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Let $X\subset \mathbb P^n$ be a projective variety of degree $d$, and suppose $O=[0,0,\cdots,0,1] \notin X$. Then we can define a morphism: $$\phi:X\to \mathbb P^{n-1}$$ by sending a point $[x_0,\cdots,x_n]\in X$to $[x_0,\cdots,x_{n-1}]$. Let $Y$ denote the image scheme of $\phi$, then can we prove $\deg(X)=\deg(Y)$?(the \textbf{degree} of a variety is defined in Hartshorne's textbook, Chapter 1) If not, will it be true when $\dim (X)=1$?

If we set $R_n=k[x_0,\cdots,x_n]$ and $X=V_+(I)$, then the ideal defining $Y$ is $I\cap R_{n-1}$. So if we can prove $R_n/I$ and $R_{n-1}/(I\cap R_{n-1})$ have the same Hilbert polynomial, it will be done. But how? Could you provide some hints for me?

Any help is appreciated. Thanks.

Richard
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    You cannot expect them to have the same Hilbert polynomial. The cuspidal curve $V(Y^2Z-X^3)$ has the Hilbert polynomial $3n$. On the other had this is the image of the Veronose embedding of $\mathbb{P}^1$ in $\mathbb{P}^3$ which has Hilbert polynomial $3n+1$. – Kapil Mar 18 '22 at 12:53
  • It's degree $d$ when $X$ is a hypersurface, and degree one when codimension is at least two. See https://math.stackexchange.com/questions/3741881/generic-linear-projection-with-textcodim-ge-2-is-birational-geometric-proo – AG learner Mar 18 '22 at 15:30
  • @AGlearner Thanks for your help. But I can't find a proof there, could you provide a reference for this result? – Richard Mar 19 '22 at 05:28

1 Answers1

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When $X$ is a hypersurface of degree $d$, then $\phi$ is surjective, so $X\to \phi(X)$ has degree $d$. To see this, for any $p=[x_0,\ldots,x_{n-1}]\in \mathbb P^{n-1}$, the line $L$ as span of $p$ and $O$ will either intersects $X$ at $d$ points (with multiplicity) or will be contained in $X$.

When $X$ has codimension $r\ge 2$, and when the projection center $O$ is in general position (as KReiser pointed it out), $\phi$ is birational onto its image (see here and the links provided in the first comment in the link). In this case, $\deg\phi(X)=\deg X$. This is because by choosing a general linear subspace $H\subseteq \mathbb P^{n-1}$ of dimension $r-1$ complementary to $X$, the intersection $H\cap \phi(X)$ is transverse and consists of $d$ distinct points $p_1,\ldots, p_d$. Note we can choose $H$ so that these $d$ points lie in the open locus where $\phi$ is isomorphic. Therefore, the cone $\tilde{H}$ as span of $H$ and $O$ has dimension $r$ and $\tilde{H}$ intersects $X$ transversely at $d$ distinct points $\phi^{-1}(p_1),\ldots, \phi^{-1}(p_d)$.

When $X$ has codimension $r\ge 2$ and the projection center $O$ is not in general position, however, $\phi$ is not generaically 1-to-1 map and the degree of $\phi(X)$ can be smaller than $X$. For example, $X$ is a intersection of a hypersurface $Y$ of degree $d$ and a hyperplane $H$ in $\mathbb P^n$. If $O\in H\setminus Y$, then $X\to \phi(X)$ is a degree $d$. This is just a "cone" of the codimension one setting.

AG learner
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