Let $\mathbb{Q}$($\zeta_n$) be some cyclotomic field, where $\zeta_n$ is a n-th root of unity.  I already managed to show that $\mathbb{Q}$($\zeta_n$) is an Galois extension, but now i struggle to show that  $[\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \varphi(n)$, where $\varphi(n)$ is Euler's totient function.
61149
I studied the cyclotomic extension using Fraleigh's text.   To prove that Galois group of the $n$th cyclotomic extension has order $\phi(n)$( $\phi$ is the Euler's phi function.), the writer assumed, without proof, that $n$th cyclotomic polynomial $\Phi_n(x)$ is irreducible over $\mathbb{Q}$.  I know that for n=p, p is prime, $\Phi_n(x)$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion.  But I don't know how $\Phi_n(x)$ is irreducible over $\mathbb{Q}$ when n is not prime.
75109
Possible Duplicate:          Consider the field extension $\mathbb{Q}(\sqrt2)$. I want to show that $\sqrt5 \notin \mathbb{Q}(\sqrt2)$. If this were not the case, then we could write $\sqrt5 = a + b\sqrt2$ for $a,b\in\mathbb{Q}$. However, I do not see the contradiction here. Is there a better/easier way to prove this?
90131
Would you please help me to prove this problem.  Let $G=\mathrm{Hol}(C_8)$. And let $f(x)\in \mathbb{Q}[x]$ be an irreducible polynomial of degree $8$, $L$ a splitting field of $f(x)$ over $\mathbb{Q}$ and suppose that $\mathrm{Gal}(L/\mathbb{Q})\cong G$.     Show that there is a subgroup of $S_8$ isomorphic to $G$.   I am reviewing for my test and i find this good exercice    Thanks.
91654
Let $K$ be a field and $f \in K[x]$ a polynomial of degree $n&gt;0$. Assume that $E$ is an extension of $K$ such that $f$ can be expressed as a product of linear factors over $E$:  $$f(x)=\gamma (x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_n).$$  I want to show that $\lvert K(\alpha_1,\dots,\alpha_2) : K \rvert$ is a divisor of $n!$. I also want to show that if $\lvert K(\alpha_1,\dots,\alpha_n): K \geq (n-1)!$ then $f$ is irreducible over $K$ or $f$ already has a root in $K$.  The problem has a hint. It says: "Use induction on $n=\deg f$. If $f=gh$ in $K[x]$ with $\deg g =r$ and $\deg h=n-r$, then $r!(n-r)!$ is a divisor of $n!$.  Any idea for solving this problem would be appreciated.
114096
Proposition 7.13 of Neukirch's ANT states that for a primitive $p^{m}$-th root of unity $\zeta$ ($p$ prime) the extension $\mathbb{Q}_{p}(\zeta)/\mathbb{Q}_{p}$ is totally ramified of degree $(p-1)p^{m-1}$. In the proof he shows that the polynomial $\phi(X)=X^{(p-1)p^{m-1}}+X^{(p-2)p^{m-1}}+\cdots+1$ is the minimal polynomial of $\zeta$ over $\mathbb{Q}_p$ .  I can't see how to obtain the congruence $\phi(X)=(X^{p^m}-1)/(X^{p^{m-1}}-1)\equiv(X-1)^{p^{m-1}(p-1)}$ mod $p$ as claimed in the book.  Here, I take it that mod $p$ means mod $p\mathbb{Z}$ and not mod $p\mathbb{Z}_{p}$ ?  Also, does Eisenstein's irreducibility criterion extend to polynomials in $\mathbb{Q}_{p}[X]$ ?   Any help would be very much appreciated.
94168
Let $K$ be a field and $K(X)$ be the field of its rational functions.  Now let $\phi \in K(X)$ be a rational function such that $K(\phi) \neq K(X)$.  Now, since $\phi$ is transcendental over $K$, $K(\phi)$ is isomorphic to $K(X)$.  Is this a correct example of a field being isomorphic to its subfield?  Are there any other examples?
4763
I want to know if there is a way to decide if a cyclotomic polynomial is irreducible over a field $\mathbb{F}_q$?
106766
Let $\omega$ be a primitive third root of unity, and $K=\mathbb{Q}(\omega,\sqrt{2})$.  I found that the degree of $[K:\mathbb{Q}]=6$. How can I find the Galois group $\operatorname{Gal}(K/\mathbb{Q})$?  After this, could anyone give me any hints for showing that $K=\mathbb{Q}(\omega\sqrt{2})$ and finding $\operatorname{min}(\omega\sqrt{2},\mathbb{Q})$?
113998
I'm having a hard time showing this:     If $K$ is an extension of $\mathbb{Q}$ with degree $m$ and $f(x)$ an irreducible polynomial over the rationals with degree $n$, such that $\gcd(m, n)=1$, then $f(x)$ is irreducible over $K$.   I have tried it by writing $f(x)=a(x)b(x)$ and then looking at the coefficients of those polynomials (some of them must belong to $K-\mathbb{Q}$ which could possibly result in a contradiction) to no success. I have no idea where to use the hypothesis of m and n being relatively prime.  Any help would be appreciated.
116701
From Berkeley Problems in Mathematics, Spring 1999, .  Let $f(x)\in \mathbb{Q}[x]$ be an irreducible polynomial of degree $n\ge 3$. Let $L$ be the splitting field of $f$, and let $\alpha\in L$ be a zero of $f$. Give that $[L:\mathbb{Q}]=n!$, prove that $\mathbb{Q}(\alpha^{4})=\mathbb{Q}(\alpha)$.   (BELOW IS NONSENSE, SHOULD BE IGNORED)  Following Gerry Myerson's advice, since $L$ is the splitting field of $f$, $L$ must be Galois over $\mathbb{Q}$. Thus the Galois group for $[L:\mathbb{Q}]=S_{n}$. Gerry now asserts that either $\mathbb{Q}(\alpha^{4})$ is of degree 2 over rationals or it must be $\mathbb{Q}$. But assuming this one can rule out the case $\mathbb{Q}$, since this would imply $\alpha$ is the root of a 4th or 2nd degree polynomial over $\mathbb{Q}$. But we know $n\ge 3$. So $\alpha$ must be the root of a 4th degree polynomial. We know $S_{n}$ has normal subgroups $A_{n}$ when $n\not=4$ and $V,A_{4}$ when $n=4$. I do not know how to proceed any further.   Now assuming $\mathbb{Q}(\alpha^{4})$ is a degree 2 abelian extension over $\mathbb{Q}$. We should have a chain of normal extensions $\mathbb{Q}(\alpha)\supset \mathbb{Q}(\alpha^{2})\supset \mathbb{Q}$. This would imply $S_{n}$ has a normal subgroup which has a normal subgroup. But we know $A_{n}$ for $n\ge 5$ are simple. So the only possibility is $\mathbb{Q}(\alpha)$ has a Galois group isomorphic to $V$. In this case $n=4$ as well. I can only proceed to here.
86276
Is it true that $[K(\zeta_n):K]=\phi(n)$ where $\zeta_n$ be primitive root of unity and K be field of char zero?  I think it should be equal to the degree of cyclotomic polynomial which has degree $\phi(n)$.  Can someone please elaborate that what is the procedure for finite fields
69153
Let $K=\mathbb{Q}(\zeta_m)$. Then if $p\nmid m$ is any odd prime, how i can show that Frobenius map is $(p,K/\mathbb{Q})(\zeta_m)=\zeta_m^p$.  We know, if $P$ is a prime above $p$   $$(p,K/\mathbb{Q})(x) \equiv x^p (\mbox{mod } P) \quad\forall x \in \mathbb{Z}_K$$  Thanks!
100654
Given the galois extension $L/K$, where $L=\mathbb{Q}(\zeta, \sqrt[3]{2})$. I've already calculated that $[L:\mathbb{Q}]=6$. I'm trying to prove that the number of intermediate fields is also 6, including $L$ and $\mathbb{Q}$ as intermediate fields. I'm told that the fundamental theorem of galois theory can be useful here, but I can't find a use for it.
71389
From Chapter VII of Lang's Algebra.  The question asks if $n\geq 6$ and $n$ is divisible by at least two primes, show that $1-\zeta$ is a unit in the ring $\mathbb{Z}[\zeta]$  I am having a hard time understanding why this is true. This is in the integral dependence chapter, but that has not given me any inspiration. I have also tried using cyclotonic polynomial to no avail  Thanks for any direction.
111450
I want to show that $\mathbb{Q}(i,\sqrt[4]{1+i} )$ is normal and thus nonabelian extension of $\mathbb{Q}$, but got stuck. The main problem is that for the extension to be normal $\root4\of{1-i}$ should also be an element. I failed to show that. Thank you for your helps. ----------------------edit--------------- My question arose from , page 5, Remark 2.7. It is given as a counterexample of Kronecker and Weber theorem: When $K=\mathbb{Q}(i)$, $K(\root4\of{1-i})$ can never be any intermediate field of $K({ \zeta  }_{ n })/K$, where ${ \zeta  }_{ n }$ is a primitive $n$-th root of unity. I first expected that above statement could be proved by showing that $K(\root4\of{1-i})$ is not abelian over  $\mathbb{Q}$ since any cyclotomic extension of K is a cyclotomic extension of $\mathbb{Q}$ thus abelian. However from your advices, it seems that it's not even normal. But I can't still understand the method you showed me.. Is there any more easy explanation for this? When $K=\mathbb{Q}(i)$, $K(\root4\of{1-i})$ can never be any intermediate field of $K({ \zeta  }_{ n })/K$, where ${ \zeta  }_{ n }$ is a primitive $n$-th root of unity. I only know about elementary field and Galois theory.
109287
For a polynomial $f=X^n+a_1X^{n-1}+\ldots+a_n \in \mathbb{Q}[X]$ we define $\varphi(f):=a_1 \in \mathbb{Q}$. Now I want to show that for the $n$th cyclotomic polynomial $\Phi_n$ it holds that $$\varphi(\Phi_n)=-\mu(n)$$ where $\mu(n)$ is the Möbius function. What I know is that $\displaystyle\Phi_n=\prod_{d|n} (X^{\frac{n}{d}}-1)^{\mu(d)}$.
14601
$\zeta_{15}$ is $15$th primitive $n$th root of unity.   Question: Find the structure of the group $Gal(\mathbb{Q}(\zeta_{15})/\mathbb{Q})$  I know that if $p$ is prime then $G=Gal(\mathbb{Q}(\zeta_{p})/\mathbb{Q})=\mathbb{Z_{p}}^*$ but when $p$ is not prime, I am not show how to solve this if not prime.  Does my required group have any relation to some cyclic group $\mathbb{Z_n}$, i.e. abelian  Would really appreciate your guidance as I have not got my head around these concepts of cyclotomic fields and Galois structure. Thanks