Comparison with singular cohomology for the cohomology of quadrics. In "Etale cohomology and the Weil Conjectures" By Freitag & Kiehl, chapter 3, paragraph 4, the following statement is made about the computation of the cohomology of quadrics:

For what follows, we need some result about the cohomology of quadrics over a separably closed base field $k$. One can for instance deduce them easily from the corresponding results on singular cohomology of quadrics over the field of complex numbers by using the comparison theorem (Chap. I §12) and the method of specialization (I, §8).

The aforementioned "specialization method" points to the existence of a specialization morphism between stalks of a sheaves given two geometric points $\eta, a$ such that $a$ is a specialization of $\eta$.
I cannot derive what they said in this chapter. To me, comparison with singular cohomology only works for schemes/varieties over the complex numbers. Yet this results are used in order to prove the Picard-Lefschetz formula and eventually to be used in the proof of the Weil's conjectures, which are about varieties over $\mathbb{F}_p$. I suspect that going from $\mathbb{C}$ to any other field is done by the mentionned "specialization method", and that checking on $\mathbb{C}$ is enough since every smooth quadrics on a separably closed field have a "normal form" that is independent on the coefficient field. But I can not derive this.
So I am asking for either a sketch of how it is done, or a referrence where the passage from $\mathbb{C}$ to any separably closed field is done explicitly. 
 A: I think that I found what the authors meant in this paragraph. Please let me know in the comments if I made any mistake in the following.
Every smooth quadric in $\mathbb{P}^{n+1}$ has a normal form as defined by the zero of one of the two following quadratics:


*

*$Q = \sum\limits_{\nu=0}^mX_\nu X_{\nu+m+1}$ if $n = 2m$

*$Q = \sum\limits_{\nu=0}^{m}X_\nu X_{\nu+m+1} + x_{n+1}^2$ if $n = 2m+1$
The second case can only arise in characteristic different from $2$. Suppose that $Q$ denote one of these two. Consider the subvariety of $\mathbb{P}_{\mathbb{Z}}^{n+1}$ defined by $Q$. It is proper and smooth (edit: in the case of the second polynomial, there might be a problem about smoothness over $(2)$, in this case, one could simply work "away" from this point and still get a smooth proper morphism). 
For any separably closed field $k$, one has a geometric point of $\mathbb{P}_{\mathbb{Z}}^{n+1}$ above the prime number corresponding to its characteristic. And $\mathbb{C}$ can be considered as a geometric point above the generic point of $\mathbb{P}_{\mathbb{Z}}^{n+1}$.
Since $f: X \to \mathbb{P}^{n+1}_{\mathbb{Z}}$ is proper smooth, for any constant sheaf $\Lambda_{X}$ on ${X}$, one has that the specialization morphism $(R^qf_*\Lambda_X)_{\mathbb{C}} \to (R^qf_*\Lambda_X)_{k}$ is an isomorphism (edit: if $\Lambda$ has order prime to the characteristic of $k$). 
One has $(R^qf_*\Lambda_X)_{\mathbb{C}} = H^q(X\times_{\mathbb{P}^{n+1}_{\mathbb{Z}}}\mathrm{Spec}(\mathbb{C}), \Lambda)$ and $(R^qf_*\Lambda_X)_{k} = H^q(X\times_{\mathbb{P}^{n+1}_{\mathbb{Z}}}\mathrm{Spec}(k), \Lambda)$ by proper base change. 
But $X\times_{\mathbb{P}^{n+1}_{\mathbb{Z}}}\mathrm{Spec}(k)$ is the zero set of $Q$ in $\mathbb{P}_k^{n+1}$ so it is indeed enough to compute the cohomology of quadrics over $\mathbb{C}$, and the comparison theorem can be used to do that.
