Two disjoint number fields $K$, $L$ such that $(\mathrm{disc}({\cal O}_K), \mathrm{disc}({\cal O}_L))\neq 1$ but ${\cal O}_L{\cal O}_K={\cal O}_{KL}$ I know that if two disjoint number field $K$, $L$ are such that $(\operatorname{disc}(\mathcal{O}_K), \operatorname{disc}(\mathcal{O}_L))= 1$ then $\mathcal{O}_L\mathcal{O}_K=\mathcal{O}_{KL}$. It is true the converse or there are some counterexamples?
EDIT. Is pretty standard notation but with $\mathcal{O}_K$ I denote the integer closure of $\mathbb{Z}$ in $K$, if $K$ is a finite extension of $\mathbb{Q}$. Then $\operatorname{disc}(\mathcal{O}_K)$ is just the discriminant of $\mathcal{O}_K$.
 A: The inclusion $\mathcal{O}_{LK} \subset d^{-1} \mathcal{O}_K \mathcal{O}_L$ in the other answer doesn't seem to say that (when $d \ne 1$) that $\mathcal{O}_{LK} \ne \mathcal{O}_K \mathcal{O}_L$, but indeed this is true; they can never be equal if $p$ ramifies in both fields.
Suppose that $\mathcal{O}$ is the ring of integers of a number field.
Let $\mathfrak{p}$ be a prime, and let $\mathfrak{p}^e$ divide $p$. (This can only happen for $e > 1$ if $\mathfrak{p}$ is ramified.)
Then
$$\mathcal{O}/\mathfrak{p}^e = k[x]/x^e,$$
where $k$ is the residue field of $\mathfrak{p}$. This is a relatively easy property of local fields (More generally, the ring of integers has the form $W(k)[\pi]$ where $W(k)$ is the Witt vetors of $k$ and $\pi$ is a uniformizer.) In particular, the tangent space of this ring has dimension $0$ if $e = 1$ and $1$ if $e > 1$. Now let
$$p = \prod_{i=1}^{r} \mathfrak{p}^{e_i}_i.$$
We consequently have
$$\mathcal{O}/p = \prod \mathcal{O}/\mathfrak{p}^{e_i}_i \simeq \prod k_i[x_i]/x^{e_i}_i$$
is a direct sum of local rings whose tangent space over the residue field has dimension either $0$ or $1$, with $1$ occurring if and only if $p$ is ramified.

We may assume that $[L:\mathbf{Q}][K:\mathbf{Q}] = [KL:\mathbf{Q}]$.
If $\mathcal{O}_L \mathcal{O}_K = \mathcal{O}_{LK}$, then there is an
isomorphism:
$$\mathcal{O}_L \otimes \mathcal{O}_K \rightarrow \mathcal{O}_{KL}$$
under multiplication. Hence there is an isomorphism
$$\mathcal{O}_L/p \otimes \mathcal{O}_K/p \rightarrow \mathcal{O}_{KL}/p.$$
(The above isomorphism means this is surjective, and both rings have the same length so surjective implies injective.) Now suppose that the discriminant of $L$ and $K$ is divisible by $p$. This is equivalent to saying that $p$ is ramified in $L$ and $K$. In particular, there are primes of ramification index $a > 1$ in $L$ and $b > 1$ in $K$. Hence, in the LHS, there is a factor of the form:
$$k[x]/x^a \otimes k[y]/y^b = k[x,y]/(x^a,y^b).$$
This has a tangent space of dimension two, and so cannot occur in the RHS as explained above. Hence it can never happen.
