1
$\begingroup$

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..

$\endgroup$
  • 3
    $\begingroup$ For $x\not \in \overline{F}$ then $\{ \frac{1}{x-a}, a \in F\}$ is $F$-linearly independent. $\endgroup$ – reuns Nov 24 '17 at 8:18
  • $\begingroup$ @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? $\endgroup$ – josephz Nov 24 '17 at 8:35
  • $\begingroup$ @josephz See math.stackexchange.com/questions/2245705/… $\endgroup$ – Phil. Z Nov 24 '17 at 8:47
  • $\begingroup$ @Phil.Z Thank! That does help a lot! $\endgroup$ – josephz Nov 24 '17 at 8:54
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @josephz Note the mistake I corrected regarding the result of applying the evaluation homomorphism. It's a bit longer than just $b_i$. $\endgroup$ – Arthur Nov 24 '17 at 8:58
  • $\begingroup$ Ah, yeah, I noticed this. Thanks! $\endgroup$ – josephz Nov 24 '17 at 9:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.