Asal Beag Dubh
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 3m comment Help to undertand the meaning of bounded family of surface Wikipedia (even in English) has lots of gaps. This is a specialised technical term, that is not even defined in standard algebraic geometry books such as Hartshorne. 2h comment Help to undertand the meaning of bounded family of surface Bounded family is a technical term (not the same as the usual meaning of bounded) with a precise meaning in algebraic geometry, and famille limit\'ee is the French translation of that term. 2h comment Help to undertand the meaning of bounded family of surface Anyway, I think it is best for you to ask your teacher directly about these things. There are only so many questions that can be answered on this website. 2h comment Help to undertand the meaning of bounded family of surface I don't really understand your last question; in particular, a family does not consist of a single surface or a finite number of surfaces, but rather one surface for each point in the parameter space. 4h reviewed Approve question about theorem references (who made it, year, etc.) 7h comment Help to undertand the meaning of bounded family of surface Actually after the statement of Proposition 1.7 it explains exactly what is meant ("Cela signifie..."): there is a variety $S$ and a morphism $X \rightarrow S$ such that each surface in the class under discussion is isomorphic to one of the fibres $X_s$. Just to emphasise, bounded refers to the fact that the "parameter space" $S$ for these surface is of finite type, rather than being, say, some scheme with infinitely many components. 1d comment About the pluricanonical map of a surface of general type. Without knowing anything, note that if $2K_X$ gives a birational map except in certain cases, then so must $4K_X$. Apr15 comment Proof of $\mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(a,b)$ is ample $\iff$ $a,b >0$. OK, now it seems complete! Apr15 comment Proof of $\mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(a,b)$ is ample $\iff$ $a,b >0$. OK, but now there is a little gap, because you don't deal with the case $a=0$. That was what my comment was meant to address. Apr15 revised Is the Godeaux surface irrational? added 37 characters in body Apr15 answered Is the Godeaux surface irrational? Apr15 comment Proof of $\mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(a,b)$ is ample $\iff$ $a,b >0$. Your argument is pretty much correct. The only part I don't like is when you say "As for the case when a=0, the global sections of $O_{P^1}$ is a 1-dimensional vector space, thus a rational map to $P^n$ can not be defined." In fact sections of $O_{P^1}$ give a perfectly good map to the projective space $\mathbf P^0$. Instead, I would simply argue that the restriction of an ample line bundle to a subvariety is still ample. Since $O_{P^1}$ isn't ample on $\mathbf P^1$, that proves what you want. Apr15 revised locally path connectedness edited tags Apr14 reviewed Approve Is the Godeaux surface irrational? Apr14 comment Is the Godeaux surface irrational? A surface of general type cannot be ruled. Do you know why the Godeaux surface is of general type? (By the way, the questions in the title and in the body do not match.) Apr14 comment Number of fibres under finite non-flat morphism No, if $Y$ is smooth (even normal) then every fibre has at most $d$ points. This is in Shafarevich Chapter II. Apr14 comment Surface constructed using curves Draw a picture of what this surface looks like. Where are $E$ and $F$ in your picture? How many points of intersection do they have? Apr12 answered $D$ is effective iff $f^*D$ is effective? Apr9 comment global section of some divisor Do you know how to express $K_X$ in terms of $K_{\mathbf P^2}$ and the exceptional divisor? If so, you should be able to simplify $K_X+D'$ helpfully. Apr9 comment Picard group of projective space Let me say something you may already know. If $D$ is an irreducible subvariety of codimension 1, it is the zero locus of a homogeneous irreducible polynomial $F$ of degree $n$ say. Then $F/x_0^n$ is a rational function whose divisor is $D-H$ where $H$ is the hyperplane $\{x_0=0\}$. Apply this to each term appearing in a divisor $\sum_i n_i D_i$, and you are done. Is that the kind of answer you want?