3,680 reputation
831
bio website nullteilerfrei.de
location Germany
age 29
visits member for 3 years, 2 months
seen 8 hours ago

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


Aug
15
comment Number of roots of two polynomials
@Tom: I am sorry, I am unfamiliar with that theorem, or at least with the name. What does it state?
Aug
14
comment Number of roots of two polynomials
Dear @Tom, you still get the bound $\deg(f)\cdot\deg(g)$ which is sharp in general. In your example, you could conclude that there are at most $\max(d_1,d_2)^2$ many points in the intersection of the curves defined by $f$ and $g$.
Aug
14
comment Ideal of affine piece of a projective variety
@user169368: No problem, glad I could help!
Aug
14
comment Ideal of affine piece of a projective variety
That said, this does not feel too strong at all: A projective algebraic set is just a finite union of varieties, so you could easily deduct the statement topologically for projective algebraic sets if you knew it was true for varieties.
Aug
14
comment Ideal of affine piece of a projective variety
If a projective algebraic set is just the vanishing of any homogeneous ideal (i.e. the vanishing of a certain number of homogeneous polynomials) then yes, this is in no way dependent on the fact that $V$ is irreducible.
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.
Jun
26
comment Two nonassociated functions defining the same hypersurface?
Indeed, this works! Thanks!
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
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
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?
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.
May
26
comment How to find the lengths of the shortest paths in a directed graph in $O(m)$ steps?
I think you can simply use BFS.
May
21
comment Differentials on a curve
Is $G$ finite? Or is it an algebraic group? In general, it is not clear how $C/G$ has the structure of a variety.
Apr
23
comment Ring of Formal Power Series Over a Field is a Local Ring
@fretty: +1 for "everyday power series".
Apr
19
comment Kid's homework: 4 equations 5 unknowns? Going crazy!
There is no unique solution to that system of equations even if you limit yourself only to the natural numbers. I am posting just to confirm your observation, but concur with @naslundx.
Apr
8
comment Do rational functions separate points?
@Cantlog: That's really great and a clear, elegant proof at that. Why don't you post it as an answer, I'd accept in a heartbeat.
Apr
7
comment Do rational functions separate points?
@AsalBeagDubh: Well. I wanted to avoid it, but if you have a solution for quasi-projective $X$, I'd be curious, too.