Lemma 110.35.1. There exists a local ring $R$ with a unique prime ideal and a nonzero ideal $I \subset R$ which is a flat $R$-module

## 110.35 Zero dimensional local ring with nonzero flat ideal

In [Lazard] and [Autour] there is an example of a zero dimensional local ring with a nonzero flat ideal. Here is the construction. Let $k$ be a field. Let $X_ i, Y_ i$, $i \geq 1$ be variables. Take $R = k[X_ i, Y_ i]/(X_ i - Y_ i X_{i + 1}, Y_ i^2)$. Denote $x_ i$, resp. $y_ i$ the image of $X_ i$, resp. $Y_ i$ in this ring. Note that

in this ring. The ring $R$ has only one prime ideal, namely $\mathfrak m = (x_ i, y_ i)$. We claim that the ideal $I = (x_ i)$ is flat as an $R$-module.

Note that the annihilator of $x_ i$ in $R$ is the ideal $(x_1, x_2, x_3, \ldots , y_ i, y_{i + 1}, y_{i + 2}, \ldots )$. Consider the $R$-module $M$ generated by elements $e_ i$, $i \geq 1$ and relations $e_ i = y_ i e_{i + 1}$. Then $M$ is flat as it is the colimit $\mathop{\mathrm{colim}}\nolimits _ i R$ of copies of $R$ with transition maps

Note that the annihilator of $e_ i$ in $M$ is the ideal $(x_1, x_2, x_3, \ldots , y_ i, y_{i + 1}, y_{i + 2}, \ldots )$. Since every element of $M$, resp. $I$ can be written as $f e_ i$, resp. $h x_ i$ for some $f, h \in R$ we see that the map $M \to I$, $e_ i \to x_ i$ is an isomorphism and $I$ is flat.

**Proof.**
See discussion above.
$\square$

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