I am currently reading Takeshi Saito's book "Fermat's Last Theorem: The Proof". In the proof of Proposition 8.12, there is a part that I do not understand clearly. For my problem in particular, I am not sure that all the context is actually needed in order to understand what is going on. However, just to be sure, let me write down what we are doing.
The statement is the following.
Let $S$ be a scheme, let $E$ be a smooth curve over $S$, let $N\geq 1$ be an integer. Suppose $X$ is a closed subscheme of $E$ that is finite flat of finite presentation over $S$ of degree $N$. For sections $P_1,\ldots ,P_N:S\rightarrow X$, the following are equivalent:
1. $P_1,\ldots ,P_N$ form a full set of sections of $X$.
2. The following equality of effective Cartier divisors holds: $$X=\sum_{i=1}^N[P_i]$$
Actually, my problem lies in the easiest part of the proof, that is $2.\Rightarrow 1.$ Here is how the proof of the book goes.
We may assume that $S=\operatorname{Spec}(R)$ is affine. For $i=1,\ldots ,N$, we let $\mathcal{I}_i$ be the defining sheaf of ideals of the effective Cartier divisor $[P_i]$ of $E$. By the equality of divisors $X=\sum_{i=1}^N[P_i]$, the finitely generated free $\mathcal{O}_S$-module $\mathcal{O}_X$ is a successive extension of the invertible $\mathcal{O}_S$-modules $\prod_{j=1}^{i-1}\mathcal{I}_j/\prod_{j=1}^{i}\mathcal{I}_j$. For any element $f\in \Gamma (X,\mathcal{O})$, the multiplication-by-$f$ map of $\mathcal{O}_X$ induces the multiplication-by-$f(P_i)$ map of $\prod_{j=1}^{i-1}\mathcal{I}_j/\prod_{j=1}^{i}\mathcal{I}_j$. Hence, we have $\operatorname{Norm}_{X/S}(f)=\prod_{i=1}^Nf(P_i)$.
That is the proof. Now, the part of it that bugs me is the one I wrote in italic: what does "a successive extension of $\mathcal{O}_S$-modules" actually mean?
I understand that $\mathcal{O}_X=\mathcal{O}_E/\prod_{j=1}^{N}\mathcal{I}_j$ (or, to be more exact, this is rather $i_{\star}\mathcal{O}_X$ where $i$ denotes the closed immersion of $X$ in $E$). What does "successive extension" then mean? Should we exploit the algebraic fact that if $L\subset M\subset N$ are modules, then $(N/L)/(M/L)=N/M$?
If so, how does this writing allows us to make sense of the following argument about the multiplication by $f$ endomorphism?
I thank you very much for your explanations.
Yes, he probably means the following exact sequences of $\mathcal{O}_X$ modules, though they are written as if $\mathcal{O}_E$ modules, justified by the quotient by $\displaystyle\prod_{j = 1}^N\mathcal{I}_j$: $$ 0\to \mathcal{I}_1/\prod_{j = 1}^N\mathcal{I}_j\to \mathcal{O}_X = \mathcal{O}_E/\prod_{j = 1}^N\mathcal{I}_j\to \mathcal{O}_E/\mathcal{I}_1\to 0, \\ 0\to \mathcal{I}_1\mathcal{I}_2/\prod_{j = 1}^N\mathcal{I}_j\to \mathcal{I}_1/\prod_{j = 1}^N\mathcal{I}_j\to \mathcal{I}_1/\mathcal{I}_1\mathcal{I}_2\to 0, \\ \vdots \\ 0\to \prod_{j = 1}^{N-1}\mathcal{I}_j/\prod_{j = 1}^N\mathcal{I}_j\to \prod_{j = 1}^{N-2}\mathcal{I}_j/\prod_{j = 1}^N\mathcal{I}_j\to \prod_{j = 1}^{N-2}\mathcal{I}_j/\prod_{j = 1}^{N-1}\mathcal{I}_j\to 0. $$ The sequences remain exact after being pushed on $S$ because $g:X\to S$ is affine so that the higher direct images of the leftmost terms vanish. As the rightmost terms pushed on $S$ can be identified with the pullback by the closed immersion defining cartier divisors (e.g. if $i:X \to E$, then $\mathcal{I}_1/\mathcal{I}_1\mathcal{I}_2 = g_*P_2{}_*P_2^*i^*\mathcal{I}_1\cong P_2^*i^*\mathcal{I}_1$), they are invertible modules on $S$ so that the sequences split on $S$.
As $f$ acts on $\displaystyle\prod_{j = 1}^k\mathcal{I}_j/\displaystyle\prod_{j = 1}^{k+1}\mathcal{I}_j$ as a multiplication by $f(P_{k+1})$ (cf. see also the above example), the norm of $f$ can just be calculated as the product of them by the splitting, just like linear algebra.