral" elements are not a subring. There are at least two ways to understand this.</p> <p>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 $\pmatrix{1 &amp; 0 \cr 0 &amp; 1}$, $\pmatrix{i &amp; 0 \cr 0 &amp; -i}$, $\pmatrix{0 &amp; 1 \cr -1 &amp; 0}$, and $\pmatrix{0 &amp; i \cr i &amp; 0}$, conjugating by diagonal elements $\pmatrix{a+bi &amp; 0 \cr 0 &amp; a-bi}$ where $a\pm bi$ have some odd Gaussian prime factors <em>not</em> in common will move us out of the Hurwitz integers.</p> <p>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 <em>not</em> a unique such maximal subring. Approximating this globally gives the kind of counterexample in the previous paragraph.</p>