Yeah. My son and I visited him at the hospital in Texas over the weekend and said goodbye. He's in hospice care, so no more antibiotics, feeding tube, dialysis, etc.
Suppose I have a category with all pullbacks. Consider the pullback $X \times_B Y$, and suppose I have an automorphism $Y \to Y$. Do I necessarily get an automorphism $X \times_B Y \to X \times_B Y$?
@TheChaz: Hey, I'm not the one who brought up the guess-and-check method. Skull's question was itself about the method. || I recall when I was in elementary school and we were taught the guess-and-check method, I caught on to the formulas behind them (this is before we had the symbols to articulate algebra or formulas), so I would just put the answer down pat, and my teacher marked these wrong because I didn't guess! :/
Unfortunately, it seems you cannot understand profinite groups without a good background in general topology. I don't know what "Hausdorff" or "Tychonoff" mean, argh!
@anon Hausdorff: any two distinct points admit disjoint neighborhoods. Tychonoff: usually (Hausdorff +) for every point $x$ outside a closed set $A$ there is a continuous function to the unit interval such that $f(x) = 1$ and $f|_A = 0$.
@Zhen: Wait, all profinite groups are ultimately limits of inverse systems?! @t.b. I was reading about the "Tychonoff topology" given to a direct product of spaces.
So here's the problem statement: Prove that the automorphisms of the rational function field $k(t)$ which fix $k$ are precisely the fractional linear transformations determined by $ t\mapsto\frac{at+b}{ct+d}$ for $a,b,c,d\in k,ad-bc\neq0$.
I've established that if $\sigma$ is an automorphism that fixes $k$, then $\sigma(t)=\frac{p(t)}{q(t)}$ (in lowest terms). And that $\deg\{p\}+\deg\{q\}\geq1$.
@tb Thanks for that. I have already consulted that thread. The thread develops several facts that are very useful for this exercise. However, unless I missed it, Arturo does not make mention of the requirement that $ad-bc\neq0$.
If $ad = bc$ then $(ax+b)c = acx + ad = a(cx + d)$ so the fraction $\frac{ax+b}{cx+d} = \frac{a}{c}$.
(assuming that $c \neq 0$)
@DavidK what's wrong with those? you can easily write down inverses for both (invert the matrices $\left(\begin{smallmatrix} 0 & b \\ c & d\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix} a & b \\ 0 & d\end{smallmatrix}\right)$.
@tb Ok, I just ran through the 4 possibilities in turn: $a=0$, $b=0$, $c=0$ and $d=0$. Each results in a contradiction. I guess I was just hoping for a cleaner way than brute force.
@anon Just so I'm clear about something, this implies that $[k(t):k(\sigma(x))]=1$, right?
t.b.'s case works when $c\ne0$. If $c=0$, then $d\ne0$ (division by 0), along with $ad-bc=0$ implies $a=0$, giving $b/d$ as the ratio. @David: It's an automorphism, so $k(t)=\sigma k(t) = k(\sigma(t))$ (I think...), so yes.
I wonder if there's any chance of finding Kleiman's "Algebraic cycles and the Weil conjectures" online. It seems to be an article in a book rather than a journal...
@tb Btw, I am not sure we were talking about the same thing there: My day was spoiled because I couldn't make sense of your comment about homomorphisms on $S^1$. Not because of your "this is what you get when". : )
In mind, I think, when trying to keep an high rep user away, the consensus of the community must have been a priority and not a mod who is almost unaware of our community acting on a flag.
No, I am not against Robjohn or something, but I am not letting down Asaf, We want him back here.
@robjohn On a completely different note: maybe this is silly, but I was wondering if there was a way of implementing that MathJaX bookmark of yours as an addition to Skype chat. That would be a tremendously useful and convenient thing to have. Do you think that might be feasible without too much effort?
@robjohn exactly my thinking :) I haven't really thought about it (and I don't think I would be able to implement it even if it were very easy) but I thought I might ask.
