4,279 reputation
834
bio website nullteilerfrei.de
location Germany
age 29
visits member for 3 years, 6 months
seen 3 hours ago

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


Aug
29
comment Definition of multiplication in Grothendieck ring
You should add Zhen Lin's comment to your answer: The problem is really that I treated $\mathscr G$ as part of the resolution. It is not enough to just start the sum at $1$, one needs to look at a different complext $\mathscr P_\bullet$ than I did.
Aug
29
comment Definition of multiplication in Grothendieck ring
@ZhenLin: Yea, I think I see now. I should replace $\mathscr P_0$ and $d_1$ by $0$ in my notation. I will get roughly the same, but will end up with $[\mathscr F]\cdot\sum_{i=1}^n (-1)^{i+1} [\mathscr R_i]$ which is precisely $[\mathscr F]\cdot[\mathscr G]$.
Aug
28
comment Definition of multiplication in Grothendieck ring
Hm. I actually thought about this? But starting the sum at $i=1$ doesn't seem to change anything, since $[\mathscr T_0]=[\mathscr K_0]-[\mathscr I_1]=[\mathscr P_0]-[\mathscr P_0]=0$ in my notation.
Aug
28
asked Definition of multiplication in Grothendieck ring
Aug
23
revised Maximal tori in $SO(n,\mathbb{C})$
edited tags
Aug
23
answered Maximal tori in $SO(n,\mathbb{C})$
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
revised Number of roots of two polynomials
added 16 characters in body
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
answered Number of roots of two polynomials
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
answered Ideal of affine piece of a projective variety
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