Can a polynomial have an isolated local minimum at a transcendental point? Let $f:\mathbb{R}^n \to \mathbb{R}$ be a polynomial with coefficients in $\mathbb{Q}$. Is it possible for there to be a point $\textbf{a} \in \mathbb{R}^n$ with all transcendental coordinates such that $\textbf{a}$ is an isolated local minimum of f? By isolated I mean that there exists a neighborhood of $\textbf{a}$ such that for every $\textbf{b}$ in the neighborhood, either $f(\textbf{a}) < f(\textbf{b})$ or $\textbf{a} = \textbf{b}$. In particular, I am thinking about the case when $f$ is non-negative and $f(\textbf{a}) = 0$.
It seems to me like such a local minimum shouldn't be possible. For $n = 1$ it is not possible. I am pretty sure it is not possible for $n=2$. For example, if $f = (x - 2y)^2$, then $f$ has a minimum at $(2\pi,\pi)$, but it is not isolated. I can not figure out how to prove this for $n>2$ and I can't find any counterexamples. I have tried various ideas from calculus and for the case when $f(\textbf{a}) = 0$, I have tried to make arguments about the dimension of the variety not being zero. I'm not very familiar with Algebraic Geometry though so maybe this idea doesn't work. Any thoughts would be appreciated.
 A: The answer is no; the coordinates of any isolated local minimum must be algebraic.
There's a nice argument I'd like to be able to use involving quotienting by the Jacobian ideal which I think is what Tabes suggests in the comments, but the problem is that the critical locus might be positive-dimensional over $\mathbb{C}$. Instead we can argue as follows. There are a collection of fields called real closed fields that can be defined in several equivalent ways, and we need that

*

*the real algebraic numbers $\mathbb{R} \cap \overline{\mathbb{Q}}$ are a real closed field, and that

*every real closed field satisfies the same first-order sentences in the language of fields as $\mathbb{R}$.

The latter might not seem so impressive until you know that inequalities such as $x \le y$ are expressible as first-order sentences: over a real closed field this condition is equivalent to $\exists z : y - x = z^2$. We can also express $x < y$ as the conjunction of $x \le y$ and $x \neq y$.
In particular, it follows that the claim that there exists a point $x = (x_1, \dots x_n)$ which is an isolated local minimum of $f$ can be expressed in the first-order language of fields! (Here we crucially need that the coefficients of $f$ are rational so we can write them all down in the first-order language of fields.) Namely, it's equivalent to the existence of $\epsilon > 0$ such that for all points $y = (y_1, \dots y_n)$ such that $\sum (x_i - y_i)^2 < \epsilon$ we have that either $f(y) > f(x)$ or $y = x$.
Hence if this sentence is true over $\mathbb{R}$ it's true over any real closed field and so true over the real algebraic numbers. But we can say more: if $f$ has an isolated local minimum $a = (a_1, \dots a_n)$ then we can express the claim that $f$ has an isolated local minimum near $a$ as a first-order sentence, by picking a sufficiently small $\delta > 0$ and finding rational upper and lower bounds $r_i \in (a_i - \delta, r), s_i \in (r, a_i + \delta)$ and adding to the above sentence the conditions that $r_i \le x_i \le s_i$. If we pick $\delta$ small enough so that, over $\mathbb{R}$, $a$ is the only isolated local minimum satisfying these bounds, then the existence of $a$ over $\mathbb{R}$ implies the existence of an isolated local minimum near $a$ over any real closed field and in particular over the real algebraic numbers, which must be $a$ itself. So $a$ has only algebraic coordinates.
