Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Assume that I have an affine hypersurface $X =V(f)\subset \mathbb{C}^4$ of degree $d$ with an isolated singularity of multiplicity $m$ at the origin $o=(0,0,0,0)$. Let $$f:=f_m + f_{m+1}+ \cdots +f_d$$ be the decomposition of $f$ into homogeneous pieces and suppose, in addition, that the singularity at $o$ is ordinary, that is the tangent cone $C=V(f_m) \subset \mathbb{C}^4$ is a cone over a smooth surface of degree $m$ in $\mathbb{P}^3$.

Simple examples show that, in general, the germ $(X, o)$ is not isomorphic to the germ $(C, o)$. For instance, take $$f=x_0^4+4x_0^3x_1+6x_0^2x_1^2+4x_0x_1^3+x_2^4+4x_2^3x_3+6x_2^2x_3^2+4x_2x_3^3+x_0^5+x_0x_1^4$$ $$+x_0^4x_2+7x_1^4x_2+3x_0x_2^4-x_2^5+4x_0x_3^4+11x_2x_3^4,$$ so that $$f_4=x_0^4+4x_0^3x_1+6x_0^2x_1^2+4x_0x_1^3+x_2^4+4x_2^3x_3+6x_2^2x_3^2+4x_2x_3^3.$$ Then, denoting with $\tau$ the Tjurina number, by using Singular one finds $$\tau(f, 0)=71, \quad \tau(f_4, 0)=81,$$ so the two germs are not isomorphic. This implies that the (analytic) local rings $\mathcal{O}^{an}_{X,0}, \quad \mathcal{O}^{an}_{C,0}$ are not isomorphic, see for instance [De Jong-Pfister, Local analytic geometry, page 120].

Now my question is:

Are at least the completed rings $\widehat{\mathcal{O}}^{an}_{X,0}, \quad \widehat{\mathcal{O}}^{an}_{C,0}$ isomorphic? If the answer is yes, what is a reference? If the answer is no, what is a counterexample?

My apologies in advance to the experts of singularity theory if this question turns out to be a trivial one.

EDIT. Jason Starr's comments suggest that the two completed rings should not be isomorphic. Let me then ask a weaker form of my question:

(1) Assume that the local ring of the tangent cone at the vertex $\mathcal{O}^{an}_{C,0}$ is a UFD. Can one conclude that the same is true for $\mathcal{O}^{an}_{X,0}$?

(2) Assume that the completed local ring of the tangent cone at the vertex $\widehat{\mathcal{O}}^{an}_{C,0}$ is a UFD. Can one conclude that the same is true for $\widehat{\mathcal{O}}^{an}_{X,0}$?

share|improve this question
    
I am pretty certain that Mike Artin's theory of algebraization of formal moduli implies that the completed local rings are not isomorphic. If the completed local rings are isomorphic, Artin proves that the Henselized local rings are isomorphic. However, the local rings of germs of analytic functions factor through the Henselized local rings. –  Jason Starr Dec 6 '12 at 18:43
    
In previous comment: "algebraization of formal moduli" --> "approximation of complete local rings". –  Jason Starr Dec 6 '12 at 20:09
    
@Jason. Thank you very much for your comments, I will think about them. Actually, I realized that in my situation I do not really need that the two rings are isomorphic, but only that they are both UFD. I edited the question accordingly. Please be so kind and let me know if you have any reference or suggestion in that direction. –  Francesco Polizzi Dec 7 '12 at 9:46
    
@Francesco: I am probably not the best person to ask. The best result I can find is Lemme 3.16, p. 133, SGA 2, Exp. XI (immediately following Grothendieck's proof of the Samuel Conjecture). Combined with the Rees algebra / deformation to the tangent cone (whose generic fiber, essentially, will be $X$ and whose closed fiber will be $C$), this gives the following: if $C$ is parafactorial and has depth $\geq 3$, and if also $X$ has depth $\geq 2$, then also $X$ is parafactorial. –  Jason Starr Dec 7 '12 at 15:08
    
@Jason: Thank you, I will check this. At any rate, since $X$ and $C$ are complete intersection (hence Cohen-Macauley) of dimension $3$ the condition about depth is satisfied. Since the singularity is isolated, if I can prove that $X$ is parafactorial, it seems to me that I can conclude that it is factorial by SGA2, Exp. XI, Corollaire 3.9. Or I missing something? –  Francesco Polizzi Dec 7 '12 at 17:35
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.