Lemma 10.134.4 (00S2): Jacobi-Zariski sequence

Lemma 10.134.4 (Jacobi-Zariski sequence). Let $A \to B \to C$ be ring maps. Choose a presentation $\alpha : A[x_ s, s \in S] \to B$ with kernel $I$ . Choose a presentation $\beta : B[y_ t, t \in T] \to C$ 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

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

with exact rows. We get the following exact sequence of homology groups

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

of $C$ -modules extending the sequence of Lemma 10.131.7. If $\text{Tor}_1^ B(\Omega _{B/A}, C) = 0$ and $\text{Tor}_2^ B(\Omega _{B/A}, C) = 0$ , then $H_1(\mathop{N\! L}\nolimits _{B/A} \otimes _ B C) = H_1(L_{B/A}) \otimes _ B C$ .

Proof. The precise definition of the maps is omitted. The exactness of the top row follows as the $\text{d}x_ s$ , $\text{d}y_ t$ form a basis for the middle module. The map $\gamma$ factors

$A[x_ s, y_ t] \to B[y_ t] \to C$

with surjective first arrow and second arrow equal to $\beta$ . Thus we see that $K \to J$ is surjective. Moreover, the kernel of the first displayed arrow is $IA[x_ s, y_ t]$ . Hence $I/I^2 \otimes C$ surjects onto the kernel of $K/K^2 \to J/J^2$ . Finally, we can use Lemma 10.134.2 to identify the terms as homology groups of the naive cotangent complexes.

The final assertion is a statement in homological algebra. Recall that $\mathop{N\! L}\nolimits _{B/A} = (N^{-1} \to N^0)$ is a two term complex of $B$ -modules with $N^0$ free and cohomology modules $H^0 = \Omega _{B/A}$ and $H^{-1} = H_1(L_{B/A})$ . Write $M \subset N^0$ for the image of the differential. If $\text{Tor}_1^ B(H^0, C) = 0$ , then we have an exact sequence

$0 \to M \otimes _ B C \to N^0 \otimes _ B C \to N^0 \otimes _ B C \to 0$

Since $N^0$ is free, we also see that $\text{Tor}_2^ B(H^0, C) = \text{Tor}_1^ B(M, C)$ . Hence if $\text{Tor}_2^ B(H^0, C) = 0$ then we also have an exact sequence

$0 \to H^{-1} \otimes _ B C \to N^{-1} \otimes _ B C \to M \otimes _ B C \to 0$

Putting everything together we see that if $\text{Tor}_1^ B(H^0, C) = 0$ and $\text{Tor}_2^ B(H^0, C) = 0$ , then $H^{-1} \otimes _ B C$ is the kernel of $N^{-1} \otimes _ B C \to N^0 \otimes _ B C$ as desired. $\square$

Comments (9)

Comment #694 by Keenan Kidwell on June 16, 2014 at 19:59

In the third line up from the bottom, the map $J/J^2\to K/K^2$ should go in the opposite direction.

Comment #1718 by Yogesh More on December 03, 2015 at 18:56

I couldn't immediately see why $IA[x_s, y_t]$ being the kernel of the first map implies that $I/I^2 \otimes C$ surjects onto the kernel of $K/K^2 \to J/J^2$ , and in case anyone else is wondering, here is one explanation why: $IA[x_s, y_t]$ is the therefore the kernel of $K \to J$ , and hence tensoring the exact sequence $0 \to IA[x_s, y_t] \to K \to J \to 0$ by $C$ gives an exact sequence $IA[x_s, y_t] \otimes C \to K \otimes C \to J \otimes C \to 0$ . Since $C\cong A[x_s, y_t]/K$ , we have $K \otimes C \cong K/K^2$ , and similarly $J \otimes C \cong J/J^2$ so our exact sequence is $IA[x_s, y_t] \otimes C \to K/K^2 \to J/J^2 \to 0$ .

Comment #8327 by Et on April 19, 2023 at 07:59

For the final assertion, would we not also want $Tor_{2}^{B}(\Omega_{B/A},C)=0$ ? Let P be a presentation of B over A, with kernel I. The condition $Tor_{1}^{B}(\Omega_{B/A},C)=0$ ensures we have an injection $\text{Im}\frac{I}{I^{2}}\bigotimes_{B}C\rightarrow\Omega_{P/A}\bigotimes_{A}C$ , so that to compute $H^{1}(NL_{B/A}\bigotimes_{B}C)=H^{1}(L_{B/A})\bigotimes_{B}C$ It's enough to show the sequence $0\rightarrow H^{1}(L_{B/A})\bigotimes_{B}C\rightarrow\frac{I}{I^{2}}\bigotimes_{B}C\rightarrow\text{Im}\frac{I}{I^{2}}\bigotimes_{B}C\rightarrow0$ is exact. The obstruction here is $Tor_{1}^{B}(\text{Im}\frac{I}{I^{2}},C)$ . The tor sequence associateded to the exact sequence $0\rightarrow\text{Im}\frac{I}{I^{2}}\rightarrow\Omega_{P/A}\bigotimes_{A}B\rightarrow\Omega_{B/A}\rightarrow0$ shows that we have a surjection $Tor_{2}^{B}(\Omega_{B/A},C)\rightarrow Tor_{1}^{B}(\text{Im}\frac{I}{I^{2}},C)\rightarrow0$ and hence $Tor_{2}^{B}(\Omega_{B/A},C)=0$ is the necessary condition to get what we need.

Comment #8332 by Johan on April 20, 2023 at 11:51

@#8327. Yes, very good for finding this mistake! Thanks! I will fix this the next time I go through all the comments. I checked all the places in the stacks project where this gets used and in the places where we use the vanishing of the tor it comes from projectivity of $\Omega$ .

Comment #8474 by Et on June 09, 2023 at 07:06

Suggestion: since this proposition uses multiple rings, maybe add a subscript for the tensor products indicating over what they are taken

Comment #8942 by Stacks project on April 11, 2024 at 10:04

Thanks and fixed here.

Comment #8988 by Zheng Yang on April 15, 2024 at 17:05

As noted above, is there supposed to be a condition on $\text{Tor}_2^B(\Omega_{B/A}, C)$ ?

Assuming this, we can use the Tor spectral sequence (Tag 061Z) for $NL_{B/A}$ tensored by $\otimes_B C$ to conclude the result in the last sentence.

Comment #8990 by Stacks project on April 16, 2024 at 16:52

@#8988: the edit about the Tor-2 has already been made as you can see when you click on the link I gave. I will correspondingly update the website when all the comments have been addressed, see also this page

Comment #9806 by Manolis C. Tsakiris on October 16, 2024 at 22:48

There is a typo in the first displayed short exact sequence in the proof: the last term should be $H^0\otimes_B C$ instead of $N^0 \otimes_B C$ .

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