Lemma 15.22.10 (0539)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 15: More on Algebra
Section 15.22: Torsion free modules
Lemma 15.22.10 (cite)

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 (7)

Comment #2329 by Guignard on December 12, 2016 at 14:58

Two typos : - " $f_i \in A$ " should be " $a_i \in A$ ". - " $A$ is torsion free" should be " $M$ is torsion free".

Comment #2400 by Johan on February 17, 2017 at 01:23

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 July 25, 2020 at 22:17

In the last paranthetical following "explicitly," it should read $a_{11} = 0$ and $a_{1j} = -a_j'$ for $j> 1$ .

As it is currently written, with $Y$ the vector of $a_i$ and $L$ the matrix of $a_{ij}$ , you get $Y^t L = (-a_1, 0, 0, \ldots)$

Comment #5650 by Johan on November 14, 2020 at 16:42

Thanks and fixed here.

Comment #7452 by Hao Peng on June 24, 2022 at 04:14

I noticed that after minor change this lemma holds for a Bezout domain too:

Let $\sum a_ix_i=0$ where $a_i\in R^*, x_i\in M$ , then consider $(a_1, \ldots, a_n)=(a)$ , thus $a_i=ab_i$ for some $b_i\in R$ , and $\sum c_ib_i=1$ for some $c_i\in R$ . Because $R$ is a domain, $\sum_i b_ix_i=0$ , so we can take $\overrightarrow{y}=A\overrightarrow{x}$ , where $A=1-\overrightarrow{c}\overrightarrow{b}^t$ . Then $\overrightarrow{b}^tA=\overrightarrow{b}^t-\overrightarrow{b}^t\overrightarrow{c}\overrightarrow{b}^t=0$ .

Comment #7453 by Hao Peng on June 24, 2022 at 04:24

Sorry, in the argument above, change " $R$ is a domain" to " $M$ is torsion-free".

Comment #7605 by Stacks Project on August 14, 2022 at 07:42

Yes, this is good. I'm going to leave it for now. If we every need Bezout domains we'll add this (maybe with a new tag).

There are also:

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

Comments (7)

Post a comment