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 $n\geq 3$.
Best Answer
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 $A$.