Nonflat completions

examples.tex

@@ -460,6 +460,148 @@ \section{The category of complete modules is not abelian}
 \end{proof}
 
 
+\section{Nonflat completions}
+\label{section-nonflat}
+
+\noindent
+In this section we give some examples of completions which are not exact.
+
+\begin{lemma}
+\label{lemma-countable-fg-tensor}
+Let $R$ be a ring. Let $M$ be an $R$-module which is countable.
+Then $M$ is a finite $R$-module if and only if
+$M \otimes_R R^\mathbf{N} \to M^\mathbf{N}$ is surjective.
+\end{lemma}
+
+\begin{proof}
+If $M$ is a finite module, then the map is surjective by Algebra, Proposition
+\ref{algebra-proposition-fg-tensor}. Conversely, assume the map is surjective.
+Let $m_1, m_2, m_3, \ldots$ be an enumeration of the elements of $M$.
+Let $\sum_{j = 1, \ldots, m} x_j \otimes a_j$ be an element of the
+tensor product mapping to the element $(m_n) \in M^\mathbf{N}$. Then
+we see that $x_1, \ldots, x_m$ generate $M$ over $R$ as in the proof of
+Algebra, Proposition \ref{algebra-proposition-fg-tensor}.
+\end{proof}
+
+\begin{lemma}
+\label{lemma-countable-fp-tensor}
+Let $R$ be a countable ring. Let $M$ be a countable $R$-module. Then $M$
+is finitely presented if and only if the canonical map
+$M \otimes_R R^\mathbf{N} \to M^\mathbf{N}$ is an isomorphism.
+\end{lemma}
+
+\begin{proof}
+If $M$ is a finitely presented module, then the map is an isomorphism
+by Algebra, Proposition \ref{algebra-proposition-fp-tensor}. Conversely,
+assume the map is an isomorphism. By Lemma \ref{lemma-countable-fg-tensor}
+the module $M$ is finite. Choose a surjection $R^{\oplus m} \to M$ with
+kernel $K$. Then $K$ is countable as a submodule of $R^{\oplus m}$.
+Arguing as in the proof of Algebra, Proposition
+\ref{algebra-proposition-fp-tensor} we see that
+$K \otimes_R R^\mathbf{N} \to K^\mathbf{N}$ is surjective.
+Hence we conclude that $K$ is a finite $R$-module by
+Lemma \ref{lemma-countable-fg-tensor}.
+Thus $M$ is finitely presented.
+\end{proof}
+
+\begin{lemma}
+\label{lemma-countable-coherent}
+Let $R$ be a countable ring. Then $R$ is coherent if and only if
+$R^\mathbf{N}$ is a flat $R$-module.
+\end{lemma}
+
+\begin{proof}
+If $R$ is coherent, then $R^\mathbf{N}$ is a flat module by
+Algebra, Proposition \ref{algebra-proposition-characterize-coherent}.
+Assume $R^\mathbf{N}$ is flat. Let $I \subset R$ be a finitely
+generated ideal. To prove the lemma we show that $I$ is finitely
+presented as an $R$-module. Namely, the map
+$I \otimes_R R^\mathbf{N} \to R^\mathbf{N}$ is
+injective as $R^\mathbf{N}$ is flat and its image is
+$I^\mathbf{N}$ by Lemma \ref{lemma-countable-fg-tensor}.
+Thus we conclude by Lemma \ref{lemma-countable-fp-tensor}.
+\end{proof}
+
+\noindent
+Let $R$ be a countable ring. Observe that $R[[x]]$ is isomorphic to
+$R^\mathbf{N}$ as an $R$-module. By Lemma \ref{lemma-countable-coherent}
+we see that $R \to R[[x]]$ is flat if and only if $R$ is coherent.
+There are plenty of noncoherent countable rings, for example
+$$
+R = k[y, z, a_1, b_1, a_2, b_2, a_3, b_3, \ldots]/
+(a_1 y + b_1 z, a_2 y + b_2 z, a_3 y + b_3 z, \ldots)
+$$
+where $k$ is a countable field. This ring is not coherent because
+the ideal $(y, z)$ of $R$ is not a finitely presented $R$-module.
+Note that $R[[x]]$ is the completion of $R[x]$ by the principal
+ideal $(x)$.
+
+\begin{lemma}
+\label{lemma-completion-polynomial-ring-not-flat}
+There exists a ring such that the completion $R[[x]]$ of $R[x]$
+at $(x)$ is not flat over $R$ and a fortiori not flat over $R[x]$.
+\end{lemma}
+
+\begin{proof}
+See discussion above.
+\end{proof}
+
+\noindent
+Next, we will construct an example where the completion of a localization
+is nonflat. To do this consider the ring
+$$
+R = k[y, z, a_1, a_2, a_3, \ldots]/(ya_i, a_i a_j)
+$$
+Denote $f \in R$ the residue class of $z$. We claim the ring map
+\begin{equation}
+\label{equation-nonflat}
+R[[x]] \longrightarrow R_f[[x]]
+\end{equation}
+isn't flat. Let $I$ be the kernel of $y : R[[x]] \to R[[x]]$. A typical
+element $g$ of $I$ looks like $g = \sum g_{n, m} a_mx^n$
+where $g_{n, m} \in k[z]$ and for a given $n$ only a finite number of
+nonzero $g_{n, m}$. Let $J$ be the kernel of $y : R_f[[x]] \to R_f[[x]]$.
+We claim that $J \not = I R_f[[x]]$. Namely, if this were true then we
+would have
+$$
+\sum z^{-n} a_n x^n = \sum\nolimits_{i = 1, \ldots, m} h_i g_i
+$$
+for some $m \geq 1$, $g_i \in I$, and $h_i \in R_f[[x]]$. Say
+$h_i = \bar h_i \bmod (y, a_1, a_2, a_3, \ldots)$
+with $\bar h_i \in k[z, 1/z][[x]]$.  Looking at the coefficient of
+$a_n$ and using the description of the elements $g_i$ above we would get
+$$
+z^{-n} x^n = \sum \bar h_i \bar g_{i, n}
+$$
+for some $\bar g_{i, n} \in k[z][[x]]$. This would mean that
+all $z^{-n}x^n$ are contained in the finite $k[z][[x]]$-module
+generated by the elements $\bar h_i$. Since $k[z][[x]]$ is Noetherian
+this implies that the $R[z][[x]]$-submodule of $k[z, 1/z][[x]]$
+generated by $1, z^{-1}x, z^{-2}x^2, \ldots$ is finite. By
+Algebra, Lemma \ref{algebra-lemma-characterize-integral-element}
+we would conclude that $z^{-1}x$ is integral over $k[z][[x]]$
+which is absurd. On the other hand,
+if (\ref{equation-nonflat}) were flat, then we would
+get $J = IR_f[[x]]$ by tensoring the exact sequence
+$0 \to I \to R[[x]] \xrightarrow{y} R[[x]]$ with $R_f[[x]]$.
+
+\begin{lemma}
+\label{lemma-nonflat-completion-localization}
+There exists a ring $A$ complete with respect to a principal ideal $I$
+and an element $f \in A$ such that the $I$-adic completion
+$A_f^\wedge$ of $A_f$ is not flat over $A$.
+\end{lemma}
+
+\begin{proof}
+Set $A = R[[x]]$ and $I = (x)$ and observe that $R_f[[x]]$
+is the completion of $R[[x]]_f$.
+\end{proof}
+
+
+
+
+
+
 \section{Regular sequences and base change}
 \label{section-regular-base-change}