Let $A$ be an integral domain and $K$ its field of fractions. If $M$ is a non-zero fractional ideal of $A$, then $$N=\{x \in K : xM \subseteq A\}$$ is also a fractional ideal of $A$.
The proof I am trying to follow is given in Dummit and Foote.
I agree that it is easy to check that $N$ is an $A$-submodule of $K$.
The next part of the proof goes as follows:
"By definition, there exists some $d \in A\setminus\{0\}$ such that $dM\subseteq A$ and so $M$ contains non-zero elements of $A$."
I really can't see how this follows - any help would be greatly appreciated. I'm sure I'm just being dense but it's driving me crazy!
$M\ne 0$ and thus there is $x\in M$, $x\ne 0$. What about $dx$? (In fact, you don't need $dM\subset A$ in order to find a non-zero element in $M$ which is also in $A$: write $x=a/b$ with $a,b\ne 0$, and notice that $bx=a\in M\cap A$.)
Edit. In order to conclude that $N$ is a fractional ideal note that $(dx)N\subseteq A$, and we are done.