The Stacks project

Lemma 109.14.2. There exists a local homomorphism of local rings $A \to B$ and a regular sequence $x, y$ in the maximal ideal of $B$ such that $B/(x, y)$ is flat over $A$, but such that the images $\overline{x}, \overline{y}$ of $x, y$ in $B/\mathfrak m_ AB$ do not form a regular sequence, nor even a Koszul-regular sequence.

Proof. Set $A = k[z]_{(z)}$ and let $B = (k[x, y, z] \oplus E)_{(x, y, z, E)}$. Since $x, y, z$ is a regular sequence in $B$, see proof of Lemma 109.14.1, we see that $x, y$ is a regular sequence in $B$ and that $B/(x, y)$ is a torsion free $A$-module, hence flat. On the other hand, there exists a nonzero element $\delta \in B/\mathfrak m_ AB = B/zB$ which is annihilated by $\overline{x}, \overline{y}$. Hence $H_2(K_\bullet (B/\mathfrak m_ AB, \overline{x}, \overline{y})) \not= 0$. Thus $\overline{x}, \overline{y}$ is not Koszul-regular, in particular it is not a regular sequence, see More on Algebra, Lemma 15.30.2. $\square$

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Your email address will not be published.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 0641. Beware of the difference between the letter 'O' and the digit '0'.