# formal completion

When I study formal completion and formal schemes, on p.194 of Hartshorne's "Algebraic Geometry", he said "However, one should note that the local rings of $\hat{X}$ are in general not complete."

But I can't see the reason for that the local rings of $\hat{X}$ are in general not complete.

Could someone explains this for me, thanks.

• Let $A=k[[x,y]]$. This is complete with respect to $m=(x,y)$.Now,localize at theprime $p=(x)$. Note that $x/y \in m_p$,but$\sum_{n=1}^{\infty} (x/y)^n \notin A_p$ – the L Feb 23 '11 at 17:18
• For which formal scheme $\hat X$ and which point $x\in\hat X$ is this $A_p$ the local ring? – user2035 Feb 25 '11 at 10:05
• I see your point, the ideal $p$ is not open... Thanks! – the L Feb 25 '11 at 10:57
• Even for open $p$, the local ring of $\mathrm{Spf}(A)$ at $p$ is not $A_p$, but rather the direct limit of the completions of $A_f$ for $f\notin p$. – user2035 Feb 25 '11 at 12:35

Consider the completion of $\mathrm{Spec}(\mathbb Q[U,V])$ along $U=0$, i.e., $\hat X=\mathrm{Spf}(\mathbb Q[[U]]\{V\})$, and let $x$ be the point $V=0$. Then $U^k/(V-k)$ for $k=1,2,\dots$ is a sequence in $\mathcal O_{\hat X,x}$ converging $U$-adically to $0$, but the corresponding series does not converge. Each element in $\mathcal O_{\hat X,x}$ comes from an open subset of $\hat X$, so is regular outside a finite subset.