# Can splitting field be generated by one root?

Say $$f$$ is an irreducible polynomial over a field $$F$$, and $$\alpha$$ is one of its roots, then is $$F(\alpha)$$ a splitting field for $$f$$? I tried to find some counterexample, but I failed.

• Sometimes a single root generates a splitting field (e.g., $f(x) = x^2 - 2$ over $\mathbb{Q}$), and sometimes it does not (e.g., $f(x) = x^3 - 2$). – FredH Feb 24 at 12:13

A related concept is that of a normal extension. A normal extension $$K/k$$ is an algebraic extension such that every irreducible polynomial in $$k[X]$$ that has a root in $$K$$ is decomposed as the product of linear factors in $$K$$. It can be shown that splitting fields are in fact normal, finite extensions.
As FredH pointed out, the splitting field of $$f(x)=x^3-2$$ over $$\mathbb{Q}$$ is $$\mathbb{Q}(\sqrt[3]{2},\omega)=\mathbb{Q}(\sqrt[3]{2},i\sqrt{3})$$ where $$w=\frac{-1+i\sqrt{3}}{2}$$ is a third root of unity, i.e. $$\omega^3=1$$. So, $$\mathbb{Q}(\sqrt[3]{2})$$ is not the splitting field of $$f$$. Meanwhile, $$\mathbb{Q}(\sqrt[3]{2})$$ is not a normal extension of $$\mathbb{Q}$$. However, for $$f(x)=x^2+1$$, the splitting field is $$\mathbb{Q}(i)$$.