Bourbaki

Exercise 3, § 4

Commutative Algebra, III, 256

Bourbaki marks this one as harder than the rest.

Let AA be a local ring, m\mathfrak{m} its maximal ideal, k=A/mk = A/\mathfrak{m} its residue field and f:A→kf : A \to k the canonical homomorphism. Show that the two following conditions are equivalent:

(H) For every monic polynomial P∈A[X]P \in A[X] and every decomposition of f(P)∈k[X]f(P) \in k[X] as a product f(P)=Q.Q′f(P) = Q.Q' of relatively prime monic polynomials, there exist two monic polynomials Q,Q′Q, Q' in A[X]A[X] such that f(Q)=Q‾f(Q) = \overline{Q}, f(Q′)=Q‾′f(Q') = \overline{Q}' and P=QQ′P = QQ'.

(C) Every commutative algebra over AA which is a finitely generated AA-module is a direct composition of AA-algebras which are local rings.

(To prove that (H) implies (C), argue as in Proposition 8 of no. 6, using Exercise 2 (a). To see that (C) implies (H), show first that, for every commutative AA-algebra BB which is a finitely generated AA-module, every decomposition of B/mBB/\mathfrak{m}B as a direct sum of two ideals is necessarily of the form b/mb⊕b′/mb′b/\mathfrak{m}b \oplus b'/\mathfrak{m}b', where B=b⊕b′B = b \oplus b' is a decomposition of BB as a direct sum of two ideals; then use Exercise 2 (b).)

A local ring satisfying conditions (H) and (C) is called Henselian. Every complete Hausdorff local ring is Henselian. If AA is Henselian and BB is a commutative AA-algebra which is a local ring and a finitely generated AA-module, then BB is Henselian.

Solution

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Transcribed from the printed volume. Not translated.