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Let $L$ be the splitting field of a degree $n$ irreducible monic polynomial with integer coefficients over $\mathbb Q$. Let the $n$ roots $\{r_i\}$, then $L = \mathbb Q(r_1,\dots, r_n)$.

Suppose I have computed the roots to a high degree of accuracy with a computer. How could I test if a permutation of the roots was a valid field automorphism?

Kenta S
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  • Arbitrary precision isn't sufficient, you need to know exactly what the roots are to discover which permutations are automorphisms. This is because many roots are irrational numbers, like the roots of $x^2-2$. – CyclotomicField Aug 04 '20 at 16:26
  • @CyclotomicField Surely there will be numerical expressions which come out as zero for valid automorphisms and nonzero for invalid ones, and then calculating to enough accuracy should enable me to differentiate between them? –  Aug 04 '20 at 16:56
  • Nope, you sure can't. Algebraic numbers are dense in the reals so no matter how accurate it's not enough. – CyclotomicField Aug 04 '20 at 17:07
  • I understand your point about density but you can often put a lower bound on an error, so if R(x1,x2,x3) = 0 for a valid permutation of the roots but expected to be > 10^-3 for an invalid one you would only need to calculate R to 10^-4 places to perform the test –  Aug 04 '20 at 17:25
  • Given that the PSLQ Algorithm can be used "experimentally" to find integer linear dependence relations between irrational numbers, for example in order to find expressions for Euler sums in terms of known constants, surely it must be a lot easier to do what this question is asking for. Of course, "experimental" results must then be proved. (I know nothing about the subject myself.) – Calum Gilhooley Aug 04 '20 at 17:43
  • @CalumGilhooley the PSLQ algorithm assumes you know the roots. If you're using rational approximations but don't know the roots it is no longer useful. It's much easier to identify the automorphisms when the roots are given and in many cases you can do so by inspection. – CyclotomicField Aug 04 '20 at 17:48
  • @CyclotomicField I wasn't suggesting that the PSLQ algorithm itself could be used to solve this problem; I was just using it as an example of something apparently more complicated that can in fact be done. (But I'm comparing two things of which I have no knowledge, so I'd rather not get into a long discussion. I could very easily have completely misunderstood.) – Calum Gilhooley Aug 04 '20 at 17:54
  • I have the original polynomial and I can calculate the roots to whatever precision is required. –  Aug 04 '20 at 17:57
  • @CalumGilhooley if we know the roots then finding linear dependence among them reduces the complexity significantly. For example you can express $\sqrt{2}$ using just two integers, but it would be impossible to store all the digits in memory using Newton's method if you wanted to identify the roots in the first place. If you consider $x^2+a$, the Galois group when $a=0$ is trivial, but if $a=1/100000000000000000$ then the automorphism $x \rightarrow -x$ is non-trivial. Tiny variations must be detectable. – CyclotomicField Aug 04 '20 at 18:21
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    @CyclotomicField The assumption here is that the polynomial has integer coefficients. By Vieta relations (or simply by expanding $\prod_i(x-x_i)$) we get approximate coefficients of the polynomial. Knowing that they are integers we then recover the polynomial. There may be difficulties if some of the roots are huge, when a tiny error in the other roots may throw the calculation off. But knowing the approximate magnitudes of the zeros, we can calculate a threshold for required accuracy for recovery of the exact polynomial. – Jyrki Lahtonen Aug 04 '20 at 22:28
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    Anyway, I don't know of a method for calculating the Galois group using approximate zeros. The obstructions to an arbitraty permutation coming from an automorphism come from relations satisfied by the zeros, and PSLQ may work, but I'm not familiar with it. – Jyrki Lahtonen Aug 04 '20 at 22:31

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