I was reading section 2 in EGA II and I have an issue understanding a little step.
It appears in several propositions, so I will present one instance here. Page 22, proof of Proposition (2.1.9). Since the theorem and proof are somewhat lengthy to copy here, I am only copying the exact sentence, and you can find the full context in the book here.
Notons que, si pour un entier $d\ge n_0$, $f\in S_d$ n'appartient pas à $\mathfrak p_d$, alors, si $\mathfrak p$ existe, $\mathfrak p_m$, pour $m<n_0$, este nécessairement égal à l'ensemble des $x\in S_m$ tels que $f^rx\in\mathfrak p_{m+rd}$, sauf pour un nombre fini de valeurs de $r$."
Now, for the life of me, I cannot understand what's with the except for finitely many values of $r$. Maybe it's a language misunderstanding or I am missing something obvious.
Here's my reasoning: if $\mathfrak p$ exists (si $\mathfrak p$ existe) with the required properties then, since $d\ge n_0$, $f\notin\mathfrak p_d$ is implies $f\notin\mathfrak p$ (as for $n\ge n_0$, $\mathfrak p\cap S_n=\mathfrak p_n$). Also $m+rd\ge n_0$, hence $\mathfrak p_{m+rd}=\mathfrak p\cap S_{m+rd}$, so $f^rx\in\mathfrak p_{m+rd}\subset\mathfrak p$, so for any $r$, $f^rx\in\mathfrak p$ is equivalent to $x\in\mathfrak p$, since $\mathfrak p$ is prime and $f\notin\mathfrak p$; so the statement is true for all positive $r$.
I apologize in advance if I am missing something silly. I tried to ignore this and read on, but the argument appeared a few more times later (second time being only a few lines below).