What kinds of fields do exist that are non-perfect?

Question

This question has already been asked twice in a similar manner (Non-Perfect Fields and Examples of fields which are not perfect). In both cases, the standard answer found in introductory textbooks was given. This is, the function fields $\mathbb{F}_q (t)$ for some finite field are not perfect.

I couldn't think of any other way to construct a non-perfect fields (except adding more variables to get $\mathbb{F}_q (t,s^q)$ or some variatons of that). I asked myself wether there are other ways to obtain non-perfect fields and in what situations these might show up.

My question thus is:

Which interesting examples of non-perfect fields do you know?

Edit. I understand that every non-perfect field must have $\mathbb{F}_p$ as it's prime field. As it cannot be algebraic, it must be transcendental. The answers and comments so far suggest that this already tells us how to get all non-perfect fields. But nonetheless: Looking at $\mathbb{Q}(i)$ we obtain a highly transcendental extension $\mathbb{C}$ in a "non-boring" way. By this I mean that we don't just add variables to get some kind of function field, but we use an analytic construction, involving some kind of metric etc. Thus, adding variables is not the only way to get transcendental extensions. This makes me wonder wether there are similar (maybe more algebraic) methods that produce transcendental extension of finite prime fields that are not perfect.

Answer 1

Fields of characteristic $0$ are perfect, and a field of characteristic $p$ contains $\mathbb{F}_p$. Algebraic extensions of $\mathbb{F}_p$ are perfect, so non-perfect extensions must contain some transcendental element $t$, i.e. they contain $\mathbb{F}_p(t)$. In this sense $\mathbb{F}_p(t)$ is the "universal" example of a non-perfect field.

Notice, however, that not every field containing $\mathbb{F}_p(t)$ must be non-perfect, just consider its perfect closure $\mathbb{F}_p(t,t^{1/p},t^{1/p^2},\dotsc)$.

If $K$ is any field of characteristic $p$, then $K(t)$ is not perfect, since $t^{1/p} \notin K(t)$.

You can also impose relations. For example, $\mathbb{F}_3(u,v)[t] / \langle t^2 + ut + v \rangle$ is not perfect.

Answer 2

It's been a while since this question was asked, but consider this question and answer.

In your edit you mentioned that if we extend fields not algebraically but by completion we may get other examples. Any completion of a global field of characteristic $0$, i.e. a number field, has characteristic $0$ and is thus perfect. This is how we get $\mathbb{C}$, as you mentioned.

The only global fields of characteristic $p>0$ are the $\mathbb{F}_q(t)$, for a power $q$ of $p$. The valuations on such field come from either the degree function or the irreducible polynomials. If we pass to the completion, in any case, we obtain a field isomorphic to the field of formal Laurent series $\mathbb{F}_q((t))$. This is now a local field, in fact, it is complete with respect to a discrete valuation.

$\mathbb{F}_q((t))$ does obviously contain $\mathbb{F}_q(t)$, so it has a chance to be non perfect. Notice that $t$ generates the maximal ideal of the valuation ring $\mathbb{F}_q[[t]]$ (formal power series). In particular, this means that $X^p-t \in \mathbb{F}_q[[t]][X]$ is irreducible, else a root of this polynomial lies in $\mathbb{F}_q[[t]]$ and has stricly smaller valuation than $t$. This means the Frobenius is not an isomorphism, so $\mathbb{F}_q((t))$ is not perfect.

I guess this is how you get a somewhat interesting example of a non-perfect field, one which is not an algebraic extension of $\mathbb{F}_q(t)$.