The Stacks project

Proof. Omitted. $\square$

Proof. Omitted. $\square$

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 $R'$. $\square$

Proof. Omitted. $\square$

Proof. Omitted. Hint: Use Lemma 15.22.3 and Algebra, Lemma 10.23.1. $\square$

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 $R^{\oplus r} = R e_1/b \oplus \ldots \oplus Re_ r/b$. $\square$

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 10.157.2. 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 10.63.3 and 10.63.7). This is an embedded associated prime which contradicts the assumption. $\square$

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. $\square$

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

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

Proof. (For the parenthetical remark in the statement of the lemma, see Algebra, Lemma 10.120.15.) Proof of (1). By Lemma 15.22.6 and Algebra, Lemma 10.39.18 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 10.120.17) we win by Lemma 15.22.10.

Proof of (2). Follows from Algebra, Lemma 10.78.2 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 $M' \subset A^{\oplus n - 1}$ be the image of the projection onto the last $n - 1$ summands of $A^{\oplus n}$. 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 $A$ of $A^{\oplus n}$. By induction we see that $M$ is an extension of finite free $A$-modules, whence finite free. $\square$

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