All Questions
Tagged with genus algebraic-curves
7 questions
17
votes
1
answer
471
views
Existence of space curves of given genus and degree
In Hartshorne's Algebraic Geometry Chapter IV, Section 6, he summarizes known results on the existence of smooth space curves of degree $d$ and genus $g$ for $g\le 12$ and $d \le 10$. He shows the ...
1
vote
0
answers
174
views
Derivation for genus-degree formula from algebraic functions field theory
This is a copy of my question from math.stackexchange: https://math.stackexchange.com/questions/4517289/derivation-for-genus-degree-formula-from-algebraic-functions-field-theory. I didn't get any ...
1
vote
0
answers
220
views
Degree and genus of projected curve
Let $C\subset\mathbb{P}^n$ be a normal curve over an algebraically closed field of characteristic $0$. Assume that $C$ is not contained in any hyperplane. We may assume that $P=[0:\cdots:0:1]$ is on $...
6
votes
2
answers
671
views
Simple proof that the arithmetic genus is non-negative
I take an irreducible and reduced closed curve $C\subseteq \mathbb{P}^n$, defined over an algebraically closed field $k$ and define the arithmetic genus $p_a(C)$ as the integer such that the Hilbert ...
1
vote
0
answers
117
views
Tangent Bundle of reducible genus one curves
I need to know what can be said in general about the tangent bundle of reducible curves over complex numbers with arithmetic genus one, say $I_N$.
As far as I know for any Simpson semistable torsion ...
0
votes
2
answers
644
views
Rationality of curve does not depend on base change
By a curve I mean an integral one-dimensional scheme of finite type over a spectrum of a field.
Let $C$ be a curve over an arbitrary field $k$. It's probably a very well known fact, that $C$ is ...
0
votes
1
answer
443
views
Example of projective variety that do not contain algebraic curves of genus strictly greater to $1$
Does exist a smooth, complex, projective variety $X$ of dimension $d\geq2$ such that $X$ does not contain smooth, complex, projective curves of wichever genus $g\geq2$?
Answer by Bertie: No, it does ...