2

I know that an algebraically closed field $K$ cannot have an finite dimensional proper field extension. But can there be a division ring containing $K$ such that $[D:K]<\infty $.?

blabla
  • 1,114

2 Answers2

6

Sure. For instance, the quaternions contain $\mathbb{C}$ as a subring and are 2-dimensional over $\mathbb{C}$.

However, this is impossible if you require $K$ to be central in $D$, since then for any $x\in D$, the subring generated by $x$ and $K$ is commutative. The fact that $K$ is algebraically closed then implies that $x\in K$, so $D=K$.

Eric Wofsey
  • 342,377
  • Quaternions aren't a division ring over complex numbers. See here https://math.stackexchange.com/questions/2090305/quaternion-hc – Hinko Pih Pih May 14 '22 at 16:44
  • @HinkoPihPih: I'm talking about the usual quaternions over $\mathbb{R}$, not the quaternions over $\mathbb{C}$. (The quaternions over $\mathbb{R}$ are not a division algebra over $\mathbb{C}$, since $\mathbb{C}$ is not central in them. They are still a division ring that contains $\mathbb{C}$ as a subring, though.) – Eric Wofsey May 14 '22 at 20:18
  • Oh, I see what you meant. I assumed that $D$ should be an extension of K, because of the notation $[D:K]$. – Hinko Pih Pih May 17 '22 at 07:28
0

Take $t \in D - K$ and look at the collection of elements $\frac{1}{t-a}$ for $a \in K$. The claim is that these must all be linearly independent over $K$. Indeed, a linear combination of these being zero would force $t$ to be algebraic over $K$ which you know can't happen for algebraically closed $K$. In particular, any division algebra $D$ over $K$ has dimension at least $|K|$.

Nate
  • 11,785