Bourbaki

Proposition 2

Tag 0531, permanent. Cite it as [Bourbaki, Tag 0531].

Commutative Algebra, chapter VII, § 4. Modules over integrally closed Noetherian domains

Let M be a lattice of V and M, a sub-A-module of V. If there exist two elements x, y of K∗K^* such that xM⊂M1⊂yMxM \subset M_1 \subset yM, M1M_1 is a lattice of V; conversely, if M1M_1 is a lattice of V, there exist two non-zero elements a, b of A such that aM⊂M,⊂b−1MaM \subset M, \subset b^{-1}M.

If L1,L2L_1, L_2 are two free lattices of V such that L,⊂M⊂L,L, \subset M \subset L,, the relations xM⊂M1⊂yMxM \subset M_1 \subset yM imply xL1⊂M,⊂yL2xL_1 \subset M, \subset yL_2 and xL1xL_1 and yL2yL_2 are free lattices; conversely, if M, is a lattice and (ei)1≤i≤n(e_i)_{1 \leq i \leq n} a basis of L, over A, the relation KM, = V implies the existence of x=a/s∈K∗x = a/s \in K^* (where a and s are non-zero elements of A) such that xei∈M,xe_i \in M, for all ii, whence xM⊂xL2⊂M1xM \subset xL_2 \subset M_1 and a fortiori aM⊂M,aM \subset M,; exchanging the roles of M and M, it can be similarly shown that there exists b≠0b \neq 0 in A such that bM1⊂MbM_1 \subset M.

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

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Cites

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Transcribed from the printed volume. Not translated.