The Stacks project

Lemma 15.22.10. Let $A$ be a valuation ring. An $A$-module $M$ is flat over $A$ if and only if $M$ is torsion free.

Proof. The implication “flat $\Rightarrow $ torsion free” is Lemma 15.22.9. For the converse, assume $M$ is torsion free. By the equational criterion of flatness (see Algebra, Lemma 10.39.11) we have to show that every relation in $M$ is trivial. To do this assume that $\sum _{i = 1, \ldots , n} a_ i x_ i = 0$ with $x_ i \in M$ and $a_ i \in A$. After renumbering we may assume that $v(a_1) \leq v(a_ i)$ for all $i$. Hence we can write $a_ i = a'_ i a_1$ for some $a'_ i \in A$. Note that $a'_1 = 1$. As $M$ is torsion free we see that $x_1 = - \sum _{i \geq 2} a'_ i x_ i$. Thus, if we choose $y_ i = x_ i$, $i = 2, \ldots , n$ then

\[ x_1 = \sum \nolimits _{j \geq 2} -a'_ j y_ j, \quad x_ i = y_ i, (i \geq 2)\quad 0 = a_1 \cdot (-a'_ j) + a_ j \cdot 1 (j \geq 2) \]

shows that the relation was trivial (to be explicit the elements $a_{ij}$ are defined by setting $a_{11} = 0$, $a_{1j} = -a'_ j$ for $j > 1$, and $a_{ij} = \delta _{ij}$ for $i, j \geq 2$). $\square$

Comments (4)

Comment #2329 by Guignard on

Two typos : - "" should be "". - " is torsion free" should be " is torsion free".

Comment #2400 by on

THanks, fixed here.

If you want your name on the list of contributors, please give at least one first name.

Comment #5423 by Badam Baplan on

In the last paranthetical following "explicitly," it should read and for .

As it is currently written, with the vector of and the matrix of , you get

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  • 2 comment(s) on Section 15.22: Torsion free modules

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