# The local rings of $xy=0$ and $xy+x^3+y^3=0$ are not isomorphic, but have isomorphic completions?

I know that if you have a commutative local ring $R$, and you take its completion $\widehat{R}$ the inverse limit of the $R/\mathfrak{m}^i$, you get another local ring. However, nonisomorphic local rings might have isomorphic completions. I'm a little mystified by an example.

Assume everything takes place over an algebraically closed field. Consider the two curves $xy=0$ and $xy+x^3+y^3=0$. I'll denote these two curves by $C_1$ and $C_2$, respectively, and let $O=(0,0)$. I suspect there is some bizarre behavior near the origin.

Is there a nice way to see that the corresponding local rings $\mathcal{O}_{O,C_1}$ and $\mathcal{O}_{O,C_2}$ are not isomorphic, but that their completions $\widehat{\mathcal{O}}_{O,C_1}$ and $\widehat{\mathcal{O}}_{O,C_2}$ are indeed isomorphic? Thank you.

• The second ring is an integral domain, so they can't be isomorphic. Commented Dec 11, 2014 at 23:45
• By the way, the standard example is $y^2 = x^3 + x^2$. Maybe try that local computation first?
– Hoot
Commented Dec 12, 2014 at 0:31
• I think step one is showing that $xy+x^3+y^3$ factors into $(x+\dots)(y+\dots)$ over the completion. Commented Dec 12, 2014 at 6:08

It is well known in the algebraic geometry of curves that at a node the completion is isomorphic to $K[[X,Y]]/(XY)$ (sometimes taken as the definition). Your curve $x^3+xy+y^3=0$ has a node at $(0,0)$ and this can be easily verified; see this in order to learn what to do.
If you want a purely algebraic approach, then follow the suggestion of Hugh Thomas and try to write $X^3+XY+Y^3$ in $K[[X,Y]]$ as $(X+f_2+f_3+\cdots)(Y+g_2+g_3+\cdots)$ with $f_i,g_i$ homogeneous of degree $i$.