Technically, an element u in K is purely inseparable, with polynomial x-u, but we usually require a multiplicity > 1, so that q(x) is nontrivial, and u does not belong to K. In other words, a purely inseparable element is inseparable.
The field extension F/K is purely inseparable if every u in F-K is purely inseparable. Let's try to characterize such an extension.
Let u be purely inseparable over K. We already know K cannot have characteristic 0, nor can K be finite, so K is a transcendental extension of Zp.
If q(u) = 0, then q is (x-u)e for some exponent e. (Remember that - really means + when p = 2.) Here e is the "common multiplicity" of the single root u. Write e as mpr, where p does not divide m. We can raise to the p power first, remembering that (a+b)p = ap+bp. Hence q can be rewritten as x(pr)-u(pr), all raised to the m.
Think of q as (a-b)m and apply the binomial theorem. Consider the term mbam-1. Remember, q(x) has coefficients in K, hence mb lies in K, hence b lies in K. So a-b represents a polynomial that is already in K. We can't raise this to a higher power and find an irreducible polynomial. Therefore m = 1, and q = x(pr)-u(pr).
In summary, u is purely inseparable if it is the pr root of v for some v in K. And F is a purely inseparable extension if every u has some r such that u to the pr lies in K.
Now for the converse. Let F/K be a field extension such that every u has some r such that u to the pr lies in K. Let q(x) = x(pr)-u(pr). Now q is a polynomial in K with root u. It can be rewritten as (x-u)(pr). The same root u appears over and over again to build q. If q is irreducible then u is purely inseparable and we are done. If q can be factored, use one of the irreducible factors. It is still made of copies of x-u, and u remains purely inseparable. Since this holds for all u in F, F/K is purely inseparable.
It is interesting to watch this fail when K is finite. The frobenius homomorphism xp is an automorphism that can be reversed. The pth root of anything in K is already in K. There are no purely inseparable elements over a finite field. However, when K is infinite the map xp may not be onto, and we have an opportunity for a purely inseparable extension.
There is another characterization of a purely inseparable extension, but first we need a lemma.
Let u be an inseparable element over the base field K. Again, K is some transcendental extension of Zp. Let q be the irreducible polynomial with root u. Remember that u has multiplicity > 1 in q. Let r be the derivative of q. Compute gcd(q,r) to find all the repeated factors. We know there are some, since u is inseparable. However, a gcd would divide q, whence q could not be irreducible. This apparent contradiction is resolved only when r drops to 0 in the field K.
The derivative of q is 0, so all the exponents on the terms of q(x) are divisible by p. Divide these exponents by p and replace u with up. Note that u is still a root of q. Repeat this process as long as u remains inseparable. In conclusion, we can raise any inseparable element to the pr power, for some r, and get a separable element.
With this lemma in hand, a purely inseparable extension is an algebraic extension with all its separable elements in K. One direction is obvious, so assume the separable elements are all in K. Every u in F-K is inseparable, and by the above lemma, we can raise it to a power and find a separable element, which lies in K. Hence every u satisfies u(pr) in K for some r, and F is purely inseparable.
When raising u±v, u*v, or u/v to the p power, we can rais u and v to the p, then apply the arithmetic operator. Therefore, if u and v are purely inseparable, there is a sufficiently large r that shows u+v is purely inseparable. The same for u*v and u/v, hence the field generated by purely inseparable elements is purely inseparable. Also, the set of purely inseparable elements in the extension F/K forms a field, a purely inseparable extension of K inside F.
If F is purely inseparable over K, and u is purely inseparable over F, then u is the root of something in F, which is the root of something in K, hence u is purely inseparable over K. The composition of purely inseparable extensions is purely inseparable.
If K is countable, such as Zp(x,y,z), (polynomials in x y and z with coefficients mod p), adjoin the pth roots of elements in K, one by one, and repeat, to create a complete purely inseparable extension of K. If K is uncountable, the complete purely inseparable extension exists, and is a subset of the closure of K. As such, it is unique up to isomorphism. This fact relies on the axiom of choice.