Re: cyclotomic field is galois

On Jun 29, 9:47 pm, Tonico <Tonic...@xxxxxxxxx> wrote:
On Jun 29, 9:10 pm, charles.cado...@xxxxxxxxxxxxxx wrote:



On Jun 29, 7:14 pm, magi...@xxxxxxxxxxxxxxxxx (Arturo Magidin) wrote:

In article <d2edcbf3-0034-43f1-b6a9-a470e99fe...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,

<charles.cado...@xxxxxxxxxxxxxx> wrote:
On Jun 29, 6:51 pm, Tonico <Tonic...@xxxxxxxxx> wrote:

[...]

Perhaps it's only me but I can't see why the simple argument of a
cyclotomic extension being the splitting field of (a cyclotomic)
polynomial is less elementary than messing around with fixed fields of
automorphism: how the latter is "basic algebra" but the former isn't?
yes the concept of elementariness is highly subjective, you're right.
how would you prove then that the splitting field of a cyclotomic
polynomial is galois?
thanks!

That depends on what you consider to be the "definition" of "Galois
extension" and what it is you consider to be "derived equivalences."
Many books ->define<- a Galois extension to be the splitting field of
a (set of separable) polynomials, so "is the splitting field of a
cyclotomic polynomial" ->means<- it is Galois. No futher argument
required.

On the other hand, others define "Galois extension" differently (e.g.,
that the relevant automorphism group have the same order, when finite,
as the degree of the extension), and then ->prove<- that this is
equivalent to some of the other definitions. Without knowing what it
is you consider basic and what you consider derived, it is impossible
to give an answer that will satisfy you.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

Ok, I'll try to explain better what is my problem, sorry if I've been
confusing so far. Let's say I'm given the polynomial P(z) = z^n - 1
over Q. Call r = exp(2*pi*i/n) and consider the field extension Q(r).
Now I associate to this field extension Q(r)\supset Q the group of
automorphisms of Q(r) fixing Q. This is the Galois group of the
extension. I want to show that any element which is kept fixed by all
the Galois group is necessarily in Q.
Thanks-

************************************************************

Alright. then I think you could proceed as follows:

1) Since the set of all roots of unit of order a fixed n is a cyclic
group, it is clear Q(r) is the splitting field of P(z) over Q;

2) Since char Q = 0 the extension is separable;

3) Clearly the extension is finite and thus algebraic

4) Finally, by (1), Q(r) contains all the roots of the minimal
polynomial of r over Q and thus Q(r)/Q is a normal extension.

As Arturo said, if we already know the extension is separable, being
the splitting field of some polynomials over some given field is
enough to conclude the extension is Galois. But you can also prove it
directly by (2)-(4).

Regards
Tonio

Thanks for the reply. Your argument is perfectly clean but
unfortunately is not what I had in mind. Basically I wanted to avoid
the theorem which says that a finite normal separable extension is
Galois. This theorem is powerful in that it allows to bypass the
knowledge of the automorphisms of the extension. In my case, since I
know precisely what is the Galois group, I wanted to verify in some
way that it had sufficiently many automorphisms to force the fixed
field to be Q. I am now afraid this could be more time-consuming than
just facing the general Galois theory.
Thank you again for your insight and patience.
.

Relevant Pages

  • Re: cyclotomic field is galois
    ... how would you prove then that the splitting field of a cyclotomic ... That depends on what you consider to be the "definition" of "Galois ... extension" and what it is you consider to be "derived equivalences." ... a polynomials, so "is the splitting field of a ...
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  • Re: cyclotomic field is galois
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