Lemma 10.45.2. A field $k$ is perfect if and only if it is a field of characteristic $0$ or a field of characteristic $p > 0$ such that every element has a $p$th root.
Proof. The characteristic zero case is clear. Assume the characteristic of $k$ is $p > 0$. If $k$ is perfect, then all the field extensions where we adjoin a $p$th root of an element of $k$ have to be trivial, hence every element of $k$ has a $p$th root. Conversely if every element has a $p$th root, then $k = k^{1/p}$ and every field extension of $k$ is separable by Lemma 10.44.2. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)