Consider the system of polynomial equations $$\left\{ \begin{aligned} P_1(x_1,x_2,\dots,x_n)&=0 \\ P_2(x_1,x_2,\dots,x_n)&=0 \\ &\;\; \vdots \\ P_m(x_1,x_2,\dots,x_n)&=0 \end{aligned} \right. \tag{1}\label{eq1}$$ where $P_1,P_2,\dots,P_m$ are polynomials in $n$ variables with integer coefficients. Let $\mathcal S \subset \Bbb C^n$ be the set of its solutions. We call $\eqref{eq1}$ nontrivial on the unknown $x_i$ if we can only obtain a finite number of distinct values of $x_i$ from $\eqref{eq1}$, in other words, if the projection $X_i=\mathrm{proj}_{i}(\mathcal S)$ of $\mathcal S$ onto the $i$th component (axis) of $\mathbb C^n$ is a finite set.

Is it true that if $\eqref{eq1}$ is nontrivial on $x_i$, then any value from $X_i$ is an algebraic number?


Yes, this is closely related to elimination, and building the elimination ideal. See for instance

Wikipedia :: Gröbner basis :: Elimination

In our case, let $J=(P_1,P_2,\dots,P_m)$ be the ideal in $$\Bbb Q[x_i;\ x_1,\dots,x_{i-1},\ x_{i+1},\dots,x_n]$$ generated by the given polynomials. Construct the elimination ideal $J^i$ in $$\Bbb Q[x_i]$$ after eliminating all other variables $x_1,\dots,x_{i-1},\ x_{i+1},\dots,x_n$. This realizes the projection to the $i$.th component in algebraic geometry, $$ \Bbb Q[x_i]/J^i \to \Bbb Q[x_i;x_1,\dots,x_{i-1},\ x_{i+1},\dots,x_n]/J \ , $$ which comes from the inclusion $$ \Bbb Q[x_i] \to \Bbb Q[x_i;x_1,\dots,x_{i-1},\ x_{i+1},\dots,x_n] \ , $$ and the fact that $J^i\subseteq J$.

(A point in $\mathcal S$ is a map $*=\operatorname{Spec}Q\to \operatorname{Spec}\Bbb Q[x_i;x_1,\dots,x_{i-1},\ x_{i+1},\dots,x_n]/J$, i.e. a map between rings in opposite direction, $\Bbb Q\leftarrow\Bbb Q[x_i;x_1,\dots,x_{i-1},\ x_{i+1},\dots,x_n]/J$. Now precompose with the above map.)

The main point is that elimination works over any field, here over $\Bbb Q$. (So we do not even need to pass to an algebraically closed field, e.g. $\bar{\Bbb Q}$.)

(Example: A particular case of elimination is building the resultant, for this construction we do not need to enlarge the base field.)

So the projected points are algebraic, one can moreover algorithmically get the polynomial generator(s) of $J^i$ in a given situation.

Note: The condition $(1)$ guarantees that the projection is a true $0$-dimensional variety (tested by using geometrical, $\Bbb C$-rational points). (So we do not have the problems like the ones in MO :: 86099/projections-of-real-algebraic-curves...)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.