Is a complete local ring determined by its values in local fields?

Question

Let $A$ be a complete, Noetherian, local ring with finite residue field of characteristic $p$. If $F$ is a non-Archimedean local field, then we will denote the ring of integers of $F$ by $\mathcal{O}_F$. Let $I$ denote the intersection of kernels of all (local) morphisms $A\to \mathcal{O}_F$ where $F$ runs over all non-Archimedean local field of zero or $p$.

Question: Is $I=(0)$?

In other words, is a complete local ring determined by its values in local fields? Any comments and reference would be appreciated. The answer to the question is Yes when $A$ is flat over $\mathbb{Z}_p$ and reduced, cf. Corollary 2.3 Serre’s modularity conjecture (II).

Edit: As pointed out in the comment, I should assume that $A$ is reduced.

Answer 1

The paper you cite itself cites Corollary 10.5.8 of EGA 4, part III. Corollary 10.5.9 says the points of dimension $1$ in $\operatorname{Spec} A$ are dense in $\operatorname{Spec} A - \mathfrak m$. Each point of dimension $1$ corresponds to a homomorphism to a complete local ring of dimension $1$, whose field of fractions is a local field, and the intersection of the kernels corresponds to an ideal whose vanishing set is dense, which if $A$ is reduced must be the zero ideal.

See also 10.5.10 which explains how to use these local rings to form a basis for the topology on $A$.