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Next: 6. The Weyl algebra Up: Standard bases, syzygies and Previous: 4. On Schreyer's method

5. Zariski's question, Milnor numbers and multiplicities

The generalization of Buchberger's algorithm presented in this paper has many applications. We just mention the computation of Milnor numbers, Tjurina numbers, local multiplicities, Buchsbaum-Rim and Polar multiplicities, first and second order deformations of isolated singularities, projections of families with affine fibres onto a local base space and, of course, all the usual ideal theoretic operations in a factorring Loc;SPMlt;K[x]/I such as intersection, ideal quotient and decision about ideal or radical membership. For further applications see [AMR]. Here we shall only explain how its implementation in SINGULAR helped to find a partial answer to Zariski's multiplicity question.

Let $f \in {\Bbb C}\{x_1, \ldots, x_n\} = {\Bbb C}\{x\}$, $f = \sum\, c_\alpha
x^\alpha$, f(0) = 0, be a not constant convergent powerseries and mult $(f) = \min \{\vert\alpha\vert \mid c_\alpha
\not= 0\}$ the multiplicity of f (for the general definition of multiplicity see the end of this chapter). Let $B \subseteq {\Bbb C}^n$ be a sufficiently small ball with centre 0 and $X = f^{-1} (0) \cap B$, the hypersurface singularity defined by f. If $g \in {\Bbb C}\{x\}$ is another powerseries and $Y = g^{-1} (0) \cap B$, then f and g (or X and Y) are called topologically equivalent if there exists a homeomorphism $h : (B,0) \to (B,0)$ such that h(X) = Y. The topological type of f is its class with respect to topological equivalence.

Zariski asked in 1971 (cf. [Z]) whether two complex hypersurface singularities f and g with the same topological type have the same multiplicity.

Zariski's question (usually called Zariski's conjecture) is, in general, unsettled but the answer is known to be yes in the following special cases:

-
n = 2, that is for plane curve singularities (Zariski, Lê Dung Trang),
-
f is semiquasihomogeneous and g is a deformation of f (Greuel [Gr], O'Shea [OS]).

Recall that f is called semiquasihomogeneous if there exists an analytic change of coordinates and positive weights such that the sum of terms of smallest weighted degree has an isolated singularity.

The two series of examples fta, b, c and gta, b, c, d, e in Chapter 3 were actually constructed to find a counter example to Zariski's conjecture. The idea is as follows: let

\begin{displaymath}f_t(x) = f(x) + tf_1(x) + t^2 f_2(x) + \ldots
\end{displaymath}

be a deformation of f(x). Let

\begin{displaymath}\mu(f_t) = \dim_{\Bbb C}{\Bbb C}\{x\}/\left(\frac{\partial f_...
...partial x_1}, \ldots,
\frac{\partial f_t}{\partial x_n}\right)
\end{displaymath}

denote the Milnor number of ft which we assume to be finite for t = 0 (then it is finite for t close to 0). Then, if the topological type of ft is independent of t, the Milnor number $\mu(f_t)$ is independent of t (for t sufficiently close to 0). The converse is also known to be true if $n \not= 3$. Hence, if $\mu(f_t)$ is constant but mult(ft) is not, we get a counter example (at least if $n \not= 3$).

For the above mentioned series ftabc and gtabcde the multiplicity is not constant. For t = 0 both series have non-degenerate Newton diagram and one can show that $\mu(f^{abc}_0) - \mu(f^{abc}_t) \le 12$ and $\mu(g_0^{abcde}) - \mu(g_t^{abcde}) \le 4$ $(t \not= 0$ and small and some restrictions on a, b, c, d, e). Since $\mu$ is several hundred or even several thousand there seemed to be a good chance for $\mu(f_t)$ or $\mu(g_t)$ to be constant. Using SINGULAR we were able to compute many of these Milnor numbers but neither $\mu(f_t)$ nor $\mu(g_t)$ were constant. (Actually, in most cases we obtained $\mu(f_0) - \mu(f_t) = 6$ and $\mu(g_0) - \mu(g_t) = 2$.) None of the existing computer algebra systems were able to compute the standard basis of the ideal of partials of ft respectively gt for relevant cases. (Only the system Macaulay was able to do some cases with small a, b, c using Lazard's method but it needed hours or days, whereas SINGULAR needed seconds or minutes. In these cases the success of SINGULAR was mainly due to the HCtest.) The failure to find a counter example led to the following positive result which shows that the families ft and gt can never be a counter example.

Let ft(x) be a (1-parameter) holomorphic family of isolated hypersurface singularities, that is $0 \in {\Bbb C}^n$ is an isolated critical point of ft for each t close to $0 \in {\Bbb C}$. The polar curve of such a family is the curve singularity in ${\Bbb C}^n \times {\Bbb C}$ defined by the ideal $(\partial f_t/\partial x_1, \ldots, \partial f_t/\partial x_n) \subset
{\Bbb C}\{x,t\}$.

Lemma 5..1   Let ft be a family of isolated hypersurface singularities. Let $H \cong
{\Bbb C}^{n-1}$ be a hyperplane through 0 such that formation of the polar curve is compatible with restriction to H. That is: polar curve $(f_t \mid H) =$ polar curve $(f_t) \cap H$. Then

\begin{displaymath}\mu(f_t) = \mbox{ constant } \Rightarrow \mu(f_t\vert H) = \mbox{ constant}.
\end{displaymath}

Proof: We may assume that $H = \{x_n = 0\}$ and then the polar curve(ft|H) is given by $(\partial f_t/\partial x_1, \ldots, \partial
f_t/\partial x_{n-1}, x_n)$ while polar curve $(f_t) \cap H$ is given by $(\partial
f_t/\partial x_1, \ldots, \partial f_t/\partial x_n, x_n)$. Hence, the assumption is equivalent to $\partial f_t/\partial x_n \in (\partial
f_t/\partial x_1, \ldots, \partial f_t/\partial x_{n-1}, x_n)$.

We shall use the valuation test for $\mu$-constant by Lê and Saito ([LS]):
$\mu(f_t) =$ constant $\Leftrightarrow$ for any holomorphic curve $\gamma : ({\Bbb C},0) \to ({\Bbb C}^n \times {\Bbb C},0)$ we have val $(\partial f_t/\partial
t(\gamma(s))) \ge \min\{\mbox{val}(\partial f_t/\partial x_i (\gamma(s)))$, $i
= 1, \ldots, n\}$. Moreover, this is equivalent to ``$\ge$'' replaced by ``>''. (val denotes the natural valuation with respect to s.)

Now let $\gamma (s)$ be any curve in $H = \{x_n = 0\}$. Then $\partial f_t/\partial x_n \in (\partial
f_t/\partial x_1, \ldots, \partial f_t/\partial x_{n-1}, x_n)$ implies that val $(\partial f_t/\partial x_n(\gamma(s)) \ge \min
\{\mbox{val}(\partial f_t/\partial x_i(\gamma (s)))$, $i = 1, \ldots, n-1\}$.

Applying the valuation test to ft and to $f_t\mid H$, the result follows.

Proposition 5..2   Let $f_t(x_1, \ldots, x_n) = g_t(x_1, \ldots, x_{n-1}) + x_n^2 h_t (x_1,
\ldots, x_n)$ be a family of isolated hypersurface singularities. Let g0 be semiquasihomogeneous or let n = 3. If the topological type of ft is constant then the multiplicity of gt is constant (for t close to 0).

Proof: Since ft has an isolated singularity we may add terms of sufficiently high degree without changing the analytic type of ft. If n = 3 we may replace gt by gt(x1, x2) + x1N + x2N, N sufficiently big, which has an isolated singularity and the same multiplicity as gt(x1, x2). Hence, in any case we may assume that gt has an isolated singularity. Applying the preceding lemma to the hyperplane $\{x_n =
0\}$ we obtain $\mu(g_t)$ constant. But since Zariski's conjecture is true for plane curve singularities and for deformations of semiquasihomogeneous singularities ([Gr], [OS]), mult(gt) is constant.

The Milnor number $\mu(f)$ of an isolated singularity can be computed as the number of monomials in $K[x_1, \ldots, x_n]/ L(I)$ where I is the leading ideal of $(\partial f/\partial x_1, \ldots, \partial f/\partial x_n)$ with respect to any ordering > such that xi < 1, $i = 1, \ldots, n$. This follows from the following Corollary 5.4.

The reason why standard bases can be applied to compute certain invariants of algebraic varieties or singularities (given in terms of submodules $I \subset
K[x]^r$), is that for any monomial ordering on K[x]r we have:

a)
(Loc ;SPMlt; K[x])r/I is a (flat) deformation of Loc ;SPMlt;K[x]r/L(I) (as we shall show below) which implies that certain invariants behave semicontinuously or even continuously during the deformation.

b)
For a monomial ideal these invariants can be computed combinatorially (but one needs extra algorithms for the actual computation).

In order to show a) we make the following construction:

Let $g_1, \ldots, g_q$ be a standard basis of $I \subset
K[x]^r = \sum_{i=1, \ldots, r} K[x]e_i$. Any monomial $x^\alpha e_k$ may be identified with the point $(\alpha_1, \ldots, \alpha_n, 0, \ldots, 1, \ldots
0) \in {\Bbb N}^{n+r}$. For a weight vector $w = (w_1, \ldots, w_{n+r}) \in
{\Bbb Z}^{n+r}$ we define

\begin{displaymath}w_1 \alpha_1 + \cdots + w_n \alpha_n + w_{n+k}
\end{displaymath}

to be the weighted degree of $x^\alpha e_k$. Let inw(f) the initial term of $f \in K[x]^r$, that is the sum of terms (monomial times coefficient) of f with maximal weighted degree and inw(I) the submodule generated by all inw(f), $f \in I$.

It is not difficult to see that there exists a weight vector $w \in {\Bbb Z}^{n+r}$ (indeed almost all w will do) such that in w(gi) = c(gi) L(gi), $i = 1,
\ldots, q$, and, moreover, in w(I) = L(I).

We choose such a w and shall now construct the deformation from L(I) to I:
For $f \in K[x]^r$ we can write $f = f_p + f_{p-1} + f_{p-2} + \cdots$ such that the weighted degree of each monomial of $f_\nu$ is $\nu$. Let t be one extra variable and put

\begin{displaymath}\tilde{f}(x,t) = f_p(x) + tf_{p-1}(x) + t^2f_{p-2} (x) + \cdots \in K[x,t]^r.
\end{displaymath}

Let $\tilde{I} \subset K[x,t]^r$ be the submodule generated by all $\tilde{f}$, $f \in I$. On K[x,t]r we choose the product ordering with lex- on K[t]: $x^\alpha t^p e_k < x^\beta t^q e_l$ if p > q or if p = q and $x^\alpha
e_k < x^\beta e_l$.

With respect to this ordering we have $L(\tilde{f}) = L(f)$ and, moreover, $L(\tilde{I}) = L(I)K[t]$. In particular, $\tilde{g}_1, \ldots, \tilde{g}_q$ is a standard basis of $\tilde{I}$.

Let $R:= \mbox{ Loc}_< K[x]$, $S:= \mbox{ Loc}_<K[x,t]$ and K(t) the quotient field of K[t].

Proposition 5..3   If $I \not= R^r$ then $S^r/\tilde{I}S$ is a faithfully flat K[t](t)-module with ``special fibre''

\begin{displaymath}(S^r/\tilde{I}S) \otimes_{K[t]_{(t)}} K \cong R^r/L(I)R
\end{displaymath}

and ``generic fibre''

\begin{displaymath}(S^r/\tilde{I}S) \otimes_{K[t]_{(t)}} K(t) \cong
R^r/IR \otimes_K K(t).
\end{displaymath}

Proof: The statements regarding the special and the generic fibres are easy. If $I \not= R^r$ then the support of $S^r/\tilde{I}S$ is surjective over Spec K[t](t) and hence ([AK], V, Proposition 2.4) it remains to show that t is a non-zero divisor of $S^r/\tilde{I}S$. Let $f \in S^r$ and $tf \in \tilde{I}S$. By Corollary 1.11 we have

\begin{displaymath}(1 + g) tf = \sum \xi_i \tilde{g}_i, L((1+g)tf) = L(tf) = tL(f),
\end{displaymath}

and $L(\xi_i \tilde{g}_i) \le L(tf)$ if $\xi_i \not= 0$. Hence, tL(f) is equal to $L(\xi_i)L(g_i)$ for some i. Then $L(\xi_i)$ is divisible by t and therefore also $\xi_i$, by definition of > on K[x,t]r. Subtracting $\xi_i
\tilde{g}_i$ on both sides and arguing as before, we see that all $\xi_i$ are divisible by t. Therefore, $(1+g) f \in \tilde{I}$ and $f \in \tilde{I}S$.

Corollary 5..4   Let either < be a wellordering or Rr/IR a finite dimensional K-vector space. Then the monomials in $K[x]^r \backslash L(I)$ represent a K-basis of Rr/IR.

Proof: If < is a wellordering, the monomials not in L(I) are a basis of the free module $S^r/\tilde{I}S$ (Theorem of Macaulay cf. [E]), hence the result. In general, it is easy to see that these monomial are linear independent modulo IR. (Use a standard basis of I and Corollary 1.11.)

Hence, if Rr/IR is finite dimensional, there are only finitely many monomials in $K[x]^r \backslash L(I)$. The proposition implies that $S^r/\tilde{I}S$ is K[t](t)-free with these monomials as basis, hence they also generate Rr/IR.

Remark 5..5   In general, the monomials not in L(I) are not a basis of Loc;SPMlt;K[x]/I. Take, for example, K[x] with lex- and I = (0). Then Loc ;SPMlt;K[x] = K[x](x) is not K-generated by monomials. If < is a wellordering, then $S^r/\tilde{I}S$ is even free over K[t] (cf. [E]).

Corollary 5..6   For any module ordering $\dim R^r/IR = \dim K[x]^r/L(I)$ where dim denotes the Krull dimension.

Proof: I = Rr implies L(I) = K[x]r, hence we may assume $I \not= R^r$. Faithful flatness implies that $\dim R^r/IR = \dim R^r/L(I)R$ ([AK], V, Proposition 2.10), hence the result.


Let us finish with a final remark about multiplicites in the local case:

Consider the local ring R = K[x](x) with maximal ideal $(x) = (x_1,
\ldots, x_n)$ and M = Rr/IR a finitely generated R-module, where I is given as a submodule of K[x]r by finitely many generators. Consider

\begin{displaymath}gr\, M = \sum_{i \ge 0} (x)^i M/(x)^{i+1} M,
\end{displaymath}

which is a graded module over $gr\, R = K[x]$. The formal power series

\begin{displaymath}S_M(T) = \sum_{i \ge 0} \dim_K (M/(x)^{i+1} M) T^i,
\end{displaymath}

is called the Hilbert-Samuel series and the function $s_M(i) =
\dim_K(M/(x)^{i+1} M)$ the Hilbert-Samuel function of M (with respect to the maximal ideal).
If $h_{gr\,M}$ denotes the Hilbert function of the graded module $gr\, M$ we have (cf. [Ma]) $h_{gr\,M}(i) = \dim_K (x)^iM/(x)^{i+1}M$ and a polynomial $hp_{gr\,M} \in {\Bbb Q}[t]$ (the Hilbert polynomial of $gr\, M$) of degree d-1 of the form

\begin{displaymath}hp_{gr\,M}(t) = \frac{e_{gr\,M}}{(d-1)!} t^{d-1} + \mbox{ (terms of lower
degree)}
\end{displaymath}

such that $hp_{gr\,M}(i) = h_{gr\,M}(i)$ for i sufficiently big. $e_{gr\,M}$ is called the degree of the graded module $gr\, M$ whilst d is equal to the Krull dimension of $gr\, M$.

It follows that

\begin{displaymath}s_M(i) - s_M(i-1) = h_{gr\,M}(i)
\end{displaymath}

and there exists a polynomial $sp_M \in {\Bbb Q}[t]$ of degree d such that

\begin{displaymath}sp_M(t) = \frac{e_M}{d!} t^d + (\mbox{terms of lower degree}),
\end{displaymath}

where $d = \dim\, M = \dim\, gr\, M$ and $e_M = e_{gr\,M}$ is called the (Samuel) multiplicity of M. The following proposition follows now easily.

Proposition 5..7   Let < be a degree ordering (cf. Chapter 1) on the monomials of K[x] such that $w_i = \mbox{ degree }(x_i) = -1$ for $i = 1, \ldots, n$ which is extended to a module ordering on K[x]r arbitrarily. Let M = Rr/IR as above and L(I) the leading ideal of I. Then the Hilbert function $h_{gr\,M}$ coincides with the Hilbert function hK[x]r/L(I) of the graded module K[x]r/L(I). In particular, $\dim\, M = \dim\,K[x]^r/L(I)$ and $e_M = \mbox{degree}(K[x]^r/L(I))$. Consequently, $\dim\,M$ and eM can be computed from a standard basis of I with repect to an ordering as above and a Hilbert polynomial algorithm.

Remark 5..8   If one is only interested in the dimension and the multiplicity it is not necessary to compute the full Hilbert series. A fast, direct algorithm for computing this is implemented in SINGULAR. Moreover, for computing dimensions we can use the fast lex- ordering by Corollary 5.6 (but, of course, not to compute multiplicities or Hilbert functions).


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