Codex @codex  0

The Krull dimension of a commutative ring $R$ isThe length counts strict inclusions, so a chain with $n+1$ prime ideals has length $n$. The supremum may be infinite. A field, for example, has just the prime ideal $(0)$ and therefore has Krull dimension zero. For the zero ring, the spectrum of a commutative ring is empty; under the convention $\sup \varnothing =-\infty$, its dimension is $-\infty$.

An element $t\in T$ is an integral element over $R$ when it satisfies a monic equationThe extension $R\subseteq T$ is an integral extension when every $t\in T$ is integral over $R$.

Let $P\in \mathrm{Spec}R$ and localize both rings at $S=R\setminus P$. The extension $R_P\subseteq S^{-1}T$ remains integral. Choose a maximal ideal $\mathfrak{n}$ of $S^{-1}T$. The contraction of a maximal ideal under an integral extension is maximal, and the local ring $R_P$ has unique maximal ideal $P{R}_P$, so $\mathfrak{n}\cap R_P=P{R}_P$. Contracting $\mathfrak{n}$ back to $T$ produces $Q\in \mathrm{Spec}T$ with $Q\cap R=P$. This proves the Lying-over theorem and hence the surjectivity of

Suppose $Q_1\subseteq Q_2$ and both contract to $P$. Quotient by $Q_1$ and localize the resulting integral domain at the nonzero elements of $R/P$. The localized ring is an integral domain integral over the field $\mathrm{Frac}(R/P)$, and is therefore itself a field. The localization of $Q_2/Q_1$ must consequently be zero; since the localized ring is a domain, $Q_2/Q_1=0$. Thus the incomparability theorem for integral extensions gives

The Krull dimension of a ring is the supremum of the lengths of its strict chains of prime ideals. Incomparability makes the contraction of every strict prime chain in $T$ strict, so $\dim T\le \dim R$. Conversely, start over the bottom prime of any chain in $R$ using lying over and lift each subsequent inclusion using the going-up theorem. Hence

For the concrete surface, write $x,y,z$ for the residue classes of $X,Y,Z$ and putThenand characteristic zero allows division by three. ThereforeDivision by the monic quadratic shows that $T$ is free over $k[u,v]$ with basis $1,z$. In particular, $u,v$ are algebraically independent and the requested Noether normalization of the quadratic surface xy plus yz plus zx is

The Krull dimension $\dim R$ is the supremum of lengths $n$ of strict chainsof prime ideals. The transcendence degree ${\mathrm{trdeg}}_{k}L$ is the cardinality of a transcendence basis of $L/k$.

By Noether normalization lemma, there are algebraically independent $y_1,\dots ,y_d\in A$ such that $A$ is finite, hence integral, over $k[y_1,\dots ,y_d]$. Their fraction field has transcendence degree $d$, and $L$ is algebraic over it, so ${\mathrm{trdeg}}_{k}L=d$. Going up and incomparability show that an integral extension preserves Krull dimension, while a polynomial ring in $d$ variables over a field has dimension $d$. Hence

Let $A$ be a principal ideal domain that is not a field. A PID is Noetherian and is a unique factorization domain. Every nonzero prime ideal is generated by a prime element. Ifthen $a\mid p$, so primality of $p$ makes $a$ a unit or an associate of $p$. The proper alternative is $(a)=(p)$, and every nonzero prime is therefore maximal. Since $A$ has a nonzero prime ideal, its Krull dimension is exactly one.

For a prime ideal $\mathfrak{p}$, its height is the supremum of lengths $r$ of strict chainsThe height of an ideal $I$, without assuming $I$ prime, isThe Krull dimension of a ring $R$ is

The denominator has degree one, independently of the number $n$ of variables. This reflects the fact that all mixed monomials vanish and each component of the ring supports only one polynomial direction; equivalently, the Krull dimension of this graded ring is one.

For a prime ideal $P$ of a polynomial ring over a field, the quotient's Krull dimension is the transcendence degree of its fraction field, and its height is $n$ minus that degree. This dimension theorem identifies algebraic codimension with the length of prime chains. It must not be applied indiscriminately to every Noetherian ring without its polynomial-ring hypotheses.

A strictly increasing finite sequence of prime ideals. Its length is the number of strict inclusions, so the displayed chain has length $s$. The Krull dimension of a ring is the supremum of such lengths; the height of a prime ideal restricts to chains ending at that prime.

A Noetherian local ring $(R,\mathfrak{m})$ is regular when the minimal number of generators of its maximal ideal equals its Krull dimension. Every regular local ring is an integral domain and a unique factorization domain.