constructible numbers
Kent Holing
Is it an easy way to prove that if alpha is an element in E and constuctible then the degree of alpha (over Q) is less or equal to 2?
Robert Israel
Since alpha is constructible, its minimal polynomial over Q
must have degree a power of 2. In this case alpha is contained in
E, where [E:Q] = 4, so that minimal polynomial is quadratic or
quartic. But if it was quartic, [Q(alpha):Q] = 4, which would imply
Q(alpha) = E, and then the members of E are all constructible.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Kent Holing
Kent Holing
The Galois group is either S4 or A4. The A4 case is easy. If there is a quadratic field extension of Q in E, say L, then [L:Q] = 2 and [E:L] = 6. This means that the subgroub H of G fixing L, i.e. H = G(E/L) must be of order 6. But A4 has no subgroup of order 6 so there is no intermediate quadratic field between Q and E. This means that the only constructible elements in E when G = A4 are the rationals.
Anybody, give the details for the case G = S4?
quasi
Then [E:Q] = 24.
Let L = {x in E | x is constructible}. Then L is a subfield of E.
Let r be one of the roots of the given quartic. The degree of r over L
can't be 1 or 2, otherwise r would be constructible, and the degree of
r over L can't be 3, otherwise one of the other roots would have
degree 1 over L (and hence be constructible). Therefore the degree of
r over L is 4, which implies [E:L] is a multiple of 4.
But [L:Q] is a power of 2, and [E:Q] = 24, so [E:L] is a multiple of
3.
But then [E:L] is a multiple of 12, hence [L:Q] = 1 or 2.
quasi
quasi
In fact, the same proof works for the case G = A4.
In that case, [E:Q] = 12, but by the same argument as above, we get
that [E:L] is a multiple of 12, hence [E:L] = 12, so L = Q.
quasi
Kent Holing
By the Galois correspondence and knowledge of the subgroups of S4 we easily see that L is the fixed field of A4, i.e.
G(E/L) = A4 so [E:L] = |A4| = 12, and therefore [E:Q] = 2.
So, L = Q if G = A4 and [E:L] = 2 (the fixed field of A4) if G = S4.
quasi
>quasi wrote:
>> Kent Holing wrote:
>By the Galois correspondence and knowledge of the subgroups
>of S4 we easily see that L is the fixed field of A4,
Shorter? Seems like a lot of your proof is hidden in the above.
Let's see the details. How do you know L is the fixed field of A4?
How does your argument depend on the fact that the polynomial is
quartic? If the polynomial was degree n instead of degree 4, what
aspect of your argument would not apply?
Once you supply the details, assuming your argument actually works at
all for the case n=4, I suspect it will be just as long, if not
longer, and moreover, less elementary.
>i.e.G(E/L) = A4
>so [E:L] = |A4| = 12, and therefore [E:Q] = 2.
In the line above you meant [L:Q] = 2, not [E:Q] = 2.
>So, L = Q if G = A4 and [E:L] = 2 (the fixed field of A4)
>if G = S4.
Above you meant either [E:L] = 12 or [L:Q] = 2 (I'm not sure which you
intended), not [E:L] = 2.
quasi
quasi
Ok, this has been proved now.
So keeping n even, let's go past n=4 and ask the same question.
What's the first n, if any, where it breaks?
More precisely ...
Let n be an even positive integer, let f in Q[x] be an irreducible
polynomial of degree n, and let E be the splitting field of f in the
algebraic closure of Q.
Let L = {x in E | x is constructible}.
Questions ...
If for some root r of f, (Q(r) intersect L) = Q,
must [L:Q] = 1 or 2?
If yes, can we improve the conclusion to [L:Q] = 1?
quasi
Kent Holing
Of course I meant [L:Q] = 2 since [E:L] = |A_4| = 12.
Sorry about that.
quasi
>Let n be an even positive integer, let f in Q[x] be an irreducible
>polynomial of degree n, and let E be the splitting field of f in the
>algebraic closure of Q.
>
>Let L = {x in E | x is constructible}.
>
>Questions ...
>
>If for some root r of f, (Q(r) intersect L) = Q,
>must [L:Q] = 1 or 2?
>
>If yes, can we improve the conclusion to [L:Q] = 1?
Might as well include odd n as well.
Ok, so how about this ...
Let f in Q[x] be irreducible of degree n, and let E be the splitting
field of f in the algebraic closure of Q.
Let L = {x in E | x is constructible}.
Prove or disprove:
If either (1) n is odd, or (2) n is even and,
for some root r of f, (Q(r) intersect L) = Q,
then [L:Q] = 1 or 2.
quasi
achille
In general, I think you can't.
For n >= 5, the Galois group of generic polynomial
of degree n is either A_n or S_n depends on whether
sqrt(D) in Q or not (where D is the discriminant).
Since sqrt(D) is constructible, [L:Q] is at least 2
when Gal(E/Q) = S_n.
AP
>Note that [E:Q]is either 12 or 24 since the Galois group of the quartic either A4 or S4.
why not Klein's group ou D_4?
(note : there exist r with degre(r)=4 and r not constructible )
achille
Hmm.. maybe I have make a statement much stronger than
what is true. The corrected version:
For n >= 5, if we pick a random polynomial of degree
n over Q, the Galois group is almost always S_n or
sometimes A_n. A necessary condition for it to be A_n
instead of S_n is sqrt(D) \in Q.
If sqrt(D) \notin Q, [L:Q] is at least 2 because sqrt(D)
\in E and D \in Q. Since one can show that L is a normal
extension of Q with [L:Q] a power of 2. Gal(E/L) is a
normal subgroup of Gal(E/Q) with [Gal(E/Q):Gal(E/L)]
= [L:Q] = a power of 2.
Corollary:
[1] sqrt(D) \notin Q
=> Gal(E/Q) cannot be A_n.
[2] sqrt(D) \notin Q AND Gal(E/Q) is S_n
=> Gal(E/L) is A_n
=> [L:Q] = 2 AND L = Q(sqrt(D)).