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Could anyone tell me how to show for any positive integer n, there exists a normal extension of rational number field of degree n?

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Please try math.stackexchange.com. – Chandan Singh Dalawat Nov 27 2011 at 12:59
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 This is actually simple if you are willing to remain in the realm of abelian extensions. Since $Q(\nu_n)/Q$ has Galois group isomorphic to $Z/NZ^*$, you should look for an integer $N$ such that this group surjects onto a group of order $n$. Which really amounts to require that $n$ divide the order of $Z/NZ^*$. And you are done! – Tommaso Centeleghe Nov 27 2011 at 13:02
Could you tell me what is Vn and what is Z/NZ*? Thanks. – Zhouzhou Nov 27 2011 at 13:10
by $\nu_n$ I meant to write $\mu_N$, which you can think of as a primitive N-th root of unity. $Z/NZ^∗$ is quite standard notation for the multiplicative group of the ring of integers $Z$ modulo $N$. If you are not familiar with these notations you can learn more in a basic algebra book touching some Galois theory as well! – Tommaso Centeleghe Nov 27 2011 at 13:31
the size is $n-1$ only if $n$ is prime. you should google Euler totient function for the general case. – Marc Palm Nov 27 2011 at 13:44

closed as too localized by Felipe Voloch, Chandan Singh Dalawat, Franz Lemmermeyer, Leonid Positselski, Joe Silverman Nov 27 2011 at 13:34

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