# Could different polynomials induce the same splitting field?

Let $\alpha := e^{2\pi i/6}$ be the sixth root of unity. Is $\mathbb Q(\alpha)$ both the splitting field of $x^6 - 1$ and $x^2 - x + 1$?

• The two second degree factors of $x^6-1$, one of which equals your $x^2-x-1$, have the same discriminant, $-3$. Consequently $x^6-1$ has splitting field $\mathbb{Q}(\sqrt{-3})$, which is also the splitting field of $x^2-x-1$. – Erik Vesterlund Jul 28 '13 at 20:19
• Note that of the two $x^2-x+1$ is irreducible over $\mathbb Q$. Sometimes it is important to test for irreducibility before applying apparently general theorems about extensions. – Mark Bennet Jul 28 '13 at 20:52
• Possibly also interesting is that $\mathbb{Q}(\alpha)$ is also the splitting field of $(3x-2)^6-1$. – user14972 Jul 28 '13 at 21:28

Hint: Alll sixth roots of unity are powers of $\alpha$. And $x^6-1=(x^3+1)(x^3-1)$.
• and it factors as $x^6 - 1 = (x-1)(x+1)(x^2+x+1)(x^2-x+1)$ into irreducible factors, but what does this tell me regarding my question. – StefanH Jul 28 '13 at 21:20
• I had left it at $x^3+1$. The two roots of this are primitive $6$-th roots of unity. So one of them is $alpha$. So we have shown that the splitting fields of the two polynomials you were given are both $\mathbb{Q}(\alpha)$. – André Nicolas Jul 28 '13 at 23:51
Clearly, $x^6-1$ splits over ${\bf Q}(\alpha)$, as does $x^2-x+1$. Furthermore, $\alpha$ is a root of $x^6-1$, so it must belong to its splitting field.
What are roots of $x^2-x+1$? Can you produce a sixth root of unity from one of them?
• the roots of $x^2 - x + 1$ are the primitve sixth roots of unity, and so they generate all roots of $x^6 - 1$. – StefanH Jul 28 '13 at 21:19