By Andy R. Magid

ISBN-10: 1482208059

ISBN-13: 9781482208054

**The Separable Galois idea of Commutative earrings, moment Edition** offers a whole and self-contained account of the Galois concept of commutative earrings from the perspective of specific type theorems and utilizing exclusively the concepts of commutative algebra. in addition to updating approximately each outcome and rationalization, this version includes a new bankruptcy at the thought of separable algebras.

The ebook develops the suggestion of commutative separable algebra over a given commutative ring and explains tips on how to build an similar classification of profinite areas on which a profinite groupoid acts. It explores how the relationship among the kinds depends upon the development of an appropriate separable closure of the given ring, which in flip relies on yes notions in profinite topology. The e-book additionally discusses the right way to deal with jewelry with infinitely many idempotents utilizing profinite topological areas and different methods.

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**Additional resources for The Separable Galois Theory of Commutative Rings, Second Edition**

**Sample text**

It follows that neither does its subring R/M ⊗R S, which means the latter is a finite product of field extensions. 3, ✐ ✐ ✐ ✐ ✐ ✐ 2014/5/19 — 17:41 ✐ ✐ 26 CHAPTER 1. 11 we have that SP is a projective SP /RP bimodule. 12 shows that S is a projective S/R bimodule. 6 (for every prime ideal P of R, SP is a finitely generated and projective RP algebra) will be called locally finitely generated projective. We are now ready to define separable ring extension: Definition 3. A commutative R algebra S is separable if S is a projective S/R bimodule.

Thus the identity, as an S point of S, comes from an A point of S, which is what we needed to show surjectivity. 7 we obtain the following characterization of separability, at least in the finitely generated projective case. 12. Let S be an R algebra which is a finitely generated and projective R module. Then S is a separable R algebra if and only if for every square zero extension p : A → T of R algebras the corresponding map SpecR (S)(T ) → SpecR (S)(A) is surjective. ✐ ✐ ✐ ✐ ✐ ✐ 2014/5/19 — 17:41 ✐ ✐ 34 CHAPTER 1.

In particular, if R has no idempotents except 0 and 1 then R is an R module direct summand of S. Proof. Let I = Ker(R → S). Then S is also an R/I module, and since HomR (S, ·) = HomR/I (S, ·), S is a projective R/I module, as well as finitely generated. 17 implies that R/I is a direct sum- mand of S, and since S is R projective, so is R/I. Thus R → R/I splits, which implies that I is generated by an idempotent e. If R has no idempotents except 0 and 1, then e = 0 so R/I = R is a direct summand of S.

### The Separable Galois Theory of Commutative Rings, Second Edition by Andy R. Magid

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