Suppose $X \subset \mathbb{P}_k^n$ is a smooth, projective curve over an algebraically closed field $k$ of degree $d$ . In this case, degree of $X$ is defined as the leading coefficient of $P_X$, where $P_X$ is the Hilbert polynomial of $X$.

I guess Hartshorne using following fact without proof:

Under certain condition, the projection morphism $\phi$ from $O \notin X$ to some lower dimension space $\mathbb{P}_k^m$ gives a curve $\phi(X)$, and $deg X= deg\ \phi(X)$. I don't know which condition is needed to make this possible (the condition for $X \cong \phi(X)$ is given in the book), and more importantly, why they have the same degree.

One thing giving me trouble is the definition of degree by Hilbert polynomial. Is it possible to define it more geometrically?


The geometric way to understand the degree is as follows:

If $H$ is a general hyperplane in $\mathbb{P}^n$ then $H$ intersects $X$ in a finite number of points, and this number is the degree of $X$.

Note that we had to use the word general; see I.7 of Hartshorne for a discussion of why this is a tough notion to use.

Now, $\phi(X)$ is either a point or a curve (II Prop. 6.8 Hartshorne), so let's assume it's not a point. Next, $\phi$ being an isomorphism is equivalent to it being of degree $1$ (i.e. 1 pre-image over every point), and the degree of $\phi$ is the degree of the field extension $[k(X): k(\phi(X)]$ ( II.6 of Hartshorne). It's not always true that $\phi$ is an isomorphism (imagine a plane curve in $\mathbb{P}^n$ for large $n$ and $O$ is in the same plane).

To be of degree $1$ we need that $O$ lies on no secant or tangent line of $X$, and as long as $n \geq 4$ we always have such a point (IV Prop 3.4, 3.5 Hartshorne).

  • 2
    $\begingroup$ Thank you for your explanation of degree, that makes a lot sense! However, as for the condition $\phi$ preserve the degree, I think $\phi$ is a isomorphism alone is not enough (consider a plane conic which is always isomorphism to $\mathbb{P}^1$). $\endgroup$ – Li Zhan Jul 15 '12 at 7:25

Chris Dionne already explained the condition for $\phi$ to be an isomorphism from $X$ to $\phi(X)$. Suppose $\phi$ is a projection to $Q=\mathbb P^{n-1}$. Let $H'$ be a hyperplane in $Q$, then the schematic pre-image $\phi^{-1}Q=H\setminus \{ O\}$ where $H$ is a hyperplane of $P:=\mathbb P^n$ passing through $O$. Now $$O_P(H)|_X =(O_P(H)|_{P\setminus \{ O \}})|_X \simeq (\phi^*O_Q(H'))|_X \simeq (\phi|{}_X)^*(O_Q(H')|_{\phi(X)})$$ hence $$ \deg X=\deg O_P(H)|_X= \deg O_Q(H')|_{\phi(X)}=\deg \phi(X).$$


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