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The following problem came up in something I am working on. It has a really elementary statement but I couldn't crack it in a couple of hours of thinking about it. It isn't clear to me if I am being dumb or I don't have the right tools to understand it. Here it goes, given the ring $\mathbb{C}[x_1,\ldots ,x_n]$ and n functions $f_i(x_1,\ldots, x_n)$ (quasihomogeneous if it matters) such that the ring $\mathbb{C}[x_1,\ldots,x_n]/(f_i)$ is a finite dimensional local artin algebra concentrated at the origin.

Now fix an integer d and consider $\mathbb{C}[x_1,\ldots,x_n,y_1,\ldots y_n]$ with the function $ \widetilde{f}(x_i,y_i)=(\sum f_i(x_1,\ldots x_n)y_i)+p(y_1,\ldots,y_n)$ for polynomials p only depending on y's whose lowest order terms have total degree at least d. Can we find a p such that this function $\widetilde{f}$ has only isolated singularities?

The answer is yes, for d=1 I believe. I am actually interested in knowing if for any collection of f's as above there is some d>1 for which we can solve the problem. If this turns out to be impossible, a nice clean condition where we can find such a p would be great too. This problem seems open to brute force attacks, geometric interpretations, algebraic thinking (like maybe Grobner bases ?) so I am hoping someone will be able to know exactly what is going on. Thanks.

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Are you sure it is an elementary question, given that no-one has given any answer yet? :-) More seriously, is d arbitrary or is it related to your original data? –  Thierry Zell Apr 3 '11 at 4:21
What do you mean by "singularities"? –  Tom Goodwillie Apr 3 '11 at 4:32
@Thierry d is arbitrary but the existence of a single d and a corresponding p for any collection of f's as above is what I am interested in. @Tom a critical point of $\tilde{f}$, that is a point in C^2n where the 2n partial derivatives simultaneously vanish. –  Daniel Pomerleano Apr 3 '11 at 6:03
@Thierry meant to say "a single d>1" –  Daniel Pomerleano Apr 3 '11 at 6:05
For a brute force but elementary approach: I would just pick $p = \sum y_i^{d}$. You have to show that the Jacobian ideal generated by the partial derivatives of $\tilde f$ has finite colength. It will suffice to show that all high powers of each $y_i$ and $f_i$ are in $J$. The Euler identity: $$ \sum_j \frac{dF}{dx_j} x_j = (deg F)F $$ (when $F$ homogenous) may be helpful. –  Hailong Dao Apr 3 '11 at 7:51
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