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?
More generally, what does intermediate field look like corresponding a subgroup of S_n?
On Nov 19, 1:50 pm, Kusanagi <Kusan...@hotmail.com> wrote:
> 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.
On Nov 19, 1:50 pm, Kusanagi <Kusan...@hotmail.com> wrote:
> 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 .)
> On Nov 19, 1:50 pm, Kusanagi <Kusan...@hotmail.com> > wrote:
> > 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
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.
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)?
On Nov 19, 11:10 pm, Kusanagi <Kusan...@hotmail.com> asked about 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.
> On Nov 19, 11:10 pm, Kusanagi <Kusan...@hotmail.com> > asked about > 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
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.
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.
> 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)?
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).
> > 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).
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