Ariyan Javanpeykar
Reputation
845
Next privilege 1,000 Rep.
Create tags
 Apr14 comment Surface constructed using curves @dario Compute $E^2$ by writing it as $p^{-1}(y) \cdot p^{-1}(y)$ with $y$ some point in $F$ and $p$ the projection $E\times F\to F$. Mar26 comment Higher dimensional analogue for Riemann Hurwitz formula The setting in the higher-dimensional case is similar to the case of curves: you start with a reasonable map $f:X\to Y$ (say proper with finite fibres) and you want to relate topological invariants of $X$ to topological invariants of $Y$. Unfortunately, one of the points made in the answers to the questions you link to is that you should forget about "numerical relations between topological invariants" and rather work with equalities of line bundles (or divisors). So maybe try to learn a bit about canonical bundles and understand the meaning of Riemann-Hurwitz: $K_X = f*K_Y + R$. Mar17 comment Zariski dense implies classically dense? Dear @DustanLevenstein , I will to translate the proof for you if I find the time today. Mar17 revised Zariski dense implies classically dense? added 32 characters in body Mar17 answered Zariski dense implies classically dense? Feb20 revised (Reference Request) Desingularization of Fibrations deleted 142 characters in body Feb20 revised (Reference Request) Desingularization of Fibrations Fixed a typo Feb20 comment (Reference Request) Desingularization of Fibrations Dear @Paul you are completely right. You need an $n$th root of $t$. I was probably thinking of $X_2$ when writing this. Feb7 awarded Yearling Jan28 comment Properties of fibers of a morphism of varieties @Dubious I miswrote some things in my last comment as well. If $X$ is irreducible and generically non-reduced, then the locus of smoothness is empty. Of course, you can have an irreducible variety with only one non-reduced local ring. Then the locus of smoothness is non-empty (if $X$ has more than one point). Jan24 comment Number of elements in fiber @PeterWalker What might bother you is the definition of $\deg f$ is $f$ is not finite but only quasi-finite. In general, by $\deg f$ one denotes the degree of the corresponding extension of function fields. This makes sense for etale morphisms of integral schemes $X\to Y$ as etale morphisms are dominant and quasi-finite. Jan24 comment Families of Elliptic Curves See the books Modular functions of One Variable for some Tables, especially part IV and V. Jan24 comment Scheme over S and morphisms @Grobber Consider for all complex numbers $t$ the conic $x^2 + y^2 = tz^2$. You have probably a good idea of how these conics look like for all small enough real numbers $t$. For $t=0$, you get the union of two lines (namely, $x+iy =0$ and $x-iy = 0$). It was an important innovation of Grothendieck to realize this intuitive picture in the category of schemes as the relative morphism $X =$ Proj$(R[x,y,z]/(x^2 + y^2 - tz^2])\to$ Spec $R = S$, where $R$ is a ring and $t$ is an element of $R$. (In the example: $R$ would correspond to $\mathbb C[t]$.) Jan21 comment Cubic hypersurface of singular conics Dear @user198182 , I can't write all the details now, but could this hypersurface be defined by the Jacobi matrix of the "universal" conic? For a conic given by $a= (a_0:\ldots:a_5)$ to be singular it is necessary and suffices for the Jacobi matrix of the associated quadric to be of rank zero. If $f_{a}(x,y,z)$ is the associated quadric, then this translates into the vanishing of the discriminant of $f_{a}$ probably. I wrote this in a hurry, so probably someone else could correct/complement this comment. Jan20 comment Question on Algebraic Hartogs Lemma for locally Noetherian normal schemes But now use that $\cap_{\mathfrak p, ht = 1} A_{\mathfrak p} = A$, to see that it is the identity. Or am I missing something? Jan20 comment Question on Algebraic Hartogs Lemma for locally Noetherian normal schemes @enoughsaid05 The (injective) morphism $A\to A_{\mathfrak p}$ corresponds to the morphism Spec $A_{\mathfrak p}\to X =$ Spec $A$. For all $\mathfrak p$, it factorizes through $\mathcal O_X(U)$, as Spec $A_{\mathfrak p} \to X$ factorizes through $U$ (by assumption). Therefore, we have $$A \to \mathcal O_X(U) \to A_{\mathfrak p}$$ for all $\mathfrak p$ of height $1$. But, then we get $$A\to \mathcal O_X(U)\to \cap_{\mathfrak p, ht =1} A_{\mathfrak p}$$ where the map $A\to \cap_{\mathfrak p, ht = 1} A_{\mathfrak p}$ is injective (clearly). Jan20 comment Morphism $f\colon X\to S$ is proper iff $f^{-1}(V_j)\to V_j$ is proper for some open cover $\{V_j\}$ of $S$? (Lemma 28.42.3 of Stacks Project) Dear @GeorgesElencwajg , thank you very much for your kind words. I have to say that I truly enjoy your many answers and questions on MSE and MO. Best, Ariyan Jan20 answered Morphism $f\colon X\to S$ is proper iff $f^{-1}(V_j)\to V_j$ is proper for some open cover $\{V_j\}$ of $S$? (Lemma 28.42.3 of Stacks Project) Jan19 comment Reference request: Zero set of global section What if you take $(f,\ldots,f)\in \oplus_{i=1}^n\Gamma(\mathbb P^n,\mathcal O(1)) \subset \Gamma(\mathbb P^n,\mathcal O(1)^{\oplus n})$? Jan19 answered Jacobian of a Riemann surface and double unramified covers