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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.2011-12-17
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    If the Galois group of a polynomial has order 2 then what can you say about the degree of the polynomial?2011-12-17
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    Do _all_ of the roots need to be irrational? It seems like this would be hard to arrange for $n = 3$.2011-12-17
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    What have *you* tried to do to solve this problem?2011-12-17
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    $(x^2+2)^{n/2}$ or something similar for $n$ even. if $n$ is odd, the poly has a real (irrational) root.2011-12-17
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    Equivalent to: For each n even find a polynomial P in Q [x] with degree P = n ,distinct roots not rational, such that the Galois group of its field of decomposition is isomorphic to Z2.2011-12-17
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    yoyo, Why?, Explain how to deduce the expression2011-12-17
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    There's not much to explain in yoyo's answer. The only roots of, say, $(x^2-2)^{100}$ are $\pm\sqrt2$, they aren't rational, the field they generate has degree 2 over the rationals, so Galois group the group of order 2.2011-12-17
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    yeah, if n is odd?2011-12-17
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    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.2011-12-17

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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 $n\ge2$).2013-01-22
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    @GerryMyerson fixed.2013-01-22