Non-Monogenic Quadratic Extensions Fix a quadratic extension of global fields $L/K$; i.e., $[L:K]=2$. Must there exist some $\alpha\in\mathcal O_L$ such that $\mathcal O_L=\mathcal O_K[\alpha]$?
Of course, the answer is positive for $K=\mathbb Q$, but I suspect the answer is negative in general. If the statement is false, is there some reasonably large (e.g., infinite) family of global fields $K$ for which the statement is true?
 A: An infinite set of examples can be found at https://kconrad.math.uconn.edu/blurbs/gradnumthy/notfree.pdf.
In all of these examples, $\mathcal O_K$ is not a PID.
When $\mathcal O_K$ is a PID, $\mathcal O_L$ is monogenic as an $\mathcal O_K$-module because $[L:K] = 2$: let’s show there is an $\mathcal O_K$-basis containing $1$, and writing that basis as $\{1,\alpha\}$ makes $\mathcal O_L = \mathcal O_K[\alpha]$. (Such reasoning breaks down for extensions of degree greater than $2$, and there are many examples of non-monogenic extensions of degree $3$ when $\mathcal O_K$ is a PID, in fact in the simplest case when $\mathcal O_K = \mathbf Z$.)
There is some $\mathcal O_K$-basis $\{\alpha_1,\alpha_2\}$ of $\mathcal O_L$ since $\mathcal O_K$ is a PID and $L/K$ is quadratic. Relative to this basis, we can write
$$
1 = a_1\alpha_1 + a_2\alpha_2
$$
where $a_1$ and $a_2$ are in $\mathcal O_K$. That equation implies a common factor of $a_1$ and $a_2$ in $\mathcal O_K$ is a unit in $\mathcal O_L$, and thus in $\mathcal O_K$, so $a_1$ and $a_2$ are relatively prime in $\mathcal O_K$. Thus
$$
a_1b_1+a_2b_2= 1
$$
for some $b_1$ and $b_2$ in $\mathcal O_K$.
Set
$$
\begin{pmatrix}
a_1&a_2\\-b_2&b_1
\end{pmatrix}
\begin{pmatrix}
\alpha_1\\ \alpha_2
\end{pmatrix}=
\begin{pmatrix}
1\\ \alpha
\end{pmatrix}.
$$
That $2\times 2$ matrix with entries in $\mathcal O_K$ has determinant $1$, so $1$ and $\alpha$ have the same $\mathcal O_K$-span as $\alpha_1$ and $\alpha_2$. Hence
$$
\mathcal O_L = \mathcal O_K + \mathcal O_K\alpha = \mathcal O_K[\alpha].
$$
A: Try $$O_K=\Bbb{Z}[\sqrt{-5}],\qquad L= K(\sqrt2)$$
$$O_L = O_K \oplus \frac{\sqrt2}2 I, \qquad I=2O_K+(1+\sqrt{-5})O_K$$
$O_L/O_K\cong I \not\cong O_K$ as $O_K$-module, so $O_L$ can't be monogenic.
