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Let V be a analytic set of $C^n$, $I(V)$ is the sheaf of ideals of V (the sheaf whose stalks are ideals defining germs of V at its points). Since $I(V)$ is a coherent analytic sheaf, we see that in a neighborhood D of a point p of V there are holomorphic function $g_1,g_2,\ldots,g_t$ such that $V\cap D=$ {$g_1=g_2=\cdots=g_t=0$}. Let $k=\dim_p V$.

If p is a regular point of V, evidently we can choose $t=n-k$.

If V is hypersurface we can choose $t=1=n-k$.

My question is: in general, can we choose $t=n-k$, or equivalently, an analytic is locally intersection of (n-k) hypersurface?

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    $\begingroup$ Giving $V$ its reduced analytic structure, the question is asking whether the local ring at $p$ is a complete intersection ring (due to general facts about Cohen-Macaulay and complete intersection rings; 17.4(iii)(3) and 21.2(ii) in Matsumura's "Commutative Ring Theory" book). It is equivalent to the completion of the local ring being a c.i. ring (21.2(i) in Matsumura). Thus, taking an reduced affine algebraic $\mathbf{C}$-scheme having a non-c.i. point gives (after analytification) a counterexample (by using excellence of analytic and algebraic local rings to track reducedness). $\endgroup$
    – BCnrd
    Commented May 11, 2010 at 23:28

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Let $V \subset {\mathbb{C}}^6$ be the set defined by $$ \operatorname{rank} \begin{bmatrix}z_1 & z_2 & z_3 \\\ z_4 & z_5 & z_6 \end{bmatrix} \leq 1 . $$ Then dimension of $V$ is 4, but near the origin you need at least 3 holomorphic functions to define $V$ (the three $2\times 2$ subdeterminants being zero). That is, $V$ is not a set-theoretic complete intersection.

There is an extra hickup in this. The minimal number of germs of holomorphic functions necessary to define the set (the germ of the set) need not be the same as the number of germs of holomorphic functions necessary to define the ideal. I do not have an example offhand for this.

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  • $\begingroup$ Det? Don't you mean rank <= 1? $\endgroup$
    – Thierry Zell
    Commented Aug 20, 2010 at 10:54
  • $\begingroup$ yes, duh .... I mixed up in my head the 3 equations (the three 2x2 determinants) and the condition ... sorry about that $\endgroup$
    – Jiri Lebl
    Commented Aug 29, 2010 at 19:07
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I have found the answer of my question in Chirka's book "complex analytic sets", section 5.7, page 64.

He wrote that "in general, the answer is negative". However, he didn't give an example for this statement.

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    $\begingroup$ Since c.i. implies Cohen-Macaulay, a non-normal surface which is regular in codimension 1 is not c.i. at the non-normal points (due to Serre's homological normality criterion). So scrunching up a smooth surface at a point does the job. For example, create a singularity at the origin in the affine plane: ${\rm{Spec}}(A)$ for $A \subset \mathbf{C}[x_1,x_2,x_3]$ the subring of $f$ satisfying $f \equiv f(0,0) \bmod (x_1,x_2,x_3)^2$ (i.e., $A = \mathbf{C}[x_i x_j]_ {1 \le i, j \le 3}$). $\endgroup$
    – BCnrd
    Commented May 12, 2010 at 14:44

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