The Stacks project

Lemma 15.38.1. Let $(R, \mathfrak m) \to (S, \mathfrak n)$ be a local homomorphism of local rings. The following are equivalent

  1. $R \to S$ is formally smooth in the $\mathfrak n$-adic topology,

  2. for every solid commutative diagram

    \[ \xymatrix{ S \ar[r] \ar@{-->}[rd] & A/J \\ R \ar[r] \ar[u] & A \ar[u] } \]

    of local homomorphisms of local rings where $J \subset A$ is an ideal of square zero, $\mathfrak m_ A^ n = 0$ for some $n > 0$, and $S \to A/J$ induces an isomorphism on residue fields, a dotted arrow exists which makes the diagram commute.

If $S$ is Noetherian these conditions are also equivalent to

  1. same as in (2) but only for diagrams where in addition $A \to A/J$ is a small extension (Algebra, Definition 10.141.1).

Proof. The implication (1) $\Rightarrow $ (2) follows from the definitions. Consider a diagram

\[ \xymatrix{ S \ar[r] \ar@{-->}[rd] & A/J \\ R \ar[r] \ar[u] & A \ar[u] } \]

as in Definition 15.37.1 for the $\mathfrak m$-adic topology on $R$ and the $\mathfrak n$-adic topology on $S$. Pick $m > 0$ with $\mathfrak n^ m(A/J) = 0$ (possible by continuity of maps in diagram). Consider the subring $A'$ of $A$ which is the inverse image of the image of $S$ in $A/J$. Set $J' = J$ viewed as an ideal in $A'$. Then $J'$ is an ideal of square zero in $A'$ and $A'/J'$ is a quotient of $S/\mathfrak n^ m$. Hence $A'$ is local and $\mathfrak m_{A'}^{2m} = 0$. Thus we get a diagram

\[ \xymatrix{ S \ar[r] \ar@{-->}[rd] & A'/J' \\ R \ar[r] \ar[u] & A' \ar[u] } \]

as in (2). If we can construct the dotted arrow in this diagram, then we obtain the dotted arrow in the original one by composing with $A' \to A$. In this way we see that (2) implies (1).

Assume $S$ Noetherian. The implication (1) $\Rightarrow $ (3) is immediate. Assume (3) and suppose a diagram as in (2) is given. Then $\mathfrak m_ A^ n J = 0$ for some $n > 0$. Considering the maps

\[ A \to A/\mathfrak m_ A^{n - 1}J \to \ldots \to A/\mathfrak mJ \to A/J \]

we see that it suffices to produce the lifting if $\mathfrak m_ A J = 0$. Assume $\mathfrak m_ A J = 0$ and let $A' \subset A$ be the ring constructed above. Then $A'/J'$ is Artinian as a quotient of the Artinian local ring $S/\mathfrak n^ m$. Thus it suffices to show that given property (3) we can find the dotted arrow in diagrams as in (2) with $A/J$ Artinian and $\mathfrak m_ A J = 0$. Let $\kappa $ be the common residue field of $A$, $A/J$, and $S$. By (3), if $J_0 \subset J$ is an ideal with $\dim _\kappa (J/J_0) = 1$, then we can produce a dotted arrow $S \to A/J_0$. Taking the product we obtain

\[ S \longrightarrow \prod \nolimits _{J_0 \text{ as above}} A/J_0 \]

Clearly the image of this arrow is contained in the sub $R$-algebra $A'$ of elements which map into the small diagonal $A/J \subset \prod _{J_0} A/J$. Let $J' \subset A'$ be the elements mapping to zero in $A/J$. Then $J'$ is an ideal of square zero and as $\kappa $-vector space equal to

\[ J' = \prod \nolimits _{J_0 \text{ as above}} J/J_0 \]

Thus the map $J \to J'$ is injective. By the theory of vector spaces we can choose a splitting $J' = J \oplus M$. It follows that

\[ A' = A \oplus M \]

as an $R$-algebra. Hence the map $S \to A'$ can be composed with the projection $A' \to A$ to give the desired dotted arrow thereby finishing the proof of the lemma. $\square$

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