So we know that $\mathbb{C}$ itself is algebraically closed. But I was thinking if there maybe is an algebraically closed field which contains $\mathbb{C}$ strictly, so that it is not equal to $\mathbb{C}$.
I considered the quaternions, as discussed in Is the set of quaternions $\mathbb{H}$ algebraically closed? but couldn't really figure out wether it was algebraically closed or not. So do the quaternions fulfill my statement or is there another field? Or can someone prove the statement is false?
As was mentioned in the comments, the algebraical closure of $\mathbb{C}(x)$ provides an example.
If you don't mind using the axiom of choice, then there is a result that there is precisely one algebraically closed field of characteristic zero of each uncountable cardinality, see e.g. https://kconrad.math.uconn.edu/blurbs/zorn2.pdf , Theorem 6.