Transcendental solutions to constrained polynomial optimization problems?

Question

Can an optimization problem in which the objective and constraints are all polynomials with rational coefficients have a solution involving transcendental values?

Answer 1

I'm assuming what you mean is the following:

Let $f(x_1,\cdots,x_n)$ and $g(x_1,\cdots,x_n)$ be polynomials with rational coefficients, and let $(a_1,\cdots,a_n)$ be the point in $\mathbb{R}^n$ where $f$ is minimized, given that $g(a_1,\cdots,a_n)=0$. Is it possible that not all of $(a_1,\cdots,a_n)$ is algebraic?

As A.Γ. pointed out, it's possible that the set that reaches the minimum is a line, in which case they can certainly be transcendental. I am henceforth assuming that there are finitely many points where the minimum is reached; a corollary of this is that all of the points reaching global minima satisfy $\nabla f = \lambda \nabla g$ for some real $\lambda$ (Lagrange multipliers).

We can consider $a_1,\cdots,a_n,\lambda$ as $n+1$ variables, with the $n+1$ conditions that

$$\frac{\partial f}{\partial x_i}\bigg|_{\mathbb{a}}=\lambda \frac{\partial g}{\partial x_i}\bigg|_{\mathbb{a}}$$

for all $1\leq i\leq n$, and $g(\mathbb{a})=0$. These are all polynomial equations in the predefined $n+1$ variables, and (by our assumptions) it is well-behaved. Algebraically, for some polynomials with rational (we may presume integer) coefficients $P_1,\cdots,P_{n+1}$, we can represent our equations as $P_k(a_1,\cdots,a_n,\lambda)=0$ for all $1\leq k\leq n+1$.

We will prove, by induction on $m$, that a system of $m$ polynomial equations with rational coefficients in $m$ variables will, if it has only finitely many solutions, have only algebraic ones (it might have none).

Base case: If $m=1$ this is just the definition of an algebraic number; either there are finitely many algebraic solutions or the polynomial is the zero polynomial.

Inductive step: Assume one has a set of $m+1$ polynomial equations in $m+1$ variables. Let those variables be $x_1,\cdots,x_{m+1}$, and WLOG let the $(m+1)$-th equation depend on $x_{m+1}$ (if not, we may simply reorder the equations). Each of our $m+1$ equations can be viewed as a polynomial equation in $x_{m+1}$ with coefficients rational-coefficiented polynomials in $x_1,\cdots,x_m$.

Recall that two polynomials have a common root iff their resultant is $0$. Since this resultant is a rational-coefficiented polynomial in terms of the coefficients, we get that for any two polynomials in our list, their resultant (when viewed as a polynomial in $x_{m+1}$) being $0$ is a polynomial equation in $x_1,\cdots,x_m$. Thus, there exists a simultaneous solution for $x_{m+1}$ iff the polynomial equations determined by the resultant of $P_i$ and $P_{m+1}$ (for all $1\leq i \leq m$) are all solved (they are $m$ equations in $x_1,\cdots,x_m$).

So, by the inductive hypothesis, either there are infinitely many solutions, or all solutions result in $x_1,\cdots,x_m$ all being algebraic. In addition, since there was nothing special in the ordering of our variables, $x_2,\cdots,x_{m+1}$ are all algebraic. Thus, all of the variables are algebraic, finishing the inductive step.