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?
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 $$ \mathcal O_L = \mathcal O_K + \mathcal O_K\alpha = \mathcal O_K[\alpha]. $$