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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
deleted 90 characters in body
Apr
8
comment 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
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Oct
9
answered Factorisation of local quaternionic zeta functions
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accepted Infinite dimensional simple algebras of finite degree
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Mar
12
comment Totally real points on curves
Cool, thank you very much!
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11
asked Totally real points on curves
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11
answered Explicit isomorphism for quaternion algebras over $\mathbb{Q}$?
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