# Let $D$ be a countably dimensional division algebra over an uncountable algebraically closed field $F$. Why $D=F$?

Problem. Suppose that $F$ is an uncountable algebraically closed field and $D$ is a division $F$-algebra. If $\dim_F D$ is countable, then $D=F$.

For each $x\in D$, $F(x)$ is a field extension of $F$ and therefore $[F(x)\colon F]\leqslant\aleph_0$. Since $F$ is algebraically closed, if $F(x)$ is a proper extension, it must be transcendental. Thus I think if the uncountableness of $F$ implies that the degree of any transcendental simple extension must be uncountable, then the problem is done, and this is where I am stuck. So I would like to ask whether I am in a right way, and what to do for the rest of the proof? Thanks in advance..

• For $x\not \in \overline{F}$ then $\{ \frac{1}{x-a}, a \in F\}$ is $F$-linearly independent. – reuns Nov 24 '17 at 8:18
• @reuns Thanks! I have thought about this before asking, but when I tried to prove their $F$-linear independence I am still stuck. So could you please elaborate a little? – josephz Nov 24 '17 at 8:35
• @josephz See math.stackexchange.com/questions/2245705/… – Phil. Z Nov 24 '17 at 8:47
• @Phil.Z Thank! That does help a lot! – josephz Nov 24 '17 at 8:54

Elaborating on the linear independence of the set described in the comments above, namely for some fixed $x\notin F$, $X = \left\{\frac{1}{x-a}\mid a\in F\right\}$. Note that since $x$ is trancendental over $F$, we may as well assume that $F(x)$ is the field of rational functions with coefficients in $F$ and variable $x$.
Take some linear combination of the elements of $X$ that equals $0$. It is of the form $$0 = \frac{b_1}{x-a_1} + \frac{b_2}{x-a_2}+\cdots\frac{b_n}{x-a_n}$$ where all the $a_i$ are distinct. Multiply this by $\prod_{i = 1}^n(x-a_i)$, and we get \begin{align} 0 = {}&b_1(x-a_2)(x-a_3)\cdots(x-a_n) \\ &{}+ b_2(x-a_1)(x-a_3)\cdots(x-a_n)\\ &\vdots\\ &{}+b_n(x-a_1)(x-a_2)\cdots(x-a_{n-1}) \end{align} Since this is an element in the subring of polynomials of the field of rational functions, we have evaluation homomorphisms $h_a: F[x]\to F$, defined by $h_a(f) = f(a)$. Now note what happens when we apply $h_{a_i}$ to the element above. It becomes $0 = b_i(a_i-a_1)\cdots(a_i-a_n)$ (where the term $(a_i-a_i)$ is skipped). Since all the $a_i$ were distinct, this means that $b_i = 0$, and thus the linear combination above is trivial (all coefficients $b_i$ are $0$), so $X$ is linearly dependent.
• @josephz Note the mistake I corrected regarding the result of applying the evaluation homomorphism. It's a bit longer than just $b_i$. – Arthur Nov 24 '17 at 8:58