The Stacks project

Lemma 86.4.2. Let $A$ be a Noetherian ring and let $I$ be an ideal. Let $B$ be a finite type $A$-algebra.

  1. If $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$, then $B^\wedge $ satisfies the equivalent conditions of Lemma 86.4.1.

  2. If $B^\wedge $ satisfies the equivalent conditions of Lemma 86.4.1, then there exists $g \in 1 + IB$ such that $\mathop{\mathrm{Spec}}(B_ g)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$.

Proof. Assume $B^\wedge $ satisfies the equivalent conditions of Lemma 86.4.1. The naive cotangent complex $\mathop{N\! L}\nolimits _{B/A}$ is a complex of finite type $B$-modules and hence $H^{-1}$ and $H^0$ are finite $B$-modules. Completion is an exact functor on finite $B$-modules (Algebra, Lemma 10.96.2) and $\mathop{N\! L}\nolimits ^\wedge _{B^\wedge /A}$ is the completion of the complex $\mathop{N\! L}\nolimits _{B/A}$ (this is easy to see by choosing presentations). Hence the assumption implies there exists a $c \geq 0$ such that $H^{-1}/I^ nH^{-1}$ and $H^0/I^ nH^0$ are annihilated by $I^ c$ for all $n$. By Nakayama's lemma (Algebra, Lemma 10.19.1) this means that $I^ cH^{-1}$ and $I^ cH^0$ are annihilated by an element of the form $g = 1 + x$ with $x \in IB$. After inverting $g$ (which does not change the quotients $B/I^ nB$) we see that $\mathop{N\! L}\nolimits _{B/A}$ has cohomology annihilated by $I^ c$. Thus $A \to B$ is étale at any prime of $B$ not lying over $V(I)$ by the definition of étale ring maps, see Algebra, Definition 10.142.1.

Conversely, assume that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$. Then for every $a \in I$ there exists a $c \geq 0$ such that multiplication by $a^ c$ is zero $\mathop{N\! L}\nolimits _{B/A}$. Since $\mathop{N\! L}\nolimits ^\wedge _{B^\wedge /A}$ is the derived completion of $\mathop{N\! L}\nolimits _{B/A}$ (see Lemma 86.3.1) it follows that $B^\wedge $ satisfies the equivalent conditions of Lemma 86.4.1. $\square$


Comments (0)


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Your email address will not be published.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 0ALQ. Beware of the difference between the letter 'O' and the digit '0'.