Is there an algebraic closed field which contains the complex field $\mathbb{C}$ strictly?

Question

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?

The quaternions are not a field (at least usually you require a field to be commutative). — Julian Kuelshammer, Aug 19 '19 at 08:15
Look at https://en.m.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras). This strictly limits finite dimension extensions. Are you considering infinite dimension extensions? — badjohn, Aug 19 '19 at 08:17
Okay, so we can drop the quaternions, but is there another field then? — Belgium_Physics, Aug 19 '19 at 08:18
Any field has an algebraic closure. Consider the algebraic closure of $\mathbb C(x)$. — Wojowu, Aug 19 '19 at 08:20

score 3 · Accepted Answer · answered Aug 19 '19 at 08:26

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.

Is there an algebraic closed field which contains the complex field $\mathbb{C}$ strictly?

1 Answers1