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.

  • 6
    $\begingroup$ 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., $f(x) = x^3 - 2$). $\endgroup$ – 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 $\mathbb{Q}(i)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.