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May
15
answered Weighted projective space
May
13
answered Dimension of linear system of divisor of two points on curve of genus greater than 2
May
13
comment Dimension of linear system of divisor of two points on curve of genus greater than 2
Ah ok, sorry, I thought it was the dimension of the vector space, not of the projective one, I will delete my comment.
May
12
comment Solving exercise 1.10 in Silverman's AEC
You can reverse the process: basically first you find a rational point on $V_p$ and then you show that the projection from that point is an isomorphism with $\mathbb{P}^1$.
May
11
comment Does an inseparable extension have a purely inseparable element?
Thanks! I wrote my answer above after some explicit computations and I was wondering about a more abstract approach.
May
10
answered Does an inseparable extension have a purely inseparable element?
Apr
20
awarded  Popular Question
Mar
18
comment Prove the Inverse of a Nonconstant Harmonic Function is Unbounded
I think that $u^{-1}(c)$ here means the preimage: $u^{-1}(c)=\{z\in \mathbb{C}\, |\, u(z)=c\}$
Feb
23
comment The complement $\mathbb{P}_k^2 \setminus C$ of a plane curve $C$ is affine.
Yes that's what I mean. I also have posted a full answer below to be more clear
Feb
23
answered The complement $\mathbb{P}_k^2 \setminus C$ of a plane curve $C$ is affine.
Feb
23
comment The complement $\mathbb{P}_k^2 \setminus C$ of a plane curve $C$ is affine.
Well, you know that $\mathbb{P}^n_k\setminus H \cong \mathbb{A}^n_k$ and then you have a closed subset of affine space that is by definition affine.
Feb
23
comment The complement $\mathbb{P}_k^2 \setminus C$ of a plane curve $C$ is affine.
Using the Veronese embedding corresponding to the degree of the curve you can reduce to the situation of $\mathbb{P}^n$ minus an hyperplane.
Feb
16
answered Relating the cardinality to surjectivity
Feb
16
comment Can a product of a Stein manifold and a compact manifold be again Stein?
You're welcome!
Feb
16
comment Can a product of a Stein manifold and a compact manifold be again Stein?
I think that the answer should be no, as a closed submanifold of a Stein manifold is Stein, and $Y$ is a closed submanifold of $ X \times Y$.
Feb
16
awarded  Supporter
Dec
20
awarded  Caucus
Dec
15
comment Flexes of cubic curve
@evinda: I wrote an answer that hopefully explains enough. Sorry that it took me so long and tell me if you would like further clarifications.
Dec
15
answered Flexes of cubic curve
Dec
9
comment Flexes of cubic curve
sorry I have to go now, I will write a more extensive explanation later. Anyway you can try to have a look at wikipedia: en.wikipedia.org/wiki/Projective_space