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In Borceaux and Janelidze's Galois Theories, a construction of the Pierce spectrum is given. It is the poset of ideals in a Boolean ring. It's construction is reminiscent of the Zariski spectrum in commutative ring theory, however I've not seen it used elsewhere.

Q. Is the Pierce spectrum important elsewhere in mathematics?

Mozibur Ullah
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    It's used by some people in ring theory. For a boolean ring it is the Zariski spectrum. For a began ring it plays the role of the central primitive idempotents in an Artinian ring in that it encodes direct product decompositions by representing the end in a sheaf but not a sheaf of local rings. If your ring is von Neumann regular it can be used to represent your ring as a sheaf of directly indecomposable rings. You should see Pierce's memoir https://www.ams.org/books/memo/0070. – Benjamin Steinberg Dec 01 '20 at 02:53
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    I believe Magid uses it in his galois theory for commutative rings. We use it in https://arxiv.org/abs/1906.06952 to represent skew inverse semigroup rings as convolution algebras on groupoids – Benjamin Steinberg Dec 01 '20 at 02:56
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    It is the Zariski spectrum (of a Boolean ring)! – Qiaochu Yuan Dec 01 '20 at 03:19
  • Actually if the ring is von Neumann regular you get a sheaf of fields. For a general ring you get it represented as the global sections of a sheaf of directly indecomposable rings over a compact totally disconnected space – Benjamin Steinberg Dec 01 '20 at 03:48
  • @BenjaminSteinberg "began ring" ? – David Roberts Dec 01 '20 at 05:13
  • @BenjaminSteinberg I have a feeling auto-correct interfered more than once in your first comment. I couldn't parse "representing the end in a sheaf..." – Todd Trimble Dec 01 '20 at 12:07
  • @ToddTrimble, this is nearly as bad as when I was writing about the cosmological dimension of groups. I meant representing the ring as the global sections of a sheaf of indecomposable rings. When you have finitely many central idempotents you can do this via a direct product but otherwise you need to use sheaves and the Pierce spectrum is how to do it – Benjamin Steinberg Dec 01 '20 at 19:08
  • @Qiaochu Yuan: I should have spotted that. Thanks for pointing this out. – Mozibur Ullah Dec 02 '20 at 17:44

1 Answers1

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For Boolean rings, the Pierce spectrum coincides with the Zariski spectrum and is one of the functors implementing the Stone duality between Boolean algebras and compact totally disconnected Hausdorff spaces and Stonean duality between complete Boolean algebras and compact extremally disconnected Hausdorff spaces.

Restricting further to complete Boolean algebras of projections of commutative von Neumann algebras produces an equivalence of categories of commutative von Neumann algebras and compact strictly localizable enhanced measurable spaces, i.e., a version of the Gelfand duality in the setting of measure theory.

Dmitri Pavlov
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