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 ?
What is the Galois group of $x^n + (x-1)^n $?
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2J. Milne's notes on Galois theory are, in my opinion, quite a good material i$f$ you're a beginner on the topic. They both covers the theory and the practice, giving many example and a $f$ew tricks on how to calculate Galois groups of polinomials over the rationals. Here's the link http://www.jmilne.org/math/CourseNotes/ft.html – 2012-09-19
1 Answers
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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4The Galois group of a reducible polynomial is still the Galois group of its splitting field. – 2012-09-19