Proof. There exists an \(f\in R^\circ\) such that \(R'_f\) is flat over \(R_f\), so \((R'^\wedge)_f\) is flat over \((R^\wedge)_f\). As \(f\) is a nonzerodivisor in both \(R^\wedge\) and \(R'^\wedge\) we get the desired result, cf. stacks? Tags 00ON and0337. ◻

Proof. Let \(x'\in R'^\circ\) be not in the fraction field \(K\) of \(R\). Let \(f_{x'}(T)=\sum a_i T^i\) be the monic minimal polynomial of \(x'\) over \(K\), and let \(d=\deg f_{x'}\). Then for \(x\in R^\circ\) the monic minimal polynomial of \(xx'\) over \(K\) is \(f_{xx'}(T)=\sum x^{d-i}a_i T^i.\) Therefore we may choose \(x\) so that \(f_{xx'}\in R[T]\), so \(R[xx']\cong R[T]/f_{xx'}(T)\) is flat over \(R\). Inductively we can find our \(R''\). ◻

Proof. By Lemma 10 we may assume \(R\) and \(R'\) has the same fraction field. Let \(\mathfrak{p}'\) be a minimal prime divisor of \(xR'\) and let \(\mathfrak{p}=\mathfrak{p}'\cap R\).

Let \(b\in R^\circ\) be such that \(bR'\subseteq R\) and that \(b\in\mathfrak{p}\). As \(\operatorname{ht}(\mathfrak{p}')=1\), \(\mathfrak{p}'^nR'_{\mathfrak{p}'}\subseteq bR'_{\mathfrak{p}'}\) for some \(n\). We can therefore find an \(s\in R'\setminus\mathfrak{p}'\) such that \(s\mathfrak{p}'^n\subseteq bR'\subseteq R\), and consequently \(bs^t\mathfrak{p}'^{nt}\subseteq b^{t+1}R'\subseteq b^{t}R\) for all \(t\). If \(\mathfrak{p}\not\in\operatorname{Ass}_{R}(R/xR)\), then \(\operatorname{depth}R_\mathfrak{p}\geq 2\), so \(\mathfrak{p}\not\in\operatorname{Ass}_{R}(R/bR)\), and we can take \(c\in\mathfrak{p}\) a nonzerodivisor on \(R/bR\), thus a nonzerodivisor on \(R/b^{t}R\) for all \(t\). As \(bs^t\mathfrak{p}'^{nt}\subseteq b^{t}R\), we have \(c^{nt}bs^t\in b^{t}R\), so \(bs^t\in b^tR\) as \(bs^t\in bR'\subseteq R\). Then \(b\in b^tR'_{\mathfrak{p}'}\) for all \(t\), contradiction. Therefore \(\mathfrak{p}\in\operatorname{Ass}_R(R/xR)\). ◻

Proof. If \(S=R^\nu\), then this is part of Nagata-local?. The general case follows from the fact \(\operatorname{Spec}(R^\nu)\to \operatorname{Spec}(S)\) is surjective. ◻

Proof. By Theorem 12, we can take a finite subalgebra \(R'\subseteq S\) so that \(a\in R'\) and \(\mathfrak{q}\) is the only prime ideal of \(S\) above \(\mathfrak{p}'=\mathfrak{q}\cap R'\). If \(\mathfrak{p}'\) were not a minimal prime divisor of \(aR'\), then we can find primes \(\mathfrak{p}_0'\subsetneq\mathfrak{p}'\) in \(R'\) containing \(a\). We can then find primes \(\mathfrak{q}_0\subsetneq\mathfrak{q}_1\) of \(S\) lying above \(\mathfrak{p}_0'\subsetneq\mathfrak{p}'\); they automatically contain \(a\). By uniqueness, \(\mathfrak{q}_1=\mathfrak{q}\), contradicting the minimality of \(\mathfrak{q}\). Therefore \(\mathfrak{p}'\) is a minimal prime divisor of \(aR'\), so \(\operatorname{ht}(\mathfrak{p}')=1\) as \(R'\) is Noetherian. This is [MtoA:ht], and we get [MtoA:Ass] by Lemma 11. ◻

Proof. We may assume \(a\in R\). Finiteness follows from Lemma 13[MtoA:Ass] and Theorem 12. For each \(\mathfrak{q}\), Lemma 13[MtoA:ht] gives a map \(R'_{\mathfrak{q}\cap R'}\to S_\mathfrak{q}\), so \(S_\mathfrak{q}\) is a \(1\)-dimensional Noetherian ring by the theorem of Krull–Akizuki Nagata-local?. ◻

Proof. That [S2ass:S2] implies [S2ass:ass] is clear as minimal prime divisors of and the set of zero divisors modulo \(aS\) can be read off of a primary decomposition, cf. AMcommalg?.

Assume [S2ass:ass]. Let \(z=x/y\in \bigcap_{\mathfrak{q}\in\operatorname{Spec}_1(S)}S_\mathfrak{q}\) where \(x,y\in S^\circ\). Then \((xS:_S yS)\) is not contained in any minimal prime of \(yS\), as they are of height \(1\) (Theorem 14). Therefore \((xS:_S yS)\) contains a nonzerodivisor on \(S/yS\) by prime avoidance, so \(x\in yS,z\in S\).

Finally, assume [S2ass:intersect]. Then \(aS=\bigcap_{\mathfrak{q}\in\operatorname{Spec}_1(S)} aS_\mathfrak{q}\) for all \(a\in S^\circ\), so \(aS=\bigcap_{\mathfrak{q}\in\operatorname{Spec}_1(S)} (aS_\mathfrak{q}\cap S)\). By Theorem 14, all but finitely many of the ideals \(aS_\mathfrak{q}\cap S\) are \(S\), and the others are \(\mathfrak{q}\)-primary. This gives a primary decomposition of \(aS\). ◻

Proof. [S2loc:localize] is trivial as primary decompositions localize, whereas [S2loc:localcheck] follows from Lemma 16. ◻

Proof. Let \(a\in S^\circ\); we may assume \(a\in S_\alpha\) for all \(\alpha\). If \(\mathfrak{p}_\alpha\) is a minimal prime divisor of \(aS_\alpha\), then there exists a minimal prime divisor \(\mathfrak{p}\) of \(aS\) above \(\mathfrak{p}_\alpha\), as \(\operatorname{Spec}(S)\to\operatorname{Spec}(S_\alpha)\) is surjective. Now for a \(b\in S\) not in any minimal prime divisor of \(aS\), which we may assume in \(S_\alpha\) for all \(\alpha\), we have \(b\) not in any minimal prime divisor of \(aS_\alpha\). Therefore \(b\) is a nonzerodivisor on \(S_\alpha/aS_\alpha\) by Lemma 16. Consequently, \(b\) a nonzerodivisor on \(S/aS=\operatorname{colim}_\alpha S_\alpha/aS_\alpha\), so \(S\) is \((S_2)\) by Lemma 16. ◻

Proof. We will show \(W^{-1}(S^{n\sigma})=(W^{-1}S)^{n\sigma}\); the corresponding identity \(W^{-1}(S^{\sigma})=(W^{-1}S)^{\sigma}\) follows as localization commutes with finite intersections.

The inclusion \(W^{-1}(S^{n\sigma})\subseteq (W^{-1}S)^{n\sigma}\) is clear. For the other inclusion, let \(z=x/y\in (W^{-1}S)^{n\sigma}\) where \(x,y\in S^\circ\). Let \(\mathfrak{q}_1,\ldots,\mathfrak{q}_m,\mathfrak{q}_{m+1},\ldots,\mathfrak{q}_n\) be the minimal prime divisors of \(yS\) (there are only finitely many, Theorem 14), ordered in a way that \(\mathfrak{q}_1,\ldots,\mathfrak{q}_m\) are disjoint from \(W\) and \(\mathfrak{q}_{m+1},\ldots,\mathfrak{q}_n\) are not. Then \(z\in S_{\mathfrak{q}_j}\) for all \(1\leq j\leq m\), and \(z\in S_{\mathfrak{q}}\) for all \(\mathfrak{q}\in\operatorname{Spec}_1(S)\setminus\{\mathfrak{q}_1,\ldots,\mathfrak{q}_m,\mathfrak{q}_{m+1},\ldots,\mathfrak{q}_n\}\). As \(S_{\mathfrak{q}_j}\) is a \(1\)-dimensional Noetherian local ring (Theorem 14), we can take \(w\in W\) so that \(w\in yS_{\mathfrak{q}_j}\) for all \(m+1\leq j\leq n\). Then \(wz=wx/y\) is in \(S^{n\sigma}\). ◻

Proof. For every \(\mathfrak{q}\in\operatorname{Spec}_1(S)\), we have \(S\subseteq S'\subseteq S_\mathfrak{q}\), so \(S'_\mathfrak{q}=S_\mathfrak{q}\). Therefore the family of rings defining the intersection of \(S'^{n\sigma}\) contains that of \(S^{n\sigma}\), giving the \((-)^{n\sigma}\) case.

If \(S'\subseteq S^{\sigma}\), then \(S'^{\nu}=S^{\nu}\), which gives the \((-)^{\sigma}\) case. ◻

Proof. By Lemmas 20 and 17, we may assume \(R\) is local. Let \(K\) be the fraction field of \(R\).

Let \(0=Q_1\cap\ldots\cap Q_m\cap Q_{m+1}\cap\ldots\cap Q_n\) be a shortest primary decomposition of \(0\) in \(R^\wedge\), ordered in a way that \(P_j=\sqrt{Q_j}\) is a minimal prime of \(R^\wedge\) for \(1\leq j\leq m\) and not for \(m+1\leq j\leq n\). Let \(S_j=R^\wedge/Q_j\) for \(1\leq j\leq m\), \(S=\prod_{j=1}^m S_j\). Let \(L_j=(S_j)_{P_j}\), so \(L=\prod_{j=1}^m L_j\) is the total fraction ring of \(S\).

Every \(a\in R^\circ\) is a nonzerodivisor on \(R^\wedge,R^{n\sigma}\otimes_R R^\wedge\), and \(S\). Therefore there is a commutative diagram \[\begin{tikzcd} R^\wedge \arrow[hookrightarrow,r] \arrow[d] &[0.5em] R^{n\sigma}\otimes_R R^\wedge \arrow[hookrightarrow,r] \arrow[d] &[0.5em] K\otimes_R R^\wedge\arrow[d] \\ S \arrow[hookrightarrow,r] & T \arrow[hookrightarrow,r] & L \end{tikzcd}\] where \(T=\prod_{j=1}^m T_j\) and each \(T_j\) denote the image of \(R^{n\sigma}\otimes_R R^\wedge\) in \(L_j\). Note that the kernel of \(K\otimes_R R^\wedge\to L\) is \(\bigcap_{j=1}^m Q_j(K\otimes_R R^\wedge)\), which is nilpotent, and is zero if \(R^\wedge\) is \((S_1)\).

Observe that for every \(\mathfrak{p}\in\operatorname{Spec}_{1}(R)\), we have \(R_\mathfrak{p}=(R^{n\sigma})_\mathfrak{p}\). As \(\operatorname{OS}_2^\wedge({R})=\emptyset\), we see that for every \(P\in\operatorname{Spec}(R^\wedge)\) so that \(\operatorname{ht}(P/P_j)=1\) for some \(1\leq j\leq m\), we have \((R^\wedge)_{P}=(R^{n\sigma}\otimes_R R^\wedge)_{P}\). Therefore the sub-\(S_j\)-algebra \(T_j\) of \(L_j\) satisfies \((S_j)_{P}=(T_j)_{P}\) for all \(P\in \operatorname{Spec}_1(S_j)\).

By the lemma, each \(T_j\) is finite over \(S_j\), so \(T\) is finite over \(S\). The kernel of \(R^{n\sigma}\otimes_R R^\wedge\to T\) is nilpotent, and is zero if \(R^\wedge\) is \((S_1)\). Therefore \(R^{n\sigma}\otimes_R R^\wedge\) is integral over \(R^\wedge\), and is finite over \(R^\wedge\) if \(R^\wedge\) is \((S_1)\). As \(R\to R^\wedge\) is faithfully flat, we get [NS2:integral] and [NS2:formalS1], cf. stacks? Tags 02L9 and02LA.

Let \(S\) be an arbitrary subalgebra of \(R^{n\sigma}=R^\sigma\). Then \(S^{n\sigma}=R^{n\sigma}\) by Lemmas 21 and 28 below. Therefore \(S\) is \((S_2)\) if and only if \(S=R^{n\sigma}\). This gives [NS2:S2][NS2:S2unique]. ◻

Proof. The “in particular” assertion follows at once as \(S'^{n\sigma}=\bigcap_{\mathfrak{q}'\in\operatorname{Spec}_1(S')} S'_{\mathfrak{q}'}\supseteq \bigcap_{\mathfrak{q}'\in\operatorname{Spec}_1(S')} R_{\mathfrak{q}'\cap R}\supseteq R^{n\sigma}\).

By Lemma 13, we may assume \(S'=R'\). We may assume \(R\) local, so \(R'\) is semilocal. By Lemma 9 every minimal prime of \(R'^\wedge\) lies above a minimal prime of \(R^\wedge\).

Let \(P'\) be a minimal prime divisor of \(\mathfrak{q}'R'^\wedge\), so \(\operatorname{ht}(P')=1\) and \(P'\cap R'=\mathfrak{q}'\). Take \(P_0'\in\operatorname{Min}(R'^\wedge)\) contained in \(P'\), so \(\operatorname{ht}(P'/P_0')=1\). As \(R^\wedge\) is universally catenary, we have \(\operatorname{ht}(P'\cap R^\wedge/P_0'\cap R^\wedge)=1\) by the dimension formula stacks? Tag02IJ. Therefore either \(\operatorname{ht}(P'\cap R)=1\) or \(P'\cap R\in\operatorname{OS}_2^\wedge({R})\). As \(P'\cap R=\mathfrak{q}'\cap R\) this proves the lemma. ◻

Proof. Let \(S'\) be a \(S\)-subalgebra of \(R^\nu\). Let \(\mathfrak{q}'\in \operatorname{Spec}_1(S')\). Then \(\operatorname{ht}(\mathfrak{q}'\cap R)=1\) by Lemma 28, so \(\operatorname{ht}(\mathfrak{q}'\cap S)=1\). Therefore \(S^{n\sigma}\subseteq S'^{n\sigma}\). Apply this to \(S'=R^\nu\), noting that \((R^\nu)^{n\sigma}=R^\nu\) as \(R^\nu\) is a Krull domain Nagata-local?, we see [NS2c:integral] holds.

Now apply the inclusion \(S^{n\sigma}\subseteq S'^{n\sigma}\) and Lemma 21 to a \(S\)-subalgebra \(S'\) of \(S^{n\sigma}=S^\sigma\). We get \(S^{n\sigma}= S'^{n\sigma}\), giving [NS2c:S2][NS2c:S2unique]. ◻

Proof. As \(\operatorname{depth}R\geq 1\), \(R\) is a subring of \(\mathcal{O}(D(\mathfrak{m}))\). Let \(y\in R+_\mathfrak{m}M\). Then \(y\mathfrak{m}\subseteq R\). If \(y\mathfrak{m}=R\), then we can write \(1=yt\) for some \(t\in \mathfrak{m}\), so \(a=ayt\in tR\) for all \(a\in\mathfrak{m}\), and \(\mathfrak{m}=tR\) is principal, contradiction. Therefore \(y\mathfrak{m}=\mathfrak{m}\), so \(y\) is integral over \(R\) by stacks? Tag0B5T and the fact \(\mathfrak{m}\) contains a nonzerodivisor on \(R\), which is automatically a nonzerodivisor on \(\mathcal{O}(D(\mathfrak{m}))\). ◻

Proof. By standard theory of limits, particularly stacks? Tags 01ZO and081E, we may assume \(W\) is generated by a single element \(f\). Let \(T\) be a finite \((W^{-1}U)\)-modification of \(W^{-1}R\). Then \(T\) and \(\mathcal{O}_U\) glue to a coherent \(\mathcal{O}_{V}\)-algebra \(\mathcal{A}\), where \(V=U\cup D(f)\). Let \(j:V\to \operatorname{Spec}(A)\) be the open immersion, and let \(\mathcal{A}'\) be the integral closure of \(\mathcal{O}_{\operatorname{Spec}(A)}\) in \(j_*\mathcal{A}\). Then \(j^*\mathcal{A}'=\mathcal{A}\), and we can find a coherent subalgebra \(\mathcal{A}''\) of \(\mathcal{A}'\) such that \(j^*\mathcal{A}''=\mathcal{A}\), which corresponds to a \(U\)-modification \(R''\) of \(R\) that satisfies \(W^{-1}R''=T\). Alternatively, one can use Zariski’s Main Theorem, stacks? Tag05K0. ◻

Proof. Let \(M_1\) be the socle of \(H^1_\mathfrak{m}(R)\), which is finite as \(H^1_\mathfrak{m}(R)\) is Artinian Residues-Duality?. The ring \(R_1:=R[\mathfrak{m};M_1]\) is then finite over \(R\) by Lemma 34. As \(\mathfrak{m}\in \operatorname{Spec}(R)\) is p-unibranch, \(R_1\) is a local ring, and we denote its maximal ideal by \(\mathfrak{m}_1\). We know \(\dim R_1=\dim R\geq 2\), therefore the same argument applies to the ring \(R_1\), and from Discussion 32 we see \(R[\mathfrak{m};M_2]\) is finite over \(R\) for \(M_2=(M_1^a:\mathfrak{m})\). Inductively we see \(R[\mathfrak{m};H^1_\mathfrak{m}(R)]\) is integral over \(R\), and so is its subring \(R[\mathfrak{m};M]\). ◻

Proof. The proof is similar to the proof of Lemma 28.

Let \(P'\in\operatorname{OS}_2^\wedge({R'})\). Take \(P_0'\in\operatorname{Min}(R'^\wedge)\) contained in \(P'\) with \(\operatorname{ht}(P'/P_0')=1\). As \(R^\wedge\) is universally catenary, we have \(\operatorname{ht}(P'\cap R^\wedge/P_0'\cap R^\wedge)=1\) by the dimension formula stacks? Tag02IJ. By Lemma 9 \(P_0'\cap R^\wedge\in\operatorname{Min}(R^\wedge)\). We have \(\operatorname{ht}(P'\cap R)\geq \operatorname{ht}(P'\cap R')>1\), by the fact \(R\to R'\) is integral and by the definition of \(\operatorname{OS}_2^\wedge({R'})\). Therefore \(P'\cap R^\wedge\in \operatorname{OS}_2^\wedge({R}).\) ◻

Proof. This follows from Theorem 41 as \(R^\wedge\) has only finitely many minimal primes. ◻

Proof. By Lemmas 40 and 42 it suffices to find a finite subalgebra \(R'\) of \(R^\sigma\) for each \(P\in\operatorname{OS}_2^\wedge({R})\) such that no FONSIs of \(R'\) are above \(P\). Fix a \(P\in\operatorname{OS}_2^\wedge({R})\) and let \(\mathfrak{p}=P\cap R\in\operatorname{Spec}(R)\), so we have \(\operatorname{ht}(\mathfrak{p})>1\).

By Theorem 12, there exists a finite \(D(\mathfrak{p})\)-modification \(R_1\) of \(R\) such that all preimages of \(\mathfrak{p}\) in \(\operatorname{Spec}(R_1)\) are p-unibranch. We know \(R_1\subseteq R^\sigma\) as \(\operatorname{ht}(\mathfrak{p})>1\). Next, we take a finite \(D(\mathfrak{p}R_1)\)-modification \(R_2\) of \(R_1\), so that the number of maximal ideals of \((R_2^\wedge)_{P}\) (i.e. the number of preimages of \(P\) in \(\operatorname{Spec}(R_2^\wedge)\)) is maximal among all possible \(R_2\); to see this is achievable, note that \((R_2^\wedge)_{\operatorname{red}}\) is finite birational over \((R_1^\wedge)_{\operatorname{red}}\), so \(((R_2^\wedge)_{\operatorname{red}})_{P}\) is contained in the normalization of \(((R_1^\wedge)_{\operatorname{red}})_{P}\), which is finite as a complete Noetherian local ring is Nagata stacks? Tag0335, so the number of maximal ideals of \((R_2^\wedge)_{P}\) is bounded. Note that we still have \(R_2\subseteq R^\sigma\) and that all preimages of \(\mathfrak{p}\) in \(\operatorname{Spec}(R_2)\) are p-unibranch. We will show no FONSIs of \(R_2\) are above \(P\).

Assume that there exists a \(P_2\in \operatorname{OS}_2^\wedge({R_2})\) above \(P\). Let \(\mathfrak{p}_2=P_2\cap R_2\), so \(\mathfrak{p}_2\) lies above \(\mathfrak{p}\), therefore is p-unibranch; and \(\operatorname{ht}(\mathfrak{p}_2)>1\) as \(P_2\in \operatorname{OS}_2^\wedge({R_2})\). In particular \(T:=(R_2)_{\mathfrak{p}_2}\) and \(A:=(R_2^\wedge)_{P_2}\) have depth at least \(1\).

The punctured spectrum \(U\) of the ring \(A\) is disconnected (Remark 25), therefore \(\mathcal{O}(U)\) has a nontrivial idempotent \(e\). In the notations of Discussion 30, there exists a finite submodule \(N\subseteq H^1_{P_2A}(A)\) such that \(A+_{P_2A}N=A[P_2A;N]=A[e]\). As \(T\to A\) is flat and as \(H^0_{\mathfrak{p}_2T}(T)=0\), we have \(H^1_{P_2A}(A)=H^0_{P_2A}(H^1_{\mathfrak{p}_2T}(T)\otimes_T A)\) (Discussion 33). Therefore there exists a finite submodule \(M\subseteq H^1_{\mathfrak{p}_2T}(T)\) such that \(N\subseteq M\otimes_T A\). Note that \(\mathfrak{p}_2T\in\operatorname{Spec}(T)\) is p-unibranch (Lemma 37), so the ring \(T[\mathfrak{p}_2T;M]\) is finite over \(T\) (Lemma 38), and therefore a finite \(D(\mathfrak{p}_2T)\)-modification. We can then find a finite \(D(\mathfrak{p}_2)\)-modification \(R_3\) of \(R_2\), which is automatically a finite \(D(\mathfrak{p}R_1)\)-modification of \(R_1\), such that \((R_3)_{\mathfrak{p}_2}=T[\mathfrak{p}_2T;M]\) (Lemma 37). Then \((R_3^\wedge)_{P_2}=T[\mathfrak{p}_2T;M]\otimes_T A=A[\mathfrak{p}_2A;M\otimes_T A]\supseteq A[\mathfrak{p}_2A;N]=A[P_2A;N]=A[e]\), where we used compatibilities in Discussions 30 and 33. In particular, \((R_3^\wedge)_{P_2}\) is not local, so \(P_2\) has more than \(1\) preimages in \(\operatorname{Spec}(R_3^\wedge)\), so \(P\) has more preimages in \(\operatorname{Spec}(R_3^\wedge)\) than in \(\operatorname{Spec}(R_2^\wedge)\), contradicting maximality. ◻

Proof. The “in particular” statement follows from Lemma 20.

That \(\operatorname{OS}_2^\wedge({R})=\emptyset\) implies \(R^{n\sigma}\) being integral over \(R\) is Theorem 26[NS2:integral]. Assume \(\operatorname{OS}_2^\wedge({R})\neq\emptyset\). We show \(R^{n\sigma}\) is not integral over \(R\). By Lemma 20 we may assume \(R\) is local. Let \(R'\) be as in Theorem 43. Let \(P\in \operatorname{OS}_2^\wedge({R})\) and let \(P_0\in\operatorname{Min}(R^\wedge)\) be contained in \(P\) so that \(\operatorname{ht}(P/P_0)=1\). Write \(\mathfrak{p}=P\cap R\), so \(\operatorname{ht}(\mathfrak{p})>1\).

Note that \(R^\wedge\to R'^\wedge\) is finite injective. Let \(P_0'\in\operatorname{Min}(R'^\wedge)\) be above \(P_0\). Let \(P'\in V(P_0')\) be above \(P\). As \(R^\wedge\) is universally catenary, we have \(\operatorname{ht}(P'/P_0')=\operatorname{ht}(P/P_0)=1\). As \(\operatorname{OS}_2^\wedge({R'})=\emptyset\), we have \(\operatorname{ht}(\mathfrak{p}')\leq 1\), where \(\mathfrak{p}'=P'\cap R'\). As \(\mathfrak{p}'\cap R=\mathfrak{p}\neq0\) we have \(\operatorname{ht}(\mathfrak{p}')=1\).

Let \(\Sigma\) be the finite set of all preimages of \(\mathfrak{p}\) in \(\operatorname{Spec}(R')\) and let \(\Sigma_0=\Sigma\setminus\{\mathfrak{p}'\}\). Let \(W'=R'\setminus \cup\Sigma_0\). We claim that \(W'^{-1}R'\subseteq (R_\mathfrak{p})^{n\sigma}\). This tells us \((R_\mathfrak{p})^{n\sigma}\) is not integral over \(R_\mathfrak{p}\) (as \(\operatorname{Spec}(W'^{-1}R')\to \operatorname{Spec}(R'_\mathfrak{p})\) is not surjective), and Lemma 20 gives \(R^{n\sigma}\) is not integral over \(R\).

To see the claim, let \(\mathfrak{q}\in\operatorname{Spec}_1(R)\) be contained in \(\mathfrak{p}\). We need to show \(W'^{-1}R'\subseteq R_\mathfrak{q}\). As \(\operatorname{ht}(\mathfrak{p})>1\), we can find \(x\in\mathfrak{p}\setminus\mathfrak{q}\). Let \(\beta\in W'^{-1}R'\). Since \(\operatorname{ht}(\mathfrak{p}')=1\), we know the fraction field of \(R'_{\mathfrak{p}'}\), which is also the fraction field of \(R\) and \(R'\), is equal to \(R'_{\mathfrak{p}'}[\frac{1}{x}]\). Therefore \(x^N\beta\in R'_{\mathfrak{p}'}\) for some \(N\), so \(x^N\beta\in R'_{\mathfrak{p}}\) as \(\beta\in W'^{-1}R'\). As \(R'\subseteq R^\sigma\), we have \(R_\mathfrak{q}=R'_{\mathfrak{q}}\). Thus \(x^N\beta\in R_\mathfrak{q}\), so \(\beta\in R_\mathfrak{q}\) as \(x\not\in\mathfrak{q}\). ◻

Proof. By Lemmas 20 and 17, we may assume \(R\) is local. By Theorem 43 we can find a finite \(R'\subseteq R^\sigma\) so that \(\operatorname{OS}_2^\wedge({R'})=\emptyset\). Let \(\mathcal{R}\) be the set of all finite \(R'\)-subalgebras of \(R^\sigma\). For every \(R''\in\mathcal{R}\) we have \(R''^{\sigma}\subseteq R^{\sigma}\) by Lemma 21, so \(R^{\sigma}=\bigcup_{R''\in\mathcal{R}}R''^{\sigma}\), and this union is filtered by Lemmas 40 and 28. By Lemma 40 and Theorem 26, each \(R''^{\sigma}\) is \((S_2)\). We conclude by Lemma 18. ◻

Proof. By Lemmas 20 and 17, we may assume \(R\) is local. Let \(\Sigma=\{\mathfrak{q}\in\operatorname{Spec}_1(S)\mid \operatorname{ht}(\mathfrak{q}\cap R)>1\}\). Then by Lemmas 28 and 42 and Theorem 12, \(\Sigma\) is finite. We may therefore find a finite subalgebra \(R'\) of \(S\) such that \(\operatorname{ht}(\mathfrak{q}\cap R')=1\) for all \(\mathfrak{q}\in\Sigma\), Lemma 13. For all \(\mathfrak{q}\in\operatorname{Spec}_1(S)\setminus\Sigma\), \(\operatorname{ht}(\mathfrak{q}\cap R)=1\), so \(\operatorname{ht}(\mathfrak{q}\cap R')=1\) as \(R\to R'\) is integral. Therefore \(\operatorname{ht}(\mathfrak{q}\cap R')=1\) for all \(\mathfrak{q}\in\operatorname{Spec}_1(S)\), consequently \(\operatorname{ht}(\mathfrak{q}\cap R'')=1\) for all \(\mathfrak{q}\in\operatorname{Spec}_1(S)\) and all \(R''\in \mathcal{R}:=\) the set of all finite \(R'\)-subalgebras of \(S\). As in the proof of Corollary 29, we see \(S^{n\sigma}=\bigcup_{R''\in\mathcal{R}}R''^{n\sigma}\), so \(S^{\sigma}=\bigcup_{R''\in\mathcal{R}}R''^{\sigma}\) as \(S^\nu=R''^\nu=R^\nu\) for all \(R''\in\mathcal{R}\). Again, as in the proof of Corollary 29, this union is filtered, and we conclude by Lemma 18 and Theorem 45. ◻

Proof. Let \(R\) be a semi-Nagata local ring. We need to show the fibers of \(R\to R^\wedge\) are \((S_1)\). This is enough by Remark 52.

By Noetherian induction we may assume this is true for all proper quotients of \(R\). If \(R\) were not an integral domain we are done, so we may assume \(R\) is an integral domain. Then there exist a finite inclusion \(R\subseteq R'\) of integral domains so that \(R'\) is \((S_2)\).

Let \(x\in R^\circ\) be a noninvertible element. Then \(R'/xR'\) is \((S_1)\). As the fibers of \(R/xR\to (R/xR)^\wedge\) are \((S_1)\) by the induction hypothesis, so are the fibers of \(R'/xR'\to (R'/xR')^\wedge\). Therefore \((R'/xR')^\wedge\) is \((S_1)\) stacks? Tag0339, so \(R'^\wedge\) is \((S_1)\) EGA4_2?. As \(R\subseteq R'\) is a finite inclusion of Noetherian semilocal domains we see \(R^\wedge\) is \((S_1)\) (Lemma 9). ◻

Proof. By Theorem 43 we can find a finite subalgebra \(R'\) of \(R^\sigma\) so that \(\operatorname{OS}_2^\wedge({R'})=\emptyset\). \(R'^\wedge\) is \((S_1)\) by Lemma 9. Therefore \(R'^\sigma\) is a finite \(R'\)-algebra inside \(R^\sigma\) (Lemma 21) that is \((S_2)\) (Theorem 26). ◻

Proof. Lemma 53 gives [semi-Nagatalocal:semi-Nagata] implies [semi-Nagatalocal:S1ring], whereas [semi-Nagatalocal:finiteS2ify] trivially implies [semi-Nagatalocal:semi-Nagata]. Finally, to see [semi-Nagatalocal:S1ring] implies [semi-Nagatalocal:finiteS2ify], we may replace \(R\) by \(B\) and assume \(R=B\) is an integral domain. Then \(R^\wedge\) is \((S_1)\) by stacks? Tag0339, so [semi-Nagatalocal:finiteS2ify] follows from Lemma 54. ◻

Proof. EGA4_2? gives [S2-2:open]. For [S2-2:fg], it suffices to show for a finite type inclusion of Noetherian domains \(R\subseteq B\), if \(R\) is \((S_2)\), then \(B_g\) is \((S_2)\) for some \(g\in B^\circ\). To see this, we may assume \(R\to B\) is flat stacks? Tag051R, and has Cohen–Macaulay fibers stacks? Tag045U. Then \(B\) is \((S_2)\) by stacks? Tag0339. ◻

Proof. As note in Lemma 60 and before Lemma 53, [semi-Nagata:S1ring] is preserved by essentially finitely generated algebras, giving the last assertion.

To see [semi-Nagata:semi-Nagata] implies [semi-Nagata:S1ring], Lemma 53 says a semi-Nagata ring is an \((S_1)\)-ring. Therefore it suffices to show for a semi-Nagata domain \(R\), there exists an \(f\in R^\circ\) so that \(R_f\) is \((S_2)\). Let \(R\subseteq R'\) be a finite inclusion so that \(R'\) is \((S_2)\). Then any \(f\in R^\circ\) so that \(R'_f\) is flat over \(R_f\) works.

As [semi-Nagata:finiteS2ify] implies [semi-Nagata:semi-Nagata], it suffices to show [semi-Nagata:S1ring] implies [semi-Nagata:finiteS2ify]. Assume [semi-Nagata:S1ring] and assume \(R\) is an integral domain. We must show that there exists a finite subalgebra \(R'\) of \(R^\sigma\) that is \((S_2)\). Let \(U\) be the \((S_2)\) locus of \(R\), which is open by Lemma 60, and let \(\mathfrak{p}\not\in U\), so \(\operatorname{ht}(\mathfrak{p})>1\).

The local ring \(R_\mathfrak{p}\) is semi-Nagata by Theorem 55, so by Lemma 20 there exists a finite subalgebra \(R_1\) of \(R^\sigma\) such that \((R_1)_\mathfrak{p}\) is \((S_2)\). As \(R\) is \((S_2)\)-2, the \((S_2)\) locus \(U_1\) of the ring \(R_1\) is open by Lemma 60, and we have \(\bigcap_{h\in R\setminus\mathfrak{p}}D_{R_1}(f)\subseteq U_1\). As the constructible topology of \(\operatorname{Spec}(R_1)\) is compact stacks? Tag0901, we see there exists \(f_1\in R\setminus\mathfrak{p}\) such that \((R_1)_{f_1}\) is \((S_2)\). Since \(R_1\subseteq R^\sigma\) we know \(R=R_1\) over \(U\), so the image \(Z_1\) of the non-\((S_2)\) locus of \(R_1\) in \(\operatorname{Spec}(R)\) is disjoint from \(U\cup D(f_1)\). If \(Z_1\neq\emptyset\), take \(\mathfrak{q}\in Z_1\), then similarly the semilocal ring \((R_1)_\mathfrak{q}\) is semi-Nagata, and we can find \(R_2\subseteq R_1^\sigma\subseteq R^\sigma\) (Lemma 21) so that the image \(Z_2\) of the non-\((S_2)\) locus of \(R_2\) in \(\operatorname{Spec}(R)\) is disjoint from \(U\cup D(f_1)\cup D(f_2)\) and \(f_2\in R\setminus\mathfrak{q}\). As the topological space \(\operatorname{Spec}(R)\) is Noetherian, we get our desired \(R'\) after finitely many steps. ◻

Alternative proof of [semi-Nagata:S1ring] implies [semi-Nagata:finiteS2ify]. Let \(R\) be a Noetherian integral domain that satisfies [semi-Nagata:S1ring]. We want to show there exists a finite subalgebra of \(R^\sigma\) that is \((S_2)\).

Let \(\mathfrak{m}\in\operatorname{Max}(R)\). Then \(R_\mathfrak{m}\) is semi-Nagata by Theorem 55, so by Lemma 20 there exists a finite subalgebra \(R(\mathfrak{m})\) of \(R^\sigma\) such that \((R(\mathfrak{m}))_\mathfrak{m}\) is \((S_2)\).

As \(R\) is \((S_2)\)-2, the \((S_2)\) locus of the ring \(R(\mathfrak{m})\) is open (Lemma 60). As the constructible topology of \(\operatorname{Spec}(R(\mathfrak{m}))\) is compact stacks? Tag0901, there exists \(f(\mathfrak{m})\in R\setminus\mathfrak{m}\) such that \(R(\mathfrak{m})_{f(\mathfrak{m})}\) is \((S_2)\).

Take finitely many \(\mathfrak{m}_1,\ldots,\mathfrak{m}_n\) so that \(D(f(\mathfrak{m}_i))\;(1\leq i\leq n)\) cover \(\operatorname{Spec}(R)\) and let \(R_1\) be the \(R\)-algebra generated by all \(R(\mathfrak{m}_i)\). Then \(R(\mathfrak{m}_i)_{f(\mathfrak{m}_i)}\subseteq (R_1)_{f(\mathfrak{m}_i)}\), so by Remark 25 and Lemma 40 we have \(\operatorname{OS}_2^\wedge({R_1})=\emptyset\). As \(R_1^\sigma\subseteq R^\sigma\) (Lemma 21) we may replace \(R\) by \(R_1\) to assume \(\operatorname{OS}_2^\wedge({R})=\emptyset\). Apply the same construction again, we see from Theorem 26[NS2:S2unique] (and Lemma 20) that \(R(\mathfrak{m}_i)_{f(\mathfrak{m}_i)}=R^\sigma_{f(\mathfrak{m}_i)}\) for all \(i\), so \(R^\sigma\) is finite over \(R\). ◻

Proof. EGA4_2? tells us a Cohen–Macaulay ring is an \((S_1)\)-ring, and EGA4_2? tell us a Cohen–Macaulay ring is \((S_2)\)-2. ◻

Proof. Immediate from Theorem 26[NS2:S2unique]. ◻

Proof. We will use Theorem 61[semi-Nagata:S1ring].

Let \(U\) be the locus where \(M\) is \((S_2)\). Then \(U\) is open (Lemma 60), and contains \(\Sigma\) as \(\Sigma=\operatorname{Spec}_1(R)\) and \(\operatorname{Ass}(M)=\operatorname{Min}(R)\). Let \(j:U\to \operatorname{Spec}(R)\) be the canonical open immersion. It is now clear that \(N=j_*j^*M\).

Let \(\mathfrak{p}_0\) be a minimal prime of \(R\), and let \(\overline{R}=R/\mathfrak{p}_0\). By stacks? Tag0BK3 to show \(N\) is finite it suffices to show for \(\mathfrak{p}\in \operatorname{Spec}(R)\setminus U\) containing \(\mathfrak{p}_0\), every \(P_0\in\operatorname{Ass}(\overline{R}^\wedge_{\mathfrak{p}})\) satisfies \(\dim (\overline{R}^\wedge_\mathfrak{p}/P_0)>1\).

We know \(\overline{R}^\wedge_\mathfrak{p}\) is \((S_1)\) as \(\overline{R}\) is \((S_1)\) with \((S_1)\) formal fibers. Moreover, for every \(P_0\in \operatorname{Min}(\overline{R}^\wedge_\mathfrak{p})\), we have \(\dim (\overline{R}^\wedge_\mathfrak{p}/P_0)>1\) as \(\operatorname{ht}(\mathfrak{p})>1\) and as \(\operatorname{OS}_2^\wedge({\overline{R}_\mathfrak{p}})=\emptyset\) (Corollary 44). Thus for all \(P_0\in \operatorname{Ass}(\overline{R}^\wedge_\mathfrak{p})\), we have \(\dim (\overline{R}^\wedge_\mathfrak{p}/P_0)>1\). Therefore \(N\) is finite. It is \((S_2)\) by for example EGA4_2?. ◻

Proof. Fix a strictly functorial \(\mathbf{P}\)-assignment \(A\mapsto\mathfrak{c}(A)\) on \(\mathcal{D}^\mathbf{Q}_\mathbf{P}\).

Let \(R\) be a ring with an ideal \(I\) that satisfy the assumptions. For every \(R\)-algebra \(S\) we denote by \(S^*\) the \((IS)\)-adic completion of \(S\). By induction, we may assume

1. the theorem holds when the dimension of \(R\) is strictly smaller.

It suffices to show for every \(R\)-algebra \(B\) that is an integral domain, \((B^\circ)^{-1}B^\wedge\) satisfies \(\mathbf{P}\). By [Nsmr:Q-ify] we may assume \(B\) satisfies \(\mathbf{Q}\). Replace \(R\) by \(B\) we may assume

1. \(R\) is an integral domain that satisfies \(\mathbf{Q}\).

By [Nsmr:Qring], [PisOpen], and the fact a complete local ring is a \(\mathbf{P}\)-ring we see

1. \(R^\wedge\in \mathcal{D}^\mathbf{Q}_\mathbf{P}\).

By [Nsmr:indOnDim], we have

1. \((R_\mathfrak{p})^*\) is a \(\mathbf{P}\)-ring for all \(\mathfrak{p}\in\operatorname{Spec}(R)\setminus\operatorname{Max}(R)\).

Write \(C=\mathfrak{c}(R^\wedge)\neq 0\), and for every \(n\in\mathbf{Z}_{\geq 1}\), \(C_n=C+I^nR^\wedge\), \(\mathfrak{a}_n=C_n\cap R\). We will show \(C_n=\mathfrak{a}_nR^\wedge\) for all \(n\), which implies \(C\cap R\neq 0\) by Rotthaus-qe-semilocal-lifting?, which then tells us the generic fiber \((R^\circ)^{-1}R^{\wedge}\) is \(\mathbf{P}\), as \(V(C)=\operatorname{Spec}(R^\wedge)\setminus U_\mathbf{P}(R^\wedge)\).

By consideration of a primary decomposition of \(C_n\), and by the fact flat base change commutes with finite intersections, it suffices to show for a primary ideal \(Q\) containing \(C_n\) we have \(C_n\subseteq (Q\cap R)R^\wedge\). If \(\sqrt{Q}\) is maximal then this is trivial as \(Q=(Q\cap R)R^\wedge\). Therefore we may assume \(\sqrt{Q}\) is not maximal. Let \(\mathfrak{p}=\sqrt{Q}\cap R=\sqrt{Q\cap R}\in\operatorname{Spec}(R)\setminus\operatorname{Max}(R)\). As \(R\to R^\wedge\) is flat and as \(Q\cap R\) is \(\mathfrak{p}\)-primary, every prime divisor of \((Q\cap R)R^\wedge\) is above \(\mathfrak{p}\). Therefore it suffices to show \(C_n (R^\wedge)_\mathfrak{p}\subseteq (Q\cap R)(R^\wedge)_\mathfrak{p}\). In the remainder of the proof we show \(C_n (R^\wedge)_\mathfrak{p}=\mathfrak{a}_n (R^\wedge)_\mathfrak{p}\) for all \(\mathfrak{p}\in\operatorname{Spec}(R)\setminus\operatorname{Max}(R)\), which is enough as \(Q\cap R\supseteq C_n\cap R=\mathfrak{a}_n\).

Consider the commutative diagram of rings \[\begin{CD} R@>>> R^\wedge\\ @VVV @VVV\\ R_\mathfrak{p}@>{f_\mathfrak{p}}>> (R^\wedge)_\mathfrak{p}\\ @V{g_\mathfrak{p}}VV @V{g^\wedge_\mathfrak{p}}VV\\ (R_\mathfrak{p})^* @>{f^*_\mathfrak{p}}>> ((R^\wedge)_\mathfrak{p})^*.\\ \end{CD}\] We know \((R^\wedge)_\mathfrak{p}\) is (quasi-)excellent as \(R^\wedge\) is complete, therefore \(((R^\wedge)_\mathfrak{p})^*\) is quasi-excellent formal-lifting-excellence-Gabber?. In particular, both \((R^\wedge)_\mathfrak{p}\) and \(((R^\wedge)_\mathfrak{p})^*\) are \(\mathbf{P}\)-rings and have open \(\mathbf{P}\)-locus. By [Nsmr:locPring] \((R_\mathfrak{p})^*\) is a local \(\mathbf{P}\)-ring, hence its \(\mathbf{P}\)-locus is open by [PisOpen], cf. stacks? Tag02JY.

We know \(f_\mathfrak{p}\) and \(g_\mathfrak{p}\) are \(\mathbf{Q}\)-maps and \(g^\wedge_\mathfrak{p}\) is a \(\mathbf{P}\)-map by [Nsmr:Qring][Nsmr:locPring] and stacks? Tag0BK9. The map \(f^*_\mathfrak{p}\) is faithfully flat stacks? Tag0AGW, and for every \(\mathfrak{Q}\in V(I(R_\mathfrak{p})^*)\), the fiber of \(f^*_\mathfrak{p}\) over \(\mathfrak{Q}\) is the same as the formal fiber of \(R\) over \(\mathfrak{Q}\cap R\in V(I)\), which is geometrically \(\mathbf{P}\) by [Nsmr:quotPring]. As every maximal ideal of \(((R^\wedge)_\mathfrak{p})^*\) contains \(I((R^\wedge)_\mathfrak{p})^*\), we see from [PGroLocalizes] that \(f^*_\mathfrak{p}\) is a \(\mathbf{P}\)-map. Consequently, if we remove \(R\) and \(R_\mathfrak{p}\), then the diagram above is a diagram in \(\mathcal{D}^\mathbf{Q}_\mathbf{P}\) (cf. [Nsmr:RhatisQ]). Therefore \(C((R^\wedge)_\mathfrak{p})^*=\mathfrak{c}(((R^\wedge)_\mathfrak{p})^*)=\mathfrak{c}((R_\mathfrak{p})^*)((R^\wedge)_\mathfrak{p})^*\).

Now, let \(\mathfrak{b}=(\mathfrak{c}((R_\mathfrak{p})^*)+I^n(R_\mathfrak{p})^*)\cap R_\mathfrak{p}\), so \(\mathfrak{b}(R_\mathfrak{p})^*=\mathfrak{c}((R_\mathfrak{p})^*)+I^n(R_\mathfrak{p})^*\) as \((R_\mathfrak{p})^*\) is the \(I\)-adic completion of \(R_\mathfrak{p}\). Then we have \(\mathfrak{b}((R^\wedge)_\mathfrak{p})^*=C_n((R^\wedge)_\mathfrak{p})^*\), so \(\mathfrak{b}(R^\wedge)_\mathfrak{p}=C_n(R^\wedge)_\mathfrak{p}\) as both sides contain \(I^n (R^\wedge)_\mathfrak{p}\). Contract to \(R\) we see \(\mathfrak{b}=\mathfrak{a}_n R_\mathfrak{p}\), so \(\mathfrak{a}_n (R^\wedge)_\mathfrak{p}=C_n(R^\wedge)_\mathfrak{p}\), as desired. ◻

Proof. Let \(\mathfrak{c}(-)\) on \(\mathcal{A}^\mathbf{Q}_\mathbf{P}\) be a given strictly functorial \(\mathbf{P}\)-assignment.

For \(A\in \mathcal{D}^\mathbf{Q}_{\mathbf{P}}\), let \(\mathfrak{p}_1,\ldots,\mathfrak{p}_n\;(n\geq 0)\) be the generic points of \(\operatorname{Spec}(A)\setminus U_\mathbf{P}(A)\). If a desired extension exists, then it must satisfy \(\mathfrak{c}(A)=\bigcap_i(\mathfrak{c}(A_{\mathfrak{p}_i})\cap A)\) as \(\mathfrak{c}(A)\) has no embedded prime divisors. Furthermore, we must have \(\mathfrak{c}(A_{\mathfrak{p}_i})=\mathfrak{c}(A^\wedge_{\mathfrak{p}_i})\cap A_{\mathfrak{p}_i}\), as the completion map \(A_{\mathfrak{p}_i}\to A^\wedge_{\mathfrak{p}_i}\) is in \(\mathcal{D}^\mathbf{Q}_\mathbf{P}\) (i.e. a \(\mathbf{P}\)-map) by condition [DisPring] in Definition 68. As \(\mathfrak{p}_i\) is a generic point of \(\operatorname{Spec}(A)\setminus U_\mathbf{P}(A)\) we see \(U_\mathbf{P}(A_{\mathfrak{p}_i})=D(\mathfrak{p}_iA_{\mathfrak{p}_i})\), therefore \(U_\mathbf{P}(A^\wedge_{\mathfrak{p}_i})=D(\mathfrak{p}_iA^\wedge_{\mathfrak{p}_i})\), in other words \(A^\wedge_{\mathfrak{p}_i}\in \mathcal{A}^\mathbf{Q}_\mathbf{P}\). This shows the uniqueness of the extension; we must have \(\mathfrak{c}(A)=\bigcap_i\mathfrak{c}(A^\wedge_{\mathfrak{p}_i})\cap A\).

It remains to verify that \(\mathfrak{c}(A):=\bigcap_i\mathfrak{c}(A^\wedge_{\mathfrak{p}_i})\cap A\) is indeed a strictly functorial \(\mathbf{P}\)-assignment; by construction it has no embedded prime divisors as each \(\mathfrak{c}(A^\wedge_{\mathfrak{p}_i})\cap A\) is \(\mathfrak{p}_i\)-primary. It is clear that \(V(\mathfrak{c}(A))=\operatorname{Spec}(A)\setminus U_\mathbf{P}(A)\). We have \(\mathfrak{c}(A)\neq 0\) as \(\mathfrak{c}(A)=A\) when \(n=0\), and \(\mathfrak{c}(A)_{\mathfrak{p}_1}=\mathfrak{c}(A^\wedge_{\mathfrak{p}_1})\cap A_{\mathfrak{p}_1}\neq 0\) when \(n>0\), as \(\mathfrak{c}(A^\wedge_{\mathfrak{p}_1})\) is nonzero and \((\mathfrak{p}_1A^\wedge_{\mathfrak{p}_1})\)-primary.

It remains to show for \(\varphi:A\to B\) in \(\mathcal{D}^\mathbf{Q}_\mathbf{P}\), we have \(\mathfrak{c}(A)B=\mathfrak{c}(B)\), where \(\varphi\) is omitted in the notation. Let \(\mathfrak{q}_{ij}\;(1\leq j\leq m_i)\) be the minimal prime divisors of \(\mathfrak{p}_i B\), where \(m_i\geq 0\). As \(\varphi\) is a \(\mathbf{P}\)-map, \(\operatorname{Spec}(\varphi)^{-1}(U_\mathbf{P}(A))=U_\mathbf{P}(B)\), so \(\mathfrak{q}_{ij}\;(1\leq j\leq m_i,1\leq i\leq n)\) are exactly the generic points of \(\operatorname{Spec}(B)\setminus U_\mathbf{P}(B)\). Moreover, as \(\mathbf{P}\) implies \((S_1)\), we see for \(\mathfrak{c}_i:=\mathfrak{c}(A^\wedge_{\mathfrak{p}_i})\cap A\), \(\operatorname{Ass}_{B}(B/\mathfrak{c}_iB)=\{\mathfrak{q}_{ij}\mid 1\leq j\leq m_i\}\). Therefore it suffices to show \(\mathfrak{c}_iB_{\mathfrak{q}_{ij}}=\mathfrak{c}(B^\wedge_{\mathfrak{q}_{ij}})\cap B_{\mathfrak{q}_{ij}}\), and as both sides are \(\mathfrak{q}_{ij}\)-primary, passing to the completion we see it suffices to show \(\mathfrak{c}(A^\wedge_{\mathfrak{p}_i})B^\wedge_{\mathfrak{q}_{ij}}=\mathfrak{c}(B^\wedge_{\mathfrak{q}_{ij}})\). We know \(A^\wedge_{\mathfrak{p}_i},B^\wedge_{\mathfrak{q}_{ij}}\in {^d\mathcal{A}^\mathbf{Q}_\mathbf{P}}\) for \(d:=\operatorname{ht}(\mathfrak{p}_i)=\operatorname{ht}(\mathfrak{q}_{ij})\), therefore, as our \(\mathfrak{c}(-)\) is strictly functorial on \(\mathcal{A}^\mathbf{Q}_\mathbf{P}\), it suffices to show \(A^\wedge_{\mathfrak{p}_i}\to B^\wedge_{\mathfrak{q}_{ij}}\) is a \(\mathbf{P}\)-map. By [PGroLocalizes] it suffices to show \(\kappa(\mathfrak{p}_i)\to (B/\mathfrak{p}_iB)^\wedge_{\mathfrak{q}_{ij}}\) is a \(\mathbf{P}\)-map. This follows from the fact \(\varphi:A\to B\) is a \(\mathbf{P}\)-map and the fact \(B/\mathfrak{p}_iB\), a quotient of \(B\in\mathcal{D}^\mathbf{Q}_\mathbf{P}\), is a \(\mathbf{P}\)-ring. ◻

Proof. It is clear that \(\mathfrak{c}(A)\) is \(\mathfrak{m}_A\)-primary and \(\mathfrak{c}(-)\) is strictly functorial, as the closed fiber of all maps in \(\mathcal{A}_{\mathbf{P}}\) have dimension \(0\). As an Artinian ring is \((S_1)\) we see \(^0\mathcal{A}^\mathbf{Q}_\mathbf{P}=\emptyset\), so \(\dim A>0\) and any \(\mathfrak{m}_A\)-primary ideal is nonzero. ◻

Proof. Again, as an Artinian ring is \((S_2)\) the \(\mathfrak{m}_A\)-primary ideal \(\mathfrak{c}(A)\) is nonzero. It remains to show for \(\varphi:A\to B\) in \(\mathcal{A}^\mathbf{Q}_\mathbf{P}\), we have \(\mathfrak{c}(A)B=\mathfrak{c}(B)\).

We know \(\varphi\) has \((S_1)\) fibers and its closed fiber has dimension \(0\). For \(Q\in\Sigma_1\), \(B/QB\) is therefore \(1\)-dimensional and \((S_1)\), so all prime divisors \(\mathfrak{P}\) of \(QB\) are such that \(\dim(B/\mathfrak{P})=1\). If we can show for all \(Q\in\Sigma_2\) and all prime divisors \(\mathfrak{P}\) of \(QB\) (which are automatically minimal), we have \(\dim(B/\mathfrak{P})>1\), then it will follow that \(A_1\otimes_A B=B_1\) and \(A_2\otimes_A B=B_2\), so \(A_2'\otimes_A B=B_2'\) and \(\mathfrak{c}(A)B=\mathfrak{c}(B)\).

We have a commutative diagram \[\begin{CD} A_2@>>> A_2'\\ @VVV @VVV\\ A_2\otimes_A B@>>> A_2'\otimes_A B \end{CD}\] of rings. The ring \(A_2'\otimes_A B\) is \((S_2)\) (as \(A_2'\) and the fibers of \(\varphi\) are) and universally catenary, hence locally equidimensional EGA4_2?. Let \(\mathfrak{N}'\in\operatorname{Max}(A_2'\otimes_A B)\). Then \(\mathfrak{N}'\cap (A_2\otimes_A B)\) is the maximal ideal \(\mathfrak{N}\) of the local ring \(A_2\otimes_A B\), so \(\mathfrak{N}'\cap A_2\) is the maximal ideal of \(A_2\), so \(\mathfrak{N}'\cap A_2'\in\operatorname{Max}(A_2')\). By flatness \(\operatorname{ht}(\mathfrak{N}')\geq \operatorname{ht}(\mathfrak{N}'\cap A_2')>1\). As \(\mathfrak{N}'\) was arbitrary, a similar discussion as the case of \(A_2\) tells us for all \(\mathfrak{Q}_0\in\operatorname{Min}(A_2\otimes_A B)\) we have \(\dim((A_2\otimes_A B)/\mathfrak{Q}_0)>1\), as desired. ◻

Proof. By Discussion 78 \(H^{\operatorname{depth}A}_{\mathfrak{m}_A}(A)\) is of finite length, so \(\mathfrak{c}(A)\) is \(\mathfrak{m}_A\)-primary. Again, as an Artinian ring is \((S_k)\) we see \(\mathfrak{c}(A)\neq 0\). For all \(\varphi:A\to B\) in \(\mathcal{A}^\mathbf{Q}_\mathbf{P}\), the closed fiber of \(\varphi\) has dimension \(0\), so we have \(\operatorname{depth}A=\operatorname{depth}B\) stacks? Tag0337. This shows \(\mathfrak{c}(A)B=\mathfrak{c}(B)\). ◻

Proof. We have seen \(\mathfrak{c}(A)\) is \(\mathfrak{m}_A\)-primary and nonzero. Strict functoriality follows immediately from the fact Fitting ideals commute with base change stacks? Tag07ZA and that for \(\varphi:A\to B\) in \(\mathcal{A}^\mathbf{Q}_\mathbf{P}\) and a normalized dualizing complex \(\omega\) of \(A\), \(\omega\otimes^L_A B\) is a normalized dualizing complex of \(B\). See for example Lyu-dual-complex-lift?, note \(\dim A=\dim B\). ◻

Proof. Apply Discussion 84 to the triangle \[\begin{CD} L_{A/\mathbf{Z}}@>>> L_{A/R}@>>> (L_{R/\mathbf{Z}}\otimes^L_R A)[1]@>>> +1, \end{CD}\] we get an exact sequence \[\begin{CD} H@>>> C^a(L_{A/\mathbf{Z}})@>>> C^a(L_{A/R})@>>> P@>>> 0. \end{CD}\] where \(P=C^{a+1}(L_{R/\mathbf{Z}}\otimes^L_R A)\), \(H=H^{a}(L_{R/\mathbf{Z}}\otimes^L_R A)\). Since \(R\) is lci, \(L_{R/\mathbf{Z}}\) has tor-amplitude in \([-1,0]\) avramov-ci?, so \(H=0\). Moreover, every flat \(A\)-module has projective dimension \(\leq \dim A\) Raynaud?. Therefore \(L_{R/\mathbf{Z}}\) has projective-amplitude in \([-\dim A-1,0]\), so \(P\) is projective, and we get \(C^a(L_{A/\mathbf{Z}})\oplus P\cong C^a(L_{A/R})\). It remains to observe \(C^a(L_{A/R})\) is finite up to projective summands, as \(L_{A/R}\in D_{Coh}(A)\) stacks? Tag08PZ. ◻

Proof. Take the same exact sequence as in the proof of Lemma 85. We have \(H=0\) and \(P\) is flat. Thus \(C^a(L_{A/\mathbf{Z}})\) is flat if and only \(C^a(L_{A/R})\) is flat, if and only if \(R\to A\) is lci (Briggs-Iyenger-Cotangent-Complex? and avramov-ci?), if and only if \(A\) is lci avramov-ci?. ◻

Proof. We first show [Fit:wdf] for finite-by-flat modules. Let \(M,M'\) be two finite submodules of a given module \(X\) so that \(X/M\) and \(X/M'\) are flat. By Lazard’s Theorem stacks? Tag058G, we can write \(X/M=\operatorname{colim}_\alpha L_\alpha\), where the colimit is filtered and \(L_\alpha\) are finite free. Let \(X_\alpha=X\times_{X/M}L_\alpha\), so we have a commutative diagram with exact rows \[\begin{CD} 0@>>> M@>>> X_\alpha@>>> L_\alpha@>>> 0\\ @. @| @VVV @VVV @.\\ 0@>>> M@>>> X@>>> X/M@>>> 0. \end{CD}\] We have \(X=\operatorname{colim}_\alpha X_\alpha\) as filtered colimits commute with finite limits. Therefore the inclusion \(M'\subseteq X\) factors through some \(X_\alpha\) stacks? Tag0G8P. By stacks? Tag058M \(M'\subseteq X\) is pure (i.e. universally injective), so \(M'\to X_\alpha\) is also pure, thus split stacks? Tag058L. On the other hand \(X_\alpha\cong M\oplus L_\alpha\) as \(A\)-modules as \(L_\alpha\) is free. We conclude that \(M'\) is isomorphic to a direct summand of \(M\oplus F\) where \(F\) is a finite free \(A\)-module, and by symmetry \(M\) is isomorphic to a direct summand of \(M'\oplus F'\) where \(F'\) is a finite free \(A\)-module. If \(A\) is complete, it follows from Krull–Schmidt Leuschke-Wiegand-Cohen-Macaulay-Modules? that \(M\oplus P\cong M'\oplus P'\) for some finite free \(A\)-modules \(P,P'\). For a general \(A\) the same is true by Leuschke-Wiegand-Cohen-Macaulay-Modules?. Therefore the Fitting invariants of \(M\) and \(M'\) are the same stacks? Tag07ZA.

Given an inclusion of modules \(X\subseteq Y\) with flat quotient, if \(M\) is a finite submodule of \(X\) so that \(X/M\) is flat, then \(M\) is a finite submodule of \(Y\) so that \(Y/M\) is flat, as \(Y/M\) is an extension of \(X/M\) by \(Y/X\). This gives [Fit:cokerFlatSameFit] in the finite-by-flat case.

Next, let \(X\) be a pseudo-finite-by-flat module and \(C,D\) be flat modules so that \(X\oplus C\) and \(X\oplus D\) are both finite-by-flat. Then \(X\oplus C\oplus D\) is finite-by-flat and has the same Fitting invariant as \(X\oplus C\) and \(X\oplus D\) by [Fit:cokerFlatSameFit] for finite-by-flat modules. Therefore the Fitting invariants of \(X\oplus C\) and \(X\oplus D\) are the same, which is [Fit:wdf] for \(X\).

Finally, given an inclusion of modules \(X\subseteq Y\) with flat quotient, if \(C\) is a flat module, then we have an inclusion of modules \(X\oplus C\subseteq Y\oplus C\) with flat quotient. This gives [Fit:cokerFlatSameFit] in the pseudo-finite-by-flat case. ◻

Proof. We can find a complete regular local ring \(R\) and a surjective ring map \(R\to A\). By Theorem 90, \(\mathfrak{c}(A)\) is well-defined and \(0\neq \mathfrak{c}(A)\neq A\). To show \(\mathfrak{c}(-)\) is strictly functorial, let \(\varphi:A\to B\) be in \({^d\mathcal{A}^\mathbf{Q}_\mathbf{P}}\). Apply Discussion 84 to \[\begin{CD} L_{A/\mathbf{Z}}\otimes^L_A B@>>> L_{B/\mathbf{Z}}@>>> L_{B/A}@>>> +1, \end{CD}\] we get an exact sequence \[\begin{CD} H@>>> C^a(L_{A/\mathbf{Z}})\otimes_A B@>>> C^a(L_{B/\mathbf{Z}})@>>> C@>>> 0 \end{CD}\] where \(H=H^{a-1}(L_{B/A})\) and \(C=C^a(L_{B/A})\). As \(\varphi\) is lci and as \(a\leq -1\) we have \(H=0\) and \(C\) flat (in fact projective since \(a=-d-2\), cf. proof of Lemma 85), so Lemma 89[Fit:cokerFlatSameFit] gives \(\mathfrak{c}(A)B=\mathfrak{c}(B)\).

It remains to show \(\mathfrak{c}(A)\) is \(\mathfrak{m}_A\)-primary. When \(d=0\) this is trivial, so we assume \(d>0\). Let \(P^\bullet\) be a complex of finite free modules that satisfies \(P^{>-1}=0\) and represents \(L_{A/R}\) stacks? Tags 08PZ and08QF. As seen in Lemma 85 \(\mathfrak{c}(A)\) is the Fitting invariant of \(C^a(P^\bullet)\). We will show \(M:=C^a(P^\bullet)\) is finite flat of constant rank, say \(r\), on the punctured spectrum of \(A\). Then by stacks? Tag07ZD, for \(\mathfrak{a}=\operatorname{Fit}_j(M)\;(j<r)\), we have \(\mathfrak{a}_\mathfrak{p}=0\) for all \(\mathfrak{p}\in\operatorname{Spec}(A)\setminus \{\mathfrak{m}_A\}\), so \(\mathfrak{a}=0\) as \(\operatorname{depth}A\geq 1\); therefore \(\mathfrak{c}(A)=\operatorname{Fit}_r(M)\), and \(\mathfrak{c}(A)_\mathfrak{p}=A_\mathfrak{p}\).

We know \(M\) is finite flat on the punctured spectrum of \(A\) which is lci. If \(d\geq 2\), then \(\operatorname{depth}A\geq 2\), as \(A\) is \((S_2)\). Therefore the punctured spectrum of \(A\) is connected stacks? Tag0BLR, so the rank is constant. We may now assume \(d=1\).

Let \(I=\ker(R\to A)\), \(\mathfrak{p}\in V(I)\setminus\operatorname{Max}(R)\). The complex \((\tau_{\geq a} P^\bullet)_\mathfrak{p}\) represents \((I/I^2)_\mathfrak{p}[1]\) stacks? Tag08SJ, as \(A_\mathfrak{p}\) is lci. Computing Euler characteristic, we see \((-1)^a\operatorname{rank}M_\mathfrak{p}+ \sum_{i=a+1}^{-1} (-1)^i\operatorname{rank}P^i=-\operatorname{rank}(I/I^2)_\mathfrak{p}\). We know \(\operatorname{rank}(I/I^2)_\mathfrak{p}=\dim R_\mathfrak{p}-\dim A_\mathfrak{p}\), as \(I_\mathfrak{p}\) is generated by a regular sequence. As \(d=1\) and as \(R\) is a catenary domain, we have \(\dim R_\mathfrak{p}=\dim R-1,\dim A_\mathfrak{p}=0\), independent of the choice of \(\mathfrak{p}\). Therefore, \(\operatorname{rank}M_\mathfrak{p}\) is independent of the choice of \(\mathfrak{p}\), as desired. ◻

Proof. First consider the case \(\mathbf{P}\)=“\((S_1)\),” as every Noetherian ring is \((S_0)\). Let \(\mathbf{Q}\) be the trivial property. A strictly functorial \(\mathbf{P}\)-assignment on \(\mathcal{D}_\mathbf{P}\) exists, Lemmas 73 and 74. The assumptions [Nsmr:Qring] and [Nsmr:Q-ify] in Theorem 70 are trivial, and we conclude.

Next, consider the case \(\mathbf{P}\)=“\((S_2)\).” Let \(\mathbf{Q}\)=“\((S_1)\).” A strictly functorial \(\mathbf{P}\)-assignment on \(\mathcal{D}^\mathbf{Q}_\mathbf{P}\) exists, Lemmas 73 and 76. The assumption [Nsmr:Q-ify] in Theorem 70 is trivial as a domain in \((S_1)\), whereas [Nsmr:Qring] follows from the case \(\mathbf{P}\)=“\((S_1)\),” and we conclude.

Finally, consider the case \(\mathbf{P}\)=“\((S_k)\)” \((k\geq 3)\), “Cohen–Macaulay,” “Gorenstein,” or “lci.” Let \(\mathbf{Q}\)=“\((S_2)\).” A strictly functorial \(\mathbf{P}\)-assignment on \(\mathcal{D}^\mathbf{Q}_\mathbf{P}\) exists, Lemma 73 and Lemma 79, Remark 80, Lemma 82, and Theorem 92. The assumption [Nsmr:Qring] in Theorem 70 follows from the case \(\mathbf{P}\)=“\((S_2)\),” and [Nsmr:Q-ify] follows from Theorem 55 (or Macaulay-Cesnavi?), and we conclude. ◻

Proof. By Noetherian induction, we may assume \(A\) is an integral domain and that \(A/I\to B/IB\) is a \(\mathbf{P}_k\)-map for all nonzero ideals \(I\subseteq A\). Let \(K\) be the fraction field of \(A\). We need to show \(B\otimes_A K\) is a geometrically \(\mathbf{P}_k\) \(K\)-algebra. Let \(\mathfrak{q}\in\operatorname{Spec}(B)\) be above \(0\in\operatorname{Spec}(A)\) and of height \(\leq k\). By [PisPointwise], it suffices to show \(B_\mathfrak{q}\) is a geometrically \(\mathbf{P}\) \(K\)-algebra.

Let \(\underline{x}\) be a system of parameters of \(A\), and let \(\mathfrak{Q}\) be a minimal divisor of \(\mathfrak{q}+(\underline{x})B\). Then \(\operatorname{ht}(\mathfrak{Q}/\mathfrak{q})\leq \dim A\), so \(\operatorname{ht}(\mathfrak{Q})\leq \operatorname{ht}(\mathfrak{q})+\dim A\) as \(B\) is catenary and equidimensional. By stacks? Tag00ON we have \(\operatorname{ht}(\mathfrak{Q}/\mathfrak{m}B)\leq \operatorname{ht}(\mathfrak{q})\leq k\), \(\mathfrak{m}\) being the maximal ideal of \(A\). Since \(B/\mathfrak{m}B\) is geometrically \(\mathbf{P}_k\) by [RkGro:closedFiber], we see \(B_\mathfrak{Q}/\mathfrak{m}B_\mathfrak{Q}\) is geometrically \(\mathbf{P}\). By [PGroLocalizes], \(A\to B_\mathfrak{Q}\) is a \(\mathbf{P}\)-map, hence \(B_\mathfrak{Q}\otimes_A K\) is geometrically \(\mathbf{P}\) over \(K\), thus so is the localization \(B_\mathfrak{q}\otimes_A K=B_\mathfrak{q}\) by [PisPointwise]. ◻

Proof. Note that \(\mathbf{P}\) satisfies [PisSing]–[PisOpen] in Discussion 66, see Discussion 96. Similar to the proof of Theorem 70 we may assume \(R\) is an integral domain and the theorem holds for all \(R\) of smaller dimension. If \(\dim R\leq 1\) then either \(I=0\) or \(R\) is complete, so we may assume \(\dim R\geq 2\). By [NsmrP0:P-ify-codim-1] we may assume \(U_{\mathbf{P}}(R)\neq\{0\}\). Let \(0\neq \mathfrak{P}\in U_{\mathbf{P}}(R)\) and let \(P\in\operatorname{Min}(\mathfrak{P}R^\wedge)\). By induction hypothesis \(R/\mathfrak{P}\) is a \(\mathbf{P}_0\)-ring, so \((R^\wedge)_P/\mathfrak{P}(R^\wedge)_P\) is \(\mathbf{P}\). Thus \((R^\wedge)_P\) is \(\mathbf{P}\) by [Pascends], so \(U_\mathbf{P}(R^\wedge)\neq\emptyset\).

Let \(C=\mathfrak{c}(R^\wedge)\) where \(\mathfrak{c}(-)\) is as in Discussion 100. Notice that \(C\neq 0\) as \(U_\mathbf{P}(R^\wedge)\neq\emptyset\). The rest of the proof is identical to that of Theorem 70, with [PGroLocalizes] replaced by Theorem 98. We only need to show that for every prime ideal \(M\) of \(T=((R^\wedge)_\mathfrak{p})^*\), \(T_M\) is catenary and equidimensional (cf. Remark 99), and we only need to check maximal \(M\).

As \(R^\wedge\) is a quotient of a regular ring, so are \(S:=(R^\wedge)_\mathfrak{p}\) and \(T\), thus we only need to check \(T_M\) is equidimensional. As \(M\) is maximal the map \(S_{M\cap S}\to T_M\) induces an isomorphism of completions. If \(S_{M\cap S}\) is equidimensional, so is its completion stacks? Tag0AW3, thus so is \(T_M\) stacks? Tag0AW4. As \(S\) is a localization of \(R^\wedge\) it now suffices to show \((R^\wedge)_Q\) is equidimensional for all \(Q\in\operatorname{Spec}(R^\wedge)\). Again, as \(R^\wedge\) is catenary it suffices to check maximal \(Q\), for which equidimensionality follows from stacks? Tag0AW3 as \(R\) is a universally catenary domain by assumption [NsmrP0:UC]. ◻

Proof. Let \(A\) be any Noetherian ring with \(U_{\mathbf{P}_1}(A)\) open. As an Artinian ring is Cohen–Macaulay it is clear that there exists finitely many primes \(\mathfrak{p}_1,\ldots,\mathfrak{p}_m\in\operatorname{Spec}_1(A)\cap \operatorname{Ass}(A)\) so that \(U_{\mathbf{P}_1}(A)=\operatorname{Spec}(A)\setminus V(\mathfrak{p}_1\cap\ldots\cap\mathfrak{p}_m)\). Therefore, we can similarly take \(\mathfrak{c}(A)=\operatorname{Ann}H^0_{\mathfrak{b}}(A)\) where \(\mathfrak{b}=\mathfrak{p}_1\cap\ldots\cap\mathfrak{p}_m\), so \(V(\mathfrak{c}(A))=\operatorname{Spec}(A)\setminus U_{\mathbf{P}_1}(A)\) and and \(\mathfrak{c}(A)B=\mathfrak{c}(B)\) for all \(\mathbf{P}_1\)-maps \(A\to B\). Note that we automatically have \(\mathfrak{c}(A)\neq 0\).

The rest of the proof is now verbatim to that of Theorem 70, with [PGroLocalizes] replaced by Theorem 98, and with equidimensionality guaranteed by [NsmrCM1:UC] as seen in the proof of Theorem 101. ◻