2
$\begingroup$

This is one of the theorems in Field Arithmetic written by M. Fried and M. Jarden as following which is Proposition 13.4.6 in that monograph:

Every field K has a regular extension F which is PAC and Hilbertian

The first part of the proofs goes as follows:

Wellorder the (absolutely irreducible) varieties of dimension at least 1 that are defined over K in a transfinite sequence $\{ V_\alpha | \alpha < m \}$ for some ordinal m. Use transfinite induction to define, for each $\beta < m$, a function field $F_\beta$ for $V_\beta$ which is algebraically independent from $\prod_{\alpha <\beta}F_\alpha$ (the composite of the $F_\alpha$'s with $\alpha < \beta$) over K. Since $F_\beta$ is a regular extension of K, the field $\prod_{\alpha \leq \beta}F_\alpha$ is a proper finitely generated regular extension of $\prod_{\alpha <\beta}F_\alpha$ (Corollary 2.6.8(a)). It follows that $K_1=\prod_{\alpha <m}F_\alpha$ is Hilbertian (Lemma 13.4.5) and regular over K (Lemma 2.6.5(d)).Moreover, every variety defined over $K$ has a $K_1$-rational point.

Note that varieties considered here are all absolutely irreducible. But I have some problems about the proof above:

(1) Is the set $\{ V_\alpha | \alpha < m \}$ contains all varieties? If not, is it possible that I take some particular variety Y of dimension at least 1 as our $V_1$?($V_1$ means the first variety in$\{ V_\alpha | \alpha < m \}$ )

(2) Why every variety defined over $K$ has a $K_1$-rational point?

Any hints or suggestions are welcomed!

Improve this question
$\endgroup$
2
  • $\begingroup$ Effectively you can choose $V_1$ to be a particular (absolutely / geometrically) irreducible variety $W$. This is equivalent to running through the proof with the field $K$ replaced by the regular extension field $K(W)$. For your second question, I completely agree that this is not obvious. The construction that I know for $K_1$ uses Zorn's lemma for the partially ordered set of chains of regular field extensions, but they need not be extensions of the form $\prod_{\alpha<\beta} F_\alpha \to \prod_{\alpha \leq \beta} F_\alpha$. $\endgroup$
    – Jason Starr
    Mar 22, 2018 at 9:05
  • $\begingroup$ @JasonStarr Thanks for your comments! Any suggestion for second question ? $\endgroup$
    – Max CYLin
    Mar 26, 2018 at 15:46

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.