# Galois group of x^10+x^5+1

Originally published at 狗和留美者不得入内. You can comment here or there.

This was a problem from an old qualifying exam, that I solved today, with a few pointers. First of all, is it reducible? It actually is. Note that $x^{15} - 1 = (x^5-1)(x^{10}+x^5+1) = (x^3-1)(x^{12} + x^9 + x^6 + x^3 + 1)$. $1 + x + x^2$, as a prime element of $\mathbb{Q}[x]$ that divides not $x^5-1$ must divide the polynomial, the Galois group of which we are looking for. The other factor of it corresponds to the multiplicative group of $\mathbb{F}_{15}$, which has $8$ elements. Seeing that it has $3$ elements of order $2$ and $4$ elements of order $4$ and is abelian, it must be $C_2 \times C_4$. Thus, the answer is $C_2 \times C_2 \times C_4$.

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