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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?

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2 Answers 2

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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.

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Thank you very much! –  John Nov 27 '11 at 19:27

Hint: Look at cyclotomic fields. Any intermediate field of an abelian extension is normal.

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Could you give me more details? I still can't work it out. Thank you. –  John Nov 27 '11 at 17:15
    
Sorry, @user20193, I was off air. Fortunately Arturo has given you the necessary details. –  Jyrki Lahtonen Nov 27 '11 at 20:09
    
Thank you any way! Your hint also helps a lot, I just stuck at finding subgroup of cyclotomic field of required size. –  John Nov 27 '11 at 20:20

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