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Definition : Let $K$ be a field. If $\alpha$ is an algebraic element over K, such that $\alpha^n \in K $ and such that $x^n-\alpha^n \in K[x] $ is irreducible over K. Then we call $ K(\alpha)$ a simple radical and irreducible extension. A field extension $ L / K$ is said to be solvable by simple and irreducible radicals, if there exist a simple radical and irreducible extension $K(\alpha)$ such that $ K(\alpha) / L $.

Problem:

Let $\zeta_n$ be a primitive n-rooth of unity. Prove that the field extension $ \Bbb Q(\zeta_n) $ is solvable by simple and irreducible radicals.

And then give a field tower of simple and irreducible radicals extensions $\Bbb Q = E_0 \subset E_1 \subset ... \subset E_n $ such that $ \zeta_{47}\in E_n $

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I think nobody will ever be able to prove the first assertion unless you edit it appropriately. –  Makoto Kato Nov 25 '12 at 0:34

2 Answers 2

Note that you are not asked to prove that there is a chain of simple irreducible radical extensions $\Bbb Q \subset E_1 \subset E_2 \subset \ldots \subset \Bbb Q(\zeta_n)$, but that there is a chain $\Bbb Q \subset \ldots \subset L$ such that $\Bbb Q(\zeta_n) \subset L$.

If a normal extension $K \subset L$ has Galois Group isomorphic to $\Bbb Z/p \Bbb Z$, then $K(\zeta_p) \subset L(\zeta_p)$ is a normal simple irreducible radical extension :

$K \subset K(\zeta_p)$ is a normal extension, thus $Gal_{K(\zeta_p)}(L(\zeta_p))$ and $Gal_L(L(\zeta_p))$ are normal subgroups of $Gal_K(L(\zeta_p))$ and there is a well-defined restriction map $Gal_K(L(\zeta_p)) \to Gal_K(L) \times Gal_K(K(\zeta_p))$. This map is injective because $L$ and $\zeta_p$ generate $L(\zeta_p)$. Furthermore, the composition of this map with either projection has to be surjective. Since $|Gal_K(L)|=p $ and $|Gal_K(K(\zeta_p))|$ are coprime, the only subgroups of their product are products of subgroups, so the map has to be surjective. Thus it is an isomorphism.
let $x \in L \setminus K$. Then $K(x) = L$. Let $\sigma$ be a generator for $Gal_K(L) = Gal_{K(\zeta_p)}(L(\zeta_p))$, and look at $y = \sum_{k=0}^{p-1} \zeta_p^{-k} \sigma^k(x) \in L(\zeta_p)$. Then, $\sigma(y) = \sum_{k=0}^{p-1} \zeta_p^{-k} \sigma^{k+1}(x) = \zeta_p y$.
Thus $\sigma(y^p) = \sigma(y)^p = \zeta_p^p y^p = y^p$. Since $y^p$ is fixed by $\sigma$, $y^p \in K(\zeta_p)$.
Since $X^p - y^p = \prod(X - \zeta_p^k y) = \prod(X - \sigma^k(y))$, this polynomial must be irreducible over $K(\zeta_p)$ (there is exactly one orbit of the action $Gal_{K(\zeta_p)}(L(\zeta_p))$ on its roots)
Finally, $K(\zeta_p,y) = L(\zeta_p)$ because there is an obvious inclusion and their degree over $K(\zeta_p)$ are the same.

Next, if $K \subset L$ is a normal simple irreducible radical extension of degree $p$, and $K \subset K'$ is a normal extension, then $K' \subset K'L$ is still a normal simple irreducible radical extension, of degree $1$ or $p$. The polynomial $X^p - y^p$ can't suddenly become reducible unless all the $y$ are in $K'$, because $\Bbb Z/p\Bbb Z$ has no nontrivial subgroup.

With this, you can show that if $K \subset L$ and $L \subset M$ are solvable by simple irreducible radical extensions, then so is $K \subset R$ (just add all the roots one after the other).

Finally, with an induction argument, you can finally show that since $K \subset K(\zeta_p)$ is abelian of degree dividing $p-1$, it is solvable, and since $K(\zeta_p) \subset L(\zeta_p)$ is solvable, $K \subset L(\zeta_p)$ is solvable, which implies that $K \subset L$ is too. And by decomposing any abelian extension into cyclic prime extensions, you obtain that any abelian extension is solvable.


Now we apply this to $\Bbb Q \subset \Bbb Q(\zeta_{47})$. Since its Galois group is isomorphic to $\Bbb Z / 2 \Bbb Z \times \Bbb Z / 23 \Bbb Z$, we need to add $\zeta_2$ (which is already there, it's $-1$) and $\zeta_{23}$. For this one, we need $\zeta_{11}$, which needs $\zeta_5$, which needs $\zeta_4$.

The resulting chain (showing the degrees of the extensions) is : $\Bbb Q \subset^2 \Bbb Q(\sqrt{-1}) \subset^2 \Bbb Q(\sqrt{-1},\sqrt 5) \subset^2 \Bbb Q(\zeta_{20}) \subset^2 \Bbb Q(\zeta_{20}, \sqrt{-11}) \subset^5 \Bbb Q(\zeta_{220}) \subset^2 \Bbb Q(\zeta_{220}, \sqrt{-23}) \\ \subset^{11} \Bbb Q(\zeta_{5060}) \subset^2 \Bbb Q(\zeta_{5060}, \sqrt{-47}) \subset^{23} \Bbb Q(\zeta_{237820}) \supset \Bbb Q(\zeta_{47})$.

Theoretically, you can compute explicitly at each step what are the things you are taking $n$th roots of and express everyone in terms of radicals, though it gets messy really fast. (I only explicited the quadratic subfields of the $\mathbb Q(\zeta_p)$)

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now I realize the galois groups aren't always this nice, so this needs more detail. –  mercio Nov 25 '12 at 9:02

Let $n$ be fixed, and $\zeta:=\zeta_n$. Then $\zeta,\zeta^2,..,\zeta^n=1$ are all the $n$th roots of unity, and $\zeta^m$ is primitive iff $\gcd(m,n)=1$. It also follows that $\Bbb Q(\zeta)$ is the splitting field of $x^n-1$, hence it is a normal field extension.

The relation $\phi(s)=s$ between automorphisms $\phi$ and field elements $s$ establishes the Galois connection, in which intermediate fields $\Bbb Q\subseteq E\subseteq\Bbb Q(\zeta)$ correspond to subgroups of $Aut(\Bbb Q(\zeta):{\Bbb Q})$.

Now $\zeta$ can be mapped by an automorphism to any primitive root of unity, i.e. to any $\zeta^m$ where $m$ is coprime to $n$, and then $\zeta^k$ is mapped to $\zeta^{km}$, so $m$ uniquely determines the automorphism, and in fact this automorphism group is isomorphic to $(\Bbb Z_n^*,\cdot)$ where $$\Bbb Z_n^* =\{m\in\Bbb Z_n\mid \gcd(m,n)=1\}$$ and it has $\varphi(n)$ elements and commutative. So, by the classification of finite(-ly generated) Abelian groups, we have that it is a direct product of cyclic groups, and basically these will give the intermediate simple radical extensions.

In the given example, when $n=47$, as it is a prime, $\varphi(47)=46=2\cdot 23$.

So, $\Bbb Z_2\hookrightarrow \Bbb Z_{46}\cong Aut(\Bbb Q(\zeta):\Bbb Q)$, generated by the automorphism $\zeta\mapsto \zeta^{-1}$. This fixes, for example $\eta:=\zeta+\zeta^{-1}$. Hence, $E_1:=\Bbb Q(\eta)$ is an intermediate field. Show that $\zeta$ has order 2 over $\Bbb Q(\eta)$ and hence (expressable by a) simple radical element.

The quotient group $\Bbb Z_{23}=\Bbb Z_{46}/\Bbb Z_{2}$ corresponds to the Galois group $Aut(\Bbb Q(\eta):\Bbb Q)$...

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And $ E_1 = {\Bbb Q}\left( \eta \right) $is a simple radical and irreducible over $\Bbb Q$. I have to find and $n$ such that $\eta^n \in \Bbb Q$ since the order of $\eta$ over $\Bbb Q $ is $46$ that implies that if that $n$ exist then $ 46 \le n$ , but I'm not sure if this is true :/ . Maybe considering some subfield before of $E_1$ –  Daniel Nov 18 '12 at 2:19
    
Well, $n$ must be $23$, and not necessarily $\eta$ itself is the $23$th root of a rational, but generates the same $E_1$ field. Try to prove that any field extension of prime order is a simple radical extension. –  Berci Nov 18 '12 at 23:51

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