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From: Marc Bogaerts on 22 Jan 2010 13:55
I tried to have a deeper insight in the splitting field of a polynomial
of the form f(x)=x^3-p*x-q for rationals p and q. First of all I
considered the companion matrix of this polynomial:
0 0 q
1 0 p
0 1 0
Let's call it M. M satisfies M^3-p*M-q*I where I is the 3 x 3 identity
matrix.

This matrix generates an 3 dimensional extension F of the rationals,
where the elements are of the form a + b*M + c*M^2. I don't know if this
is the splitting field of the polynomial or not. This depends on the
fact if the quotient of f(x) by x-M has roots in F or not. Formally I
can write this quotient as g(x) = x^2 + M*x + M^2 - q*I, a polynomial
with coefficients in F.
If there are no roots of g(x) in F I do the same with this polynomial by
taking the companion matrix over F. We get a matrix N as a block matrix:

0 -M^2+q*I
I -M

with 3 x 3 matrices as coefficients. Now N is also a root of f(x), and
generates a two dimensional extension G over F, that is also a six
dimensional extension over the rationals. The third "root" can be found
as P = -M - N (where M is the diagonal block-matrix) since the second
degree term of f(x) is zero. Let us now look at what the Galois group
may be. If we take as a basis of the field G the elements I, M, M^2, N,
N*M, N*M^2 then we can express the automorphism that maps:
I -> I
M -> M
N -> P = - M - N
P -> N
as a block - matrix R :
I -M
0 -I

We can check that R^2 = 1, R*M*R^-1 = M , R*N*R^-1 = P and R*P*R^-1 = N.

The automorphism S that fixes P but permutes M and N is a little harder
to calculate; the result is:

1 0 q 0 0 -p
0 0 0 1 0 0
0 0 -1 0 0 0

0 1 0 0 0 0
0 0 -1 0 1 0
0 0 0 0 0 -1

We see that R and S generate the symmetric group S_3. R and S do not
commute. If we look at R*S - S*R we see that it contains nonzero
elements that do not depend on p neither q so that the galois group is
never abelian. We conclude that galois group is non abelian if the field
F is not the splitting field of f(x). If F is the splitting field of
f(x) then g(x) has a root N in F and the Galois group contains maximally
3 elements so it must be the cyclic group C_3. As an example we can take
the polynomial f(x) = x^3 - 3*x + 1, then N = M^2 - 2*I is a root of
g(x) = x^2 + M*x + M^2-q in F.
In the general case the discriminant of g(x) is D = -3*M^2 - 4*q, if the
root exists in F then there is a matrix T with coefficients in F such
that T^2 = D. The determinant of D is 4*q^3 - 27*p^2, which is precisely
the discriminant of f(x). If D is the square of a matrix T then its
determinant must be the square of a rational. I know the reverse is true
but I don't know how to prove it.
From: Achava Nakhash, the Loving Snake on 23 Jan 2010 11:37
On Jan 22, 10:55 am, Marc Bogaerts <mbg-dot-ni...(a)gmail.com> wrote:
> I tried to have a deeper insight in the splitting field of a polynomial
> of the form f(x)=x^3-p*x-q for rationals p and q. First of all I
> considered the companion matrix of this polynomial:
> 0   0   q
> 1   0   p
> 0   1   0
> Let's call it M. M satisfies M^3-p*M-q*I where I is the 3 x 3 identity
> matrix.
>
> This matrix generates an 3 dimensional extension F of the rationals,
> where the elements are of the form a + b*M + c*M^2. I don't know if this
> is the splitting field of the polynomial or not. This depends on the
> fact if the quotient of f(x) by x-M has roots in F or not. Formally I
> can write this quotient as g(x) = x^2 + M*x + M^2 - q*I, a polynomial
> with coefficients in F.
> If there are no roots of g(x) in F I do the same with this polynomial by
> taking the companion matrix over F. We get a matrix N as a block matrix:
>
> 0  -M^2+q*I
> I  -M
>
> with 3 x 3 matrices as coefficients. Now N is also a root of f(x), and
> generates a two dimensional extension G over F, that is also a six
> dimensional extension over the rationals. The third "root" can be found
> as P = -M - N (where M is the diagonal block-matrix) since the second
> degree term of f(x) is zero. Let us now look at what the Galois group
> may be. If we take as a basis of the field G the elements I, M, M^2, N,
> N*M, N*M^2 then we can express the automorphism that maps:
> I -> I
> M -> M
> N -> P = - M - N
> P -> N
> as a block - matrix R :
> I -M
> 0 -I
>
> We can check that R^2 = 1, R*M*R^-1 = M , R*N*R^-1 = P and R*P*R^-1 = N.
>
> The automorphism S that fixes P but permutes M and N is a little harder
> to calculate; the result is:
>
> 1  0  q    0  0  -p
> 0  0  0    1  0  0
> 0  0  -1   0  0  0
>
> 0  1  0    0  0  0
> 0  0  -1   0  1  0
> 0  0  0    0  0  -1
>
> We see that R and S generate the symmetric group S_3. R and S do not
> commute. If we look at R*S - S*R we see that it contains nonzero
> elements that do not depend on p neither q so that the galois group is
> never abelian. We conclude that galois group is non abelian if the field
> F is not the splitting field of f(x). If F is the splitting field of
> f(x) then g(x) has a root N in F and the Galois group contains maximally
> 3 elements so it must be the cyclic group C_3. As an example we can take
> the polynomial f(x) = x^3 - 3*x + 1, then N = M^2 - 2*I is a root of
> g(x) = x^2 + M*x + M^2-q in F.
> In the general case the discriminant of g(x) is D = -3*M^2 - 4*q, if the
> root exists in F then there is a matrix T with coefficients in F such
> that T^2 = D. The determinant of D is 4*q^3 - 27*p^2, which is precisely
> the discriminant of f(x). If D is the square of a matrix T then its
> determinant must be the square of a rational. I know the reverse is true
> but I don't know how to prove it.

I haven't had a chance to review what you have done here, but here is
what I came up with a few minutes ago in order to prove that the
splitting field of any irreducible cubic equation with rational
coefficients is of degree 3 if and only if the discriminant D of the
polynomial is a square in Q.

First of all, the definition of discriminant is that

D = ((x1 - x2)*(x1 - x3)*(x2 - x3))^2

where x1, x2, and x3 are the 3 roots of the polynomial.

It is easy to see that D is a symmetric function in the roots and so
must be a rational number. In your case it must be an integer, since
the polynomial is monic and has integer coefficients. It is easy
enough to work out the value of D in terms of the coefficients of the
polynomial, but I shall spare myself the details. They can easily be
looked up in a lot of algebra books and no doubt on line as well. You
get what you expcet to get, the standard result.

Now if D is a square in Q, the (x1 - x2)*(x1 - x3)*(x2 - x3) must also
be in Q and hence invariant under the action of the entire Galois
group. Since this quantity is turned into its negative by all
transpositions of the 3 roots, the Galois group cannot contain any
transpositions and so must have order 3. In other words the splitting
field must have degree 3.

Now suppose the splitting field has degree 3. The only subgroup of
S3 of degree 3 is A3, which has order 4 and contains no
transpositions. It is then easy to verify that
(x1 - x2)*(x1 - x3)*(x2 - x3) is invariant under the Galois group and
so is in Q. Hence D, the square of this quantity, is a square in Q.

The idea of embedding number fields in rings of matrices (and your
construction works for any number field) has been around a long time
and can be quite useful. I used it myself in my thesis, although I
don't remmeber how. One thing that it buys you is that the number
field now acts on the ring of all matrices, and no dbout a great deal
can be learned that way. I know there is literature on this, but I
recall it being rather difficult, and I am not familiar with what you
get. There is a book by Irving Reiner called something like
"Hypercoomplex Numbers" which deals with it.


Regards,
achava
From: Marc Bogaerts on 24 Jan 2010 11:46
Achava Nakhash, the Loving Snake wrote:
> On Jan 22, 10:55 am, Marc Bogaerts <mbg-dot-ni...(a)gmail.com> wrote:
....
>> In the general case the discriminant of g(x) is D = -3*M^2 - 4*q, if the
>> root exists in F then there is a matrix T with coefficients in F such
>> that T^2 = D. The determinant of D is 4*q^3 - 27*p^2, which is precisely
>> the discriminant of f(x). If D is the square of a matrix T then its
>> determinant must be the square of a rational. I know the reverse is true
>> but I don't know how to prove it.
>
> I haven't had a chance to review what you have done here, but here is
> what I came up with a few minutes ago in order to prove that the
> splitting field of any irreducible cubic equation with rational
> coefficients is of degree 3 if and only if the discriminant D of the
> polynomial is a square in Q.
>
> First of all, the definition of discriminant is that
>
> D = ((x1 - x2)*(x1 - x3)*(x2 - x3))^2
>
> where x1, x2, and x3 are the 3 roots of the polynomial.
>
> It is easy to see that D is a symmetric function in the roots and so
> must be a rational number. In your case it must be an integer, since
> the polynomial is monic and has integer coefficients. It is easy
> enough to work out the value of D in terms of the coefficients of the
> polynomial, but I shall spare myself the details. They can easily be
> looked up in a lot of algebra books and no doubt on line as well. You
> get what you expcet to get, the standard result.
>
> Now if D is a square in Q, the (x1 - x2)*(x1 - x3)*(x2 - x3) must also
> be in Q and hence invariant under the action of the entire Galois
> group. Since this quantity is turned into its negative by all
> transpositions of the 3 roots, the Galois group cannot contain any
> transpositions and so must have order 3. In other words the splitting
> field must have degree 3.
>
> Now suppose the splitting field has degree 3. The only subgroup of
> S3 of degree 3 is A3, which has order 4 and contains no
> transpositions. It is then easy to verify that
> (x1 - x2)*(x1 - x3)*(x2 - x3) is invariant under the Galois group and
> so is in Q. Hence D, the square of this quantity, is a square in Q.
>
This is perfectly correct (except that I presume that you meant that A3
has order 3 instead of 4). But what I would have liked to find is that
in the second degree equation that one obtains by splitting of the root
M and that has coefficients in Q[M] the "determinant" squared D^2 is a 3
x 3 matrix and that it is not obvious how to find D. In the example of
the polynomial f(x) = x^3 - 3*x +1 whe have the matrix D^2 :

[ [ 12, 3, 0 ],
[ 0, 3, 3 ],
[ -3, 0, 3 ] ]

and it is not obvious that D equals

[ [ -4, -2, -1 ],
[ 1, 2, 1 ],
[ 2, 1, 2 ] ]

using only techniques

> The idea of embedding number fields in rings of matrices (and your
> construction works for any number field) has been around a long time
> and can be quite useful.

That is indeed the whole idea behind it. It can be generalized to more
general polynomials and gives an insight into the working of the Galois
group as the automorphisms can also be expressed as matrices.

> I used it myself in my thesis, although I
> don't remmeber how. One thing that it buys you is that the number
> field now acts on the ring of all matrices, and no dbout a great deal
> can be learned that way. I know there is literature on this, but I
> recall it being rather difficult, and I am not familiar with what you
> get. There is a book by Irving Reiner called something like
> "Hypercoomplex Numbers" which deals with it.
>
>
HAND,

Marc.
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