The Stacks Project


Tag 00UO

Chapter 10: Commutative Algebra > Section 10.144: Formally unramified maps

Lemma 10.144.2. Let $R \to S$ be a ring map. The following are equivalent:

  1. $R \to S$ is formally unramified,
  2. the module of differentials $\Omega_{S/R}$ is zero.

Proof. Let $J = \mathop{\rm Ker}(S \otimes_R S \to S)$ be the kernel of the multiplication map. Let $A_{univ} = S \otimes_R S/J^2$. Recall that $I_{univ} = J/J^2$ is isomorphic to $\Omega_{S/R}$, see Lemma 10.130.13. Moreover, the two $R$-algebra maps $\sigma_1, \sigma_2 : S \to A_{univ}$, $\sigma_1(s) = s \otimes 1 \bmod J^2$, and $\sigma_2(s) = 1 \otimes s \bmod J^2$ differ by the universal derivation $\text{d} : S \to \Omega_{S/R} = I_{univ}$.

Assume $R \to S$ formally unramified. Then we see that $\sigma_1 = \sigma_2$. Hence $\text{d}(s) = 0$ for all $s \in S$. Hence $\Omega_{S/R} = 0$.

Assume that $\Omega_{S/R} = 0$. Let $A, I, R \to A, S \to A/I$ be a solid diagram as in Definition 10.144.1. Let $\tau_1, \tau_2 : S \to A$ be two dotted arrows making the diagram commute. Consider the $R$-algebra map $A_{univ} \to A$ defined by the rule $s_1 \otimes s_2 \mapsto \tau_1(s_1)\tau_2(s_2)$. We omit the verification that this is well defined. Since $A_{univ} \cong S$ as $I_{univ} = \Omega_{S/R} = 0$ we conclude that $\tau_1 = \tau_2$. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 38725–38733 (see updates for more information).

    \begin{lemma}
    \label{lemma-characterize-formally-unramified}
    Let $R \to S$ be a ring map.
    The following are equivalent:
    \begin{enumerate}
    \item $R \to S$ is formally unramified,
    \item the module of differentials $\Omega_{S/R}$ is zero.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Let $J = \Ker(S \otimes_R S \to S)$ be the kernel of
    the multiplication map. Let $A_{univ} = S \otimes_R S/J^2$. Recall
    that $I_{univ} = J/J^2$ is isomorphic to $\Omega_{S/R}$, see
    Lemma \ref{lemma-differentials-diagonal}. Moreover, the two $R$-algebra maps
    $\sigma_1, \sigma_2 : S \to A_{univ}$, $\sigma_1(s) = s \otimes 1 \bmod J^2$,
    and $\sigma_2(s) = 1 \otimes s \bmod J^2$ differ by the
    universal derivation $\text{d} : S \to \Omega_{S/R} = I_{univ}$.
    
    \medskip\noindent
    Assume $R \to S$ formally unramified.
    Then we see that $\sigma_1 = \sigma_2$.
    Hence $\text{d}(s) = 0$ for all $s \in S$.
    Hence $\Omega_{S/R} = 0$.
    
    \medskip\noindent
    Assume that $\Omega_{S/R} = 0$. Let $A, I, R \to A, S \to A/I$
    be a solid diagram as in Definition \ref{definition-formally-unramified}.
    Let $\tau_1, \tau_2 : S \to A$ be two dotted arrows making the
    diagram commute. Consider the $R$-algebra map $A_{univ} \to A$
    defined by the rule $s_1 \otimes s_2 \mapsto \tau_1(s_1)\tau_2(s_2)$.
    We omit the verification that this is well defined. Since $A_{univ} \cong S$
    as $I_{univ} = \Omega_{S/R} = 0$ we conclude that $\tau_1 = \tau_2$.
    \end{proof}

    Comments (0)

    There are no comments yet for this tag.

    Add a comment on tag 00UO

    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 lower-right corner).

    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 box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?