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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.

If $I$ is an $\mathfrak{m}$-primary ideal generated by $n>0$ elements in a Noetherian local ring and $L=\mathrm{length}(R/I)$, then $I^j/I^{j+1}$ is a quotient of $\left(\genfrac{}{}{0ex}{}{j+n-1}{n-1}\right)$ copies of $R/I$. Summing their lengths gives the displayed bound. Since $I\subseteq \mathfrak{m}$, it also bounds $\mathrm{length}(R/{\mathfrak{m}}^t)$. Together with the prime-chain lower bound for local length, localization at a minimal prime proves the Krull height theorem.

The height of a prime ideal is the supremum of lengths of strict prime chains ending at it, equivalently the Krull dimension of its localization at a prime ideal. The Krull height theorem bounds the height of a minimal prime over an $n$-generated ideal by $n$.

For closed irreducible subvarieties $Y,Z$ of a smooth $d$-dimensional algebraic variety $X$, every irreducible component of their intersection has dimension at least $\dim Y+\dim Z-d$. Locally the diagonal of $X\times X$ is cut out by $d$ parameters. Intersecting it with $Y\times Z$ and applying the Krull height theorem gives the bound. The three-dimensional affine quadric cone shows that the same lower bound can fail when $X$ is singular.

A formal power series $f={\sum }_{n\ge 0}{a}_n{x}^n$ is a unit exactly when its constant coefficient is a unit in $A$. If $fg=1$, constant coefficients give $a_0{b}_0=1$. Conversely, for invertible $a_0$, setThen every positive-degree coefficient of $fg$ is zero and its constant coefficient is one, so $g=\sum b_n{x}^n$ is an inverse. This is a formal coefficient calculation and requires no convergence.

Since $A$ is a local ring, its nonunits are exactly $\mathfrak{m}$. The nonunits of $B$ are consequently the series with $a_0\in \mathfrak{m}$, which are exactly $I=\mathfrak{m}B+xB$: such a series is $a_0+xg$, and conversely an element of $I$ has constant coefficient in $\mathfrak{m}$. The map $B\to k$ taking the constant coefficient modulo $\mathfrak{m}$ is surjective with kernel $I$, so $I$ is a maximal ideal. Every proper ideal consists of nonunits and hence lies in $I$, proving that $I$ is the unique maximal ideal. ThusThis proves the description of formal power series over a local ring.

To compute the cotangent space of a formal power series ring, expandDefine the $k$-linear mapChanging a lift $a$ by ${\mathfrak{m}}^2$ or a lift $b$ by $\mathfrak{m}$ changes the image by $I^2$, so the map is well-defined. Every series in $I$ is congruent modulo $x^{2}B\subseteq I^2$ to its first two coefficients, proving surjectivity. If $a+bx\in I^2$, its constant coefficient lies in ${\mathfrak{m}}^2$ and its linear coefficient lies in $\mathfrak{m}$, so both input classes are zero. This proves injectivity. ThereforeFinally, part (a) gives generators $a_1,\dots ,a_d$ of $\mathfrak{m}$, so $I=(a_1,\dots ,a_d,x)$. Use the Krull height theorem in its lecture form: in a Noetherian ring, a prime minimal over an ideal generated by $r$ elements has height at most $r$. The problem allows Noetherianity of $B$ to be assumed. Apply the theorem to $I$ itself, whose height is the Krull dimension of the local ring $B$. Hence $\boxed{\dim B=\mathrm{ht}(I)\le d+1}$.

Use the usual convention that a variety is an irreducible variety. For a quasi-projective algebraic set $T$, its algebraic dimension is the supremum of the lengths $r$ of strict chainsof nonempty irreducible closed subsets of $T$. It is the maximum of the algebraic dimensions of its irreducible components. On an affine algebraic set, the correspondence between irreducible closed subsets and prime ideals reverses inclusion, so this is the Krull dimension of the coordinate ring. On a quasi-projective algebraic set, it is the supremum of the Krull dimensions of the coordinate rings of its affine open subsets. At a closed point $x$, the Krull dimension of the local ring measures chains through $x$.

Here is a closed-point dimension lemma for affine domains that avoids transcendence degree. Put $A=k[X]$. By Noether normalization, there is a integral extensionThe number of variables is $d$ because integral extensions preserve Krull dimension and $\dim B=d$. For any maximal ideal $\mathfrak{m}$ of $A$, its contraction $\mathfrak{n}$ to $B$ is a maximal ideal. Since $k$ is an algebraically closed field, $\mathfrak{n}=(z_1-c_1,\dots ,z_d-c_d)$ has height of a prime ideal $d$. The going-down theorem applies because $B$ is an integrally closed domain and $A$ is a integral domain. It lifts a length-$d$ chain below $\mathfrak{n}$ to one below $\mathfrak{m}$. HenceThe opposite inequality follows from $\dim A=d$.

For the nonempty open subset $U$, choose a nonempty principal open subset $D(g)\subset U$ and a closed point $x\in D(g)$. Every prime ideal below ${\mathfrak{m}}_x$ avoids $g$, so the preceding chain survives in the localization $A_g$. Consequently $\dim D(g)\ge d$. Conversely, any chain of irreducible closed subsets in $U$ gives a chain of the same length after taking closures in $X$: intersecting those closures with $U$ recovers the original subsets. Therefore

For the principal hypersurface dimension lemma, let $P$ be a minimal prime ideal over $(f)$. Since $f\ne 0$ in the integral domain $A$, $P\ne (0)$. The Krull principal ideal theorem gives $\mathrm{ht}P=1$. Choose a closed point $x$ on $V(P)$ lying on none of the other finitely many irreducible components of $V(f)$. Such a point exists because those other components cut out proper closed subsets of the irreducible variety $V(P)$, and closed points are dense. Set $R=A_{{\mathfrak{m}}_x}$. Then $\dim R=d$ andWrite $s=\dim R/fR$. Choose a system of parameters ${\bar{y}}_1,\dots ,{\bar{y}}_s$ in $R/fR$ and lift it to $R$. The ideal $(f,y_1,\dots ,y_s)$ has radical equal to the maximal ideal of $R$. The Krull height theorem yields $d\le s+1$. On the other hand, any chain of prime ideals containing $f$ can be extended strictly at the bottom by the zero prime ideal of the integral domain $R$, giving $s\le d-1$. Thus $s=d-1$. Since $R/PR$ is a localization of $A/P$, $\dim A/P\ge d-1$. Extending a chain in $A/P$ by $(0)⊊P$ also gives $\dim A/P\le d-1$. Hence every component has the required dimension:This argument uses Noether normalization, going-down theorem, and the Krull height theorem, never the dimension from the function field theorem. Irreducibility is essential: if the word variety were instead allowed to mean an arbitrary reducible affine algebraic set, neither assertion would hold without extra hypotheses. For example, a disjoint union of an affine plane and an affine line has an open component of smaller algebraic dimension; a function equal to $1$ on the plane and a coordinate on the line has a nonempty zero set of algebraic dimension $0$.

The height of a prime ideal is the supremum of the lengths of strict chains of prime ideals ending at $P$:The last equality follows from the prime ideal correspondence for localization. Here is a length-growth proof of the Krull height theorem that makes the bound explicit.

We first establish a prime-chain lower bound for local length. For a Noetherian local ring $(B,\mathfrak{m})$ admitting a prime chain of length $d$, put $H_B(t)=\mathrm{length}(B/{\mathfrak{m}}^t)$. Thenfor some $c>0$. For $d=0$, use $H_B(t)\ge 1$. For $d>0$, quotient by the first prime ideal of the chain; this only decreases $H_B$ and reduces us to a local integral domain $B$ withTake $0\ne a\in P_1$ and put $C=B/(a)$. It has a prime chain of length $d-1$, so induction bounds $H_C(t)$ below by a positive multiple of $t^{d-1}$.

The Artin-Rees lemma applied to $(a)\subseteq B$ gives, for a fixed integer $c_0\ge 1$,Because $a$ is a non-zero-divisor, the first term in the exact sequencehas length at least $H_B(t-c_0)$. HenceIterate this inequality about $t/(2c_0)$ times, while its arguments remain at least $t/2$. This gives the desired positive multiple of $t^d$. The particular Artin-Rees lemma inclusion used here follows directly from finite generation: the Rees algebra ${⨁}_{j\ge 0}{\mathfrak{m}}^j{t}^j$ is Noetherian, by the Hilbert basis theorem, and its graded submodule ${⨁}_j((a)\cap {\mathfrak{m}}^j)t^j$ has finitely many homogeneous generators. A bound on their degrees gives $c_0$.

Now localize at the given minimal prime ideal $P$. Write $B=R_P$, $\mathfrak{m}=P{R}_P$, and $J=I{R}_P$. The only prime ideal of $B$ containing $J$ is $\mathfrak{m}$, so $\sqrt{J}=\mathfrak{m}$. Since $\mathfrak{m}$ is finitely generated, some ${\mathfrak{m}}^q\subseteq J$. Thus $B/J$ has finite module length, say $L$. The ideal $J$ is generated by at most $n$ elements.

For $n>0$, products of these generators give a surjection from $\left(\genfrac{}{}{0ex}{}{j+n-1}{n-1}\right)$ copies of $B/J$ onto $J^j/J^{j+1}$. Summing the lengths gives the generator bound for primary-ideal lengthSince $J\subseteq \mathfrak{m}$, we have $J^t\subseteq {\mathfrak{m}}^t$, and thereforeThe prime-chain lower bound for local length now excludes any chain of length greater than $n$. ConsequentlyFor $n=0$, minimality over the zero ideal says that $P$ is a minimal prime ideal, so its height is zero. This also handles that boundary case.

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 $\mathbb{Q}$.

For a prime ideal $\mathfrak{p}$, its height is the supremum of the lengths of strict chains of prime ideals ending at $\mathfrak{p}$. For a proper ideal $I$,This is the height of an ideal.

The Krull height theorem states that if $R$ is Noetherian, $I=(a_1,\dots ,a_m)$, and $\mathfrak{p}$ is minimal over $I$, then

We prove it by induction on $m$. The case $m=1$ is the Krull principal ideal theorem. For the induction step, let ${\mathfrak{q}}_1,\dots ,{\mathfrak{q}}_s$ be the finitely many minimal primes over $(a_2,\dots ,a_m)$ that lie below $\mathfrak{p}$. By induction each has height at most $m-1$; if one equals $\mathfrak{p}$, we are done.

Otherwise, suppose $\mathfrak{p}$ has finite height $d$ and choose a chainwhose first nonminimal term ${\mathfrak{p}}_1$ is contained in none of the ${\mathfrak{q}}_i$. Such a chain is obtained by prime avoidance and the principal ideal theorem: a three-term segment can be replaced by a prime minimal over ${\mathfrak{p}}_0+(b)$ for an element $b$ avoiding the finitely many unwanted primes. Choose

The prime $\mathfrak{p}$ is minimal over $(b,a_2,\dots ,a_m)$. Otherwise a prime strictly between some ${\mathfrak{q}}_i$ and $\mathfrak{p}$ would show that $\mathfrak{p}/(a_2,\dots ,a_m)$ has height at least two, although it is minimal over the principal ideal generated by $a_1$; this contradicts the principal ideal theorem. In $R/(b)$, the prime $\mathfrak{p}/(b)$ is therefore minimal over an ideal generated by $m-1$ elements, so induction bounds its height by $m-1$. The $d-1$ strict inclusions from ${\mathfrak{p}}_1/(b)$ to $\mathfrak{p}/(b)$ giveand hence $d\le m$. If the height were infinite, the same argument applied to arbitrarily long finite chains would give the same fixed bound, which is impossible. This completes the proof.

A prime chain of length $s$ in a Noetherian local ring forces its maximal-ideal length function to be at least $c{t}^s$ for large $t$. Quotient by the initial prime to reduce to a domain, choose a nonzero $a$ in the next prime, and use the Artin-Rees lemma to obtain $H_R(t)\ge H_R(t-c_0)+H_{R/(a)}(t)$. Induction on chain length and summing about $t/(2c_0)$ terms prove the bound. This yields $\dim R\le d(R)$ without assuming the Krull height theorem.

For an irreducible variety which is an affine variety of dimension $d$ and a nonzero nonunit regular function $f$, each irreducible component of $V(f)$ has dimension $d-1$. Choose a closed point on just the component in question. In its local ring $R$, $\sqrt{(f)}$ is that component's prime ideal. A system of parameters of $R/(f)$, lifted and supplemented by $f$, generates a maximal-primary ideal; the Krull height theorem gives $\dim R/(f)\ge d-1$. Extending any chain above $(f)$ by the zero prime of the domain gives the reverse inequality. This proof does not identify dimension with transcendence degree.

In a Noetherian local ring $(R,\mathfrak{m})$ of dimension $d$, a list $x_1,\dots ,x_d\in \mathfrak{m}$ whose generated ideal has radical $\mathfrak{m}$. Such lists exist, by successively avoiding the finitely many top-dimensional minimal primes. No list shorter than $d$ can generate a maximal-primary ideal, by the Krull height theorem.