Proof.
It is clear that we have the implications (1) \Rightarrow (2) and (2) \Rightarrow (4). Recall that a domain which is J-1 is J-0. Hence we also have the implications (1) \Rightarrow (3) and (3) \Rightarrow (4).
Let R \to S be a finite type ring map and let's try to show S is J-1. By Lemma 15.47.3 it suffices to prove that S/\mathfrak q is J-0 for every prime \mathfrak q of S. In this way we see (2) \Rightarrow (1).
Assume (4). We will show that (2) holds which will finish the proof. Let R \to S be a finite type ring map with S a domain. Let \mathfrak p = \mathop{\mathrm{Ker}}(R \to S). Let K be the fraction field of S. There exists a diagram of fields
\xymatrix{ K \ar[r] & K' \\ \kappa (\mathfrak p) \ar[u] \ar[r] & L \ar[u] }
where the horizontal arrows are finite purely inseparable field extensions and where K'/L is separable, see Algebra, Lemma 10.42.4. Choose R' \subset L as in (4) and let S' be the image of the map S \otimes _ R R' \to K'. Then S' is a domain whose fraction field is K', hence S' is J-0 by Lemma 15.47.5 and our choice of R'. Then we apply Lemma 15.47.4 to see that S is J-0 as desired.
\square
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