hcju/stacksqa huggingface dataset:
The implications (1) $\Rightarrow$ (2) and (3) $\Rightarrow$ (1) hold in general, see Lemmas \ref{lemma-normal-ring-integrally-closed} and \ref{lemma-finite-product-normal}. \medskip\noindent Let $\mathfrak p_1, \ldots, \mathfrak p_n$ be the minimal primes of $R$. By Lemmas \ref{lemma-reduced-ring-sub-product-fields} and \ref{lemma-total-ring-fractions-no-embedded-points} we have $Q(R) = R_{\mathfrak p_1} \times \ldots \times R_{\mathfrak p_n}$, and by Lemma \ref{lemma-minimal-prime-reduced-ring} each factor is a field. Denote $e_i = (0, \ldots, 0, 1, 0, \ldots, 0)$ the $i$th idempotent of $Q(R)$. \medskip\noindent If $R$ is integrally closed in $Q(R)$, then it contains in particular the idempotents $e_i$, and we see that $R$ is a product of $n$ domains (see Sections \ref{section-connected-components} and \ref{section-tilde-module-sheaf}). Each factor is of the form $R/\mathfrak p_i$ with field of fractions $R_{\mathfrak p_i}$.  By Lemma \ref{lemma-finite-product-integral-closure} each map $R/\mathfrak p_i \to R_{\mathfrak p_i}$ is integrally closed.  Hence $R$ is a finite product of normal domains.
Closure under addition is clear from the characterization of Lemma \ref{lemma-characterize-integral-ideal}. Any element $s \in S$ which is integral over $R$ corresponds to the degree $0$ element $s$ of $S[x]$ which is integral over $A$ (because $R \subset A$). Hence we see that multiplication by $s$ on $S[x]$ preserves the property of being integral over $A$, by Lemma \ref{lemma-integral-closure-is-ring}.
Let $L/K$ be a field extension such that we can write $x^m + b_{m - 1} x^{m - 1} + \ldots + b_0 = \prod_{i = 1}^m (x - \beta_i)$ with $\beta_i \in L$. See Fields, Section \ref{fields-section-splitting-fieds}. Each $\beta_i$ is integral over $R_0$. Since each $a_i$ is a homogeneous polynomial in $\beta_1, \ldots, \beta_m$ we deduce the same for the $a_i$ (use Lemma \ref{lemma-integral-closure-is-ring}). \medskip\noindent Choose $c_0, \ldots, c_{m - n - 1} \in K$ such that $$ \begin{matrix} x^m + b_{m - 1} x^{m - 1} + \ldots + b_0 =  \\ (x^n + a_{n - 1}x^{n - 1} + \ldots + a_0) (x^{m - n} + c_{m - n - 1}x^{m - n - 1}+ \ldots + c_0). \end{matrix} $$ By part (1) the elements $c_i$ are integral over $R$. Consider the integral extension $$ R \subset R' = R[c_0, \ldots, c_{m - n - 1}] \subset K $$ By Lemmas \ref{lemma-integral-overring-surjective} and \ref{lemma-surjective-spec-radical-ideal} we see that $R \cap \sqrt{(b_0, \ldots, b_{m - 1})R'} = \sqrt{(b_0, \ldots, b_{m - 1})R}$. Thus we may replace $R$ by $R'$ and assume $c_i \in R$. Dividing out the radical $\sqrt{(b_0, \ldots, b_{m - 1})}$ we get a reduced ring $\overline{R}$. We have to show that the images $\overline{a}_i \in \overline{R}$ are zero. And in $\overline{R}[x]$ we have the relation $$ \begin{matrix} x^m = x^m + \overline{b}_{m - 1} x^{m - 1} + \ldots + \overline{b}_0 = \\ (x^n + \overline{a}_{n - 1}x^{n - 1} + \ldots + \overline{a}_0) (x^{m - n} + \overline{c}_{m - n - 1}x^{m - n - 1}+ \ldots + \overline{c}_0). \end{matrix} $$ It is easy to see that this implies $\overline{a}_i = 0$ for all $i$. Indeed by Lemma \ref{lemma-minimal-prime-reduced-ring} the localization of $\overline{R}$ at a minimal prime $\mathfrak{p}$ is a field and $\overline{R}_{\mathfrak p}[x]$ a UFD. Thus $f = x^n + \sum \overline{a}_i x^i$ is associated to $x^n$ and since $f$ is monic $f = x^n$ in $\overline{R}_{\mathfrak p}[x]$. Then there exists an $s \in \overline{R}$, $s \not\in \mathfrak p$ such that $s(f - x^n) = 0$.  Therefore all $\overline{a}_i$ lie in $\mathfrak p$ and we conclude by Lemma \ref{lemma-reduced-ring-sub-product-fields}.
Let $P = x^m + b_{m-1} x^{m-1} + \ldots + b_0$ be a polynomial with coefficients in $R$ such that $P(g) = 0$. Let $Q = x^n + a_{n-1}x^{n-1} + \ldots + a_0$ be the minimal polynomial for $g$ over the fraction field $K$ of $R$. Then $Q$ divides $P$ in $K[x]$. By Lemma \ref{lemma-polynomials-divide} we see the $a_i$ are integral over $R$. Since $R$ is normal this means they are in $R$.
The first statement of the lemma is a particular case of the second, so it is clearly enough to prove the latter. Let $R \to R'$ be a flat ring map, and $M'$ a flat $R'$-module. We need to prove that $M'$ is a flat $R$-module. Let $N_1 \to N_2 \to N_3$ be an exact complex of $R$-modules. Then, the complex $R' \otimes_R N_1 \to R' \otimes_R N_2 \to R' \otimes_R N_3$ is exact (since $R'$ is flat as an $R$-module), and so the complex $M' \otimes_{R'} \left(R' \otimes_R N_1\right) \to M' \otimes_{R'} \left(R' \otimes_R N_2\right) \to M' \otimes_{R'} \left(R' \otimes_R N_3\right)$ is exact (since $M'$ is a flat $R'$-module). Since $M' \otimes_{R'} \left(R' \otimes_R N\right) \cong \left(M' \otimes_{R'} R'\right) \otimes_R N \cong M' \otimes_R N$ for any $R$-module $N$ functorially (by Lemmas \ref{lemma-tensor-with-bimodule} and \ref{lemma-flip-tensor-product}), this complex is isomorphic to the complex $M' \otimes_R N_1 \to M' \otimes_R N_2 \to M' \otimes_R N_3$, which is therefore also exact. This shows that $M'$ is a flat $R$-module. Tracing this argument backwards, we can show that if $R \to R'$ is faithfully flat, and if $M'$ is faithfully flat as an $R'$-module, then $M'$ is faithfully flat as an $R$-module.
Let $\mathfrak a \subset R$ be a finitely generated ideal. By Lemma \ref{lemma-flat} it suffices to show that $\mathfrak a \otimes_R M \to M$ is injective. We can find an $i \in I$ and a finitely generated ideal $\mathfrak a' \subset R_i$ such that $\mathfrak a = \mathfrak a'R$. Then $\mathfrak a = \colim_{i' \geq i} \mathfrak a'R_{i'}$. Hence the map $\mathfrak a \otimes_R M \to M$ is the colimit of the maps $$ \mathfrak a'R_{i'} \otimes_{R_{i'}} M \longrightarrow M $$ which are all injective by assumption. Since $\otimes$ commutes with colimits and since colimits over $I$ are exact by Lemma \ref{lemma-directed-colimit-exact} we win.
By Lemma \ref{lemma-flat-base-change} we see that if $M$ is flat then $M'$ is flat. For the converse, suppose that $M'$ is flat. Let $N_1 \to N_2 \to N_3$ be an exact sequence of $R$-modules. We want to show that $N_1 \otimes_R M \to N_2 \otimes_R M \to N_3 \otimes_R M$ is exact. We know that $N_1 \otimes_R R' \to N_2 \otimes_R R' \to N_3 \otimes_R R'$ is exact, because $R \to R'$ is flat. Flatness of $M'$ implies that $N_1 \otimes_R R' \otimes_{R'} M' \to N_2 \otimes_R R' \otimes_{R'} M' \to N_3 \otimes_R R' \otimes_{R'} M'$ is exact. We may write this as $N_1 \otimes_R M \otimes_R R' \to N_2 \otimes_R M \otimes_R R' \to N_3 \otimes_R M \otimes_R R'$. Finally, faithful flatness implies that $N_1 \otimes_R M \to N_2 \otimes_R M \to N_3 \otimes_R M$ is exact.
Let $N \to N'$ be an injection of $R$-modules. By the flatness of $S \to S'$ we have $$ \Ker(N \otimes_R M \to N' \otimes_R M) \otimes_S S' = \Ker(N \otimes_R M' \to N' \otimes_R M') $$ If $M$ is flat over $R$, then the left hand side is zero and we find that $M'$ is flat over $R$ by the second characterization of flatness in Lemma \ref{lemma-flat}. If $M'$ is flat over $R$ then we have the vanishing of the right hand side and if in addition $S \to S'$ is faithfully flat, this implies that $\Ker(N \otimes_R M \to N' \otimes_R M)$ is zero which in turn shows that $M$ is flat over $R$.
We will use the criterion that a module $N$ is flat if for every ideal $I \subset R$ the map $N \otimes_R I \to N$ is injective, see Lemma \ref{lemma-flat}. Consider an ideal $I \subset R$. Consider the diagram $$ \begin{matrix} 0 & \to & M' & \to & M & \to & M'' & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M'\otimes_R I & \to & M \otimes_R I & \to & M''\otimes_R I & \to & 0 \end{matrix} $$ with exact rows. This immediately proves the first assertion. The second follows because if $M''$ is flat then the lower left horizontal arrow is injective by Lemma \ref{lemma-flat-tor-zero}.
Assume $M$ faithfully flat and $N \not = 0$. By Lemma \ref{lemma-easy-ff} the nonzero map $1 : N \to N$ induces a nonzero map $M \otimes_R N \to M \otimes_R N$, so $M \otimes_R N \not = 0$. Thus (1) implies (2). The imlpications (2) $\Rightarrow$ (3) $\Rightarrow$ (4) are immediate. \medskip\noindent Assume (4). Suppose that $N_1 \to N_2 \to N_3$ is a complex and suppose that $N_1 \otimes_R M \to N_2\otimes_R M \to N_3\otimes_R M$ is exact. Let $H$ be the cohomology of the complex, so $H = \Ker(N_2 \to N_3)/\Im(N_1 \to N_2)$. To finish the proof we will show $H = 0$. By flatness we see that $H \otimes_R M = 0$. Take $x \in H$ and let $I = \{f \in R \mid fx = 0 \}$ be its annihilator. Since $R/I \subset H$ we get $M/IM \subset H \otimes_R M = 0$ by flatness of $M$. If $I \not =  R$ we may choose a maximal ideal $I \subset \mathfrak m \subset R$. This immediately gives a contradiction.
Let us prove the last statement of the lemma. In the proof we will use repeatedly that localization is exact and commutes with tensor product, see Sections \ref{section-localization} and \ref{section-tensor-product}. \medskip\noindent Suppose $R \to A$ is a ring map, and $M$ is an $A$-module. Assume that $M_{\mathfrak m}$ is a flat $R_{\mathfrak p}$-module for all maximal ideals $\mathfrak m$ of $A$ (with $\mathfrak p = R \cap \mathfrak m$). Let $I \subset R$ be an ideal. We have to show the map $I \otimes_R M \to M$ is injective. We can think of this as a map of $A$-modules. By assumption the localization $(I \otimes_R M)_{\mathfrak m} \to M_{\mathfrak m}$ is injective because $(I \otimes_R M)_{\mathfrak m} = I_{\mathfrak p} \otimes_{R_{\mathfrak p}} M_{\mathfrak m}$. Hence the kernel of $I \otimes_R M \to M$ is zero by Lemma \ref{lemma-characterize-zero-local}. Hence $M$ is flat over $R$. \medskip\noindent Conversely, assume $M$ is flat over $R$. Pick a prime $\mathfrak q$ of $A$ lying over the prime $\mathfrak p$ of $R$. Suppose that $I \subset R_{\mathfrak p}$ is an ideal. We have to show that $I \otimes_{R_{\mathfrak p}} M_{\mathfrak q} \to M_{\mathfrak q}$ is injective. We can write $I = J_{\mathfrak p}$ for some ideal $J \subset R$. Then the map $I \otimes_{R_{\mathfrak p}} M_{\mathfrak q} \to M_{\mathfrak q}$ is just the localization (at $\mathfrak q$) of the map $J \otimes_R M \to M$ which is injective. Since localization is exact we see that $M_{\mathfrak q}$ is a flat $R_{\mathfrak p}$-module. \medskip\noindent This proves (7) and (6). The other statements follow in a straightforward way from the last statement (proofs omitted).
By Lemma \ref{lemma-flat-localization} the local ring map $R_{\mathfrak p'} \to S_{\mathfrak q'}$ is flat. By Lemma \ref{lemma-local-flat-ff} this local ring map is faithfully flat. By Lemma \ref{lemma-ff-rings} there is a prime mapping to $\mathfrak p R_{\mathfrak p'}$. The inverse image of this prime in $S$ does the job.
By Lemma \ref{lemma-colimit-flat} we see that $S$ is flat. Let $\mathfrak m \subset R$ be a maximal ideal. By Lemma \ref{lemma-ff-rings} none of the rings $S_i/\mathfrak m S_i$ is zero. Hence $S/\mathfrak mS = \colim S_i/\mathfrak mS_i$ is nonzero as well because $1$ is not equal to zero. Thus the image of $\Spec(S) \to \Spec(R)$ contains $\mathfrak m$ and we see that $R \to S$ is faithfully flat by Lemma \ref{lemma-ff-rings}.
Set $I = \text{Ann}_R(m)$. By definition there is an exact sequence $0 \to I \to R \to M$ where the map $R \to M$ sends $f$ to $fm$. Using flatness we obtain an exact sequence $0 \to I \otimes_R S \to S \to M \otimes_R S$ which proves the first assertion. If $m_1, \ldots, m_n$ is a set of generators of $M$ then $\text{Ann}_R(M) = \bigcap \text{Ann}_R(m_i)$. Similarly $\text{Ann}_S(M \otimes_R S) = \bigcap \text{Ann}_S(m_i \otimes 1)$. Set $I_i = \text{Ann}_R(m_i)$. Then it suffices to show that $\bigcap_{i = 1, \ldots, n} (I_i S) = (\bigcap_{i = 1, \ldots, n} I_i)S$. This is Lemma \ref{lemma-flat-intersect-ideals}.
We will show that $V(I) = \text{Supp}(M)$. \medskip\noindent Suppose $\mathfrak p \in \text{Supp}(M)$. Then $M_{\mathfrak p} \not = 0$. Choose an element $m \in M$ whose image in $M_\mathfrak p$ is nonzero. Then the annihilator of $m$ is contained in $\mathfrak p$ by construction of the localization $M_\mathfrak p$. Hence a fortiori $I = \text{Ann}(M)$ must be contained in $\mathfrak p$. \medskip\noindent Conversely, suppose that $\mathfrak p \not \in \text{Supp}(M)$. Then $M_{\mathfrak p} = 0$. Let $x_1, \ldots, x_r \in M$ be generators. By Lemma \ref{lemma-localization-colimit} there exists an $f \in R$, $f\not\in \mathfrak p$ such that $x_i/1 = 0$ in $M_f$. Hence $f^{n_i} x_i = 0$ for some $n_i \geq 1$. Hence $f^nM = 0$ for $n = \max\{n_i\}$ as desired.
Let $\mathfrak p \in \text{Supp}(M)$. By Nakayama's lemma (Lemma \ref{lemma-NAK}) we see that $$ M \otimes_R \kappa(\mathfrak p) = M_\mathfrak p/\mathfrak p M_\mathfrak p $$ is a nonzero $\kappa(\mathfrak p)$ vector space. Hence for every prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$ we see that $$ (M \otimes_R R')_{\mathfrak p'}/\mathfrak p' (M \otimes_R R')_{\mathfrak p'} = (M \otimes_R R') \otimes_{R'} \kappa(\mathfrak p') = M \otimes_R \kappa(\mathfrak p) \otimes_{\kappa(\mathfrak p)} \kappa(\mathfrak p') $$ is nonzero. This implies $\mathfrak p' \in \text{Supp}(M \otimes_R R')$. For the converse, if $\mathfrak p' \subset R'$ is a prime lying over an arbitrary prime $\mathfrak p \subset R$, then $$ (M \otimes_R R')_{\mathfrak p'} = M_\mathfrak p \otimes_{R_\mathfrak p} R'_{\mathfrak p'}. $$ Hence if $\mathfrak p' \in \text{Supp}(M \otimes_R R')$ lies over the prime $\mathfrak p \subset R$, then $\mathfrak p \in \text{Supp}(M)$.
We may replace $M$ by $Rm \subset M$. Then (1) $\text{Ann}(m) = \text{Ann}(M)$ and (2) $x$ does not map to zero in $M_\mathfrak p$ if and only if $\mathfrak p \in \text{Supp}(M)$. The result now follows from Lemma \ref{lemma-support-closed}.
Choose a presentation $$ R^{\oplus m} \longrightarrow R^{\oplus n} \longrightarrow M \to 0 $$ Let $A \in \text{Mat}(n \times m, R)$ be the matrix of the first map. By Nakayama's Lemma \ref{lemma-NAK} we see that $$ M_{\mathfrak p} \not = 0 \Leftrightarrow M \otimes \kappa(\mathfrak p) \not = 0 \Leftrightarrow \text{rank}(A \bmod \mathfrak p) < n. $$ Hence, if $I$ is the ideal of $R$ generated by the $n \times n$ minors of $A$, then $\text{Supp}(M) = V(I)$. Since $I$ is finitely generated, say $I = (f_1, \ldots, f_t)$, we see that $\Spec(R) \setminus V(I)$ is a finite union of the standard opens $D(f_i)$, hence quasi-compact.
The functors $M \mapsto M_{\mathfrak p}$ are exact. This immediately implies all but the first assertion. For the first assertion we need to show that $M_\mathfrak p \not = 0$ and $I \subset \mathfrak p$ implies $(M/IM)_{\mathfrak p} = M_\mathfrak p/IM_\mathfrak p \not = 0$. This follows from Nakayama's Lemma \ref{lemma-NAK}.
Suppose that $\mathfrak p \subset \mathfrak p' \subset R$ and $\mathfrak q' \subset S$ lies over $\mathfrak p'$. As $\varphi$ is open, for every $g \in S$, $g \not \in \mathfrak q'$ we see that $\mathfrak p$ is in the image of $D(g) \subset \Spec(S)$. In other words $S_g \otimes_R \kappa(\mathfrak p)$ is not zero. Since $S_{\mathfrak q'}$ is the directed colimit of these $S_g$ this implies that $S_{\mathfrak q'} \otimes_R \kappa(\mathfrak p)$ is not zero, see Lemmas \ref{lemma-localization-colimit} and \ref{lemma-tensor-products-commute-with-limits}. Hence $\mathfrak p$ is in the image of $\Spec(S_{\mathfrak q'}) \to \Spec(R)$ as desired.
We give two proofs. \medskip\noindent First proof. Let $\mathfrak p \subset R$ be a prime ideal such that the corresponding point of $\Spec(R)$ is in the closure of $T$. This means that for every $f \in R$, $f \not \in \mathfrak p$ we have $D(f) \cap T \not = \emptyset$. Note that $D(f) \cap T$ is the image of $\Spec(S_f)$ in $\Spec(R)$. Hence we conclude that $S_f \not = 0$. In other words, $1 \not = 0$ in the ring $S_f$. Since $S_{\mathfrak p}$ is the directed colimit of the rings $S_f$ we conclude that $1 \not = 0$ in $S_{\mathfrak p}$. In other words, $S_{\mathfrak p} \not = 0$ and considering the image of $\Spec(S_{\mathfrak p}) \to \Spec(S) \to \Spec(R)$ we see there exists a $\mathfrak p' \in T$ with $\mathfrak p' \subset \mathfrak p$. As we assumed $T$ closed under specialization we conclude $\mathfrak p$ is a point of $T$ as desired. \medskip\noindent Second proof. Let $I = \Ker(R \to S)$. We may replace $R$ by $R/I$. In this case the ring map $R \to S$ is injective. By Lemma \ref{lemma-injective-minimal-primes-in-image} all the minimal primes of $R$ are contained in the image $T$. Hence if $T$ is stable under specialization then it contains all primes.
It is a general fact that specializations lift along a closed map of topological spaces, see Topology, Lemma \ref{topology-lemma-closed-open-map-specialization}. Hence the second condition implies the first. \medskip\noindent Assume that going up holds for $R \to S$. Let $V(I) \subset \Spec(S)$ be a closed set. We want to show that the image of $V(I)$ in $\Spec(R)$ is closed. The ring map $S \to S/I$ obviously satisfies going up. Hence $R \to S \to S/I$ satisfies going up, by Lemma \ref{lemma-going-up-down-composition}. Replacing $S$ by $S/I$ it suffices to show the image $T$ of $\Spec(S)$ in $\Spec(R)$ is closed. By Topology, Lemmas \ref{topology-lemma-open-closed-specialization} and \ref{topology-lemma-lift-specializations-images} this image is stable under specialization. Thus the result follows from Lemma \ref{lemma-image-stable-specialization-closed}.
First proof. The first assertion follows from Lemma \ref{lemma-image-stable-specialization-closed} combined with Lemma \ref{lemma-constructible-is-image}. The second follows because the complement of a constructible set is constructible (see Topology, Lemma \ref{topology-lemma-constructible}), the first part of the lemma and Topology, Lemma \ref{topology-lemma-open-closed-specialization}. \medskip\noindent Second proof. Since $\Spec(R)$ is a spectral space by Lemma \ref{lemma-spec-spectral} this is a special case of Topology, Lemma \ref{topology-lemma-constructible-stable-specialization-closed}.
Let $\mathfrak p \subset R$ be in the image of the south-west arrow. This means (Lemma \ref{lemma-in-image}) that $$ (S' \otimes_k R)_f \otimes_R \kappa(\mathfrak p) = (S' \otimes_k \kappa(\mathfrak p))_f $$ is not the zero ring, i.e., $S' \otimes_k \kappa(\mathfrak p)$ is not the zero ring and the image of $f$ in it is not nilpotent. The ring map $S' \otimes_k \kappa(\mathfrak p) \to S \otimes_k \kappa(\mathfrak p)$ is injective. Hence also $S \otimes_k \kappa(\mathfrak p)$ is not the zero ring and the image of $f$ in it is not nilpotent. Hence $(S \otimes_k R)_f \otimes_R \kappa(\mathfrak p)$ is not the zero ring. Thus (Lemma \ref{lemma-in-image}) we see that $\mathfrak p$ is in the image of the south-east arrow as desired.
Let $f \in S \otimes_k R$. It suffices to prove that the image of the standard open $D(f)$ is open. Let $S' \subset S$ be a finite type $k$-subalgebra such that $f \in S' \otimes_k R$. The map $R \to S' \otimes_k R$ is flat and of finite presentation, hence the image $U$ of $\Spec((S' \otimes_k R)_f) \to \Spec(R)$ is open by Proposition \ref{proposition-fppf-open}. By Lemma \ref{lemma-same-image} this is also the image of $D(f)$ and we win.
The meaning of the statement is as follows. Let $\mathfrak p \subset \mathfrak p' \subset R$ be primes. Let $\mathfrak q' \subset S$ be a prime $\mathfrak q' \in \text{Supp}(N)$ Then there exists a prime $\mathfrak q \subset \mathfrak q'$, $\mathfrak q \in \text{Supp}(N)$ lying over $\mathfrak p$. As $N$ is flat over $R$ we see that $N_{\mathfrak q'}$ is flat over $R_{\mathfrak p'}$, see Lemma \ref{lemma-flat-localization}. As $N_{\mathfrak q'}$ is finite over $S_{\mathfrak q'}$ and not zero since $\mathfrak q' \in \text{Supp}(N)$ we see that $N_{\mathfrak q'} \otimes_{S_{\mathfrak q'}} \kappa(\mathfrak q')$ is nonzero by Nakayama's Lemma \ref{lemma-NAK}. Thus $N_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p')$ is also not zero. We conclude from Lemma \ref{lemma-ff} that $N_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p)$ is nonzero. Let $J \subset S_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p)$ be the annihilator of the finite nonzero module $N_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p)$. Since $J$ is a proper ideal we can choose a prime $\mathfrak q \subset S$ which corresponds to a prime of $S_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p)/J$. This prime is in the support of $N$, lies over $\mathfrak p$, and is contained in $\mathfrak q'$ as desired.
This lemma is only interesting when the characteristic of $k$ is $p > 0$. Choose $x_1, \ldots, x_r$ a transcendence basis of $K$ over $k$. As $K$ is finitely generated over $k$ the extension $k(x_1, \ldots, x_r) \subset K$ is finite. Let $k(x_1, \ldots, x_r) \subset K_{sep} \subset K$ be the subextension found in Fields, Lemma \ref{fields-lemma-separable-first}. If $K = K_{sep}$ then we are done. We will use induction on $d = [K : K_{sep}]$. \medskip\noindent Assume that $d > 1$. Choose a $\beta \in K$ with $\alpha = \beta^p \in K_{sep}$ and $\beta \not \in K_{sep}$. Let $P = T^d + a_1T^{d - 1} + \ldots + a_d$ be the minimal polynomial of $\alpha$ over $k(x_1, \ldots, x_r)$. Let $k \subset k'$ be a finite purely inseparable extension obtained by adjoining $p$th roots such that each $a_i$ is a $p$th power in $k'(x_1^{1/p}, \ldots, x_r^{1/p})$. Such an extension exists; details omitted. Let $L$ be a field fitting into the diagram $$ \xymatrix{ K \ar[r] & L \\ k(x_1, \ldots, x_r) \ar[u] \ar[r] & k'(x_1^{1/p}, \ldots, x_r^{1/p}) \ar[u] } $$ We may and do assume $L$ is the compositum of $K$ and $k'(x_1^{1/p}, \ldots, x_r^{1/p})$. Let $k'(x_1^{1/p}, \ldots, x_r^{1/p}) \subset L_{sep} \subset L$ be the subextension found in Fields, Lemma \ref{fields-lemma-separable-first}. Then $L_{sep}$ is the compositum of $K_{sep}$ and $k'(x_1^{1/p}, \ldots, x_r^{1/p})$. The element $\alpha \in L_{sep}$ is a zero of the polynomial $P$ all of whose coefficients are $p$th powers in $k'(x_1^{1/p}, \ldots, x_r^{1/p})$ and whose roots are pairwise distinct. By Fields, Lemma \ref{fields-lemma-pth-root} we see that $\alpha = (\alpha')^p$ for some $\alpha' \in L_{sep}$. Clearly, this means that $\beta$ maps to $\alpha' \in L_{sep}$. In other words, we get the tower of fields $$ \xymatrix{ K \ar[r] & L \\ K_{sep}(\beta) \ar[r] \ar[u] & L_{sep} \ar[u] \\ K_{sep} \ar[r] \ar[u] & L_{sep} \ar@{=}[u] \\ k(x_1, \ldots, x_r) \ar[u] \ar[r] & k'(x_1^{1/p}, \ldots, x_r^{1/p}) \ar[u] \\ k \ar[r] \ar[u] & k' \ar[u] } $$ Thus this construction leads to a new situation with $[L : L_{sep}] < [K : K_{sep}]$. By induction we can find $k' \subset k''$ and $L \subset L'$ as in the lemma for the extension $k' \subset L$. Then the extensions $k \subset k''$ and $K \subset L'$ work for the extension $k \subset K$. This proves the lemma.
By Lemma \ref{lemma-limit-argument} we may assume that $R$ is of finite type over $k$. Then $R$, as a reduced Noetherian ring, embeds into a finite product of fields (see Lemmas \ref{lemma-total-ring-fractions-no-embedded-points}, \ref{lemma-Noetherian-irreducible-components}, and \ref{lemma-minimal-prime-reduced-ring}). Hence we may assume $R$ is a finite product of fields. In this case it follows from Definition \ref{definition-geometrically-reduced} that $R \otimes_k S$ is reduced.