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What is the Galois group of $x^n + (x-1)^n $ over the rationals in terms of the integer $n$ ? In case that is too hard , what is it for the first 20 integers ?

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A minor question , what is preferred 'group of f(x)' or 'group of f(x)=0' ? I seem to think the first at least that is what I read all the time. Maybe both are correct ? –  mick Sep 19 '12 at 14:50
    
I suppose you mean its Galois Group over the rationals $\mathbb Q$, but you should specify it. Because you know, we could consider its Galois group on every finite field $\mathbb F_q$ as well. –  Niccolò Sep 19 '12 at 15:38
    
I am not so familiar with Galois groups over finite fields , plz tell me more or give good link for beginners. I will edit. –  mick Sep 19 '12 at 15:42
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J. Milne's notes on Galois theory are, in my opinion, quite a good material if you're a beginner on the topic. They both covers the theory and the practice, giving many example and a few tricks on how to calculate Galois groups of polinomials over the rationals. Here's the link jmilne.org/math/CourseNotes/ft.html –  Niccolò Sep 19 '12 at 15:45

1 Answer 1

First of all notice that $f(x):=x^n+(x-1)^n=0$ is equivalent to $(\frac{x-1}{x})^n+1=0$. The roots $f$ are therefore of the form $x=1/(1-y)$ where $y$ are the roots of $y^n+1=0$, so the Galois group is the same as for the polynomial $x^n+1$.

I'm not sure what "Galois group of a polynomial" is when the polynomial is reducible, so let me determine the Galois group for every irreducible factor. For every odd divisor $k$ of $n$ there is one such a factor, its roots are the $2n/k$-th primitive roots of $1$. The Galois group is thus the multiplicative group $(\mathbb{Z}/(2n/k))^*$. (you might mean the product of these groups over all $k$'s, as that's the automorphism group of he ring $\mathbb{Q}[x]/(f)$)

edit (to take account of Qiaochu's comment): the splitting field of $f$ is $\mathbb{Q}(e^{\pi i/n})$, so the Galois group is $(\mathbb{Z}/(2n))^*$.

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The Galois group of a reducible polynomial is still the Galois group of its splitting field. –  Qiaochu Yuan Sep 19 '12 at 17:15

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