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Let $X\subseteq\mathbb{P}^n\times\mathbb{P}^m\subseteq \mathbb P^{(n+1)(m+1)-1}$ be a projective variety of dimension $p$ and degree $d$ defined over an algebraically closed field $k$, where $\mathbb{P}^n\times\mathbb{P}^m\subseteq \mathbb P^{(n+1)(m+1)-1}$ via the Segre embedding. Let $X'\subseteq \mathbb{P}^n$ be the projection of $X$ on $\mathbb{P}^n$, which we will assume having dimension equal to that of $X$. What can we say about the degree $d'$ of $X'$? Is there some formula linking the degrees $d$ and $d'$. Since an analytical charcaterization is also good for my purposes, feel free to assume $k=\mathbb C$.

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    $\begingroup$ What do you mean by degree $d$? With respect to what ample line bundle? $\endgroup$ – R. van Dobben de Bruyn Oct 9 '18 at 23:17
  • $\begingroup$ @R.vanDobbendeBruyn I mean the classical one: Number of points in the intersection of $X$ with $p$-hyperplanes in general position. $\endgroup$ – Vincenzo Zaccaro Oct 9 '18 at 23:50
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    $\begingroup$ @VincenzoZaccaro: Your question is ill-posed. Without specifying more details, there is no such thing as a hyperplane in $\mathbb{P}^n \times \mathbb{P}^m$. Subvarieties such as $X$ come with bidegrees, not degrees. $\endgroup$ – Ozob Oct 10 '18 at 3:01
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    $\begingroup$ Let $h_1,h_2$ be the hyperplane class in $CH^1(\mathbb{P}^n)$ and $CH^1(\mathbb{P}^m)$, and $q_1,q_2$ the two projections. Then $d=(q_1^*h_1+q_2^*h_2)^p$, and $d'=q_1^*h_1^p$. Obviously you cannot say more than $d'\leq d$. $\endgroup$ – abx Oct 10 '18 at 4:10
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    $\begingroup$ One can say more if $X$ is irreducible. Would you be willing to add that hypothesis? Or do you need more general results, for possibly reducible varieties? $\endgroup$ – Zach Teitler Nov 10 '18 at 13:35
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As abx says in his comment, all you can say is $d' \le d$. Here are some explicit examples with $m = 2$ and $n = 1$.

(1) To see that $d'$ can be arbitrarily smaller than $d$, fix a line $L \subseteq \mathbb{P}^2$ and set $X = L \times S$, where $S \subseteq \mathbb{P}^1$ is a finite set of points. Then $X$ has degree $\lvert S\rvert$ while has $X' = L$ has degree 1.

(2) To see that we can also have $d' = d$, take a degree $d$ curve $C \subseteq \mathbb{P}^2$ and set $X = C \times \{*\}$ for some point $* \in \mathbb{P}^1$. Then $X$ has degree $d$, and $X' = C$ also has degree $d$.

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