Bourbaki

Lemma 3

Tag 04F0, permanent. Cite it as [Bourbaki, Tag 04F0].

Commutative Algebra, chapter III, § 2. General results on filtered rings and modules

Let AA be a complete Hausdorf topological ring, in which there exists a fundamental system G\mathscr{G} of neighbourhoods of 0 consisting of additive subgroups of AA. (i) For all x∈Ax \in A such that lim⁡n→mxn=0\lim_{n \to m} x^n = 0, 1−x1 - x is invertible in AA and its inverse is equal to ∑n=0∞xn\sum_{n=0}^\infty x^n. (ii) Let aa be a two-sided ideal of AA such that lim⁡n→∞xn=0\lim_{n \to \infty} x^n = 0 for all x∈ax \in a. For an element y of A\mathbf{A} to be invertible, it is necessary and sufficient that its class mod. a be invertible in A/a\mathbf{A}/a; in particular a is contained in the Jacobson radical of A\mathbf{A}.

(i) As

(1−x)(1+x+⋯+xn)=(1+x+⋯+xn)(1−x)=1−xn+1, (1 - x)(1 + x + \cdots + x^n) = (1 + x + \cdots + x^n)(1 - x) = 1 - x^{n+1},

it all amounts to proving that the series with general term xnx^n is convergent in A\mathbf{A}; now, by hypothesis, for every neighbourhood V∈SV \in \mathfrak{S} of 0 in A\mathbf{A}, there exists an integer p>0p > 0 such that xn∈Vx^n \in V for all n≥pn \geq p. We conclude that

xp+xp+1+…+xq∈V x^p + x^{p+1} + \ldots + x^q \in V

for all q≥pq \geq p and our assertion then follows from Cauchy's criterion (General Topology, Chapter III, § 5, no. 2, Theorem 1).

(ii) Suppose that there exists y′∈Ay' \in \mathbf{A} such that yy′≡1yy' \equiv 1 (mod. a) and y′y≡1y'y \equiv 1 (mod. a). The hypothesis on a implies, by (i), that yy′yy' and y′yy'y are invertible in A\mathbf{A} and hence yy is invertible in A\mathbf{A}. In particular, every x∈ax \in a is such that 1−x1 - x is invertible in A\mathbf{A} and, as a is a two-sided ideal of A\mathbf{A}, it is contained in the Jacobson radical of A\mathbf{A} (Algebra, Chapter VIII, § 6, no. 3, Theorem 1).

Having established this lemma, it is sufficient to apply it to the topological ring A^\hat{\mathbf{A}} and the ideal m^\hat{\mathfrak{m}}, as, for all x∈m^x \in \hat{\mathfrak{m}}, xn∈(m^)n⊂(mn)∧x^n \in (\hat{\mathfrak{m}})^n \subset (\mathfrak{m}^n)^{\wedge} and the sequence (xn)(x^n) therefore tends to 0.

If we take A=Z\mathbf{A} = \mathbf{Z}, every maximal ideal of Z\mathbf{Z} is of the form pZp\mathbf{Z} where pp is prime. The ring of pp-adic numbers Zp\mathbf{Z}_p is then a local ring of which pZpp\mathbf{Z}_p is the maximal ideal (Corollary 2 to Proposition 16) and whose residue field is isomorphic to Z/pZ=Fp\mathbf{Z}/p\mathbf{Z} = \mathbf{F}_p, and Z(p)\mathbf{Z}_{(p)} with the pZ(p)p\mathbf{Z}_{(p)}-adic topology is identified with a topological subring of Z\mathbf{Z}, containing Z\mathbf{Z}.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.