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We know that all finite fields are perfect (fields with char $p$). Also fields with char 0 (infinite fields) are perfect. Then what are the fields that are not perfect?

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By your remarks, it has to be an infinite field of characteristic $p$. The first such thing that comes to mind, $\mathbb{F}_p(T)$, turns out to work (why?). – Cam McLeman Feb 7 '12 at 12:16
sorry, I want to ask an example of a field that is not perfect.- Madhav Bapat – Madhav Bapat Feb 7 '12 at 12:16
(You can edit your post to reflect your comment). – Cam McLeman Feb 7 '12 at 12:18
To Cam McLeman,Thanks for suggestion- Bapat – Madhav Bapat Feb 7 '12 at 12:31
May I ask you to post an answer to your own question? In such a way you will be sure that your intuition is correct and the question will not remain in the "unanswered" category forever! If you don´t have time to do that just let it know to someone who can answer. Thank you and welcome to Math.Se! – Giovanni De Gaetano Feb 7 '12 at 13:43

Example of non-perfect field: $\,\mathbb F_p(T)=\,$ the field of rational functions in an unknown (transcendental element) $\,T\,$ .

Why? The polynomial $\,f(x)=x^p-T\in\mathbb F_p(T)[x]\,$ is

$\,(1)\,\,$ irreducible: Apply Eisenstein's Criterion in the UFD $\,\mathbb F_p[T]\subset \mathbb F_p(T)\,$ and the prime $\,T\,$ in it

$\,(2)\,\,$ Let $\,\alpha\,$ be some root of $\,f(x)\,$ in some field extension, then $$\alpha^p=T\Longrightarrow x^p-\alpha^p=(x-\alpha)^p\in\mathbb F_p[T]$$and thus $\,\alpha\,$ is the unique root of $\,f(x)\,$, what makes this irreducible polynomial as inseparable as one could ever hope and, thus, the field $\,\mathbb F_p(T)\,$ is non-perfect.

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