## 10.141 Smooth ring maps in the Noetherian case

Definition 10.141.1. Let $\varphi : B' \to B$ be a ring map. We say $\varphi$ is a small extension if $B'$ and $B$ are local Artinian rings, $\varphi$ is surjective and $I = \mathop{\mathrm{Ker}}(\varphi )$ has length $1$ as a $B'$-module.

Clearly this means that $I^2 = 0$ and that $I = (x)$ for some $x \in B'$ such that $\mathfrak m' x = 0$ where $\mathfrak m' \subset B'$ is the maximal ideal.

Lemma 10.141.2. Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime ideal of $S$ lying over $\mathfrak p \subset R$. Assume $R$ is Noetherian and $R \to S$ of finite type. The following are equivalent:

1. $R \to S$ is smooth at $\mathfrak q$,

2. for every surjection of local $R$-algebras $(B', \mathfrak m') \to (B, \mathfrak m)$ with $\mathop{\mathrm{Ker}}(B' \to B)$ having square zero and every solid commutative diagram

$\xymatrix{ S \ar[r] \ar@{-->}[rd] & B \\ R \ar[r] \ar[u] & B' \ar[u] }$

such that $\mathfrak q = S \cap \mathfrak m$ there exists a dotted arrow making the diagram commute,

3. same as in (2) but with $B' \to B$ ranging over small extensions, and

4. same as in (2) but with $B' \to B$ ranging over small extensions such that in addition $S \to B$ induces an isomorphism $\kappa (\mathfrak q) \cong \kappa (\mathfrak m)$.

Proof. Assume (1). This means there exists a $g \in S$, $g \not\in \mathfrak q$ such that $R \to S_ g$ is smooth. By Proposition 10.138.13 we know that $R \to S_ g$ is formally smooth. Note that given any diagram as in (2) the map $S \to B$ factors automatically through $S_{\mathfrak q}$ and a fortiori through $S_ g$. The formal smoothness of $S_ g$ over $R$ gives us a morphism $S_ g \to B'$ fitting into a similar diagram with $S_ g$ at the upper left corner. Composing with $S \to S_ g$ gives the desired arrow. In other words, we have shown that (1) implies (2).

Clearly (2) implies (3) and (3) implies (4).

Assume (4). We are going to show that (1) holds, thereby finishing the proof of the lemma. Choose a presentation $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. This is possible as $S$ is of finite type over $R$ and therefore of finite presentation (see Lemma 10.31.4). Set $I = (f_1, \ldots , f_ m)$. Consider the naive cotangent complex

$\text{d} : I/I^2 \longrightarrow \bigoplus \nolimits _{j = 1}^ m S\text{d}x_ j$

of this presentation (see Section 10.134). It suffices to show that when we localize this complex at $\mathfrak q$ then the map becomes a split injection, see Lemma 10.137.12. Denote $S' = R[x_1, \ldots , x_ n]/I^2$. By Lemma 10.131.11 we have

$S \otimes _{S'} \Omega _{S'/R} = S \otimes _{R[x_1, \ldots , x_ n]} \Omega _{R[x_1, \ldots , x_ n]/R} = \bigoplus \nolimits _{j = 1}^ m S\text{d}x_ j.$

Thus the map

$\text{d} : I/I^2 \longrightarrow S \otimes _{S'} \Omega _{S'/R}$

is the same as the map in the naive cotangent complex above. In particular the truth of the assertion we are trying to prove depends only on the three rings $R \to S' \to S$. Let $\mathfrak q' \subset R[x_1, \ldots , x_ n]$ be the prime ideal corresponding to $\mathfrak q$. Since localization commutes with taking modules of differentials (Lemma 10.131.8) we see that it suffices to show that the map

10.141.2.1
$$\label{algebra-equation-target-map} \text{d} : I_{\mathfrak q'}/I_{\mathfrak q'}^2 \longrightarrow S_{\mathfrak q} \otimes _{S'_{\mathfrak q'}} \Omega _{S'_{\mathfrak q'}/R}$$

coming from $R \to S'_{\mathfrak q'} \to S_{\mathfrak q}$ is a split injection.

Let $N \in \mathbf{N}$ be an integer. Consider the ring

$B'_ N = S'_{\mathfrak q'} / (\mathfrak q')^ N S'_{\mathfrak q'} = (S'/(\mathfrak q')^ N S')_{\mathfrak q'}$

and its quotient $B_ N = B'_ N/IB'_ N$. Note that $B_ N \cong S_{\mathfrak q}/\mathfrak q^ NS_{\mathfrak q}$. Observe that $B'_ N$ is an Artinian local ring since it is the quotient of a local Noetherian ring by a power of its maximal ideal. Consider a filtration of the kernel $I_ N$ of $B'_ N \to B_ N$ by $B'_ N$-submodules

$0 \subset J_{N, 1} \subset J_{N, 2} \subset \ldots \subset J_{N, n(N)} = I_ N$

such that each successive quotient $J_{N, i}/J_{N, i - 1}$ has length $1$. (As $B'_ N$ is Artinian such a filtration exists.) This gives a sequence of small extensions

$B'_ N \to B'_ N/J_{N, 1} \to B'_ N/J_{N, 2} \to \ldots \to B'_ N/J_{N, n(N)} = B'_ N/I_ N = B_ N = S_{\mathfrak q}/\mathfrak q^ NS_{\mathfrak q}$

Applying condition (4) successively to these small extensions starting with the map $S \to B_ N$ we see there exists a commutative diagram

$\xymatrix{ S \ar[r] \ar[rd] & B_ N \\ R \ar[r] \ar[u] & B'_ N \ar[u] }$

Clearly the ring map $S \to B'_ N$ factors as $S \to S_{\mathfrak q} \to B'_ N$ where $S_{\mathfrak q} \to B'_ N$ is a local homomorphism of local rings. Moreover, since the maximal ideal of $B'_ N$ to the $N$th power is zero we conclude that $S_{\mathfrak q} \to B'_ N$ factors through $S_{\mathfrak q}/(\mathfrak q)^ NS_{\mathfrak q} = B_ N$. In other words we have shown that for all $N \in \mathbf{N}$ the surjection of $R$-algebras $B'_ N \to B_ N$ has a splitting.

Consider the presentation

$I_ N \to B_ N \otimes _{B'_ N} \Omega _{B'_ N/R} \to \Omega _{B_ N/R} \to 0$

coming from the surjection $B'_ N \to B_ N$ with kernel $I_ N$ (see Lemma 10.131.9). By the above the $R$-algebra map $B'_ N \to B_ N$ has a right inverse. Hence by Lemma 10.131.10 we see that the sequence above is split exact! Thus for every $N$ the map

$I_ N \longrightarrow B_ N \otimes _{B'_ N} \Omega _{B'_ N/R}$

is a split injection. The rest of the proof is gotten by unwinding what this means exactly. Note that

$I_ N = I_{\mathfrak q'}/ (I_{\mathfrak q'}^2 + (\mathfrak q')^ N \cap I_{\mathfrak q'})$

By Artin-Rees (Lemma 10.51.2) we find a $c \geq 0$ such that

$S_{\mathfrak q}/\mathfrak q^{N - c}S_{\mathfrak q} \otimes _{S_{\mathfrak q}} I_ N = S_{\mathfrak q}/\mathfrak q^{N - c}S_{\mathfrak q} \otimes _{S_{\mathfrak q}} I_{\mathfrak q'}/I_{\mathfrak q'}^2$

for all $N \geq c$ (these tensor product are just a fancy way of dividing by $\mathfrak q^{N - c}$). We may of course assume $c \geq 1$. By Lemma 10.131.11 we see that

$S'_{\mathfrak q'}/(\mathfrak q')^{N - c}S'_{\mathfrak q'} \otimes _{S'_{\mathfrak q'}} \Omega _{B'_ N/R} = S'_{\mathfrak q'}/(\mathfrak q')^{N - c}S'_{\mathfrak q'} \otimes _{S'_{\mathfrak q'}} \Omega _{S'_{\mathfrak q'}/R}$

we can further tensor this by $B_ N = S_{\mathfrak q}/\mathfrak q^ N$ to see that

$S_{\mathfrak q}/\mathfrak q^{N - c}S_{\mathfrak q} \otimes _{S'_{\mathfrak q'}} \Omega _{B'_ N/R} = S_{\mathfrak q}/\mathfrak q^{N - c}S_{\mathfrak q} \otimes _{S'_{\mathfrak q'}} \Omega _{S'_{\mathfrak q'}/R}.$

Since a split injection remains a split injection after tensoring with anything we see that

$S_{\mathfrak q}/\mathfrak q^{N - c}S_{\mathfrak q} \otimes _{S_{\mathfrak q}} (02HU) = S_{\mathfrak q}/\mathfrak q^{N - c}S_{\mathfrak q} \otimes _{S_{\mathfrak q}/\mathfrak q^ N S_{\mathfrak q}} (I_ N \longrightarrow B_ N \otimes _{B'_ N} \Omega _{B'_ N/R})$

is a split injection for all $N \geq c$. By Lemma 10.74.1 we see that (10.141.2.1) is a split injection. This finishes the proof. $\square$

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).