Let $\mathbb{F}$ be an algebraically closed field. Let $X$ be an affine algebraic set, $x \in X$ and $M_x$ the maximal ideal in $\mathbb{F}[X]$ of functions which vanish at $x$. I want to show that the Krull dimension of the ring $\mathbb{F}[X]_{M_x}$ is equal to the maximum of the dimensions of the irreducible components of $X$ which contain $x$.
What I've done so far: I know that the Krull dimension of $\mathbb{F}[X]_{M_x}$ is just the height of the maximal ideal $M_x$ in $\mathbb{F}[X]$. Since a chain of prime ideals in $M_x$ corresponds to a chain of irreducible algebraic sets in $X$ each containing $x$, the Krull dimension of $\mathbb{F}[X]_{M_x}$ is equal to the longest chain of irreducible algebraic sets in $X$, $\textit{each}$ of which contains $x$.
What I want to say is that, if an irreducible algebraic set $X_i$ contains $x$ then I can construct a chain of irreducible subsets $\{x\} = Y_0 \subsetneq Y_1 \subsetneq \cdots \subsetneq Y_{\text{dim}(X_i)} = X_i$. If I could show this then the irreducible component which contains $x$ of largest dimension which have dimension equal to the longest chain of irreducible algebraic sets in $X$, each of which contain $x$.
Another lemma that, if shown, would solve the problem is the following: Let $X$ and $Y$ be irreducible such that $Y \subset X$ and $\dim(Y) < dim(X) - 1$. Then there exists an irreducible $Z$ such that $Y \subsetneq Z \subsetneq X$.
Any advice on either of these fronts would be super useful.