Let $R \to S$ be a ring map. Denote $R[S]$ the polynomial ring whose variables are the elements $s \in S$. Let's denote $[s] \in R[S]$ the variable corresponding to $s \in S$. Thus $R[S]$ is a free $R$-module on the basis elements $[s_1] \ldots [s_ n]$ where $s_1, \ldots , s_ n$ ranges over all unordered sequences of elements of $S$. There is a canonical surjection
whose kernel we denote $I \subset R[S]$. It is a simple observation that $I$ is generated by the elements $[s + s'] - [s] - [s']$, $[s][s'] - [ss']$ and $[r] - r$. According to Lemma 10.131.9 there is a canonical map
whose cokernel is canonically isomorphic to $\Omega _{S/R}$. Observe that the $S$-module $\Omega _{R[S]/R} \otimes _{R[S]} S$ is free on the generators $\text{d}[s]$.
Definition 10.134.1. Let $R \to S$ be a ring map. The naive cotangent complex $\mathop{N\! L}\nolimits _{S/R}$ is the chain complex (10.134.0.2) with $I/I^2$ placed in (homological) degree $1$ and $\Omega _{R[S]/R} \otimes _{R[S]} S$ placed in degree $0$. We will denote $H_1(L_{S/R}) = H_1(\mathop{N\! L}\nolimits _{S/R})$1 the homology in degree $1$.
\[ \mathop{N\! L}\nolimits _{S/R} = \left(I/I^2 \longrightarrow \Omega _{R[S]/R} \otimes _{R[S]} S\right) \]
Before we continue let us say a few words about the actual cotangent complex (Cotangent, Section 92.3). Given a ring map $R \to S$ there exists a canonical simplicial $R$-algebra $P_\bullet $ whose terms are polynomial algebras and which comes equipped with a canonical homotopy equivalence
The cotangent complex $L_{S/R}$ of $S$ over $R$ is defined as the chain complex associated to the simplicial module
The naive cotangent complex as defined above is canonically isomorphic to the truncation $\tau _{\leq 1}L_{S/R}$ (see Homology, Section 12.15 and Cotangent, Section 92.11). In particular, it is indeed the case that $H_1(\mathop{N\! L}\nolimits _{S/R}) = H_1(L_{S/R})$ so our definition is compatible with the one using the cotangent complex. Moreover, $H_0(L_{S/R}) = H_0(\mathop{N\! L}\nolimits _{S/R}) = \Omega _{S/R}$ as we've seen above.
Let $R \to S$ be a ring map. A presentation of $S$ over $R$ is a surjection $\alpha : P \to S$ of $R$-algebras where $P$ is a polynomial algebra (on a set of variables). Often, when $S$ is of finite type over $R$ we will indicate this by saying: “Let $R[x_1, \ldots , x_ n] \to S$ be a presentation of $S/R$”, or “Let $0 \to I \to R[x_1, \ldots , x_ n] \to S \to 0$ be a presentation of $S/R$” if we want to indicate that $I$ is the kernel of the presentation. Note that the map $R[S] \to S$ used to define the naive cotangent complex is an example of a presentation.
Note that for every presentation $\alpha $ we obtain a two term chain complex of $S$-modules
Here the term $I/I^2$ is placed in degree $1$ and the term $\Omega _{P/R} \otimes S$ is placed in degree $0$. The class of $f \in I$ in $I/I^2$ is mapped to $\text{d}f \otimes 1$ in $\Omega _{P/R} \otimes S$. The cokernel of this complex is canonically $\Omega _{S/R}$, see Lemma 10.131.9. We call the complex $\mathop{N\! L}\nolimits (\alpha )$ the naive cotangent complex associated to the presentation $\alpha : P \to S$ of $S/R$. Note that if $P = R[S]$ with its canonical surjection onto $S$, then we recover $\mathop{N\! L}\nolimits _{S/R}$. If $P = R[x_1, \ldots , x_ n]$ then will sometimes use the notation $I/I^2 \to \bigoplus _{i = 1, \ldots , n} S\text{d}x_ i$ to denote this complex.
Suppose we are given a commutative diagram
of rings. Let $\alpha : P \to S$ be a presentation of $S$ over $R$ and let $\alpha ' : P' \to S'$ be a presentation of $S'$ over $R'$. A morphism of presentations from $\alpha : P \to S$ to $\alpha ' : P' \to S'$ is defined to be an $R$-algebra map
such that $\phi \circ \alpha = \alpha ' \circ \varphi $. Note that in this case $\varphi (I) \subset I'$, where $I = \mathop{\mathrm{Ker}}(\alpha )$ and $I' = \mathop{\mathrm{Ker}}(\alpha ')$. Thus $\varphi $ induces a map of $S$-modules $I/I^2 \to I'/(I')^2$ and by functoriality of differentials also an $S$-module map $\Omega _{P/R} \otimes S \to \Omega _{P'/R'} \otimes S'$. These maps are compatible with the differentials of $\mathop{N\! L}\nolimits (\alpha )$ and $\mathop{N\! L}\nolimits (\alpha ')$ and we obtain a map of naive cotangent complexes
It is often convenient to consider the induced map $\mathop{N\! L}\nolimits (\alpha ) \otimes _ S S' \to \mathop{N\! L}\nolimits (\alpha ')$.
In the special case that $P = R[S]$ and $P' = R'[S']$ the map $\phi : S \to S'$ induces a canonical ring map $\varphi : P \to P'$ by the rule $[s] \mapsto [\phi (s)]$. Hence the construction above determines canonical(!) maps of chain complexes
associated to the diagram (10.134.1.1). Note that this construction is compatible with composition: given a commutative diagram
we see that the composition of
is the map $\mathop{N\! L}\nolimits _{S/R} \to \mathop{N\! L}\nolimits _{S''/R''}$ given by the outer square.
It turns out that $\mathop{N\! L}\nolimits (\alpha )$ is homotopy equivalent to $\mathop{N\! L}\nolimits _{S/R}$ and that the maps constructed above are well defined up to homotopy (homotopies of maps of complexes are discussed in Homology, Section 12.13 but we also spell out the exact meaning of the statements in the lemma below in its proof).
Lemma 10.134.2. Suppose given a diagram (10.134.1.1). Let $\alpha : P \to S$ and $\alpha ' : P' \to S'$ be presentations.
There exists a morphism of presentations from $\alpha $ to $\alpha '$.
Any two morphisms of presentations induce homotopic morphisms of complexes $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha ')$.
The construction is compatible with compositions of morphisms of presentations (see proof for exact statement).
If $R \to R'$ and $S \to S'$ are isomorphisms, then for any map $\varphi $ of presentations from $\alpha $ to $\alpha '$ the induced map $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha ')$ is a homotopy equivalence and a quasi-isomorphism.
In particular, comparing $\alpha $ to the canonical presentation (10.134.0.1) we conclude there is a quasi-isomorphism $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits _{S/R}$ well defined up to homotopy and compatible with all functorialities (up to homotopy).
Proof. Since $P$ is a polynomial algebra over $R$ we can write $P = R[x_ a, a \in A]$ for some set $A$. As $\alpha '$ is surjective, we can choose for every $a \in A$ an element $f_ a \in P'$ such that $\alpha '(f_ a) = \phi (\alpha (x_ a))$. Let $\varphi : P = R[x_ a, a \in A] \to P'$ be the unique $R$-algebra map such that $\varphi (x_ a) = f_ a$. This gives the morphism in (1).
Let $\varphi $ and $\varphi '$ morphisms of presentations from $\alpha $ to $\alpha '$. Let $I = \mathop{\mathrm{Ker}}(\alpha )$ and $I' = \mathop{\mathrm{Ker}}(\alpha ')$. We have to construct the diagonal map $h$ in the diagram
where the vertical maps are induced by $\varphi $, $\varphi '$ such that
Consider the map $\varphi - \varphi ' : P \to P'$. Since both $\varphi $ and $\varphi '$ are compatible with $\alpha $ and $\alpha '$ we obtain $\varphi - \varphi ' : P \to I'$. This implies that $\varphi , \varphi ' : P \to P'$ induce the same $P$-module structure on $I'/(I')^2$, since $\varphi (p)i' - \varphi '(p)i' = (\varphi - \varphi ')(p)i' \in (I')^2$. Also $\varphi - \varphi '$ is $R$-linear and
Hence the induced map $D : P \to I'/(I')^2$ is a $R$-derivation. Thus we obtain a canonical map $h : \Omega _{P/R} \otimes _ P S \to I'/(I')^2$ such that $D = h \circ \text{d}$. A calculation (omitted) shows that $h$ is the desired homotopy.
Suppose that we have a commutative diagram
and that
$\alpha : P \to S$,
$\alpha ' : P' \to S'$, and
$\alpha '' : P'' \to S''$
are presentations. Suppose that
$\varphi : P \to P'$ is a morphism of presentations from $\alpha $ to $\alpha '$ and
$\varphi ' : P' \to P''$ is a morphism of presentations from $\alpha '$ to $\alpha ''$.
Then it is immediate that $\varphi ' \circ \varphi : P \to P''$ is a morphism of presentations from $\alpha $ to $\alpha ''$ and that the induced map $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha '')$ of naive cotangent complexes is the composition of the maps $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha ')$ and $\mathop{N\! L}\nolimits (\alpha ') \to \mathop{N\! L}\nolimits (\alpha '')$ induced by $\varphi $ and $\varphi '$.
In the simple case of complexes with 2 terms a quasi-isomorphism is just a map that induces an isomorphism on both the cokernel and the kernel of the maps between the terms. Note that homotopic maps of 2 term complexes (as explained above) define the same maps on kernel and cokernel. Hence if $\varphi $ is a map from a presentation $\alpha $ of $S$ over $R$ to itself, then the induced map $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha )$ is a quasi-isomorphism being homotopic to the identity by part (2). To prove (4) in full generality, consider a morphism $\varphi '$ from $\alpha '$ to $\alpha $ which exists by (1). The compositions $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha ') \to \mathop{N\! L}\nolimits (\alpha )$ and $\mathop{N\! L}\nolimits (\alpha ') \to \mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha ')$ are homotopic to the identity maps by (3), hence these maps are homotopy equivalences by definition. It follows formally that both maps $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha ')$ and $\mathop{N\! L}\nolimits (\alpha ') \to \mathop{N\! L}\nolimits (\alpha )$ are quasi-isomorphisms. Some details omitted. $\square$
Lemma 10.134.3. Let $A \to B$ be a polynomial algebra. Then $\mathop{N\! L}\nolimits _{B/A}$ is homotopy equivalent to the chain complex $(0 \to \Omega _{B/A})$ with $\Omega _{B/A}$ in degree $0$.
Proof. Follows from Lemma 10.134.2 and the fact that $\text{id}_ B : B \to B$ is a presentation of $B$ over $A$ with zero kernel. $\square$
The following lemma is part of the motivation for introducing the naive cotangent complex. The cotangent complex extends this to a genuine long exact cohomology sequence. If $B \to C$ is a local complete intersection, then one can extend the sequence with a zero on the left, see More on Algebra, Lemma 15.33.6.
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 with exact rows. We get the following exact sequence of homology groups 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 $H_1(\mathop{N\! L}\nolimits _{B/A} \otimes _ B C) = H_1(L_{B/A}) \otimes _ B C$.
\[ \xymatrix{ 0 \ar[r] & \Omega _{A[x_ s]/A} \otimes C \ar[r] & \Omega _{A[x_ s, y_ t]/A} \otimes C \ar[r] & \Omega _{B[y_ t]/B} \otimes C \ar[r] & 0 \\ & I/I^2 \otimes C \ar[r] \ar[u] & K/K^2 \ar[r] \ar[u] & J/J^2 \ar[r] \ar[u] & 0 } \]
\[ H_1(\mathop{N\! L}\nolimits _{B/A} \otimes _ B C) \to H_1(L_{C/A}) \to H_1(L_{C/B}) \to C \otimes _ B \Omega _{B/A} \to \Omega _{C/A} \to \Omega _{C/B} \to 0 \]
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
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
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
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. $\square$
Remark 10.134.5. Let $A \to B$ and $\phi : B \to C$ be ring maps. Then the composition $\mathop{N\! L}\nolimits _{B/A} \to \mathop{N\! L}\nolimits _{C/A} \to \mathop{N\! L}\nolimits _{C/B}$ is homotopy equivalent to zero. Namely, this composition is the functoriality of the naive cotangent complex for the square Write $J = \mathop{\mathrm{Ker}}(B[C] \to C)$. An explicit homotopy is given by the map $\Omega _{A[B]/A} \otimes _ A B \to J/J^2$ which maps the basis element $\text{d}[b]$ to the class of $[\phi (b)] - b$ in $J/J^2$.
\[ \xymatrix{ B \ar[r]_\phi & C \\ A \ar[r] \ar[u] & B \ar[u] } \]
Lemma 10.134.6. Let $A \to B$ be a surjective ring map with kernel $I$. Then $\mathop{N\! L}\nolimits _{B/A}$ is homotopy equivalent to the chain complex $(I/I^2 \to 0)$ with $I/I^2$ in degree $1$. In particular $H_1(L_{B/A}) = I/I^2$.
Proof. Follows from Lemma 10.134.2 and the fact that $A \to B$ is a presentation of $B$ over $A$. $\square$
Lemma 10.134.7. Let $A \to B \to C$ be ring maps. Assume $A \to C$ is surjective (so also $B \to C$ is). Denote $I = \mathop{\mathrm{Ker}}(A \to C)$ and $J = \mathop{\mathrm{Ker}}(B \to C)$. Then the sequence is exact.
\[ I/I^2 \to J/J^2 \to \Omega _{B/A} \otimes _ B B/J \to 0 \]
Proof. Follows from Lemma 10.134.4 and the description of the naive cotangent complexes $\mathop{N\! L}\nolimits _{C/B}$ and $\mathop{N\! L}\nolimits _{C/A}$ in Lemma 10.134.6. $\square$
Lemma 10.134.8 (Flat base change). Let $R \to S$ be a ring map. Let $\alpha : P \to S$ be a presentation. Let $R \to R'$ be a flat ring map. Let $\alpha ' : P \otimes _ R R' \to S' = S \otimes _ R R'$ be the induced presentation. Then $\mathop{N\! L}\nolimits (\alpha ) \otimes _ R R' = \mathop{N\! L}\nolimits (\alpha ) \otimes _ S S' = \mathop{N\! L}\nolimits (\alpha ')$. In particular, the canonical map is a homotopy equivalence if $R \to R'$ is flat.
\[ \mathop{N\! L}\nolimits _{S/R} \otimes _ S S' \longrightarrow \mathop{N\! L}\nolimits _{S \otimes _ R R'/R'} \]
Proof. This is true because $\mathop{\mathrm{Ker}}(\alpha ') = R' \otimes _ R \mathop{\mathrm{Ker}}(\alpha )$ since $R \to R'$ is flat. $\square$
Lemma 10.134.9. Let $R_ i \to S_ i$ be a system of ring maps over the directed set $I$. Set $R = \mathop{\mathrm{colim}}\nolimits R_ i$ and $S = \mathop{\mathrm{colim}}\nolimits S_ i$. Then $\mathop{N\! L}\nolimits _{S/R} = \mathop{\mathrm{colim}}\nolimits \mathop{N\! L}\nolimits _{S_ i/R_ i}$.
Proof. Recall that $\mathop{N\! L}\nolimits _{S/R}$ is the complex $I/I^2 \to \bigoplus _{s \in S} S\text{d}[s]$ where $I \subset R[S]$ is the kernel of the canonical presentation $R[S] \to S$. Now it is clear that $R[S] = \mathop{\mathrm{colim}}\nolimits R_ i[S_ i]$ and similarly that $I = \mathop{\mathrm{colim}}\nolimits I_ i$ where $I_ i = \mathop{\mathrm{Ker}}(R_ i[S_ i] \to S_ i)$. Hence the lemma is clear. $\square$
Lemma 10.134.10. If $S \subset A$ is a multiplicative subset of $A$, then $\mathop{N\! L}\nolimits _{S^{-1}A/A}$ is homotopy equivalent to the zero complex.
Proof. Since $A \to S^{-1}A$ is flat we see that $\mathop{N\! L}\nolimits _{S^{-1}A/A} \otimes _ A S^{-1}A \to \mathop{N\! L}\nolimits _{S^{-1}A/S^{-1}A}$ is a homotopy equivalence by flat base change (Lemma 10.134.8). Since the source of the arrow is isomorphic to $\mathop{N\! L}\nolimits _{S^{-1}A/A}$ and the target of the arrow is zero (by Lemma 10.134.6) we win. $\square$
Lemma 10.134.11. Let $S \subset A$ is a multiplicative subset of $A$. Let $S^{-1}A \to B$ be a ring map. Then $\mathop{N\! L}\nolimits _{B/A} \to \mathop{N\! L}\nolimits _{B/S^{-1}A}$ is a homotopy equivalence.
Proof. Choose a presentation $\alpha : P \to B$ of $B$ over $A$. Then $\beta : S^{-1}P \to B$ is a presentation of $B$ over $S^{-1}A$. A direct computation shows that we have $\mathop{N\! L}\nolimits (\alpha ) = \mathop{N\! L}\nolimits (\beta )$ which proves the lemma as the naive cotangent complex is well defined up to homotopy by Lemma 10.134.2. $\square$
Lemma 10.134.12. Let $A \to B$ be a ring map. Let $g \in B$. Suppose $\alpha : P \to B$ is a presentation with kernel $I$. Then a presentation of $B_ g$ over $A$ is the map extending $\alpha $ and sending $x$ to $1/g$. The kernel $J$ of $\beta $ is generated by $I$ and the element $f x - 1$ where $f \in P$ is an element mapped to $g \in B$ by $\alpha $. In this situation we have
$J/J^2 = (I/I^2)_ g \oplus B_ g (f x - 1)$,
$\Omega _{P[x]/A} \otimes _{P[x]} B_ g = \Omega _{P/A} \otimes _ P B_ g \oplus B_ g \text{d}x$,
$\mathop{N\! L}\nolimits (\beta ) \cong \mathop{N\! L}\nolimits (\alpha ) \otimes _ B B_ g \oplus (B_ g \xrightarrow {g} B_ g)$
\[ \beta : P[x] \longrightarrow B_ g \]
Hence the canonical map $\mathop{N\! L}\nolimits _{B/A} \otimes _ B B_ g \to \mathop{N\! L}\nolimits _{B_ g/A}$ is a homotopy equivalence.
Proof. Since $P[x]/(I, fx - 1) = B[x]/(gx - 1) = B_ g$ we get the statement about $I$ and $fx - 1$ generating $J$. Consider the commutative diagram
with exact rows of Lemma 10.134.4. The $B_ g$-module $\Omega _{B[x]/B} \otimes B_ g$ is free of rank $1$ on $\text{d}x$. The element $\text{d}x$ in the $B_ g$-module $\Omega _{P[x]/A} \otimes B_ g$ provides a splitting for the top row. The element $gx - 1 \in (gx - 1)/(gx - 1)^2$ is mapped to $g\text{d}x$ in $\Omega _{B[x]/B} \otimes B_ g$ and hence $(gx - 1)/(gx - 1)^2$ is free of rank $1$ over $B_ g$. (This can also be seen by arguing that $gx - 1$ is a nonzerodivisor in $B[x]$ because it is a polynomial with invertible constant term and any nonzerodivisor gives a quasi-regular sequence of length $1$ by Lemma 10.69.2.)
Let us prove $(I/I^2)_ g \to J/J^2$ injective. Consider the $P$-algebra map
sending $x$ to $1/f$. Since $J$ is generated by $I$ and $fx - 1$ we see that $\pi (J) \subset (I/I^2)_ f = (I/I^2)_ g$. Since this is an ideal of square zero we see that $\pi (J^2) = 0$. If $a \in I$ maps to an element of $J^2$ in $J$, then $\pi (a) = 0$, which implies that $a$ maps to zero in $I_ f/I_ f^2$. This proves the desired injectivity.
Thus we have a short exact sequence of two term complexes
Such a short exact sequence can always be split in the category of complexes. In our particular case we can take as splittings
This works because $\text{d}(fx - 1) = x\text{d}f + f \text{d}x = g(g^{-2}\text{d}f + \text{d}x)$ in $\Omega _{P[x]/A} \otimes B_ g$. $\square$
Lemma 10.134.13. Let $A \to B$ be a ring map. Let $S \subset B$ be a multiplicative subset. The canonical map $\mathop{N\! L}\nolimits _{B/A} \otimes _ B S^{-1}B \to \mathop{N\! L}\nolimits _{S^{-1}B/A}$ is a quasi-isomorphism.
Proof. We have $S^{-1}B = \mathop{\mathrm{colim}}\nolimits _{g \in S} B_ g$ where we think of $S$ as a directed set (ordering by divisibility), see Lemma 10.9.9. By Lemma 10.134.12 each of the maps $\mathop{N\! L}\nolimits _{B/A} \otimes _ B B_ g \to \mathop{N\! L}\nolimits _{B_ g/A}$ are quasi-isomorphisms. The lemma follows from Lemma 10.134.9. $\square$
Lemma 10.134.14. Let $R$ be a ring. Let $A_1 \to A_0$, and $B_1 \to B_0$ be two term complexes. Suppose that there exist morphisms of complexes $\varphi : A_\bullet \to B_\bullet $ and $\psi : B_\bullet \to A_\bullet $ such that $\varphi \circ \psi $ and $\psi \circ \varphi $ are homotopic to the identity maps. Then $A_1 \oplus B_0 \cong B_1 \oplus A_0$ as $R$-modules.
Proof. Choose a map $h : A_0 \to A_1$ such that
Similarly, choose a map $h' : B_0 \to B_1$ such that
A trivial computation shows that
This shows that both matrices on the right hand side are invertible and proves the lemma. $\square$
Lemma 10.134.15. Let $R \to S$ be a ring map of finite type. For any presentations $\alpha : R[x_1, \ldots , x_ n] \to S$, and $\beta : R[y_1, \ldots , y_ m] \to S$ we have as $S$-modules where $I = \mathop{\mathrm{Ker}}(\alpha )$ and $J = \mathop{\mathrm{Ker}}(\beta )$.
\[ I/I^2 \oplus S^{\oplus m} \cong J/J^2 \oplus S^{\oplus n} \]
Lemma 10.134.16. Let $R \to S$ be a ring map of finite type. Let $g \in S$. For any presentations $\alpha : R[x_1, \ldots , x_ n] \to S$, and $\beta : R[y_1, \ldots , y_ m] \to S_ g$ we have as $S_ g$-modules where $I = \mathop{\mathrm{Ker}}(\alpha )$ and $J = \mathop{\mathrm{Ker}}(\beta )$.
\[ (I/I^2)_ g \oplus S^{\oplus m}_ g \cong J/J^2 \oplus S_ g^{\oplus n} \]
Proof. Let $\beta ' : R[x_1, \ldots , x_ n, x] \to S_ g$ be the presentation of Lemma 10.134.12 constructed starting with $\alpha $. Then we know that $\mathop{N\! L}\nolimits (\alpha ) \otimes _ S S_ g$ is homotopy equivalent to $\mathop{N\! L}\nolimits (\beta ')$. We know that $\mathop{N\! L}\nolimits (\beta )$ and $\mathop{N\! L}\nolimits (\beta ')$ are homotopy equivalent by Lemma 10.134.2. We conclude that $\mathop{N\! L}\nolimits (\alpha ) \otimes _ S S_ g$ is homotopy equivalent to $\mathop{N\! L}\nolimits (\beta )$. Finally, we apply Lemma 10.134.15. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (13)
Comment #170 by Kiran Kedlaya on
Comment #172 by Johan on
Comment #692 by Keenan Kidwell on
Comment #1716 by Yogesh More on
Comment #1756 by Johan on
Comment #1943 by Maarten Derickx on
Comment #1999 by Johan on
Comment #5555 by minsom on
Comment #5739 by Johan on
Comment #7374 by Yiming TANG on
Comment #7577 by Stacks Project on
Comment #9024 by Zheng Yang on
Comment #9192 by Stacks project on