Galois Theory/Splitting FieldsDate: 03/07/2002 at 18:27:21 From: Julius Subject: Galois Theory / Splitting Fields Determine the splitting field of the polynomial x^p - x - a over F_p where a is not equal to zero and a is an element of F_p. Show explicitly that the Galois group is cyclic. [Hint: Show alpha |--> (alpha + 1) is an automorphism.] Can you help? Date: 03/07/2002 at 22:06:48 From: Doctor Pete Subject: Re: Galois Theory / Splitting Fields Hi Julius, I believe I can help you on this one. Recall that the splitting field of a polynomial f(x) in F[x] is an extension K of F in which f factors completely into linear factors, and in a sense K is the smallest such extension, as f cannot be expressed as a product of linear factors in any proper subfield of K. (Basically, this second condition means that if K' is an extension containing K, then this is not a splitting field for f since the "smaller" extension K splits f before K' does.) Thus, if our field is Fp, the field of characteristic p, let f(x) = x^p - x - a where a is nonzero, and suppose b is a root: f(b) = b^p - b - a = 0. Now consider f(b+1) = (b+1)^p - (b+1) - a = b^p + 1^p - b - 1 - a = b^p - b - a = 0. (That (b+1)^p = b^p + 1^p in Fp is due to the Frobenius endomorphism, and can be explicitly proved using the Binomial Theorem on the left-hand side and noting that all terms except the first and last are multiples of p and thus equal to zero.) Therefore, if b is a root, then b+1 is also a root of f. It follows that the set {b, b+1, b+2, ..., b+p-1} are the p roots of f, and at this point it should be obvious that the extension Fp(b) splits f completely into linear factors, and no smaller extension does so; hence Fp(b) is the splitting field of f. To find Gal(f(x)), we must first show that f(x) is separable, from which it follows that Fp(b)/Fp is Galois. But f is separable since it has no repeated roots. (Alternatively we can see this by noting that the formal derivative D(f(x)) = px^(p-1) - 1 = -1, and hence has no roots, thus f(x) and D(f(x)) share no roots and f is separable.) So, now having shown Fp(b)/Fp is Galois, Gal(f(x)) = Aut(Fp(b)/Fp), and it is important to note that there are only p permutations (not p! ; why?), of the form s(k) : b -> b+k, 0 <= k <= p-1. This much is quite clear just from looking at the set of roots; and furthermore, each permutation admits an element of Aut(Fp(b)/Fp) so is an element of Gal(f(x)). Therefore we have specified all the elements of the Galois group can see that the map s(1) is a generator of this Galois group. Therefore, we finally have Gal(f(x)) = <s : s^p = 1> ~= Z/pZ (the symbol ~= is meant to be the isomorphism symbol.) - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ Date: 03/08/2002 at 09:04:34 From: Julius Subject: Galois Theory / Splitting Fields Many thanks to Dr. Pete for a very clear response. It was just what I was hoping for. I guess the trick was to notice that if b is a root of f(x), then b+1 is also a root of f(x). Thanks again. Julius |
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