4,378 reputation
934
bio website nullteilerfrei.de
location Germany
age 29
visits member for 3 years, 7 months
seen yesterday

I'm a PhD student with research interests in Algebraic Geometry, Algebraic Complexity Theory and Geometric Invariant Theory.


Aug
13
comment Rank, degree and slope of a general coherent sheaf
Well, frankly I am unsure. But a compact Kähler manifold is a projective variety iff it is a Hodge variety, that much I happened to pick up. In this case, your definition of degree should coincide with "the degree" of this projective variety, but you cannot formulate the latter notion without a reference to the embedding, which as I said, is not unique. So maybe your notion of degree is not what I am used to.
Aug
12
comment Rank, degree and slope of a general coherent sheaf
There is no reasonable definition for degree unless, maybe, with respect to a fixed, very ample invertible sheaf, i.e. you fix a projective embedding. My reason for saying this is simply the fact that you can change the degree of a projective variety by embedding it differently, for instance using Veronese embedding.
Aug
4
comment Finite surjective morphism of smooth varieties is flat
If you need a reference, this is Remark 3.11 in Chapter 4 of Qing Liu's book Algebraic Geometry and Arithmetic Curves.
Jul
8
comment Integral dependence of coordinate ring
I don't think he shows that. First, $S(Y)_{(x_i)}$ does not properly contain $S(Y)$ and it is not clear how $S(Y)$ should act (by scalar multiplication) on $S(Y)_{(x_i)}$. Observe that the latter are only the fractions in degree zero! Second, the ring $S(Y)_{x_i}$ is a good example of an extension of $S(Y)$ which is not integral. Intuitively, this is because you added a new fraction, which is against the spirit of remaining integral. I can also write a formal proof if you like.
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jun
26
comment Two nonassociated functions defining the same hypersurface?
Indeed, this works! Thanks!
Jun
26
accepted Two nonassociated functions defining the same hypersurface?
Jun
9
comment What are the one-parameter subgroups of GL?
Nice! The density argument is really elegant, also thanks for answering such an old question.
Jun
9
accepted What are the one-parameter subgroups of GL?
Jun
3
comment Two nonassociated functions defining the same hypersurface?
It does work, but I realized what bothers me about it and slightly changed my question: The variety $X$ is not irreducible in this case and I am wondering if it will still hold under that condition.
Jun
3
revised Two nonassociated functions defining the same hypersurface?
added 12 characters in body
Jun
3
comment Two nonassociated functions defining the same hypersurface?
This yields an affine example, which is already nice: Take the Neil Parabola with coordinate ring $\mathbb C[x,y]/(x^2-y^3)$ and the functions $g=x$ and $f=y$. They both define the same point, they are not associated and irreducible. And you are right, it is because they are both not prime, the minimal prime ideal over both of them is $(x,y)$. However, this argument does not work well after homogenization to $\mathbb C[x,y,z]/(zx^2-y^3)$. Any ideas? Is the projective case different or am I looking at the wrong example?
Jun
3
revised Two nonassociated functions defining the same hypersurface?
deleted 31 characters in body
Jun
3
revised Two nonassociated functions defining the same hypersurface?
deleted 12 characters in body
Jun
3
revised Two nonassociated functions defining the same hypersurface?
edited tags
Jun
3
asked Two nonassociated functions defining the same hypersurface?
Jun
2
awarded  Yearling
May
29
asked Looking for an example for a very particular kind of cone
May
27
comment How to find the lengths of the shortest paths in a directed graph in $O(m)$ steps?
You're right, but now that I think about it we have to be more specific. Do you want the lengths of all shortest paths for all pairs of vertices? In this case, you need $\mathcal O(n^2)$ steps anyway, because you have quadraticly many values to compute.