# Linearly independent set of a free module on a field of fractions of an integral domain

Can anyone check my proof and let me know if I made some mistakes, please?

Let $R$ be an integral domain, let $K$ be its field of fraction Let $M=R^{\oplus A}$ be a free module on $R$ with basis $A$. View $M$ as a subset of $K^{\oplus A}$. Let $S$ be a subset of M.

Q : Show that if $S$ is linearly independent on $R$, then so it is on $K$.

I tried the following:[contrapositive]

Assume $S$ is NOT linearly independent on $K$, then there exist $s_1,\ldots,s_n$ in $S$, and $a_1,\ldots, a_n$ in $K$ not all of the $a_i$'s are zeros such that $a_1s_1+\cdots+a_ns_n=0$. Write every $a_i$ as a fraction i.e. $\displaystyle a_i= \frac{u_i }{v_i}$ where $u_i$ and $v_i$ are in $R$ and $v_i\neq 0$ .

Let $d= v_1v_2\ldots v_n$, then $d\neq 0$ because $R$ is integral domain. Multiply by $d$ $$d (a_1 s_1+\cdots +a_n s_n)=0$$ and then get $$c_1s_1+\cdots+c_ns_n=0$$ where $\displaystyle c_i=d \frac{u_i}{v_i}\in R$ and not all of the $c_i$'s are zeros. This will imply $S$ is linearly independent on $R$. Hence the statement is true.

Thank you.

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Yes, your argument is correct. In fact it works more generally to show that if $S$ is an $R$-linearly independent subset of any $R$-module $M$, then $S \otimes 1$ is a $K$-linearly independent subset of the $K$-module $M \otimes_R K$. That is, you do not need $M$ to be a free $R$-module.