## Sharp effective Nullstellensatz.(English)Zbl 0682.14001

Let K be any field and n, $$d_ 1,...,d_ k$$ be natural numbers. Let $$N(n,\underline d)=\min \{s| \quad for$$ any homogeneous polynomials $$\phi_ 1,...,\phi_ k\in K[X_ 0,...,X_ n]$$ such that $$\deg (\phi_ i)=d_ i$$, we have $$(rad(I))^ s\subset I$$ where $$I=(\phi_ 1,...,\phi_ k)\}.$$
Main theorem: If we assume $$d_ 1\geq d_ 2\geq...\geq d_ k$$ and $$d_ i\neq 2$$ (for $$all\quad i),$$ then N(n,ḏ)$$=d_ 1...d_ k$$ if $$k\leq n$$, N(n,ḏ)$$=d_ 1...d_{n-1}\cdot d_ k$$ if $$k>n>1$$, N(n,ḏ)$$=d_ 1+d_ k-1$$ if $$k>n=1.$$
Corollary. Let $$f_ 1,...,f_ k, h\in K[X_ 1,...,X_ n]$$ and assume that h vanishes on all common zeros of $$f_ 1,...,f_ k$$ in the algebraic closure of K. Let $$d_ i=\deg (f_ i)\neq 2$$ (for $$all\quad i).$$ Then one can find $$g_ 1,...,g_ k\in K[X_ 1,...,X_ n]$$ and a natural number s such that $$\sum g_ if_ i =h^ s$$, $$s\leq N(n,\underline d)$$, $$\deg (g_ if_ i)\leq (1+\deg (h))\cdot N(n,\underline d)$$. In particular, if $$h=1$$ (so that $$f_ 1,...,f_ k$$ have no common zeros), we can choose $$g_ i$$ such that $$\sum g_ if_ i =1$$ and $$\deg (g_ if_ i)\leq N(n,\underline d).$$
This estimate is the best possible. The condition $$d_ i\neq 2$$ is technical, and the author expects that it is not necessary. The proof uses elementary methods of algebraic geometry including local cohomology, and works in all characteristics.
Reviewer: H.Matsumura

### MSC:

 14A05 Relevant commutative algebra 13F20 Polynomial rings and ideals; rings of integer-valued polynomials

### Keywords:

Nullstellensatz; local cohomology
Full Text: