The formation of the naive cotangent complex commutes with localization at an element.

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

$\beta : P[x] \longrightarrow B_ g$

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

1. $J/J^2 = (I/I^2)_ g \oplus B_ g (f x - 1)$,

2. $\Omega _{P[x]/A} \otimes _{P[x]} B_ g = \Omega _{P/A} \otimes _ P B_ g \oplus B_ g \text{d}x$,

3. $\mathop{N\! L}\nolimits (\beta ) \cong \mathop{N\! L}\nolimits (\alpha ) \otimes _ B B_ g \oplus (B_ g \xrightarrow {g} B_ g)$

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

$\xymatrix{ 0 \ar[r] & \Omega _{P/A} \otimes B_ g \ar[r] & \Omega _{P[x]/A} \otimes B_ g \ar[r] & \Omega _{B[x]/B} \otimes B_ g \ar[r] & 0 \\ & (I/I^2)_ g \ar[r] \ar[u] & J/J^2 \ar[r] \ar[u] & (gx - 1)/(gx - 1)^2 \ar[r] \ar[u] & 0 }$

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

$\pi : P[x] \to (P/I^2)_ f = P_ f/I_ f^2$

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

$0 \to \mathop{N\! L}\nolimits (\alpha ) \otimes _ B B_ g \to \mathop{N\! L}\nolimits (\beta ) \to (B_ g \xrightarrow {g} B_ g) \to 0$

Such a short exact sequence can always be split in the category of complexes. In our particular case we can take as splittings

$J/J^2 = (I/I^2)_ g \oplus B_ g (fx - 1)\quad \text{and}\quad \Omega _{P[x]/A} \otimes B_ g = \Omega _{P/A} \otimes B_ g \oplus B_ g (g^{-2}\text{d}f + \text{d}x)$

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 $\Omega _{P[x]/A} \otimes B_ g$. $\square$

Comment #701 by Keenan Kidwell on

In the fourth sentence of the statement of the lemma, $1/b$ should be $1/g$.

Comment #703 by on

Thank you very much for all the comments. As of now, the last batch of fixes can be found in this commit. I intend to update the stacks project later today so then you should be able to see the changes reflected online as well.

Comment #850 by Bhargav Bhatt on

Suggested slogan: The formation of the naive cotangent complex commutes with localization at an element.

Comment #1514 by Rob Roy on

There's an error in the argument for injectivity of $(I/I^2)_g \rightarrow (J/J^2)$, because in the short exact sequence $0 \rightarrow I[x] \rightarrow J \rightarrow (fx-1) \rightarrow 0$, $(fx-1)$ really represents a non-free quotient of $(fx-1)$ (annihilated by $I[x]$); so the last sentence of the proof is false. Here's a different argument. It is enough to show that $\alpha ( I \cap J^2)=0$ where $\alpha$ is the obvious map $I \rightarrow I_f/I_f^2$. So consider the extension of $\alpha$ to the P-algebra map $\pi: P[x] \rightarrow P_f/I_f^2$ given by $\pi(x)=1/f$, and note that $\pi(J^2)=\pi(J)^2=\pi(I[x])^2=0$, since $\pi(fx-1)=0$ implies $\pi(J)=\pi(I[x])$.

Comment #1531 by on

Thanks very much. There is a generalization of this proof (the correct one you give) in the proof of Lemma 15.33. Here is the commit.

Comment #1951 by Brian Conrad on

Perhaps mention that $fx - 1 \in B[x]$ is not a zero-divisor (easy calculation beginning at constant terms), so $(fx-1)/(fx-1)^2$ is a free module (of rank 1) over $B[x]/(fx-1)= B_f$. This freeness over $B_f$ is implicitly used in the use of the present result in the proof of Tag 07CF.

Comment #2006 by on

OK, I added your remark and I tried to explain the proof better one more time, see here. But I think there is a curse on this lemma which prevents it from being explained clearly.

Comment #7193 by Paolo on

Shouldn't point (1) say $J/J^2 = (I/I^2)_g \oplus B_g(gx-1)/(gx-1)^2$?

Comment #7195 by on

The $fx - 1$ in part (1) means the element of $J/J^2$ which is the image of $fx - 1 \in J$ and the $B_g$ in front of it means that this element gives a free summand; see also discussion in the proof of the lemma. OTOH $(gx - 1)/(gx - 1)^2$ is a quotient module of $J/J^2$ so I think it would be less clear to write that here.

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