Hilbert’s 11th problem which demands that we ‘classify quadratic forms over algebraic number fields’ has been of interest to me and I would like to know what makes it partially resolved currently. Or in other words, what does the progress made in recent times lack? What equation(s) does one have to solve in particular to provide full solutions to the problem?
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6$\begingroup$ It would be helpful to remind the rest of us what is Hilbert's 11th problem. Those who aren't actively studying a particular Hilbert problem have trouble remembering which is which. $\endgroup$– Dan FoxCommented Feb 19, 2022 at 11:08
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$\begingroup$ @DanFox I have edited the question. I hope I have made clear enough. Thanks $\endgroup$– A. J.Commented Feb 19, 2022 at 11:27
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2$\begingroup$ I think this is a good question. The article I posted as a comment to the OP's other question (jtnb.centre-mersenne.org/article/JTNB_2003__15_1_33_0.pdf) says that the work it discusses "essentially resolves the remaining open case of Hilbert's 11th problem." So exactly what is left is unclear to me, even though on en.wikipedia.org/wiki/Hilbert%27s_problems it is indeed listed as only "partially resolved." $\endgroup$– Sam HopkinsCommented Feb 19, 2022 at 13:48
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$\begingroup$ @SamHopkins I recommend a contemporary survey by Schulze-Pillot math.uni-sb.de/ag/schulze/Preprints/talca_survey.pdf see also arxiv.org/search/… $\endgroup$– Will JagyCommented Feb 19, 2022 at 17:43
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$\begingroup$ isn’t the issue that the bounds one gets are ineffective (Siegel zeroes)? in my nonexpert understanding you get a main term for the number of rep.s of $N$ which is something like the value of a Dirichlet $L$-function at 1 (maybe assoc. to the quad. char. with disc. the discriminant of your form, maybe I’m implicitly imagining it’s diag.) times the usual local const.s (which at infinity gives the expected $N^{1/2}$) plus an error term which you effectively upper bound via a subconvex estimate (which saves a small power), and so you need $L(1,\chi)$ to not be too small, whence Siegel zeroes. $\endgroup$– alpogeCommented Feb 19, 2022 at 22:10
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