= Solution
Choose an <integral basis> $u_1,\ldots,u_d$ of the nonzero <ideal> $\mathfrak a$. Its rational span is $L$, since it contains a nonzero integer multiple of $\mathcal O_L$. The condition $x\mathfrak a\subseteq\mathfrak a$ gives integers $m_{ij}$ with $xu_j=\sum_i m_{ij}u_i$. Thus multiplication by $x$ on $L$ has an integer <matrix> $M$ in this basis.
Its <characteristic polynomial> $P(t)=\det(tI-M)$ is monic with integer coefficients. By the <Cayley-Hamilton theorem>, $P(M)=0$. On the <field> $L$, this is multiplication by $P(x)$, so $P(x)=0$. Hence $x$ is an <algebraic integer>, and
$$
\boxed{x\in\mathcal O_L}.
$$
This is the <determinant trick> for an <ideal> preserved by multiplication.
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