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Let $X$ be an irreducible algebraic variety and suppose that $L$ is a linear space defined by the linear forms $l_1,l_2,\ldots,l_k$. I want to study $L\cap X$. I would like to know whether the following argument is correct and I would appreciate very much a reference:

The linear sections of $X$ are all birational equivalent and the birational map between $X \cap V(l_1)$ and $X \cap V(l')$ will preserve also the degree. So the dimension and the degree of $X\cap L$ will be the same of the dimension and the degree of $X\cap L'$, with $L'$ defined by $l',l_2,\ldots,l_k$.

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  • $\begingroup$ In general the linear sections are not birational equivalent. As a silly example, take as $X$ a smooth quadric surface in $\mathbb{P}^3$ and consider a linear section given by a smooth conic and one given by two incident lines. $\endgroup$ Sep 26, 2014 at 9:33
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    $\begingroup$ The part after "So" is more or less right: the dimension will be constant in a dense open subset of the space of linear forms, and all those varieties will have the same degree. But the statement about birational equivalence is totally false. For instance, the plane sections of a smooth cubic surface over $\mathbf C$ are not all isomorphic (=birational). Do you really want the "birational" part, or just equality of degree and dimension? $\endgroup$
    – user5117
    Sep 26, 2014 at 9:33
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    $\begingroup$ For a less trivial example, take the $d$-ple Veronese embedding $\mathbb{P}^2 \subset \mathbb{P}^N$. The linear sections of the image correspond to the linear system of curves of degree $d$ in $\mathbb{P}^2$, and two general curves in this system are not birationally equivalent as soon as $d \geq 3$. $\endgroup$ Sep 26, 2014 at 9:36
  • $\begingroup$ Yes, but both have dimension 1 and degree 2, so I can say that I obtain something of the same degree and dimension. Can I use this in some way? (I am referring to your first comment) $\endgroup$
    – user46071
    Sep 26, 2014 at 9:37
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    $\begingroup$ @FrancescoPolizzi: but the OP wants to intersect $X$ with a linear subspace of arbitrary codimension. Hence my babble about dense open sets, etc. $\endgroup$
    – user5117
    Sep 26, 2014 at 9:48

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