Are fractional ideals usual ideals? I'm going through some notes, and have the following definition:

Let $K$ be a number field. Then $ \mathfrak{a} \subset K$ is a fractional ideal if there exists a non-zero $c \in K$ such that $c\mathfrak{a} \subset \mathcal O_K$ is an ideal.

I'm concerned that this is unclearly stated; specifically, shouldn't it specify that $\mathfrak{a}$ is an ideal of $K$? If $\mathfrak{a}$ is just any subset of $K$, then I can't prove the lemma that gives the correspondence between fractional ideals and finitely generated $\mathcal O_K$ modules.
Thanks
 A: Well, no. $K$ is a field, so it has no non-trivial ideals. But $\mathfrak{a}$ is an $\mathcal{O}_K$-submodule of $K$ (this follows from the fact that $c \mathfrak{a}$ is an ideal).
A: Just as a complement to the other answers: the field $K$ is a ring, and hence a module over itself.  Thus it is also an $\mathcal O_K$-module in a natural way (just using the inclusion of $\mathcal O_K$ in $K$).  A fractional ideal is a certain kind of $\mathcal O_K$-submodule of $K$, namely one that is finitely generated over $\mathcal O_K$.  (Note that $K$ itself doesn't have this property).  
Now, one can prove that an $\mathcal O_K$ submodule $\mathfrak a$ of $K$ is finitely generated if and only if $c\mathfrak a \subset \mathcal O_K$ for 
some non-zero $c \in K$.  
Also, one can prove that a subset $\mathfrak a$ of $K$ is an $\mathcal O_K$-submodule and only if $c \mathfrak a$ is an $\mathcal O_K$-submodule of $K$ for some (equivalently,
any) non-zero $c \in K$.  
Finally, the $\mathcal O_K$-submodules of $\mathcal O_K$ itself are precisely the ideals in $\mathcal O_K$.
Putting these last three conditions  together, one obtains the alternative characterization of fractional ideals given in your question.
A: $K$ is a field.  One of the basic facts about fields is that they only have two ideals: $(0)$ and the whole ring $K$.  (Exericse: if you haven't seen this before, prove it!).
So $\mathfrak{a}$ is not, in general, an ideal of $K$.  It is not closed under multiplication by elements of $K$.  It is also not an ideal of $\mathcal{O}_K$ because it may contain elements of $K$ that are not in $\mathcal{O}_K$.  However, it is closed under addition and multiplication by elements of $\mathcal{O}_K$, so it is an $\mathcal{O}_K$-module.
A: Let $R$ be a Dedekind domain and $K$ its field of fractions. Let $M$ be an $R$ - submodule of $K$. Then the following conditions are equivalent


*

*$M$ is a finitely generated $R$ - submodule of $K$.

*There exists $z \in R$ such that $zM$ is an integral ideal of $R$.

*$M = \alpha I$ for some  $\alpha \in K$ and $I$ an integral ideal of $R$.


$1 \implies 2 : $ 
Suppose $M$ is generated by $x_1,\ldots,x_n \in K$. Then if we let $z$ to be equal to the product of the denominators of the $x_i$ we have $zx_i = y_i$ for some $y_i \in R$. It now follows that $$zM =  I$$
where $I = (y_1,\ldots,y_n)$. 
$2 \implies 3:$ If there is $z \in R$ so that $zM$ is an integral ideal $I$ of $R$ then considering $z$ now as an element of $K$ we have $M = z^{-1}I$.
$3 \implies 1:$ This comes from the more general fact that because $R$ is Noetherian, $M$ must be finitely generated.
