Torsion (Free) Elements and Fractional Modules

In this section, we study the relationship between torsion-free finitely generated $R$-modules $M$ and their embeddings $FM$ into $F=\mathrm{Frac}(R)$.

This leads to the concept of fractional modules.

Torsion Elements and Torsion Free Elements

Definition

Let $R$ be a ring, and let $M$ be a $R$-module. $m\in M$ is said to be

• a torsion element if ${\mathrm{Ann}}_R(m)\ne 0$
• a torsion free element if ${\mathrm{Ann}}_R(m)=0$.

Example.

• $\mathbb{Z}$-module $\mathbb{Q}$ has no torsion element but $0$.
• $R$ is commutative,
  • ${}_{R}R$ has no torsion element but $0$ iff $R$ is an integral domain;
  • if $I$ is a proper ideal, all elements of $R$-module $R/I$ are torsion;
  • $R/I$ has no torsion element but $0$, iff $I$ is prime.

Lemma

If $R$ is an integral domain, the subset of $M$

$$\mathrm{tor}(M)=\{m\in M:{\mathrm{Ann}}_R(m)\ne 0\}$$

is a submodule of $M$, which is called the torsion submodule of $M$.

Definition

We say $M$ is a torsion module if $M=\mathrm{tor}(M)$, and $M$ is torsion free if $\mathrm{tor}(M)=0$.

Examples.

• $\mathbb{Z}$-modules: $\mathbb{Z},\mathbb{Q},\mathbb{Z}/2\mathbb{Z}$, $\mathrm{tor}(\mathbb{Q})=0$ and $\mathrm{tor}(\mathbb{Z}/2\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}$
• For an integral domain $R$ and a $R$-module $M$, there is $\mathrm{tor}(M^{\ast })=0$, where $M^{\ast }={\mathrm{Hom}}_R(M,R)$.

Proposition

Let $R$ be an integral domain, then its free modules are torsion free.

Proposition

Let $R$ be an integral domain. Any finitely generated torsion free $R$-module is isomorphic to a submodule of a free $R$-module of finite rank.

\begin{proof} Suppose $M$ is generated by non-zero elements $x_1,\cdots ,x_n$. Up to reordering, assume $x_1,\cdots ,x_r$ is a maximal linearly independent subset of $x_1,\cdots ,x_n$. If $r=n$, then $M$ is free. Now assume that $r<n$. Consider $M_0=R{x}_1+\cdots +R{x}_r=R{x}_1\oplus \cdots \oplus R{x}_r$. For each $j⩾r+1$, there exists nonzero ${\lambda }_j\in R$ such that ${\lambda }_j{x}_j\in M_0$. Let $\lambda ={\lambda }_{r+1}\cdots {\lambda }_m$. Then $\lambda M\subseteq M_0$. Consider $M\to M_0,m\mapsto \lambda m$, which is injective. Therefore, $M$ is isomorphic to a free $R$-module of finite rank. \end{proof}

Torsion-Free Modules and Fractional Ideals

Let $R$ be an integral domain, and let $F=\mathrm{Frac}(R)$ be the fraction field of $R$. The following proposition shows that any torsion-free $R$-module $M$ can be embedded into a vector space over $F$. This new setting, in which division is always possible, motivates the generalization of an ideal to a fractional ideal.

Proposition

Define $FM=F{\otimes }_{R}M$, then $FM\simeq S^{-1}M$ with $S=R\setminus \{0\}$. The kernel of $M\to FM,m\mapsto 1\otimes m$ is $\mathrm{tor}(M)$.

\begin{proof} If $1\otimes m=0$, then $m=0\in \mathrm{tor}(M)$. For any $m\in \mathrm{tor}(M)$, there exists nonzero $r\in R$ such that $rm=0$ and then $1\otimes m=r^{-1}r\otimes m=r^{-1}\otimes 0=0$. Therefore, $\ker (M\to FM)=\mathrm{tor}(M)$. \end{proof}

Corollary

• $F{\otimes }_R\mathrm{tor}(M)=0$
• TFAE:
  • the natural $R$-morphism $M\to FM,m\mapsto 1\otimes m$ is injective
  • $M$ is torsion free.

Corollary

If $M$ is finitely generated projective $R$-module, then $M\to FM,m\mapsto 1\otimes m$ is injective.

\begin{proof} Since $M$ is a finitely generated projective $R$-module by ^ajkzkw, $M$ is a direct summand of $R^n$ for some $n$. Then $M$ is torsion free by ^m4tylw and so $M\to FM$ is injective. \end{proof}

Definition

The rank of a finitely generated torsion free $R$-module is the dimension of the $F$-space $FM$.

Remark. A finitely generated torsion free $R$-module $M$ of rank $1$ is isomorphic to a finitely generated $R$-submodule of $F$.

• Indeed, if $FM$ is a $1$-dimensional vector space over $F$, then $FM=Fe$. Define $I=\{\lambda \in F:\lambda e\in M\}$, then $I$ is a $R$-submodule of $F$, and we have $M=Ie$.
• In other words, a finitely generated torsion-free $R$-module of rank $1$ can be viewed as a fractional ideal of $R$, that is, as a finitely generated $R$-submodule of its field of fractions $F$.

Let $M$ be a finitely generated torsion free $R$-module. By ^952afe, $M\to FM,m\mapsto 1\otimes m$ injective. So $M$ can be viewed as an $R$-submodule of $FM$. Given an $F$-basis $(e_i{)}_{1⩽i⩽r}$ of $FM$, assume $M$ is generated by $m_1,\cdots ,m_n$ as $R$-module. Then each $m_i$ can be written as

$$m_i=\frac{s_1}{t_1}{e}_1+\cdots +\frac{s_r}{t_r}{e}_r,$$

where $s_i\in R$ and $t_i\in R\setminus \{0\}$. There exists $\lambda \in R\setminus \{0\}$ such that $\lambda M\subseteq R{e}_1\oplus \cdots \oplus R{e}_r$. Additionally, $M$ generated $FM$. This motivates the following definition:

Definition

Let $V$ be an $F$-vector space of dimension $r$. A fractional $R$-module in $V$ is an $R$-submodule $M\subseteq V$ such that

• $M$ generates $V$;
• given any basis $(e_i{)}_{1⩽i⩽r}$ of $V$, there exists $\lambda \in R$ with $\lambda \ne 0$ such that $\lambda M\subseteq {\oplus }_{i=1}^{r}R{e}_i$.

In the case $V=F$, fractional modules are called fractional ideals of $R$. It is an $R$-submodule $I$ of $F$ such that there exists nonzero $\lambda \in R$ such that $\lambda I\subseteq R$. Hence $\lambda I$ is an ideal of $R$. Note that

• every finitely generated $R$-submodule of $F$ is a fractional ideal (if it is not finitely generated, there may not exist $\lambda \in R$ such that $\lambda I\subseteq R$).
• if $R$ is Noetherian, then these submodules of $F$ are all fractional ideals. That is, when $R$ is Noetherian, finitely generated $R$-module $⟺$ fractional ideal.

the proof of the second statement

Let $R$ be a Noetherian integral domain and $F=\mathrm{Frac}(R)$. Let $I$ be an $R$-submodule of $F$. We want to show that $I$ is a fractional ideal if and only if $I$ is finitely generated.

($⟹$) Assume $I$ is finitely generated. Let $I={\sum }_{k=1}^{n}R{x}_k$ for some $x_k\in I$. Since $x_k\in F$, we can write $x_k=p_k/q_k$ where $p_k,q_k\in R$ and $q_k\ne 0$. Let $\lambda =q_1{q}_2\dots q_n$. Then $\lambda \in R$ and $\lambda \ne 0$. For each $k$, $\lambda x_k=(q_1\dots q_{k-1}{q}_{k+1}\dots q_n)p_k\in R$. Then $\lambda I={\sum }_{k=1}^{n}R(\lambda x_k)\subseteq R$. By definition, $I$ is a fractional ideal.

($⟸$) Assume $I$ is a fractional ideal. By definition, $I$ is an $R$-submodule of $F$ and there exists $\lambda \in R\setminus \{0\}$ such that $\lambda I\subseteq R$. Let $J=\lambda I$. Note that $J$ is an $R$-submodule of $R$, which means $J$ is an ideal of $R$. Since $R$ is a Noetherian ring, the ideal $J$ must be finitely generated. Let $J={\sum }_{k=1}^{n}R{y}_k$ for some $y_k\in J$. Since $y_k\in J=\lambda I$, we can write $y_k=\lambda x_k$ for some $x_k\in I$. We claim $I={\sum }_{k=1}^{n}R{x}_k$.

Take any $x\in I$. Then $\lambda x\in J$. So, $\lambda x={\sum }_{i=1}^n{r}_i{y}_i$ for some $r_i\in R$. Substituting $y_i=\lambda x_i$, we get $\lambda x={\sum }_{i=1}^n{r}_i(\lambda x_i)=\lambda ({\sum }_{i=1}^n{r}_i{x}_i)$. Since $\lambda \ne 0$ and $F$ is a field, we can multiply by $1/\lambda$ to get $x={\sum }_{i=1}^n{r}_i{x}_i$. This shows $I\subseteq {\sum }_{k=1}^{n}R{x}_k$. The inclusion ${\sum }_{k=1}^{n}R{x}_k\subseteq I$ is clear since each $x_k\in I$ and $I$ is an $R$-module. Thus, $I={\sum }_{k=1}^{n}R{x}_k$, which means $I$ is finitely generated.