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Consider the local rings

$$R = \mathbb{C}[[x,y,z,w]]/\langle xyz+xyw+xzw+yzw\rangle$$

and

$$S = \mathbb{C}[[x,y,z,w]]/\langle xyz+xyw+xzw+yzw+xyzw\rangle.$$

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.

Some history: In a previous post (Two rings...are they isomorphic?), I asked a version of this question with one fewer variable, and Bjorn Poonen pointed out that the rings are isomorphic because there is only one kind of rational double point. I think this was essentially an accident, hence the new post.

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    $\begingroup$ Your rings have different dimensions and hence are not isomorphic. Is there perhaps a typo in the question? $\endgroup$
    – Steven Landsburg
    Commented Nov 12, 2014 at 14:22
  • $\begingroup$ Ack, sorry! I introduced some typos while translating between the previous version and this one. Both are supposed to be local rings at the origin on singular hypersurfaces in $\mathbb{C}^4$. The first hypersurface is homogeneous of degree 3, while the second has an extra term of degree 4. Hence the first hypersurface is isomorphic to the tangent cone (at the origin) of the second hypersurface. $\endgroup$
    – Nicholas Proudfoot
    Commented Nov 12, 2014 at 14:49
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    $\begingroup$ Anyway, already for isolated singularities, it is not true in general that the analytic (or formal) type of the singularity is the same of the type of the tangent cone. For instance, take $$\mathbb{C}[[x, y, z, w]]/(x^3+y^3+z^3+w^3), \quad \mathbb{C}[[x, y, z, w]]/(x^3+y^3+z^3+w^3+xyzw).$$ Then these two germs are not isomorphic: in fact they have the same Milnor number ($16$) but different Tjurina number ($16$ and $15$, respectively). $\endgroup$
    – Francesco Polizzi
    Commented Nov 12, 2014 at 15:02
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    $\begingroup$ That's correct. I don't think that there will be any general statement that gets me the isomorphism for free. If the rings are isomorphic (which I hope that they are), I think it will involve some trick that is specific to this example. (Of course, I hope that it will generalize to other similar examples that arise in the problem that I'm considering). $\endgroup$
    – Nicholas Proudfoot
    Commented Nov 12, 2014 at 15:07
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    $\begingroup$ @joro: I'm a little confused; does it make sense to talk about the zeta series when you are dealing with a local ring? I thought that the zeta function had to do with counting points, and these guys each have only one point. (I definitely agree that, if they were quotients of polynomial rings, these two rings would not be isomorphic!) $\endgroup$
    – Nicholas Proudfoot
    Commented Nov 13, 2014 at 16:01

3 Answers 3

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Consider the map $\varphi:S\to R$ given by $$\varphi(x) = \frac{4x}{4-x},$$ and similarly for $y$, $z$, and $w$. One needs to check that this is well-defined; indeed, $\varphi$ maps $xyz+xyw+xzw+yzw+xyzw$ to $$\frac{4^4(xyz+xyw+xzw+yzw)}{(4-x)(4-y)(4-z)(4-w)}.$$ Since $\varphi$ obviously induces an isomorphism on the associated graded of the filtration by powers of the maximal ideal, it is itself an isomorphism.
More explicitly, the inverse homomorphism $\psi:R\to S$ is given by $$\psi(x) = \frac{4x}{4+x},$$ and similarly for $y$, $z$, and $w$.

Note also that this generalizes to any number of variables.

Thanks David and Vladimir for your answers; that was really helpful for getting me going!

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    $\begingroup$ @WłodzimierzHolsztyński: While of course one can overdo this, I think that what the OP does here is totally fine, and I see no reason to criticize him for it (implicitly or otherwise). $\endgroup$
    – Andy Putman
    Commented Nov 16, 2014 at 7:32
  • $\begingroup$ @WłodzimierzHolsztyński: I agree with Andy: I believe that the OP is quite right to have handled it like this once he figured out a complete solution (in case hearing this from one of the people who wrote incomplete answers above is more convincing). $\endgroup$
    – Vladimir Dotsenko
    Commented Nov 16, 2014 at 14:19
  • $\begingroup$ @NicholasProudfoot: this is a very nice argument, albeit somewhat specific for the problem you wanted to solve. Do you now have a feeling as to whether the condition that arose in both the approach of David Speyer and the approach I outlined (absence of $x_i^n$) can be maintained inductively if it was the case in the beginning? $\endgroup$
    – Vladimir Dotsenko
    Commented Nov 16, 2014 at 14:23
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    $\begingroup$ In the more general situation that I care about, there will be a number of relations like this, with coefficients, each involving some subset of the variables. I realized that if I need to construct an isomorphism that deals with each of the equations simultaneously, it would be really helpful if $\varphi(x_i)$ were a power series in only $x_i$. Furthermore, because of the symmetry, I should try the same power series in each variable. Once I restricted the problem this much, it was easy to work out the first few terms when $n=3$ and guess the general solution. $\endgroup$
    – Nicholas Proudfoot
    Commented Nov 16, 2014 at 16:42
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    $\begingroup$ It turns out that there's an even easier isomorphism that works in any characteristic: just send $x_1$ to $\frac{x_1}{1+x_1}$, and send every other $x_i$ to itself! (This is the map from $R$ to $S$.) $\endgroup$
    – Nicholas Proudfoot
    Commented Nov 19, 2014 at 0:19
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This is very similar to Vladimir Dotsenko's approach.

The natural approach (at least, the one both he and I took) is to make a sequence of changes of variable, of the form $$\begin{array}{rcl} w_{n} &=& w_{n+1}+a_{n+1}(w_{n+1},x_{n+1},y_{n+1},z_{n+1}) \\ x_{n} &=& x_{n+1}+b_{n+1}(w_{n+1},x_{n+1},y_{n+1},z_{n+1}) \\ y_{n} &=& y_{n+1}+c_{n+1}(w_{n+1},x_{n+1},y_{n+1},z_{n+1}) \\ z_{n} &=& z_{n+1}+d_{n+1}(w_{n+1},x_{n+1},y_{n+1},z_{n+1}) \\ \end{array}$$ where $(a_n, b_n, c_n, d_n)$ have degree $n$ so that the polynomial $w_1 x_1 y_1 + w_1 x_1 z_1 + w_1 y_1 z_1 + x_1 y_1 z_1 + w_1 x_1 y_1 z_1$ becomes $w_n x_n y_n + w_n x_n z_n + w_n y_n z_n + x_n y_n z_n + \Delta$ for $\Delta$ of higher and higher degree.

Thus, the key computation is the following:

If we have $wxy+wxz+wyz+xyz+\Delta$, with $\Delta$ in degrees $\geq n$, when can we make a change of variables which eliminates the degree $n$ part of $\Delta$? The answer is the following: if and only if $\Delta$ has no $w^n$, $x^n$, $y^n$ or $z^n$ term.

Proof: Let $\Delta_n$ be the degree $n$ part of $\Delta$. We can eliminate it if and only if $\Delta_n$ is of the form $$(wx+wy+xy) d + (wx+wz+xz) c + (wy+wz+yz) b+ (wx+wy+xy) a$$ for $a$, $b$, $c$, $d$ of degree $n-2$. In other words, if and only if $\Delta_n$ is in the ideal $$I:= \langle wx+wy+xy, wx+wz+xz, wy+wz+yz, wx+wy+xy \rangle.$$ It is obvious that $I$ is contained in $$J: = \langle wx, wy, wz, xy, xz, wz \rangle.$$ I verified by explicit linear algebra that the degree $3$ piece of $I$ has dimension $31$, as does the degree $3$ piece of $J$. So $I$ and $J$ are equal in degree $3$, and hence in all higher degrees. The degree $n$ part of $J$ is precisely the polynomials with no $w^n$, $x^n$, $y^n$ or $z^n$ term.

Thus, to win, we must show that we can keep clearing away the lowest remaining terms without creating $w^n$, $x^n$, $y^n$ or $z^n$ in higher degrees. It isn't clear to me whether or not this is possible.

Further thoughts: the germ $wxy+wxz+wyz+xyz$ is singular along $x=y=z=0$ and the permutations thereof, and likewise for $WXY+WXZ+WYZ+XYX+WXYZ$. It seems to me that this should imply that the change of variables should take $x=y=z=0$ to $X=Y=Z=0$, and the same for the other four combinations of coordinates. This should mean that $W-w$ is in the ideal generated by $\langle w, xy, xz, yz \rangle$. Imposing these restrictions on $a$, $b$, $c$, $d$ gives a new ideal $I'$, which is only $19$ dimensional in degree $4$ (so much less than $J$.) But $wxyz$ is in $I'$, so we can still make the first change of variable.

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  • $\begingroup$ This idea of comparing $I$ and $J$ is absolutely lovely. I was too concentrated on the symmetry argument to spot it. I still think that my argument can be pushed further to justify absence of pure $n$-th powers, maybe I will have a clean reasoning in the morning :) $\endgroup$
    – Vladimir Dotsenko
    Commented Nov 12, 2014 at 22:38
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This is a bit too long for a comment.

Let me denote by $L_k=\sum_i x_i^{k+1}\frac{\partial\phantom{x_i}}{\partial x_i}$ the operators via which vector fields on a line act on polynomials in several variables (via the diagonal embedding). I noticed that if you take polynomials in $n$ variables, then the elementary symmetric function $e_n$ can be obtained as a combination of $L_1(e_{n-1})$ and $(x_1+\cdots+x_n)L_0(e_{n-1})$. Indeed, $L_1(e_{n-1})$ is the monomial symmetric function $m_{2,1^{n-1}}$, and $(x_1+\cdots+x_n)L_0(e_{n-1})=(x_1+\cdots+x_n)(n-1)e_{n-1}$ is $(n-1)(m_{2,1^{n-1}}+ne_n)$. Therefore, $e_n=\frac{1}{n(n-1)}((x_1+\cdots+x_n)L_0-(n-1)L_1)(e_{n-1})$, so $e_n$ can be obtained as the action of a vector field (infinitesimal diffeomorphisms) on $e_{n-1}$.

This means that in the question as stated you can at least use the action of diffeomorphisms to eliminate $x_1x_2x_3x_4$ at the expense of arising higher terms (I intentionally performed the computation for any $n$ since you want it to be generalizable).

Now specifically for $n=4$: for the next step, we have some terms of degree $5$ to kill. By construction, these terms are symmetric polynomials so they all live in a vector space of dimension $6$ (of symmetric polynomials in $4$ variables of degree $5$). At the same time, if we take "obvious" infinitesimal diffeomorphisms, that is vector fields which are combinations of $L_i$'s with coefficients being symmetric polynomials, which increase degree by $2$, then they form a $7$-dimensional space spanned by $L_2$, $e_1L_1$, $e_2L_0$, $e_1^2L_0$, $e_3L_{-1}$, $e_1e_2L_{-1}$, $e_1^3L_{-1}$. I think it is 7-dimensional, but I did not check very carefully if these are linearly independent. I also did not check carefully what happens when we apply this to $e_3$, but my feeling is that the only thing which is not hit by the action of these is scalar multiples of $\sum_i x_i^5$, and these will not appear in the infinitesimal action killing $x_1x_2x_3x_4$ either.

My impression is that this then can propagate: at each step the only thing that is not hit by appropriate infinitesimal diffeomorphisms consists of scalar multiples of $\sum_i x_i^k$, and these never have to be eliminated.

I am sorry for not filling in the details, but I have to finish a lot of stuff today (this lovely question already distracted me more than it should have!)

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    $\begingroup$ Thanks, this is a great idea! In fact, I would be happy producing an etale map from Spec(R) to Spec(S). Since etale is equivalent to inducing an isomorphism on tangent cones, any homomorphism from R to S of the form $x_i\mapsto x_i + \text{higher order}$ would do the trick. When $n=2$, your approach of sending $x_i$ to $x_i + \frac{1}{n(n-1)} e_1 x_i - \frac{1}{n}x_i^2$ does this perfectly on the first try. When $n=3$, this doesn't quite work; a second step is required. But the idea seems very promising! $\endgroup$
    – Nicholas Proudfoot
    Commented Nov 12, 2014 at 22:00

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