Degree of a Segre variety (without Hilbert polynomial)

Question

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?

Answer 1

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.