## 109.75 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 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 [§6.7, 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 37.22.5. Part (2) was proven in ; 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 109.75.1. 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 $f \mapsto f^ p - f$.

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).