Lemma 110.15.1. There exists a local ring $R$ and a regular sequence $x, y, z$ (in the maximal ideal) such that there exists a nonzero element $\delta \in R/zR$ with $x\delta = y\delta = 0$.

## 110.15 Regular sequences and base change

We are going to construct a ring $R$ with a regular sequence $(x, y, z)$ such that there exists a nonzero element $\delta \in R/zR$ with $x\delta = y\delta = 0$.

To construct our example we first construct a peculiar module $E$ over the ring $k[x, y, z]$ where $k$ is any field. Namely, $E$ will be a push-out as in the following diagram

where the rows are short exact sequences (we dropped the outer zeros due to typesetting problems). Another way to describe $E$ is as

where $(f, g) \sim (f', g')$ if and only if there exists a $h \in k[x, y, z, y^{-1}]$ such that

We claim: (a) $x : E \to E$ is injective, (b) $y : E/xE \to E/xE$ is injective, (c) $E/(x, y)E = 0$, (d) there exists a nonzero element $\delta \in E/zE$ such that $x\delta = y\delta = 0$.

To prove (a) suppose that $(f, g)$ is a pair that gives rise to an element of $E$ and that $(xf, xg) \sim 0$. Then there exists a $h \in k[x, y, z, y^{-1}]$ such that $xf + xh \in yk[x, y, z, x^{-1}]$ and $xg - zh \in yzk[x, y, z]$. We may assume that $h = \sum a_{i, j, k}x^ iy^ jz^ k$ is a sum of monomials where only $j \leq 0$ occurs. Then $xg - zh \in yzk[x, y, z]$ implies that only $i > 0$ occurs, i.e., $h = xh'$ for some $h' \in k[x, y, z, y^{-1}]$. Then $(f, g) \sim (f + xh', g - zh')$ and we see that we may assume that $g = 0$ and $h = 0$. In this case $xf \in yk[x, y, z, x^{-1}]$ implies $f \in yk[x, y, z, x^{-1}]$ and we see that $(f, g) \sim 0$. Thus $x : E \to E$ is injective.

Since multiplication by $x$ is an isomorphism on $\frac{k[x, y, z, x^{-1}, y^{-1}]}{yk[x, y, z, x^{-1}]}$ we see that $E/xE$ is isomorphic to

and hence multiplication by $y$ is an isomorphism on $E/xE$. This clearly implies (b) and (c).

Let $e \in E$ be the equivalence class of $(1, 0)$. Suppose that $e \in zE$. Then there exist $f \in k[x, y, z, x^{-1}, y^{-1}]$, $g \in k[x, y, z, y^{-1}]$, and $h \in k[x, y, z, y^{-1}]$ such that

This is impossible: the monomial $1$ cannot occur in $zf$, nor in $xh$. On the other hand, we have $ye = 0$ and $xe = (x, 0) \sim (0, -z) = z(0, -1)$. Hence setting $\delta $ equal to the congruence class of $e$ in $E/zE$ we obtain (d).

**Proof.**
Let $R = k[x, y, z] \oplus E$ where $E$ is the module above considered as a square zero ideal. Then it is clear that $x, y, z$ is a regular sequence in $R$, and that the element $\delta \in E/zE \subset R/zR$ gives an element with the desired properties. To get a local example we may localize $R$ at the maximal ideal $\mathfrak m = (x, y, z, E)$. The sequence $x, y, z$ remains a regular sequence (as localization is exact), and the element $\delta $ remains nonzero as it is supported at $\mathfrak m$.
$\square$

Lemma 110.15.2. 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 110.15.1, 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.
$\square$

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