Lemma 110.37.1. There exists a local ring $A$, a finite type ring map $A \to B$ and a prime $\mathfrak q$ lying over $\mathfrak m_ A$ such that $B_{\mathfrak q}$ is flat over $A$, and for any element $g \in B$, $g \not\in \mathfrak q$ the ring $B_ g$ is neither finitely presented over $A$ nor flat over $A$.
110.37 Finite type, not finitely presented, flat at prime
Let $k$ be a field. Consider the local ring $A_0 = k[x, y]_{(x, y)}$. Denote $\mathfrak p_{0, n} = (y + x^ n + x^{2n + 1})$. This is a prime ideal. Set
\[ A = A_0[z_1, z_2, z_3, \ldots ]/(z_ n z_ m, z_ n(y + x^ n + x^{2n + 1})) \]
Note that $A \to A_0$ is a surjection whose kernel is an ideal of square zero. Hence $A$ is also a local ring and the prime ideals of $A$ are in one-to-one correspondence with the prime ideals of $A_0$. Denote $\mathfrak p_ n$ the prime ideal of $A$ corresponding to $\mathfrak p_{0, n}$. Observe that $\mathfrak p_ n$ is the annihilator of $z_ n$ in $A$. Let
\[ C = A[z]/(xz^2 + z + y)[\frac{1}{2zx + 1}] \]
Note that $A \to C$ is an étale ring map, see Algebra, Example 10.137.8. Let $\mathfrak q \subset C$ be the maximal ideal generated by $x$, $y$, $z$ and all $z_ n$. As $A \to C$ is flat we see that the annihilator of $z_ n$ in $C$ is $\mathfrak p_ nC$. We compute
\begin{align*} C/\mathfrak p_ n C & = A_0[z]/(xz^2 + z + y, y + x^ n + x^{2n + 1})[1/(2zx + 1)] \\ & = k[x]_{(x)}[z]/(xz^2 + z - x^ n - x^{2n + 1})[1/(2zx + 1)] \\ & = k[x]_{(x)}[z]/(z - x^ n) \times k[x]_{(x)}[z]/(xz + x^{n + 1} + 1)[1/(2zx + 1)] \\ & = k[x]_{(x)} \times k(x) \end{align*}
because $(z - x^ n)(xz + x^{n + 1} + 1) = xz^2 + z - x^ n - x^{2n + 1}$. Hence we see that $\mathfrak p_ nC = \mathfrak r_ n \cap \mathfrak q_ n$ with $\mathfrak r_ n = \mathfrak p_ nC + (z - x^ n)C$ and $\mathfrak q_ n = \mathfrak p_ nC + (xz + x^{n + 1} + 1)C$. Since $\mathfrak q_ n + \mathfrak r_ n = C$ we also get $\mathfrak p_ nC = \mathfrak r_ n \mathfrak q_ n$. It follows that $\mathfrak q_ n$ is the annihilator of $\xi _ n = (z - x^ n)z_ n$. Observe that on the one hand $\mathfrak r_ n \subset \mathfrak q$, and on the other hand $\mathfrak q_ n + \mathfrak q = C$. This follows for example because $\mathfrak q_ n$ is a maximal ideal of $C$ distinct from $\mathfrak q$. Similarly we have $\mathfrak q_ n + \mathfrak q_ m = C$ for $n \not= m$. At this point we let
\[ B = \mathop{\mathrm{Im}}(C \longrightarrow C_{\mathfrak q}) \]
We observe that the elements $\xi _ n$ map to zero in $B$ as $xz + x^{n + 1} + 1$ is not in $\mathfrak q$. Denote $\mathfrak q' \subset B$ the image of $\mathfrak q$. By construction $B$ is a finite type $A$-algebra, with $B_{\mathfrak q'} \cong C_{\mathfrak q}$. In particular we see that $B_{\mathfrak q'}$ is flat over $A$.
We claim there does not exist an element $g' \in B$, $g' \not\in \mathfrak q'$ such that $B_{g'}$ is of finite presentation over $A$. We sketch a proof of this claim. Choose an element $g \in C$ which maps to $g' \in B$. Consider the map $C_ g \to B_{g'}$. By Algebra, Lemma 10.6.3 we see that $B_ g$ is finitely presented over $A$ if and only if the kernel of $C_ g \to B_{g'}$ is finitely generated. But the element $g \in C$ is not contained in $\mathfrak q$, hence maps to a nonzero element of $A_0[z]/(xz^2 + z + y)$. Hence $g$ can only be contained in finitely many of the prime ideals $\mathfrak q_ n$, because the primes $(y + x^ n + x^{2n + 1}, xz + x^{n + 1} + 1)$ are an infinite collection of codimension 1 points of the 2-dimensional irreducible Noetherian space $\mathop{\mathrm{Spec}}(k[x, y, z]/(xz^2 + z + y))$. The map
\[ \bigoplus \nolimits _{g \not\in \mathfrak q_ n} C/\mathfrak q_ n \longrightarrow C_ g, \quad (c_ n) \longrightarrow \sum c_ n \xi _ n \]
is injective and its image is the kernel of $C_ g \to B_{g'}$. We omit the proof of this statement. (Hint: Write $A = A_0 \oplus I$ as an $A_0$-module where $I$ is the kernel of $A \to A_0$. Similarly, write $C = C_0 \oplus IC$. Write $IC = \bigoplus Cz_ n \cong \bigoplus (C/\mathfrak r_ n \oplus C/\mathfrak q_ n)$ and study the effect of multiplication by $g$ on the summands.) This concludes the sketch of the proof of the claim. This also proves that $B_{g'}$ is not flat over $A$ for any $g'$ as above. Namely, if it were flat, then the annihilator of the image of $z_ n$ in $B_{g'}$ would be $\mathfrak p_ nB_{g'}$, and would not contain $z - x^ n$.
As a consequence we can answer (negatively) a question posed in [Part I, Remarques (3.4.7) (v), GruRay]. Here is a precise statement.
Proof. See discussion above. $\square$
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Comment #1153 by John on