Bourbaki

Lemma 1

Tag 04MT, permanent. Cite it as [Bourbaki, Tag 04MT].

Commutative Algebra, chapter V, § 1. Notion of an integral element

Let AA be a ring, RR an algebra (not necessarily commutative) over AA and xx an element of RR. Let MM be a faithful module over A[x]A[x] which is a finitely generated AA-module. If qq is an ideal of AA such that xM⊂qMxM \subset qM, then xx is a root of a monic polynomial with coefficients in AA all ∉\notin whose coefficients other than the dominant coefficient belong to qq.

Let (ui)1≤i≤n(u_i)_{1 \leq i \leq n} be a finite family of elements of MM such that M=∑i=1nAuiM = \sum_{i=1}^n Au_i. For all ii, there exists by hypothesis a finite family (qij)1≤j≤n(q_{ij})_{1 \leq j \leq n} of elements of qq such that

xui=∑j=1nqijujfor1≤i≤n. xu_i = \sum_{j=1}^n q_{ij}u_j \quad \text{for} \quad 1 \leq i \leq n.

Therefore (Algebra, Chapter III, § 8) if dd is the determinant of the matrix (qij−δijx)(q_{ij} - \delta_{ij}x) with elements in A[x]A[x] (δij\delta_{ij} denoting the Kronecker index), then dui=0du_i = 0 for all ii and hence dM=0dM = 0; as MM is assumed to be a faithful A[x]A[x]-module, necessarily d=0d = 0. This means that xx is a root of the polynomial det⁡(qij−δijX)\det(q_{ij} - \delta_{ij}X) in A[X]A[X] which, up to a sign, is a monic polynomial whose coefficientsother than the dominant coefficient belong to qq.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

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Cites

Transcribed from the printed volume. Not translated.