It is clear that a Noetherian universally Japanese ring is universally Nagata (i.e., condition (2) holds). Let $R$ be a Nagata ring. We will show that any finitely generated $R$-algebra $S$ is Nagata. This will prove the proposition. \medskip\noindent Step 1. There exists a sequence of ring maps $R = R_0 \to R_1 \to R_2 \to \ldots \to R_n = S$ such that each $R_i \to R_{i + 1}$ is generated by a single element. Hence by induction it suffices to prove $S$ is Nagata if $S \cong R[x]/I$. \medskip\noindent Step 2. Let $\mathfrak q \subset S$ be a prime of $S$, and let $\mathfrak p \subset R$ be the corresponding prime of $R$. We have to show that $S/\mathfrak q$ is N-2. Hence we have reduced to the proving the following: (*) Given a Nagata domain $R$ and a monogenic extension $R \subset S$ of domains then $S$ is N-2. \medskip\noindent Step 3. Let $R$ be a Nagata domain and $R \subset S$ a monogenic extension of domains. Let $R \subset R'$ be the integral closure of $R$ in its fraction field. Let $S'$ be the subring of the fraction field of $S$ generated by $R'$ and $S$. As $R'$ is finite over $R$ (by the Nagata property) also $S'$ is finite over $S$. Since $S$ is Noetherian it suffices to prove that $S'$ is N-2 (Lemma \ref{lemma-finite-extension-N-2}). Hence we have reduced to proving the following: (**) Given a normal Nagata domain $R$ and a monogenic extension $R \subset S$ of domains then $S$ is N-2. \medskip\noindent Step 4: Let $R$ be a normal Nagata domain and let $R \subset S$ be a monogenic extension of domains. Suppose the induced extension of fraction fields of $R$ and $S$ is purely transcendental. In this case $S = R[x]$. By Lemma \ref{lemma-polynomial-ring-N-2} we see that $S$ is N-2. Hence we have reduced to proving the following: (**) Given a normal Nagata domain $R$ and a monogenic extension $R \subset S$ of domains inducing a finite extension of fraction fields then $S$ is N-2. \medskip\noindent Step 5. Let $R$ be a normal Nagata domain and let $R \subset S$ be a monogenic extension of domains inducing a finite extension of fraction fields $L/K$. Choose an element $x \in S$ which generates $S$ as an $R$-algebra. Let $L \subset M$ be a finite extension of fields. Let $R'$ be the integral closure of $R$ in $M$. Then the integral closure $S'$ of $S$ in $M$ is equal to the integral closure of $R'[x]$ in $M$. Also the fraction field of $R'$ is $M$ and $R \subset R'$ is finite (by the Nagata property of $R$). This implies that $R'$ is a Nagata ring (Lemma \ref{lemma-quasi-finite-over-nagata}). To show that $S'$ is finite over $S$ is the same as showing that $S'$ is finite over $R'[x]$. Replace $R$ by $R'$ and $S$ by $R'[x]$ to reduce to the following statement: (***) Given a normal Nagata domain $R$ with fraction field $K$, and $x \in K$, the ring $S \subset K$ generated by $R$ and $x$ is N-1. \medskip\noindent Step 6. Let $R$ be a normal Nagata domain with fraction field $K$. Let $x = b/a \in K$. We have to show that the ring $S \subset K$ generated by $R$ and $x$ is N-1. Note that $S_a \cong R_a$ is normal. Hence by Lemma \ref{lemma-characterize-N-1} it suffices to show that $S_{\mathfrak m}$ is N-1 for every maximal ideal $\mathfrak m$ of $S$. \medskip\noindent With assumptions as in the preceding paragraph, pick such a maximal ideal and set $\mathfrak n = R \cap \mathfrak m$. The residue field extension $\kappa(\mathfrak n) \subset \kappa(\mathfrak m)$ is finite (Theorem \ref{theorem-nullstellensatz}) and generated by the image of $x$. Hence there exists a monic polynomial $f(X) = X^d + \sum_{i = 1, \ldots, d} a_iX^{d -i}$ with $f(x) \in \mathfrak m$. Let $K \subset K''$ be a finite extension of fields such that $f(X)$ splits completely in $K''[X]$. Let $R'$ be the integral closure of $R$ in $K''$. Let $S' \subset K''$ be the subring generated by $R'$ and $x$. As $R$ is Nagata we see $R'$ is finite over $R$ and Nagata (Lemma \ref{lemma-quasi-finite-over-nagata}). Moreover, $S'$ is finite over $S$. If for every maximal ideal $\mathfrak m'$ of $S'$ the local ring $S'_{\mathfrak m'}$ is N-1, then $S'_{\mathfrak m}$ is N-1 by Lemma \ref{lemma-characterize-N-1}, which in turn implies that $S_{\mathfrak m}$ is N-1 by Lemma \ref{lemma-finite-extension-N-2}. After replacing $R$ by $R'$ and $S$ by $S'$, and $\mathfrak m$ by any of the maximal ideals $\mathfrak m'$ lying over $\mathfrak m$ we reach the situation where the polynomial $f$ above split completely: $f(X) = \prod_{i = 1, \ldots, d} (X - a_i)$ with $a_i \in R$. Since $f(x) \in \mathfrak m$ we see that $x - a_i \in \mathfrak m$ for some $i$. Finally, after replacing $x$ by $x - a_i$ we may assume that $x \in \mathfrak m$. \medskip\noindent To recapitulate: $R$ is a normal Nagata domain with fraction field $K$, $x \in K$ and $S$ is the subring of $K$ generated by $x$ and $R$, finally $\mathfrak m \subset S$ is a maximal ideal with $x \in \mathfrak m$. We have to show $S_{\mathfrak m}$ is N-1. \medskip\noindent We will show that Lemma \ref{lemma-criterion-analytically-unramified} applies to the local ring $S_{\mathfrak m}$ and the element $x$. This will imply that $S_{\mathfrak m}$ is analytically unramified, whereupon we see that it is N-1 by Lemma \ref{lemma-analytically-unramified-easy}. \medskip\noindent We have to check properties (1), (2), (3)(a) and (3)(b). Property (1) is trivial. Let $I = \Ker(R[X] \to S)$ where $X \mapsto x$. We claim that $I$ is generated by all linear forms $aX - b$ such that $ax = b$ in $K$. Clearly all these linear forms are in $I$. If $g = a_d X^d + \ldots a_1 X + a_0 \in I$, then we see that $a_dx$ is integral over $R$ (Lemma \ref{lemma-make-integral-trivial}) and hence $b := a_dx \in R$ as $R$ is normal. Then $g - (a_dX - b)X^{d - 1} \in I$ and we win by induction on the degree. As a consequence we see that $$ S/xS = R[X]/(X, I) = R/J $$ where $$ J = \{b \in R \mid ax = b \text{ for some }a \in R\} = xR \cap R $$ By Lemma \ref{lemma-normal-domain-intersection-localizations-height-1} we see that $S/xS = R/J$ has no embedded primes as an $R$-module, hence as an $R/J$-module, hence as an $S/xS$-module, hence as an $S$-module. This proves property (2). Take such an associated prime $\mathfrak q \subset S$ with the property $\mathfrak q \subset \mathfrak m$ (so that it is an associated prime of $S_{\mathfrak m}/xS_{\mathfrak m}$ -- it does not matter for the arguments). Then $\mathfrak q$ is minimal over $xS$ and hence has height $1$. By the sequence of equalities above we see that $\mathfrak p = R \cap \mathfrak q$ is an associated prime of $R/J$, and so has height $1$ (see Lemma \ref{lemma-normal-domain-intersection-localizations-height-1}). Thus $R_{\mathfrak p}$ is a discrete valuation ring and therefore $R_{\mathfrak p} \subset S_{\mathfrak q}$ is an equality. This shows that $S_{\mathfrak q}$ is regular. This proves property (3)(a). Finally, $(S/\mathfrak q)_{\mathfrak m}$ is a localization of $S/\mathfrak q$, which is a quotient of $S/xS = R/J$. Hence $(S/\mathfrak q)_{\mathfrak m}$ is a localization of a quotient of the Nagata ring $R$, hence Nagata (Lemmas \ref{lemma-quasi-finite-over-nagata} and \ref{lemma-nagata-localize}) and hence analytically unramified (Lemma \ref{lemma-local-nagata-domain-analytically-unramified}). This shows (3)(b) holds and we are done.