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Let $S \subseteq \mathbb{P}^n$ be a smooth projective surface with given embedding in projective space. Moreover, let $X$ be another smooth surface and let there be a map $\pi: X \rightarrow S$ that is finite of degree 2. So if we count multiplicities, every point in $S$ has two preimages in $X$.

Question 1: Is the sheaf $\pi^*(\mathcal{O}_S(1))$ again very ample? Alternatively, if we denote by $H$ a hyperplane section on $S$, is the pullback divisor $\pi^*H$ on $X$ again a hyperplane section of some embedding of $X$ in some projective space?

Question 1(b): If this is not true, can you give a counterexample?

Question 2: If this is not true, would it be true if we replace $\mathcal{O}_S(1)$ by $\mathcal{O}_S(k)$ for some $k \in \mathbb{N}$?

If necessary, everything can be over $\mathbb{C}$.

I think a part of my confusion arises from the following: Hartshorne defines the Serre sheaf of some $\operatorname{Proj}(T)$, but this depends on the graded ring $T$. Question 3: Is it true that being able to write a projective variety as the $\operatorname{Proj}$ of some ring is equivalent by having an explicit embedding in some projective space? Then, is it true that without a given embedding in projective space of some projective variety $X$, the sheaf $\mathcal{O}_X(1)$ is not even defined?

As you can tell, i am struggling with the concepts of twisting sheaf, hyperplane sections and projective embeddings. So any elaboration on these concepts would be immensely appreciated, even if it does not contain an answer to the above questions.

Thanks a lot!

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By the way, the definition of a very ample invertible sheaf that i know/use is that its space of sections defines an embedding into projective space (whose dimension is the dimension of the space of global sections minus one). Please correct me if this is not right ok? –  Joachim Jan 11 '13 at 11:44
    
This definition is correct. An equivalent one is there is an embedding $ i : X\to \mathbb P^n$ such that the given invertible sheaf is isomorphic to $i^*O_{\mathbb P^n}(1)$. –  user18119 Jan 11 '13 at 15:23

1 Answer 1

up vote 3 down vote accepted

Question (1): no.

Example: let $S=\mathbb P^1$ and let $\pi: X\to S$ be a morphism of degree $2$ with $X$ of positive genus (e.g. $X$ is a hyperelliptic curve). Then $\pi^*O_S(1)\simeq O_X(D)$ for some effective divisor $D$ of degree $2$ over $X$. The space of sectios of $D$ has dimension $2$ (use Riemann-Roch or direct computations). So $\pi^*O_{S}(1)$ can't define an immersion to $\mathbb P^n$ with $n\ge 2$. And you don't have immersion to $\mathbb P^1$.

Question (2): yes. It is relatively easy to show that $\pi^*O_S(1)$ is ample. As $X$ is quasi-projective, some power $(\pi^*O_S(1))^{\otimes r}=\pi^*(O_S(r))$ becomes very ample.

Question (3): It depends on what do you call "Proj of some ring". If you can write $X$ as the projective of a homogeneous $k$-algebra $A$, then by definition $A=k[T_0,\dots, T_n]/I$ for some homogeneous ideal $I$. So $X$ is embedded into $\mathbb P^n$ as the closed subvariety $V_+(I)$. But if you write $X$ as the projective of some general graded $k$-algebra, you don't have directly an embedding into some projective space.

Conversely, if you are given an embedding $i : X\to \mathbb P^n$. Let $\mathcal I$ be the sheaf of ideals on $\mathbb P^n$ defining $i(X)$, let $$ I=\oplus_{d\ge 0} H^0(\mathbb P^n, \mathcal I(d)).$$ This is a homogeneous ideal of $k[T_0, \dots, T_n]$ and $i$ induces an isomorphism $X\simeq V_+(I)$. So $X=\mathrm{Proj}(A)$ with $A=k[T_0, \dots, T_n]/I$.

Finally, $O_X(1)$ is not intrinsic. It depends on the choice of an embedding of $X$ in some projective space. Typically, when we fix an $O_X(1)$, then $O_X(r)$ for any $r\ge 1$ is an $O_X(1)$ for a different embedding into a different projective space.

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