The Stacks project

108.71 Flat maps are not directed limits of finitely presented flat maps

The goal of this section is to give an example of a flat ring map which is not a filtered colimit of flat and finitely presented ring maps. In [gabber-nonexcellent] it is shown that if $A$ is a nonexcellent local ring of dimension $1$ and residue characteristic zero, then the (flat) ring map $A \to A^\wedge $ to its completion is not a filtered colimit of finite type flat ring maps. The example in this section will have a source which is an excellent ring. We encourage the reader to submit other examples; please email stacks.project@gmail.com if you have one.

For the construction, fix a prime $p$, and let $A = \mathbf{F}_ p[x_1, \ldots , x_ n]$. Choose an absolute integral closure $A^+$ of $A$, i.e., $A^+$ is the integral closure of $A$ in an algebraic closure of its fraction field. In [ยง6.7, HHBigCM] it is shown that $A \to A^+$ is flat.

We claim that the $A$-algebra $A^+$ is not a filtered colimit of finitely presented flat $A$-algebras if $n \geq 3$.

We sketch the argument in the case $n = 3$, and we leave the generalization to higher $n$ to the reader. It is enough to prove the analogous statement for the map $R \to R^+$, where $R$ is the strict henselization of $A$ at the origin and $R^+$ is its absolute integral closure. Observe that $R$ is a henselian regular local ring whose residue field $k$ is an algebraic closure of $\mathbf{F}_ p$.

Choose an ordinary abelian surface $X$ over $k$ and a very ample line bundle $L$ on $X$. The section ring $\Gamma _*(X, L) = \bigoplus _ n H^0(X,L^ n)$ is the coordinate ring of the affine cone over $X$ with respect to $L$. It is a normal ring for $L$ sufficiently positive. Let $S$ denote the henselization of $\Gamma _*(X, L)$ at vertex of the cone. Then $S$ is a henselian Noetherian normal domain of dimension $3$. We obtain a finite injective map $R \to S$ as the henselization of a Noether normalization for the finite type $k$-algebra $\Gamma _*(X, L)$. As $R^+$ is an absolute integral closure of $R$, we can also fix an embedding $S \to R^+$. Thus $R^+$ is also the absolute integral closure of $S$. To show $R^+$ is not a filtered colimit of flat $R$-algebras, it suffices to show:

  1. If there exists a factorization $S \to P \to R^+$ with $P$ flat and finite type over $R$, then there exists a factorization $S \to T \to R^+$ with $T$ finite flat over $R$.

  2. For any factorization $S \to T \to R^+$ with $S \to T$ finite, the ring $T$ is not $R$-flat.

Indeed, since $S$ is finitely presented over $R$, if one could write $R^+ = \mathop{\mathrm{colim}}\nolimits _ i P_ i$ as a filtered colimit of finitely presented flat $R$-algebras $P_ i$, then $S \to R^+$ would factor as $S \to P_ i \to R^+$ for $i \gg 0$, which contradicts the above pair of assertions. Assertion (1) follows from the fact that $R$ is henselian and a slicing argument, see More on Morphisms, Lemma 37.21.5. Part (2) was proven in [BhattSmallCMMod]; for the convenience of the reader, we recall the argument.

Let $U \subset \mathop{\mathrm{Spec}}(S)$ be the punctured spectrum, so there are natural maps $X \leftarrow U \subset \mathop{\mathrm{Spec}}(S)$. The first map gives an identification $H^1(U, \mathcal{O}_ U) \simeq H^1(X, \mathcal{O}_ X)$. By passing to the Witt vectors of the perfection and using the Artin-Schreier sequence1, this gives an identification $H^1_{\acute{e}tale}(U, \mathbf{Z}_ p) \simeq H^1_{\acute{e}tale}(X, \mathbf{Z}_ p)$. In particular, this group is a finite free $\mathbf{Z}_ p$-module of rank $2$ (since $X$ is ordinary). To get a contradiction assume there exists an $R$-flat $T$ as in (2) above. Let $V \subset \mathop{\mathrm{Spec}}(T)$ denote the preimage of $U$, and write $f : V \to U$ for the induced finite surjective map. Since $U$ is normal, there is a trace map $f_*\mathbf{Z}_ p \to \mathbf{Z}_ p$ on $U_{\acute{e}tale}$ whose composition with the pullback $\mathbf{Z}_ p \to f_*\mathbf{Z}_ p$ is multiplication by $d = \deg (f)$. Passing to cohomology, and using that $H^1_{\acute{e}tale}(U, \mathbf{Z}_ p)$ is nontorsion, then shows that $H^1_{\acute{e}tale}(V, \mathbf{Z}_ p)$ is nonzero. Since $H^1_{\acute{e}tale}(V, \mathbf{Z}_ p) \simeq \mathop{\mathrm{lim}}\nolimits H^1_{\acute{e}tale}(V, \mathbf{Z}/p^ n)$ as there is no $R^1\mathop{\mathrm{lim}}\nolimits $ interference, the group $H^1(V_{\acute{e}tale},\mathbf{Z}/p)$ must be non-zero. Since $T$ is $R$-flat we have $\Gamma (V, \mathcal{O}_ V) = T$ which is strictly henselian and the Artin-Schreier sequence shows $H^1(V, \mathcal{O}_ V) \neq 0$. This is equivalent to $H^2_\mathfrak m(T) \neq 0$, where $\mathfrak m \subset R$ is the maximal ideal. Thus, we obtain a contradiction since $T$ is finite flat (i.e., finite free) as an $R$-module and $H^2_\mathfrak m(R) = 0$. This contradiction proves (2).

Lemma 108.71.1. There exists a commutative ring $A$ and a flat $A$-algebra $B$ which cannot be written as a filtered colimit of finitely presented flat $A$-algebras. In fact, we may either choose $A$ to be a finite type $\mathbf{F}_ p$-algebra or a $1$-dimensional Noetherian local ring with residue field of characteristic $0$.

Proof. See discussion above. $\square$

[1] Here we use that $S$ is a strictly henselian local ring of characteristic $p$ and hence $S \to S$, $f \mapsto f^ p - f$ is surjective. Also $S$ is a normal domain and hence $\Gamma (U, \mathcal{O}_ U) = S$. Thus $H^1_{\acute{e}tale}(U, \mathbf{Z}/p)$ is the kernel of the map $H^1(U, \mathcal{O}_ U) \to H^1(U, \mathcal{O}_ U)$ induced by $f \mapsto f^ p - f$.

Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0ATE. Beware of the difference between the letter 'O' and the digit '0'.