Let $R$ be a Dedekind ring, $K$ its quotient field. If $J$ is a fractional $R$-ideal in $K$ then I want to show that $KJ=K$, so that it's a full $R$-lattice in $K$.
Since $J$ is non-zero, we can choose $c\in J-\{0\}$. Then $1=c/c\in KJ$, so $K\subset KJ\subset K$.
Is this it? Sorry for the stupid question but I'm asking because I haven't used any of the properties of a fractional ideal, except that $J$ is a non-zero subset of $K$.
If one considers $K$ as an $R$-module, and $J\subset R$ a non-zero ideal, then $JK=K$ for the same reason as you noticed.