The Stacks project

108.33 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 ├ętale ring map, see Algebra, Example 10.136.8. 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.

Lemma 108.33.1. 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 $A$.

Proof. See discussion above. $\square$

Comments (1)

Comment #1153 by John on

The command \romannumeral does not seem to compile in the online version. This seems a global issue, not just of this tag. (I suspect you might already be aware of this, but just in case)

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 05G1. Beware of the difference between the letter 'O' and the digit '0'.