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Take a field $k$. If $k'/k$ is a field extension of degree $p$, it is known that there are many possibilities for the isomorphism class of $k'$. See https://math.stackexchange.com/questions/353928/does-mathbbf-px-has-only-finitely-many-extension-of-a-given-degree/354804#354804 for example. But the examples given in Keith Conrad's post are all separable. Could you have more than one possibility for the isomorphism class of $k'$ such that $k'/k$ is a purely inseparable extension of degree $p$?

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  • $\begingroup$ You mean isomorphism as $K$-algebras? (i.e. as extensions of $K$?) $\endgroup$ – YCor Jul 29 '15 at 18:07
  • $\begingroup$ Trivial example: $K(t,u^{1/p})\supset K(t,u)\subset K(t^{1/p},u)$, where $p$ is the characteristic of $K$. $\endgroup$ – YCor Jul 29 '15 at 18:09
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    $\begingroup$ If you want field isomorphism (less natural), replace $K(t,u)$ with $\bigcup_{n\ge 0}K(t^{1/q^n},u)$, where $q$ is coprime to $p$. $\endgroup$ – YCor Jul 29 '15 at 18:10
  • $\begingroup$ @YCor. That's great thanks. We were aware of the first example, but the second is what we were after. Do you happen to know a good reference where this sort of stuff is discussed? $\endgroup$ – David Stewart Jul 29 '15 at 19:21
  • $\begingroup$ No idea where this could be written. It sounds quite artificial, so possibly nowhere. $\endgroup$ – YCor Jul 29 '15 at 19:28
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Suppose $k$ is finitely generated over a perfect field $k_0$ of characteristic $p$ (e.g. $k_0 = \mathbf{F}_p$), with transcendence degree $d > 0$. We'll make infinitely many such pairwise non-isomorphic extensions whenever $d> 1$. (Of course, $k^{1/p}$ is the unique such extension when $d=1$, as is well-known but will also drop out from the considerations below.)

By perfectness of $k_0$, there exists a separating transcendence basis $\{x_1, \dots, x_d\}$ over $k_0$, which is to say that $k$ is separable (of finite degree) over its subfield $K = k_0(x_1, \dots, x_d)$. You're looking for isomorphism classes of degree-$p$ sub extensions of $k^{1/p}$ over $k$. By separability of $k/K$ we have $$k^{1/p} = kK^{1/p} = k \otimes_K K^{1/p}.$$ (Indeed, to show the containment $kK^{1/p} \subset k^{1/p}$ is an equality it is the same to show that the containment $Kk^p \subset k$ is an equality. But $k$ is separable over $Kk^p$ since $Kk^p$ is intermediate to the extension $k/K$ that is separable, and it is clearly also purely inseparable over $Kk^p$ since $Kk^p$ is intermediate to the extension $k/k^p$ that is purely inseparable.)

It follows that for any subfield $F \subset K^{1/p}$ of degree $p$ over $K$ we see that $kF = k \otimes_K F$ is a subfield of $k^{1/p}$ of degree $p$ over $k$ (obviously purely inseparable as such). By consideration of separable and inseparable field degrees, the subfield $F$ is uniquely determined by the extension $kF/K$ as its unique subfield that is purely inseparable of degree $p$ over $K$. Thus, $kF_1 \simeq kF_2$ over $k$ implies $kF_1 \simeq kF_2$ over $K$ implies $F_1 \simeq F_2$ over $K$. Hence, we get an injection from the set of isomorphism classes of degree-$p$ purely inseparable extensions of $K = k_0(x_1, \dots, x_d)$ into the analogous such set of extensions of $k$.

Provided that $d > 1$, there are infinitely many such isomorphism classes in a sense we will soon make precise. Consideration of $p$-bases and their relationship with differential bases in characteristic $p$ (se Matsumura's book "Commutative Ring Theory", for example) shows that if $a \in K - K^p$ and $F := K(a^{1/p})$ then the kernel of the natural map $\Omega^1_{K/k_0} \rightarrow \Omega^1_{F/k_0}$ is the $K$-line spanned by ${\rm{d}}a$. But $\Omega^1_{K/k_0}$ is $d$-dimensional over $K$ and even just letting $a$ vary through elements of the form $x_1 + \sum_{j>1} c_j^p x_j$ with $c_2, \dots, c_d \in K$ makes the ${\rm{d}}a$'s pairwise linearly independent over $K$ in $\Omega^1_{K/k_0}$, so this gives a collection of such distinct $K$-lines parameterized by a $(d-1)$-dimensional vector space over $K^p$.

So we get a set of pairwise non-isomorphic purely inseparable degree-$p$ extensions of $k$ parameterized by a $(d-1)$-dimensional vector space over $K^p$. As Donald Trump would say, that is huuuuuge$^{\sf{TM}}$ (when $d > 1$).

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