$\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\R}{\mathbf R}$ $\newcommand{\rank}{\text{rank}}$ $\newcommand{\reg}{\text{reg}}$ $\newcommand{\sing}{\text{sing}}$
Definitions
Definition. Let $X$ be a topological space. The dimension of $X$ is defined as the supremum of the length of chains $$ \emptyset\neq X_0\subsetneq X_1 \subsetneq \cdots \subsetneq X_d\subseteq X $$ or irreducible (Recall that a topological space $X$ is said to be irreducible if any two distinct open sets in $X$ intersect. closed subspaces of $X$). (The length of the above chain in $d$ and not $d+1$). The dimension of $X$ is written as $\dim(X)$.
Definition. A real algebraic variety is a subset $V$ of $\R^n$ which occurs as the common zero set of a collection of polynomials in $\R[x_1 , \ldots, x_n]$. The dimension of a real algebraic variety $V$ is defined as the dimension of $V$ when $V$ is endowed with the Zariski topology.
Definition. Let $V$ be an affine algebraic variety in $\R^n$. Let $I(V)$ denote the set of all the polynomials in $n$ variables which vanish on each point of $V$. Writing $Df_p$ to denote the derivative of a polynomial $f$ at the point $p$, we note that the set $\set{Df_p:\ f\in I(V)}$ is a real vector space, and is a subspace of $(\R^n)^*$. It is therefore finite dimensional. The rank of $V$ at $p$, written $\rank_p(V)$, is defined as the dimension of this vector space. The rank of $V$ is defined as $$ \rank(V) = \max_{p\in V}\rank_p(V) $$ A point $p\in V$ is said to be a simple point if $\rank_p(V)=\rank(V)$. A simple point may also be referred to as a non-singular point. The set of all the simple points of $V$ will be denoted by $\reg(V)$. A point in $V$ which is not simple is called singular.
It is shown in Whitney's Elementary Structure of Real Algebraic Varieties, Annals of Math., 1957 that if $p\in V$ is a simple point of a real algebraic variety $V\subseteq \R^n$ having rank $r$, then the set of simple points of $V$ constitute a smooth submanifold of $\R^n$ of dimension $n-r$.
Questions
My question is the following:
Question. Let $X\subseteq \R^n$ be an irreducible real algebraic variety and let $p\in X$ be a simple point of $X$. Let $f\in \R[x_1 , \ldots, x_n]$ be a polynomial which vanishes on a Euclidean neighborhood of $p$ in $X$. Is it necessary that $f$ vanishes on the whole of $X$?
I asked a similar question here and accepted the answer by @KReiser. Based on the helper's response it seems that the answer to the above question is 'yes'.
I did not fully understand the reasoning at that point since I was largely ignorant of the terminology used by the helper and of algebraic geometry in general. However, I now am trying to understand the answer step by step. The helper starts his/her reasoning as follows:
Let $Z$ be the zero set of $f$. Then $\dim Z\geq \dim X$.
I am stuck here itself. Of course, we need to make use of the hypothesis that $Z$ contains a Euclidean neighborhood of $p$ in $V$ Can somebody explain why the last assertion is true?
Yes, this is correct, and the proof is as given in the linked question. An aside: the term "simple point" is vastly less common in the literature than "smooth point", and I would recommend sticking with "smooth point".
Recall that $\dim U = \dim X$ was assumed, so it remains to explain why $\dim (f=0)\geq \dim U$. This is an instance of a general fact: If $S\subset T$ is an inclusion, then the Krull dimension of $T$ must be at least that of $S$. Proof: given a chain of proper inclusions of irreducible closed subsets of $S$, the closures of these sets again form a chain of proper inclusions of irreducible closed subsets of $T$ (why? if the inclusion of the closures is nonproper, then the inclusion of the original sets is nonproper, and the closure of an irreducible space is again irreducible).
Edit to respond to comments: There's some wrangling introduced here by the fact that $U$ is not in the same category as $X$: the best we can do is consider $U$ as an analytic space, and then apply analytic variety machinery to $U$ and $X(\Bbb R)$, note where that machinery matches the algebraic world, then go from there. The Krull dimension of an analytic space should be defined by chains of analytically irreducible analytic closed sets (this means these subsets cannot be written as the union of two other closed analytic spaces which are both strictly subsets of our given space). On the other hand, if $x\in X$ is a smooth closed point, the Krull dimension of $X(\Bbb R)$ at $x$ is the same as the Krull dimension of $X$ at $x$, since these can both be described by the number of generators for the maximal ideal of the local rings of $x$ in each of these varieties, and since $x$ is a smooth point, the local rings are $\Bbb R\{x_1,\cdots,x_n\}$ and $\Bbb R[x_1,\cdots,x_n]_{(x_1,\cdots,x_n)}$ which both have $x_1,\cdots,x_n$ as generating sets for their maximal ideals.