Definition 15.22.1. Let $R$ be a domain. Let $M$ be an $R$-module.

We say an element $x \in M$ is

*torsion*if there exists a nonzero $f \in R$ such that $fx = 0$.We say $M$ is

*torsion free*if the only torsion element of $M$ is $0$.

In this section we discuss torsion free modules and the relationship with flatness (especially over dimension 1 rings).

Definition 15.22.1. Let $R$ be a domain. Let $M$ be an $R$-module.

We say an element $x \in M$ is

*torsion*if there exists a nonzero $f \in R$ such that $fx = 0$.We say $M$ is

*torsion free*if the only torsion element of $M$ is $0$.

Let $R$ be a domain and let $S = R \setminus \{ 0\} $ be the multiplicative set of nonzero elements of $R$. Then an $R$-module $M$ is torsion free if and only if $M \to S^{-1}M$ is injective. In other words, if and only if the map $M \to M \otimes _ R K$ is injective where $K = S^{-1}R$ is the fraction field of $R$.

Lemma 15.22.2. Let $R$ be a domain. Let $M$ be an $R$-module. The set of torsion elements of $M$ forms a submodule $M_{tors} \subset M$. The quotient module $M/M_{tors}$ is torsion free.

**Proof.**
Omitted.
$\square$

Lemma 15.22.3. Let $R$ be a domain. Let $M$ be a torsion free $R$-module. For any multiplicative set $S \subset R$ the module $S^{-1}M$ is a torsion free $S^{-1}R$-module.

**Proof.**
Omitted.
$\square$

Lemma 15.22.4. Let $R \to R'$ be a flat homomorphism of domains. If $M$ is a torsion free $R$-module, then $M \otimes _ R R'$ is a torsion free $R'$-module.

**Proof.**
If $M$ is torsion free, then $M \subset M \otimes _ R K$ is injective where $K$ is the fraction field of $R$. Since $R'$ is flat over $R$ we see that $M \otimes _ R R' \to (M \otimes _ R K) \otimes _ R R'$ is injective. Since $M \otimes _ R K$ is isomorphic to a direct sum of copies of $K$, it suffices to see that $K \otimes _ R R'$ is torsion free. This is true because it is a localization of $R'$.
$\square$

Lemma 15.22.5. Let $R$ be a domain. Let $0 \to M \to M' \to M'' \to 0$ be a short exact sequence of $R$-modules. If $M$ and $M''$ are torsion free, then $M'$ is torsion free.

**Proof.**
Omitted.
$\square$

Lemma 15.22.6. Let $R$ be a domain. Let $M$ be an $R$-module. Then $M$ is torsion free if and only if $M_\mathfrak m$ is a torsion free $R_\mathfrak m$-module for all maximal ideals $\mathfrak m$ of $R$.

**Proof.**
Omitted. Hint: Use Lemma 15.22.3 and Algebra, Lemma 10.22.1.
$\square$

Lemma 15.22.7. Let $R$ be a domain. Let $M$ be a finite $R$-module. Then $M$ is torsion free if and only if $M$ is a submodule of a finite free module.

**Proof.**
If $M$ is a submodule of $R^{\oplus n}$, then $M$ is torsion free. For the converse, assume $M$ is torsion free. Let $K$ be the fraction field of $R$. Then $M \otimes _ R K$ is a finite dimensional $K$-vector space. Choose a basis $e_1, \ldots , e_ r$ for this vector space. Let $x_1, \ldots , x_ n$ be generators of $M$. Write $x_ i = \sum (a_{ij}/b_{ij}) e_ j$ for some $a_{ij}, b_{ij} \in R$ with $b_{ij} \not= 0$. Set $b = \prod _{i, j} b_{ij}$. Since $M$ is torsion free the map $M \to M \otimes _ R K$ is injective and the image is contained in $R^{\oplus r} = R e_1/b \oplus \ldots \oplus Re_ r/b$.
$\square$

Lemma 15.22.8. Let $R$ be a Noetherian domain. Let $M$ be a nonzero finite $R$-module. The following are equivalent

$M$ is torsion free,

$M$ is a submodule of a finite free module,

$(0)$ is the only associated prime of $M$,

$(0)$ is in the support of $M$ and $M$ has property $(S_1)$, and

$(0)$ is in the support of $M$ and $M$ has no embedded associated prime.

**Proof.**
We have seen the equivalence of (1) and (2) in Lemma 15.22.7. We have seen the equivalence of (4) and (5) in Algebra, Lemma 10.151.2. The equivalence between (3) and (5) is immediate from the definition. A localization of a torsion free module is torsion free (Lemma 15.22.3), hence it is clear that a $M$ has no associated primes different from $(0)$. Thus (1) implies (5). Conversely, assume (5). If $M$ has torsion, then there exists an embedding $R/I \subset M$ for some nonzero ideal $I$ of $R$. Hence $M$ has an associated prime different from $(0)$ (see Algebra, Lemmas 10.62.3 and 10.62.7). This is an embedded associated prime which contradicts the assumption.
$\square$

Lemma 15.22.9. Let $R$ be a domain. Any flat $R$-module is torsion free.

**Proof.**
If $x \in R$ is nonzero, then $x : R \to R$ is injective, and hence if $M$ is flat over $R$, then $x : M \to M$ is injective. Thus if $M$ is flat over $R$, then $M$ is torsion free.
$\square$

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 $i, j \geq 2$). $\square$

Lemma 15.22.11. Let $A$ be a Dedekind domain (for example a discrete valuation ring or more generally a PID).

An $A$-module is flat if and only if it is torsion free.

A finite torsion free $A$-module is finite locally free.

A finite torsion free $A$-module is finite free if $A$ is a PID.

**Proof.**
Proof of (1). Since a PID is a Dedekind domain (Algebra, Lemma 10.119.15), it suffices to prove this for Dedekind domains. By Lemma 15.22.6 and Algebra, Lemma 10.38.19 it suffices to check the statement over $A_\mathfrak m$ for $\mathfrak m \subset A$ maximal. Since $A_\mathfrak m$ is a discrete valuation ring (Algebra, Lemma 10.119.17) we win by Lemma 15.22.10.

Proof of (2). Follows from Algebra, Lemma 10.77.2 and (1).

Proof of (3). Let $A$ be a PID and let $M$ be a finite torsion free module. By Lemma 15.22.7 we see that $M \subset A^{\oplus n}$ for some $n$. We argue that $M$ is free by induction on $n$. The case $n = 1$ expresses exactly the fact that $A$ is a PID. If $n > 1$ let $M' \subset R^{\oplus n - 1}$ be the image of the projection onto the last $n - 1$ summands of $R^{\oplus n}$. Then we obtain a short exact sequence $0 \to I \to M \to M' \to 0$ where $I$ is the intersection of $M$ with the first summand $R$ of $R^{\oplus n}$. By induction we see that $M$ is an extension of finite free $R$-modules, whence finite free. $\square$

Lemma 15.22.12. Let $R$ be a domain. Let $M$, $N$ be $R$-modules. If $N$ is torsion free, so is $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$.

**Proof.**
Choose a surjection $\bigoplus _{i \in I} R \to M$. Then $\mathop{\mathrm{Hom}}\nolimits _ R(M, N) \subset \prod _{i \in I} N$.
$\square$

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