Symmetric functions, Galois groups, and intermediate fields
Kusanagi
Then
[F(x_1, x_2, ... ,x_n) : F(s_1, s_2, ...,s_n)] = n! with Galois group S_n.
I am looking for intermediate fields corresponding each subgroup of Galois group of S_n with small order.
For instance, S_n has a cyclic subgroup C_n of order n. By the fundamental theorem of Galois theory, this subgroup corresponds to the intermediate field between F(x_1, x_2, ... ,x_n) and F(s_1, s_2, ...,s_n). What does this intermediate field look like?
More generally, what does intermediate field look like corresponding a subgroup of S_n?
Thanks.
Arturo Magidin
> Let F(x_1, x_2, ... , x_n) be the splitting field over F(s_1, s_2, ... , s_n) of the polynomial of degree n, where x_1, x_2,..,x_n are indeterminates of the general polynomial of degree n and s_1, s_2, .. , s_n are elementary symmetric functions for indeterminates x_1, x_2,..., x_n(http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial#Examples)
>
> Then
>
> [F(x_1, x_2, ... ,x_n) : F(s_1, s_2, ...,s_n)] = n! with Galois group S_n.
>
> I am looking for intermediate fields corresponding each subgroup of Galois group of S_n with small order.
>
> For instance, S_n has a cyclic subgroup C_n of order n. By the fundamental theorem of Galois theory, this subgroup corresponds to the intermediate field between F(x_1, x_2, ... ,x_n) and F(s_1, s_2, ...,s_n). What does this intermediate field look like?
The Fundamental Theorem tells you exactly what it is.
The action of S_n on F(x_1,...,x_n) is the obvious one: if sigma is a
permutation, then sigma(x_i) = x_{sigma(i)}. Given a subgroup H of
S_n, the field that corresponds to H is the subfield of F(x_1,...,x_n)
of all elements that are fixed by each element of H.
A cyclic subgroup of order n is conjugate to the one generated by the
permutation (1,2,3,...,n). So you want to know what is the fixed field
of (1,2,3,...,n). This will be precisely those elements of F
(x_1,...,x_n) which map to themselves under the permutation
(1,2,3,...,n). So for example, it will contain x_1x_2 + x_2x_3+ ... +
x_(n-1)x_n + x_nx_1, which is not a symmetric polynomial in the x_i
(hence not an element of F(s_1,..,s_n)) (it is not fixed by the
permutation (1,2), for example).
> More generally, what does intermediate field look like corresponding a subgroup of S_n?
Take the generators of the subgroup, find the polynomials in
x_1,..,x_n that are fixed by the generators.
--
Arturo Magidin
dave.rusin
> Let F(x_1, x_2, ... , x_n) be the splitting field over F(s_1, s_2,
> ... , s_n) of the polynomial of degree n, where x_1, x_2,..,x_n
> are indeterminates of the general polynomial of degree n and s_1,
> s_2, .. , s_n are elementary symmetric functions for
> indeterminates x_1, x_2,...,x_n
> I am looking for intermediate fields corresponding each subgroup
> of Galois group of S_n with small order.
> For instance, S_n has a cyclic subgroup C_n of order n. By the
> fundamental theorem of Galois theory, this subgroup corresponds to
> the intermediate field between F(x_1, x_2, ... ,x_n) and
> F(s_1, s_2, ...,s_n). What does this intermediate field look like?
I think you mean something like this: when n=3, the big field is
F(x,y,z) and the small field is F(r,s,t) where r=x+y+z,
s=xy+yz+zx, and t=xyz, that is, these are rational functions of
x,y,z which are invariant under all of S_3, and any invariant
rational function is a rational function of these three.
Now you want the intermediate field K of rational functions
invariant under the three-cycle (123) (but not necessarily invariant
under the transposition (12) ). This field K includes F(r,s,t) but
is (by Galois theory) a quadratic extension thereof: you should be
able to recognize K as F(r,s,t)[ X^(1/2) ] for some rational
function X of r,s,t .
Indeed in this particular case (or more generally when looking for
the intermediate field K associated to the alternating subgroup A_n
of S_n ), it is known how to construct X using the idea of the
discriminant: if you form the product
D = (x-y)(y-z)(x-z)
(i.e. D = prod_{i<j} (x_i - x_j) in F(x_1, x_2, ..., x_n) ) then
this expression is not itself invariant under transpositions, but
its square X = D^2 is. So this is the thing you want to adjoin the
square root of to get the quadratic extension of F(r,s,t) . You can
figure out how to express this S_3-invariant expression explicitly:
X = -t*(27*t+4*r^3) + 18*t*r*s + r^2*s^2 - 4*s^3 .
But for more general subgroups of S_n it can take a little more
work to identify a generating set of invariants, and then to find
algebraic relations among them, and then (if desired) to find a
minimal set of generating invariants.
In general the buzzword you need is "invariant theory": you are
looking for invariants for the action of subgroups of S_n on the
function field F(x_1, ..., x_n). One can also ask for the
invariants of the action on the polynomial ring F[x_1, ..., x_n] ;
that's harder because in general that invariant ring is not a pure
polynomial extension of F. (I'm pretty sure that's true even for
A_3 , as above: you can't find three algebraically independent
polynomials which generate the same subring of F[x,y,z] that is
generated by r, s, t, and D .)
dave
Kusanagi
Let D = (x-y)(y-z)(x-z) such that D = prod_{i<j} (x_i - x_j) in F(x_1, x_2, ..., x_n); let X=D^2 and consider E=F(r,s,t)[ X^(1/2) ].
Once we square D, it seems like every permutation is fixed.
identity : fixed
transpositions: fixed
3-cycles: fixed
Is it true that Gal(E:F(r,s,t)) = S_3? If this is wrong, could you explain what I am confused with?
Another questions is
Is it possible to find an intermediate field E explicitly such that Gal(E/F(r,s,t)) = Z_2, Z_3, and A_3 (alternating group of degree 3), where E is a subfield of F(x,y,z)?
Many thanks.
dave.rusin
identifying intermediate fields in the Galois correspondence,
in a specific case.
Recap: inside the function field F(x,y,z) we identify the subfield
F(r,s,t) of symmetric functions, where r=x+y+z, s=xy+yz+zx, t=xyz.
Then F(x,y,z) is a Galois extension of F(r,s,t) with galois group
S_3.
> Let D = (x-y)(y-z)(x-z) [...]
> let X=D^2 and consider E=F(r,s,t)[ X^(1/2) ].
> Once we square D, it seems like every permutation is fixed.
You mean: every permutation in S_3 fixes X=D^2. That's correct.
> Is it true that Gal(E:F(r,s,t)) = S_3?
No, you jumped the gun. You need the small field to be fixed
(element-wise) by the group, yes. But you also need the large field
to be fixed setwise by the group (it is), and you need to know that
everything not in the small field is moved by some element of the
group (it is), and you need every element of the group to move
_something_ in the big field (not true here).
Indeed, look at the definition of E: it's a quadratic extension of
F(r,s,t), hence a Galois extension with a Galois group of order 2.
Your mistake was to observe only that F(r,s,t) is fixed by all
of S_3, so that there is a homomorphism of S_3 _into_ the
Galois group you seek; but that homomorphism has a kernel, because
all of the alternating subgroup of S_3 fixes not only F(r,s,t)
but the larger field E as well.
> Is it possible to find an intermediate field E explicitly such
> that Gal(E/F(r,s,t)) = Z_2, Z_3, and A_3 (alternating group of
> degree 3), where E is a subfield of F(x,y,z)?
Um, what is "Z_3" if not A_3 ?
I guess you know about the Galois correspondence: the subgroups
of S_3 are in one-to-one correspondence with the intermediate
fields between F(x,y,z) and F(r,s,t). In this particular case
we can describe the correspondence explicitly. Here is a table
showing the different subgroups H of S_3 and the corresponding
subfields of F(x,y,z) :
H = {1} corresponds to F(x,y,z) itself
H = (12) corresponds to F( x+y, xy, z)
H = (13) corresponds to F( x+z, xz, y)
H = (23) corresponds to F( y+z, yz, x)
H = (123) corresponds to E = F( r,s,t, D )
H = S_3 corresponds to F( r, s, t)
You might want to check all the features of the Galois
correspondence: that H1 < H2 implies E1 > E2; that H1=normal
iff E / F(r,s,t) is Galois; and most importantly that in
each case the subfield consists precisely of all the elements
of F(x,y,z) that are invariant under the action of H.
dave
Kusanagi
Kusanagi
Hello, do you have any example of an irreducible polynomial in F(r, s, t) that splits in the field F( x+y, xy, z)? A field F( x+y, xy, z) here is invariant under the action of H=<(1,2)>, as shown in the previuos reply.
Thank you.
Kusanagi
Arturo Magidin
>
> Hello, do you have any example of an irreducible polynomial in F(r, s, t) that splits in the field F( x+y, xy, z)? A field F( x+y, xy, z) here is invariant under the action of H=<(1,2)>, as shown in the previuos reply.
If such a polynomial existed, then F(x+y,xy,z) would be Galois over F
(r,s,t). By the Fundamental Theorem of Galois Theory, this will occur
if and only if the subgroup H is normal in Gal(F(x,y,z)/F(r,s,t))=S_3.
But H is *not* normal in S_3, so no such polynomial can exist.
--
Arturo Magidin
Kusanagi
Oh, that's right.
Any example of an irreducible polynomial in F(r, s, t) that splits in the field F(r,s,t, D)?
A_3=Z_3=Gal(F(r,s,t,D)/F(r,s,t)) is normal in S_3=Gal(F(x,y,z)/F(r,s,t)) where D = (x-y)(y-z)(x-z).
Thanks.
Kusanagi
Arturo Magidin
> > On Nov 23, 9:37 pm, Kusanagi <Kusan...@hotmail.com>
> > wrote:
>
> > > Hello, do you have any example of an irreducible
> > polynomial in F(r, s, t) that splits in the field F(
> > x+y, xy, z)? A field F( x+y, xy, z) here is invariant
> > under the action of H=<(1,2)>, as shown in the
> > previuos reply.
>
> > If such a polynomial existed, then F(x+y,xy,z) would
> > be Galois over F
> > (r,s,t). By the Fundamental Theorem of Galois Theory,
> > this will occur
> > if and only if the subgroup H is normal in
> > Gal(F(x,y,z)/F(r,s,t))=S_3.
> > But H is *not* normal in S_3, so no such polynomial
> > can exist.
>
> > --
> > Arturo Magidin
>
> Oh, that's right.
> Any example of an irreducible polynomial in F(r, s, t) that splits in the field F(r,s,t, D)?
X^2 - D^2. Note that D^2 is in F(r,s,t), as was noted before, but D is
not. Since the two roots of this polynomial are D and -D, and neither
is in F(r,s,t), the polynomial is irreducible, but trivially splits in
F(r,s,t,D).
--
Arturo Magidin