dhy
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The set $S$ gives rise to a subscheme (which let's also denote by $S$) of $\mathbb{P}^1_{\mathbb{Z}},$ because relatively a prime pair $(x,y)$ corresponds to a section of $\mathbb{P}^1_{\mathbb{Z}}\... View answer 33 votes The "conjecture" in algebraic geometry that all rationally connected varieties are unirational comes to mind. It's usually thrown around as a way of saying, "See, we know so little about what ... View answer Accepted answer 21 votes Really this is mostly just consolidating what has been (implicitly) said in the comments and cleaning it up a bit (e.g. using the Chow ring instead of singular cohomology), but might as well make it ... View answer 20 votes The Tate extension. Let$k$be a field, and let$V$be the space$k((t))$be the space of Laurent series with coefficients in$k$, considered as a topological vector space. If we write$\operatorname{...

No, take $f=b^4-a^3c$ and $g=ac^3.$ Now the variety $V(b^4-a^3c,y^2-ac^3)$ being reducible is equivalent to the ideal $(b^4-a^3c,y^2-ac^3)$ being prime. But we have $(ya-b^2c)(ya+b^2c)=y^2a^2-b^4c^2=a^... View answer 13 votes Here is an argument that Alice wins for odd$n$. Label the points$1$to$n.$Lemma: If at any point Alice has marked$k$points in a row and no other points, Bob has marked the two points ... View answer 13 votes Consider a$4\times 2k+1$rectangle. Label its points by$(x,y)$, with$x$from$0$to$3$and$y$from$0$to$2k.$Remove from the rectangle the points$(1,y)$and$(2,y)$for$y$odd. I claim that ... View answer 11 votes An update from 2018: There has been some recent work on a heuristic suggesting that there are infinitely many elliptic curves of every rank$<21$but only finitely many of rank$> 21$(it's ... View answer 10 votes One place where$(\infty,2)$-categories shows up is the geometric Langlands program. (As in David Ben-Zvi's comment, this is again related to the TFT example.) Indeed, local geometric Langlands is ... View answer Accepted answer 10 votes As discussed in the comments, I don't see how to extract the desired matrix from the original question (about spanning the vector space). However, the matrix being nonzero IS equivalent to the ... View answer Accepted answer 9 votes This equation has no solutions when$d$is odd. (EDIT: See below for the general case.) Actually, for$d$odd, there are no triples$(F_0,F_1,F_2)$with$F_1\cdot F_2-F_0^2$a multiple of$x$, let ... View answer Accepted answer 8 votes Here is one way to see it, via classifying$G$-invariant radical ideals. (This has the bonus that it implicitly describes the boundary.) Lemma:$G$-invariant ideals$I$of$\mathbb{C}[G/U]$are in ... View answer 8 votes Some comments: It is not necessarily true that chiral algebras are essentially conformal vertex algebras, as chiral algebras are allowed to vary over the curve in a way that vertex algebras are not. ... View answer Accepted answer 7 votes Let's see what the data of a diagram \begin{matrix} \partial\Delta^{n-2}&{\to}&X_{/f}\\ \downarrow &&\downarrow\\ \Delta^{n-2}&{\to}&X_{/y}\times_{S_{/p(y)}}S_{p(f)}\end{... View answer 7 votes The answer to 1) is that this is a special case of a broader phenomenon for symplectic resolutions (though I think some features are specific to the Grothendieck-Springer case.) For instance you have ... View answer 7 votes Lusztig's conjecture in modular representation theory, which would describe the characters of simple$G_1$-modules for$p$larger than the Coxeter number, was generally believed to be true for a long ... View answer 7 votes Here's an argument that one should expect that the two numbers gotten this way must be transcendental. Really, what I am showing is that the locus of$(a,b)$in$\mathbb{R}\times\mathbb{R}$where one ... View answer Accepted answer 7 votes There are lots of applications of toric varieties to singularities, e.g., the proof of the weak factorization theorem in characteristic zero. (Indeed, the name of the linked paper is "Torification and ... View answer Accepted answer 6 votes (In characteristic zero) Flatness implies the other two definitions; integrability and formal lifting are very weak conditions (in fact if I haven't made a mistake I think this notion of integrability ... View answer Accepted answer 6 votes I think one way of explaining this is via quadratic forms. The usual correspondence sends$X$to the quadratic form$X(v)=vXv^t,v$a row vector. But in characteristic$2$, this formula simplifies ... View answer 5 votes (Written on my phone - apologies for any typos.) A few comments: a) First, as to the source of the braided monoidal structure on the Kazhdan-Lusztig category. The category of integrable affine Lie ... View answer 5 votes All perfect complexes on an open$U$of a smooth variety$X$extend to perfect complexes on$X$. In fact, I believe a much more general statement is true: For an open substack$U$of a separated ... View answer 5 votes From the Lie algebra inclusion$\mathfrak{g}\rightarrow \Gamma(T_G)$of right invariant vector fields we get a map$U(\mathfrak{g})\rightarrow\Gamma(\mathcal{D}_G).$This induces a map$U(\mathfrak{g})...

(Expanding my comments into an answer for more visibility.) By Larsen-Lunts $K_0(\operatorname{Var}_k)/[\mathbb{A}^1]$ is the free abelian group on stable birational equivalence classes. It thus ...

(This should really be a comment on S. Carnahan's answer, but don't have enough reputation.) All coherent sheaves on a Riemann surface split into the direct sum of a torsion sheaf and a vector bundle....

The automorphisms $a_1,a_2,a_3$ are part of a $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$ action on $F_k$, which in turn induces an action on the space $H^0(\Omega^1_{F_k})$ of holomorphic ...

Maybe let me try to synthesize my comments into an answer. All of this is contained in Raskin's beautiful paper arxiv.org/abs/1611.04937 on Whittaker categories. Convention: We work here in the ...

Your map (assuming that the determinant is nonzero for all elements of the grassmannian) is a projection of the Plucker embedding. This tells you that the degree of the corresponding map of complex ...

I think the answer is no. Here is an argument. Degenerate your $8$ points so that there are $2$ $5$-tuples that lie in $3$-planes $A,B$. The degree $3$ rational curves through the first $5$-tuple ...
This statement isn't true; as you've currently defined everything, your $X_i$ are infinite dimensional (in fact they are unions of connected components.) You want instead to take the $X_i$ unions of \$\...