hcju/stacksqa huggingface dataset:
The map $\Pic(X) \to \Pic(X_{\overline{k}})$ is injective by Varieties, Lemma \ref{varieties-lemma-change-fields-pic}. By Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology} $\dim_k H^1(X, \mathcal{O}_X)$ equals $\dim_{\overline{k}} H^1(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})$. Hence we may assume $k$ is algebraically closed. \medskip\noindent Let $X_{red}$ be the reduction of $X$. Then the surjection $\mathcal{O}_X \to \mathcal{O}_{X_{red}}$ induces a surjection $H^1(X, \mathcal{O}_X) \to H^1(X, \mathcal{O}_{X_{red}})$ because cohomology of quasi-coherent sheaves vanishes in degrees $\geq 2$ by Cohomology, Proposition \ref{cohomology-proposition-vanishing-Noetherian}. Since $X_{red} \to X$ induces an isomorphism on irreducible components over $\overline{k}$ and an isomorphism on $\ell$-torsion in Picard groups (Picard Schemes of Curves, Lemma \ref{pic-lemma-torsion-descends}) we may replace $X$ by $X_{red}$. In this way we reduce to Proposition \ref{proposition-torsion-picard-reduced-proper}.
Assume $I \not = 0$. Write $I = (f_1, \ldots, f_r)$. As $A$ is a UFD (More on Algebra, Lemma \ref{more-algebra-lemma-regular-local-UFD}) we can write $f_i = fg_i$ where $f$ is the gcd of $f_1, \ldots, f_r$. Thus the gcd of $g_1, \ldots, g_r$ is $1$ which means that there is no height $1$ prime ideal over $g_1, \ldots, g_r$. Since $\dim(A) = 2$ this implies that $V(g_1, \ldots, g_r) = \{\mathfrak m\}$, i.e., $\mathfrak m = \sqrt{(g_1, \ldots, g_r)}$. \medskip\noindent Assume $A/I$ reduced, i.e., $I$ radical. If $f$ is a unit, then since $I$ is radical we see that $I = \mathfrak m$. If $f \in \mathfrak m$, then we see that $f^n$ maps to zero in $A/I$. Hence $f \in I$ by reducedness and we conclude $I = (f)$. \medskip\noindent Assume $A/I$ has depth $1$. Then $\mathfrak m$ is not an associated prime of $A/I$. Since the class of $f$ modulo $I$ is annihilated by $g_1, \ldots, g_r$, this implies that the class of $f$ is zero in $A/I$. Thus $I = (f)$ as desired.
Assume (3). Then $A^\wedge$ is reduced because $ax^2 + bxy + cy^2$ is either irreducible or a product of two nonassociated prime elements. Hence $A \subset A^\wedge$ is reduced. It follows that (1) is true. \medskip\noindent Assume (1). Then $A$ cannot be Artinian, since it would not be reduced because $\mathfrak m \not = (0)$. Hence $\dim(A) \geq 1$, hence $\text{depth}(A) \geq 1$ by Algebra, Lemma \ref{algebra-lemma-criterion-reduced}. On the other hand $\dim(A) = 2$ implies $A$ is regular which contradicts the existence of $q$ by Algebra, Lemma \ref{algebra-lemma-regular-graded}. Thus $\dim(A) \leq 1$ and we conclude $\text{depth}(A) = 1$ by Algebra, Lemma \ref{algebra-lemma-bound-depth}. It follows that (2) is true. \medskip\noindent Assume (2). Since the depth of $A$ is the same as the depth of $A^\wedge$ (More on Algebra, Lemma \ref{more-algebra-lemma-completion-depth}) and since the other conditions are insensitive to completion, we may assume that $A$ is complete. Choose $\kappa \to A$ as in More on Algebra, Lemma \ref{more-algebra-lemma-lift-residue-field}. Since $\dim_\kappa(\mathfrak m/\mathfrak m^2) = 2$ we can choose $x_0, y_0 \in \mathfrak m$ which map to a basis. We obtain a continuous $\kappa$-algebra map $$ \kappa[[x, y]] \longrightarrow A $$ by the rules $x \mapsto x_0$ and $y \mapsto y_0$. Let $q$ be the class of $ax_0^2 + bx_0y_0 + cy_0^2$ in $\text{Sym}^2_\kappa(\mathfrak m/\mathfrak m^2)$. Write $Q(x, y) = ax^2 + bxy + cy^2$ viewed as a polynomial in two variables. Then we see that $$ Q(x_0, y_0) = ax_0^2 + bx_0y_0 + cy_0^2 = \sum\nolimits_{i + j = 3} a_{ij} x_0^iy_0^j $$ for some $a_{ij}$ in $A$. We want to prove that we can increase the order of vanishing by changing our choice of $x_0$, $y_0$. Suppose that $x_1, y_1 \in \mathfrak m^2$. Then $$ Q(x_0 + x_1, y_0 + y_1) = Q(x_0, y_0) + (2ax_0 + by_0)x_1 + (bx_0 + 2cy_0)y_1 \bmod \mathfrak m^4 $$ Nondegeneracy of $Q$ means exactly that $2ax_0 + by_0$ and $bx_0 + 2cy_0$ are a $\kappa$-basis for $\mathfrak m/\mathfrak m^2$, see discussion preceding the lemma. Hence we can certainly choose $x_1, y_1 \in \mathfrak m^2$ such that $Q(x_0 + x_1, y_0 + y_1) \in \mathfrak m^4$. Continuing in this fashion by induction we can find $x_i, y_i \in \mathfrak m^{i + 1}$ such that $$ Q(x_0 + x_1 + \ldots + x_n, y_0 + y_1 + \ldots + y_n) \in \mathfrak m^{n + 3} $$ Since $A$ is complete we can set $x_\infty = \sum x_i$ and $y_\infty = \sum y_i$ and we can consider the map $\kappa[[x, y]] \longrightarrow A$ sending $x$ to $x_\infty$ and $y$ to $y_\infty$. This map induces a surjection $\kappa[[x, y]]/(Q) \longrightarrow A$ by Algebra, Lemma \ref{algebra-lemma-completion-generalities}. By Lemma \ref{lemma-reduced-quotient-regular-ring-dim-2} the kernel of $k[[x, y]] \to A$ is principal. But the kernel cannot contain a proper divisor of $Q$ as such a divisor would have degree $1$ in $x, y$ and this would contradict $\dim(\mathfrak m/\mathfrak m^2) = 2$. Hence $Q$ generates the kernel as desired.
By Lemma \ref{lemma-fitting-ideal-well-defined} we may change our system of coordinates and the choice of generators during the proof. \medskip\noindent If (1) holds, then we may change coordinates such that $x_1, \ldots, x_{n - 2}$ map to zero in $A/I$ and $A/I = k[[x_{n - 1}, x_n]]/(a x_{n - 1}^2 + b x_{n - 1}x_n + c x_n^2)$ for some nondegenerate quadric $a x_{n - 1}^2 + b x_{n - 1}x_n + c x_n^2$. Then we can explicitly compute to show that both (2) and (3) are true. \medskip\noindent Assume the $(n - 1) \times (n - 1)$ minors of the matrix $(\partial f_j/\partial x_i)$ generate $I + \mathfrak m_A$. Suppose that for some $i$ and $j$ the partial derivative $\partial f_j/\partial x_i$ is a unit in $A$. Then we may use the system of parameters $f_j, x_1, \ldots, x_{i - 1}, \hat x_i, x_{i + 1}, \ldots, x_n$ and the generators $f_j, f_1, \ldots, f_{j - 1}, \hat f_j, f_{j + 1}, \ldots, f_m$ of $I$. Then we get a regular system of parameters $x_1, \ldots, x_n$ and generators $x_1, f_2, \ldots, f_m$ of $I$. Next, we look for an $i \geq 2$ and $j \geq 2$ such that $\partial f_j/\partial x_i$ is a unit in $A$. If such a pair exists, then we can make a replacement as above and assume that we have a regular system of parameters $x_1, \ldots, x_n$ and generators $x_1, x_2, f_3, \ldots, f_m$ of $I$. Continuing, in finitely many steps we reach the situation where we have a regular system of parameters $x_1, \ldots, x_n$ and generators $x_1, \ldots, x_t, f_{t + 1}, \ldots, f_m$ of $I$ such that $\partial f_j/\partial x_i \in \mathfrak m_A$ for all $i, j \geq t + 1$. \medskip\noindent In this case the matrix of partial derivatives has the following block shape $$ \left( \begin{matrix} I_{t \times t} & * \\ 0 & \mathfrak m_A \end{matrix} \right) $$ Hence every $(n - 1) \times (n - 1)$-minor is in $\mathfrak m_A^{n - 1 - t}$. Note that $I \not = \mathfrak m_A$ otherwise the ideal of minors would contain $1$. It follows that $n - 1 - t \leq 1$ because there is an element of $\mathfrak m_A \setminus \mathfrak m_A^2 + I$ (otherwise $I = \mathfrak m_A$ by Nakayama). Thus $t \geq n - 2$. We have seen that $t \not = n$ above and similarly if $t = n - 1$, then there is an invertible $(n - 1) \times (n - 1)$-minor which is disallowed as well. Hence $t = n - 2$. Then $A/I$ is a quotient of $k[[x_{n - 1}, x_n]]$ and Lemma \ref{lemma-reduced-quotient-regular-ring-dim-2} implies in both cases (2) and (3) that $I$ is generated by $x_1, \ldots, x_{n - 2}, f$ for some $f = f(x_{n - 1}, x_n)$. In this case the condition on the minors exactly says that the quadratic term in $f$ is nondegenerate, i.e., $A/I$ is as in Lemma \ref{lemma-nodal-algebraic}.
If $x$ is a closed point of $X$, then $y$ is too (look at residue fields). But conversely, this need not be the case, i.e., it can happen that a closed point of $Y$ maps to a nonclosed point of $X$. However, in this case $y$ cannot be a node. Namely, then $X$ would be geometrically unibranch at $x$ (because $x$ would be a generic point of $X$ and $\mathcal{O}_{X, x}$ would be Artinian and any Artinian local ring is geometrically unibranch), hence $Y$ is geometrically unibranch at $y$ (Varieties, Lemma \ref{varieties-lemma-geometrically-unibranch-and-change-of-fields}), which means that $y$ cannot be a node by Lemma \ref{lemma-nodal}. Thus we may and do assume that both $x$ and $y$ are closed points. \medskip\noindent Choose algebraic closures $\overline{k}$, $\overline{K}$ and a map $\overline{k} \to \overline{K}$ extending the given map $k \to K$. Using the equivalence of (1) and (3) in Lemma \ref{lemma-nodal} we reduce to the case where $k$ and $K$ are algebraically closed. In this case we can argue as in the proof of Lemma \ref{lemma-nodal} that the geometric number of branches and $\delta$-invariants of $X$ at $x$ and $Y$ at $y$ are the same. Another argument can be given by choosing an isomorphism $k[[x_1, \ldots, x_n]]/(g_1, \ldots, g_m) \to \mathcal{O}_{X, x}^\wedge$ of $k$-algebras as in Varieties, Lemma \ref{varieties-lemma-complete-local-ring-structure-as-algebra}. By Varieties, Lemma \ref{varieties-lemma-base-change-complete-local-ring} this gives an isomorphism $K[[x_1, \ldots, x_n]]/(g_1, \ldots, g_m) \to \mathcal{O}_{Y, y}^\wedge$ of $K$-algebras. By definition we have to show that $$ k[[x_1, \ldots, x_n]]/(g_1, \ldots, g_m) \cong k[[s, t]]/(st) $$ if and only if $$ K[[x_1, \ldots, x_n]]/(g_1, \ldots, g_m) \cong K[[s, t]]/(st) $$ We encourage the reader to prove this for themselves. Since $k$ and $K$ are algebraically closed fields, this is the same as asking these rings to be as in Lemma \ref{lemma-nodal-algebraic}. Via Lemma \ref{lemma-fitting-ideal} this translates into a statement about the $(n - 1) \times (n - 1)$-minors of the matrix $(\partial g_j/\partial x_i)$ which is clearly independent of the field used. We omit the details.
By Lemma \ref{lemma-node-field-extension} we may base change to the algebraic closure of $k$. Then the residue fields of $x$ and $y$ are $k$. Hence the map $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{Y, y}^\wedge$ is an isomorphism (for example by \'Etale Morphisms, Lemma \ref{etale-lemma-characterize-etale-completions} or More on Algebra, Lemma \ref{more-algebra-lemma-flat-unramified}). Thus the lemma is clear.
We urge the reader to find their own proof of this lemma; what follows is just putting together earlier results and may hide what is really going on. \medskip\noindent Assume (2). Since $Z \to \Spec(k)$ is quasi-finite (Morphisms, Lemma \ref{morphisms-lemma-unramified-quasi-finite}) we see that the residue fields of points $x \in Z$ are finite over $k$ (as well as separable) by Morphisms, Lemma \ref{morphisms-lemma-residue-field-quasi-finite}. Hence each $x \in Z$ is a closed point of $X$ by Morphisms, Lemma \ref{morphisms-lemma-algebraic-residue-field-extension-closed-point-fibre}. The local ring $\mathcal{O}_{X, x}$ is Cohen-Macaulay by Algebra, Lemma \ref{algebra-lemma-lci-CM}. Since $\dim(\mathcal{O}_{X, x}) = 1$ by dimension theory (Varieties, Section \ref{varieties-section-algebraic-schemes}), we conclude that $\text{depth}(\mathcal{O}_{X, x})) = 1$. Thus $x$ is a node by Lemma \ref{lemma-nodal}. If $x \in X$, $x \not \in Z$, then $X \to \Spec(k)$ is smooth at $x$ by Divisors, Lemma \ref{divisors-lemma-d-fitting-ideal-omega-smooth}. \medskip\noindent Assume (1). Under this assumption $X$ is geometrically reduced at every closed point (see Varieties, Lemma \ref{varieties-lemma-geometrically-reduced-upstairs}). Hence $X \to \Spec(k)$ is smooth on a dense open by Varieties, Lemma \ref{varieties-lemma-geometrically-reduced-dense-smooth-open}. Thus $Z$ is closed and consists of closed points. By Divisors, Lemma \ref{divisors-lemma-d-fitting-ideal-omega-smooth} the morphism $X \setminus Z \to \Spec(k)$ is smooth. Hence $X \setminus Z$ is a local complete intersection by Morphisms, Lemma \ref{morphisms-lemma-smooth-syntomic} and the definition of a local complete intersection in Morphisms, Definition \ref{morphisms-definition-syntomic}. By Lemma \ref{lemma-nodal} for every point $x \in Z$ the local ring $\mathcal{O}_{Z, x}$ is equal to $\kappa(x)$ and $\kappa(x)$ is separable over $k$. Thus $Z \to \Spec(k)$ is unramified (Morphisms, Lemma \ref{morphisms-lemma-unramified-over-field}). Finally, Lemma \ref{lemma-nodal} via part (3) of Lemma \ref{lemma-nodal-algebraic}, shows that $\mathcal{O}_{X, x}$ is a complete intersection in the sense of Divided Power Algebra, Definition \ref{dpa-definition-lci}. However, Divided Power Algebra, Lemma \ref{dpa-lemma-check-lci-agrees} and Morphisms, Lemma \ref{morphisms-lemma-local-complete-intersection} show that this agrees with the notion used to define a local complete intersection scheme over a field and the proof is complete.
The Gorenstein assertion follows from Lemma \ref{lemma-nodal-lci} and Duality for Schemes, Lemma \ref{duality-lemma-gorenstein-lci}. Or you can use that it suffices to check after passing to the algebraic closure (Duality for Schemes, Lemma \ref{duality-lemma-gorenstein-base-change}), then use that a Noetherian local ring is Gorenstein if and only if its completion is so (by Dualizing Complexes, Lemma \ref{dualizing-lemma-flat-under-gorenstein}), and then prove that the local rings $k[[t]]$ and $k[[x, y]]/(xy)$ are Gorenstein by hand. \medskip\noindent To see that $X$ is geometrically reduced, it suffices to show that $X_{\overline{k}}$ is reduced (Varieties, Lemmas \ref{varieties-lemma-perfect-reduced} and \ref{varieties-lemma-geometrically-reduced}). But $X_{\overline{k}}$ is a nodal curve over an algebraically closed field. Thus the complete local rings of $X_{\overline{k}}$ are isomorphic to either $\overline{k}[[t]]$ or $\overline{k}[[x, y]]/(xy)$ which are reduced as desired.
Since $X$ and $Y$ are reduced and equidimensional of dimension $1$, we see that $Y$ is the scheme theoretic union of a subset of the irreducible components of $X$ (in a reduced ring $(0)$ is the intersection of the minimal primes). Let $y \in Y$ be a closed point. If $y$ is in the smooth locus of $X \to \Spec(k)$, then $y$ is on a unique irreducible component of $X$ and we see that $Y$ and $X$ agree in an open neighbourhood of $y$. Hence $Y \to \Spec(k)$ is smooth at $y$. If $y$ is a node of $X$ but still lies on a unique irreducible component of $X$, then $y$ is a node on $Y$ by the same argument. Suppose that $y$ lies on more than $1$ irreducible component of $X$. Since the number of geometric branches of $X$ at $y$ is $2$ by Lemma \ref{lemma-nodal}, there can be at most $2$ irreducible components passing through $y$ by Properties, Lemma \ref{properties-lemma-number-of-branches-irreducible-components}. If $Y$ contains both of these, then again $Y = X$ in an open neighbourhood of $y$ and $y$ is a node of $Y$. Finally, assume $Y$ contains only one of the irreducible components. After replacing $X$ by an open neighbourhood of $x$ we may assume $Y$ is one of the two irreducble components and $Z$ is the other. By Properties, Lemma \ref{properties-lemma-number-of-branches-irreducible-components} again we see that $X$ has two branches at $y$, i.e., the local ring $\mathcal{O}_{X, y}$ has two branches and that these branches come from $\mathcal{O}_{Y, y}$ and $\mathcal{O}_{Z, y}$. Write $\mathcal{O}_{X, y}^\wedge \cong \kappa(y)[[u, v]]/(uv)$ as in Remark \ref{remark-trivial-quadratic-extension}. The field $\kappa(y)$ is finite separable over $k$ by Lemma \ref{lemma-nodal} for example. Thus, after possibly switching the roles of $u$ and $v$, the completion of the map $\mathcal{O}_{X, y} \to \mathcal{O}_{Y, Y}$ corresponds to $\kappa(y)[[u, v]]/(uv) \to \kappa(y)[[u]]$ and the completion of the map $\mathcal{O}_{X, y} \to \mathcal{O}_{Y, Y}$ corresponds to $\kappa(y)[[u, v]]/(uv) \to \kappa(y)[[v]]$. The scheme theoretic intersection of $Y \cap Z$ is cut out by the sum of their ideas which in the completion is $(u, v)$, i.e., the maximal ideal. Thus (2)(a) and (2)(b) are clear. Finally, (2)(c) holds: the completion of $\mathcal{O}_{Y, y}$ is regular, hence $\mathcal{O}_{Y, y}$ is regular (More on Algebra, Lemma \ref{more-algebra-lemma-completion-regular}) and $\kappa(y)/k$ is separable, hence smoothness in an open neighbourhood by Algebra, Lemma \ref{algebra-lemma-separable-smooth}.
Recall that the formation of $\text{Sing}(f)$ commutes with base change, see Divisors, Lemma \ref{divisors-lemma-base-change-and-fitting-ideal-omega}. Thus the lemma follows from Lemma \ref{lemma-nodal-lci}, Morphisms, Lemma \ref{morphisms-lemma-syntomic-flat-fibres}, and Morphisms, Lemma \ref{morphisms-lemma-unramified-etale-fibres}. (We also use the trivial Morphisms, Lemmas \ref{morphisms-lemma-syntomic-locally-finite-presentation} and \ref{morphisms-lemma-syntomic-flat}.)
This is true because the base change of a syntomic morphism is syntomic (Morphisms, Lemma \ref{morphisms-lemma-base-change-syntomic}), the base change of a morphism of relative dimension $1$ has relative dimension $1$ (Morphisms, Lemma \ref{morphisms-lemma-base-change-relative-dimension-d}), the formation of $\text{Sing}(f)$ commutes with base change (Divisors, Lemma \ref{divisors-lemma-base-change-and-fitting-ideal-omega}), and the base change of an unramified morphism is unramified (Morphisms, Lemma \ref{morphisms-lemma-base-change-unramified}).
By Morphisms, Lemma \ref{morphisms-lemma-set-points-where-fibres-lci} we find that there is such an open where $f$ is syntomic. Hence we may assume that $f$ is a syntomic morphism. In particular $f$ is a Cohen-Macaulay morphism (Duality for Schemes, Lemmas \ref{duality-lemma-lci-gorenstein} and \ref{duality-lemma-gorenstein-CM-morphism}). Thus $X$ is a disjoint union of open and closed subschemes on which $f$ has given relative dimension, see Morphisms, Lemma \ref{morphisms-lemma-flat-finite-presentation-CM-fibres-relative-dimension}. This decomposition is preserved by arbitrary base change, see Morphisms, Lemma \ref{morphisms-lemma-base-change-relative-dimension-d}. Discarding all but one piece we may assume $f$ is syntomic of relative dimension $1$. Let $\text{Sing}(f) \subset X$ be the closed subscheem defined by the first fitting ideal of $\Omega_{X/S}$. There is a maximal open subscheme $W \subset \text{Sing}(f)$ such that $W \to S$ is unramified and its formation commutes with base change (Morphisms, Lemma \ref{morphisms-lemma-set-points-where-fibres-unramified}). Since also formation of $\text{Sing}(f)$ commutes with base change (Divisors, Lemma \ref{divisors-lemma-base-change-and-fitting-ideal-omega}), we see that $$ U = (X \setminus \text{Sing}(f)) \cup W $$ is the maximal open subscheme of $X$ such that $f|_U : U \to S$ is at-worst-nodal of relative dimension $1$ and that formation of $U$ commutes with base change.
Observe that $g$ is flat and locally of finite presentation as a composition of morphisms which are flat and locally of finite presentation (use Morphisms, Lemmas \ref{morphisms-lemma-etale-locally-finite-presentation}, \ref{morphisms-lemma-etale-flat}, \ref{morphisms-lemma-composition-finite-presentation}, and \ref{morphisms-lemma-composition-flat}). Thus it suffices to prove the fibres have at-worst-nodal singularities. This follows from Lemma \ref{lemma-node-etale-local} (and the fact that the composition of an \'etale morphism and a smooth morphism is smooth by Morphisms, Lemmas \ref{morphisms-lemma-etale-smooth-unramified} and \ref{morphisms-lemma-composition-smooth}).
Observe that $g$ is flat and locally of finite presentation as a composition of morphisms which are flat and locally of finite presentation (use Morphisms, Lemmas \ref{morphisms-lemma-etale-locally-finite-presentation}, \ref{morphisms-lemma-etale-flat}, \ref{morphisms-lemma-composition-finite-presentation}, and \ref{morphisms-lemma-composition-flat}). Thus it suffices to prove the fibres of $g$ have at-worst-nodal singularities. This follows from Lemma \ref{lemma-node-over-separable-extension} and the analogous result for smooth points.
One direction we have seen in Lemma \ref{lemma-nodal-family-precompose-etale}. For the other direction, observe that being locally of finite presentation, flat, or to have relative dimension $1$ is \'etale local on the source (Descent, Lemmas \ref{descent-lemma-locally-finite-presentation-fppf-local-source}, \ref{descent-lemma-flat-fpqc-local-source}, and \ref{descent-lemma-dimension-at-point}). Taking fibres we reduce to the case where $S$ is the spectrum of a field. In this case the result follows from Lemma \ref{lemma-node-etale-local} (and the fact that being smooth is \'etale local on the source by Descent, Lemma \ref{descent-lemma-smooth-smooth-local-source}).
One direction we have seen in Lemma \ref{lemma-base-change-nodal-family}. For the other direction, observe that being locally of finite presentation, flat, or to have relative dimension $1$ is fpqc local on the target (Descent, Lemmas \ref{descent-lemma-descending-property-locally-finite-presentation}, \ref{descent-lemma-descending-property-flat}, and Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-after-base-change}). Taking fibres we reduce to the case where $S$ is the spectrum of a field. In this case the result follows from Lemma \ref{lemma-node-field-extension} (and the fact that being smooth is fpqc local on the target by Descent, Lemma \ref{descent-lemma-descending-property-smooth}).
By Limits, Lemma \ref{limits-lemma-descend-syntomic} there exists an $i$ such that $f_i$ is syntomic. Then $X_i = \coprod_{d \geq 0} X_{i, d}$ is a disjoint union of open and closed subschemes such that $X_{i, d} \to Y_i$ has relative dimension $d$, see Morphisms, Lemma \ref{morphisms-lemma-syntomic-relative-dimension}. Because of the behaviour of dimensions of fibres under base change given in Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-after-base-change} we see that $X \to X_i$ maps into $X_{i, 1}$. Then there exists an $i' \geq i$ such that $X_{i'} \to X_i$ maps into $X_{i, 1}$, see Limits, Lemma \ref{limits-lemma-limit-contained-in-constructible}. Thus $f_{i'} : X_{i'} \to Y_{i'}$ is syntomic of relative dimension $1$ (by Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-after-base-change} again). Consider the morphism $\text{Sing}(f_{i'}) \to Y_{i'}$. We know that the base change to $Y$ is an unramified morphism. Hence by Limits, Lemma \ref{limits-lemma-descend-unramified} we see that after increasing $i'$ the morphism $\text{Sing}(f_{i'}) \to Y_{i'}$ becomes unramified. This finishes the proof.