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Apr 13, 2017 at 12:57 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jun 15, 2014 at 22:17 comment added Will Jagy Cool. Thanks, Noam. That does look much better. I found my degree eight on that website you mentioned, by putting in $221^4.$ Actually, the first time it thought i meant 2214 and complained. i decided it was better to multiply out the number.
Jun 15, 2014 at 22:14 comment added Noam D. Elkies The repeated root mod $101$ is an example of my warning that "a polynomial generating [the Hilbert class field] $H$ might have to err on the first few primes" if $H$ has no generator $x$ such that ${\bf Z}[x]$ is the full ring of integers of $H$. A better choice for this purpose is $x^8 + 34 x^6 + 83 x^4 + 34 x^2 + 1$, which has spurious repeated roots only mod $2$ and $3$ (and as it happens no linear factors modulo either of these primes). Then $x+1/x$ generates a quartic isomorphic with the one you found with coefficients $1,1,1,2,4$.
Jun 15, 2014 at 20:57 comment added Will Jagy For 221, the quartic $$ x^4 + x^3 + x^2 + 2 x + 4 $$ behaves well
Jun 15, 2014 at 20:47 comment added Will Jagy For the very similar discriminant 221 and $x^2 + 13 x y - 13 y^2, $ I was surprised to find that $$ f(x) = x^8 + x^6 - 4 x^5 - 38 x^4 - 2 x^3 + 123 x^2 -34 x + 17, $$ has a repeated root $\pmod {101},$ although still linear factors, seven of them with one of them squared. Is that allowed? After that it is always 8 roots. This time the primes are $ 17,101,103,127,179,251,263,373,433,$
Jun 15, 2014 at 18:55 history edited Noam D. Elkies CC BY-SA 3.0
Further explanation and alternatives
Jun 14, 2014 at 17:18 vote accept Will Jagy
Jun 14, 2014 at 5:03 history answered Noam D. Elkies CC BY-SA 3.0