The Stacks project

Lemma 10.137.7. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c) = R[x_1, \ldots , x_ n]/I$ be a standard smooth algebra. Then

  1. the ring map $R \to S$ is smooth,

  2. the $S$-module $\Omega _{S/R}$ is free on $\text{d}x_{c + 1}, \ldots , \text{d}x_ n$,

  3. the $S$-module $I/I^2$ is free on the classes of $f_1, \ldots , f_ c$,

  4. for any $g \in S$ the ring map $R \to S_ g$ is standard smooth,

  5. for any ring map $R \to R'$ the base change $R' \to R'\otimes _ R S$ is standard smooth,

  6. if $f \in R$ maps to an invertible element in $S$, then $R_ f \to S$ is standard smooth, and

  7. the ring $S$ is a relative global complete intersection over $R$.

Proof. Consider the naive cotangent complex of the given presentation

\[ (f_1, \ldots , f_ c)/(f_1, \ldots , f_ c)^2 \longrightarrow \bigoplus \nolimits _{i = 1}^ n S \text{d}x_ i \]

Let us compose this map with the projection onto the first $c$ direct summands of the direct sum. According to the definition of a standard smooth algebra the classes $f_ i \bmod (f_1, \ldots , f_ c)^2$ map to a basis of $\bigoplus _{i = 1}^ c S\text{d}x_ i$. We conclude that $(f_1, \ldots , f_ c)/(f_1, \ldots , f_ c)^2$ is free of rank $c$ with a basis given by the elements $f_ i \bmod (f_1, \ldots , f_ c)^2$, and that the homology in degree $0$, i.e., $\Omega _{S/R}$, of the naive cotangent complex is a free $S$-module with basis the images of $\text{d}x_{c + j}$, $j = 1, \ldots , n - c$. In particular, this proves $R \to S$ is smooth.

The proofs of (4) and (6) are omitted. But see the example below and the proof of Lemma 10.136.9.

Let $\varphi : R \to R'$ be any ring map. Denote $S' = R'[x_1, \ldots , x_ n]/(f_1^\varphi , \ldots , f_ c^\varphi )$ where $f^\varphi $ is the polynomial obtained from $f \in R[x_1, \ldots , x_ n]$ by applying $\varphi $ to all the coefficients. Then $S' \cong R' \otimes _ R S$. Moreover, the determinant of Definition 10.137.6 for $S'/R'$ is equal to $g^\varphi $. Its image in $S'$ is therefore the image of $g$ via $R[x_1, \ldots , x_ n] \to S \to S'$ and hence invertible. This proves (5).

To prove (7) it suffices to show that $S \otimes _ R \kappa (\mathfrak p)$ has dimension $n - c$ for every prime $\mathfrak p \subset R$. By (5) it suffices to prove that any standard smooth algebra $k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ over a field $k$ has dimension $n - c$. We already know that $k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a local complete intersection by Lemma 10.137.5. Hence, since $I/I^2$ is free of rank $c$ we see that $k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ has dimension $n - c$, by Lemma 10.135.4 for example. $\square$


Comments (2)

Comment #2824 by Dario Weißmann on

Typos in the proof of (7):

  • is not introduced as a prime of .

  • "Hence, since is free of rank we see that it dimension ..." Maybe something like "Hence, since is free of rank we see that our standard smooth algebra has dimension ..."?


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 00T7. Beware of the difference between the letter 'O' and the digit '0'.