Let $ R$ be a Noetherian normal domain, and let $ R^+$ denote the absolute integral closure of $ R$ , i.e. the directed union of all modulefinite extension domains. I’m trying to understand what “separable” should mean for elements of $ R^+$ . It’s hard to say what “should mean” means, exactly, but let’s suppose we want a definition
 That would allow us to factor a modulefinite extension $ R\subseteq S \subseteq R^+$ into a ‘separable’ part $ R \subseteq R^{\text{sep} }_S$ followed by a purely inseparable part $ R^{\text{sep} }_S\subseteq S$ (it’s much clearer what purely inseparable ‘should mean’),
 That agrees with the usual definition when $ R$ is a field, and
 That reduces to (or is not unimaginably stronger than) checking geometric reducedness for a certain fiber of the associated map $ \mathbb{A} ^1_S \to \mathbb{A}^1_R$ (as is true in the field case).
Anyway, here’s a fairly naive guess at a ‘correct’ definition: Let’s call $ \alpha \in R^+$ separable over $ R$ if the compositum $ R[\alpha]\subseteq R^+$ has the property that $ S \otimes_R R[\alpha]$ is reduced for any modulefinite extension domain $ S$ of $ R$ . We can write $ R[\alpha]$ as $ R[X]/P_\alpha$ where $ P_\alpha$ is prime, and under mild hypotheses on $ R$ (e.g. excellence), the height of $ P_\alpha$ must be $ 1$ . Either way, we are asking about the reducedness of $ S[X]/P_\alpha S[X]$ for varying $ S$ . This is of course equivalent to checking:

$ (S_1)$ : $ S[X]/P_\alpha S[X]$ has no embedded primes.

$ (R_0)$ : The local rings at the minimal primes of $ S[X]/P_\alpha S[X]$ are fields.
In trying to get my hands on this a bit more concretely, I’ve been trying to answer the following questions:
Q1: If $ R$ is a normal (excellent) domain, will the $ (S_1)$ condition automatically be satisfied, regardless of $ S$ and $ \alpha$ ?
Q2: In the same setting, is the $ (R_0)$ condition equivalent to reducedness for the fiber $ \kappa(P_\alpha)\otimes_{R[X]} S[X]$ of $ \mathbb{A} ^1_S \to \mathbb{A}^1_R$ over $ P_\alpha$ ?
The answer to both questions is ‘Yes’ if $ R$ is a field: In this case, $ S[X]/P_\alpha S[X]$ is torsionfree and modulefinite over the field $ R[X]/P_\alpha$ . Being zerodimensional, it’s associated primes need to be minimal (answering Q1), and being torsionfree over a domain implies that the fiber of $ R[X]/P_\alpha \to S[X]/P_\alpha S[X]$ over $ (0)$ is the product of the local rings at all the minimal primes of $ S[X]/P_\alpha S[X]$ (answering Q2). I emphasize torsionfreeness here because a hypothesis on $ R$ which would force $ S[X]/P_\alpha S[X]$ to be torsionfree over $ R[X]/P_\alpha$ for any $ S$ and $ \alpha$ would answer Q2.
Anyway, I’m somewhat more worried about Q1 than I am about Q2. You can localize the issue of embedded primes in a straightforward way:
Suppose that $ Q\in \text{Spec}(S[X])$ is an an embedded prime of the expansion $ P_A S[X]$ , and let $ P$ denote the contraction of $ Q$ to $ R[X]$ (which contains $ P_\alpha$ ). We are in a situation where we can use Going Down: since $ R$ is normal, so is $ R[X]$ , and $ S[X]$ is a modulefinite domain extension of $ R[X]$ . The upshot is that if we replace $ R[X]$ and $ S[X]$ by their localizations at $ P$ and $ Q$ , respectively, then the induced map on Spec is still surjective. In particular, we get a local injection
$ $ \frac{R[X]_P}{P_\alpha R[X]_P} \hookrightarrow \frac{S[X]_Q}{P_\alpha S[X]_Q}$ $
I was a little tempted to try to show that this general situation is just not possible:
$ (A,\mathfrak{m},K)\hookrightarrow (B,\mathfrak{n},L)$ is a local injection from a local domain to a local ring of depth $ 0$ , with both rings at least $ 1$ dimensional.
but you could in principle have something like this ($ k$ is a field):
$ $ k[[x]] \hookrightarrow \frac{(k[[x]])[\epsilon]}{(x\epsilon, \epsilon^2)}$ $
I’m not sure whether a scenario like this could actually happen just by looking at expansions of primes along $ R[X]\hookrightarrow S[X]$ and localizing. So, it seems like you’d either need you use the additional structure in our situation more carefully (I’m not clearly seeing how, and it seems like it could be tricky) or you’d need impose stronger conditions on $ R$ (maybe asking for $ R$ to be CohenMacaulay?). Of course, the other obvious thing to ask is simply
Q3: Is there a better notion of ‘separable’ to use in $ R^+$ when $ R$ is not a field?
(For the record, I’m most interested in the case where $ R$ is a complete local domain of positive characteristic, thinking of $ R^+$ as an enlargement in which I can find sequences of elements (getting small with respect to some $ \mathbb{Q}$ valuation) to use for tight closure tests over $ R$ . I mention this because there’s a large amount of extra structure on $ R^+$ as an $ R$ algebra, namely that it’s big CohenMacaulay, although I’m not sure whether this is relevant.)