4,006 reputation
832
bio website nullteilerfrei.de
location Germany
age 29
visits member for 3 years, 5 months
seen 1 hour ago

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


1d
answered All the $k\times k$ minors determines the matrix?
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
22
revised Rational functions on varieties
added 6 characters in body
Oct
21
answered Reference request for algebraic Peter-Weyl theorem?
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
17
answered A question about Klaus Hulek algebraic geometry (regarding Noether normalization)
Oct
15
comment Divisibility of polynomials in a subfield of a field.
+1 and yes, I admit I chose an easy path.
Oct
15
answered Why $\mathbb Z/ 2 \mathbb Z$ is not a free module?
Oct
15
answered Divisibility of polynomials in a subfield of a field.
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 $G_m$.
Oct
1
answered How to show that $GL_n/U$ is birationally isomorphic to $B^-$?
Sep
30
awarded  Explainer
Sep
28
answered Lang Category Theory
Sep
25
accepted Rational Points, classical versus modern notion
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.
Sep
21
comment Can we say “commutative ring = field”?
clearly, that was supposed to be a "but". thx =)