# Computation of completion of a local ring

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 $$n\geq 3$$.

• 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? – Youngsu Feb 13 at 16:23
• stacks.math.columbia.edu/tag/0C0S Here it states that it is sufficient for it contain any field – Notone Feb 13 at 16:48
• 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). – Youngsu Feb 13 at 21:09

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 $$A$$.

• 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! – Notone Feb 14 at 15:47