# Degree of the Segre product of lines

Does anybody know whether there exists a proof by induction (or at least a proof that does not use Hilbert polynomials) that the degree of the Segre variety product of $n$ lines is $n!$ ?

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The Segre embedding is defined by the line bundle $\ L:=\mathcal{O}_{\mathcal{P}^1}(1)\boxtimes\ldots \boxtimes \mathcal{O}_{\mathcal{P}^1}(1)\$ on $(\mathbb{P}^1)^n$, therefore its degree is $L^n=(p_1^*h+\ldots +p_n^*h)^n$, where $h$ is the class of a point in $\mathrm{Pic}(\mathbb{P}^1)$ and $p_i$ the $i$-th projection. The only nonzero term in this expression is $p_1^*h\cdot \ldots \cdot p_n^*h=1$, with coefficient $n!$.

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Let me give a proof by induction.

For $n=2$ we have $\mathbb{P}^1 \times \mathbb{P}^1$ embedded as a quadric in $\mathbb{P}^3$, so the claim is true in this case.

Now let $X_n$ be the Segre embedding of the product of $n$ lines $\mathbb{P}^1 \times \ldots \times \mathbb{P}^1$. An easy computation using the definition of Segre embedding shows that there is a hyperplane section of $X_n$ given by the union of $n$ copies of $X_{n-1}$, one for each copy of $\mathbb{P}^1$ (for instance, there is a hyperplane section of the quadric $X_2$ given by the union of two lines).

By the induction assumption, $X_{n-1}$ has degree $(n-1)!$. So the degree of $X_n$ is $$n \cdot(n-1)! = n!.$$

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Algebraic geometry is not my topic and my problem is that in any proof I have ever read about the degree of a Segre variety is that the subspace chosen is never really "general", at least for me. I mean, you are assuming that there is at least one linear equation defining the subspace of codimension $n$ that is a simple tensor (of the dual space). Is it always the case? Is it not possible that there exists a subspace of codimension $n$ intersecting the variety in a finite number of points and such that the space of linear equations defining it does not contain a simple tensor? –  user46071 Jun 13 '14 at 11:21