My question arises from the reading of Katz & Mazur's book "Arithmetic Moduli of Elliptic Curves", which can be found here. Given an arbitrary scheme $S$, an elliptic curve over $S$ is a proper smooth curve $E/S$ with geometrically connected fibers all of genus $1$, together with a section $"0"$.

By Abel's theorem, we know that such a curve comes with a commutative $S$-group-scheme structure. In theorem 2.3.1 (page 73 in the book, page 42 in the pdf), we would like to show the following statement about the behaviour of the "multiplication by $N$" homomorphism.

Let $N\geq 1$ be an integer. Then the $S$-homomorphism "multiplication by $N$" $$[N]: E\rightarrow E$$ is finite locally free of rank $N^2$. If $N$ is invertible on $S$, its kernel $E[N]$ is finite étale over $S$, locally for the étale topology on $S$ isomorphic to $\mathbb{Z}/N\mathbb{Z}\times \mathbb{Z}/N\mathbb{Z}$.

As for the proof of this theorem, we begin by a reduction argument "to the universal case", which is precisely what I don't understand. It is stated the following way.

Zariski locally on $S$, $E$ is given by a smooth Weierstrass cubic in $\mathbb{P}^2_S$, with origin $(0,1,0)$, and any smooth Weierstrass cubic in $P^2_S$ is an elliptic curve over $S$ with origin $(0,1,0)$. So by reduction to the universal case, we may syppose that $S$ is the open set $U$ in $$\operatorname{Spec}(\mathbb{Z}[a_1,a_2,a_3,a_4,a_6])$$ over which the cubic $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$ is smooth.

I do not understand what the so-called "universal case" means and what role it plays... Are we trying to build a morphism $S\rightarrow \operatorname{Spec}(\mathbb{Z}[a_1,a_2,a_3,a_4,a_6])$ identifying $S$ with $U$ and then considering the base change? Is it what is meant here? But as far as I can tell, if $E$ is described by a smooth cubic $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$ in $\mathbb{P}^2_S$ with coefficients $a_i\in \Gamma(S, \mathcal{O})$, the resulting map $S\rightarrow \operatorname{Spec}(\mathbb{Z}[a_1,a_2,a_3,a_4,a_6])$ indeed has image in $U$, but does not identify $S$ with $U$...