Normal extension of any degree Could any one tell me how to show that for any positive integer n, there exists a normal extension of rational number field of degree n?
 A: Hint: Look at cyclotomic fields. Any intermediate field of an abelian extension is normal.
A: The splitting field of $x^n-1$ over $\mathbb{Q}$ is known as the $n$th cyclotomic field. It is known that the extension is abelian (that is, the Galois group is abelian). In fact, the extension has degree $\varphi(n)$ (Euler's totient function), and the Galois group is isomorphic to the multiplicative group of units modulo $n$.
By the Fundamental Theorem of Galois Theory, any subextension $K\subseteq L \subseteq M$ of an abelian extension $M/K$ is necessarily normal; moreover, as in any normal extension, the degree of $L$ over $K$ equals the index of the subgroup $\mathrm{Gal}(M/K)$ of all automorphisms that fix $L$ pointwise.
It is an easy result in group theory that abelian groups satisfy the converse of Lagrange's Theorem: if $G$ is abelian, $|G|=n$, and $d$ is a positive integer that divides $n$, then $G$ has a subgroup of order $d$.
Therefore, to find a normal extension of $\mathbb{Q}$ of degree $m$ it suffices to find an integer $n$ such that $m$ divides $\varphi(n)$: take the cyclotomic extension given by the splitting field of $x^n-1$; it has degree $\varphi(n)$ and abelian Galois group. Write $\varphi(n)=mk$, then find a subgroup of order $k$. The fixed field of the subgroup of order $k$ will have degree $m$ over $\mathbb{Q}$.
As an overkill, then, given a positive integer $n$, take the splitting field of $x^{n^2}-1$. Since $\varphi(n^2) = n\varphi(n)$, this will do. 
