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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.

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    $\begingroup$ 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
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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 $\mathbb{Q}(i)$.

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