bio | website | math.dartmouth.edu/~jvoight |
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location | Dartmouth College | |
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Apr 28 |
awarded | Nice Question |
Apr 9 |
comment |
Noncommutative group of invertible ideals of a ring
@FernandoMuro The base ring $R$ is assumed to be commutative: it is a domain. So the field of fractions means what it usually does. (There are notions of rings of quotients for noncommutative rings, but they don't play a role here.) |
Apr 9 |
comment |
Noncommutative group of invertible ideals of a ring
@Aurel: conventions differ, but you might ask only that $\mathcal{O}$ is contained in the left and right orders; if $I$ is invertible, then equality holds, so this doesn't matter for the question. (@FernandoMuro: It is also equivalent to take a fractional ideal $I$ to be a $\mathcal{O}$-sub-bimodule of $K \otimes_R \mathcal{O}$ of the form $I=cJ$ where $c \in K^\times$ and $J \subseteq \mathcal{O}$ is a two-sided ideal. Hence the ``fractional''.) |
Apr 8 |
revised |
Noncommutative group of invertible ideals of a ring
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Apr 8 |
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Noncommutative group of invertible ideals of a ring
Thanks Pace! (Something happened in the copy and paste, ugh.) The definition of invertible is what it must be to make this into a group: a fractional ideal $I$ is invertible if there exists a fractional ideal $J$ such that $JI=IJ=\mathcal{O}$. JV |
Apr 8 |
revised |
Noncommutative group of invertible ideals of a ring
deleted 48 characters in body |
Apr 8 |
asked | Noncommutative group of invertible ideals of a ring |
Oct 10 |
comment |
Factorisation of local quaternionic zeta functions
What is $q$? It better be $q^n=\mathrm{nrd}(I)$: the reduced norm is not always a square. |
Oct 9 |
awarded | Yearling |
Oct 9 |
answered | Factorisation of local quaternionic zeta functions |
Sep 10 |
awarded | Popular Question |
Jul 2 |
awarded | Curious |
Jun 4 |
awarded | Nice Question |
Apr 9 |
accepted | Infinite dimensional simple algebras of finite degree |
Apr 7 |
asked | Infinite dimensional simple algebras of finite degree |
Mar 12 |
comment |
Totally real points on curves
Cool, thank you very much! |
Mar 11 |
asked | Totally real points on curves |
Mar 11 |
answered | Explicit isomorphism for quaternion algebras over $\mathbb{Q}$? |
Mar 10 |
awarded | Yearling |
Mar 9 |
awarded | Revival |