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If p, q and r are pairwise coprime the generalized Fermat curve is embedded in the weighted projective space ${\mathbb P}(qr, pr, pq) $ which is isomorphic to a straight ${\mathbb P}^2$ via $(x:y:z)\m …

My question is related to this question, but I'm looking for something a bit more explicit. Let $S$ be a smooth surface over $\mathbb C$, fix a point $s\in S$ and take the blow up $\beta \colon S' \r …

Interesting question. I think the answer is yes, let me try to prove it. As you noticed, the ideal sheaf sequence shows that $h^1(D)=0$ is equivalent to the fact that $H^0({\mathcal O}_D)$ is 1-dimen …

Of course not. If $L$ is very ample, $D_1$ and $D_2$ are two hyperplane sections for some embedding in a projective space. Therefore their intersection is at most codimension 2 in $X$, intersection o …

If the genus of the fibre is not 0, by Theorem 2.1 in Serrano's paper "Fibrations on algebraic surfaces" any isotrivial fibration is birational to $(A \times B)/G \rightarrow B/G$ where $G$ is a finit …

As abx note in the comments, you miswrote the equations. Still the singular point remains. The point is that the "reduced" data works well algebraically, as indeed you can deduce $L_3$ from the othe …

Xiao Gang is taking a configuration of six lines in the plane with 4 triple points $x,z_1,z_2,z_3$ and three double points $y_1,y_2,y_3$. He considers a general quartic through the seven points which …

I think the answer to the question as it is stated is negative: if I do understand correctly the question I can prove that, if $G$ is abelian and both curves $C_i/G$ are rational, this is never true …

(1) Yes. Consider a net of reducible quadrics (here the dimension of the projective space is not relevant): since all the associate quadratic forms have rank at most 2, and since they vary linearly, …

I may have misunderstood something. However my impression is that, since you are asking $Y'$ smooth, the answer is negative. The simplest counterexamples are probably compact Riemann surfaces, since b …

1) is correct, 2) is not. Indeed, if $i(C)=C$, then the map $C \rightarrow C'$ is a double cover, and $f^*{\mathcal O}_Y(C')={\mathcal O}_X(C)$ since in a neighbourhood of a general point of $C$ the …

This is meant to be an integration to Yusuf answer. Consider a section $$ \stackrel{\rightarrow}{\alpha}= \begin{pmatrix} \alpha_1\\ \vdots\\ \alpha_r \end{pmatrix} \in {\mathbb C}^r\cong H^0({\math …

abx has already pointed out that your divisors are not the canonical divisors. Still, your (not-so-clear) question was maybe slighlty different, on "how to find the equation"? If I understand your qu …

By the comments of abx you only need an answer to question $3$. You get easily an answer if you assume the results in the classical book of Beauville on algebraic surfaces. Namely, if $S$ is uniruled …

A fine classification of the quartic surfaces that are not normal is in Tohsuke URABE, "Classification of Non-normal Quartic Surfaces", TOKYO J. MATH. VOL. 9, No. 2, 1986, 265-295