Is the zero set of a finite collection of polynomials defined on a submanifold of $R^n$ either the entire manifold or of measure zero?

Question

My question is related to this one.

I'd like to explore this result in the context of polynomials defined on a smooth, properly embedded (thus closed) $m$-dimensional submanifold $M \subset R^n$. That is, I have a finite collection of polynomials defined on $M$ (say the determinants of submatrices whose entries are functions of the points of $M$) and would like to investigate the subset of $M$ (call it $Z$) on which all those functions vanish.

My approach is to think of $Z$ as the intersection of a finite number of manifolds; i.e. the zero sets of each of the determinant functions, and $M$.

Question: Is it correct to say that either $Z$ is all of $M$ or is a subset of $M$ of measure zero? If that is true then, since $Z$ is clearly closed, can we say that $M \setminus Z$ is either empty or open-dense?

I realize there is a question of what measure to use on $M$, I would guess $m$-dimensional Hausdorff measure, but I haven't gotten quite that far yet.

I've searched around for answers to this question (as you can see from my link) but I know things can become delicate when dealing with submanifolds, so really any pointers to resources addressing this question would be appreciated. I wouldn't have asked if I thought the question was trivial, but I guess it's possible, in which case I apologize.

Edit: Moishe Kohan has kindly pointed out that the concept of real analytic manifold as it pertains to $M$ is relevant here. I will look into this further.

Edit$^2$: I have found this post on MO, go figure. For anyone who has interest in this question, the fact that analytic manifolds and zeros of polynomials are related is fascinating. Then again, having locally an expansion logically connects these manifolds to polynomials. At least that is my immediate impression. There is another post which discusses a question closer to mine, but still no cigar. A key may be if I can compose chart coordinates with real analytic functions and get real analytic functions then my problem reduces to $R^n$. Thanks are also due to Severin Schraven for a counterexample that illustrates the difficulty with real analytic manifolds.

Edit$^3$: Severin Schraven has made the suggestion that the implicit function theorem coupled with this post will give me what I'm looking for (at least when $M$ is connected, or on each connected component of $M$). Will look into it.

Update: As you can see, Moishe Kohan has graciously provided a detailed proof, which I have marked as the Answer. From my perspective, the key seems to be (Severin Schraven also pointed this out) that when $M$ is an analytic manifold the coordinate maps are analytic so that my question can be reduced, in one way or another, to questions about real analytic functions on domains in $R^n$. This is crucial since there are existing results (e.g. Identity Theorem) available for real analytic fuctions in this context.

I do, however, have some questions about that proof which I will post over the next day or two in the comment section.

Answer 1

To convert my comments to an answer. First, suppose that your submanifold $M^m\subset \mathbb R^n$ is real-analytic, i.e. is locally a graph of a real-analytic vector-function. (Equivalently, locally, $M$ is the image of a real-analytic embeddings from open subsets of $\mathbb R^m$ to $\mathbb R^n$. Equivalently, locally, near each point $p\in M$, the submanifold $M$ is the regular level set of real-analytic function $U\to \mathbb R^{n-m}$, defined on a neighborhood $U$ of $p$.) Then for every real-analytic function $f: \mathbb R^n\to \mathbb R^k$ and for every component $C$ of $M$, either $f$ is identically zero on $C$ or $f^{-1}(0)\cap C$ has measure zero in $C$. Here one can take any reasonable measure on $M$, for instance, the one given by a Riemannian metric on $M$ or the $m$-dimensional Hausdorff measure restricted to $M$: All "reasonable" measures will be absolutely continuous with respect to each other. To prove the dichotomy, it suffices to argue in local real-analytic coordinates on $M$, $\psi: V\to M$, $V$ is open in $\mathbb R^m$. Namely, it suffices to check that $h=f\circ \psi$ satisfies $h^{-1}(0)$ has zero Lebesgue measure. (One can use the Lebesgue measure since on $\psi(U)$ again, the push-forward of the Lebesgue measure will be "reasonable".) The composition of real-analytic functions is again real-analytic. If for one of the coordinate functions $\psi$ the composition $h$ will be identically zero, then $f$ will vanish on $\psi(V)$. By real-analyticity, it will follow that $f$ is identically zero on $C$. Thus, it suffices to observe that each nonconstant real-analytic function $h$ defined on an open connected subset of $\mathbb R^m$ has zero measure of the zero-level set $h^{-1}(0)$. This was explained many times on this site, see for instance here or here, etc...

Now, if $M$ is merely smooth, then the $m$-dimensional measure of $f^{-1}(0)\cap C$ need not be zero even if $f$ is a linear function. For instance, note that for every closed subset $A\subset \mathbb R$ there is a smooth function $g: \mathbb R\to \mathbb R$ such that $g^{-1}(0)=A$. (This was explained many times on this site, but it's a good exercise to find such a function for, say, $A=(-\infty, 0]$.) Now, let $M$ denote the graph of $g$: It is a smooth 1-dimensional submanifold in $\mathbb R^2$. Take $f(x,y)=y$. Then $f^{-1}(0)\cap M=A\times \{0\}$. Take $A\subset \mathbb R$, a proper closed subset which has positive 1-dimensional measure. Then $A$ will have positive measure in $M$ (with respect to any "reasonable" measure on $M$). Since $g$ is nonconstant, so is $f|_A$. That's a counter-example.