I'm studying a Galois Theory module in my undergraduate maths course, and the following theorem has come up:
Let $\mathbb{K} \subseteq \mathbb{F}$ be a field extension where
(a) $\mathbb{K}$ is finite
or
(b) $\mathbb{K}$ is infinite and $\mathbb{K} \subseteq \mathbb{F}$ is separable.
Then $\mathbb{F} = \mathbb{K}(\theta)$ for some $\theta \in \mathbb{F}$
This seems to make sense to me, but then I thought of the field of algebraic integers over $\mathbb{Q}$.
If the theorem is true, it suggests that we should have some element in the field of algebraic integers $\mathbb{O}_k$ whose minimal polynomial over $\mathbb{Q}$ has a repeated root.
If this were not the case, the field extension would be separable, and $\mathbb{O}_k$ would, according to the theorem, be a simple extension of the field $\mathbb{Q}$, which is obviously isn't, because $\mathbb{O}_k$ has infinite dimension as a vector space over $\mathbb{Q}$.
Can anyone provide me with an example of such an element of $\mathbb{O}_k$?
I believe you make a confusion, the result that you seem to quote is the primitive element theorem which says that if an extension is finite and separable, it is generated by an element.
As you point out neither infinite algebraic infinite extensions are generated by an element nor every finite algebraic extension. For the last part of this assertion, see the counterexample section in the link.
https://en.wikipedia.org/wiki/Primitive_element_theorem