I need some help with the following problem:

Let $f_1,\dots,f_k\in \mathbb{Z}[X_1,\dots,X_n]$ be polynomials without any common zero in $\mathbb{C}^n$. Then there are $g_1,\dots,g_k\in \mathbb{Z}[X_1,\dots,X_n]$ s.t. $\sum_1^kg_if_i\in\mathbb{Z}-\{0\}$.

What I have is the following:

Let $k$ be an algebraic closure of $\mathbb{Q}$. Consider the ideal $\mathfrak{a}=(f_1,\dots,f_k)\subset k[X_1,\dots,X_n]$. Since every proper ideal $I$ is contained in some maximal ideal and there is a natural bijection between $Z(I)$ and $\operatorname{MaxSpec}(k[X_1,\dots,X_n])\cap V(I)$ from $Z(\mathfrak{a})=\emptyset$ it follows that $\mathfrak{a}=(1)$. So there are $g_i\in k[X_1,\dots,X_n]$ s.t. $\sum_1^kg_if_i=1$. Now I want to prove that the $g_i$ actually lie in $\mathbb{Q}$ so that I can multiply through with the common denominator. Let $L$ be a finite and normal extension of $\mathbb{Q}$ containing the coefficients of the $g_i$. If $\sigma$ is any element of $\operatorname{Gal}(L/\mathbb{Q})$ then $$\sum_{i=1}^kg_i^\sigma f_i=\sum_{i=1}^kg_i^\sigma f_i^\sigma=(\sum_{i=1}^kg_if_i)^\sigma=1=\sum_{i=1}^kg_if_i$$ I would like to conclude from this that $g_i=g_i^\sigma$ and that therefore we have $g_i \in \mathbb{Q}$ but I'm not sure if this is OK.

I would appreciate any help and suggestions :)


1 Answer 1


What about considering $$\sum_{i=1}^k\left(\sum_{\sigma\in G}g_i^\sigma\right) f_i=|G| \sum_{i=1}^kg_if_i?$$


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