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For a regular local ring of embedding dimension $e$, a minimal maximal-ideal generating set induces the displayed polynomial-algebra isomorphism. It is surjective in every degree. A nonzero homogeneous kernel relation would bound the cumulative Hilbert function by $O(t^{e-1})$, contradicting the prime-chain lower bound for local length at $\dim R=e$. The graded ring is a domain; the Krull intersection theorem then shows the original ring is a domain by multiplying nonzero initial forms.

For a local ring $(A,\mathfrak{m})$ with residue field $k$, put $B=A[[x]]$ and $I=\mathfrak{m}B+xB$. There is a natural $k$-vector space isomorphism $I/I^2\cong (\mathfrak{m}/{\mathfrak{m}}^2)\oplus k$. The inverse sends $(\bar{a},\bar{b})$ to the class of $a+bx$. It is well-defined because changing $a$ by ${\mathfrak{m}}^2$ or $b$ by $\mathfrak{m}$ changes the result by $I^2$. It is surjective since every $f\in I$ is congruent modulo $x^{2}B\subseteq I^2$ to its constant and linear terms. As $I^2={\mathfrak{m}}^{2}B+x\mathfrak{m}B+x^{2}B$, those coefficients of an element of $I^2$ lie in ${\mathfrak{m}}^2$ and $\mathfrak{m}$, respectively. This proves injectivity. Thus adjoining one formal variable increases finite embedding dimension by one.

Three coordinate axes in affine three-space have coordinate ideal $(xy,xz,yz)$; their common point has embedding dimension three. Three distinct coplanar lines meeting at one point have embedding dimension two there. An isomorphism preserves the unique point lying on all three irreducible components and preserves the algebraic cotangent space, so these two configurations are not isomorphic despite identical component incidence.

Let $(A,\mathfrak{m})$ be a regular local ring of Krull dimension $d$. Its formal power series ring $B=A[[x]]$ is Noetherian and local, with maximal ideal $I=\mathfrak{m}B+(x)$ and embedding dimension $d+1$. Since $I$ is generated by $d+1$ elements, the Krull height theorem gives $\dim B\le d+1$. For the reverse inequality choose a strict prime chain ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_d=\mathfrak{m}$ in $A$. Finite generation of each ${\mathfrak{p}}_i$ gives $B/{\mathfrak{p}}_{i}B\cong (A/{\mathfrak{p}}_i)[[x]]$, an integral domain, so each extended ideal is prime. The inclusions remain strict by contraction to $A$, and $\mathfrak{m}B⊊I$ adds one more step since $x\notin \mathfrak{m}B$. This proves $\dim B=d+1$, equal to its embedding dimension, and hence regularity.

For a Noetherian local ring $(R,\mathfrak{m})$, let $H_R(t)=\mathrm{length}(R/{\mathfrak{m}}^t)$. Its eventual Hilbert-Samuel polynomial has degree $d(R)$, equivalently the pole order at $1$ of the Hilbert series of ${\mathrm{gr}}_{\mathfrak{m}}R$. This is the local length-growth invariant, not the embedding dimension. The prime-chain lower bound for local length proves $\dim R\le d(R)$; the full local dimension theorem gives equality.

Since $A$ is a regular local ring, $\dim A=d$ is finite. Choose a chain of prime ideals of length $d$,Such a chain exists because the supremum defining this finite integer Krull dimension is attained; a maximal-length chain in a local ring ends at its maximal ideal.

For each $i$, the ideal ${\mathfrak{p}}_i$ is finitely generated because $A$ is Noetherian. This ensures that its extension ${\mathfrak{p}}_{i}B$ consists precisely of series all of whose coefficients lie in ${\mathfrak{p}}_i$: if ${\mathfrak{p}}_i=(b_1,\dots ,b_r)$, write each coefficient as ${\sum }_j{b}_j{c}_{jn}$ and gather them into the finitely many series ${\sum }_n{c}_{jn}{x}^n$. Thus coefficient reduction givesThe coefficient ring $A/{\mathfrak{p}}_i$ is an integral domain, and so is its formal power series ring: the first nonzero coefficient of a product is the product of the first nonzero coefficients of its factors. Hence each ${\mathfrak{p}}_{i}B$ is prime. Also ${\mathfrak{p}}_{i}B\cap A={\mathfrak{p}}_i$, so the strict inclusions persist upon extension. Finally, $x\notin \mathfrak{m}B$, since its image in $B/\mathfrak{m}B\cong k[[x]]$ is nonzero. We obtain the prime chainof length $d+1$. Therefore $\dim B\ge d+1$. Together with part (b),Its embedding dimension is also $d+1$, by part (b), so $B$ is itself a regular local ring. This proves that formal power series preserve regular local rings.

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.

Put $u=X_1-1$ and $v=X_2-1$. Evaluation at $(1,1)$ is surjective, andso its kernel is the prime ideal $\boxed{P=(u,v)}$. In particular $P/P^2$ is naturally an $R/P=\mathbb{Z}$-module, generated by the classes of $u,v$.

For integers $a,b$, including negative integers, reduction modulo $P^2$ givesFor negative powers, this follows from $(1+u{)}^{-1}\equiv 1-u$ and $(1+v{)}^{-1}\equiv 1-v$; the mixed product $uv$ vanishes modulo $P^2$. Therefore a Laurent polynomial $f$ satisfiesThe formal derivatives obey the product rule, so $D_i(ab)=\theta (a)D_i(b)+\theta (b)D_i(a)$. If $a,b\in P$, both terms vanish. Hence both $D_i$ vanish on $P^2$ and define a mapIt sends $[u]$ to $(1,0)$ and $[v]$ to $(0,1)$, and the displayed expansion gives its inverse. Thus the conormal module has the explicit basis

Let $A=R_P$ be the localization at a prime ideal, with maximal ideal $\mathfrak{m}=P{R}_P$. Every nonzero integer lies outside $P$ and is therefore inverted. The residue field isIndeed there is an explicit identificationinverting $1+u$ and $1+v$ has no effect in this local ring, and a polynomial denominator lies outside $(u,v)$ precisely when its constant term is nonzero. Conversely, clearing rational denominators turns every such fraction into a fraction from $R_P$.

For the Krull dimension, the prime ideal correspondence for localization preserves the strict chainof prime ideals in $R$. They are prime because the successive quotients are the integral domains $R$, $\mathbb{Z}[X_2,X_2^{-1}]$, and $\mathbb{Z}$. They remain distinct after localization at a prime ideal because they are all contained in $P$. Hence $\dim A\ge 2$.

For the upper bound, $R$ is a Noetherian ring by the Hilbert basis theorem and Localization of a Noetherian ring. Since $P$ is generated by two elements, the Krull height theorem gives $\mathrm{ht}P\le 2$. The prime ideal correspondence for localization identifies $\dim R_P$ with $\mathrm{ht}P$, so

A Noetherian local ring $(B,\mathfrak{n})$ with residue field $\kappa$ is a regular local ring ifBy the Nakayama lemma, the dimension on the left is the minimal number of generators of $\mathfrak{n}$; this is also called the embedding dimension.

The localization of a module is exact and commutes with quotients and products of ideals. Applying it to the already computed conormal module gives the cotangent space of a local ringIts dimension is two, equal to $\dim A$. Therefore $\boxed{R_P\text{ is a regular local ring}}$. Notice that $P/P^2$ has a basis over $\mathbb{Z}$, whereas the localized cotangent space of a local ring has a basis over the residue field $\mathbb{Q}$.

In the Noetherian local ring $A=k[[u,v]]$ with maximal ideal $\mathfrak{m}=(u,v)$, the embedding dimension is two but ${\mathfrak{m}}^r$ needs $r+1$ generators for every $r\ge 1$. The degree-$r$ monomials $u^r,u^{r-1}v,\dots ,v^r$ generate ${\mathfrak{m}}^r$, and their images form a basis of ${\mathfrak{m}}^r/{\mathfrak{m}}^{r+1}$, so no smaller generating list exists. These ideals are $\mathfrak{m}$-primary: if $ab\in {\mathfrak{m}}^r$ and $a\notin {\mathfrak{m}}^r$, then $b$ cannot be a unit; hence $b\in \mathfrak{m}$ and $b^r\in {\mathfrak{m}}^r$. Consequently embedding dimension bounds the minimum over all maximal-primary ideals, because the maximal ideal itself is eligible, but does not bound the number of generators of each such ideal.