The Stacks project

Proof. Choose $f_ j \in A[x_1, \ldots , x_ n]$ with $\alpha (f_ j) = \beta (y_ j)$ and $g_ i \in A[y_1, \ldots , y_ m]$ with $\beta (g_ i) = \alpha (x_ i)$. Then we get a commutative diagram

Note that the kernel $K$ of $A[x_ i, y_ j] \to B$ is equal to $K = (I, y_ j - f_ j) = (J, x_ i - f_ i)$. In particular, as $I$ is finitely generated by Lemma 15.32.2 we see that $J = K/(x_ i - f_ i)$ is finitely generated too.

Pick a prime $\mathfrak q \subset B$. Since $I/I^2 \oplus B^{\oplus m} = J/J^2 \oplus B^{\oplus n}$ (Algebra, Lemma 10.134.15) we see that

Pick $p_1, \ldots , p_ t \in I$ which map to a basis of $I/I^2 \otimes \kappa (\mathfrak q) = I \otimes _{A[x_ i]} \kappa (\mathfrak q)$. Pick $q_1, \ldots , q_ s \in J$ which map to a basis of $J/J^2 \otimes \kappa (\mathfrak q) = J \otimes _{A[y_ j]} \kappa (\mathfrak q)$. So $s + n = t + m$. By Nakayama's lemma there exist $h \in A[x_ i]$ and $h' \in A[y_ j]$ both mapping to a nonzero element of $\kappa (\mathfrak q)$ such that $I_ h = (p_1, \ldots , p_ t)$ in $A[x_ i, 1/h]$ and $J_{h'} = (q_1, \ldots , q_ s)$ in $A[y_ j, 1/h']$. As $I$ is Koszul-regular we may also assume that $I_ h$ is generated by a Koszul regular sequence. This sequence must necessarily have length $t = \dim I/I^2 \otimes _ B \kappa (\mathfrak q)$, hence we see that $p_1, \ldots , p_ t$ is a Koszul-regular sequence by Lemma 15.30.15. As also $y_1 - f_1, \ldots , y_ m - f_ m$ is a regular sequence we conclude

is a Koszul-regular sequence in $A[x_ i, y_ j, 1/h]$ (see Lemma 15.30.13). This sequence generates the ideal $K_ h$. Hence the ideal $K_{hh'}$ is generated by a Koszul-regular sequence of length $m + t = n + s$. But it is also generated by the sequence

of the same length which is thus a Koszul-regular sequence by Lemma 15.30.15. Finally, by Lemma 15.30.14 we conclude that the images of $q_1, \ldots , q_ s$ in

form a Koszul-regular sequence generating $J_{h''}$. Since $h''$ is the image of $hh'$ it doesn't map to zero in $\kappa (\mathfrak q)$ and we win. $\square$

Proof. Let $S = R[x_1, \ldots , x_ n]/I$ be a presentation. Pick $h_ j \in R[x_1, \ldots , x_ n]$ mapping to $g_ j$ in $S$. Then $R[x_1, \ldots , x_ n, x_{n + 1}]/(I, x_{n + 1}h_ j - 1)$ is a presentation of $S_{g_ j}$. Hence $I_ j = (I, x_{n + 1}h_ j - 1)$ is a Koszul-regular ideal in $R[x_1, \ldots , x_ n, x_{n + 1}]$. Pick a prime $I \subset \mathfrak q \subset R[x_1, \ldots , x_ n]$. Then $h_ j \not\in \mathfrak q$ for some $j$ and $\mathfrak q_ j = (\mathfrak q, x_{n + 1}h_ j - 1)$ is a prime ideal of $V(I_ j)$ lying over $\mathfrak q$. Pick $f_1, \ldots , f_ r \in I$ which map to a basis of $I/I^2 \otimes \kappa (\mathfrak q)$. Then $x_{n + 1}h_ j - 1, f_1, \ldots , f_ r$ is a sequence of elements of $I_ j$ which map to a basis of $I_ j \otimes \kappa (\mathfrak q_ j)$. By Nakayama's lemma there exists an $h \in R[x_1, \ldots , x_ n, x_{n + 1}]$ such that $(I_ j)_ h$ is generated by $x_{n + 1}h_ j - 1, f_1, \ldots , f_ r$. We may also assume that $(I_ j)_ h$ is generated by a Koszul regular sequence of some length $e$. Looking at the dimension of $I_ j \otimes \kappa (\mathfrak q_ j)$ we see that $e = r + 1$. Hence by Lemma 15.30.15 we see that $x_{n + 1}h_ j - 1, f_1, \ldots , f_ r$ is a Koszul-regular sequence generating $(I_ j)_ h$ for some $h \in R[x_1, \ldots , x_ n, x_{n + 1}]$, $h \not\in \mathfrak q_ j$. By Lemma 15.30.14 we see that $I_{h'}$ is generated by a Koszul-regular sequence for some $h' \in R[x_1, \ldots , x_ n]$, $h' \not\in \mathfrak q$ as desired. $\square$

Proof. Recall that the homology groups $H_ i(K_\bullet (f_\bullet ))$ are annihilated by the ideal $(f_1, \ldots , f_ c)$. Hence it suffices to show that $H_ i(K_\bullet (f_\bullet ))_\mathfrak q$ is zero for all primes $\mathfrak q \subset R[x_1, \ldots , x_ n]$ containing $(f_1, \ldots , f_ c)$. This follows from Algebra, Lemma 10.136.12 and the fact that a regular sequence is Koszul regular (Lemma 15.30.2). $\square$

Proof. Assume (1). Then $R \to S$ is flat by definition. By Algebra, Lemma 10.136.15 and Lemma 15.33.3 we see that it suffices to show a relative global complete intersection is a local complete intersection homomorphism which is Lemma 15.33.4.

Assume (2). A local complete intersection is of finite presentation because a Koszul-regular ideal is finitely generated. Let $R \to k$ be a map to a field. It suffices to show that $S' = S \otimes _ R k$ is a local complete intersection over $k$, see Algebra, Definition 10.135.1. Choose a prime $\mathfrak q' \subset S'$. Write $S = R[x_1, \ldots , x_ n]/I$. Then $S' = k[x_1, \ldots , x_ n]/I'$ where $I' \subset k[x_1, \ldots , x_ n]$ is the image of $I$. Let $\mathfrak p' \subset k[x_1, \ldots , x_ n]$, $\mathfrak q \subset S$, and $\mathfrak p \subset R[x_1, \ldots , x_ n]$ be the corresponding primes. By Definition 15.32.1 exists an $g \in R[x_1, \ldots , x_ n]$, $g \not\in \mathfrak p$ and $f_1, \ldots , f_ r \in R[x_1, \ldots , x_ n]_ g$ which form a Koszul-regular sequence generating $I_ g$. Since $S$ and hence $S_ g$ is flat over $R$ we see that the images $f'_1, \ldots , f'_ r$ in $k[x_1, \ldots , x_ n]_ g$ form a $H_1$-regular sequence generating $I'_ g$, see Lemma 15.31.4. Thus $f'_1, \ldots , f'_ r$ map to a regular sequence in $k[x_1, \ldots , x_ n]_{\mathfrak p'}$ generating $I'_{\mathfrak p'}$ by Lemma 15.30.7. Applying Algebra, Lemma 10.135.4 we conclude $S'_{gg'}$ for some $g' \in S$, $g' \not\in \mathfrak q'$ is a global complete intersection over $k$ as desired. $\square$

Proof. The only thing to prove is the injectivity of the map $I/I^2 \otimes C \to K/K^2$. By assumption the ideal $J$ is Koszul-regular. Hence we have $IA[x_ s, y_ j] \cap K^2 = IK$ by Lemma 15.32.5. This means that the kernel of $K/K^2 \to J/J^2$ is isomorphic to $IA[x_ s, y_ j]/IK$. Since $I/I^2 \otimes _ A C = IA[x_ s, y_ j]/IK$ by right exactness of tensor product, this provides us with the desired injectivity of $I/I^2 \otimes _ A C \to K/K^2$. $\square$

Proof. Follows from Lemma 15.33.6 and Algebra, Lemma 10.134.9. $\square$

Proof. Since $A^ h$ is a filtered colimit of étale algebras over $A$ we see that $\mathop{N\! L}\nolimits _{A^ h/A}$ is an acyclic complex by Algebra, Lemma 10.134.9 and Algebra, Definition 10.143.1. The same is true for $B^ h/B$. Using the Jacobi-Zariski sequence (Algebra, Lemma 10.134.4) for $A \to A^ h \to B^ h$ we find that $\mathop{N\! L}\nolimits _{B^ h/A} \to \mathop{N\! L}\nolimits _{B^ h/A^ h}$ induces isomorphisms on cohomology groups. Moreover, an étale ring map is a local complete intersection as it is even a global complete intersection, see Algebra, Lemma 10.143.2. By Lemma 15.33.7 we get a six term exact Jacobi-Zariski sequence associated to $A \to B \to B^ h$ which proves that $\mathop{N\! L}\nolimits _{B/A} \otimes _ B B^ h \to \mathop{N\! L}\nolimits _{B^ h/A}$ induces isomorphisms on cohomology groups. This finishes the proof in the case of the map on henselizations. The case of strict henselization is proved in exactly the same manner. $\square$