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Let $X$ be smooth over a field $k$, and let $x$ be a $k$-rational point of local dimension $d$. Put $R={\mathcal{O}}_{X,x}$, whose residue field is $k$. A chosen regular system of parameters gives the displayed completion isomorphism by sending $T_i$ to $t_i$. The associated graded ring of a regular local ring is the polynomial ring on their initial forms. Successively removing the homogeneous initial form of each remainder gives a unique formal series for every local function. The Krull intersection theorem embeds the local ring in its completion. The power-series characterization is also given in Stacks Project Lemma 10.160.10.

Put $R={\mathcal{O}}_{X,x}$, $\mathfrak{m}={\mathfrak{m}}_x$ and $d=\dim R$. Smoothness makes $R$ a regular local ring, with residue field $k$. A list of local parameters $t_1,\dots ,t_d$ is a regular system of parameters: its images form a basis of the cotangent space of a local ring $\mathfrak{m}/{\mathfrak{m}}^2$. Set $J=(t_1,\dots ,t_d)$. The basis condition says $\mathfrak{m}=J+{\mathfrak{m}}^2$, so the finitely generated module $\mathfrak{m}/J$ equals $\mathfrak{m}(\mathfrak{m}/J)$. The Nakayama lemma gives

We now construct the requested formal power series, rather than assert that a local function has a convergent expansion. The associated graded ring of a regular local ring isIndeed these initial forms generate the graded ring, since the $t_i$ generate $\mathfrak{m}$. The Dimension theorem for Noetherian local rings gives $\dim {\mathrm{gr}}_{\mathfrak{m}}R=d$. A nonzero kernel in the surjection from the polynomial ring would make its quotient have dimension at most $d-1$, so that kernel is zero.

For $f\in R$, choose $P_0\in k$ to be its residue at $x$. Inductively suppose homogeneous polynomials $P_0,\dots ,P_{n-1}$ have been chosen such thatThe graded-ring isomorphism gives a unique homogeneous polynomial $P_n(T_1,\dots ,T_d)$ of degree $n$ representing the class of $r_n$ in ${\mathfrak{m}}^n/{\mathfrak{m}}^{n+1}$. Subtracting $P_n(t)$ makes the next remainder lie in ${\mathfrak{m}}^{n+1}$. HenceThis equality is interpreted in the adic completion of a module $\hat{R}=\underleftarrow{\lim }R/{\mathfrak{m}}^n$. The same induction applies to every compatible sequence of residues in this inverse limit, so evaluation $T_i\mapsto t_i$ defines an isomorphism $k[[T_1,\dots ,T_d]]\cong \hat{R}$. It is injective because the lowest nonzero homogeneous term of a series has nonzero initial form. The Krull intersection theorem gives ${\bigcap }_n{\mathfrak{m}}^n=0$, so $R$ embeds in its completion and the series determines $f$. This is the formal expansion at a smooth point; it depends on the chosen regular system of parameters, and does not claim that the infinite sum converges inside $R$. If $d=0$, the local ring is just $k$, and only the constant term is needed.

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.