determine the galois group of p(x) ... over Q?
Daniel C. Bastos
(x^2 - 2)(x^2 - 3)(x^2 - 5).
Over which field? I'm assuming Q.
Solution. I first look for a splitting field. The polynomial
splits in Q(sqrt{2}, sqrt{3}, sqrt{5}).
Now, let s be in G(Q(sqrt{2}, sqrt{3}, sqrt{5})/Q). To
determine s, we must know s(sqrt{2}), s(sqrt{3}),
s(sqrt{5})? If so, why?
The possibilities for s(sqrt{3}) are the roots of the
polynomial? Only the roots? I believe so, but I haven't
clearly understood why yet.
Daniel C. Bastos
solved the exercise, but now I'm working on a second part
which I am not completely tuned yet.
In article <20060424232927.000007d4@Saturn>,
Daniel C. Bastos wrote:
> Exercise. Determine the Galois group of
>
> (x^2 - 2)(x^2 - 3)(x^2 - 5).
>
> Over which field? I'm assuming Q.
>
> Solution. I first look for a splitting field. The polynomial
> splits in Q(sqrt{2}, sqrt{3}, sqrt{5}).
Let s be arbitrary in G(Q(sqrt{2}, sqrt{3}, sqrt{5})/Q), and
let r = sqrt{2}. Then, the splitting field must also split
r's minimal polynomial which is r^2 - 2 = 0, and so s must
take sqrt{2} to sqrt{2} or - sqrt{2}. Let n = sqrt{3}, so s
must take sqrt{3} to sqrt{3} or - sqrt{3}. Let m = sqrt{5},
so s must take sqrt{5} to sqrt{5} or - sqrt{5}.
So, the automorphisms are
sqrt{2} sqrt{3} sqrt{5}
1 sqrt{2} sqrt{3} sqrt{5}
s sqrt{2} sqrt{3} -sqrt{5}
t sqrt{2} -sqrt{3} sqrt{5}
st sqrt{2} -sqrt{3} -sqrt{5}
g -sqrt{2} sqrt{3} sqrt{5}
gs -sqrt{2} sqrt{3} -sqrt{5}
gt -sqrt{2} -sqrt{3} sqrt{5}
gst -sqrt{2} -sqrt{3} -sqrt{5}
So the galois group is determined, and it is completely
determined by s: sqrt{5} to -sqrt{5}, t: sqrt{3} to -sqrt{3}
and g: sqrt{2} to - sqrt{2}.
Now if the above is all correct, then I would work on the
following exercise.
Exercise. Determine all subfields of the splitting field,
found in the previous exercise, and build a field tree.
The term ``field tree'' is being given by myself; what I
mean is one of those diagrams of extension fields.
Solution. We already have the trivial subfields; they will
be at the top and bottom of the field tree. So, we're only
looking for the fixed fields that are contained in the
splitting field found above.
Let u = sqrt{2}, v = sqrt{3}, w = sqrt{5}. A basis for the
splittig field is
a + bu + cv + dw + e(uv) + f(uw) + g(vw) + h(uwv).
Now I seem to lack a nice way to go on. I started
considering subfields like H = {1, s} which moves w to -w,
and fixes Q(u,v). Then I considered H = {e, t} which moves v
to -v, and fixes Q(u,w). So, I tried to find a pattern on
the discovery and I started to draw a tree like this
Q(u,v,w)
Q(u,v) Q(u,w) Q(u,vw) Q(v,w) Q(v,uw) Q(w,uv)
I'm missing the arrows connecting the fields. I'm also
missing some knowledge to be sure that I've written the
second level completely. Here are my questions:
I wrote 6 subfields for the second level. By ``second
level'' I mean subfields in which we adjoin two roots. How
would I know that 6 is the total? Is 6 the total?
We don't write the subfields more than once in any of level,
right? So once I write ``Q(sqrt{2}) = Q(u)'' in the third
level, then I shouldn't write it anywhere else.
Paul Sperry
<dba...@yahoo.com.br> wrote:
Good for you. I was just about to post a hint or two but you didn't
need it.
The Galois group is isomorphic to Z_2 x Z_2 x Z_2. In fact, with your
notation The Galois group is <s><t><g>
> Now if the above is all correct, then I would work on the
> following exercise.
>
> Exercise. Determine all subfields of the splitting field,
> found in the previous exercise, and build a field tree.
[...]
It might be easier to get the subgroup lattice of Z_2 x Z_2 x Z_2,
convert that to the subgroup lattice of the Galois group and then find
the fixed fields.
--
Paul Sperry
Columbia, SC (USA)
Daniel C. Bastos
Paul Sperry wrote:
I think I don't know how to get the lattice of any group. I
see that the Fundamental Theorem would allow me to more
easily link the two things, but I am not sure why the
lattice is built and what it means. What do the arrows mean?
Paul Sperry
<dba...@yahoo.com.br> wrote:
What I had in mind is surely not elegant.
First, for Z_2 x Z_2 x Z_2, calculate all of the subgroups more or less
by brute force. For example {(0, 1, 0), (1, 0, 1)} generates
{(0, 1, 0), (1, 0, 1), (1, 1, 1), (0, 0, 0)}. I get 17 distinct
subgroups - I may have missed some.
Now associate (1, 0, 0) with s, (0, 1, 0) with t and (0, 0, 1) with g.
The above says that {t, sg} generates {t, sg, stg, 1}. My reason for
using Z_2 x Z_2 x Z_2 is just because the arithmetic is more familiar.
Finally, figure out what is left fixed by all four of t, sg, stg, and
1. Note sqrt(3)sqrt(5) is fixed although neither sqrt(3) nor sqrt(5)
is. By my calculations the fixed field is Q(sqrt(10)).
Finally note that the fixed fields reverse the ordering of the groups:
the larger the group, the smaller the fixed field. So, the largest
group, G(Q,K), gives the smallest field: Q. The smallest group, {1},
gives the largest field: K.
Unless there is a cleverer way to do all of this, it is a tedious
problem. Good luck.
Daniel C. Bastos
Paul Sperry wrote:
17? Hmm, I don't work so easily with Z_2 x Z_2 x Z_2, but I
think that, besides the trivial subfields, these will be
subfields of Q(sqrt{2}, sqrt{3}, sqrt{5}):
Q(sqrt{2},sqrt{3}), Q(sqrt{2},sqrt{5}), Q(sqrt{3}, sqrt{5}),
Q(sqrt{2},sqrt{15}), Q(sqrt{6},sqrt{5}), Q(sqrt{3},sqrt{10})
Also, these will be subfields: Q(sqrt{2}), Q(sqrt{3}),
Q(sqrt{5}), Q(sqrt{6}), Q(sqrt{10}), Q(sqrt{15}),
Q(sqrt{30}).
With the trivial ones, I get 15 total, but you get at least
17 subgroups and we should have a one-to-one and onto
correspondence, but I can't seem to see how another subfield
of Q(sqrt{2},sqrt{3},sqrt{5}) wouldn't be the same as one of
the above.
> Now associate (1, 0, 0) with s, (0, 1, 0) with t and (0, 0, 1) with g.
> The above says that {t, sg} generates {t, sg, stg, 1}. My reason for
> using Z_2 x Z_2 x Z_2 is just because the arithmetic is more familiar.
>
> Finally, figure out what is left fixed by all four of t, sg, stg, and
> 1. Note sqrt(3)sqrt(5) is fixed although neither sqrt(3) nor sqrt(5)
> is. By my calculations the fixed field is Q(sqrt(10)).
Oh, I see something here that I hadn't seen before for some
reason. *The* fixed field is the field that is fixed by all
automorphisms of a Galois group. Why would it be Q(sqrt{10])
though?
Here's how I'm trying to answer this. I look at the bases
and consider the automorphisms. So here's what I get for s
s(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
+ f sqrt{10} + g sqrt{15} + h sqrt{30}) =
a + b sqrt{2} + c sqrt{3} - d sqrt{5} + e sqrt{6}
- f sqrt{10} - g sqrt{15} - h sqrt{30}),
because s moves sqrt{5} to -sqrt{5}. But I already see a
problem here. I moved sqrt{10}, didn't I? Here's why I wrote
- f sqrt{10}:
s(f sqrt{10}) = s(f sqrt{2} sqrt{5})
= s(f) s(sqrt{2}) s(sqrt{5})
= (f sqrt{2})(-sqrt{5})
= - f sqrt{10}.
> Finally note that the fixed fields reverse the ordering of the groups:
> the larger the group, the smaller the fixed field. So, the largest
> group, G(Q,K), gives the smallest field: Q. The smallest group, {1},
> gives the largest field: K.
Right. I agree.
[...]
Paul Sperry
<dba...@yahoo.com.br> wrote:
In fact, I counted one twice.
> 17? Hmm, I don't work so easily with Z_2 x Z_2 x Z_2, but I
> think that, besides the trivial subfields, these will be
> subfields of Q(sqrt{2}, sqrt{3}, sqrt{5}):
>
> Q(sqrt{2},sqrt{3}), Q(sqrt{2},sqrt{5}), Q(sqrt{3}, sqrt{5}),
> Q(sqrt{2},sqrt{15}), Q(sqrt{6},sqrt{5}), Q(sqrt{3},sqrt{10})
>
> Also, these will be subfields: Q(sqrt{2}), Q(sqrt{3}),
> Q(sqrt{5}), Q(sqrt{6}), Q(sqrt{10}), Q(sqrt{15}),
> Q(sqrt{30}).
>
> With the trivial ones, I get 15 total, but you get at least
> 17 subgroups and we should have a one-to-one and onto
> correspondence, but I can't seem to see how another subfield
> of Q(sqrt{2},sqrt{3},sqrt{5}) wouldn't be the same as one of
> the above.
>
> > Now associate (1, 0, 0) with s, (0, 1, 0) with t and (0, 0, 1) with g.
> > The above says that {t, sg} generates {t, sg, stg, 1}. My reason for
> > using Z_2 x Z_2 x Z_2 is just because the arithmetic is more familiar.
> >
> > Finally, figure out what is left fixed by all four of t, sg, stg, and
> > 1. Note sqrt(3)sqrt(5) is fixed although neither sqrt(3) nor sqrt(5)
> > is. By my calculations the fixed field is Q(sqrt(10)).
>
> Oh, I see something here that I hadn't seen before for some
> reason. *The* fixed field is the field that is fixed by all
> automorphisms of a Galois group. Why would it be Q(sqrt{10])
> though?
Arithmetic malfunction.
> Here's how I'm trying to answer this. I look at the bases
> and consider the automorphisms. So here's what I get for s
>
> s(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
> + f sqrt{10} + g sqrt{15} + h sqrt{30}) =
>
> a + b sqrt{2} + c sqrt{3} - d sqrt{5} + e sqrt{6}
> - f sqrt{10} - g sqrt{15} - h sqrt{30}),
>
> because s moves sqrt{5} to -sqrt{5}. But I already see a
> problem here. I moved sqrt{10}, didn't I? Here's why I wrote
> - f sqrt{10}:
>
> s(f sqrt{10}) = s(f sqrt{2} sqrt{5})
> = s(f) s(sqrt{2}) s(sqrt{5})
> = (f sqrt{2})(-sqrt{5})
> = - f sqrt{10}.
Here, in no particular order, are the non-trivial subgroups of
Z_2 x Z_2 x Z_2. To save typing, I've left (0, 0, 0) out of all of
them. I've associated (1, 0 ,0) with g, (0, 1, 0) with t and
(0, 0, 1) with s and give the fixed field - note the one you left out.
(1, 0, 0) <-> Q(sqrt(3), sqrt(5))
(0, 1, 0) <-> Q(sqrt(2), sqrt(5))
(0, 0, 1) <-> Q(sqrt(2), sqrt(3))
(1, 1, 0) <-> Q(sqrt(5), sqrt(6))
(1, 0, 1) <-> Q(sqrt(3), sqrt(10))
(0, 1, 1) <-> Q(sqrt(2), sqrt(15))
(1, 1, 1) <-> Q(sqrt(6), sqrt(10)) <-------------
(1, 0, 0), (0, 1, 0), (1, 1, 0) <-> Q(sqrt(5))
(1, 0, 0), (0, 0, 1), (1, 0, 1) <-> Q(sqrt(3))
(1, 0, 0), (0, 1, 1), (1, 1, 1) <-> Q(sqrt(15))
(0, 1, 1), (0, 0, 1), (0, 1, 0) <-> Q(sqrt(2))
(0, 1, 0), (1, 0, 1), (1, 1, 1) <-> Q(sqrt(10))
(0, 0, 1), (1, 1, 1), (1, 1, 0) <-> Q(sqrt(6))
(1, 1, 0), (1, 0, 1), (0, 1, 1) <-> Q(sqrt(30))
> > Finally note that the fixed fields reverse the ordering of the groups:
> > the larger the group, the smaller the fixed field. So, the largest
> > group, G(Q,K), gives the smallest field: Q. The smallest group, {1},
> > gives the largest field: K.
>
> Right. I agree.
>
> [...]
--
Daniel C. Bastos
Paul Sperry wrote:
You chose these mappings wisely, right? We know that the map
s moves only sqrt{5}, so you assigned (0, 0, 1) to s because
if we add an arbitrary element of Z_2 x Z_2 x Z_2 to (0, 0,
1), then the only component of the 3-tuple that will move is
the last one --- sort of moving sqrt{5}. Similarly for t and
g. Is this how you chose them?
> (1, 0, 0) <-> Q(sqrt(3), sqrt(5))
> (0, 1, 0) <-> Q(sqrt(2), sqrt(5))
> (0, 0, 1) <-> Q(sqrt(2), sqrt(3))
> (1, 1, 0) <-> Q(sqrt(5), sqrt(6))
> (1, 0, 1) <-> Q(sqrt(3), sqrt(10))
> (0, 1, 1) <-> Q(sqrt(2), sqrt(15))
> (1, 1, 1) <-> Q(sqrt(6), sqrt(10)) <-------------
> (1, 0, 0), (0, 1, 0), (1, 1, 0) <-> Q(sqrt(5))
> (1, 0, 0), (0, 0, 1), (1, 0, 1) <-> Q(sqrt(3))
> (1, 0, 0), (0, 1, 1), (1, 1, 1) <-> Q(sqrt(15))
> (0, 1, 1), (0, 0, 1), (0, 1, 0) <-> Q(sqrt(2))
> (0, 1, 0), (1, 0, 1), (1, 1, 1) <-> Q(sqrt(10))
> (0, 0, 1), (1, 1, 1), (1, 1, 0) <-> Q(sqrt(6))
> (1, 1, 0), (1, 0, 1), (0, 1, 1) <-> Q(sqrt(30))
What would now be the fixed field? In my computations here I
find that the only fixed field by each map is just Q. Here's
each of them; it's easy to check because soon enough we find
that all radicals are getting moved.
s(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
+ f sqrt{10} + g sqrt{15} + h sqrt{30}) =
a + b sqrt{2} + c sqrt{3} - d sqrt{5} + e sqrt{6}
- f sqrt{10} - g sqrt{15} - h sqrt{30}),
t(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
+ f sqrt{10} + g sqrt{15} + h sqrt{30}) =
a + b sqrt{2} - c sqrt{3} + d sqrt{5} - e sqrt{6}
+ f sqrt{10} - g sqrt{15} - h sqrt{30}),
st(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
+ f sqrt{10} + g sqrt{15} + h sqrt{30}) =
a + b sqrt{2} - c sqrt{3} - d sqrt{5} - e sqrt{6}
- f sqrt{10} + g sqrt{15} + h sqrt{30},
g(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
+ f sqrt{10} + g sqrt{15} + h sqrt{30}) =
a - b sqrt{2} + c sqrt{3} + d sqrt{5} - e sqrt{6}
- f sqrt{10} + g sqrt{15} - h sqrt{30},
gs(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
+ f sqrt{10} + g sqrt{15} + h sqrt{30}) =
a - b sqrt{2} + c sqrt{3} - d sqrt{5} - e sqrt{6}
+ f sqrt{10} - g sqrt{15} + h sqrt{30},
gst(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
+ f sqrt{10} + g sqrt{15} + h sqrt{30}) =
a - b sqrt{2} - c sqrt{3} - d sqrt{5} + e sqrt{6}
+ f sqrt{10} + g sqrt{15} - h sqrt{30}.
So the fixed field of G(Q(sqrt{2},sqrt{3},sqrt{5})/Q) is Q.
I think I understand now why ``the'' fixed field; we have
only one because in the end whatever components didn't get
moved are adjoined in the defined fixed field of the Galois
group to form one single fixed field of each automorphism of
the Galois group.
Paul Sperry
<dba...@yahoo.com.br> wrote:
[...]
> > Here, in no particular order, are the non-trivial subgroups of
> > Z_2 x Z_2 x Z_2. To save typing, I've left (0, 0, 0) out of all of
> > them. I've associated (1, 0 ,0) with g, (0, 1, 0) with t and
> > (0, 0, 1) with s and give the fixed field - note the one you left out.
>
> You chose these mappings wisely, right? We know that the map
> s moves only sqrt{5}, so you assigned (0, 0, 1) to s because
> if we add an arbitrary element of Z_2 x Z_2 x Z_2 to (0, 0,
> 1), then the only component of the 3-tuple that will move is
> the last one --- sort of moving sqrt{5}. Similarly for t and
> g. Is this how you chose them?
You can associate (1, 0, 0), (0, 1, 0), (0, 0, 1) with any three
independent automorphisms. I picked the associations just to make it
easy to remember what was associated with what and what the associated
automorphism moved.
> > (1, 0, 0) <-> Q(sqrt(3), sqrt(5))
> > (0, 1, 0) <-> Q(sqrt(2), sqrt(5))
> > (0, 0, 1) <-> Q(sqrt(2), sqrt(3))
> > (1, 1, 0) <-> Q(sqrt(5), sqrt(6))
> > (1, 0, 1) <-> Q(sqrt(3), sqrt(10))
> > (0, 1, 1) <-> Q(sqrt(2), sqrt(15))
> > (1, 1, 1) <-> Q(sqrt(6), sqrt(10)) <-------------
> > (1, 0, 0), (0, 1, 0), (1, 1, 0) <-> Q(sqrt(5))
> > (1, 0, 0), (0, 0, 1), (1, 0, 1) <-> Q(sqrt(3))
> > (1, 0, 0), (0, 1, 1), (1, 1, 1) <-> Q(sqrt(15))
> > (0, 1, 1), (0, 0, 1), (0, 1, 0) <-> Q(sqrt(2))
> > (0, 1, 0), (1, 0, 1), (1, 1, 1) <-> Q(sqrt(10))
> > (0, 0, 1), (1, 1, 1), (1, 1, 0) <-> Q(sqrt(6))
> > (1, 1, 0), (1, 0, 1), (0, 1, 1) <-> Q(sqrt(30))
>
> What would now be the fixed field? In my computations here I
> find that the only fixed field by each map is just Q. Here's
> each of them; it's easy to check because soon enough we find
> that all radicals are getting moved.
>
> s(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
> + f sqrt{10} + g sqrt{15} + h sqrt{30}) =
>
> a + b sqrt{2} + c sqrt{3} - d sqrt{5} + e sqrt{6}
> - f sqrt{10} - g sqrt{15} - h sqrt{30}),
>
So, what gets fixed by s has d = f = g = h = 0. In other words:
a + b*sqrt(2) + c*sqrt(3 + e*sqrt(6) which are the elements of
Q(sqrt(2), sqrt(3)). [ See (0, 0, 1) above.]
> t(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
> + f sqrt{10} + g sqrt{15} + h sqrt{30}) =
>
> a + b sqrt{2} - c sqrt{3} + d sqrt{5} - e sqrt{6}
> + f sqrt{10} - g sqrt{15} - h sqrt{30}),
>
> st(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
> + f sqrt{10} + g sqrt{15} + h sqrt{30}) =
>
> a + b sqrt{2} - c sqrt{3} - d sqrt{5} - e sqrt{6}
> - f sqrt{10} + g sqrt{15} + h sqrt{30},
>
> g(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
> + f sqrt{10} + g sqrt{15} + h sqrt{30}) =
>
> a - b sqrt{2} + c sqrt{3} + d sqrt{5} - e sqrt{6}
> - f sqrt{10} + g sqrt{15} - h sqrt{30},
>
> gs(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
> + f sqrt{10} + g sqrt{15} + h sqrt{30}) =
>
> a - b sqrt{2} + c sqrt{3} - d sqrt{5} - e sqrt{6}
> + f sqrt{10} - g sqrt{15} + h sqrt{30},
>
> gst(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
> + f sqrt{10} + g sqrt{15} + h sqrt{30}) =
>
> a - b sqrt{2} - c sqrt{3} - d sqrt{5} + e sqrt{6}
> + f sqrt{10} + g sqrt{15} - h sqrt{30}.
>
> So the fixed field of G(Q(sqrt{2},sqrt{3},sqrt{5})/Q) is Q.
Yes. Q(sqrt(2), sqrt(3), sqrt(5)) is normal (it is a splitting field)
> I think I understand now why ``the'' fixed field; we have
> only one because in the end whatever components didn't get
> moved are adjoined in the defined fixed field of the Galois
> group to form one single fixed field of each automorphism of
> the Galois group.
I'm not sure what you mean here but I'm not convinced that you've got
the idea yet. We have _16_ distinct fixed fields corresponding to the
16 subgroups of the Galois group. In other words, there are 16 fields
between Q and Q(sqrt(2), sqrt(3), sqrt(5)) (inclusive).
Daniel C. Bastos
Paul Sperry wrote:
Okay, sure. The three of them (the independent ones) square
to the identity, so we could associate them differently; for
example, let (0,0,1) be t instead of s, but I think I
understood your choice: we placed sqrt{5} at the last column,
so you assigned (0,0,1) to s because s moves sqrt{5}.
Okay, s fixes Q(sqrt{2}, sqrt{3}). Oh, I see, this is how we
figure the correspondence. By ``see (0, 0, 1) above'' you
probably mean: notice that the subgroup {(0, 0, 0), (0, 0, 1)}
corresponds to the subfield Q(sqrt{2}, sqrt{3}) because
(0, 0, 1) is assigned to s and s fixes Q(sqrt{2}, sqrt{3}),
and the identity evidently fixes anything.
So, in summary: we take each subgroup and we look at their
corresponding automorphisms; whatever subfield the
automorphisms fix, we assign a link between the fixed
subfield and the subgroup we're working with.
So, the more automorphisms we have, the smaller will be the
corresponding fixed field, as suggests your correspondence
above --- and as the fundamental theorem of galois theory
claims.
After I finished this message I realize that I wasn't
completely aware of the definition of fixed field. I
initially thought that term ``the fixed field'' was always
meant to be used with the whole Galois group, and so I
thought that fixed field had slightly different meanings in
different contexts.
Definition. Let H be a group of automorphisms of a field
D. The subfield E of D of all elements that are fixed by
every automorphism in H is called the fixed field of H.
So, {1, s} is just as good of a group as
G(Q(sqrt{2}, sqrt{3}, sqrt{5})/Q),
so Q(sqrt{2}, sqrt{3}) is the fixed field of {1, s} just
like Q is the fixed field of
G(Q(sqrt{2}, sqrt{3}, sqrt{5})/Q).
Because of the number of different fixed fields that we see
which varies with the subgroups we deal with, I was thinking
that the article ``the'' used for ``fixed field'' had to be
meant for one single fixed field, and somehow I assumed that
this single fixed field was the one correspondent to the
whole Galois group, but actually each group G, subgroup of
G(E/F), has one unique fixed field.
>
> > t(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
> > + f sqrt{10} + g sqrt{15} + h sqrt{30}) =
> >
> > a + b sqrt{2} - c sqrt{3} + d sqrt{5} - e sqrt{6}
> > + f sqrt{10} - g sqrt{15} - h sqrt{30}),
> >
> > st(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
> > + f sqrt{10} + g sqrt{15} + h sqrt{30}) =
> >
> > a + b sqrt{2} - c sqrt{3} - d sqrt{5} - e sqrt{6}
> > - f sqrt{10} + g sqrt{15} + h sqrt{30},
> >
> > g(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
> > + f sqrt{10} + g sqrt{15} + h sqrt{30}) =
> >
> > a - b sqrt{2} + c sqrt{3} + d sqrt{5} - e sqrt{6}
> > - f sqrt{10} + g sqrt{15} - h sqrt{30},
> >
> > gs(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
> > + f sqrt{10} + g sqrt{15} + h sqrt{30}) =
> >
> > a - b sqrt{2} + c sqrt{3} - d sqrt{5} - e sqrt{6}
> > + f sqrt{10} - g sqrt{15} + h sqrt{30},
> >
> > gst(a + b sqrt{2} + c sqrt{3} + d sqrt{5} + e sqrt{6}
> > + f sqrt{10} + g sqrt{15} + h sqrt{30}) =
> >
> > a - b sqrt{2} - c sqrt{3} - d sqrt{5} + e sqrt{6}
> > + f sqrt{10} + g sqrt{15} - h sqrt{30}.
> >
> > So the fixed field of G(Q(sqrt{2},sqrt{3},sqrt{5})/Q) is Q.
>
> Yes. Q(sqrt(2), sqrt(3), sqrt(5)) is normal (it is a splitting field)
Right.
> > I think I understand now why ``the'' fixed field; we have
> > only one because in the end whatever components didn't get
> > moved are adjoined in the defined fixed field of the Galois
> > group to form one single fixed field of each automorphism of
> > the Galois group.
>
> I'm not sure what you mean here but I'm not convinced that you've got
> the idea yet. We have _16_ distinct fixed fields corresponding to the
> 16 subgroups of the Galois group. In other words, there are 16 fields
> between Q and Q(sqrt(2), sqrt(3), sqrt(5)) (inclusive).
Right. I hadn't. Fixed fields are linked to subgroups of a
Galois group, so anytime we speak of a fixed field, we're
talking about *the* fixed *of* some subgroup of a Galois
group. I'm confident that I have the right idea now. Do you
see a problem in my words somewhere? Thanks a lot for your
help.
Paul Sperry
<dba...@yahoo.com.br> wrote:
[...]
I think you've got it now.