I have a simple and basic question concerning degree of projective curves and I'm referring to something I've read on Miranda's book, Algebraic curves and Riemann Surfaces, Chapter VII, 3. The Degree of Projective Curves.

Suppose that $X\subset \mathbb P^n$ is a smooth nondegenerate curve such that $\text{deg}(X)=d$. This means that given any hyperplane $H\subset \mathbb P^n$, the hyperplane divisor $D:=\text{div}(H)$ on $X$ has exactly degree $d$. Take the complete linear system $|D|$, we want to conclude that $\dim |D|\geq n$. So, let $Q$ be the set of hyperplane divisors on $X$: $$Q:=\{\text{div}(L)\in \text{Div}(X): L \text{ is an hyperplane}\}.$$

Now Miranda says that $Q$ is indeed a subsystem of the complete linear system $|D|$. It's clear that $Q\subseteq |D|$, but I don't understand why $Q$ is a subsystem. To see that it is a subsystem, I should show that there exists a linear subspace $V$ of $L(D)$ such that $Q=S(\mathbb P(V))$ where $$S\colon \mathbb P(L(D))\to |D|, [f]\mapsto \text{div}(f)+D$$ is the standard bijection between these two spaces. Who is $V$? Is there a more direct way to say that $Q$ is a subsystem? I'm sure this is really simple, but I'm quite new to Algebraic Geometry so if someone could help me I would appreciate a lot.


To avoid notational confusion: by $\mathbb PW^\vee$ I mean the set of one-dimensional linear subspaces of $W$, namely the quotient $(W\setminus\{0\})/k^\times$.

If I am not mistaken, in your notation $D=H\cap X$, a hyperplane section. Your $V$ is then the image of the restriction map $\rho:H^0(\mathbb P^n,\mathscr O_{\mathbb P^n}(1))\to H^0(X,\mathscr O_X(1))$. This map $\rho$ takes a degree one form $h$ on $\mathbb P^n$ (i.e. a linear homogeneous polynomial $h=\sum_{0\leq i\leq n} a_ix_i$) to its restriction to $X$.

The inclusion $V\subset H^0(X,\mathscr O_X(1))$ gives a surjection $H^0(X,\mathscr O_X(1))^\vee\to V^\vee$ and hence a closed immersion $$\mathbb PV^\vee\subset \mathbb PH^0(X,\mathscr O_X(1))^\vee,$$ which is your $Q\subset |D|$.

Added. Disregard all the above! Now, $|D|$ is the projective space whose points correspond to effective divisors on $X$ which are linearly equivalent to $D=H\cap X$. It is the projectivisation of (the dual of) $L(D)$. Now, in $L(D)$ you can consider the subspace $V\subset L(D)$ consisting of elements arising as restrictions of homogeneous degree one forms $$h=\sum_{0\leq i\leq n} a_ix_i.$$ Note that such forms (which live on $\mathbb P^n$) are exactly those whose zero loci are hyperplanes $L\subset\mathbb P^n$. But this says exactly that $Q$ is the projectivisation of (the dual) of $V$.

  • $\begingroup$ Thank you for the answer. But I don't understand what you are saying because my knowledge in algebraic geometry is very little. I don't know what is the map you're talking about. Isn't there a simpler way to prove it? $\endgroup$ – batman Nov 22 '14 at 13:58
  • $\begingroup$ Claretta, I hope now it is clearer. Let me know if not. $\endgroup$ – Brenin Nov 22 '14 at 14:45
  • $\begingroup$ I still have few doubts, but I think it is just me. For example, $h \in L(D) \iff div(h)\geq -D$. How can I say that this is true? $\endgroup$ – batman Nov 22 '14 at 15:16
  • $\begingroup$ You do not have to check it, because $V$ already lives inside $L(D)$ by definition. Please ask all that troubles you, I will do my best to make my answer understandable. $\endgroup$ – Brenin Nov 22 '14 at 16:58
  • $\begingroup$ In $L(D)$ we have all meromorphic functions with poles bounded by $D$. So you take $V$, which is made by linear equations. I don't understand why you're saying that $V$ is already in $L(D)$. And also if it were, I don't understand your conclusion when you say "$Q$ is the projectivisation on $V$". Thanks again and sorry for my delay in answering. $\endgroup$ – batman Nov 23 '14 at 9:40

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