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The Stacks project

Lemma 10.81.1. Let M be an R-module. The following are equivalent:

  1. M is flat.

  2. If f: R^ n \to M is a module map and x \in \mathop{\mathrm{Ker}}(f), then there are module maps h: R^ n \to R^ m and g: R^ m \to M such that f = g \circ h and x \in \mathop{\mathrm{Ker}}(h).

  3. Suppose f: R^ n \to M is a module map, N \subset \mathop{\mathrm{Ker}}(f) any submodule, and h: R^ n \to R^{m} a map such that N \subset \mathop{\mathrm{Ker}}(h) and f factors through h. Then given any x \in \mathop{\mathrm{Ker}}(f) we can find a map h': R^ n \to R^{m'} such that N + Rx \subset \mathop{\mathrm{Ker}}(h') and f factors through h'.

  4. If f: R^ n \to M is a module map and N \subset \mathop{\mathrm{Ker}}(f) is a finitely generated submodule, then there are module maps h: R^ n \to R^ m and g: R^ m \to M such that f = g \circ h and N \subset \mathop{\mathrm{Ker}}(h).

Proof. That (1) is equivalent to (2) is just a reformulation of the equational criterion for flatness1. To show (2) implies (3), let g: R^ m \to M be the map such that f factors as f = g \circ h. By (2) find h'': R^ m \to R^{m'} such that h'' kills h(x) and g: R^ m \to M factors through h''. Then taking h' = h'' \circ h works. (3) implies (4) by induction on the number of generators of N \subset \mathop{\mathrm{Ker}}(f) in (4). Clearly (4) implies (2). \square

[1] In fact, a module map f : R^ n \to M corresponds to a choice of elements x_1, x_2, \ldots , x_ n of M (namely, the images of the standard basis elements e_1, e_2, \ldots , e_ n); furthermore, an element x \in \mathop{\mathrm{Ker}}(f) corresponds to a relation between these x_1, x_2, \ldots , x_ n (namely, the relation \sum _ i f_ i x_ i = 0, where the f_ i are the coordinates of x). The module map h (represented as an m \times n-matrix) corresponds to the matrix (a_{ij}) from Lemma 10.39.11, and the y_ j of Lemma 10.39.11 are the images of the standard basis vectors of R^ m under g.

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