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Let $X=\mathrm{Spec}(\mathbb{R}[a,b]/(a^2+b^2+1))$ and consider the closed point $p=(a)$. I would like to compute the completion of $\mathcal{O}_{X,p}$ w.r.t. to its maximal ideal $\mathfrak m$.

Since $(\mathcal{O}_{X,p},\mathfrak{m}) $ is a noetherian regular local ring of dimension one with residue field $\mathbb{C}$, so will be $(\widehat{\mathcal{O}}_{X, \, p},\widehat{\mathfrak m})$ and by Cohen's structure theorem this should yield that $\widehat{\mathcal{O}}_{X, \, p}\cong \mathbb{C}[[t]]$ (I copied the argument from this MO answer).

I tried to write out the explicit isomorphism, but I fail to see how to construct a map $\mathbb{C}[[t]]\to (\mathbb{R}[a,b]/(a^2+b^2+1))_\mathfrak m)/a^n=\mathcal{O}_{X,p}/\mathfrak m^n$ for $n\geq 3$.

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    $\begingroup$ The argument you copied assumes that the ring contains its residue field. How do you show that the completion contains $\mathbb{C}$? Or is there a different argument? $\endgroup$ – Youngsu Feb 13 at 16:23
  • $\begingroup$ stacks.math.columbia.edu/tag/0C0S Here it states that it is sufficient for it contain any field $\endgroup$ – Notone Feb 13 at 16:48
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    $\begingroup$ Thank you. I did not know about this fact in detail, and the reference is certainly nice. I think @jgon's answer below explains the proof in the reference (the part where one needs to use formal smoothness to construct the inclusion). $\endgroup$ – Youngsu Feb 13 at 21:09
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The trick is to use Hensel's lemma.

Let $A=\widehat{\mathcal{O}}_{X,p}$, so I can save myself some typing. $\newcommand\mm{\mathfrak{m}}$

Based on the first two cases we generally expect that $i\mapsto b$ (ish) and $t\mapsto a$.

The first step is to find the actual root $u\in A$ of $x^2+1$ which satisfies $u\equiv b\pmod{\mm}$.

We are guaranteed such a thing by Hensel's lemma, but I suspect you'd like a procedure to compute it.

The way Hensel's lemma works is we inductively find solutions to the equation mod $\mm^n$. Here's the first few examples.

$b$ is a solution mod $\mm$, and $b$ is still a solution mod $\mm^2$. Now we want to find $b_2\in\mm^2/\mm^3$ with $(b+b_2)^2+1\equiv 0 \pmod{\mm^3}$. Consider now $$0=(b+b_2)^2+1\equiv b^2+2bb_2+1\pmod{\mm^3}. $$ Rearranging, and using that $2b$ is invertible (since we localized at $(a)$), we get $$b_2 = -\frac{b^2+1}{2b}+\mm_3.$$ Then similarly we have $$b_3 = \frac{-(b+b_2)^2}{2b} +\mm^4,$$ $$b_4 = \frac{-(b+b_2+b_3)^2}{2b} + \mm^5,$$ and so on. Ultimately we get $c = b+b_2+b_3+\cdots$, which is a valid element of the completion (since $b_n\in \mm^n/\mm^{n+1}$ for all $n$), and $c^2+1=0$.

This gives us the map $\Bbb{C}\to A$.

Sending $t\to a$, we should get the desired isomorphism $\Bbb{C}[[t]]$ to $A$.

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  • $\begingroup$ Ah, I didn't think of Hensel's lemma. Indeed I wouldn't like the explicit procedure, I was more interested in knowing how I would find such an element. Thanks very much! $\endgroup$ – Notone Feb 14 at 15:47

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