hcju/stacksqa huggingface dataset:
By More on Algebra, Lemma \ref{more-algebra-lemma-points-of-fibre-product} we have $T' = T \amalg_S S'$ as topological spaces, i.e., the diagram is a pushout in the category of topological spaces. Next, consider the map $$ ((i')^\sharp, (f')^\sharp) : \mathcal{O}_{T'} \longrightarrow i'_*\mathcal{O}_T \times_{g_*\mathcal{O}_S} f'_*\mathcal{O}_{S'} $$ where $g = i' \circ f = f' \circ i$. We claim this map is an isomorphism of sheaves of rings. Namely, we can view both sides as quasi-coherent $\mathcal{O}_{T'}$-modules (use Schemes, Lemmas \ref{schemes-lemma-push-forward-quasi-coherent} for the right hand side) and the map is $\mathcal{O}_{T'}$-linear. Thus it suffices to show the map is an isomorphism on the level of global sections (Schemes, Lemma \ref{schemes-lemma-equivalence-quasi-coherent}). On global sections we recover the identification $B' \to B \times_A A'$ from statement of the lemma (this is how we chose $B'$). \medskip\noindent Let $X$ be a scheme. Suppose we are given morphisms of schemes $m' : S' \to X$ and $n : T \to X$ such that $m' \circ i = n \circ f$ (call this $m$). We get a unique map of topological spaces $n' : T' \to X$ compatible with $m'$ and $n$ as $T' = T \amalg_S S'$ (see above). By the description of $\mathcal{O}_{T'}$ in the previous paragraph we obtain a unique homomorphism of sheaves of rings $$ (n')^\sharp : \mathcal{O}_X \longrightarrow (n')_*\mathcal{O}_{T'} = m'_*\mathcal{O}_T \times_{m_*\mathcal{O}_T} n_*\mathcal{O}_S $$ given by $(m')^\sharp$ and $n^\sharp$. Thus $(n', (n')^\sharp)$ is the unque morphism of ringed spaces $T' \to X$ compatible with $m'$ and $n$. To finish the proof it suffices to show that $n'$ is a morphism of schemes, i.e., a morphism of locally ringed spaces. \medskip\noindent Let $t' \in T'$ with image $x \in X$. We have to show that $\mathcal{O}_{X, x} \to \mathcal{O}_{T', t'}$ is local. If $t' \not \in T$, then $t'$ is the image of a unique point $s' \in S'$ and $\mathcal{O}_{T', t'} = \mathcal{O}_{S', s'}$. Namely, $S' \setminus S \to T' \setminus T$ is an isomorphism of schemes as $B' \to A'$ induces an isomorphism $\Ker(B' \to B) = \Ker(A' \to A)$. If $t'$ is the image of $t \in T$, then we know that the composition $\mathcal{O}_{X, x} \to \mathcal{O}_{T', t'} \to \mathcal{O}_{T, t}$ is local and we conclude also.
We first construct $Y'$ as a ringed space. Namely, as topological space we take $Y' = Y$. Denote $f' : X' \to Y'$ the map of topological spaces which equals $f$. As structure sheaf $\mathcal{O}_{Y'}$ we take the right hand side of the equation of the lemma. To see that $Y'$ is a scheme, we have to show that any point has an affine neighbourhood. Since the formation of the fibre product of sheaves commutes with restricting to opens, we may assume $Y$ is affine. Then $X$ is affine (as $f$ is affine) and $X'$ is affine as well (see Lemma \ref{lemma-thickening-affine-scheme}). Say $Y \leftarrow X \rightarrow X'$ corresponds to $B \rightarrow A \leftarrow A'$. Set $B' = B \times_A A'$; this is the global sections of $\mathcal{O}_{Y'}$. As $A' \to A$ is surjective with locally nilpotent kernel we see that $B' \to B$ is surjective with locally nilpotent kernel. Hence $\Spec(B') = \Spec(B)$ (as topological spaces). We claim that $Y' = \Spec(B')$. To see this we will show for $g' \in B'$ with image $g \in B$ that $\mathcal{O}_{Y'}(D(g)) = B'_{g'}$. Namely, by More on Algebra, Lemma \ref{more-algebra-lemma-diagram-localize} we see that $$ (B')_{g'} = B_g \times_{A_h} A'_{h'} $$ where $h \in A$, $h' \in A'$ are the images of $g'$. Since $B_g$, resp.\ $A_h$, resp.\ $A'_{h'}$ is equal to $\mathcal{O}_Y(D(g))$, resp.\ $f_*\mathcal{O}_X(D(g))$, resp.\ $f'_*\mathcal{O}_{X'}(D(g))$ the claim follows. \medskip\noindent It remains to show that $Y'$ is the pushout. The discussion above shows the scheme $Y'$ has an affine open covering $Y' = \bigcup W'_i$ such that the corresponding opens $U'_i \subset X'$, $W_i \subset Y$, and $U_i \subset X$ are affine open. Moreover, if $A'_i$, $B_i$, $A_i$ are the rings corresponding to $U'_i$, $W_i$, $U_i$, then $W'_i$ corresponds to $B_i \times_{A_i} A'_i$. Thus we can apply Lemmas \ref{lemma-basic-example-pushout} and \ref{lemma-pushout-fpqc-local} to conclude our construction is a pushout in the category of schemes.
Let $(V, U', \varphi)$ be an object of the fibre product category. Set $U = U' \times_{X'} X$. Note that $U \to U'$ is a thickening. Since $\varphi : V \times_Y X \to U' \times_{X'} X = U$ is an isomorphism we have a morphism $U \to V$ over $X \to Y$ which identifies $U$ with the fibre product $X \times_Y V$. In particular $U \to V$ is affine, see Morphisms, Lemma \ref{morphisms-lemma-base-change-affine}. Hence we can apply Lemma \ref{lemma-pushout-along-thickening} to get a pushout $V' = V \amalg_U U'$. Denote $V' \to Y'$ the morphism we obtain in virtue of the fact that $V'$ is a pushout and because we are given morphisms $V \to Y$ and $U' \to X'$ agreeing on $U$ as morphisms into $Y'$. Setting $G(V, U', \varphi) = V'$ gives the functor $G$. \medskip\noindent Let us prove that $G$ is a left adjoint to $F$. Let $Z$ be a scheme over $Y'$. We have to show that $$ \Mor(V', Z) = \Mor((V, U', \varphi), F(Z)) $$ where the morphism sets are taking in their respective categories. Let $g' : V' \to Z$ be a morphism. Denote $\tilde g$, resp.\ $\tilde f'$ the composition of $g'$ with the morphism $V \to V'$, resp.\ $U' \to V'$. Base change $\tilde g$, resp.\ $\tilde f'$ by $Y \to Y'$, resp.\ $X' \to Y'$ to get a morphism $g : V \to Z \times_{Y'} Y$, resp.\ $f' : U' \to Z \times_{Y'} X'$. Then $(g, f')$ is an element of the right hand side of the equation above (details omitted). Conversely, suppose that $(g, f') : (V, U', \varphi) \to F(Z)$ is an element of the right hand side. We may consider the composition $\tilde g : V \to Z$, resp.\ $\tilde f' : U' \to Z$ of $g$, resp.\ $f$ by $Z \times_{Y'} X' \to Z$, resp.\ $Z \times_{Y'} Y \to Z$. Then $\tilde g$ and $\tilde f'$ agree as morphism from $U$ to $Z$. By the universal property of pushout, we obtain a morphism $g' : V' \to Z$, i.e., an element of the left hand side. We omit the verification that these constructions are mutually inverse. \medskip\noindent To prove that $F \circ G$ is isomorphic to the identity we have to show that the adjunction mapping $(V, U', \varphi) \to F(G(V, U', \varphi))$ is an isomorphism. To do this we may work affine locally. Say $X = \Spec(A)$, $X' = \Spec(A')$, and $Y = \Spec(B)$. Then $A' \to A$ and $B \to A$ are ring maps as in More on Algebra, Lemma \ref{more-algebra-lemma-module-over-fibre-product} and $Y' = \Spec(B')$ with $B' = B \times_A A'$. Next, suppose that $V = \Spec(D)$, $U' = \Spec(C')$ and $\varphi$ is given by an $A$-algebra isomorphism $D \otimes_B A \to C' \otimes_{A'} A = C'/IC'$. Set $D' = D \times_{C'/IC'} C'$. In this case the statement we have to prove is that $D' \otimes_{B'} B \cong D$ and $D' \otimes_{B'} A' \cong C'$. This is a special case of More on Algebra, Lemma \ref{more-algebra-lemma-module-over-fibre-product}.
A quasi-inverse functor assigns to the triple $(\mathcal{G}, \mathcal{F}', \varphi)$ the fibre product $$ (V \to V')_*\mathcal{G} \times_{(U \to V')_*\mathcal{F}} (U' \to V')_*\mathcal{F}' $$ where $\mathcal{F} = (U \to U')^*\mathcal{F}'$. This works, because on affines we recover the equivalence of More on Algebra, Lemma \ref{more-algebra-lemma-relative-flat-module-over-fibre-product}. Some details omitted. \medskip\noindent Parts (a) and (b) follow from More on Algebra, Lemmas \ref{more-algebra-lemma-relative-finite-module-over-fibre-product} and \ref{more-algebra-lemma-relative-finitely-presented-module-over-fibre-product}.
Consider the commutative diagram of local rings $$ \xymatrix{ \mathcal{O}_{X', x'} & \mathcal{O}_{X, x} \ar[l] \\ \mathcal{O}_{S', s'} \ar[u] & \mathcal{O}_{S, s} \ar[l] \ar[u] } $$ Note that $\mathcal{O}_{X', x'}$ is a localization of $\mathcal{O}_{X, x} \otimes_{\mathcal{O}_{S, s}} \mathcal{O}_{S', s'}$, and that $((g')^*\mathcal{F})_{x'}$ is equal to $\mathcal{F}_x \otimes_{\mathcal{O}_{X, x}} \mathcal{O}_{X', x'}$. Hence the lemma follows from Algebra, Lemma \ref{algebra-lemma-base-change-flat-up-down}.
By Morphisms, Lemma \ref{morphisms-lemma-finite-presentation-permanence} the morphism $f$ is locally of finite presentation. Hence $U$ is open by Theorem \ref{theorem-openness-flatness}. Because we have assumed that $\mathcal{F}$ is flat over $S$ we see that Theorem \ref{theorem-criterion-flatness-fibre} implies $$ U = \{x \in X \mid \mathcal{F}_s \text{ flat at }x \text{ over }Y_s\}. $$ where $s$ always denotes the image of $x$ in $S$. (This description also works trivially when $\mathcal{F}_x = 0$.) Moreover, the assumptions of the lemma remain in force for the morphism $f' : X' \to Y'$ and the sheaf $\mathcal{F}'$. Hence $U'$ has a similar description. In other words, it suffices to prove that given $s' \in S'$ mapping to $s \in S$ we have $$ \{x' \in X'_{s'} \mid \mathcal{F}'_{s'} \text{ flat at }x' \text{ over }Y'_{s'}\} $$ is the inverse image of the corresponding locus in $X_s$. This is true by Lemma \ref{lemma-flat-locus-base-change} because in the cartesian diagram $$ \xymatrix{ X'_{s'} \ar[d] \ar[r] & X_s \ar[d] \\ Y'_{s'} \ar[r] & Y_s } $$ the horizontal morphisms are flat as they are base changes by the flat morphism $\Spec(\kappa(s')) \to \Spec(\kappa(s))$.
If $\mathcal{F}_x \otimes \kappa(x)$ is zero, then $\mathcal{F}_x = 0$ by Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) and hence $\mathcal{F}$ is zero in a neighbourhood of $x$ (Modules, Lemma \ref{modules-lemma-finite-type-stalk-zero}) and the lemma holds. Thus we may assume $\mathcal{F}_x \otimes \kappa(x)$ is not zero and we see that Theorem \ref{theorem-criterion-flatness-fibre} applies with $f = \text{id} : X \to X$. We conclude that $\mathcal{F}_x$ is flat over $\mathcal{O}_{X, x}$. Hence $\mathcal{F}_x$ is free, see Algebra, Lemma \ref{algebra-lemma-finite-flat-local} for example. Choose an open neighbourhood $x \in U \subset X$ and sections $s_1, \ldots, s_r \in \mathcal{F}(U)$ which map to a basis in $\mathcal{F}_x$. The corresponding map $\psi : \mathcal{O}_U^{\oplus r} \to \mathcal{F}|_U$ is surjective after shrinking $U$ (Modules, Lemma \ref{modules-lemma-finite-type-stalk-zero}). Then $\Ker(\psi)$ is of finite type (see Modules, Lemma \ref{modules-lemma-kernel-surjection-finite-free-onto-finite-presentation}) and $\Ker(\psi)_x = 0$. Whence after shrinking $U$ once more $\psi$ is an isomorphism.
Openness holds trivially. Let $x \in X$ mapping to $s \in S$. By Lemma \ref{lemma-flat-and-free-at-point-fibre} we see that $x$ is in our set if and only if $\mathcal{F}|_{X_s}$ is flat at $x$ over $X_s$. Clearly this is also equivalent to $\mathcal{F}$ being flat at $x$ over $X$ (because this statement is implied by freeness of $\mathcal{F}_x$ and implies flatness of $\mathcal{F}|_{X_s}$ at $x$ over $X_s$). Thus the base change statement follows from Lemma \ref{lemma-base-change-criterion-flatness-fibre} applied to $\text{id} : X \to X$ over $S$.
The base change of the factorization $Y_1 \to X_1' \to X_1$ to $X_2$ is a factorization $Y_2 \to X_2 \times_{X_1} X_1' \to X_2$ and $X_2 \times_{X_1} X_1' \to X_2$ is integral (Morphisms, Lemma \ref{morphisms-lemma-base-change-finite}). Hence we get a morphism $h : X_2' \to X_2 \times_{X_1} X_1'$ by the universal property of Morphisms, Lemma \ref{morphisms-lemma-characterize-normalization}. Observe that $X_2'$ is the relative spectrum of the integral closure of $\mathcal{O}_{X_2}$ in $f_{2, *}\mathcal{O}_{Y_2}$. If $\mathcal{A}' \subset f_{1, *}\mathcal{O}_{Y_1}$ denotes the integral closure of $\mathcal{O}_{X_2}$, then $X_2 \times_{X_1} X_1'$ is the relative spectrum of $\varphi^*\mathcal{A}'$, see Constructions, Lemma \ref{constructions-lemma-spec-properties}. By Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology} we know that $f_{2, *}\mathcal{O}_{Y_2} = \varphi^*f_{1, *}\mathcal{O}_{Y_1}$. Hence the result follows from Lemma \ref{lemma-integral-closure-smooth-pullback}.
By Descent, Lemma \ref{descent-lemma-locally-finite-nr-irred-local-fppf} every quasi-compact open of $X$ has finitely many irreducible components. Note that $X_{red} = X \times_Y Y_{red}$ as a scheme smooth over a reduced scheme is reduced, see Descent, Lemma \ref{descent-lemma-reduced-local-smooth}. Hence we may assume that $X$ and $Y$ are reduced (as the normalization of a scheme is equal to the normalization of its reduction by definition). Next, note that $X' = X \times_Y Y^\nu$ is a normal scheme by Descent, Lemma \ref{descent-lemma-normal-local-smooth}. The morphism $X' \to Y^\nu$ is smooth (hence flat) thus the generic points of irreducible components of $X'$ lie over generic points of irreducible components of $Y^\nu$. Since $Y^\nu \to Y$ is birational we conclude that $X' \to X$ is birational too (because $X' \to Y^\nu$ induces an isomorphism on fibres over generic points of $Y$). We conclude that there exists a factorization $X^\nu \to X' \to X$, see Morphisms, Lemma \ref{morphisms-lemma-normalization-normal} which is an isomorphism as $X'$ is normal and integral over $X$.
Let $\eta_1, \ldots, \eta_r$ be the generic points of the irreducible components of $X$ passing through $x$. The base change of the normalization to $\Spec(\mathcal{O}_{X, x})$ is the spectrum of the integral closure of $\mathcal{O}_{X, x}$ in $\prod \kappa(\eta_i)$. This follows from our construction of the normalization of $X$ in Morphisms, Definition \ref{morphisms-definition-normalization} and Morphisms, Lemma \ref{morphisms-lemma-integral-closure}; you can also use the description of the normalization in Morphisms, Lemma \ref{morphisms-lemma-description-normalization}. Thus we reduce to the following algebra problem. Let $A$ be a Noetherian local ring; recall that this implies the henselization $A^h$ and strict henselization $A^{sh}$ are Noetherian too (More on Algebra, Lemma \ref{more-algebra-lemma-henselization-noetherian}). Let $\mathfrak p_1, \ldots, \mathfrak p_r$ be its minimal primes. Let $A'$ be the integral closure of $A$ in $\prod \kappa(\mathfrak p_i)$. Problem: show that $A' \otimes_A A^h$, resp.\ $A' \otimes_A A^{sh}$ is constructed from the Noetherian local ring $A^h$, resp.\ $A^{sh}$ in the same manner. \medskip\noindent Since $A^h$, resp.\ $A^{sh}$ are colimits of \'etale $A$-algebras, we see that the minimal primes of $A$ and $A^{sh}$ are exactly the primes of $A^h$, resp.\ $A^{sh}$ lying over the minimal primes of $A$ (by going down, see Algebra, Lemmas \ref{algebra-lemma-flat-going-down} and \ref{algebra-lemma-minimal-prime-image-minimal-prime}). Thus More on Algebra, Lemma \ref{more-algebra-lemma-fibres-henselization} tells us that $A^h \otimes_A \prod \kappa(\mathfrak p_i)$, resp. $A^{sh} \otimes_A \prod \kappa(\mathfrak p_i)$ is the product of the residue fields at the minimal primes of $A^h$, resp.\ $A^{sh}$. We know that taking the integral closure in an overring commutes with \'etale base change, see Algebra, Lemma \ref{algebra-lemma-integral-closure-commutes-etale}. Writing $A^h$ and $A^{sh}$ as a limit of \'etale $A$-algebras we see that the same thing is true for the base change to $A^h$ and $A^{sh}$ (you can also use the more general Algebra, Lemma \ref{algebra-lemma-integral-closure-commutes-colim-smooth}).
Let $f : X \to Y$ be a smooth morphism. As $f$ is locally of finite presentation, see Morphisms, Lemma \ref{morphisms-lemma-smooth-locally-finite-presentation} the fibres $X_y$ are locally of finite type over a field, hence locally Noetherian. Moreover, $f$ is flat, see Morphisms, Lemma \ref{morphisms-lemma-smooth-flat}. Finally, the fibres $X_y$ are smooth over a field (by Morphisms, Lemma \ref{morphisms-lemma-base-change-smooth}) and hence geometrically normal by Varieties, Lemma \ref{varieties-lemma-smooth-geometrically-normal}. Thus $f$ is normal by Lemma \ref{lemma-normal}.
Let $f : X \to Y$ be a morphism of schemes. Let $\{\varphi_i : Y_i \to Y\}_{i \in I}$ be an fppf covering of $Y$. Denote $f_i : X_i \to Y_i$ the base change of $f$ by $\varphi_i$. Let $i \in I$ and let $y_i \in Y_i$ be a point. Set $y = \varphi_i(y_i)$. Note that $$ X_{i, y_i} = \Spec(\kappa(y_i)) \times_{\Spec(\kappa(y))} X_y. $$ Moreover, as $\varphi_i$ is of finite presentation the field extension $\kappa(y) \subset \kappa(y_i)$ is finitely generated. Hence in this situation we have that $X_y$ is locally Noetherian if and only if $X_{i, y_i}$ is locally Noetherian, see Varieties, Lemma \ref{varieties-lemma-locally-Noetherian-base-change}. This fact implies locality on the target. \medskip\noindent Let $\{X_i \to X\}$ be an fppf covering of $X$. Let $y \in Y$. In this case $\{X_{i, y} \to X_y\}$ is an fppf covering of the fibre. Hence the locality on the source follows from Descent, Lemma \ref{descent-lemma-Noetherian-local-fppf}.
We have $\mathcal{P}(f) = \mathcal{P}_1(f) \wedge \mathcal{P}_2(f) \wedge \mathcal{P}_3(f)$ where $\mathcal{P}_1(f)=$``the fibres of $f$ are locally Noetherian'', $\mathcal{P}_2(f)=$``$f$ is flat'', and $\mathcal{P}_3(f)=$``the fibres of $f$ are geometrically normal''. We have already seen that $\mathcal{P}_1$ and $\mathcal{P}_2$ are local in the fppf topology on the source and the target, see Lemma \ref{lemma-locally-Noetherian-fibres-fppf-local-source-and-target}, and Descent, Lemmas \ref{descent-lemma-descending-property-flat} and \ref{descent-lemma-flat-fpqc-local-source}. Thus we have to deal with $\mathcal{P}_3$. \medskip\noindent Let $f : X \to Y$ be a morphism of schemes. Let $\{\varphi_i : Y_i \to Y\}_{i \in I}$ be an fpqc covering of $Y$. Denote $f_i : X_i \to Y_i$ the base change of $f$ by $\varphi_i$. Let $i \in I$ and let $y_i \in Y_i$ be a point. Set $y = \varphi_i(y_i)$. Note that $$ X_{i, y_i} = \Spec(\kappa(y_i)) \times_{\Spec(\kappa(y))} X_y. $$ Hence in this situation we have that $X_y$ is geometrically normal if and only if $X_{i, y_i}$ is geometrically normal, see Varieties, Lemma \ref{varieties-lemma-geometrically-normal-upstairs}. This fact implies $\mathcal{P}_3$ is fpqc local on the target. \medskip\noindent Let $\{X_i \to X\}$ be a smooth covering of $X$. Let $y \in Y$. In this case $\{X_{i, y} \to X_y\}$ is a smooth covering of the fibre. Hence the locality of $\mathcal{P}_3$ for the smooth topology on the source follows from Descent, Lemma \ref{descent-lemma-normal-local-smooth}. Combining the above the lemma follows.
The equivalence of (1) and (2) is immediate from the definitions. Let $x \in X$ with $y = f(x)$. By definition $f$ is flat at $x$ if and only if $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is a flat ring map, and $X_y$ is geometrically regular at $x$ over $\kappa(y)$ if and only if $\mathcal{O}_{X_y, x} = \mathcal{O}_{X, x}/\mathfrak m_y\mathcal{O}_{X, x}$ is a geometrically regular algebra over $\kappa(y)$. Hence Whether or not $f$ is regular at $x$ depends only on the local homomorphism of local rings $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$. Thus the equivalence of (1) and (4) is clear. \medskip\noindent Recall (More on Algebra, Definition \ref{more-algebra-definition-regular}) that a ring map $A \to B$ is regular if and only if it is flat and the fibre rings $B \otimes_A \kappa(\mathfrak p)$ are Noetherian and geometrically regular for all primes $\mathfrak p \subset A$. By Varieties, Lemma \ref{varieties-lemma-geometrically-regular} this is equivalent to $\Spec(B \otimes_A \kappa(\mathfrak p))$ being a geometrically regular scheme over $\kappa(\mathfrak p)$. Thus we see that (2) implies (3). It is clear that (3) implies (5). Finally, assume (5). This implies that $f$ is flat (see Morphisms, Lemma \ref{morphisms-lemma-flat-characterize}). Moreover, if $y \in Y$, then $y \in V_j$ for some $j$ and we see that $X_y = \bigcup_{i \in I_j} U_{i, y}$ with each $U_{i, y}$ geometrically regular over $\kappa(y)$ by Varieties, Lemma \ref{varieties-lemma-geometrically-regular}. Another application of Varieties, Lemma \ref{varieties-lemma-geometrically-regular} shows that $X_y$ is geometrically regular. Hence (2) holds and the proof of the lemma is finished.
Let $f : X \to Y$ be a smooth morphism. As $f$ is locally of finite presentation, see Morphisms, Lemma \ref{morphisms-lemma-smooth-locally-finite-presentation} the fibres $X_y$ are locally of finite type over a field, hence locally Noetherian. Moreover, $f$ is flat, see Morphisms, Lemma \ref{morphisms-lemma-smooth-flat}. Finally, the fibres $X_y$ are smooth over a field (by Morphisms, Lemma \ref{morphisms-lemma-base-change-smooth}) and hence geometrically regular by Varieties, Lemma \ref{varieties-lemma-smooth-geometrically-normal}. Thus $f$ is regular by Lemma \ref{lemma-regular}.
We have $\mathcal{P}(f) = \mathcal{P}_1(f) \wedge \mathcal{P}_2(f) \wedge \mathcal{P}_3(f)$ where $\mathcal{P}_1(f)=$``the fibres of $f$ are locally Noetherian'', $\mathcal{P}_2(f)=$``$f$ is flat'', and $\mathcal{P}_3(f)=$``the fibres of $f$ are geometrically regular''. We have already seen that $\mathcal{P}_1$ and $\mathcal{P}_2$ are local in the fppf topology on the source and the target, see Lemma \ref{lemma-locally-Noetherian-fibres-fppf-local-source-and-target}, and Descent, Lemmas \ref{descent-lemma-descending-property-flat} and \ref{descent-lemma-flat-fpqc-local-source}. Thus we have to deal with $\mathcal{P}_3$. \medskip\noindent Let $f : X \to Y$ be a morphism of schemes. Let $\{\varphi_i : Y_i \to Y\}_{i \in I}$ be an fpqc covering of $Y$. Denote $f_i : X_i \to Y_i$ the base change of $f$ by $\varphi_i$. Let $i \in I$ and let $y_i \in Y_i$ be a point. Set $y = \varphi_i(y_i)$. Note that $$ X_{i, y_i} = \Spec(\kappa(y_i)) \times_{\Spec(\kappa(y))} X_y. $$ Hence in this situation we have that $X_y$ is geometrically regular if and only if $X_{i, y_i}$ is geometrically regular, see Varieties, Lemma \ref{varieties-lemma-geometrically-regular-upstairs}. This fact implies $\mathcal{P}_3$ is fpqc local on the target. \medskip\noindent Let $\{X_i \to X\}$ be a smooth covering of $X$. Let $y \in Y$. In this case $\{X_{i, y} \to X_y\}$ is a smooth covering of the fibre. Hence the locality of $\mathcal{P}_3$ for the smooth topology on the source follows from Descent, Lemma \ref{descent-lemma-regular-local-smooth}. Combining the above the lemma follows.
After replacing $X$ by an open neighborhood of $x$, there is a natural number $d$ such that all fibers of $X \to Y$ have dimension $d$ at every point, see Morphisms, Lemma \ref{morphisms-lemma-flat-finite-presentation-CM-fibres-relative-dimension}. Then $f$ is flat, locally of finite type and of relative dimension $d$. Hence the result follows from Morphisms, Lemma \ref{morphisms-lemma-rel-dimension-dimension}.
Consider the map of Noetherian local rings $$ \mathcal{O}_{Y_z, y} \to \mathcal{O}_{X_z, x} $$ and observe that its fibre is $$ \mathcal{O}_{X_z, x}/\mathfrak m_{Y_z, y}\mathcal{O}_{X_z, x} = \mathcal{O}_{X_y, x} $$ Thus the lemma this follows from Algebra, Lemma \ref{algebra-lemma-CM-goes-up}.