## 23.10 Smooth ring maps and diagonals

In this section we use the material above to characterize smooth ring maps as those whose diagonal is perfect.

Lemma 23.10.1. Let $A \to B$ be a local ring homomorphism of Noetherian local rings such that $B$ is flat and essentially of finite type over $A$. If

$B \otimes _ A B \longrightarrow B$

is a perfect ring map, i.e., if $B$ has finite tor dimension over $B \otimes _ A B$, then $B$ is the localization of a smooth $A$-algebra.

Proof. As $B$ is essentially of finite type over $A$, so is $B \otimes _ A B$ and in particular $B \otimes _ A B$ is Noetherian. Hence the quotient $B$ of $B \otimes _ A B$ is pseudo-coherent over $B \otimes _ A B$ (More on Algebra, Lemma 15.64.17) which explains why perfectness of the ring map (More on Algebra, Definition 15.82.1) agrees with the condition of finite tor dimension.

We may write $B = R/K$ where $R$ is the localization of $A[x_1, \ldots , x_ n]$ at a prime ideal and $K \subset R$ is an ideal. Denote $\mathfrak m \subset R \otimes _ A R$ the maximal ideal which is the inverse image of the maximal ideal of $B$ via the surjection $R \otimes _ A R \to B \otimes _ A B \to B$. Then we have surjections

$(R \otimes _ A R)_\mathfrak m \to (B \otimes _ A B)_\mathfrak m \to B$

and hence ideals $I \subset J \subset (R \otimes _ A R)_\mathfrak m$ as in Lemma 23.7.4. We conclude that $I/\mathfrak m I \to J/\mathfrak m J$ is injective.

Let $K = (f_1, \ldots , f_ r)$ with $r$ minimal. We may and do assume that $f_ i \in R$ is the image of an element of $A[x_1, \ldots , x_ n]$ which we also denote $f_ i$. Observe that $I$ is generated by $f_1 \otimes 1, \ldots , f_ r \otimes 1$ and $1 \otimes f_1, \ldots , 1 \otimes f_ r$. We claim that this is a minimal set of generators of $I$. Namely, if $\kappa$ is the common residue field of $R$, $B$, $(R \otimes _ A R)_\mathfrak m$, and $(B \otimes _ A B)_\mathfrak m$ then we have a map $R \otimes _ A R \to R \otimes _ A \kappa \oplus \kappa \otimes _ A R$ which factors through $(R \otimes _ A R)_\mathfrak m$. Since $B$ is flat over $A$ and since we have the short exact sequence $0 \to K \to R \to B \to 0$ we see that $K \otimes _ A \kappa \subset R \otimes _ A \kappa$, see Algebra, Lemma 10.39.12. Thus restricting the map $(R \otimes _ A R)_\mathfrak m \to R \otimes _ A \kappa \oplus \kappa \otimes _ A R$ to $I$ we obtain a map

$I \to K \otimes _ A \kappa \oplus \kappa \otimes _ A K \to K \otimes _ B \kappa \oplus \kappa \otimes _ B K.$

The elements $f_1 \otimes 1, \ldots , f_ r \otimes 1, 1 \otimes f_1, \ldots , 1 \otimes f_ r$ map to a basis of the target of this map, since by Nakayama's lemma (Algebra, Lemma 10.20.1) $f_1, \ldots , f_ r$ map to a basis of $K \otimes _ B \kappa$. This proves our claim.

The ideal $J$ is generated by $f_1 \otimes 1, \ldots , f_ r \otimes 1$ and the elements $x_1 \otimes 1 - 1 \otimes x_1, \ldots , x_ n \otimes 1 - 1 \otimes x_ n$ (for the proof it suffices to see that these elements are contained in the ideal $J$). Now we can write

$f_ i \otimes 1 - 1 \otimes f_ i = \sum g_{ij} (x_ j \otimes 1 - 1 \otimes x_ j)$

for some $g_{ij}$ in $(R \otimes _ A R)_\mathfrak m$. This is a general fact about elements of $A[x_1, \ldots , x_ n]$ whose proof we omit. Denote $a_{ij} \in \kappa$ the image of $g_{ij}$. Another computation shows that $a_{ij}$ is the image of $\partial f_ i / \partial x_ j$ in $\kappa$. The injectivity of $I/\mathfrak m I \to J/\mathfrak m J$ and the remarks made above forces the matrix $(a_{ij})$ to have maximal rank $r$. Set

$C = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ r)$

and consider the naive cotangent complex $\mathop{N\! L}\nolimits _{C/A} \cong (C^{\oplus r} \to C^{\oplus n})$ where the map is given by the matrix of partial derivatives. Thus $\mathop{N\! L}\nolimits _{C/A} \otimes _ A B$ is isomorphic to a free $B$-module of rank $n - r$ placed in degree $0$. Hence $C_ g$ is smooth over $A$ for some $g \in C$ mapping to a unit in $B$, see Algebra, Lemma 10.137.12. This finishes the proof. $\square$

Lemma 23.10.2. Let $A \to B$ be a flat finite type ring map of Noetherian rings. If

$B \otimes _ A B \longrightarrow B$

is a perfect ring map, i.e., if $B$ has finite tor dimension over $B \otimes _ A B$, then $B$ is a smooth $A$-algebra.

Proof. This follows from Lemma 23.10.1 and general facts about smooth ring maps, see Algebra, Lemmas 10.137.12 and 10.137.13. Alternatively, the reader can slightly modify the proof of Lemma 23.10.1 to prove this lemma. $\square$

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