# Two rings…are they isomorphic?

Edit: I have reverted my question to its original version (which Bjorn Pooenen answered correctly) as requested in the comments.

Consider the local rings

$$R = \mathbb{C}[[x,y,z]]/\langle xy+xz+yz\rangle$$

and

$$S = \mathbb{C}[[x,y,z]]/\langle xy+xz+yz+xyz\rangle.$$

Is $R$ isomorphic to $S$?

Some context: I am trying to understand formal neighborhoods of points on certain varieties. I expect one answer, and I'm getting a different answer. This is the first nontrivial case where the answer that I get does not obviously agree with the answer that I expect.

• What's the dimension of $\mathfrak{m}^3/\mathfrak{m}^4$ in both cases? Not an answer, just the first thing I'd try. – Daniel Litt Nov 12 '14 at 3:21
• Both rings become isomorphic after passing to the associated graded of the filtration by powers of the maximal ideal. Geometrically, this is the statement that Spec(R) and Spec(S) have isomorphic tangent cones (at their unique closed points). – Nicholas Proudfoot Nov 12 '14 at 3:37
• It would be extremely helpful to roll back the edit, accept Bjorn's answer, and create a separate question for the new pair of rings. The way it is done now makes Bjorn's answer hanging there without any context, it is very misleading. – Vladimir Dotsenko Nov 12 '14 at 8:03
• I completely agree with Vladimir Dotsenko. – Francesco Polizzi Nov 12 '14 at 8:54
• @NicholasProudfoot, please do revert the edit and ask another question. – Mariano Suárez-Álvarez Nov 12 '14 at 8:59

More generally, if $k$ is any algebraically closed field of characteristic not $2$, and $n$ is given, then all $k$-algebras of the form $k[[x_1,\ldots,x_n]]/(f_2+f_3+\cdots)$, where each $f_i$ is homogeneous of degree $i$, and $f_2$ is a nondegenerate quadratic form, are isomorphic. (I.e., there is only one kind of ordinary double point.)
You can construct an isomorphism to $k[[x_1,\ldots,x_n]]/(x_1^2+\cdots+x_n^2)$ with your bare hands as follows. First diagonalize the quadratic form to assume that $f_2=x_1^2+\cdots+x_n^2$. Let $m$ be the lowest degree monomial of degree greater than $2$. Then $m$ is divisible by some $x_i$, say $m=x_1 g$. Performing the analytic change of variable $x_1 \mapsto x_1-g/2$ eliminates $m$ at the expense of introducing new terms of even higher degree. By iterating, one can eventually eliminate all monomials of degree 3 to obtain $f_3=0$, and then $f_4=0$, etc. The partial compositions of this sequence of analytic coordinate changes converge to a single analytic coordinate change because they stabilize modulo any given power of the maximal ideal.
• Followup to comment from user52824: And in fact, the Freitag-Kiehl assumptions on $k$ are needed only to put the quadratic form $f_2$ into a standard form. For any commutative ring $k$ and any $n$, all $k$-algebras of the form $k[[x_1,y_1,\ldots,x_n,y_n]]/(\sum_{j=1}^n x_j y_j + \sum_{i \ge 3} f_i)$ (with $f_i$ homogeneous of degree $i$) are isomorphic, by the same kind of proof as the one I sketched. – Bjorn Poonen Nov 12 '14 at 4:08
• @Qfwfq: Yes, they can. Or are you talking of something different? One can always recall the standard proof. If $q$ is your non-degenerate quadratic form over a finite dimensional $\mathbb C$-vector space $V$, you can find a $v \in V$ such that $q(v) \neq 0$ (otherwise $q=0$ and is already diagonal), and even such that $q(v)=1$. The orthogonal space $W$ of $v$ for $q$ has dimension $dim V-1$ since $q$ is non-degenerate and does not contain $v$. Hence to diagonalize $q$ on $V$ it is enough to do so by $W$, which is already done by induction. – Joël Nov 12 '14 at 15:54
• @Qfwfq: not every symmetric matrix can be diagonalised by conjugation $X\mapsto A^{-1}XA$, but the action of $GL(V)$ on quadratic forms is given by $X\mapsto A^TXA$. This action allows to proceed with Joël's argument over any field of characteristic not 2 where you can extract square roots of any element. (Characteristic is needed to have a 1-to-1 correspondence between quadratic and bilinear form to talk of orthogonal complement, square roots - to ensure $q(v)=1$). – Vladimir Dotsenko Nov 12 '14 at 17:01