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

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


Dec
4
comment Order of product of abelian group
I think you have only shown that the order of $c=ab$ divides $\operatorname{lcm}(m,n)$. This does not prove that they are equal.
Dec
4
comment Order of product of abelian group
No, this is not true. from $b^n=a^{-m}$ you can not conclude $b^{n-1}=a^{-(m-1)}$ because $a^{-(m-1)}=a^{-m+1}=b^{n+1}\ne b^{n-1}$ in general. I have a feeling the order should be the least common multiple of $m$ and $n$.
Dec
3
comment How does Molien series describe polynomial invariants?
@monomorphic: You got it. Unfortunately, I am afraid I have never seen a proof of Molien's theorem and you might best be served asking this as another question, I am sure someone here can help you with that.
Dec
3
comment How does Molien series describe polynomial invariants?
@monomorphic: Alright =)
Dec
3
comment How does Molien series describe polynomial invariants?
@monomorphic: The Molien series does not really describe the polynomial invariants. I do not think it is possible to reconstruct the polynomial invariants from the Molien series. The Molien series is really just a power series whose $d$-th coefficient is the number of linearly independent polynomial invariants of degree $d$. What uniqueness are you referring to?
Dec
3
comment How to prove that $Aut(\mathbb{P}^2) \cong PSL_3 (\mathbb{C})$?
Sure thing. I left an answer for you, maybe it can help.
Dec
3
comment How to prove that $Aut(\mathbb{P}^2) \cong PSL_3 (\mathbb{C})$?
This fact (or actually, a more general statement) is known as the "Fundamental Theorem of Projective Geometry". You can find it as Theorem 1.5 in this book, which is available online. What you are looking for is the Corollary 1.7.
Nov
19
comment Vector space and algebraic closure of a field
@user142800: Yea, that part was not quite formal. The "structure" of a finite dimensional vector space over some field is given only by its dimension, because any two finite dimensional vector spaces are isomorphic. So basically I said nothing profound there, it's just elaborating on what I said formally before. But yes, I think of it much the same way you said: You keep the same basis, but now you can multiply those vectors with more scalars.
Oct
31
comment Representation theory of the general linear group over a finite prime field
I am perfectly sure what the question is, it is the one written up there. I just do not know what your definition of a representations of the algebraic group $\operatorname{GL}_n/\mathbb F_p$ is. @Stephen, you are correct.
Oct
31
comment Representation theory of the general linear group over a finite prime field
@Thomas: Most definitely not, everything is over $\mathbb F_p$.
Oct
31
comment Representation theory of the general linear group over a finite prime field
I worded my question more cautiously. I am not sure if I want the representations of the algebraic group $\operatorname{GL}_n/\mathbb F_p$, in case that somehow involves the algebraic closure of $\mathbb F_p$, I am indeed looking at the finite group $\operatorname{GL}_n(\mathbb F_p)$ acting on $\mathbb F_p$-vector spaces.
Oct
23
comment Worst-case time to copy one movie
For $N=99$ you'd only have $H_1$ and $H_2$ because $\log_{10}(N+1)=\log_{10}(100)=2$.
Oct
17
comment UFD and relatively prime elements
You are not quite sincere, there is a hint in the book: It says that $\gamma$ is the resultant of $u$ and $v$. The fact he states is a known property of resultants.
Oct
15
comment Divisibility of polynomials in a subfield of a field.
+1 and yes, I admit I chose an easy path.
Oct
11
comment Is a coherent locally free sheaf isomorphic it's dual?
No, a locally free sheaf is not in general isomorphic to its dual, not even in the very special case of line bundles. Here, the dual $\mathcal L^\ast$ of a line bundle $\mathcal L$ serves as an inverse to $\mathcal L$ in the Picard group. This might be a good case to study to understand what's going on.
Oct
5
comment Why is the multiplicative subgroup of a field an affine algebraic group?
No, I am relying on the fact that (2) is true. They are isomorphic as varieties and the group structure on $G_m$ induces a group structure on the hyperbola via this exact isomorphism.
Oct
4
comment Why is the multiplicative subgroup of a field an affine algebraic group?
Regarding 1), you just gave the description in the sentence before: Take $n=2$ and observe that $f=x_1x_2-1$ is irreducible. The ideal generated by $f$ is a prime ideal in $K[x_1,x_2]$ and its zero locus is (isomorphic to) the multiplicative group. Regarding 2), simply consider the projection $K\times K\to K$. You should be able to show that the restriction to $G$ is injective and has image $G_m$.
Sep
25
comment Rational Points, classical versus modern notion
Thanks, that's a good answer.
Sep
24
comment About the fixed part of a linear system
If your surface is nonsingular in codimension one, the answer is easy: For any $Z$ satisfying that property, all elements of your linear system considered as sections of the line bundle $\mathcal L(D)$ vanish at $Z$. Hence, $Z$ is contained in the base locus of your linear system $\mathcal S$. I'd have to ponder a bit if anything goes wrong in the singular case, but I have only ever heard the term "linear system" used in the context of "sufficiently nice" varieties.
Sep
22
comment Rational Points, classical versus modern notion
The problem with this is that I'd like to start with an $L$-scheme in the first place, not with a $K$-scheme.