algebraically closed field in a division ring?

Question

Is it possible to have $K \subset D$ where $K$ is algebraically closed field and $D$ is a division ring such that $K \subseteq Z(D)$?

Answer 1

Ben explained the key point (+1) that an algebraically closed field $K$ cannot be the center of such a division algebra $D$ that $\dim_KD<\infty$. I just want to add the argument that does not refer to Brauer groups (this is how I would prove that the Brauer group is trivial, but in case you haven't seen that argument here it is).

Assume that such $K$ and $D$, $K\subset Z(D)$, $1<\dim_KD<\infty$ existed. Let's pick an element $a\in D\setminus K$. Consider the ring $$K[a]=\bigoplus_{n=0}^\infty Ka^n\subseteq D.$$ Because $D$ is f.d. over $K$, so is $K[a]$. Because $K$ commutes with all of $D$, and the powers of $a$ clearly commute with each other, the ring $K[a]$ is commutative. Because all the elements of $D$ have inverses, $K[a]$ has no zero-divisors, so it is an integral domain. Undoubtedly you have seen the result stating that an integral domain that is f.d. over a field is itself a field. So $K[a]$ is a finite dimensional proper extension field of $K$, but algebraically closed fields don't have such extensions. A contradiction.

Answer 2

You may take the skew Laurent series $\overline{\mathbb{K(t)}}((x;\sigma))$.

Here $\sigma\in\mathcal{Gal}(\overline{\mathbb{K(t)}}/\overline{\mathbb{K}})$

It is the set of formal series $\sum_{k=-\infty}^{+\infty}a_kx^k$ (such that $\#\{k<0\ s.t. a_k\neq0\}$ is finite), endowed with a product

$$\left(\sum_{k=-\infty}^{+\infty}a_kx^k\right)\left(\sum_{h=-\infty}^{+\infty}b_hx^h\right):=\sum_{s=-\infty}^{+\infty}\left[\left(\sum_{h+k=s}a_k\sigma^k(b_h)\right)x^s\right]$$

One can show that this is indeed a division ring. But $\sigma|_\overline{\mathbb{K}}=id\longrightarrow\overline{\mathbb{K}}\subseteq\overline{\mathbb{K(t)}}((x;\sigma))$

Disclaimer

I found the skew Laurent series over a division ring on Lam, "A first course in noncommutative rings", p. 11

Answer 3

The other answer gives an example. However, note that in the text you showed us, $D$ is finite dimensional over $K$. This changes everything, since then $\mathop{Z}(D)/K$ is a finite field extension; thus, if $K$ is algebraically closed, actually $\mathop{Z}(D) = K$. So $D$ is a finite $K$-division algebra. But over an algebraically closed field, no such thing (besides $K$ itself) exists, because the Brauer group of any algebraically closed field is trivial.