Can the skew field of quaternions be seen as an extension over $\Bbb C$

Question

$\Bbb C$ is algebraically closed, does that mean there is no extension over it? Why can't we extend $\Bbb C$ like $\Bbb C(j,k)=\mathcal Q$? Or can we?

Answer 1

We can't extend $\Bbb C$ algebraically. All poynomials with coefficients in $\Bbb C$ have roots in $\Bbb C$, so any attempt at an algebraic extension would automatically become trivial. However, there are still transcendental field extensions.

The smallest one is $\Bbb C(t)$, the field of rational functions in one variable with complex coefficients. Of course, that field isn't algebraically closed, so you can do algebraic extensions the way you're used to, like $\Bbb C(t)(\sqrt{t})$. Or we can add another transcendental variable to get $\Bbb C(s, t)$.

As mentioned in the other (now deleted answer), the quaternions are, in a sense, an extension of the complex numbers. But we do not usually use that word because it's typically reserved to extensions which are fields.

Answer 2

From the fact that $\mathbb C$ is algebraically closed you can deduce that it has no field extension of finite degree. Of course, it has field extensions of infinite degree.

Since the quaternions aren't a field, they don't count here.

Answer 3

Given any extension of rings $R\subseteq S$ and element $\alpha\in S$ you can interpret $R[\alpha]$ to be the smallest subring of $S$ containing both $R$ and $\alpha$, and if $R$ and $S$ were division rings, I suppose you could interpret $R(\alpha)$ to be the smallest division ring of $S$ containing $R$ and $\alpha$.

The problem is that such extensions aren't constructible as quotients of polynomial rings like you're used to for field extensions.

How could they be? If we identify $\mathbb C=\langle 1,i\rangle\subseteq H$, then $\mathbb C[j]=\mathbb C(j)=\mathbb H$ if we interpret things as I mentioned above, but obviously $\mathbb H$ is not going to be isomorphic to a quotient of $\mathbb C[x]$ since a) it such a quotient is commutative, and b) there are no irreducibles of degree two or more to mod out by.