Determining the Galois Group of a Polynomial Anyone who considers arithmetical methods of producing random digits is, of course, in a state of sin." J. von Neumann Is there a simple way of determining the Galois group of a given polynomial such as A fairly quick and simplistic way is to check for roots of this polynomial modulo various primes, to see in which fields the polynomial splits completely into linear factors. With the exception of primes that divide the discriminant (which for this equation is –21674), the polynomial will have no repeated roots, so if there are 6 distinct values of x between 0 and p–1 such that f(x) = 0 (mod p), then f splits in Fp[x]. Now, by a theorem of Chebotarev we know the proportion of all primes relative to which f(x) splits is asymptotic to 1/order(G) where G is the Galois group of f. So, it's fairly simple to just count "splitting primes" for any given polynomial to give a very robust indication of the order of the group. For the polynomial (1) we find that f(x) splits relative to p = 1013, 3797, 6449, 6793, 7309, 9473, ... etc. In fact, of the first 4792 primes (not counting 2 and 7, of course), f(x) splits completely relative to 36 of them, which is about 1 out of 133. Clearly the order of the group is not 720 (which it would have to be if the Galois group was S6), so it must be a proper subgroup. The proper subgroups of S6 are and we can rule out the alternating group because the discriminant of an alternating group must be a positive square, whereas our discriminant is –157351936, so the only possibility that's really feasible is the group PGL(2,5) of order 120. If we are worried that the group might really be G72 (even though the "odds" of G72 having only 36 of the first 4792 primes as splitting primes are incredibly tiny), it's not hard to rule out the smaller groups just by noting how f(x) factors in various Fp[x], and recalling that these factorizations correspond to cyclic decompositions of the permutations in the respective Galois group. For the polynomial (1) we can verify that f splits into irreducible factors of the degrees shown below and of course we've seen that f splits into 1,1,1,1,1,1 relative to p = 1013, etc., so the Galois group must certainly contain the permutations with these cyclic decompositions, and this rigorously rules out any groups smaller than PGL(5,2). This still relies on the count of splitting fields, which indicates an order in the neighborhood of 133, to assure us that the order of the group isn't 720. However, if we compute the odds for a Poisson process of getting 36 splitting primes in a range where only 6 or 7 would be expected, we will be quite confident that the order is not 720. Moreover, the numbers of factorizations of the different types for the 168 primes (not counting 2 and 7) up to 1013 are and these densities also conform pretty well with Chebotarev's density theorem for the Galois group PGL(2,5), and of course the compositions {4,2}, {3,2,1}, {3,1,1,1}, and {2,1,1,1,1} are entirely absent, whereas if the group were S6 they should have asymptotic densities 90/720, 120/720, 40/720, and 15/720, respectively. Thus they should account for almost 37% of the primes, but they are completely absent. Furthermore, they are precisely the compositions that must be absent if the group is actually PGL(2,5). Thus, by merely examining f(x) in various Fp[x], the Galois group of the polynomial quickly becomes very apparent, even without slogging through a rigorous demonstration. Return to MathPages Main Menu