As you already know, we can embed the rationals inside the cuts, i.e. inside $\bf R$ by means of $r\mapsto \hat r= \{x\in\Bbb Q:x<r\}$.
It is immediate that set is a cut.
Now, suppose you have a bunch of rational cuts, $\hat r_1,\hat r_2,\ldots$.
The point is $\bigcup\limits_{i\geqslant 1} \hat r_i$ is a cut, but rarely it is a rational cut.
This is similar to this: why doesn't Cantor's diagonal argument show the naturals are uncountable?
Recall Cantor creates a string of zeros and ones from a countable collection of strings of zeros and ones, to create a real that was not in our list. Natural numbers can be thought of, in base two, as sequences of zeros and ones that eventually terminate in $0$. Now, if we have a long list of sequences that eventually zero out, there is no reason why taking the diagonal will produce a sequence of zeroes and ones that also zeroes out.
The point is that the reals work nice because they are all of the cuts, but the rationals are just a little subset. The "trick" of joining cuts to create another cut works fine when looking at all of the reals (i.e. all of the cuts) but if we look only at rationals, there is no reason the construction will escape them.
@Anthony OK, so if we have an infinite number of cuts $(A,B)$ (with the 'cut point' bounded above), we can certainly make a new cut $(\bigcup A, \bigcap B)$. Check that this partitions the rationals, that all $a \in \bigcup A$ are less than $b \in \bigcap B$.
@usukidoll The grad students being in the class means they're not good enough to get into the grad real analysis and have to take the undergrad one to catch up :P
The roots of a general polynomial over $\Bbb{Q}[X]$ of degree $n \geq 5$ is not contained in $\Bbb Q[x_1, x_2, \cdots, x_m]$ for some algebraics $x_1, x_2, \ldots,x_m$ with solvable Galois groups.
Is there any problem stating Abel Ruffini like that?
the difference between the two forms is that you replaced "splitting extension" with "extension by a given set of elements," which is superfluous in my opinion
easy is relative. (1) skim the paper to see if it's the kind of stuff you'd enjoy thinking through if you did choose to put effort into it, and (2) try following some initial segment (see what I did there) of the paper to test to what extent it's at your level
@anon Ah, but that depends on the context, of course. Extensions are more needed to here, especially transcendental ones. For example, note that a quintic can be solved in the extension $\Bbb{Q}[X, \vartheta(0|X)]$ (wink)
@BalarkaSen okay, territorial over the vocabulary I see. I suppose I am not prepared for that particular turf war, except to comment that I notice the word "hypergeometric" appears in the adjective "A-hypergeometric"
@anon You are chatroom co-owner? I would have spoken with a little more, uh, respect then if I knew that before. I have been haunted by the fear of banning by mods everytime I come to war-like position with them. Nevermind.
I figure $K[G]^\times$ will be $K[G]$ minus a set number of left/right ideals (principally generated by left/right nonunits), so in particular $K[G]$ minus a bunch of proper $K$-subspaces, which can be partitioned according to dimension. as long as tensoring up from ${\Bbb F}_p$ to ${\Bbb F}_{p^n}$ doesn't get too wild it shouldn't destabilize this family of subspaces, which would mean the size is in fact polynomial in $q=p^n$.
What a world. No TNT Qs anywhere and I am answering to one galois theory question. Nevertheless, prove that quintics cannot be solved even in the extension of the solvable numbers equipped with finite composition of algebraic field operators by equipping with the trigonometric functions and thus extending to the closure of $\mathcal {EL}$. Just for kicks.
@user41631 Yep, feel free to meet me in person for the resolution of your psychological problems. This room is a terrible place for me to do this kind of work.
