Equivalent conditions for a closed immersion of schemes

In Hartshorne, a closed immersion of schemes is defined to be a scheme morphism Φ:Y→X such that Φ is a homeomorphism onto Φ(Y), Φ(Y) is closed in |X| and (∗)Φ#:OX→Φ∗OY is surjective. Can I replace (∗) by the condition that Φ#U:OX(U):→OY(Φ−1(U)) is surjective for all affine open subsets U⊆X, or does this lead to another definition? … Read more

Exercise 6.3.N. on Ravi Vakil’s notes (on morphism from SS scheme to PnB\mathbb{P}_B^n)

I am working on Exercise 6.3.N. on Ravi Vakil’s notes (on morphism from S scheme to PnB) and I would like some assistance. The exercise states: Let B be a ring. If X is a B-scheme, and f0,…,fn are n+1 functions on X with no common zeros, then show that [f0,…,fn] gives a morphism of … Read more

canonical representation of three skew lines in P3\mathbb{P}^3

Consider three skew (non-intersecting) lines L,M,N in P3. Each line is given by two equations of the form α⊤i,jz=0,i=1,2,3,j=1,2, where z=(z0,z1,z2,z3) are the homogeneous coordinates of P3 and αi,j∈P3. What i want is to show that up to projective equivalence the equations for the three lines can be written as (this is possible according to … Read more

How does an irreducible quadric in projective space look like?

I read the answer to the following question: Quadrics are birational to projective space Here it is stated that: Over a field k of characteristic ≠2 every irreducible quadric Q⊂Pnk has equation q(x)=x0x1+x22+…+x2m=0 in suitable coordinates . Can anyone tell me why an irreducible quadric would look like this (and would not have the terms … Read more

If Y Y is irreducible set so is cl(Y)cl(Y). [duplicate]

This question already has answers here: The closure of an irreducible subset of an irreducible space is irreducible. (2 answers) Closed 6 years ago. If Y is irreducible set so is cl(Y). If cl(Y) is reducible then cl(Y)=A∪B where both A and B is closed in cl(Y). Now how do we proceed? Answer Suppose Y … Read more

If C⊆P2C \subseteq \mathbb{P}^2 is a plane curve, then genus(C)=12(d−1)(d−2)genus(C)=\frac{1}{2}(d-1)(d-2). Compare with example in the notes

In my Algebraic Geometry notes (see http://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2002/main.pdf) there is the following exercise: If C⊆P2 is a plane curve of degree d, then its arithmetic genus g(C) is equal to 12(d−1)(d−2). Compare this to example 0.1.3 (of the notes of course). In fact, the first part of the exercise is easy to solve using the Hilbert … Read more

A curve of genus $g\geq 2$ has a closed point of degree at most $2g-2$ over base field.

I am working on the following problem [R. Vakil] Exercise 19.8.B: Suppose $C$ is a curve of genus $g>1$ over a field $k$ that is not algebraically closed. Show that $C$ has a closed point of degree at most $2g-2$ over the base field. I have no idea how to do this question. This is … Read more

Morphisms induced by effective divisors on P1\mathbf P^1

This question is about the proof of Theorem V.2.17 in Hartshorne’s Algebraic Geometry. Here everything is defined over some algebraically closed field k. Define O=OP1. Let e≥0 be an integer. Then the following is what I do not understand: The proof says that when n≥e, we can have a surjective map O⊕O(−e)→O(n−e)→0 by arguing that … Read more

Real Points of resolution of singularities of $\mathrm{Spec} \mathbb{R}[x,y]/(x^2+y^2)$

Consider the scheme $X = \mathrm{Spec} \mathbb{R}[x,y]/(x^2+y^2)$. Scheme-theoretically, it’s a one dimensional scheme with one real, singular point (there are, of course, other complex points). I’m interested in finding the structure of the real points of the resolution of $X$, but I’m having trouble visualizing exactly what I get. I think I want to do … Read more

How to compute the K-group of this affine scheme?

By Bott periodicity, we know that Ktop0(S2)=Z⊕Z But S2 is defined by the equation x2+y2+z2=1. If we write this abstractly as X:=SpecR[x,y,z]/(x2+y2+z2−1), then how can we compute Kalg0(X) the K0 is the same definition as Hartshorne’s Ex II.6.10 Answer The algebraic K-theory of smooth affine or projective quadrics is quite well understood (as opposed to … Read more