$K$ a field, $A\subset K$ a subring, $M\subset K$ an $A$-submodule. Define
$$(A:\_{K}M):= \lbrace s\in K|sM\subset A\rbrace$$
Then it is easy to see that 


$$M\subset A\Longleftrightarrow A\subset (A:\_{K}M),$$


$$A\subset M\Longrightarrow (A:\_{K}M)\subset A$$


But I couldn't show the reverse implication of the second. It is true if $M$ is invertible, and I'm guessing that it is true only if $M$ is invertible. Any ideas?



Related to fractional ideals
This is not true, you are guessing right. Here is a counter-example: take $k$ a field, $A=k[X,Y]$, $K=k(X,Y)$.
Let $M=XA+YA$. Then using the fact that $A$ is a UFD one see easily that $(A:\_K M)=A$.
(indeed, let $P/Q \in (A:\_K M)$ with $P,Q \in A$, $Q \neq 0$. We have $XP/Q \in A$ so
$Q$ divides $XP$. Similarly $Q$ divides $YP$. Since $X$ and $Y$ are two non-equivalent irreducible elements of $A$, this implies that $Q|P$, hence $P/Q \in A$).


Yet $A \not \subset M$.


