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A local ring is a nonzero ring with one maximal ideal $\mathfrak{m}$. Write $\kappa =R/\mathfrak{m}$ for its residue field. Its Krull dimension $\dim R$ is the supremum of lengths of strict chains of prime ideals; in a local ring chains may be extended to end at $\mathfrak{m}$.

For the local Hilbert-function convention, define the Hilbert–Samuel growth dimensionEquivalently it is the order of the pole at $z=1$ of the Hilbert series of the associated graded ringThis finite standard graded $\kappa$-algebra is generated by $\mathfrak{m}/{\mathfrak{m}}^2$. The Hilbert-Serre theorem makes the cumulative Hilbert function eventually polynomial, which establishes the definition. For an Artinian local ring, the polynomial is constant and nonzero, so $d(R)=0$.

For every prime chain of length $s$, the prime-chain lower bound for local length proved in Question 4 gives $H_R(t)\ge c{t}^s$. A polynomial of degree $d(R)$ cannot satisfy this when $s>d(R)$. Thus the requested inequality isFor clarity, the embedding dimension is the different invariantThe equality follows from the Nakayama lemma. Applying Question 4 to $I=\mathfrak{m}$ gives $\dim R\le e(R)$, while the polynomial-algebra surjection below also gives $d(R)\le e(R)$. In particular, growth dimension and embedding dimension should be kept distinct.

A regular local ring is a Noetherian local ring with $e(R)=\dim R$. Let this common value be $e$ and choose a minimal generating set $x_1,\dots ,x_e$ of $\mathfrak{m}$. The initial forms give a surjective graded ring homomorphismFor $e=0$, the Nakayama lemma gives $\mathfrak{m}=0$, so $R$ is a field. Assume $e>0$. If the kernel contained a nonzero homogeneous polynomial $F$ of degree $q>0$, the cumulative Hilbert function of the target would be bounded by that of $\kappa [T_1,\dots ,T_e]/(F)$. Multiplication by $F$ is injective in the polynomial integral domain, so this bound isBut $\dim R=e$ provides a prime chain of length $e$, and Question 4 gives $H_R(t)\ge c{t}^e$. This is a contradiction. Hence the associated graded ring of a regular local ring isan integral domain.

Finally the Krull intersection theorem gives ${\bigcap }_j{\mathfrak{m}}^j=0$. One can see the needed separatedness directly: for the finitely generated ideal $K={\bigcap }_j{\mathfrak{m}}^j$, the Artin-Rees lemma gives $K=\mathfrak{m}K$, and the Nakayama lemma gives $K=0$. Therefore each nonzero $a\in R$ has finite $\mathfrak{m}$-adic order $r$, with nonzero initial form in ${\mathfrak{m}}^r/{\mathfrak{m}}^{r+1}$. For nonzero $a,b$, their initial forms have nonzero product in the graded integral domain. It follows that $ab\ne 0$, and indeed its order is the sum of their orders. Thus $R$ is an integral domain.