A multiplication notation Let $K$ be a finite algebraic number field and $\mathfrak{a}$ an $\mathfrak{O}_K$-submodule of $K$. Then $\mathfrak{a}$ is said to be a fractional ideal if there exists an $0\ne\alpha\in\mathfrak{O}_K$ such that $\alpha\mathfrak{a}\subset\mathfrak{O}_K$. I understand that it should be that
$\alpha\mathfrak{a}=\{\alpha x\mid x\in\mathfrak{a}\}$.
However, to which definition of "multiplication" is this due to? I know there is a multiplication of ideals, but this does not seem to be what is used. Thank you for your help.
 A: It turns out that 
$$ (\alpha)\mathfrak a = \{\alpha x | x \in \mathfrak a\},$$
where $(\alpha)\mathfrak a$ denotes the ideal-theoretic product of the principal ideal $(\alpha)$ with the $\mathfrak O_K$-submodule $\mathfrak a$. (By this, I mean the set $\{ z_1 x_1 + \dots + z_n x_n | n \in \mathbb N : z_1, \dots, z_n \in (\alpha), x_1, \dots x_n \in \mathfrak a \}$, which is another $\mathfrak O_K$-submodule  of $K$.) So the two notions of product are equivalent.
How can we show this? By definition, $(\alpha) = \{ \alpha y | y \in \mathfrak O_K \}$, so $$(\alpha)\mathfrak a = \{ \alpha y_1x_1 + \dots + \alpha y_n x_n | n \in \mathbb N, y_1, \dots y_n \in \mathfrak O_K, x_1, \dots, x_n \in \mathfrak a\}$$
But for any $ y_1, \dots y_n \in \mathfrak O_K$ and for any $x_1, \dots, x_n \in \mathfrak a$, we have $y_1x_1 + \dots +  y_n x_n = x$ for some $x\in \mathfrak a$, since $\mathfrak a$ is an $\mathfrak O_K$-submodule, by assumption. Hence we have the inclusion
$$ (\alpha) \subset \{ \alpha x | x \in \mathfrak a \}.$$
Conversely, for every $x \in \mathfrak a$, we have $\alpha x = \alpha . 1. x$, and $1$ is an element of $\mathfrak O_K$. Hence we have the reverse inclusion
$$ (\alpha) \supset\{ \alpha x | x \in \mathfrak a \},$$
which completes the proof.
