# $R$ is a domain such that for $x,y \in R$, $(x)\subseteq (y)$ or $(y)\subseteq (x)$ . Is every f.g. torsion free module over $R$ free?

Let $R$ be an integral domain such that $x,y \in R \implies (x)\subseteq (y)$ or $(y)\subseteq (x)$ ; then is it true that every finitely generated torsion free module over $R$ is free ?

I can only see that every finitely generated ideal of $R$ is principal . But I am stuck here . Please help. Thanks in advance

## 2 Answers

Yes, this is true. Suppose $M$ is a finitely generated torsion-free $R$-module. Let $K$ be the field of fractions of $R$. We may choose a basis for $M\otimes_R K$ to identify it with $K^n$ for some $n$, and since the canonical map $M\to M\otimes_R K$ is injective (since $M$ is torsion-free) we may assume $M$ is a submodule of $K^n$. Multiplying a set of generators of $M$ by a common denominator $d$, we may assume they are actually all elements of $R^n$ (since $dM\subseteq K^n$ is isomorphic to $M$). So we may actually assume $M$ is a submodule of $R^n$. Note also that since $M$ generates $K^n$ over $K$, for each $j\leq n$ there exists an element of $M$ whose $j$th coordinate is nonzero.

Now let $x_1,\dots,x_s\in R^n$ be a finite set of generators for $M$; we will write $x_{ij}$ for the $j$th coordinate of $x_i$. If $n=0$ or $s=0$ then $M=0$ and $M$ is trivially free; so let us assume $n>0$ and $s>0$. Without loss of generality, suppose that $(x_{11})$ contains $(x_{i1})$ for all $i$ (by the remark at the end of the previous paragraph, this implies $x_{11}\neq 0$). Then for each $i>1$, let $y_i=x_i-d_ix_1$ where $d_i$ is such that $d_ix_{11}=x_{i1}$ Note that $x_1,y_2,\dots,y_s$ still generate $M$, but now $y_{i1}=0$ for all $i$.

Now I claim the quotient module $M/(x_1)$ is torsion-free. Indeed, suppose $z\in M/(x_1)$ is a torsion element; say $z$ is the image of $\sum a_i y_i\in M$. Since $z$ is torsion, there is some nonzero $c\in R$ such that $cz=0$, so $c\sum a_iy_i$ is a multiple of $x_1$. But the first coordinate of $c\sum a_iy_i$ is $0$ and the first coordinate of any nonzero multiple of $x_1$ is nonzero. So this implies $c\sum a_iy_i=0$, which implies $\sum a_iy_i=0$, so $z=0$.

Thus $M/(x_1)$ is torsion free, and is generated by $s-1$ elements (namely the images of the $y_i$). By induction on $s$, we conclude that $M/(x_1)$ is a free module. The short exact sequence $0\to (x_1)\to M\to M/(x_1)\to 0$ then splits, so $M\cong (x_1)\oplus M/(x_1)$. Since $(x_1)$ is also free, this implies $M$ is free.

Let $M$ be a finitely generated torsion-free $R$-module and let $S=\{m_1,...,m_n\}$ be a generating set of $M$ such that no proper subset of $S$ generates $M$. We claim: $S$ is linearly independent, and thus $S$ is a basis for $M$.

Indeed, if $S$ were not linearly independent, then $\exists r_{i_1},...,r_{i_k} \in R\setminus \{0\}$, such that $r_{i_1}m_{i_1}+...+r_{i_k}m_{i_k}=0$, where $m_{i_1},...,m_{i_k}\in S$. By the hypothesis on $R$, we may assume w.lo.g., $(r_{i_j})\subseteq (r_{i_1}), \forall 1\le j\le k$. Then for every $j$, there is some $u_j \ne 0$ such that $r_{i_j}=u_j r_{i_1}$. Thus $r_{i_1}(m_{i_1}+u_2m_{i_2}+...+u_km_{i_k})=0$, and since $r_{i_1}\ne 0$ and $M$ is torsion-free, we get $-m_{i_1}=u_2m_{i_2}+...+u_km_{i_k}$.

Hence $M=\langle S \rangle=\langle S\setminus \{ m_{i_1} \} \rangle$, contradicting that no proper subset of $S$ generates $M$.

Thus $S$ is linearly independent.