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Galois Theory/Splitting Fields


Date: 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
    
Associated Topics:
College Modern Algebra

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