Lemma 31.22.1. Let $f : X \to S$ be a morphism of schemes. Let $i : Z \subset X$ be an immersion. Assume
$i$ is an $H_1$-regular (resp. quasi-regular) immersion, and
$Z \to S$ is a flat morphism.
31.22 Relative regular immersions
In this section we consider the base change property for regular immersions. The following lemma does not hold for regular immersions or for Koszul immersions, see Examples, Lemma 110.14.2.
Then for every morphism of schemes $g : S' \to S$ the base change $Z' = S' \times _ S Z \to X' = S' \times _ S X$ is an $H_1$-regular (resp. quasi-regular) immersion.
Proof. Unwinding the definitions and using Lemma 31.20.7 this translates into More on Algebra, Lemma 15.31.4. $\square$
This lemma is the motivation for the following definition.
Definition 31.22.2. Let $f : X \to S$ be a morphism of schemes. Let $i : Z \to X$ be an immersion.
We say $i$ is a relative quasi-regular immersion if $Z \to S$ is flat and $i$ is a quasi-regular immersion.
We say $i$ is a relative $H_1$-regular immersion if $Z \to S$ is flat and $i$ is an $H_1$-regular immersion.
We warn the reader that this may be nonstandard notation. Lemma 31.22.1 guarantees that relative quasi-regular (resp. $H_1$-regular) immersions are preserved under any base change. A relative $H_1$-regular immersion is a relative quasi-regular immersion, see Lemma 31.21.2. Please take a look at Lemma 31.22.6 (or Lemma 31.22.4) which shows that if $Z \to X$ is a relative $H_1$-regular (or quasi-regular) immersion and the ambient scheme is (flat and) locally of finite presentation over $S$, then $Z \to X$ is actually a regular immersion and the same remains true after any base change.
Lemma 31.22.3. Let $f : X \to S$ be a morphism of schemes. Let $Z \to X$ be a relative quasi-regular immersion. If $x \in Z$ and $\mathcal{O}_{X, x}$ is Noetherian, then $f$ is flat at $x$.
Proof. Let $f_1, \ldots , f_ r \in \mathcal{O}_{X, x}$ be a quasi-regular sequence cutting out the ideal of $Z$ at $x$. By Algebra, Lemma 10.69.6 we know that $f_1, \ldots , f_ r$ is a regular sequence. Hence $f_ r$ is a nonzerodivisor on $\mathcal{O}_{X, x}/(f_1, \ldots , f_{r - 1})$ such that the quotient is a flat $\mathcal{O}_{S, f(x)}$-module. By Lemma 31.18.5 we conclude that $\mathcal{O}_{X, x}/(f_1, \ldots , f_{r - 1})$ is a flat $\mathcal{O}_{S, f(x)}$-module. Continuing by induction we find that $\mathcal{O}_{X, x}$ is a flat $\mathcal{O}_{S, s}$-module. $\square$
Lemma 31.22.4. Let $X \to S$ be a morphism of schemes. Let $Z \to X$ be an immersion. Assume
$X \to S$ is flat and locally of finite presentation,
$Z \to X$ is a relative quasi-regular immersion.
Then $Z \to X$ is a regular immersion and the same remains true after any base change.
Proof. Pick $x \in Z$ with image $s \in S$. To prove this it suffices to find an affine neighbourhood of $x$ contained in $U$ such that the result holds on that affine open. Hence we may assume that $X$ is affine and there exist a quasi-regular sequence $f_1, \ldots , f_ r \in \Gamma (X, \mathcal{O}_ X)$ such that $Z = V(f_1, \ldots , f_ r)$. By More on Algebra, Lemma 15.31.4 the sequence $f_1|_{X_ s}, \ldots , f_ r|_{X_ s}$ is a quasi-regular sequence in $\Gamma (X_ s, \mathcal{O}_{X_ s})$. Since $X_ s$ is Noetherian, this implies, possibly after shrinking $X$ a bit, that $f_1|_{X_ s}, \ldots , f_ r|_{X_ s}$ is a regular sequence, see Algebra, Lemmas 10.69.6 and 10.68.6. By Lemma 31.18.9 it follows that $Z_1 = V(f_1) \subset X$ is a relative effective Cartier divisor, again after possibly shrinking $X$ a bit. Applying the same lemma again, but now to $Z_2 = V(f_1, f_2) \subset Z_1$ we see that $Z_2 \subset Z_1$ is a relative effective Cartier divisor. And so on until on reaches $Z = Z_ n = V(f_1, \ldots , f_ n)$. Since being a relative effective Cartier divisor is preserved under arbitrary base change, see Lemma 31.18.1, we also see that the final statement of the lemma holds. $\square$
Remark 31.22.5. The codimension of a relative quasi-regular immersion, if it is constant, does not change after a base change. In fact, if we have a ring map $A \to B$ and a quasi-regular sequence $f_1, \ldots , f_ r \in B$ such that $B/(f_1, \ldots , f_ r)$ is flat over $A$, then for any ring map $A \to A'$ we have a quasi-regular sequence $f_1 \otimes 1, \ldots , f_ r \otimes 1$ in $B' = B \otimes _ A A'$ by More on Algebra, Lemma 15.31.4 (which was used in the proof of Lemma 31.22.1 above). Now the proof of Lemma 31.22.4 shows that if $A \to B$ is flat and locally of finite presentation, then for every prime ideal $\mathfrak q' \subset B'$ the sequence $f_1 \otimes 1, \ldots , f_ r \otimes 1$ is even a regular sequence in the local ring $B'_{\mathfrak q'}$.
Lemma 31.22.6. Let $X \to S$ be a morphism of schemes. Let $Z \to X$ be a relative $H_1$-regular immersion. Assume $X \to S$ is locally of finite presentation. Then
there exists an open subscheme $U \subset X$ such that $Z \subset U$ and such that $U \to S$ is flat, and
$Z \to X$ is a regular immersion and the same remains true after any base change.
Proof. Pick $x \in Z$. To prove (1) suffices to find an open neighbourhood $U \subset X$ of $x$ such that $U \to S$ is flat. Hence the lemma reduces to the case that $X = \mathop{\mathrm{Spec}}(B)$ and $S = \mathop{\mathrm{Spec}}(A)$ are affine and that $Z$ is given by an $H_1$-regular sequence $f_1, \ldots , f_ r \in B$. By assumption $B$ is a finitely presented $A$-algebra and $B/(f_1, \ldots , f_ r)B$ is a flat $A$-algebra. We are going to use absolute Noetherian approximation.
Write $B = A[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$. Assume $f_ i$ is the image of $f_ i' \in A[x_1, \ldots , x_ n]$. Choose a finite type $\mathbf{Z}$-subalgebra $A_0 \subset A$ such that all the coefficients of the polynomials $f_1', \ldots , f_ r', g_1, \ldots , g_ m$ are in $A_0$. We set $B_0 = A_0[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$ and we denote $f_{i, 0}$ the image of $f_ i'$ in $B_0$. Then $B = B_0 \otimes _{A_0} A$ and
\[ B/(f_1, \ldots , f_ r) = B_0/(f_{0, 1}, \ldots , f_{0, r}) \otimes _{A_0} A. \]
By Algebra, Lemma 10.168.1 we may, after enlarging $A_0$, assume that $B_0/(f_{0, 1}, \ldots , f_{0, r})$ is flat over $A_0$. It may not be the case at this point that the Koszul cohomology group $H_1(K_\bullet (B_0, f_{0, 1}, \ldots , f_{0, r}))$ is zero. On the other hand, as $B_0$ is Noetherian, it is a finitely generated $B_0$-module. Let $\xi _1, \ldots , \xi _ n \in H_1(K_\bullet (B_0, f_{0, 1}, \ldots , f_{0, r}))$ be generators. Let $A_0 \subset A_1 \subset A$ be a larger finite type $\mathbf{Z}$-subalgebra of $A$. Denote $f_{1, i}$ the image of $f_{0, i}$ in $B_1 = B_0 \otimes _{A_0} A_1$. By More on Algebra, Lemma 15.31.3 the map
\[ H_1(K_\bullet (B_0, f_{0, 1}, \ldots , f_{0, r})) \otimes _{A_0} A_1 \longrightarrow H_1(K_\bullet (B_1, f_{1, 1}, \ldots , f_{1, r})) \]
is surjective. Furthermore, it is clear that the colimit (over all choices of $A_1$ as above) of the complexes $K_\bullet (B_1, f_{1, 1}, \ldots , f_{1, r})$ is the complex $K_\bullet (B, f_1, \ldots , f_ r)$ which is acyclic in degree $1$. Hence
\[ \mathop{\mathrm{colim}}\nolimits _{A_0 \subset A_1 \subset A} H_1(K_\bullet (B_1, f_{1, 1}, \ldots , f_{1, r})) = 0 \]
by Algebra, Lemma 10.8.8. Thus we can find a choice of $A_1$ such that $\xi _1, \ldots , \xi _ n$ all map to zero in $H_1(K_\bullet (B_1, f_{1, 1}, \ldots , f_{1, r}))$. In other words, the Koszul cohomology group $H_1(K_\bullet (B_1, f_{1, 1}, \ldots , f_{1, r}))$ is zero.
Consider the morphism of affine schemes $X_1 \to S_1$ equal to $\mathop{\mathrm{Spec}}$ of the ring map $A_1 \to B_1$ and $Z_1 = \mathop{\mathrm{Spec}}(B_1/(f_{1, 1}, \ldots , f_{1, r}))$. Since $B = B_1 \otimes _{A_1} A$, i.e., $X = X_1 \times _{S_1} S$, and similarly $Z = Z_1 \times _ S S_1$, it now suffices to prove (1) for $X_1 \to S_1$ and the relative $H_1$-regular immersion $Z_1 \to X_1$, see Morphisms, Lemma 29.25.7. Hence we have reduced to the case where $X \to S$ is a finite type morphism of Noetherian schemes. In this case we know that $X \to S$ is flat at every point of $Z$ by Lemma 31.22.3. Combined with the fact that the flat locus is open in this case, see Algebra, Theorem 10.129.4 we see that (1) holds. Part (2) then follows from an application of Lemma 31.22.4. $\square$
If the ambient scheme is flat and locally of finite presentation over the base, then we can characterize a relative quasi-regular immersion in terms of its fibres.
Lemma 31.22.7. Let $\varphi : X \to S$ be a flat morphism which is locally of finite presentation. Let $T \subset X$ be a closed subscheme. Let $x \in T$ with image $s \in S$.
If $T_ s \subset X_ s$ is a quasi-regular immersion in a neighbourhood of $x$, then there exists an open $U \subset X$ and a relative quasi-regular immersion $Z \subset U$ such that $Z_ s = T_ s \cap U_ s$ and $T \cap U \subset Z$.
If $T_ s \subset X_ s$ is a quasi-regular immersion in a neighbourhood of $x$, the morphism $T \to X$ is of finite presentation, and $T \to S$ is flat at $x$, then we can choose $U$ and $Z$ as in (1) such that $T \cap U = Z$.
If $T_ s \subset X_ s$ is a quasi-regular immersion in a neighbourhood of $x$, and $T$ is cut out by $c$ equations in a neighbourhood of $x$, where $c = \dim _ x(X_ s) - \dim _ x(T_ s)$, then we can choose $U$ and $Z$ as in (1) such that $T \cap U = Z$.
In each case $Z \to U$ is a regular immersion by Lemma 31.22.4. In particular, if $T \to S$ is locally of finite presentation and flat and all fibres $T_ s \subset X_ s$ are quasi-regular immersions, then $T \to X$ is a relative quasi-regular immersion.
Proof. Choose affine open neighbourhoods $\mathop{\mathrm{Spec}}(A)$ of $s$ and $\mathop{\mathrm{Spec}}(B)$ of $x$ such that $\varphi (\mathop{\mathrm{Spec}}(B)) \subset \mathop{\mathrm{Spec}}(A)$. Let $\mathfrak p \subset A$ be the prime ideal corresponding to $s$. Let $\mathfrak q \subset B$ be the prime ideal corresponding to $x$. Let $I \subset B$ be the ideal corresponding to $T$. By the initial assumption of the lemma we know that $A \to B$ is flat and of finite presentation. The assumption in (1) means that, after shrinking $\mathop{\mathrm{Spec}}(B)$, we may assume $I(B \otimes _ A \kappa (\mathfrak p))$ is generated by a quasi-regular sequence of elements. After possibly localizing $B$ at some $g \in B$, $g \not\in \mathfrak q$ we may assume there exist $f_1, \ldots , f_ r \in I$ which map to a quasi-regular sequence in $B \otimes _ A \kappa (\mathfrak p)$ which generates $I(B \otimes _ A \kappa (\mathfrak p))$. By Algebra, Lemmas 10.69.6 and 10.68.6 we may assume after another localization that $f_1, \ldots , f_ r \in I$ form a regular sequence in $B \otimes _ A \kappa (\mathfrak p)$. By Lemma 31.18.9 it follows that $Z_1 = V(f_1) \subset \mathop{\mathrm{Spec}}(B)$ is a relative effective Cartier divisor, again after possibly localizing $B$. Applying the same lemma again, but now to $Z_2 = V(f_1, f_2) \subset Z_1$ we see that $Z_2 \subset Z_1$ is a relative effective Cartier divisor. And so on until one reaches $Z = Z_ n = V(f_1, \ldots , f_ n)$. Then $Z \to \mathop{\mathrm{Spec}}(B)$ is a regular immersion and $Z$ is flat over $S$, in particular $Z \to \mathop{\mathrm{Spec}}(B)$ is a relative quasi-regular immersion over $\mathop{\mathrm{Spec}}(A)$. This proves (1).
To see (2) consider the closed immersion $Z \to D$. The surjective ring map $u : \mathcal{O}_{D, x} \to \mathcal{O}_{Z, x}$ is a map of flat local $\mathcal{O}_{S, s}$-algebras which are essentially of finite presentation, and which becomes an isomorphisms after dividing by $\mathfrak m_ s$. Hence it is an isomorphism, see Algebra, Lemma 10.128.4. It follows that $Z \to D$ is an isomorphism in a neighbourhood of $x$, see Algebra, Lemma 10.126.6.
To see (3), after possibly shrinking $U$ we may assume that the ideal of $Z$ is generated by a regular sequence $f_1, \ldots , f_ r$ (see our construction of $Z$ above) and the ideal of $T$ is generated by $g_1, \ldots , g_ c$. We claim that $c = r$. Namely,
\begin{align*} \dim _ x(X_ s) & = \dim (\mathcal{O}_{X_ s, x}) + \text{trdeg}_{\kappa (s)}(\kappa (x)), \\ \dim _ x(T_ s) & = \dim (\mathcal{O}_{T_ s, x}) + \text{trdeg}_{\kappa (s)}(\kappa (x)), \\ \dim (\mathcal{O}_{X_ s, x}) & = \dim (\mathcal{O}_{T_ s, x}) + r \end{align*}
the first two equalities by Algebra, Lemma 10.116.3 and the second by $r$ times applying Algebra, Lemma 10.60.13. As $T \subset Z$ we see that $f_ i = \sum b_{ij} g_ j$. But the ideals of $Z$ and $T$ cut out the same quasi-regular closed subscheme of $X_ s$ in a neighbourhood of $x$. Hence the matrix $(b_{ij}) \bmod \mathfrak m_ x$ is invertible (some details omitted). Hence $(b_{ij})$ is invertible in an open neighbourhood of $x$. In other words, $T \cap U = Z$ after shrinking $U$.
The final statements of the lemma follow immediately from part (2), combined with the fact that $Z \to S$ is locally of finite presentation if and only if $Z \to X$ is of finite presentation, see Morphisms, Lemmas 29.21.3 and 29.21.11. $\square$
The following lemma is an enhancement of Morphisms, Lemma 29.34.20.
Lemma 31.22.8. Let $f : X \to S$ be a smooth morphism of schemes. Let $\sigma : S \to X$ be a section of $f$. Then $\sigma $ is a regular immersion.
Proof. By Schemes, Lemma 26.21.10 the morphism $\sigma $ is an immersion. After replacing $X$ by an open neighbourhood of $\sigma (S)$ we may assume that $\sigma $ is a closed immersion. Let $T = \sigma (S)$ be the corresponding closed subscheme of $X$. Since $T \to S$ is an isomorphism it is flat and of finite presentation. Also a smooth morphism is flat and locally of finite presentation, see Morphisms, Lemmas 29.34.9 and 29.34.8. Thus, according to Lemma 31.22.7, it suffices to show that $T_ s \subset X_ s$ is a quasi-regular closed subscheme. This follows immediately from Morphisms, Lemma 29.34.20 but we can also see it directly as follows. Let $k$ be a field and let $A$ be a smooth $k$-algebra. Let $\mathfrak m \subset A$ be a maximal ideal whose residue field is $k$. Then $\mathfrak m$ is generated by a quasi-regular sequence, possibly after replacing $A$ by $A_ g$ for some $g \in A$, $g \not\in \mathfrak m$. In Algebra, Lemma 10.140.3 we proved that $A_{\mathfrak m}$ is a regular local ring, hence $\mathfrak mA_{\mathfrak m}$ is generated by a regular sequence. This does indeed imply that $\mathfrak m$ is generated by a regular sequence (after replacing $A$ by $A_ g$ for some $g \in A$, $g \not\in \mathfrak m$), see Algebra, Lemma 10.68.6. $\square$
The following lemma has a kind of converse, see Lemma 31.22.12.
Lemma 31.22.9. Let be a commutative diagram of morphisms of schemes. Assume $X \to S$ smooth, and $i$, $j$ immersions. If $j$ is a regular (resp. Koszul-regular, $H_1$-regular, quasi-regular) immersion, then so is $i$.
\[ \xymatrix{ Y \ar[rd]_ j \ar[rr]_ i & & X \ar[ld] \\ & S } \]
Proof. We can write $i$ as the composition
\[ Y \to Y \times _ S X \to X \]
By Lemma 31.22.8 the first arrow is a regular immersion. The second arrow is a flat base change of $Y \to S$, hence is a regular (resp. Koszul-regular, $H_1$-regular, quasi-regular) immersion, see Lemma 31.21.4. We conclude by an application of Lemma 31.21.7. $\square$
Lemma 31.22.10. Let be a commutative diagram of morphisms of schemes. Assume that $Y \to S$ is syntomic, $X \to S$ smooth, and $i$ an immersion. Then $i$ is a regular immersion.
\[ \xymatrix{ Y \ar[rd] \ar[rr]_ i & & X \ar[ld] \\ & S } \]
Proof. After replacing $X$ by an open neighbourhood of $i(Y)$ we may assume that $i$ is a closed immersion. Let $T = i(Y)$ be the corresponding closed subscheme of $X$. Since $T \cong Y$ the morphism $T \to S$ is flat and of finite presentation (Morphisms, Lemmas 29.30.6 and 29.30.7). Also a smooth morphism is flat and locally of finite presentation (Morphisms, Lemmas 29.34.9 and 29.34.8). Thus, according to Lemma 31.22.7, it suffices to show that $T_ s \subset X_ s$ is a quasi-regular closed subscheme. As $X_ s$ is locally of finite type over a field, it is Noetherian (Morphisms, Lemma 29.15.6). Thus we can check that $T_ s \subset X_ s$ is a quasi-regular immersion at points, see Lemma 31.20.8. Take $t \in T_ s$. By Morphisms, Lemma 29.30.9 the local ring $\mathcal{O}_{T_ s, t}$ is a local complete intersection over $\kappa (s)$. The local ring $\mathcal{O}_{X_ s, t}$ is regular, see Algebra, Lemma 10.140.3. By Algebra, Lemma 10.135.7 we see that the kernel of the surjection $\mathcal{O}_{X_ s, t} \to \mathcal{O}_{T_ s, t}$ is generated by a regular sequence, which is what we had to show. $\square$
Lemma 31.22.11. Let be a commutative diagram of morphisms of schemes. Assume that $Y \to S$ is smooth, $X \to S$ smooth, and $i$ an immersion. Then $i$ is a regular immersion.
\[ \xymatrix{ Y \ar[rd] \ar[rr]_ i & & X \ar[ld] \\ & S } \]
Proof. This is a special case of Lemma 31.22.10 because a smooth morphism is syntomic, see Morphisms, Lemma 29.34.7. $\square$
Lemma 31.22.12. Let be a commutative diagram of morphisms of schemes. Assume $X \to S$ smooth and $i$ and $j$ immersions. If $i$ is a Koszul-regular (resp. $H_1$-regular, quasi-regular) immersion, then so is $j$.
\[ \xymatrix{ Y \ar[rd]_ j \ar[rr]_ i & & X \ar[ld] \\ & S } \]
Proof. We will use Lemma 31.21.2 without further mention. Let $y \in Y$ be any point. Set $x = i(y)$ and set $s = j(y)$. It suffices to prove the result after replacing $X$ and $S$ by open neighbourhoods $U$ and $V$ of $x$ and $s$ and $Y$ by an open neighbourhood of $y$ in $i^{-1}(U) \cap j^{-1}(V)$.
We first prove the result for $X = \mathbf{A}^ n_ S$. After replacing $S$ by an affine open $V$ and replacing $Y$ by $j^{-1}(V)$ we may assume that $j$ is a closed immersions and $S$ is affine. Write $S = \mathop{\mathrm{Spec}}(A)$. Then $j : Y \to S$ defines an isomorphism of $Y$ to the closed subscheme $\mathop{\mathrm{Spec}}(A/I)$ for some ideal $I \subset A$. The map $i : Y = \mathop{\mathrm{Spec}}(A/I) \to \mathbf{A}^ n_ S = \mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ n])$ corresponds to an $A$-algebra homomorphism $i^\sharp : A[x_1, \ldots , x_ n] \to A/I$. Choose $a_ i \in A$ which map to $i^\sharp (x_ i)$ in $A/I$. Observe that the ideal of the closed immersion $i$ is
\[ J = (x_1 - a_1, \ldots , x_ n - a_ n) + IA[x_1, \ldots , x_ n]. \]
Set $K = (x_1 - a_1, \ldots , x_ n - a_ n)$. We claim the sequence
\[ 0 \to K/KJ \to J/J^2 \to J/(K + J^2) \to 0 \]
is split exact. To see this note that $K/K^2$ is free with basis $x_ i - a_ i$ over the ring $A[x_1, \ldots , x_ n]/K \cong A$. Hence $K/KJ$ is free with the same basis over the ring $A[x_1, \ldots , x_ n]/J \cong A/I$. On the other hand, taking derivatives gives a map
\[ \text{d}_{A[x_1, \ldots , x_ n]/A} : J/J^2 \longrightarrow \Omega _{A[x_1, \ldots , x_ n]/A} \otimes _{A[x_1, \ldots , x_ n]} A[x_1, \ldots , x_ n]/J \]
which maps the generators $x_ i - a_ i$ to the basis elements $\text{d}x_ i$ of the free module on the right. The claim follows. Moreover, note that $x_1 - a_1, \ldots , x_ n - a_ n$ is a regular sequence in $A[x_1, \ldots , x_ n]$ with quotient ring $A[x_1, \ldots , x_ n]/(x_1 - a_1, \ldots , x_ n - a_ n) \cong A$. Thus we have a factorization
\[ Y \to V(x_1 - a_1, \ldots , x_ n - a_ n) \to \mathbf{A}^ n_ S \]
of our closed immersion $i$ where the composition is Koszul-regular (resp. $H_1$-regular, quasi-regular), the second arrow is a regular immersion, and the associated conormal sequence is split. Now the result follows from Lemma 31.21.8.
Next, we prove the result holds if $i$ is $H_1$-regular or quasi-regular. Namely, shrinking as in the first paragraph of the proof, we may assume that $Y$, $X$, and $S$ are affine. In this case we can choose a closed immersion $h : X \to \mathbf{A}^ n_ S$ over $S$ for some $n$. Note that $h$ is a regular immersion by Lemma 31.22.11. Hence $h \circ i$ is a $H_1$-regular or quasi-regular immersion, see Lemma 31.21.7 (note that this step does not work in the “quasi-regular case”). Thus we reduce to the case $X = \mathbf{A}^ n_ S$ and $S$ affine we proved above.
Finally, assume $i$ is quasi-regular. After shrinking as in the first paragraph of the proof, we may use Morphisms, Lemma 29.36.20 to factor $f$ as $X \to \mathbf{A}^ n_ S \to S$ where the first morphism $X \to \mathbf{A}^ n_ S$ is étale. This reduces the problem to the the two cases (a) $X = \mathbf{A}^ n_ S$ and (b) $f$ is étale. Case (a) was handled in the second paragraph of the proof. Case (b) is handled by the next paragraph.
Assume $f$ is étale. After shrinking we may assume $X$, $Y$, and $S$ affine $i$ and $j$ closed immersions (small detail omitted). Say $S = \mathop{\mathrm{Spec}}(A)$, $X = \mathop{\mathrm{Spec}}(B)$ and $Y = \mathop{\mathrm{Spec}}(B/J) = \mathop{\mathrm{Spec}}(A/I)$. Shrinking further we may assume $J$ is generated by a quasi-regular sequence. The ring map $A \to B$ is étale, hence formally étale (Algebra, Lemma 10.150.2). Thus $\bigoplus I^ n/I^{n + 1} \cong \bigoplus J^ n/J^{n + 1}$ by Algebra, Lemma 10.150.5. Since $J$ is generated by a quasi-regular sequence, so is $I$. This finishes the proof. $\square$
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