Is $\mathcal O_L$ an $\mathcal O_K$-lattice in $L$? This is a basic question. Let $L/K$ be a finite extension of algebraic number fields and let $\mathcal O_L$ and $\mathcal O_K$ be their respective rings of integers. Is it true that $$K\otimes_{\mathcal O_K}\mathcal O_L \cong L$$
If so how can we prove it? 
Thanks in advance.
 A: Yes, this is true. First of all the map is injective. The key to proving this is the observation that every element of $K\otimes_{\mathscr{O}_K}\mathscr{O}_L$ is a pure tensor. This follows from a common denominator argument, and ultimately holds because $K\otimes_{\mathscr{O}_K}\mathscr{O}_L$ is a localization of $\mathscr{O}_L$. In fact, an arbitrary element of the tensor product can be written as $1/a\otimes b$ where $a\in\mathscr{O}_K\setminus\{0\}$ and $b\in\mathscr{O}_L$. The image of this element under the map in question is $a^{-1}b\in L$, so if this is zero, $b=0$, and $1/a\otimes b=1/a\otimes 0=0$. 
For surjectivity, note that if $x\in L$, then because $x$ is algebraic over $K$, for some non-zero $a\in \mathscr{O}_K$, $ax$ is integral over $\mathscr{O}_K$, i.e., $ax=b\in\mathscr{O}_L$. Then the element $1/a\otimes b\in K\otimes_{\mathscr{O}_K}\mathscr{O}_L$ maps to $a^{-1}b=a^{-1}(ax)=x$.
All we actually used was that $\mathscr{O}_K$ is a domain with field of fractions $K$ and that $L$ is an algebraic extension of $K$ with $\mathscr{O}_L$ the integral closure of $\mathscr{O}_K$ in $L$.
A: If we know that $\mathbb{Q} \otimes_{\mathbb{Z}} \mathcal{O}_K \cong K$ for any number field, e.g. because the ring of integers is a full rank lattice, or because any algebraic integer divides its norm and
$$ \frac{x}{y} = \frac{x (N(y) / y)}{N(y)} $$
then we can reason
$$ K \otimes_{\mathcal{O}_K} \mathcal{O}_L 
\cong
\mathbb{Q} \otimes_{\mathbb{Z}} \mathcal{O}_K \otimes_{\mathcal{O}_K} \mathcal{O}_L 
\cong \mathbb{Q} \otimes_{\mathbb{Z}} \mathcal{O}_L \cong L
$$
