## 15.32 Local complete intersection maps

We can use the material above to define a local complete intersection map between rings using presentations by (finite) polynomial algebras.

Lemma 15.32.1. Let $A \to B$ be a finite type ring map. If for some presentation $\alpha : A[x_1, \ldots , x_ n] \to B$ the kernel $I$ is a Koszul-regular ideal then for any presentation $\beta : A[y_1, \ldots , y_ m] \to B$ the kernel $J$ is a Koszul-regular ideal.

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

$\xymatrix{ A[x_1, \ldots , x_ n, y_1, \ldots , y_ m] \ar[d]^{x_ i \mapsto g_ i} \ar[rr]_-{y_ j \mapsto f_ j} & & A[x_1, \ldots , x_ n] \ar[d] \\ A[y_1, \ldots , y_ m] \ar[rr] & & B }$

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 15.31.2 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 10.132.15) we see that

$\dim J/J^2 \otimes _ B \kappa (\mathfrak q) + n = \dim I/I^2 \otimes _ B \kappa (\mathfrak q) + m.$

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 15.29.15. As also $y_1 - f_1, \ldots , y_ m - f_ m$ is a regular sequence we conclude

$y_1 - f_1, \ldots , y_ m - f_ m, p_1, \ldots , p_ t$

is a Koszul-regular sequence in $A[x_ i, y_ j, 1/h]$ (see Lemma 15.29.13). 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

$x_1 - g_1, \ldots , x_ n - g_ n, q_1, \ldots , q_ s$

of the same length which is thus a Koszul-regular sequence by Lemma 15.29.15. Finally, by Lemma 15.29.14 we conclude that the images of $q_1, \ldots , q_ s$ in

$A[x_ i, y_ j, 1/hh']/(x_1 - g_1, \ldots , x_ n - g_ n) \cong A[y_ j, 1/h'']$

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. $\square$

This lemma allows us to make the following definition.

Definition 15.32.2. A ring map $A \to B$ is called a local complete intersection if it is of finite type and for some (equivalently any) presentation $B = A[x_1, \ldots , x_ n]/I$ the ideal $I$ is Koszul-regular.

This notion is local.

Lemma 15.32.3. Let $R \to S$ be a ring map. Let $g_1, \ldots , g_ m \in S$ generate the unit ideal. If each $R \to S_{g_ j}$ is a local complete intersection so is $R \to S$.

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 15.29.15 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 15.29.14 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. $\square$

Lemma 15.32.4. Let $R$ be a ring. If $R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a relative global complete intersection, then $f_1, \ldots , f_ c$ is a Koszul regular sequence.

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 10.134.13 and the fact that a regular sequence is Koszul regular (Lemma 15.29.2). $\square$

Lemma 15.32.5. Let $R \to S$ be a ring map. The following are equivalent

1. $R \to S$ is syntomic (Algebra, Definition 10.134.1), and

2. $R \to S$ is flat and a local complete intersection.

Proof. Assume (1). Then $R \to S$ is flat by definition. By Algebra, Lemma 10.134.15 and Lemma 15.32.3 we see that it suffices to show a relative global complete intersection is a local complete intersection homomorphism which is Lemma 15.32.4.

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 10.133.1. 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 15.31.1 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 15.30.3. 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 15.29.7. Applying Algebra, Lemma 10.133.4 we conclude $S'_{gg'}$ for some $g' \in S$, $g' \not\in \mathfrak q'$ is a global complete intersection over $k$ as desired. $\square$

For a local complete intersection $R \to S$ we have $H_ n(L_{S/R}) = 0$ for $n \geq 2$. Since we haven't (yet) defined the full cotangent complex we can't state and prove this, but we can deduce one of the consequences.

Lemma 15.32.6. Let $A \to B \to C$ be ring maps. Assume $B \to C$ is a local complete intersection homomorphism. Choose a presentation $\alpha : A[x_ s, s \in S] \to B$ with kernel $I$. Choose a presentation $\beta : B[y_1, \ldots , y_ m] \to C$ with kernel $J$. Let $\gamma : A[x_ s, y_ t] \to C$ be the induced presentation of $C$ with kernel $K$. Then we get a canonical commutative diagram

$\xymatrix{ 0 \ar[r] & \Omega _{A[x_ s]/A} \otimes C \ar[r] & \Omega _{A[x_ s, y_ t]/A} \otimes C \ar[r] & \Omega _{B[y_ t]/B} \otimes C \ar[r] & 0 \\ 0 \ar[r] & I/I^2 \otimes C \ar[r] \ar[u] & K/K^2 \ar[r] \ar[u] & J/J^2 \ar[r] \ar[u] & 0 }$

with exact rows. In particular, the six term exact sequence of Algebra, Lemma 10.132.4 can be completed with a zero on the left, i.e., the sequence

$0 \to H_1(\mathop{N\! L}\nolimits _{B/A} \otimes _ B C) \to H_1(L_{C/A}) \to H_1(L_{C/B}) \to \Omega _{B/A} \otimes _ B C \to \Omega _{C/A} \to \Omega _{C/B} \to 0$

is exact.

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 15.31.5. This means that the kernel of $K/K^2 \to J/J^2$ is isomorphic to $IA[x_ s, y_ j]/IK$. Since $I/I^2 \otimes _ A C = IA[x_ s, y_ j]/IK$ this provides us with the desired injectivity of $I/I^2 \otimes _ A C \to K/K^2$ so that the result follows from the snake lemma, see Homology, Lemma 12.5.17. $\square$

Lemma 15.32.7. Let $A \to B \to C$ be ring maps. If $B \to C$ is a filtered colimit of local complete intersection homomorphisms then the conclusion of Lemma 15.32.6 remains valid.

Lemma 15.32.8. Let $A \to B$ be a local homomorphism of local rings. Let $A^ h \to B^ h$, resp. $A^{sh} \to B^{sh}$ be the induced map on henselizations, resp. strict henselizations (Algebra, Lemma 10.150.6, resp. Lemma 10.150.12). Then $\mathop{N\! L}\nolimits _{B/A} \otimes _ B B^ h \to \mathop{N\! L}\nolimits _{B^ h/A^ h}$ and $\mathop{N\! L}\nolimits _{B/A} \otimes _ B B^{sh} \to \mathop{N\! L}\nolimits _{B^{sh}/A^{sh}}$ induce isomorphisms on cohomology groups.

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 10.132.9 and Algebra, Definition 10.141.1. The same is true for $B^ h/B$. Using the Jacobi-Zariski sequence (Algebra, Lemma 10.132.4) 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 10.141.2. By Lemma 15.32.7 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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07CY. Beware of the difference between the letter 'O' and the digit '0'.