This is a sequence to this post. The SAIR competition (Round I) has been completed! 98.4% of the possible signatures were obtained, with only 39 non-solvable cases missing.
Some thoughts.
First, my timing in the last post of dissing the problem of realizing \(M_{23}\) as a Galois group was not so great. It seems to me that the delightful paper does an excellent job of combining human and AI thoughts but also clearly and concisely explaining the ideas, especially distinguishing between what is known, what is clever, and what is lucky. Nicely done!
Moving on to the competition. I thought that it would be better to get a precise sense of the difficulty by trying it myself. The approach I used was purely to tell CODEX to do 6 obvious things, but not to either look at any literature myself, not to write any code, and just to come back and complain when it failed. This quickly produced around 40,000 pairs, but then stalled. One approach that wasn’t successful at all was as follows. There were around 80,000 pairs or so could be realized as coming from the Galois closure of degree 12 extensions of quadratic fields. But alas, my suggestions for how to construct these were not taken up sensibly, and I didn’t pursue it.
Certainly my personal explorations produced no meaning mathematical content at all. The only mathematical idea I had that was not completely obvious was one I learnt entirely from David Roberts. In situations where one has a Galois extension \(L/K/\mathbf{Q}\) where \(K\) has Galois group \(G\) and \(L\) has Galois group a central extension of \(G\) of degree \(2\), then one can write \(L\) as the splitting field of a polynomial of the form \(f(x^2)\) where \(f(x)\) has splitting field \(K\) and one root of \(f(x)\) generates a field \(E\). But now, given \(g(x)\) with \(E \simeq \mathbf{Q}[x]/g(x)\), how does one find \(f(x)\)? The observation is that one can often find \(f(x)\) by applying \(\texttt{polred}\) to \(g(x)\).
That said, having done some of these experiments, it did help me appreciate what type of problem this was. It certainly seemed to be the case that real skill and knowledge working with explicit polynomials and explicit Galois theory would be genuinely useful, and simply a purely theoretical knowledge of (say) the general solvable case is not sufficient. It is no surprise then that Klüners and Malle (the leading team) were so successful.
But where does it lead us? I don’t think the conclusion is so far from my original prediction. I think there might be a new second round coming, and after that is done, it really could be the case that the only pairs remaining are \((G,r)\) where \(G = \mathrm{PSL}_2(\mathbf{F}_{23})\) and also \(G = \mathrm{PGL}_2(\mathbf{F}_{23})\) with \(r=0\) (which are hard for the same reasons), and then possibly some cases of \(M_{24}\) (say with \(r=0\)) as well. We shall see!