Let ${\Bbb C}[X_1,\ldots,X_n]$ be a $n$-variable polynomial ring over a complex number field ${\Bbb C}$. For its maximal ideal $(X_1,\ldots,X_n)$, we define the geometric regular local ring as

$R \colon= {\Bbb C}[X_1,\ldots,X_n]_{(X_1,\ldots,X_n)}$,

which is the localisation of ${\Bbb C}[X_1,\ldots,X_n]$ at $(X_1,\ldots,X_n)$. $R$ has the unique maximal ideal $(X_1,\ldots,X_n)$.

We shall consider the polynomial ring $R[X]$ over $R$ and choose $m$ Weierstrass polynomials

\begin{align} & f_1(X) = c_{1,0} + c_{1,1}X + \ldots + c_{1,e_1}X^{e_1} \\ & {\phantom{AAA}} ... \\ & f_m(X) = c_{m,0} + c_{m,1}X + \ldots + c_{m,e_{m}}X^{e_m}, \end{align} where $c_{1,0},\ldots,c_{i,j},\ldots,c_{m,e_m} \in (X_1,\ldots,X_n)$.

Suppose that ${\mathrm{GCD}}(f_1(X),\ldots,f_m(X)) = 1$.

Now for a function $F(X) \in R[X]$, we consider the following condition$\colon$

$(\sharp)$ $\quad F(\alpha) \in (f_1(\alpha),\ldots,f_m(\alpha))\phantom{A}$ for any $\alpha \in {\Bbb C}$,

where $(f_1(\alpha),\ldots,f_m(\alpha))$ is the ideal of $R$.

Q. Does $F(X) \in (f_1(X),\ldots,f_m(X))$?

  • $\begingroup$ No: $n=1, f_1=X_1X^2, f_2=X_1(X+1), F=X_1$ $\endgroup$ Oct 3 '17 at 13:41
  • $\begingroup$ Thanks. The condition GCD(f_1(X),...,f_m(X)) = 1 must be added. $\endgroup$
    – Rinmyaku
    Oct 3 '17 at 14:06

The answer to the question Q is ``no''. Take $n=m=2$, $f_1(X)=X_1X^2$, $f_2(X)=X_2$, $F(X)=X_1X$. If $\alpha\ne0$ then $F(\alpha)=\alpha X_1\in(\alpha^2X_1)\in(f_1(\alpha),f_2(\alpha))$. If $\alpha=0$ we have $F(0)=0\in(f_1(0),f_2(0))$. However, $X_1X\notin(X_1X^2,X_2)$.

Using Nullstellensatz, it is possible to prove that $F(X)\in\sqrt{(f_1(X),…,f_m(X))}$, where $\sqrt{(f_1(X),…,f_m(X))}$ denotes the radical of $(f_1(X),…,f_m(X))$. The condition $GCD(f_1(X),…,f_m(X))=1$ is not needed for that.


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