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

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

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

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

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

## Comments (7)

Comment #701 by Keenan Kidwell on

Comment #703 by Johan on

Comment #850 by Bhargav Bhatt on

Comment #1514 by Rob Roy on

Comment #1531 by Johan on

Comment #1951 by Brian Conrad on

Comment #2006 by Johan on

There are also: