Impact
~7k
people reached
- 0 posts edited
- 0 helpful flags
- 13 votes cast
Sep
25 |
accepted | Noncommutative group of invertible ideals of a ring |
Aug
1 |
comment |
Segments of Voronoi Diagrams on smooth manifolds. Are they geodesics?
Nevermind: on a small enough neighborhood around each point of the bisector, the geodesics from the point cover all possible tangent directions. In particular, if the bisector contains all geodesics between points, then it is totally geodesic. |
Aug
1 |
comment |
Segments of Voronoi Diagrams on smooth manifolds. Are they geodesics?
This paper only says that bisectors in $M$ are totally geodesic if and only if $M$ has constant curvature. This makes no difference for the original poster's question which was one-dimensional, but isn't it true that a bisector $B$ can be geodesic (the geodesic between any two distinct points in $B$ is contained in $B$) without being totally geodesic? I'm thinking, for example, of products. |
Jul
29 |
awarded | Revival |
Jul
29 |
answered | Does anyone have an electronic copy of Waldspurger's “Sur les coefﬁcients de Fourier des formes modulaires de poids demi-entier”? |
Jul
12 |
comment |
Generators of the graded ring of modular forms
Well, the result over $\mathbb{Z}[1/6N]$ implies the result over $\mathbb{Q}$ and therefore any field of characteristic $0$, so in particular it answers the original question (for the subring in even weight). More generally, one can apply flat base change; but I don't know how far we want to get into this in these comments. |
Jul
12 |
comment |
Generators of the graded ring of modular forms
You may also want to refer to DZB's Proposition 11.3.1, since extends Theorem 9.3.1 to more general base rings. JV |
Jul
12 |
comment |
Generators of the graded ring of modular forms
What are "minimal relations"? I think you just mean "relations", since you say "at most 12" anyway. JV |
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
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 |
awarded | Yearling |
Oct
9 |
answered | Factorisation of local quaternionic zeta functions |
Sep
10 |
awarded | Popular Question |
Jul
2 |
awarded | Curious |