From: balint@mathworks.com (bbalint)
Newsgroups: sci.math
Subject: Algebra Help
Date: 28 Sep 1995 13:31:58 GMT
Hi,
Let G be the subset of Aut K(x) over field K consisting of the 3
automorphisms induced by
x -> x
x -> 1/(1-x)
x -> (x-1)/x.
Determine the fixed field of G.
How do I go about finding this fixed field? Full disclosure: this
*was* a homework problem in a class I took last year. I've already
received a grade so this is just out of personal interest.
Thanks,
Bill Balint
==============================================================================
From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Algebra Help
Date: 28 Sep 1995 17:53:02 GMT
In article <44e84e$io2@turing.mathworks.com>,
bbalint wrote:
>Let G be the subset of Aut K(x) over field K consisting of the 3
>automorphisms induced by
>
>x -> x
>x -> 1/(1-x)
>x -> (x-1)/x.
>
>Determine the fixed field of G.
I'll do this more generally for actions on any field K1.
The way I usually find fixed fields is to create a lot of elements of
it, let L be the field they generate, and then note that K1 is
an L[G]-module, that is, it's a vector space over L on which G acts.
If you found enough elements for L, then K1 will be finite dimensional
over L, so you just want to look for the invariant subspace under the
action of G. If this is the span of basis vectors b_1, ..., b_n, then
the fixed field is the field extension F=L[b_1, ..., b_n]. (Note: although
the b_i may all be needed to express F as an L-module, it is often
the case that F is generated by fewer of them as an L-algebra.) If
you were very lucky the first time around, of course, you'll have F=L.
Finding things that are invariant under the action of G is easy. In
addition to traces and norms you can more generally take any symmetric
polynomial in {x, g(x), h(x), ...} where {1, g, h, ...} run over all
the elements of G (or over cosets G/Stab(x)).
In your case, you could take the sum of those three images of x to get
the fixed-field element X1=(x^3-3x+1)/[(x-1)x] . You could jump right
in here with L=K( X1 ), which wouldn't be a bad idea since L clearly
has transcendence degree 1 over K, and so K(x) will have to be
a finite-dimensional extension of L. Actually you'll find it's 6-dimensional,
though, and that the action of G on here is the sum of two permutation
modules of Z/3Z. That will give a 2-dimensional fixed subspace --clearly
the element 1 of K(x) accounts for one of these dimensions, but
there's another one too.
But I think I'd make L a little bigger first. The product of the three
images of x is 1, which is already in L, but the other symmetric
function x*g(x) + g(x)*h(x) + h(x)*x is not. It comes out to
X2 = (x^2-x-2)/x
So now let L=K( X1, X2 ); by construction it's contained in F, but
I'll show it's all of F.
Indeed, K(x) itself is of course an extension of L generated by x,
and it's even a finite-dimensional extension and, even better, I know
the minimal polynomial of x over L: it's just
X^3 - (X1) X^2 + (X2) X - (1) = 0
using the standard relation between polynomial coefficients and symmetric
functions of roots. So [K(x) : L] <=3, but since G acts non-trivially
on K(x), [K(x) : L ] >= [K(x) : F ] = 3. These inequalities force
L = F.
So the fixed field has to be the subfield of K(x) generated over K by the
elements X1=(x^3-3x+1)/[(x-1)x] and X2 = (x^2-x-2)/x.
dave
Exercise: as a byproduct of this construction, we see X2 must be quadratic
over K( X1 ). Find its minimal polynomial.
==============================================================================
From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Algebra Help
Date: 28 Sep 1995 20:13:54 GMT
In article <44e84e$io2@turing.mathworks.com>,
bbalint wrote:
>Let G be the subset of Aut K(x) over field K consisting of the 3
>automorphisms induced by
>
>x -> x
>x -> 1/(1-x)
>x -> (x-1)/x.
>
>Determine the fixed field of G.
Some idiot posted a bunch of hogwash in response, apparently a result
of miscalculating the second symmetric polynomial. That many of the
resulting comments don't hold water can be seen by reflecting for just
a little longer than it takes to hit the "send" key.
Let me try to rewrite it more sensibly.
I'll show how to find the fixed fields under the action of a gorup
G on a field K1.
The way I usually find fixed fields is to create a lot of elements of
it, let L be the field they generate, and then note that K1 is
an L[G]-module, that is, it's a vector space over L on which G acts.
If you found enough elements for L, then K1 will be finite dimensional
over L, so you just want to look for the invariant subspace under the
action of G. If this is the span of basis vectors b_1, ..., b_n, then
the fixed field is the field extension F=L[b_1, ..., b_n]. (Note: although
the b_i may all be needed to express F as an L-module, it is often
the case that F is generated by fewer of them as an L-algebra.) If
you were very lucky the first time around, of course, you'll have F=L.
Finding things that are invariant under the action of G is easy. In
addition to traces and norms you can more generally take any symmetric
polynomial in {x, g(x), h(x), ...} where {1, g, h, ...} run over all
the elements of G (or over cosets G/Stab(x)).
In your case, you could take the sum of those three images of x to get
the fixed-field element X1=(x^3-3x+1)/[(x-1)x] . You could jump right
in here with L=K( X1 ), which wouldn't be a bad idea since L clearly
has transcendence degree 1 over K, and so K(x) will have to be
a finite-dimensional extension of L. Indeed, you would find it's
3-dimensional, and that the action of G on here is a permutation
module of Z/3Z. That will give a 1-dimensional fixed subspace, clearly
spanned by the element 1 of K(x).
But rather than thinking about the representation space, look to see what
happens with some other invariant elements. The product X3 of the three
images of x is 1, which is already in L, and the other symmetric
function x*g(x) + g(x)*h(x) + h(x)*x comes out to
X2 = (x^3-3x^2+1)/[(x-1)x] = X1 - 3
So L=K( X1, X2, X3 ) is just K( X1 ). By construction L is contained
in F, but I'll show it's all of F.
Indeed, K(x) itself is of course an extension of L generated by x;
it's even a finite-dimensional extension and, even better, I know
the minimal polynomial of x over L: it's just
X^3 - (X1) X^2 + (X2) X - (X3) = 0
using the standard relation between polynomial coefficients and symmetric
functions of roots. So [K(x) : L] <=3, but since G acts non-trivially
on K(x), [K(x) : L ] >= [K(x) : F ] = 3. These inequalities force
L = F.
(Alternatively, notice that the definition of X1 can be turned around to
solve for x in terms of X1 by solving a cubic equation in x.)
So the fixed field has to be the subfield of K(x) generated over K by the
element X1=(x^3-3x+1)/[(x-1)x].
dave
That other guy should be silenced.