hcju/stacksqa huggingface dataset:
First assume $R$ is N-1. Let $R'$ be the integral closure of $R$ in its field of fractions $K$. By assumption we can find $x_1, \ldots, x_n$ in $R'$ which generate $R'$ as an $R$-module. Since $R' \subset K$ we can find $f_i \in R$ nonzero such that $f_i x_i \in R$. Then $R_f \cong R'_f$ where $f = f_1 \ldots f_n$. Hence $R_f$ is normal and we have (1). Part (2) follows from Lemma \ref{lemma-localize-N}. \medskip\noindent Assume (1) and (2). Let $K$ be the fraction field of $R$. Suppose that $R \subset R' \subset K$ is a finite extension of $R$ contained in $K$. Note that $R_f = R'_f$ since $R_f$ is already normal. Hence by Lemma \ref{lemma-openness-normal-locus} the set of primes $\mathfrak p' \in \Spec(R')$ with $R'_{\mathfrak p'}$ non-normal is closed in $\Spec(R')$. Since $\Spec(R') \to \Spec(R)$ is closed the image of this set is closed in $\Spec(R)$. For such a ring $R'$ denote $Z_{R'} \subset \Spec(R)$ this image. \medskip\noindent Pick a maximal ideal $\mathfrak m \subset R$. Let $R_{\mathfrak m} \subset R_{\mathfrak m}'$ be the integral closure of the local ring in $K$. By assumption this is a finite ring extension. By Lemma \ref{lemma-integral-closure-localize} we can find finitely many elements $r_1, \ldots, r_n \in K$ integral over $R$ such that $R_{\mathfrak m}'$ is generated by $r_1, \ldots, r_n$ over $R_{\mathfrak m}$. Let $R' = R[x_1, \ldots, x_n] \subset K$. With this choice it is clear that $\mathfrak m \not \in Z_{R'}$. \medskip\noindent As $\Spec(R)$ is quasi-compact, the above shows that we can find a finite collection $R \subset R'_i \subset K$ such that $\bigcap Z_{R'_i} = \emptyset$. Let $R'$ be the subring of $K$ generated by all of these. It is finite over $R$. Also $Z_{R'} = \emptyset$. Namely, every prime $\mathfrak p'$ lies over a prime $\mathfrak p'_i$ such that $(R'_i)_{\mathfrak p'_i}$ is normal. This implies that $R'_{\mathfrak p'} = (R'_i)_{\mathfrak p'_i}$ is normal too. Hence $R'$ is normal, in other words $R'$ is the integral closure of $R$ in $K$.
We may assume $x \not = 0$ since otherwise the lemma is trivial. Let $K$ be the fraction field of $R$. If the characteristic of $K$ is zero the lemma follows from (1), see Lemma \ref{lemma-domain-char-zero-N-1-2}. Hence we may assume that the characteristic of $K$ is $p > 0$, and we may apply Lemma \ref{lemma-domain-char-p-N-1-2}. Thus given $K \subset L$ be a finite purely inseparable field extension we have to show that the integral closure $S$ of $R$ in $L$ is finite over $R$. \medskip\noindent Let $q$ be a power of $p$ such that $L^q \subset K$. By enlarging $L$ if necessary we may assume there exists an element $y \in L$ such that $y^q = x$. Since $R \to S$ induces a homeomorphism of spectra (see Lemma \ref{lemma-p-ring-map}) there is a unique prime ideal $\mathfrak q \subset S$ lying over the prime ideal $\mathfrak p = xR$. It is clear that $$ \mathfrak q = \{f \in S \mid f^q \in \mathfrak p\} = yS $$ since $y^q = x$. Observe that $R_{\mathfrak p}$ is a discrete valuation ring by Lemma \ref{lemma-characterize-dvr}. Then $S_{\mathfrak q}$ is Noetherian by Krull-Akizuki (Lemma \ref{lemma-krull-akizuki}). Whereupon we conclude $S_{\mathfrak q}$ is a discrete valuation ring by Lemma \ref{lemma-characterize-dvr} once again. By Lemma \ref{lemma-finite-extension-residue-fields-dimension-1} we see that $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$ is a finite field extension. Hence the integral closure $S' \subset \kappa(\mathfrak q)$ of $R/xR$ is finite over $R/xR$ by assumption (2). Since $S/yS \subset S'$ this implies that $S/yS$ is finite over $R$. Note that $S/y^nS$ has a finite filtration whose subquotients are the modules $y^iS/y^{i + 1}S \cong S/yS$. Hence we see that each $S/y^nS$ is finite over $R$. In particular $S/xS$ is finite over $R$. Also, it is clear that $\bigcap x^nS = (0)$ since an element in the intersection has $q$th power contained in $\bigcap x^nR = (0)$ (Lemma \ref{lemma-intersect-powers-ideal-module-zero}). Thus we may apply Lemma \ref{lemma-finite-over-complete-ring} to conclude that $S$ is finite over $R$, and we win.
Observe that $R[[x]]$ is Noetherian by Lemma \ref{lemma-Noetherian-power-series}. Let $R' \supset R$ be the integral closure of $R$ in its fraction field. Because $R$ is N-2 this is finite over $R$. Hence $R'[[x]]$ is finite over $R[[x]]$. By Lemma \ref{lemma-power-series-over-Noetherian-normal-domain} we see that $R'[[x]]$ is a normal domain. Apply Lemma \ref{lemma-tate-japanese} to the element $x \in R'[[x]]$ to see that $R'[[x]]$ is N-2. Then Lemma \ref{lemma-finite-extension-N-2} shows that $R[[x]]$ is N-2.
As $S$ is essentially of finite type over $R$ it is Noetherian and has finitely many minimal primes $\mathfrak q_1, \ldots, \mathfrak q_m$, see Lemma \ref{lemma-Noetherian-irreducible-components}. Since $S$ is reduced we have $S \subset \prod S_{\mathfrak q_i}$ and each $S_{\mathfrak q_i} = K_i$ is a field, see Lemmas \ref{lemma-total-ring-fractions-no-embedded-points} and \ref{lemma-minimal-prime-reduced-ring}. It suffices to show that the integral closure $A_i'$ of $R$ in each $K_i$ is finite over $R$. This is true because $R$ is Noetherian and $A \subset \prod A_i'$. Let $\mathfrak p_i \subset R$ be the prime of $R$ corresponding to $\mathfrak q_i$. As $S$ is essentially of finite type over $R$ we see that $K_i = S_{\mathfrak q_i} = \kappa(\mathfrak q_i)$ is a finitely generated field extension of $\kappa(\mathfrak p_i)$. Hence the algebraic closure $L_i$ of $\kappa(\mathfrak p_i)$ in $\subset K_i$ is finite over $\kappa(\mathfrak p_i)$, see Fields, Lemma \ref{fields-lemma-algebraic-closure-in-finitely-generated}. It is clear that $A_i'$ is the integral closure of $R/\mathfrak p_i$ in $L_i$, and hence we win by definition of a Nagata ring.
Namely, assume the condition of the lemma. Let $R \to S$ be a finite type ring map with $S$ a domain. Let $L$ be a finite extension of the fraction field of $S$. Then there exists a finite ring extension $S \subset S' \subset L$ such that $L$ is the fraction field of $S'$. By assumption $S'$ is N-1, and hence the integral closure $S''$ of $S'$ in $L$ is finite over $S'$. Thus $S''$ is finite over $S$ (Lemma \ref{lemma-finite-transitive}) and $S''$ is the integral closure of $S$ in $L$ (Lemma \ref{lemma-integral-closure-transitive}). We conclude that $R$ is universally Japanese.
First note that $S$ is Noetherian as $R$ is Noetherian and a quasi-finite ring map is of finite type. Let $\mathfrak q \subset S$ be a prime ideal, and set $\mathfrak p = R \cap \mathfrak q$. Then $R/\mathfrak p \subset S/\mathfrak q$ is quasi-finite and hence we conclude that $S/\mathfrak q$ is N-2 by Lemma \ref{lemma-quasi-finite-over-Noetherian-japanese} as desired.
Let $\varphi : R \to S$ be a finite type ring map so that $S$ is a domain. Then $\varphi(f_1), \ldots, \varphi(f_n)$ generate the unit ideal in $S$. Hence if each $S_{f_i} = S_{\varphi(f_i)}$ is N-1 then so is $S$, see Lemma \ref{lemma-Japanese-local}. This proves (1). \medskip\noindent If each $R_{f_i}$ is Nagata, then each $R_{f_i}$ is Noetherian and hence $R$ is Noetherian, see Lemma \ref{lemma-cover}. And if $\mathfrak p \subset R$ is a prime, then we see each $R_{f_i}/\mathfrak pR_{f_i} = (R/\mathfrak p)_{f_i}$ is N-2 and hence we conclude $R/\mathfrak p$ is N-2 by Lemma \ref{lemma-Japanese-local}. This proves (2).
Let $R$ be a complete local Noetherian ring. Let $\mathfrak p \subset R$ be a prime. Then $R/\mathfrak p$ is also a complete local Noetherian ring, see Lemma \ref{lemma-quotient-complete-local}. Hence it suffices to show that a Noetherian complete local domain $R$ is N-2. By Lemmas \ref{lemma-quasi-finite-over-Noetherian-japanese} and \ref{lemma-complete-local-Noetherian-domain-finite-over-regular} we reduce to the case $R = k[[X_1, \ldots, X_d]]$ where $k$ is a field or $R = \Lambda[[X_1, \ldots, X_d]]$ where $\Lambda$ is a Cohen ring. \medskip\noindent In the case $k[[X_1, \ldots, X_d]]$ we reduce to the statement that a field is N-2 by Lemma \ref{lemma-power-series-over-N-2}. This is clear. In the case $\Lambda[[X_1, \ldots, X_d]]$ we reduce to the statement that a Cohen ring $\Lambda$ is N-2. Applying Lemma \ref{lemma-tate-japanese} once more with $x = p \in \Lambda$ we reduce yet again to the case of a field. Thus we win.
In this proof we will use the remarks immediately following Definition \ref{definition-analytically-unramified}. As $R \to R^\wedge$ is a faithfully flat local ring homomorphism it is injective and (1) follows. \medskip\noindent Let $\mathfrak q$ be a minimal prime of $R$, and assume $R$ is analytically unramified. Then $\mathfrak q$ is an associated prime of $R$ (see Proposition \ref{proposition-minimal-primes-associated-primes}). Hence there exists an $f \in R$ such that $\{x \in R \mid fx = 0\} = \mathfrak q$. Note that $(R/\mathfrak q)^\wedge = R^\wedge/\mathfrak q^\wedge$, and that $\{x \in R^\wedge \mid fx = 0\} = \mathfrak q^\wedge$, because completion is exact (Lemma \ref{lemma-completion-flat}). If $x \in R^\wedge$ is such that $x^2 \in \mathfrak q^\wedge$, then $fx^2 = 0$ hence $(fx)^2 = 0$ hence $fx = 0$ hence $x \in \mathfrak q^\wedge$. Thus $\mathfrak q$ is analytically unramified and (2) holds. \medskip\noindent Assume $R$ is reduced with minimal primes $\mathfrak q_1, \ldots, \mathfrak q_t$, and each $\mathfrak q_i$ is analytically unramified. Then $R \to R/\mathfrak q_1 \times \ldots \times R/\mathfrak q_t$ is injective. Since completion is exact (see Lemma \ref{lemma-completion-flat}) we see that $R^\wedge \subset (R/\mathfrak q_1)^\wedge \times \ldots \times (R/\mathfrak q_t)^\wedge$. Hence (3) is clear. \medskip\noindent Assume $R$ is analytically unramified. Let $\mathfrak p_1, \ldots, \mathfrak p_s$ be the minimal primes of $R^\wedge$. Then we see that $$ Q(R^\wedge) = R^\wedge_{\mathfrak p_1} \times \ldots \times R^\wedge_{\mathfrak p_s} $$ with each $R^\wedge_{\mathfrak p_i}$ a field as $R^\wedge$ is reduced (see Lemma \ref{lemma-total-ring-fractions-no-embedded-points}). Hence the integral closure $S$ of $R^\wedge$ in $Q(R^\wedge)$ is equal to $S = S_1 \times \ldots \times S_s$ with $S_i$ the integral closure of $R^\wedge/\mathfrak p_i$ in its fraction field. In particular $S$ is finite over $R^\wedge$. Denote $R'$ the integral closure of $R$ in $Q(R)$. As $R \to R^\wedge$ is flat we see that $R' \otimes_R R^\wedge \subset Q(R) \otimes_R R^\wedge \subset Q(R^\wedge)$. Moreover $R' \otimes_R R^\wedge$ is integral over $R^\wedge$ (Lemma \ref{lemma-base-change-integral}). Hence $R' \otimes_R R^\wedge \subset S$ is a $R^\wedge$-submodule. As $R^\wedge$ is Noetherian it is a finite $R^\wedge$-module. Thus we may find $f_1, \ldots, f_n \in R'$ such that $R' \otimes_R R^\wedge$ is generated by the elements $f_i \otimes 1$ as a $R^\wedge$-module. By faithful flatness we see that $R'$ is generated by $f_1, \ldots, f_n$ as an $R$-module. This proves (4). \medskip\noindent Part (5) is a special case of part (4).
Let $\mathfrak p_1, \ldots, \mathfrak p_t$ be the associated primes of the $R$-module $R/xR$. Since $R/xR$ has no embedded primes we see that each $\mathfrak p_i$ has height $1$, and is a minimal prime over $(x)$. For each $i$, let $\mathfrak q_{i1}, \ldots, \mathfrak q_{is_i}$ be the associated primes of the $R^\wedge$-module $R^\wedge/\mathfrak p_iR^\wedge$. By Lemma \ref{lemma-codimension-1-analytically-unramified} we see that $(R^\wedge)_{\mathfrak q_{ij}}$ is regular. By Lemma \ref{lemma-bourbaki} we see that $$ \text{Ass}_{R^\wedge}(R^\wedge/xR^\wedge) = \bigcup\nolimits_{\mathfrak p \in \text{Ass}_R(R/xR)} \text{Ass}_{R^\wedge}(R^\wedge/\mathfrak pR^\wedge) = \{\mathfrak q_{ij}\}. $$ Let $y \in R^\wedge$ with $y^2 = 0$. As $(R^\wedge)_{\mathfrak q_{ij}}$ is regular, and hence a domain (Lemma \ref{lemma-regular-domain}) we see that $y$ maps to zero in $(R^\wedge)_{\mathfrak q_{ij}}$. Hence $y$ maps to zero in $R^\wedge/xR^\wedge$ by Lemma \ref{lemma-zero-at-ass-zero}. Hence $y = xy'$. Since $x$ is a nonzerodivisor (as $R \to R^\wedge$ is flat) we see that $(y')^2 = 0$. Hence we conclude that $y \in \bigcap x^nR^\wedge = (0)$ (Lemma \ref{lemma-intersect-powers-ideal-module-zero}).
By induction on $\dim(R)$. The case $\dim(R) = 0$ is trivial. Hence we assume $\dim(R) = d$ and that the lemma holds for all Noetherian Nagata domains of dimension $< d$. \medskip\noindent Let $R \subset S$ be the integral closure of $R$ in the field of fractions of $R$. By assumption $S$ is a finite $R$-module. By Lemma \ref{lemma-quasi-finite-over-nagata} we see that $S$ is Nagata. By Lemma \ref{lemma-integral-sub-dim-equal} we see $\dim(R) = \dim(S)$. Let $\mathfrak m_1, \ldots, \mathfrak m_t$ be the maximal ideals of $S$. Each of these lies over the maximal ideal $\mathfrak m$ of $R$. Moreover $$ (\mathfrak m_1 \cap \ldots \cap \mathfrak m_t)^n \subset \mathfrak mS $$ for sufficiently large $n$ as $S/\mathfrak mS$ is Artinian. By Lemma \ref{lemma-completion-flat} $R^\wedge \to S^\wedge$ is an injective map, and by the Chinese Remainder Lemma \ref{lemma-chinese-remainder} combined with Lemma \ref{lemma-change-ideal-completion} we have $S^\wedge = \prod S^\wedge_i$ where $S^\wedge_i$ is the completion of $S$ with respect to the maximal ideal $\mathfrak m_i$. Hence it suffices to show that $S_{\mathfrak m_i}$ is analytically unramified. In other words, we have reduced to the case where $R$ is a Noetherian normal Nagata domain. \medskip\noindent Assume $R$ is a Noetherian, normal, local Nagata domain. Pick a nonzero $x \in \mathfrak m$ in the maximal ideal. We are going to apply Lemma \ref{lemma-criterion-analytically-unramified}. We have to check properties (1), (2), (3)(a) and (3)(b). Property (1) is clear. We have that $R/xR$ has no embedded primes by Lemma \ref{lemma-normal-domain-intersection-localizations-height-1}. Thus property (2) holds. The same lemma also tells us each associated prime $\mathfrak p$ of $R/xR$ has height $1$. Hence $R_{\mathfrak p}$ is a $1$-dimensional normal domain hence regular (Lemma \ref{lemma-characterize-dvr}). Thus (3)(a) holds. Finally (3)(b) holds by induction hypothesis, since $R/\mathfrak p$ is Nagata (by Lemma \ref{lemma-quasi-finite-over-nagata} or directly from the definition). Thus we conclude $R$ is analytically unramified.
Assume $R$ is Nagata and let $R \to S$ and $\mathfrak m' \subset S$ be as in (2). Then $S$ is Nagata by Lemma \ref{lemma-quasi-finite-over-nagata}. Hence the local ring $S_{\mathfrak m'}$ is Nagata (Lemma \ref{lemma-nagata-localize}). Thus it is analytically unramified by Lemma \ref{lemma-local-nagata-domain-analytically-unramified}. It is clear that (2) implies (3). \medskip\noindent Assume (3) holds. Let $\mathfrak p \subset R$ be a prime ideal and let $L/\kappa(\mathfrak p)$ be a finite extension of fields. To prove (1) we have to show that the integral closure of $R/\mathfrak p$ is finite over $R/\mathfrak p$. Choose $x_1, \ldots, x_n \in L$ which generate $L$ over $\kappa(\mathfrak p)$. For each $i$ let $P_i(T) = T^{d_i} + a_{i, 1} T^{d_i - 1} + \ldots + a_{i, d_i}$ be the minimal polynomial for $x_i$ over $\kappa(\mathfrak p)$. After replacing $x_i$ by $f_i x_i$ for a suitable $f_i \in R$, $f_i \not \in \mathfrak p$ we may assume $a_{i, j} \in R/\mathfrak p$. In fact, after further multiplying by elements of $\mathfrak m$, we may assume $a_{i, j} \in \mathfrak m/\mathfrak p \subset R/\mathfrak p$ for all $i, j$. Having done this let $S = R/\mathfrak p[x_1, \ldots, x_n] \subset L$. Then $S$ is finite over $R$, a domain, and $S/\mathfrak m S$ is a quotient of $R/\mathfrak m[T_1, \ldots, T_n]/(T_1^{d_1}, \ldots, T_n^{d_n})$. Hence $S$ is local. By (3) $S$ is analytically unramified and by Lemma \ref{lemma-analytically-unramified-easy} we find that its integral closure $S'$ in $L$ is finite over $S$. Since $S'$ is also the integral closure of $R/\mathfrak p$ in $L$ we win.
Consider the ring extension $A \subset B = A[x]/(x^p - a)$. If $a$ does not have a $p$th root in $A$, then $B$ is a domain whose completion isn't reduced. This contradicts our earlier results, as $B$ is a Nagata ring (Proposition \ref{proposition-nagata-universally-japanese}) and hence analytically unramified by Lemma \ref{lemma-local-nagata-domain-analytically-unramified}.
In the statement and in the proof below, we take the depth of $M$ as an $R$-module, the depth of $M \otimes_R N$ as an $S$-module, and the depth of $N/\mathfrak m_RN$ as an $S/\mathfrak m_RS$-module. Denote $n$ the right hand side. First assume that $n$ is zero. Then both $\text{depth}(M) = 0$ and $\text{depth}(N/\mathfrak m_RN) = 0$. This means there is a $z \in M$ whose annihilator is $\mathfrak m_R$ and a $\overline{y} \in N/\mathfrak m_RN$ whose annihilator is $\mathfrak m_S/\mathfrak m_RS$. Let $y \in N$ be a lift of $\overline{y}$. Since $N$ is flat over $R$ the map $z : R/\mathfrak m_R \to M$ produces an injective map $N/\mathfrak m_RN \to M \otimes_R N$. Hence the annihilator of $z \otimes y$ is $\mathfrak m_S$. Thus $\text{depth}(M \otimes_R N) = 0$ as well. \medskip\noindent Assume $n > 0$. If $\text{depth}(N/\mathfrak m_RN) > 0$, then we may choose $f \in \mathfrak m_S$ mapping to $\overline{f} \in S/\mathfrak m_RS$ which is a nonzerodivisor on $N/\mathfrak m_RN$. Then $\text{depth}(N/\mathfrak m_RN) = \text{depth}(N/(f, \mathfrak m_R)N) + 1$ by Lemma \ref{lemma-depth-drops-by-one}. According to Lemma \ref{lemma-mod-injective} the element $f \in S$ is a nonzerodivisor on $N$ and $N/fN$ is flat over $R$. Hence by induction on $n$ we have $$ \text{depth}(M \otimes_R N/fN) = \text{depth}(M) + \text{depth}(N/(f, \mathfrak m_R)N). $$ Because $N/fN$ is flat over $R$ the sequence $$ 0 \to M \otimes_R N \to M \otimes_R N \to M \otimes_R N/fN \to 0 $$ is exact where the first map is multiplication by $f$ (Lemma \ref{lemma-flat-tor-zero}). Hence by Lemma \ref{lemma-depth-drops-by-one} we find that $\text{depth}(M \otimes_R N) = \text{depth}(M \otimes_R N/fN) + 1$ and we conclude that equality holds in the formula of the lemma. \medskip\noindent If $n > 0$, but $\text{depth}(N/\mathfrak m_RN) = 0$, then we can choose $f \in \mathfrak m_R$ which is a nonzerodivisor on $M$. As $N$ is flat over $R$ it is also the case that $f$ is a nonzerodivisor on $M \otimes_R N$. By induction on $n$ again we have $$ \text{depth}(M/fM \otimes_R N) = \text{depth}(M/fM) + \text{depth}(N/\mathfrak m_RN). $$ In this case $\text{depth}(M \otimes_R N) = \text{depth}(M/fM \otimes_R N) + 1$ and $\text{depth}(M) = \text{depth}(M/fM) + 1$ by Lemma \ref{lemma-depth-drops-by-one} and we conclude that equality holds in the formula of the lemma.
Let $\mathfrak q$ be a prime of $S$ lying over a prime $\mathfrak p$ of $R$. By Lemma \ref{lemma-apply-grothendieck} we have $$ \text{depth}(S_{\mathfrak q}) = \text{depth}(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) + \text{depth}(R_{\mathfrak p}). $$ On the other hand, we have $$ \dim(R_{\mathfrak p}) + \dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) \geq \dim(S_{\mathfrak q}) $$ by Lemma \ref{lemma-dimension-base-fibre-total}. (Actually equality holds, by Lemma \ref{lemma-dimension-base-fibre-equals-total} but strictly speaking we do not need this.) Finally, as the fibre rings of the map are assumed $(S_k)$ we see that $\text{depth}(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) \geq \min(k, \dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}))$. Thus the lemma follows by the following string of inequalities \begin{eqnarray*} \text{depth}(S_{\mathfrak q}) & = & \text{depth}(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) + \text{depth}(R_{\mathfrak p}) \\ & \geq & \min(k, \dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q})) + \min(k, \dim(R_{\mathfrak p})) \\ & = & \min(2k, \dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) + k, k + \dim(R_\mathfrak p), \dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) + \dim(R_{\mathfrak p})) \\ & \geq & \min(k, \dim(S_{\mathfrak q})) \end{eqnarray*} as desired.
Let $\mathfrak q$ be a prime of $S$ lying over a prime $\mathfrak p$ of $R$. Assume that $\dim(S_{\mathfrak q}) \leq k$. Since $\dim(S_{\mathfrak q}) = \dim(R_{\mathfrak p}) + \dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q})$ by Lemma \ref{lemma-dimension-base-fibre-equals-total} we see that $\dim(R_{\mathfrak p}) \leq k$ and $\dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) \leq k$. Hence $R_{\mathfrak p}$ and $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ are regular by assumption. It follows that $S_{\mathfrak q}$ is regular by Lemma \ref{lemma-flat-over-regular-with-regular-fibre}.
For Noetherian rings reduced is the same as having properties $(S_1)$ and $(R_0)$, see Lemma \ref{lemma-criterion-reduced}. Thus we know $R$ and the fibre rings have these properties. Hence we may apply Lemmas \ref{lemma-Sk-goes-up} and \ref{lemma-Rk-goes-up} and we see that $S$ is $(S_1)$ and $(R_0)$, in other words reduced by Lemma \ref{lemma-criterion-reduced} again.
Observe that $R \to S$ is flat with regular fibres (see the list of results on smooth ring maps in Section \ref{section-smooth-overview}). In particular, the fibres are reduced. Thus if $R$ is Noetherian, then $S$ is Noetherian and we get the result from Lemma \ref{lemma-reduced-goes-up-noetherian}. \medskip\noindent In the general case we may find a finitely generated $\mathbf{Z}$-subalgebra $R_0 \subset R$ and a smooth ring map $R_0 \to S_0$ such that $S \cong R \otimes_{R_0} S_0$, see remark (10) in Section \ref{section-smooth-overview}. Now, if $x \in S$ is an element with $x^2 = 0$, then we can enlarge $R_0$ and assume that $x$ comes from an element $x_0 \in S_0$. After enlarging $R_0$ once more we may assume that $x_0^2 = 0$ in $S_0$. However, since $R_0 \subset R$ is reduced we see that $S_0$ is reduced and hence $x_0 = 0$ as desired.
For a Noetherian ring being normal is the same as having properties $(S_2)$ and $(R_1)$, see Lemma \ref{lemma-criterion-normal}. Thus we know $R$ and the fibre rings have these properties. Hence we may apply Lemmas \ref{lemma-Sk-goes-up} and \ref{lemma-Rk-goes-up} and we see that $S$ is $(S_2)$ and $(R_1)$, in other words normal by Lemma \ref{lemma-criterion-normal} again.
Observe that $R \to S$ is flat with regular fibres (see the list of results on smooth ring maps in Section \ref{section-smooth-overview}). In particular, the fibres are normal. Thus if $R$ is Noetherian, then $S$ is Noetherian and we get the result from Lemma \ref{lemma-normal-goes-up-noetherian}. \medskip\noindent The general case. First note that $R$ is reduced and hence $S$ is reduced by Lemma \ref{lemma-reduced-goes-up}. Let $\mathfrak q$ be a prime of $S$ and let $\mathfrak p$ be the corresponding prime of $R$. Note that $R_{\mathfrak p}$ is a normal domain. We have to show that $S_{\mathfrak q}$ is a normal domain. To do this we may replace $R$ by $R_{\mathfrak p}$ and $S$ by $S_{\mathfrak p}$. Hence we may assume that $R$ is a normal domain. \medskip\noindent Assume $R \to S$ smooth, and $R$ a normal domain. We may find a finitely generated $\mathbf{Z}$-subalgebra $R_0 \subset R$ and a smooth ring map $R_0 \to S_0$ such that $S \cong R \otimes_{R_0} S_0$, see remark (10) in Section \ref{section-smooth-overview}. As $R_0$ is a Nagata domain (see Proposition \ref{proposition-ubiquity-nagata}) we see that its integral closure $R_0'$ is finite over $R_0$. Moreover, as $R$ is a normal domain it is clear that $R_0' \subset R$. Hence we may replace $R_0$ by $R_0'$ and $S_0$ by $R_0' \otimes_{R_0} S_0$ and assume that $R_0$ is a normal Noetherian domain. By the first paragraph of the proof we conclude that $S_0$ is a normal ring (it need not be a domain of course). In this way we see that $R = \bigcup R_\lambda$ is the union of normal Noetherian domains and correspondingly $S = \colim R_\lambda \otimes_{R_0} S_0$ is the colimit of normal rings. This implies that $S$ is a normal ring. Some details omitted.