4
$\begingroup$

I'm having trouble understanding transcendental field extensions which are not simple. Specifically, I'm having a hard time figuring out how to write down a "general element" of such of an extension.

Simple extensions are nice. For example, in the simple transcendental field extension $F(x)$ a general element is a quotient of polynomials with coefficients in $F$.

But how can I understand, say, $\Bbb Q(x, \sqrt{x}, \sqrt[3]{x}, \cdots )$? What is the general element in this case? By "plugging in" $x=1$ I can intuitively see why $\sqrt{x+1}$ is not an element of this field. (The extension becomes $\Bbb Q(1, 1, \cdots,) = \Bbb Q$, but $\sqrt{2} \notin \Bbb Q$.) But how would I prove this fact for more complicated elements, say, $\sqrt{x^2 - x}$? Do I have to stick to ad hoc explanations like this, or is there a way to more generally characterize elements of extensions like this?

$\endgroup$
  • 3
    $\begingroup$ Each individual element of $\Bbb{Q}(S)$, $S$ some set, only involves finitely many elements of $S$. So proving that $\sqrt{x+1}\notin\Bbb{Q}(x,x^{1/2},\ldots)$ is equivalent to proving that $\sqrt{x+1}\notin\Bbb{Q}(x,x^{1/2},\ldots,x^{1/n})$ for no choice of $n$. It is easy to convince yourself of the fact that $\Bbb{Q}(x,x^{1/2},\ldots,x^{1/n})=\Bbb{Q}(x^{1/N})$ with $N=\operatorname{lcm}(1,2,\ldots,n)$. But, this last field is $\Bbb{Q}(t), t=x^{1/N}$, so you can turn the question into one involving polynomials in $t$. $\endgroup$ – Jyrki Lahtonen May 28 '18 at 4:03
  • $\begingroup$ (cont'd) Unique factorization of polynomials shows that $u^2=1+t^N=1+x$ has no solutions $u\in\Bbb{Q}(t)$. Therefore $\sqrt{1+x}\notin \Bbb{Q}(x,x^{1/2},x^{1/3},\ldots)$ $\endgroup$ – Jyrki Lahtonen May 28 '18 at 4:04
  • $\begingroup$ But, I don't have anything very useful to say about the general question, so leaving these as comments for now. $\endgroup$ – Jyrki Lahtonen May 28 '18 at 4:04
  • $\begingroup$ @JyrkiLahtonen Your comments were already very helpful, so thank you! $\endgroup$ – aras May 28 '18 at 4:06
2
$\begingroup$

Even when your extension is of transcendence degree one over $\Bbb Q$ but finitely generated and not simple, you’re wandering into very deep waters, since this is the topic of Algebraic Curves over $\Bbb Q$. Such a field may be studied purely algebraically, and indeed that was the approach I first saw, back in an earlier geological era. But in fact, there is geometry going on, unavoidably, and one almost always uses geometric methods, language, and intuition here. Welcome to the field!

For non-finite generation, all bets are off, and things can get overwhelmingly confusing. Instead of your infinitely-generated field, how about $\Bbb Q(t,\{\sqrt{t-n}\}_{n\in\Bbb Z})$? It’s something analogous to what you get by adjoining $\sqrt p$ to $\Bbb Q$ for all primes $p$.

Typically, people stick to finitely-generated extensions; there’s more than enough there that’s still unknown to occupy you for a lifetime.

$\endgroup$

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.