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Algebraic Geometry is not my topic of research and I am having troubles in understanding the following:

In Harris book, Algebraic Geometry, a firs course, Example 18.15, there is a proof of the degree of the Segre varieties $\Sigma_{m,n}$. The author intersects $\Sigma_{m,n} \subset \mathbb{P}(V \otimes W)$ with a subspace of codimension $m+n$ defined by simple tensors of the dual space $V^* \otimes W^*$, that is the subspace is the intersection of hyperplanes that are unions of pullbacks of each factor. The intersection contains at most $\binom{n+m}{n}$ points. I read similar proofs elsewhere. But, as far as I know, to prove that a variety of dimension $k$ has degree $d$, one has to prove that the intersection of the variety with a GENERAL subspace of codimension $k$ contains at most $d$ points. How is it possible to use such a special linear subspace and then deduce that it is true for a general one?

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You forget the last 3 lines of the proof : "As before, we have to check that the intersection of $\Sigma _{m,n}$ with the hyperplanes $H_i$ is transverse at each point; assuming this, we deduce that the degree of the Segre variety is $\binom{n+m}{n}$". Indeed, for any linear subspace $L$ of the right codimension intersecting transversally a subvariety $X\subset \mathbb{P}^N$, we have $\#\,L\cap X=\deg (X)$.

Of course checking transversality is not completely trivial; this is why the cohomological approach is simpler.

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  • $\begingroup$ Yes, of course we need transversality, what it is not clear to me is how to check the intersection with such a special linear subspace is enough to deduce what happens with the intersection with any linear subspace of the right codimension and intersecting transversally $X$. $\endgroup$
    – user46071
    Commented Aug 26, 2014 at 10:07
  • $\begingroup$ As I said, transversality is enough. Look at Bézout theorem 18.3 in Harris' book. $\endgroup$
    – abx
    Commented Aug 26, 2014 at 10:08
  • $\begingroup$ Sorry,I don't think I explained myself.Suppose we don't know the degree $d$ of $X$ but we know the dimension $k$. What I would do to find $d$ (but I am not algebraic geometer, so I have a very naive approach) to intersect $X$ with a linear space of codimension $k$ intersecting transversally $X$ and to check how many points I get. What I see in many notes, when $X$ is a Segre variety, is the intersection of $X$ with a subspace that has codimension $k$, intersects transversally $X$ but ALSO it is the intersection of hyperplanes union of pullback of hyperplanes of each factor. $\endgroup$
    – user46071
    Commented Aug 26, 2014 at 10:31
  • $\begingroup$ So it is not the generic transversal subspace, but a special one, why are we allowed to do that? $\endgroup$
    – user46071
    Commented Aug 26, 2014 at 10:31
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    $\begingroup$ Once again (probably for the last time): any linear subspace of the right codimension intersecting transversally gives the right number. See Harris, thm. 18.3. The fact that your subspace is special does not matter. $\endgroup$
    – abx
    Commented Aug 26, 2014 at 10:34

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