Krull Dimension and Height Theorems

If $\mathfrak{p}$ is minimal, there is no strict prime ideal properly contained in $\mathfrak{p}$. Therefore every strict chain ending at $\mathfrak{p}$ has length $0$, and [L1] gives $\mathrm{ht}(\mathfrak{p})=0$.

If $\mathrm{ht}(\mathfrak{p})=0$, then [L1] says no strict chain of positive length ends at $\mathfrak{p}$. In particular there is no prime ideal properly contained in $\mathfrak{p}$, so $\mathfrak{p}$ is minimal.

The two implications prove that minimal primes are exactly the primes of height zero.

By [L1], $R$ is nonzero, so $(0)⊊R$. If $ab\in (0)$, then $ab=0$, and [L1] gives $a=0$ or $b=0$. Thus $a\in (0)$ or $b\in (0)$, so [L2] shows that $(0)$ is prime.

Every ideal contains $0$, hence $(0)\subseteq \mathfrak{p}$. Therefore any chain of primes below $\mathfrak{p}$ can be extended by adjoining $(0)$ at the bottom if it is not already present.

So every prime chain below $\mathfrak{p}$ begins at $(0)$ after at most one downward extension.

By [L1], strict prime chains in $R\mathrm{/}\mathfrak{p}$ ending at $\mathfrak{q}\mathrm{/}\mathfrak{p}$ are in bijection with strict prime chains in $R$ beginning at $\mathfrak{p}$ and ending at $\mathfrak{q}$. Corresponding chains have the same length.

Applying [L2] to the prime $\mathfrak{q}\mathrm{/}\mathfrak{p}$ of the quotient ring $R\mathrm{/}\mathfrak{p}$, the height of $\mathfrak{q}\mathrm{/}\mathfrak{p}$ is exactly the supremum of the lengths of those quotient chains. Step 1.1 translates that supremum into the displayed chains in $R$.

Therefore height in the quotient is the relative chain length from $\mathfrak{p}$ to $\mathfrak{q}$.

Let $\mathfrak{q}\subseteq \mathfrak{p}$ be a minimal prime of $R$. By [L1], $\mathrm{ht}(\mathfrak{q})=0$. The quotient $R\mathrm{/}\mathfrak{q}$ is a Noetherian domain by [L2] and [L3], and then [L4] makes $A=(R\mathrm{/}\mathfrak{q}{)}_{\mathfrak{p}\mathrm{/}\mathfrak{q}}$ a Noetherian local domain.

If $\mathfrak{q}=\mathfrak{p}$, then $\mathfrak{p}$ already has height $0$ by step 1.1 and there is nothing left to prove. Assume now that $\mathfrak{q}\mathrm{\ne }\mathfrak{p}$. Since $\mathfrak{p}$ is minimal over $(x)$, the element $x$ cannot lie in $\mathfrak{q}$; otherwise the prime $\mathfrak{q}$ would also contain $(x)$ and minimality would force $\mathfrak{q}=\mathfrak{p}$. Thus the image of $x$ in $R\mathrm{/}\mathfrak{q}$, and hence in $A$, is nonzero.

Let $\mathfrak{m}$ be the maximal ideal of $A$. By [L5], primes of $A$ correspond to primes of $R$ that lie between $\mathfrak{q}$ and $\mathfrak{p}$. Because $\mathfrak{p}$ is minimal over $(x)$, the prime $\mathfrak{p}\mathrm{/}\mathfrak{q}$ is minimal over the image of $(x)$ in $R\mathrm{/}\mathfrak{q}$, and after localizing there is no smaller prime of $A$ containing $x\mathrm{/}1$. Hence $\mathfrak{m}$ is minimal over $(x\mathrm{/}1)$.

Let ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_d=\mathfrak{p}$ be any strict prime chain. If $d=0$ there is nothing to bound. If $d>0$, then ${\mathfrak{p}}_0\mathrm{\ne }\mathfrak{p}$, so minimality of $\mathfrak{p}$ over $(x)$ gives $x\mathrm{\notin }{\mathfrak{p}}_0$. By [L2]–[L4], $A_0=(R\mathrm{/}{\mathfrak{p}}_0{)}_{\mathfrak{p}\mathrm{/}{\mathfrak{p}}_0}$ is a Noetherian local domain in which the image of $x$ is nonzero; [L5] also shows that its maximal ideal is minimal over that image. The original chain induces a strict chain of length $d$ ending at this maximal ideal. Therefore a height bound of $1$ in every reduced local-domain case forces $d\le 1$. Since the original chain was arbitrary, ${\mathrm{ht}}_R(\mathfrak{p})\le 1$.

Since $\mathfrak{m}$ is both maximal and minimal among primes containing $(x)$, it is the only such prime. Hence the quotient $\bar{R}=R\mathrm{/}(x)$ is Noetherian by [L2] and has the single prime $\bar{\mathfrak{m}}$. Its nilradical is therefore $\bar{\mathfrak{m}}$, so [L3] gives ${\bar{\mathfrak{m}}}^N=0$ for some $N$. Each layer ${\bar{\mathfrak{m}}}^j\mathrm{/}{\bar{\mathfrak{m}}}^{j+1}$ is a finite-dimensional vector space over $R\mathrm{/}\mathfrak{m}$: it is finitely generated by [L2] and is annihilated by $\bar{\mathfrak{m}}$. A descending chain of ideals in $\bar{R}$ induces descending chains in these finitely many finite-dimensional layers, so all layers, and therefore the original chain, stabilize. Thus $\bar{R}$ is Artinian.

Let $\mathfrak{p}⊊\mathfrak{m}$ be prime. Then $x\mathrm{\notin }\mathfrak{p}$. For $r\ge 1$, define the symbolic power ${\mathfrak{p}}^{(r)}:={\mathfrak{p}}^r{R}_{\mathfrak{p}}\cap R\mathrm{.}$ These form a descending chain. By step 1.1, the ideals $({\mathfrak{p}}^{(r)}+(x))\mathrm{/}(x)$ in $\bar{R}$ stabilize, so for some $r$, ${\mathfrak{p}}^{(r)}+(x)={\mathfrak{p}}^{(r+1)}+(x)\mathrm{.}$ If $a\in {\mathfrak{p}}^{(r)}$, write $a=b+xc$ with $b\in {\mathfrak{p}}^{(r+1)}$. In $R_{\mathfrak{p}}$ the element $x$ is a unit, while $a-b\in {\mathfrak{p}}^r{R}_{\mathfrak{p}}$; hence $c\in {\mathfrak{p}}^r{R}_{\mathfrak{p}}\cap R={\mathfrak{p}}^{(r)}$. Therefore ${\mathfrak{p}}^{(r)}={\mathfrak{p}}^{(r+1)}+x{\mathfrak{p}}^{(r)}\mathrm{.}$

The quotient module $M={\mathfrak{p}}^{(r)}\mathrm{/}{\mathfrak{p}}^{(r+1)}$ is finitely generated by [L2], and step 2.1 says $M=xM$. Apply [L5] with $I=(x)$. It gives $(1-a)M=0$ for some $a\in (x)\subseteq \mathfrak{m}$. Fact [L6] makes $1-a$ a unit, so $M=0$ and ${\mathfrak{p}}^{(r)}={\mathfrak{p}}^{(r+1)}$. Localizing this equality at $\mathfrak{p}$ gives $(\mathfrak{p}{R}_{\mathfrak{p}}{)}^r=(\mathfrak{p}{R}_{\mathfrak{p}}{)}^{r+1}\mathrm{.}$

Apply [L5] in the local ring $R_{\mathfrak{p}}$ to the finite module $(\mathfrak{p}{R}_{\mathfrak{p}}{)}^r$ and the ideal $\mathfrak{p}{R}_{\mathfrak{p}}$. Step 3.1 says that ideal times the module is the module, so [L6] gives $(\mathfrak{p}{R}_{\mathfrak{p}}{)}^r=0$. Because $R_{\mathfrak{p}}$ is a domain, this forces $\mathfrak{p}{R}_{\mathfrak{p}}=0$, and contraction through [L4] gives $\mathfrak{p}=(0)$. Thus every prime strictly below $\mathfrak{m}$ is zero, and [L1] yields $\mathrm{ht}(\mathfrak{m})\le 1$.

By [L1], choose a minimal prime $\mathfrak{q}\subseteq \mathfrak{p}$ and pass to the Noetherian local domain $A=(R\mathrm{/}\mathfrak{q}{)}_{\mathfrak{p}\mathrm{/}\mathfrak{q}}$ whose maximal ideal is minimal over the image of $x$.

Fact [L2] applies to $A$, so its maximal ideal has height at most $1$.

The reduction packaged in [L1] identifies this with the desired bound $\mathrm{ht}(\mathfrak{p})\le 1$ upstairs in $R$.

Suppose $\dim R=0$. By [L2], the chain $(0)\subseteq \mathfrak{m}$ is a prime chain. If it were strict, [L3] would give $\dim R\ge 1$, contradiction. Hence $\mathfrak{m}=(0)$, so every nonzero element is outside the maximal ideal and is a unit by the AC-qualified [L4]. Thus $R$ is a field.

Conversely, if $R$ is a field, its unique maximal ideal is $(0)$. Therefore the only prime ideal is $(0)$, and [L3] gives $\dim R=0$.

So a Noetherian local domain has dimension zero exactly when it is a field.

By [L1], $\mathrm{ht}(\mathfrak{p})\le 1$.

By [L4], the nilradical is a finite intersection of minimal primes. Since $\mathfrak{p}$ contains that intersection, it contains one of those primes; choose $\mathfrak{q}\subseteq \mathfrak{p}$. The localization $R_{\mathfrak{q}}$ is a nonzero Noetherian local ring with only one prime, $\mathfrak{q}{R}_{\mathfrak{q}}$. Its nilradical is therefore that maximal ideal and is nilpotent by [L4]. If $x\in \mathfrak{q}$, the image $x\mathrm{/}1$ would be nilpotent in $R_{\mathfrak{q}}$, contradicting that multiplication by the nonzerodivisor $x$ remains injective after localization. Thus $x\mathrm{\notin }\mathfrak{q}$ and $\mathfrak{q}\mathrm{\ne }\mathfrak{p}$.

Apply [L2] to this $\mathfrak{q}$. In the resulting Noetherian local domain $A=(R\mathrm{/}\mathfrak{q}{)}_{\mathfrak{p}\mathrm{/}\mathfrak{q}}$, the image of $x$ is nonzero and lies in the maximal ideal. If that maximal ideal had height $0$, then [L3] would make $A$ a field, forcing $x\mathrm{/}1$ to be a unit, contradiction. Hence the maximal ideal of $A$ has height at least $1$, so $\mathfrak{p}$ has height at least $1$.

Steps 1.1 and 2.1 give $\mathrm{ht}(\mathfrak{p})=1$.

Write $J=(x_2,\dots ,x_n)$. By [L1], the family $\{{\mathfrak{p}}_1^{\mathrm{\prime }},\dots ,{\mathfrak{p}}_r^{\mathrm{\prime }}\}$ is finite. Every ${\mathfrak{p}}_i^{\mathrm{\prime }}$ is minimal among primes containing $J$. Since $\mathfrak{p}$ contains $J$ but is not one of them, $\mathfrak{p}⊈{\mathfrak{p}}_i^{\mathrm{\prime }}$ for every $i$. Finite prime avoidance [L2] therefore supplies an element of $\mathfrak{p}$ outside their union. The given chain has $d\ge 1$, so its ${\mathfrak{p}}_1$ exists.

Starting at $j=d-1$ and descending to $j=1$, keep the current strict chain and the invariant that ${\mathfrak{p}}_{j+1}$ is contained in none of the ${\mathfrak{p}}_i^{\mathrm{\prime }}$. If ${\mathfrak{p}}_j$ already has this property, leave it unchanged. Otherwise choose $c\in {\mathfrak{p}}_{j+1}\setminus {\bigcup }_i{\mathfrak{p}}_i^{\mathrm{\prime }}$ by [L2]. Since ${\mathfrak{p}}_{j-1}⊊{\mathfrak{p}}_j\subseteq {\mathfrak{p}}_i^{\mathrm{\prime }}$ for at least one $i$, the element $c$ is outside ${\mathfrak{p}}_{j-1}$. In the Noetherian quotient $A=R\mathrm{/}{\mathfrak{p}}_{j-1}$, the radical of $(\bar{c})$ is the finite intersection of its minimal primes by [L1]. The prime ${\mathfrak{p}}_{j+1}\mathrm{/}{\mathfrak{p}}_{j-1}$ contains this intersection, so by primality it contains one of those minimal primes; call its inverse image $\mathfrak{q}$. Then ${\mathfrak{p}}_{j-1}⊊\mathfrak{q}$ because $c\in \mathfrak{q}$ but $c\mathrm{\notin }{\mathfrak{p}}_{j-1}$. By [L3] the selected prime has height at most $1$ in $A$, whereas ${\mathfrak{p}}_{j+1}\mathrm{/}{\mathfrak{p}}_{j-1}$ has the two-step chain through ${\mathfrak{p}}_j\mathrm{/}{\mathfrak{p}}_{j-1}$. Thus $\mathfrak{q}⊊{\mathfrak{p}}_{j+1}$. It also contains $c$, so it is contained in none of the ${\mathfrak{p}}_i^{\mathrm{\prime }}$. Replacing ${\mathfrak{p}}_j$ by $\mathfrak{q}$ preserves the chain length and establishes the invariant one level lower.

After the finite descent in step 2.1, ${\mathfrak{p}}_1$ is contained in none of the ${\mathfrak{p}}_i^{\mathrm{\prime }}$. When $d=1$, this says ${\mathfrak{p}}_1=\mathfrak{p}$, as already established in step 1.1.

Since ${\mathfrak{p}}_1$ is not contained in the finite union ${\bigcup }_i{\mathfrak{p}}_i^{\mathrm{\prime }}$, [L2] gives $b\in {\mathfrak{p}}_1\setminus {\bigcup }_i{\mathfrak{p}}_i^{\mathrm{\prime }}$. In particular $b\in \mathfrak{p}$, so $\mathfrak{p}$ contains $(b,x_2,\dots ,x_n)$.

Let $\mathfrak{q}\subseteq \mathfrak{p}$ be prime minimal over $(b,x_2,\dots ,x_n)$. Because $\mathfrak{q}$ contains $(x_2,\dots ,x_n)$, it contains one of the minimal primes ${\mathfrak{p}}_i^{\mathrm{\prime }}$. By the choice of $b$, one has $b\mathrm{\notin }{\mathfrak{p}}_i^{\mathrm{\prime }}$, so $\mathfrak{q}\mathrm{\ne }{\mathfrak{p}}_i^{\mathrm{\prime }}$. If $\mathfrak{q}\mathrm{\ne }\mathfrak{p}$, then $\mathfrak{p}⊋\mathfrak{q}⊋{\mathfrak{p}}_i^{\mathrm{\prime }}$ is a strict chain in the quotient ring $R\mathrm{/}(x_2,\dots ,x_n)$, so the prime $\mathfrak{p}\mathrm{/}(x_2,\dots ,x_n)$ has height at least $2$. But $\mathfrak{p}$ is minimal over $(x_1,\dots ,x_n)$, hence $\mathfrak{p}\mathrm{/}(x_2,\dots ,x_n)$ is minimal over the principal ideal generated by the image of $x_1$, contradicting [L3]. Therefore $\mathfrak{q}=\mathfrak{p}$.

Hence $\mathfrak{p}$ is minimal over $(b,x_2,\dots ,x_n)$, with $b$ chosen from ${\mathfrak{p}}_1$ of a chain of the same length.

By [L1], every prime ideal of $R$ containing $(b,x_2,\dots ,x_n)$ and lying inside $\mathfrak{p}$ is equal to $\mathfrak{p}$. Hence $\mathfrak{p}\mathrm{/}(b)$ is minimal over the images of $x_2,\dots ,x_n$ in $R\mathrm{/}(b)$.

Because $b\in {\mathfrak{p}}_i$ for every $1\le i\le d$, [L2] sends the given chain to ${\mathfrak{p}}_d\mathrm{/}(b)⊋{\mathfrak{p}}_{d-1}\mathrm{/}(b)⊋\cdots ⊋{\mathfrak{p}}_1\mathrm{/}(b)\mathrm{.}$ The inclusions stay strict, because equality between consecutive quotient primes would contract back to equality between the corresponding primes of $R$.

Thus quotienting by $(b)$ leaves a prime minimal over $n-1$ generators and reduces this chosen chain by exactly one step.

If $n=1$, [L1] gives $\mathrm{ht}(\mathfrak{p})\le 1$.

Assume $n\ge 2$ and that the theorem is known for $(n-1)$-generated ideals. Suppose for contradiction that $\mathrm{ht}(\mathfrak{p})\ge n+1$. Then there exists a strict prime chain $\mathfrak{p}={\mathfrak{p}}_d⊋{\mathfrak{p}}_{d-1}⊋\cdots ⊋{\mathfrak{p}}_0$ with $d\ge n+1\ge 3$. Under the assumed AC, [L2] replaces $x_1$ by a suitable element $b\in {\mathfrak{p}}_1$ of a same-length chain for which $\mathfrak{p}$ is minimal over $(b,x_2,\dots ,x_n)$.

By [L3], the quotient prime $\mathfrak{p}\mathrm{/}(b)$ is minimal over the $(n-1)$-generated ideal generated by the images of $x_2,\dots ,x_n$, and the above chain descends to a strict chain of length $d-1\ge n$ ending at $\mathfrak{p}\mathrm{/}(b)$ in $R\mathrm{/}(b)$. This contradicts the induction hypothesis for $(n-1)$ generators.

Therefore the assumption $\mathrm{ht}(\mathfrak{p})\ge n+1$ is impossible, and $\mathrm{ht}(\mathfrak{p})\le n$.

By [L1], the maximal ideal of the local ring $R_{\mathfrak{p}}$ is $\mathfrak{p}{R}_{\mathfrak{p}}$. Let $n$ be its minimal number of generators, finite by Noetherianity. If $n=0$, the maximal ideal is zero and the local ring is a field, so its dimension is zero. If $n\ge 1$, the maximal ideal is minimal over the ideal generated by those same $n$ elements. Applying the Choice-qualified [L2] inside $R_{\mathfrak{p}}$ gives $\dim R_{\mathfrak{p}}\le n$.

By [L3], $\mathrm{ht}(\mathfrak{p})=\dim R_{\mathfrak{p}}$, so step 1.1 says $\mathrm{ht}(\mathfrak{p})\le {\mu }_{R_{\mathfrak{p}}}(\mathfrak{p}{R}_{\mathfrak{p}})$.

This is the claimed local-generator bound on height.

If $r=0$, the empty list works.

Assume $r\ge 1$. The prime ideals of height $0$ are exactly the minimal primes of $R$, hence are finite by [L1]. None of them contains $\mathfrak{a}$, because $\mathrm{ht}(\mathfrak{a})=r\ge 1$. Therefore [L2] provides $a_1\in \mathfrak{a}$ outside every height-zero prime. Any prime minimal over $(a_1)$ must then have height at least $1$, while [L3] gives height at most $1$. Thus $(a_1)$ has height exactly $1$.

Suppose $1<i\le r$ and $a_1,\dots ,a_{i-1}\in \mathfrak{a}$ have already been chosen so that $J=(a_1,\dots ,a_{i-1})$ has height exactly $i-1$. The minimal primes over $J$ are finite by [L1]. Among them, collect those of height $i-1$; none can contain $\mathfrak{a}$, because $\mathfrak{a}$ has height $r\ge i$. By [L2], choose $a_i\in \mathfrak{a}$ outside all of those primes. Any prime $\mathfrak{q}$ minimal over $J+(a_i)$ contains some minimal prime $\mathfrak{p}$ over $J$. If $\mathrm{ht}\mathfrak{q}\le i-1$, then $\mathrm{ht}\mathfrak{p}\ge i-1$ and strict containment $\mathfrak{p}⊊\mathfrak{q}$ would force $\mathrm{ht}\mathfrak{q}\ge i$; hence $\mathfrak{q}=\mathfrak{p}$ is one of the excluded height-$(i-1)$ primes, impossible since it contains $a_i$. Thus every such $\mathfrak{q}$ has height at least $i$. By [L3] it has height at most $i$, so $J+(a_i)$ has height exactly $i$.

Steps 1.1, 1.2, and 2.1 build the required list $a_1,\dots ,a_r\in \mathfrak{a}$.

By [L1] and [L2], the local ring $R_{\mathfrak{p}}$ has maximal ideal $\mathfrak{p}{R}_{\mathfrak{p}}$ and dimension $n$. Applying [L3] to the proper ideal $\mathfrak{p}{R}_{\mathfrak{p}}$ produces elements $u_1,\dots ,u_n\in \mathfrak{p}{R}_{\mathfrak{p}}$ such that $J=(u_1,\dots ,u_n)$ has height $n$. Writing each $u_i=a_i\mathrm{/}{s}_i$ with $a_i\in \mathfrak{p}$ and $s_i\mathrm{\notin }\mathfrak{p}$, and replacing $u_i$ by the associate $s_i{u}_i=a_i\mathrm{/}1$, we may assume $u_i=x_i\mathrm{/}1$ with $x_i\in \mathfrak{p}$.

Let $\mathfrak{q}$ be a prime ideal minimal over $J=(x_1\mathrm{/}1,\dots ,x_n\mathrm{/}1)$. Because $J$ has height $n$, the prime $\mathfrak{q}$ has height $n$. The maximal ideal $\mathfrak{p}{R}_{\mathfrak{p}}$ also contains $J$, so $\mathfrak{q}\subseteq \mathfrak{p}{R}_{\mathfrak{p}}$. If the inclusion were strict, a strict chain of length $n$ ending at $\mathfrak{q}$ would extend by one more step to a chain ending at $\mathfrak{p}{R}_{\mathfrak{p}}$, contradicting $\dim (R_{\mathfrak{p}})=n$. Therefore $\mathfrak{q}=\mathfrak{p}{R}_{\mathfrak{p}}$, so the maximal ideal is minimal over $J$.

Therefore $\mathfrak{p}$ becomes minimal over an ideal generated by $n$ elements after localizing at $\mathfrak{p}$.

Suppose $\sqrt{J}=\mathfrak{m}$. Since every element outside $\mathfrak{m}$ is a unit in a local ring, it acts injectively on $R\mathrm{/}J$. Also $\sqrt{J}=\mathfrak{m}$ means some power ${\mathfrak{m}}^N$ lies in $J$, hence kills $R\mathrm{/}J$. Therefore [L2] makes $J$ a $\mathfrak{m}$-primary ideal.

Conversely, if $J$ is $\mathfrak{m}$-primary, [L2] applied to the finite module $R\mathrm{/}J$ gives $\sqrt{J}=\mathfrak{m}$. Then [L1] says $(x_1,\dots ,x_d)$ is a system of parameters.

Thus parameter ideals are exactly the $d$-generated $\mathfrak{m}$-primary ideals.

By [L1], the minimal primes of $R$ form a finite set; because $d>0$, none equals $\mathfrak{m}$. Hence [L2] provides $x\in \mathfrak{m}$ outside every minimal prime.

Let $\mathfrak{p}\mathrm{/}(x)$ be a minimal prime of $R\mathrm{/}(x)$. By [L4], the prime $\mathfrak{p}$ is minimal over $(x)$ in $R$. Choose a minimal prime $\mathfrak{q}\subseteq \mathfrak{p}$. Since $x\mathrm{\notin }\mathfrak{q}$, its image in the domain $R\mathrm{/}\mathfrak{q}$ is a nonzerodivisor. The prime $\mathfrak{p}\mathrm{/}\mathfrak{q}$ is minimal over that image: an intervening prime would give a prime strictly below $\mathfrak{p}$ containing $x$. Thus [L3] gives ${\mathrm{ht}}_{R\mathrm{/}\mathfrak{q}}(\mathfrak{p}\mathrm{/}\mathfrak{q})=1$. Every chain in $R\mathrm{/}\mathfrak{p}$ extends upward to a chain in $R\mathrm{/}\mathfrak{q}$ longer by one step.

Since $\dim (R\mathrm{/}\mathfrak{q})\le d$, step 2.1 implies $\dim (R\mathrm{/}\mathfrak{p})\le d-1$. Taking the supremum over all minimal primes $\mathfrak{p}\mathrm{/}(x)$ of $R\mathrm{/}(x)$ yields $\dim (R\mathrm{/}(x))\le d-1$.

Thus one can choose a first parameter outside the minimal components, and every such choice lowers dimension by at least one.

By [L1], the maximal ideal of $R\mathrm{/}(x)$ is minimal over an ideal generated by $d-1$ elements. If $d=1$, that ideal is zero, so the maximal ideal is a minimal prime and the quotient has dimension zero. If $d\ge 2$, apply the height theorem to its $d-1\ge 1$ generators. In either case $\dim (R\mathrm{/}(x))\le d-1$.

Suppose instead that $\dim (R\mathrm{/}(x))\le d-2$. The quotient is a nonzero local ring, so this is impossible when $d=1$. For $d\ge 2$, [L2] applied in the local ring $R\mathrm{/}(x)$ provides an ideal generated by at most $d-2$ elements whose radical is the maximal ideal $\mathfrak{m}\mathrm{/}(x)$. Lifting those generators to $R$ and adjoining $x$, we obtain an ideal of $R$ generated by at most $d-1$ elements with radical $\mathfrak{m}$. Applying the height theorem to the maximal ideal of $R$ would then give $d=\dim R\le d-1$, contradiction.

Therefore $\dim (R\mathrm{/}(x))=d-1$.

If $d=0$, the local ring has no strict prime inclusion, so its maximal ideal $\mathfrak{m}$ is its only prime. Consequently $\sqrt{(0)}=\mathfrak{m}$, and the empty tuple generates $(0)$ and is a system of parameters by [L2].