The Stacks project

Lemma 86.15.6. Let $\varphi : A \to B$ and $A \to C$ be arrows of $\textit{WAdm}^{Noeth}$. Assume $\varphi $ is rig-flat and $A \to C$ adic and topologically of finite type. Then $C \to B \widehat{\otimes }_ A C$ is rig-flat.

Proof. Assume $\varphi $ is rig-flat. Let $f \in C$ be an element. We have to show that $C_{\{ f\} } \to B \widehat{\otimes }_ A C_{\{ f\} }$ is naively rig-flat. Since we can replace $C$ by $C_{\{ f\} }$ we it suffices to show that $C \to B \widehat{\otimes }_ A C$ is naively rig-flat.

If $A \to C$ is surjective or more generally if $C$ is finite as an $A$-module, then $B \otimes _ A C = B \widehat{\otimes }_ A C$ as a finite module over a complete Noetherian ring is complete, see Algebra, Lemma 10.97.1. By the usual base change for flatness (Algebra, Lemma 10.39.7) we see that naive rig-flatness of $\varphi $ implies naive rig-flatness for $C \to B \times _ A C$ in this case.

In the general case, we can factor $A \to C$ as $A \to A\{ x_1, \ldots , x_ n\} \to C$ where $A\{ x_1, \ldots , x_ n\} $ is the restricted power series ring and $A\{ x_1, \ldots , x_ n\} \to C$ is surjective. Thus it suffices to show $C \to B \widehat{\otimes }_ A B$ is naively rig-flat in case $C = A\{ x_1, \ldots , x_ n\} $. Since $A\{ x_1, \ldots , x_ n\} = A\{ x_1, \ldots , x_{n - 1}\} \{ x_ n\} $ by induction on $n$ we reduce to the case discussed in the next paragraph.

Here $C = A\{ x\} $. Note that $B \widehat{\otimes }_ A C = B\{ x\} $. We have to show that $A\{ x\} \to B\{ x\} $ is naively rig-flat. Let $\mathfrak q \subset B\{ x\} $ be a rig-closed prime ideal. We have to show that $B\{ x\} _{\mathfrak q}$ is flat over $A\{ x\} $. Set $\mathfrak p = A \cap \mathfrak q$. By Lemma 86.14.9 we can find an $f \in A$ such that $f$ maps to a unit in $B\{ x\} /\mathfrak q$ and such that the prime ideal $\mathfrak p'$ in $A_{\{ f\} }$ induced is rig-closed. Below we will use that $A_{\{ f\} }\{ x\} = A\{ x\} _{\{ f\} }$ and similarly for $B$; details omitted. Consider the diagram

\[ \xymatrix{ (B\{ x\} )_{\mathfrak q} \ar[r] & (B_{\{ f\} }\{ x\} )_{\mathfrak q'} \\ A\{ x\} \ar[r] \ar[u] & A_{\{ f\} }\{ x\} \ar[u] } \]

We want to show that the left vertical arrow is flat. The top horizontal arrow is faithfully flat as it is a local homomorphism of local rings and flat as $B_{\{ f\} }\{ x\} $ is the completion of a localization of the Noetherian ring $B\{ x\} $. Similarly the bottom horizontal arrow is flat. Hence it suffices to prove that the right vertical arrow is flat. This reduces us to the case discussed in the next paragraph.

Here $C = A\{ x\} $, we have a rig-closed prime ideal $\mathfrak q \subset B\{ x\} $ such that $\mathfrak p = A \cap \mathfrak q$ is rig-closed as well. This implies, via Lemma 86.14.4, that the intermediate primes $B \cap \mathfrak q$ and $A\{ x\} \cap \mathfrak q$ are rig-closed as well. Consider the diagram

\[ \xymatrix{ (B[x])_{B[x] \cap \mathfrak q} \ar[r] & (B\{ x\} )_{\mathfrak q} \\ (A[x])_{A[x] \cap \mathfrak q} \ar[r] \ar[u] & (A\{ x\} )_{A\{ x\} \cap \mathfrak q} \ar[u] } \]

of local homomorphisms of Noetherian local rings. By Lemma 86.14.10 the horizontal arrows define isomorphisms on completions. We already know that the left vertical arrow is flat (as $A \to B$ is naively rig-flat and hence $A[x] \to B[x]$ is flat away from the closed locus defined by an ideal of definition). Hence we finally conclude by More on Algebra, Lemma 15.43.8. $\square$


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