Bourbaki

Exercise 11, § 3

Commutative Algebra, X, p. 160

Soit AA a local ring.

a) Let nn be an integer ≥1\geq 1, u:An−1→Anu : A^{n-1} \to A^n an AA-linear mapping, d(u)\mathfrak{d}(u) the ideal generated by the minors of order n−1n-1 of uu (Exercise 8). If profA(d(u);A)≥2\mathrm{prof}_A(\mathfrak{d}(u); A) \geq 2, prove that d(u)\mathfrak{d}(u) admits the resolution

0→An−1→uAn→vd(u)→0, 0 \to A^{n-1} \xrightarrow{u} A^n \xrightarrow{v} \mathfrak{d}(u) \to 0,

where vv is induced by the homomorphism ∧n−1(tu):An→A\wedge^{n-1}(t u) : A^n \to A (apply Exercise 9).

b) Conversely, let JJ be an ideal of AA of projective dimension 1. Prove that there exist a cancellable element aa of AA, an integer nn and an AA-linear mapping u:An−1→Anu : A^{n-1} \to A^n such that J=ad(u)J = a \mathfrak{d}(u); one has profA(J;A)=2\mathrm{prof}_A(J; A) = 2 (“Hilbert-Burch theorem”): deduce from Exercise 9 that a minimal resolution of JJ is of the form 0→An−1→uAn→J0 \to A^{n-1} \xrightarrow{u} A^n \to J, with profA(d(u);A)≥2\mathrm{prof}_A(\mathfrak{d}(u); A) \geq 2, then apply a) and Exercise 1 of § 1, observing that d(u)\mathfrak{d}(u) contains a cancellable element of AA.

EXERCISES

AC X.161

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.