4,011 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.


1h
comment Representation theory of the general linear group over a finite prime field
@Thomas: Most definitely not, everything is over $\mathbb F_p$.
3h
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.
Sep
21
comment Can we say “commutative ring = field”?
clearly, that was supposed to be a "but". thx =)
Sep
20
comment Is the maximal ideal of a localization at a prime ideal principal?
Also: A good place to read about this, in my opinion, is 11.1 of David Eisenbud's book Commutative Algebra with a View (Toward Algebraic Geometry).
Sep
13
comment The greatest common divisor of homogeneous polynomials
If you choose $a_{ij}$ equal to the same polynomial $g$, the greatest common divisor of the $F_j$ will always be associated to $g$. Seems like this will heavily depend on the matrix $M$.
Sep
9
comment Local complete intersection ring
What is "the" maximal ideal of $R$? Is $R$ local or graded?
Aug
29
comment Definition of multiplication in Grothendieck ring
Well, changing that $(-1)^i$ to $(-1)^{i-1}$ is pretty much the same as losing the Tor-zero term.
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
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?