Consider a matrix $A\in \mathbb R^{n\times n}$ of indeterminates. The determinant of $A$ is a degree $n$ polynomial in the $n^2$ entries satisfying $\det A\ne0\iff A$ is nonsingular.
What about when $A\in\mathbb R^{n\times m}$ is not square? Is there is a similar low-degree polynomial in the $nm$ indeterminates that satisfies $f(x_{1,1},\ldots,x_{n,m})\ne0\iff A$ has full rank? It looks like $2\cdot \min\{n,m\}$ is achievable by letting $f=\det (A^TA)$ or $\det (AA^T)$, whichever makes sense. But what about degree $\min \{n,m\}$? Or at least $\max\{n,m\}$?
Edit1: Changed $\mathbb F$ to $\mathbb R$.
Edit2: Thought about it and I guess $\min \{n,m\}$ isn't achievable when one considers the $n\times 1$ case (not possible to test whether vector is $\mathbb 0$ with degree 1). Is $2\cdot \min\{n,m\}$ optimal in all cases of $n\ne m$?