2
$\begingroup$

I have the following two questions questions I am working on and am a little stuck :

Let $L=\mathbb{Q}(\zeta_{25})$ where $\zeta_{25}$ is the primitive $n$-th root of unity.

Prove that there are unique subfields $K_1$ and $K_2$ in $L$ such that $[L:K_1]=4$ and $[L:K_2]=5$

I am not sure if this is related but $\mathbb{\phi(25)=20} \implies$ the order of $Gal(\mathbb{Q}(\zeta_{25})/\mathbb{Q}) \simeq Z_{25 }^*=Z_{20}$ (I think)

Can we say that all there exists subfields of orders equal to the divisor of our Galois group? i.e. orders $1, 2, 4, 5, 10, 20$?

Prove that $L/K_2$ is cyclic and find $a \in K_2$ such that $L=K_2(\sqrt[5]{a })$

Need to show that $Gal(L/K_2)=Z_n$ for some $n\in \mathbb{Z}$ - is that right?

$Gal(L/K_2)=Gal(\mathbb{Q}(\zeta_{25})/K_2)$

We know that $[\mathbb{Q}(\zeta_{25}) : K_2]=5$ - is it true that the only groups of order $5$ is $Z_5$ respectively (I am not convinced)

If so what would $a$ be - some primitive root of unity, $-5$?

I hope you can help me with this as I need to improve my uunderstanding of these ideas (degres of field extensions and Galois groups) - thanks!

$\endgroup$
1
$\begingroup$

Remember:

1) A finite cyclic group of order $\;n\;$ has one unique subgroup of order $\;d\;$ for any $\;d\,\mid\,n\;$ ;

2) The group of units modulo $\;p^k\;,\;\;p\;$ a prime, is cyclic.

$\endgroup$
  • $\begingroup$ Thanks. So, since there are unique subgroups of order $4$ and $5$, then, by the Galois correspondence, there are unique subfields of orders $4$ and $5$ respectively? I am unsure what the group of units would be for $Gal(\mathbb{Q}(\zeta_{25}))/K_2$ $\endgroup$ – maths2332 Jun 2 '16 at 19:52
  • $\begingroup$ Yes for the first part, as for the second one: are you asking about the "group of units" ...of a group? $\endgroup$ – DonAntonio Jun 2 '16 at 20:38
  • $\begingroup$ I am just trying to figure out the second part of the question, why the extension is cyclic. I am not sure. I thought you suggested that this could be found by considering the group of units... although I can't figure out what to do $\endgroup$ – maths2332 Jun 2 '16 at 20:44
  • $\begingroup$ When you refer to the group of units, you are referring to the units of the ring $\Bbb Z/p^k\Bbb Z$. The cyclicity is in the Galois group, not the group of units of the (integer ring of the) field. But the statement about cyclicity here is valid only for primes $p\ne2$. $\endgroup$ – Lubin Jun 3 '16 at 2:25
0
$\begingroup$

It really helps to know what the minimal polynomial for $\zeta_{25}$ is: it’s $\Phi_{25}=X^{20}+X^{15}+X^{10}+X^5+1$. This makes sense, since the minimal polynomial fo $\zeta_5$ is $\Phi_5(X)=X^4+X^3+X^2+X+1$, and clearly any fifth root of a primitive fifth root of unity will be a primitive $25$-th root of unity.

For the rest of your question, I think Joanpemo’s answer does most of the work, except perhaps for the remarks that (1) up to isomorphism there is only one cyclic group of each order; and (2) every subgroup and every quotient group of a cyclic group is cyclic.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.