An incorrect proof
- From: Kent Holing <KHO@xxxxxxxxxxx>
- Date: Tue, 30 Aug 2005 14:00:05 +0000 (UTC)
The following is not true but where is the error in the proof below?
"Theorem":
Let p(x) = 0 be an irreducible monic polynomial of degree n with integer coefficients with Galois group G.
For x1 a root of p(x) = 0 we have that G = Zn (cyclic of order n) if and only if p(x) splits completely over Q[x1]. (The other roots of p(x) = 0
are x2, x3, ... xn.)
"Proof":
If G = Zn the splitting field of the equation has dimension n, so has Q[x1] since x1 has the minimal polynomial p(x). So the splitting field of p(x) = 0
must be Q[x1] and p(x) splits completely over Q[x1].
Opposite, if p(x) splits completely over Q[x1] then all elements of G is defined by its action on x1 which may go to x1, x2 , x3, ..., xn which show that
the number of elements of G is n. Let us show that we must have G = Zn. Define sigma in G such that it is an automorphism sending x1 to x2, x2 to x3,
x3 to x4, ... xn to x1. The order of sigma is n so G = <sigma> = Zn.
"Theorem" is not true since we have:
G = V, p(x) = -9 + 6*x + 7*x^2 - 6*x^3 + x^4 = 0 which splits as ((-u + x)*(-3 + u + x)*(-30 - u + 18*u^2 - 4*u^3 + 3*x)*(21 + u - 18*u^2 + 4*u^3 + 3*x))/9 for u a root.
Note that if n = p a prime the above "theorem" is true.
What is true is the following:
Let p(x) = 0 be an irreducible monic polynominal of degree n with integer coefficients with Galois group G.
For x1 a root of p(x) = 0 we have that G has n elements if and only if p(x) splits completely over Q[x1].
(The first part of "Theorem" above is true but not the second part.)
Kent Holing
.
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