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 10.38.11) 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 $a_{1j} = -a'_ j$ and $a_{ij} = \delta _{ij}$ for $i, j \geq 2$). $\square$


Comments (2)

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.


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0539. Beware of the difference between the letter 'O' and the digit '0'.