Bourbaki

Proposition 8

Tag 00DD, permanent. Cite it as [Bourbaki, Tag 00DD].

Algebra, VIII, p. 191, § 11. Grothendieck Groups

Let A be a principal ideal domain that is not a field, and let L be its field of fractions. There exists an isomorphism φ:R(A)→L∗/A∗\varphi : R(A)\rightarrow L^*/A^* such that we have

φ([A/aA])=aA∗(9) \varphi ([A/aA]) =aA^* \tag{9}

for every a≠0a\not= 0 in A.

Let P be a system of representatives consisting of irreducible elements of A (VII, §1, No. 3, p. 3). The maximal ideals of A are the ideals pApA for p∈Pp\in P; every simple A-module is therefore isomorphic to a unique module A/pAA/pA. Moreover (VII, §1, No. 3, p. 4, Theorem 2), the abelian group L∗/A∗L^*/A^* is free and has the family (pA∗)p∈P(pA^*)_{p\in P} as a basis. Hence there exists an isomorphism φ\varphi from R(A) to L∗/A∗L^*/A^* characterized by φ([A/pA])=pA∗\varphi ([A/pA]) =pA^* for every p∈Pp\in P.

Let a≠0a\not= 0 in A. There exist an integer r⩾0r\geqslant 0, elements p1,...,prp_1, . . . , p_r of P, and an element uu of A∗A^* such that we have a=up1⋯pra=up_1\cdots p_r. The module A/aAA/aA admits the composition series defined by

E0=A/aA,Ei=(p1⋯piA)/aA(1⩽i⩽r) E_0= A/aA,E_i= (p_1\cdots p_iA)/aA(1\leqslant i\leqslant r)

and the module Ei−1/Ei=(p1⋯pi−1A)/(p1⋯piA)E_{i-1}/E_i= (p_1\cdots p_{i-1}A)/(p_1\cdots p_iA) is isomorphic to A/piAA/p_iA. We therefore have (VIII, p. 185, Proposition 3)

φ([A/aA])=∏i=1rφ([A/piA])=p1⋯prA∗=aA∗ \varphi ([A/aA]) =\prod_{i=1}^r\varphi ([A/p_iA]) =p_1\cdots p_rA^*=aA^*

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 2023, Springer Nature. Not translated.