Show that the zeros of some polynomial in $\mathbb{Q}\left[T_1, \ldots T_r\right]$ has at point whose coordinates are in finite extension field of $Q$

Question

Suppose we are given a collection of polynomials in $r$ variables with rational coefficients: $$ f_1, \ldots f_N \in \mathbb{Q}\left[T_1, \ldots T_r\right] $$ We define the complex algebraic set $V_{\mathbb{C}} \subset \mathbb{C}^r$ by $$ V_{\mathbb{C}}=\left\{\left(a_1, \ldots a_r\right) \mid f_i\left(a_1, \ldots a_r\right)=0 \text { for all } i \text { from } 1 \text { to } N\right\} . $$ Suppose $V_{\mathbb{C}}$ is not empty. Show that there is a finite extension $K$ of $\mathbb{Q}$ and a point $$ \left(a_1, \ldots a_r\right) \in V_{\mathbb{C}} $$ with all $a_k \in K$.

This question comes from 2004 UCLA algebra qualifying.

I tried to do induction on r and strengthen the conclusion on $\mathbb{\bar{Q}}\left[T_1, \ldots T_r\right]$,where $\mathbb{\bar{Q}}$ is algebraic closure of $\mathbb{Q}$, but this force me to consider how to show there is at least one point have at least one algebric coordinate, which I can't prove.

The major difficult I deal with is how to transform all transcendental point into algebraic one. Like $xyz−1=0$ ,taking $x=π^{-1/3},y=π^{1/3}e^{2/3},z=e^{-2/3}$ , At first I try to show they can be genertated by another varible like t, but the example may a counter example.

Answer 1

As $V_\mathbb{C}$ is non-empty, we have that $(f_1,\ldots,f_N)\subseteq \mathbb{C}[T_1,\ldots,T_r]$ is not the unit ideal. Therefore $(f_1,\ldots,f_N)\subseteq \mathbb{Q}[T_1,\ldots,T_r]$ isn't the unit ideal either. Thus it is included in some maximal ideal $\mathfrak{m}\subseteq \mathbb{Q}[T_1,\ldots,T_r]$. The field $K:=\mathbb{Q}[T_1,\ldots,T_r]/\mathfrak{m}$ is then a finite extension of $\mathbb{Q}$, and the classes $T_1+\mathfrak{m},\ldots,T_r+\mathfrak{m}\in K$ satisfy the required equations.

Answer 2

You're getting a little tripped up here.

First, I claim that $(f_1,\cdots,f_N)\subset \overline{\Bbb Q}[T_1,\cdots,T_r]$ is a proper ideal - this is equivalent to $1\notin(f_1,\cdots,f_N)\subset\overline{\Bbb Q}[T_1,\cdots,T_r]$, but if $1\in (f_1,\cdots,f_N)\subset\overline{\Bbb Q}[T_1,\cdots, T_r]$, then we must have $1\in (f_1,\cdots,f_N) \subset \Bbb C[T_1,\cdots,T_r]$, which doesn't happen because $V_\Bbb C$ is nonempty.

Next, use the Nullstellensatz on the proper ideal $(f_1,\cdots,f_N)\subset \overline{\Bbb Q}[T_1,\cdots,T_r]$: there is at least one solution to the system $f_1=0,\cdots,f_N=0$ over $\overline{\Bbb Q}$, and now you're done.