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For each integer $n \geq 2$ , find a polynomial of degree $n$ with non-rational roots, whose Galois group over $\mathbb{Q}$ is $\mathbb{Z}/2\mathbb{Z}$.

Anybody can help me?

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You mean you want all of the roots to be irrational? That can't be done for $n$ odd. – Chris Eagle Dec 17 '11 at 19:37
Do all of the roots need to be irrational? It seems like this would be hard to arrange for $n = 3$. – Dylan Moreland Dec 17 '11 at 19:37
What have you tried to do to solve this problem? – Asaf Karagila Dec 17 '11 at 19:42
$(x^2+2)^{n/2}$ or something similar for $n$ even. if $n$ is odd, the poly has a real (irrational) root. – yoyo Dec 17 '11 at 19:59
Dylan is right: If $n=3$ and $a$ is an irrational root of the polynomial $f(X)$, then the splitting field $K$ of $f(X)$ is $\mathbb Q (a)$, because $K$ has degree $2 $, like the Galois group of $f(X)$.The minimal polynomial $g(X)$ of $a$ over $\mathbb Q$ divides $f(X)$ and has degree 2. Hence the quotient $f(X)/g(X) \in \mathbb Q[X]$ has degree one and has a rational root, which is also a root of $f(X)$. So no polynomial $f(X)$ satisfying your requirements exists. – Georges Elencwajg Dec 17 '11 at 21:55

Just collecting some comments together.

If $n$ is even then

$$\prod_{i=1}^{n/2}(x^2-2\cdot2^{2i})$$ works otherwise its not possible. For instance when $n=3$ a cubic polynomial with no rational roots is irreducible, so its Galois group is of order at least $3$.

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In one of the comments, OP wrote "distinct roots not rational". This example fails to have distinct roots (for $n\ge2$). – Gerry Myerson Jan 22 '13 at 5:10
@GerryMyerson fixed. – JSchlather Jan 22 '13 at 5:11

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