Lemma 90.4.2. Let $f: R \to S$ be a ring map in $\widehat{\mathcal{C}}_\Lambda $. The following are equivalent
$f$ is surjective,
the map $\mathfrak m_ R/\mathfrak m_ R^2 \to \mathfrak m_ S/\mathfrak m_ S^2$ is surjective, and
the map $\mathfrak m_ R/(\mathfrak m_\Lambda R + \mathfrak m_ R^2) \to \mathfrak m_ S/(\mathfrak m_\Lambda S + \mathfrak m_ S^2)$ is surjective.
Proof. Note that for $n \geq 2$ we have the equality of relative cotangent spaces
and similarly for $S$. Hence by Lemma 90.3.5 we see that $R_ n \to S_ n$ is surjective for all $n$. Now let $K_ n$ be the kernel of $R_ n \to S_ n$. Then the sequences
form an exact sequence of directed inverse systems. The system $(K_ n)$ is Mittag-Leffler since each $K_ n$ is Artinian. Hence by Algebra, Lemma 10.86.4 taking limits preserves exactness. So $\mathop{\mathrm{lim}}\nolimits R_ n \to \mathop{\mathrm{lim}}\nolimits S_ n$ is surjective, i.e., $f$ is surjective. $\square$
Your email address will not be published.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 (0)