Consider the polynomial \(f(x)=x^{10}+x^5+1\in\mathbb Q[x]\) with splitting field \(K\) over \(\mathbb Q\).
Determine whether \(f(x)\) is irreducible over \(\mathbb Q\) and find \([K:\mathbb Q]\).
Determine the structure of the Galois group \(\Gal(K/\mathbb Q)\).
Solution
1. The polynomial \(f\) is reducible over \(\mathbb Q\).
We have \[ x^{15}-1=(x^5-1)(x^{10}+x^5+1), \] so \[ f(x)=\frac{x^{15}-1}{x^5-1}. \] Using the cyclotomic factorization \[ x^{15}-1=\prod_{d\mid15}\Phi_d(x) =\Phi_1(x)\Phi_3(x)\Phi_5(x)\Phi_{15}(x) \] and \[ x^5-1=\Phi_1(x)\Phi_5(x), \] we obtain \[ f(x)=\Phi_3(x)\Phi_{15}(x). \] Both factors have positive degree, namely \(2\) and \(8\), so \(f\) is reducible over \(\mathbb Q\).
2. The roots of \(f\) are exactly the \(15\)th roots of unity whose orders are \(3\) or \(15\).
From \[ f(x)=\frac{x^{15}-1}{x^5-1}, \] a complex number \(\alpha\) is a root of \(f\) precisely when \(\alpha^{15}=1\) but \(\alpha^5\ne1\). The possible orders of a \(15\)th root of unity are \(1,3,5,15\); excluding those whose order divides \(5\) leaves orders \(3\) and \(15\). Equivalently, this is the union of the roots of \(\Phi_3\) and \(\Phi_{15}\) from 1.
3. If \(\zeta_{15}\) is a primitive \(15\)th root of unity, then\[ K=\mathbb Q(\zeta_{15}). \]
2. Therefore it is exactly the splitting field.
4. The degree of the splitting field is\[ [K:\mathbb Q]=\varphi(15)=8. \]
By 3, \(K=\mathbb Q(\zeta_{15})\). The degree of the \(15\)th cyclotomic extension is the degree of \(\Phi_{15}\), namely Euler’s totient \[ \varphi(15)=15\left(1-\frac13\right)\left(1-\frac15\right)=8. \]
5. The Galois group is\[ \operatorname{Gal}(K/\mathbb Q) \cong (\mathbb Z/15\mathbb Z)^\times \cong C_2\times C_4. \]
Every automorphism of the cyclotomic field \(\mathbb Q(\zeta_{15})\) is determined by \[ \zeta_{15}\longmapsto \zeta_{15}^a \] for a unique \(a\in(\mathbb Z/15\mathbb Z)^\times\), and every such \(a\) defines an automorphism. Hence \[ \operatorname{Gal}(K/\mathbb Q)\cong(\mathbb Z/15\mathbb Z)^\times. \] By the Chinese remainder theorem, \[ (\mathbb Z/15\mathbb Z)^\times \cong (\mathbb Z/3\mathbb Z)^\times\times(\mathbb Z/5\mathbb Z)^\times. \] The first factor is cyclic of order \(2\), and the second is cyclic of order \(4\). Thus the group is \(C_2\times C_4\).