Bourbaki

Exercise 5, Appendix 0

Commutative Algebra, IX, p. 85

Let A be a local ring, with residue field κ(A)\kappa(A). Let B be an inflation of A, with residue field κ(B)\kappa(B). a) If κ(B)=κ(A)\kappa(B) = \kappa(A), one has A=BA = B. b) If κ(B)\kappa(B) is algebraic over κ(A)\kappa(A), the A-algebra B is integral. c) If κ(B)\kappa(B) is of finite degree over κ(A)\kappa(A), B is a free A-module whose rank is the degree of κ(B)\kappa(B) over κ(A)\kappa(A).

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Machine translation of the English, by gpt-5.4. Not checked by a person.