Bourbaki

Example 2

Tag 074Q, permanent. Cite it as [Bourbaki, Tag 074Q].

Algebra, VII, p. 28, § 4. Finitely generated modules over a principal ideal domain

Let aa be a nonzero element of AA. The map x↦x/ax \mapsto x/a from AA into KK induces an isomorphism on quotient modules from A/(a)A/(a) onto the submodule (K/A)(a)(K/A)(a) of K/AK/A consisting of elements annihilated by aa. If MM is an AA-module annihilated by aa, or equivalently an A/(a)A/(a)-module, then the AA-module D(M)D(M) is identified with Hom⁡A/(a)(M,A/(a))\operatorname{Hom}_{A/(a)}(M, A/(a)). We leave the reader to rewrite the preceding results for this special case in the corresponding notation (cf. V, p. 86).

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

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Transcribed from the printed edition, 2003, Springer. Not translated.