Hello, i cant solve number c) in the first question. We can use sylow 1 to say e know atleast one sylow 2 group exists but then sylw 3 doesnt simplify it and also worst case scenario we have that there are 21 such groups which leads to no contradiction for the 42 elements of G.
Call K' the only extension of degree 6 over Q that you found in (b). That is Q(\xi) with \Xi a primitive 6 root of unity. We know that this extension K', contains a unique sub-extension if degree 2/Q, Q(isqrt7). Thus we obtain that L has an intermediate extension if degree 2/Q. We want to show that this us unique. It will be enough to show that any intermediate extension of degree 2 will be contained in K', then we will conclude by the unicity of the quadratic subextensiins in cyclotomic fields. So ket K be an intermediate extension of L qiadratic over Q. We have that Gal(K/Q) has order 2 so Gal(L/K) has order 21. In particular Gal(L/K) has a unique 7 sylow subgroup, which is also the 7-sylow subgroup of Gal(L/Q), which is exactly Gal(L/K'). Thus Gal(L/K)>Gal(L/K').
By the Galois correspondence K'\supseteq K as we wanted.
I hope this helps. If it is not clear do not hesitate to ask questions. (Which I will answer later this evening or tomorrow)
With my best regards
Sofia
Do you mean that xi is a 7th root of unity?
Also why does it contain a unique sub extension of order 2? If we know itbis unique from there arent we done?
Also i wasnt in class when you went through that p= 1 or 3 mod 4 theorem and i cant find it in the book
Also why does it contain a unique sub extension of order 2? If we know itbis unique from there arent we done?
Also i wasnt in class when you went through that p= 1 or 3 mod 4 theorem and i cant find it in the book
With Regards
Sejad
Yes \xi is a primitive 7 root of unity.
The mod 1,3 is a special case of Kroneker theorem and the sub extension if order two is too a consequence of that. It depends from the fact that cyclotomic extensions are abelian.
Dont we know that the degree 2 subfield comes from the fact of isomorphism in theorem 26, like between gal group of cyclotomic extension and the multiplicative group Z/nZ?
And then it will be isomorphic to a cyclic group of order 6, and then since its cyclic, using lagranges theorem we know a subgroup of order 2 exists?
But i am confused about the kronecker theorem and where and to what it is applied to
L
|
K' = Q(xi_7)
/ \
/ \
K N
\ /
\ /
Q
So N has deg 3 and K deg 2.
Well i dont know im lost
And then it will be isomorphic to a cyclic group of order 6, and then since its cyclic, using lagranges theorem we know a subgroup of order 2 exists?
But i am confused about the kronecker theorem and where and to what it is applied to
L
|
K' = Q(xi_7)
/ \
/ \
K N
\ /
\ /
Q
So N has deg 3 and K deg 2.
Well i dont know im lost
Sorry it is Kroneker Webber theorem and it a teorem about abelian extension. But it is how you said. The Galois group of a cyclotomic extenesion Q(\xi) with \xi a p-root of unity is the cyclic group of order p-1. As such it has a unique subgroup of index 2, which corresponds to a unique subestension, quadratic over Q.
In the chapter about abelian extension you can aslo see that this extesnsion is generated by
$\alpha:= \sum_{a\in \mathbb{F}_p^{*2}}\xi^a$
Thus the minimal polynomial of $\alpha$ is $(x-\alpha)(x-\overline{\alpha})$ which can be explicitly computed and then one finds that $\alpha\i\mathbb{Q}(\sqrt{p})$ or $\alpha\i\mathbb{Q}(i\sqrt{p})$ depending from the remainder class of $p$ mod 4.