**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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