If $R$ is a domain which is not a field and $M$ is an $R$-module for which $\bigcap_{r\in R\setminus \{0\}} rM \neq 0$, show $M$ is not projective. I can't seem to crack this one. I've tried assuming $x \in T := \bigcap_{r \in R\setminus\{0\}} rM$ so that for all nonzero nonunits $r \in R$ there are $x_{n} \in M$ so that $x = r^{n}x_{n}$ for every $n$. This didn't get me anywhere. I've also tried assuming $M$ is projective so that it's a direct summand of a free module $F$ and $0 \neq T \subset \bigcap_{r\in R\setminus\{0\}} rF$. I tried then to get a contradiction with from using basis of $F$ but with no luck. 
Any help would be appreciated. 
 A: I'll show the contrapositive: if $R$ is a domain which is not a field, and if $M$ is a projective $R$-module then $\displaystyle \bigcap_{r \in R \setminus \{0 \}} rM =0$. 
Observe first that it suffices to prove the claim for free modules. Indeed, if $M$ is projective then there is an embedding $i: M \to F$ where $F$ is a free module. Then
\begin{equation*}
i \Big( \displaystyle \bigcap_{r \in R \setminus \{0 \}} r  M \Big) \subset \displaystyle \bigcap_{r \in R \setminus \{0 \}} r  F.
\end{equation*}
So if the result is true for $F$, then it is also true for $M$.
Now let $F$ be a free module. Let $\{ e_{\lambda} : \lambda \in \Lambda \}$ be a basis of $F$. Let $x \in \displaystyle \bigcap_{r \in R \setminus \{0 \}} rF.$ We want to show that $x =0$. We can write $x$ in the chosen basis of $F$ as a sum (of finite support)
\begin{equation*}
x = \sum_{\lambda \in \Lambda} a_{\lambda} \, e_{\lambda}
\end{equation*}
for some coefficients $a_{\lambda} \in R$. 
We will show that $a_{\lambda}=0$ for all $\lambda$. Indeed, suppose on the contrary that there was a $\lambda_0$ such that $a_{\lambda_0} \neq 0$. Then also $(a_{\lambda_0})^2 \neq 0$ since $R$ is a domain. Then since by assumption $ x \in \bigcap_{r \in R \setminus \{0 \}} rF$ there is some $y \in F$ such that $x = (a_{\lambda_0})^2 \, y$. This implies that
\begin{equation*}
a_{\lambda_0} = (a_{\lambda_0})^2 \, b
\end{equation*}
where $b \in R$ is the coefficient of $e_{\lambda_0}$ in $y$. Since $a_{\lambda_0} \neq 0$ and $R$ is a domain, this implies that $1 = a_{\lambda_0} \, b$, hence $a_{\lambda_0}$ is a unit.
Now take any $r \in R \setminus \{0 \}$. Then we have $ x = r z$ for some $z \in F$. Comparing the coefficients of $e_{\lambda_0}$ as before, we must have
\begin{equation*}
a_{\lambda_0} = r \, c
\end{equation*}
for some $c \in R$. Since $a_{\lambda_0}$ is a unit, this implies that $r$ is also a unit. Since $r$ was arbitrary, we showed that $R$ is a field, which is a contradiction!
Hence we must have $a_{\lambda} =0$ for all $\lambda$, which means that $x=0$. Hence we conclude that $\displaystyle \bigcap_{r \in R \setminus \{0 \}} rF =0$.
