The GAP package has a function $\mathtt {GaloisType}$ that takes a polynomial as an argument and returns a number, the index of the transitive group of order the degree of the polynomial. I read somewhere that the algorithm is based on a theorem by Dedekind based on how the polynomial factors if taken modulo some primes. But this theorem only gives a postive answer about the existence of a cycle, if it finds it, but this gives no certainty about what the group actually is. I also know that there are some probability theorems concerning this problem, but I don't now what they state. What is the probability that the answer is wrong?
The determination of the Galois group of a polynomial
470 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtThere are 2 best solutions below
On
What is happening is a 2-stage process. First Dedekind's theorem is used to obtain some cycle shapes. This gives a lower bound for the Galois group.
Afterwards, polynomials that correspond to the action of the Galois group on k-sets or k-sequences are factored. (E.g. $$\prod_{i,j} (x-(\alpha_i+\alpha_j))$$ corresponds to the action on 2-sets.) This way one shows that the Galois group can't be larger. The result is (modular programming errors) proven correct.
IF you only want a probabilistic result, based on Chebotarev's theorem, the function $\mathtt {ProbabilityShapes}$ does so (and is much faster).
At the risk of shameless self-promotion, http://www.math.colostate.edu/~hulpke/talks/galoistalk.pdf are slides of a relevant talk, http://www.math.colostate.edu/~hulpke/paper/gov.pdf a survey paper.
It seems the comment field is limited, thus a second answer as example:
What you describe is quite possible (and a plausible and intelligent question):
Take $p(x)=x^3+x^2-2x-1$ (the minimal polynomial of E(7)+E(7)^6, so one can actually work with roots). This is irreducible over the rationals and has Galois group A3, so factoring modulo primes will only ever give irreducible or 3 linear factors. Now both A3 and S3 are transitive on 2-sets, so we need to (white lie! see at the end) take 2-sequences. We therefore form the polynomial whose roots are of the form $\alpha_i+2\alpha_j$. We can do so by calculating the resultant $r(x)=res_y(2^3\cdot p(y/2),p(x-y))$, and dividing off (the case $i=j$) the factor $3^3p(x/3)$.
In practice, a method based on symmetric functions is easier
produces the same result. If we factor this polynomial of degree 6, we get two factors of degree 3 -- two orbits on 2-sequences. If the initial polynomial had had Galois group S3, it would have stayed irreducible.
You can observe what GAP is doing by setting
Then GaloisType will print information about the process, e.g.
here cycle shapes leave groups 1,2 and 4 left. Action on 2-sets (SumRootsPol 2) leaves 1 and 2, and 2-sequences finally nails the group.
If you do this on $x^3+x^2-2x-1$ above you notice that the 2-sequence polynomial is never used. That is because we can distinguish $A_n$/$S_n$ by whether the discriminant (for $p$ it is $49$) is a square.