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compact Hausdorff rings are profinite

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Contents

Introduction

The goal of this article is to prove that a compact Hausdorff ring RR is a profinite ring, i.e., is canonically isomorphic as a topological ring to its profinite completion R^\widehat{R}, topologized as an inverse limit of finite rings with their discrete topologies.

Further commentary on this result can be found in a CafΓ© blog post of Tom Leinster, in connection with some work by his student Barry Devlin.

Preliminaries

All topological rings considered are T 0T_0 (see separation axiom). As is well-known from the study of uniform structures, this implies that all our rings are Hausdorff and even Tychonoff spaces. In particular, finite rings are Hausdorff and therefore discrete. If SS is a finite ring and f:Rβ†’Sf: R \to S is a continuous ring homomorphism, this means the kernel of ff must be an open ideal of RR (for us, β€œideal” always means two-sided ideal), since {0}\{0\} is open in SS.

In the converse direction, if RR is a compact ring and II is an open ideal, then the ring R/IR/I with the quotient topology coming from RR is discrete (every point is open), and compact since RR is compact, and so R/IR/I must then be finite.

Let π’ͺ(R)\mathcal{O}(R) denote the poset of open ideals of RR, ordered by inclusion. This poset is codirected by taking finite intersections of ideals. Each inclusion IβŠ†JI \subseteq J induces a surjection of finite rings R/Iβ†’R/JR/I \to R/J, and so we get a functor π’ͺ(R)β†’Ring\mathcal{O}(R) \to Ring whose (projective) limit is the profinite completion R^\widehat{R} of RR; the quotient maps proj I:Rβ†’R/Iproj_I: R \to R/I of course induce a canonical map Ο€:Rβ†’R^\pi: R \to \widehat{R}.

Proposition

The map π:R→R^\pi: R \to \widehat{R} is surjective.

Proof

Since RR is compact and R^\widehat{R} is Hausdorff, the image Ο€(R)\pi(R) is closed in R^\widehat{R}, so it suffices to show the image is a dense subspace.

So, we are to show that if UβŠ†R^U \subseteq \widehat{R} is open and inhabited, then there exists x∈Rx \in R with Ο€(x)∈U\pi(x) \in U. Let Ο€ I:R^β†’R/I\pi_I: \widehat{R} \to R/I denote a typical component map of the limit cone. A basis for the topology of R^\widehat{R} consists of inhabited finite intersections of the form U=Ο€ I 1 βˆ’1(x 1)βˆ©β€¦βˆ©Ο€ I n βˆ’1(x n)U = \pi_{I_1}^{-1}(x_1) \cap \ldots \cap \pi_{I_n}^{-1}(x_n) where each x kx_k is a point of R/I kR/I_k, so WLOG assume UU is of this form.

Let y=⟨y I⟩ I∈π’ͺ(R)∈R^y = \langle y_I \rangle_{I \in \mathcal{O}(R)} \in \widehat{R} be any point belonging to UU (so y I i=x iy_{I_i} = x_i for i=1,…,ni = 1, \ldots, n), and put J=β‹‚ i=1 nI iJ = \bigcap_{i=1}^n I_i. Then for each x ix_i the coordinate y Jy_J maps to x ix_i under the map R/Jβ†’R/I iR/J \to R/I_i, precisely because of the compatibility conditions on the coordinates imposed by the limit. Then, letting yβ€²βˆˆRy' \in R be an element that maps to y Jy_J under Rβ†’R/JR \to R/J, the element Ο€(yβ€²)\pi(y') lies in UU and we are done.

Compact rings are totally disconnected

A key technical result is the following (see this nn-Category CafΓ© comment):

Theorem

Every compact Hausdorff ring RR is totally disconnected.

Proof

Let R *R^\ast be the Pontryagin dual of (the additive group of) RR; as a space R *R^\ast is discrete. As RR and R *R^\ast are locally compact Hausdorff, they are exponentiable as spaces, and it follows quickly that the functorial maps

Hom(R,R)β†’Hom(R *,R *),Hom(R *,R *)β†’Hom(R **,R **)β‰…Hom(R,R)Hom(R, R) \to Hom(R^\ast, R^\ast), \qquad Hom(R^\ast, R^\ast) \to Hom(R^{\ast\ast}, R^{\ast\ast}) \cong Hom(R, R)

are continuous homomorphisms of topological groups. Using naturality of the isomorphism R≅R **R \cong R^{\ast\ast}, these maps are mutually inverse, and so they give an isomorphism of topological groups.

The multiplication map RΓ—Rβ†’RR \times R \to R transforms to a continuous injection i:Rβ†’Hom(R,R)β‰…Hom(R *,R *)i: R \to Hom(R, R) \cong Hom(R^\ast, R^\ast). As RR is compact and Hom(R *,R *)Hom(R^\ast, R^\ast) is Hausdorff, the injection ii maps RR homeomorphically onto its image in Hom(R *,R *)Hom(R^\ast, R^\ast), i.e., is a subspace embedding. But Hom(R *,R *)Hom(R^\ast, R^\ast) is manifestly a subspace of a product of discrete spaces and hence totally disconnected, so RR is totally disconnected as well. (In more detail: if CβŠ†RC \subseteq R is connected, then the image of CC under the composite

CβŠ†Rβ†’iHom(R *,R *)β†ͺ∏ x∈R *R *β†’proj xR *C \subseteq R \stackrel{i}{\to} Hom(R^\ast, R^\ast) \hookrightarrow \prod_{x \in R^\ast} R^\ast \stackrel{proj_x}{\;\;\to\;\;} R^\ast

is also connected and hence a one-point space for each x∈R *x \in R^\ast, so that i(C)i(C) and therefore CC are one-point spaces.)

Compact rings have enough open ideals

By Theorem , Pontryagin duality implies compact rings are totally disconnected, which readies us for a next wave of results.

Lemma

Let XX be a compact Hausdorff totally disconnected space. For each x∈Xx \in X and open neighborhood VV of xx, there exists a compact open set UU such that x∈UβŠ†Vx \in U \subseteq V.

Proof

In a compact Hausdorff space XX, the connected component of a point xx equals its quasi-component; an online proof may be found here. So if XX is also totally disconnected, then xx is also the intersection of all clopens containing it. Now let VV be an open neighborhood of xx, so β‹‚ x∈KclopenKβŠ†V\bigcap_{x \in K\; clopen} K \subseteq V and therefore Β¬VβŠ†β‹ƒ x∈KclopenΒ¬K\neg V \subseteq \bigcup_{x \in K\; clopen} \neg K. As Β¬V\neg V is compact, finitely many such clopens Β¬K 1,…,Β¬K n\neg K_1, \ldots, \neg K_n cover Β¬V\neg V, and so U≔⋂ i=1 nK nβŠ†VU \coloneqq \bigcap_{i=1}^n K_n \subseteq V. This UU is compact and open and contains xx, which completes the proof.

Lemma

Let AA be a compact Hausdorff totally disconnected abelian group. For each open neighborhood VV of 0∈A0 \in A, there exists an open subgroup OO such that OβŠ†VO \subseteq V.

Proof

By Lemma , there is a compact open WW with 0∈WβŠ†V0 \in W \subseteq V. Now let Ξ±:WΓ—Wβ†’A\alpha: W \times W \to A be the restriction of the addition operation +:AΓ—Aβ†’A+: A \times A \to A. Since Ξ±\alpha is continuous, there is for each x∈Wx \in W an open neighborhood V xΓ—U xβŠ†Ξ± βˆ’1(W)V_x \times U_x \subseteq \alpha^{-1}(W) of (x,0)(x, 0) where we may assume U xU_x is symmetric (i.e., U x=βˆ’U xU_x = -U_x). Finitely many of the V xV_x cover WW, say V x 1,…,V x nV_{x_1}, \ldots, V_{x_n}. Then putting U=U x 1βˆ©β€¦βˆ©U x nU = U_{x_1} \cap \ldots \cap U_{x_n}, we have that Ξ±(WΓ—U)βŠ†W\alpha(W \times U) \subseteq W, or W+UβŠ†WW + U \subseteq W for short. This UU is again symmetric: U=βˆ’UU = -U.

Hence if U n=U+…+UU^n = U + \ldots + U denotes the set of all sums of nn elements belonging to UU, then U nU^n is open and W+U nβŠ†WW + U^n \subseteq W by induction, whence U nβŠ†WU^n \subseteq W. Since UU is symmetric, the union O=⋃ nU nO = \bigcup_n U^n is the subgroup generated by UU; it is open and contained in WW and therefore also in VV, which completes the proof.

Lemma

Let RR be a compact Hausdorff totally disconnected ring. For each open neighborhood VV of 00, there is an open ideal II such that IβŠ†VI \subseteq V.

Proof

By Lemma , there is an open additive subgroup OβŠ†VO \subseteq V. Let I={x∈R:RxRβŠ†O}I = \{x \in R: R x R \subseteq O\}. It is evident that 0∈I0 \in I and II is an ideal of RR. Now hold x∈Ix \in I fixed. For each (a,b)∈RΓ—R(a, b) \in R \times R, continuity of multiplication implies there are open neighborhoods W (a,b)W_{(a, b)} of aa, W (a,b)β€²W_{(a, b)}' of bb, and V (a,b)V_{(a, b)} of xx such that W (a,b)V (a,b)W (a,b)β€²βŠ†OW_{(a, b)} V_{(a, b)} W_{(a, b)}' \subseteq O. By compactness of RΓ—RR \times R, finitely many sets of the form W (a,b)Γ—W (a,b)β€²W_{(a, b)} \times W_{(a, b)}' cover RΓ—RR \times R. The corresponding finite intersection β‹‚ i=1 nV (a i,b i)\bigcap_{i=1}^n V_{(a_i, b_i)} is an open neighborhood of xx that is contained in II, thus completing the proof.

Compact rings are profinite

Now for our final result.

Theorem

For RR a compact ring, the canonical map π:R→R^\pi: R \to \widehat{R} is injective.

This and Proposition taken together imply that Ο€\pi is a ring isomorphism. It is also a homeomorphism because it is a continuous bijective map from a compact space to a Hausdorff space. Therefore Ο€\pi is an isomorphism of topological rings.

Proof

As we have an inclusion R^β†ͺ∏ I∈π’ͺ(R)R/I\widehat{R} \hookrightarrow \prod_{I \in \mathcal{O}(R)} R/I of the profinite completion as projective limit, we see the kernel of Ο€\pi is the same as the kernel of

⟨proj I⟩ I∈π’ͺ(R):Rβ†’βˆ I∈π’ͺ(R)R/I\langle proj_I \rangle_{I \in \mathcal{O}(R)}: R \to \prod_{I \in \mathcal{O}(R)} R/I

which is precisely β‹‚ I∈π’ͺ(R)I\bigcap_{I \in \mathcal{O}(R)} I. Since by Lemma every neighborhood of 0∈R0 \in R contains an open ideal I∈π’ͺ(R)I \in \mathcal{O}(R), this intersection is contained in the intersection of all open neighborhoods of 00, which is {0}\{0\} since RR is Hausdorff. Thus the kernel of Ο€\pi is trivial, as was to be shown.

References

  • Tom Leinster, Holy Crap, Do You Know What A Compact Ring Is?, blog post, nn-Category CafΓ© (August 20, 2014). (link)
  • Todd Trimble, comment on Holy Crap, Do You Know What A Compact Ring Is?, blog post, nn-Category CafΓ© (August 24, 2014). (link)
  • L. Ribes and P. Zalesskii, Profinite groups, volume 40 of Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge (2000), Springer-Verlag, Berlin.

Last revised on September 11, 2016 at 19:44:09. See the history of this page for a list of all contributions to it.