The Stacks project

86.15 Rig-flat homomorphisms

In this section we define rig-flat homomorphisms of adic Noetherian topological rings.

Lemma 86.15.1. Let $\varphi : A \to B$ be a morphism in $\textit{WAdm}^{adic*}$ (Formal Spaces, Section 85.17). Assume $\varphi $ is adic. The following are equivalent:

  1. $B_ f$ is flat over $A$ for all topologically nilpotent $f \in A$,

  2. $B_ g$ is flat over $A$ for all topologically nilpotent $g \in B$,

  3. $B_\mathfrak q$ is flat over $A$ for all primes $\mathfrak q \subset B$ which do not contain an ideal of definition,

  4. $B_\mathfrak q$ is flat over $A$ for every rig-closed prime $\mathfrak q \subset B$, and

  5. add more here.

Proof. Follows from the definitions and Algebra, Lemma 10.39.18. $\square$

Definition 86.15.2. Let $\varphi : A \to B$ be a continuous ring homomorphism between adic Noetherian topological rings, i.e., $\varphi $ is an arrow of $\textit{WAdm}^{Noeth}$. We say $\varphi $ is naively rig-flat if $\varphi $ is adic, topologically of finite type, and satisfies the equivalent conditions of Lemma 86.15.1.

The example below shows that this notion does not “localize”.

Example 86.15.3. By Examples, Lemma 108.17.1 there exists a local Noetherian $2$-dimensional domain $(A, \mathfrak m)$ complete with respect to a principal ideal $I = (a)$ and an element $f \in \mathfrak m$, $f \not\in I$ with the following property: the ring $A_{\{ f\} }[1/a]$ is nonreduced. Here $A_{\{ f\} }$ is the $I$-adic completion $(A_ f)^\wedge $ of the principal localization $A_ f$. To be sure the ring $A_{\{ f\} }[1/a]$ is nonzero. Let $B = A_{\{ f\} }/ \text{nil}(A_{\{ f\} })$ be the quotient by its nilradical. Observe that $A \to B$ is adic and topologically of finite type. In fact, $B$ is a quotient of $A\{ x\} = A[x]^\wedge $ by the map sending $x$ to the image of $1/f$ in $B$. Every prime $\mathfrak q$ of $B$ not containing $a$ must lie over $(0) \subset A$1. Hence $B_\mathfrak q$ is flat over $A$ as it is a module over the fraction field of $A$. Thus $A \to B$ is naively rig-flat. On the other hand, the map

\[ A_{\{ f\} } \longrightarrow B_{\{ f\} } = (B_ f)^\wedge = B = A_{\{ f\} } /\text{nil}(A_{\{ f\} }) \]

is not flat after inverting $a$ because we get the nontrivial surjection $A_{\{ f\} }[1/a] \to A_{\{ f\} }[1/a]/\text{nil}(A_{\{ f\} }[1/a])$. Hence $A_{\{ f\} } \to B_{\{ f\} }^\wedge $ is not naively rig-flat!

It turns out that it is easy to work around this problem by using the following definition.

Definition 86.15.4. Let $\varphi : A \to B$ be a continuous ring homomorphism between adic Noetherian topological rings, i.e., $\varphi $ is an arrow of $\textit{WAdm}^{Noeth}$. We say $\varphi $ is rig-flat if $\varphi $ is adic, topologically of finite type, and for all $f \in A$ the induced map

\[ A_{\{ f\} } \longrightarrow B_{\{ f\} } \]

is naively rig-flat (Definition 86.15.2).

Setting $f = 1$ in the definition above we see that rig-flatness implies naive rig-flatness. The example shows the converse is false. However, in many situations we don't need to worry about the difference between rig-flatness and its naive version as the next lemma shows.

Lemma 86.15.5. Let $\varphi : A \to B$ be an arrow of $\textit{WAdm}^{Noeth}$. If $A/I$ is Jacobson for some (equivalently any) ideal of definition $I \subset A$ and $\varphi $ is naively rig-flat, then $\varphi $ is rig-flat.

Proof. Assume $\varphi $ is naively rig-flat. We first state some obvious consequences of the assumptions. Namely, let $f \in A$. Then $A, B, A_{\{ f\} }, B_{\{ f\} }$ are Noetherian adic topological rings. The maps $A \to A_{\{ f\} } \to B_{\{ f\} }$ and $A \to B \to B_{\{ f\} }$ are adic and topologically of finite type. The ring maps $A \to A_{\{ f\} }$ and $B \to B_{\{ f\} }$ are flat as compositions of $A \to A_ f$ and $B \to B_ f$ and the completion maps which are flat by Algebra, Lemma 10.97.2. The quotients of each of the rings $A, B, A_{\{ f\} }, B_{\{ f\} }$ by $I$ is of finite type over $A/I$ and hence Jacobson too (Algebra, Proposition 10.35.19).

Let $\mathfrak q' \subset B_{\{ f\} }$ be rig-closed. It suffices to prove that $(B_{\{ f\} })_{\mathfrak q'}$ is flat over $A_{\{ f\} }$, see Lemma 86.15.1. By Lemma 86.14.5 the primes $\mathfrak q \subset B$ and $\mathfrak p' \subset A_{\{ f\} }$ and $\mathfrak p \subset A$ lying under $\mathfrak q'$ are rig-closed. We are going to apply Algebra, Lemma 10.100.2 to the diagram

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

with $M = B_\mathfrak q$. The only assumption that hasn't been checked yet is the fact that $\mathfrak p$ generates the maximal ideal of $(A_{\{ f\} })_{\mathfrak p'}$. This follows from Lemma 86.14.8; here we use that $\mathfrak p$ and $\mathfrak p'$ are rig-closed to see that $f$ maps to a unit of $A/\mathfrak p$ (this is the only step in the proof that fails without the Jacobson assumption). Namely, this tells us that $A/\mathfrak p \to A_{\{ f\} }/\mathfrak p'$ is a finite inclusion of local rings (Lemma 86.14.4) and $f$ maps to a unit in the second one. $\square$

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. $\square$

Lemma 86.15.7. Consider a commutative diagram

\[ \xymatrix{ B \ar[r] & B' \\ A \ar[r] \ar[u]^\varphi & A' \ar[u]_{\varphi '} } \]

in $\textit{WAdm}^{Noeth}$ with all arrows adic and topologically of finite type. Assume $A \to A'$ and $B \to B'$ are flat. Let $I \subset A$ be an ideal of definition. If $\varphi $ is rig-flat and $A/I \to A'/IA'$ is étale, then $\varphi '$ is rig-flat.

Proof. Given $f \in A'$ the assumptions of the lemma remain true for the digram

\[ \xymatrix{ B \ar[r] & (B')_{\{ f\} } \\ A \ar[r] \ar[u]^\varphi & (A')_{\{ f\} } \ar[u] } \]

Hence it suffices to prove that $\varphi '$ is naively rig-flat.

Take a rig-closed prime ideal $\mathfrak q' \subset B'$. We have to show that $(B')_{\mathfrak q'}$ is flat over $A'$. We can choose an $f \in A$ which maps to a unit of $B'/\mathfrak q'$ such that the induced prime ideal $\mathfrak p''$ of $A_{\{ f\} }$ is rig-closed, see Lemma 86.14.9. To be precise, here $\mathfrak q'' = \mathfrak q' B'_{\{ f\} }$ and $\mathfrak p'' = A_{\{ f\} } \cap \mathfrak q''$. Consider the diagram

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

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 $B'_{\{ f\} }$ is the completion of a localization of the Noetherian ring $B'_ f$. Similarly the bottom horizontal arrow is flat. Hence it suffices to prove that the right vertical arrow is flat. Finally, all the assumptions of the lemma remain true for the diagram

\[ \xymatrix{ B_{\{ f\} } \ar[r] & B'_{\{ f\} } \\ A_{\{ f\} } \ar[r] \ar[u] & A'_{\{ f\} } \ar[u] } \]

This reduces us to the case discussed in the next paragraph.

Take a rig-closed prime ideal $\mathfrak q' \subset B'$ and assume $\mathfrak p = A \cap \mathfrak q'$ is rig-closed as well. This implies also the primes $\mathfrak q = B \cap \mathfrak q'$ and $\mathfrak p' = A' \cap \mathfrak q'$ are rig-closed, see Lemma 86.14.4. We are going to apply Algebra, Lemma 10.100.2 to the diagram

\[ \xymatrix{ B_\mathfrak q \ar[r] & B'_{\mathfrak q'} \\ A_\mathfrak p \ar[u] \ar[r] & A'_{\mathfrak p'} \ar[u] } \]

with $M = B_\mathfrak q$. The only assumption that hasn't been checked yet is the fact that $\mathfrak p$ generates the maximal ideal of $A'_{\mathfrak p'}$. This follows from Lemma 86.14.11. $\square$

Lemma 86.15.8. Consider a commutative diagram

\[ \xymatrix{ B \ar[r] & B' \\ A \ar[r] \ar[u]^\varphi & A' \ar[u]_{\varphi '} } \]

in $\textit{WAdm}^{Noeth}$ with all arrows adic and topologically of finite type. Assume $A \to A'$ flat and $B \to B'$ faithfully flat. If $\varphi '$ is rig-flat, then $\varphi $ is rig-flat.

Proof. Given $f \in A$ the assumptions of the lemma remain true for the digram

\[ \xymatrix{ B_{\{ f\} } \ar[r] & (B')_{\{ f\} } \\ A_{\{ f\} } \ar[r] \ar[u]^\varphi & (A')_{\{ f\} } \ar[u] } \]

(To check the condition on faithful flatness: faithful flatness of $B \to B'$ is equivalent to $B \to B'$ being flat and $\mathop{\mathrm{Spec}}(B'/IB') \to \mathop{\mathrm{Spec}}(B/IB)$ being surjective for some ideal of definition $I \subset A$.) Hence it suffices to prove that $\varphi $ is naively rig-flat. However, we know that $\varphi '$ is naively rig-flat and that $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(B)$ is surjective. From this the result follows immediately. $\square$

Finally, we can show that rig-flatness is a local property.

Lemma 86.15.9. The property $P(\varphi )=$“$\varphi $ is rig-flat” on arrows of $\textit{WAdm}^{Noeth}$ is a local property as defined in Formal Spaces, Remark 85.17.4.

Proof. Let us recall what the statement signifies. First, $\textit{WAdm}^{Noeth}$ is the category whose objects are adic Noetherian topological rings and whose morphisms are continuous ring homomorphisms. Consider a commutative diagram

\[ \xymatrix{ B \ar[r] & (B')^\wedge \\ A \ar[r] \ar[u]^\varphi & (A')^\wedge \ar[u]_{\varphi '} } \]

satisfying the following conditions: $A$ and $B$ are adic Noetherian topological rings, $A \to A'$ and $B \to B'$ are étale ring maps, $(A')^\wedge = \mathop{\mathrm{lim}}\nolimits A'/I^ nA'$ for some ideal of definition $I \subset A$, $(B')^\wedge = \mathop{\mathrm{lim}}\nolimits B'/J^ nB'$ for some ideal of definition $J \subset B$, and $\varphi : A \to B$ and $\varphi ' : (A')^\wedge \to (B')^\wedge $ are continuous. Note that $(A')^\wedge $ and $(B')^\wedge $ are adic Noetherian topological rings by Formal Spaces, Lemma 85.17.1. We have to show

  1. $\varphi $ is rig-flat $\Rightarrow \varphi '$ is rig-flat,

  2. if $B \to B'$ faithfully flat, then $\varphi '$ is rig-flat $\Rightarrow \varphi $ is rig-flat, and

  3. if $A \to B_ i$ is rig-flat for $i = 1, \ldots , n$, then $A \to \prod _{i = 1, \ldots , n} B_ i$ is rig-flat.

Being adic and topologically of finite type satisfies conditions (1), (2), and (3), see Lemma 86.11.1. Thus in verifying (1), (2), and (3) for the property “rig-flat” we may already assume our ring maps are all adic and topologically of finite type. Then (1) and (2) follow from Lemmas 86.15.7 and 86.15.8. We omit the trivial proof of (3). $\square$

Lemma 86.15.10. The property $P(\varphi )=$“$\varphi $ is rig-flat” on arrows of $\textit{WAdm}^{Noeth}$ is stable under composition as defined in Formal Spaces, Remark 85.17.14.

Proof. The statement makes sense by Lemma 86.15.9. To see that it is true assume we have rig-flat morphisms $A \to B$ and $B \to C$ in $\textit{WAdm}^{Noeth}$. Then $A \to C$ is adic and topologically of finite type by Lemma 86.11.4. To finish the proof we have to show that for all $f \in A$ the map $A_{\{ f\} } \to C_{\{ f\} }$ is naively rig-flat. Since $A_{\{ f\} } \to B_{\{ f\} }$ and $B_{\{ f\} } \to C_{\{ f\} }$ are naively rig-flat, it suffices to show that compositions of naively rig-flat maps are naively rig-flat. This is a consequence of Algebra, Lemma 10.39.4. $\square$

[1] Namely, we can find $\mathfrak q \subset \mathfrak q' \subset B$ with $a \in \mathfrak q'$ because $B$ is $a$-adically complete. Then $\mathfrak p' = A \cap \mathfrak q'$ contains $a$ but not $f$ hence is a height $1$ prime. Then $\mathfrak p = A \cap \mathfrak q$ must be strictly contained in $\mathfrak p'$ as $a \not\in \mathfrak p$. Since $\dim (A) = 2$ we see that $\mathfrak p = (0)$.

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