The Stacks project

110.8 Noncomplete quotient

Let $k$ be a field. Let

Note that in particular $z_ iz_ jt = 0$ in this ring. Any element $f$ of $R$ can be uniquely written as a finite sum

where each $f_ i \in k[t, z_ i, w_ j]$ has no terms involving the products $z_ it$ or $z_ iw_ j$. Moreover, if $f$ is written in this way, then $f \in (x^ n)$ if and only if $f_ i = 0$ for $i < n$. So $x$ is a nonzerodivisor and $\bigcap (x^ n) = 0$. Let $R^\wedge$ be the completion of $R$ with respect to the ideal $(x)$. Note that $R^\wedge$ is $(x)$-adically complete, see Algebra, Lemma 10.96.3. By the above we see that an element of $R^\wedge$ can be uniquely written as an infinite sum

where each $f_ i \in k[t, z_ i, w_ j]$ has no terms involving the products $z_ it$ or $z_ iw_ j$. Consider the element

i.e., we have $f_ n = w_ n$. Note that $f \in (t , x^ n)$ for every $n$ because $x^ mw_ m \in (t)$ for all $m$. We claim that $f \not\in (t)$. To prove this assume that $tg = f$ where $g = \sum g_ lx^ l$ in canonical form as above. Since $tz_ iz_ j = 0$ we may as well assume that none of the $g_ l$ have terms involving the products $z_ iz_ j$. Examining the process to get $tg$ in canonical form we see the following: Given any term $c m$ of $g_ l$ where $c \in k$ and $m$ is a monomial in $t, z_ i, w_ j$ and we make the following replacement

1. if the monomial $m$ does not involve any $z_ i$, then $ctm$ is a term of $f_ l$, and

2. if the monomial $m$ does involve a $z_ i$ then it is equal to $m = z_ i$ and we see that $cw_ i$ is term of $f_{l + i}$.

Since $g_0$ is a polynomial only finitely many of the variables $z_ i$ occur in it. Pick $n$ such that $z_ n$ does not occur in $g_0$. Then the rules above show that $w_ n$ does not occur in $f_ n$ which is a contradiction. It follows that $R^\wedge /(t)$ is not complete, see Algebra, Lemma 10.96.10.