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In Section 1.7 of EGA I, we have the following:

(1.7.1) Étant donné un $A$-module $M$, on appelle support de $M$ et on note $\text{Supp}(M)$ l'ensemble des idéaux premiers $\mathfrak{p}$ de $A$ tels que $M_{\mathfrak{p}} \ne 0$. Pour que $M = 0$, il faut et il suffit que $\text{Supp}(M) = \emptyset$, car si $M_{\mathfrak{p}}= 0$ pour tout $\mathfrak{p}$, l'annulateur d'un élément $x \in M$ ne peut être contenu dans aucun idéal premier de $A$, donc est $A$ tout entier.

I would translate this as follows:

Being given an $A$-module $M$, we say the {\it support} of $M$ and we write $\text{Supp}(M)$ for the set of prime ideals $\mathfrak{p}$ of $A$ such that $M_{\mathfrak{p}} \ne 0$. For $M = 0$, it is necessary and sufficient that $\text{Supp}(M) = \emptyset$, because if $M_{\mathfrak{p}} = 0$ for every $\mathfrak{p}$, the annihilator of an element $x \in M$ cannot be contained in any prime ideal of $A$, therefore $A$ is [???].

As you can see, I am stuck on the phrase "tout entier". It sounds like "completely integral", but what does this actually mean in this context?

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    $\begingroup$ "...therefore is equal to $A$." $\endgroup$
    – user26857
    Feb 13, 2018 at 20:45
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    $\begingroup$ "donc l'annulateur est A tout entier" $\endgroup$
    – Billy
    Feb 13, 2018 at 20:47

1 Answer 1

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Thanks to the comments by user26857 and Billy, I see now that it's just saying the annihilator must be all of $A$.

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