Skip to content
Permalink
Browse files

Fix idiotic error in last commit

As I was walking down the street it occured to me that what I had just
proven was wrong and in fact any nontrivial closed immersion would be a
counter example...
  • Loading branch information...
aisejohan committed Aug 17, 2019
1 parent b1e7dc9 commit afc4b7a0476d84ab094ed6bb6480cbdf8ca62f1f
Showing with 36 additions and 19 deletions.
  1. +1 −1 chow.tex
  2. +27 −12 dpa.tex
  3. +8 −6 more-morphisms.tex
@@ -11829,7 +11829,7 @@ \section{Gysin maps for diagonals}
Let $(S, \delta)$ be as in Situation \ref{situation-setup}. Let $f : X \to Y$
be a smooth morphism of schemes locally of finite type over $S$. Then the
diagonal morphism $\Delta : X \longrightarrow X \times_Y X$
is a regular immersion and in fact this characterizes when $f$ is smooth, see
is a regular immersion, see
More on Morphisms, Lemma \ref{more-morphisms-lemma-smooth-diagonal-perfect}.
Thus we have the gysin map
$$
39 dpa.tex
@@ -2294,7 +2294,7 @@ \section{Smooth ring maps and diagonals}
\begin{lemma}
\label{lemma-local-perfect-diagonal}
Let $A \to B$ be a local ring homomorphism of Noetherian local rings such that
$B$ is essentially of finite type over $A$. If
$B$ is flat and essentially of finite type over $A$. If
$$
B \otimes_A B \longrightarrow B
$$
@@ -2328,20 +2328,35 @@ \section{Smooth ring maps and diagonals}
Let $K = (f_1, \ldots, f_r)$ with $r$ minimal. We may and do assume that
$f_i \in R$ is the image of an element of $A[x_1, \ldots, x_n]$ which we
also denote $f_i$. Observe that $I$ is generated
by $f_i \otimes 1$ and $1 \otimes f_i$. We claim that this is a minimal
by $f_1 \otimes 1, \ldots, f_r \otimes 1$ and
$1 \otimes f_1, \ldots, 1 \otimes f_r$. We claim that this is a minimal
set of generators of $I$. Namely, if $\kappa$ is the common residue field
of $R$ and $B$, then we have a map
of $R$, $B$, $(R \otimes_A R)_\mathfrak m$, and $(B \otimes_A B)_\mathfrak m$
then we have a map
$R \otimes_A R \to R \otimes_A \kappa \oplus \kappa \otimes_A R$
which factors through $(R \otimes_A R)_\mathfrak m$. Restricting to
$I$ we obtain a map
$I \to K \otimes_A \kappa \oplus \kappa \otimes_A K \to
K \otimes_B \kappa \oplus \kappa \otimes_B K$.
The elements $f_i \otimes 1, 1 \otimes f_i$ map to a basis of
the target of this map which proves our claim.
which factors through $(R \otimes_A R)_\mathfrak m$. Since $B$ is
flat over $A$ and since we have the short exact sequence
$0 \to K \to R \to B \to 0$ we see that
$K \otimes_A \kappa \subset R \otimes_A \kappa$, see
Algebra, Lemma \ref{algebra-lemma-flat-tor-zero}.
Thus restricting the map
$(R \otimes_A R)_\mathfrak m \to R \otimes_A \kappa \oplus \kappa \otimes_A R$
to $I$ we obtain a map
$$
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 \ref{algebra-lemma-NAK})
$f_1, \ldots, f_r$ map to a basis of $K \otimes_B \kappa$.
This proves our claim.

\medskip\noindent
The ideal $J$ is generated by $f_i \otimes 1$ and the elements
$x_i \otimes 1 - 1 \otimes x_i$ (for the proof it suffices to
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
$$
@@ -2370,7 +2385,7 @@ \section{Smooth ring maps and diagonals}

\begin{lemma}
\label{lemma-perfect-diagonal}
Let $A \to B$ be a finite type ring map of Noetherian rings. If
Let $A \to B$ be a flat finite type ring map of Noetherian rings. If
$$
B \otimes_A B \longrightarrow B
$$
\end{proof}

\noindent
The following lemma gives a characterization of smooth morphisms in terms
of the diagonal.
The following lemma gives a characterization of smooth morphisms as flat
morphisms whose diagonal is perfect.

\begin{lemma}
\label{lemma-smooth-diagonal-perfect}
The following are equivalent
\begin{enumerate}
\item $f$ is smooth,
\item $\Delta : X \to X \times_Y X$ is a regular immersion,
\item $\Delta : X \to X \times_Y X$ is a local complete intersection morphism,
\item $\Delta : X \to X \times_Y X$ is perfect.
\item $f$ is flat and $\Delta : X \to X \times_Y X$ is a regular immersion,
\item $f$ is flat and $\Delta : X \to X \times_Y X$ is a local complete intersection morphism,
\item $f$ is flat and $\Delta : X \to X \times_Y X$ is perfect.
\end{enumerate}
\end{lemma}

\begin{proof}
Assume (1). Then the projections $X \times_Y X \to X$ are smooth by
Assume (1). Then $f$ is flat by
Morphisms, Lemma \ref{morphisms-lemma-smooth-flat}.
The projections $X \times_Y X \to X$ are smooth by
Morphisms, Lemma \ref{morphisms-lemma-base-change-smooth}. Hence the diagonal
is a section to a smooth morphism and hence a regular immersion, see
Divisors, Lemma \ref{divisors-lemma-section-smooth-regular-immersion}.

0 comments on commit afc4b7a

Please sign in to comment.
You can’t perform that action at this time.