Krull dimension

The Krull dimension of a commutative ring is the supremum of the lengths of strict chains of prime ideals.

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.

If $A$ is a finitely generated algebra over a field $k$, then there are algebraically independent elements $y_1,\dots ,y_d\in A$ such that $A$ is finite, hence integral, over $k[y_1,\dots ,y_d]$.

To normalize $k[a_1,\dots ,a_n]$ over an arbitrary field, choose a nonzero polynomial relation and a base $M$ larger than all its exponents. The substitutions $b_i=a_i-a_n^{M^i}$ give distinct top weights to its monomials, so its highest power of $a_n$ has a nonzero scalar coefficient. It becomes a monic polynomial equation, making the algebra finite over $k[b_1,\dots ,b_{n-1}]$. Induction proves the Noether normalization lemma, including over finite fields.

For a nonconstant polynomial over any field, use triangular substitutions $x_i=y_i+y_n^{B^i}$, $x_n=y_n$, where $B$ exceeds every exponent in its support. Base-$B$ weights distinguish all original monomials, so the transformed polynomial has a unique highest $y_n$ power with coefficient in $k^{\times }$. Rescale to make it monic. The monic polynomial quotient is finite free over $k[y_1,\dots ,y_{n-1}]$, yielding a finite flat projection. The nonlinear substitution works over finite fields as well as infinite ones; a generic linear-direction argument alone would not do so.

Letover a field in which $3$ is invertible. Put $u=x-z$ and $v=y-z$. ThensoDivision by this monic quadratic makes $T$ a free $k[u,v]$-module with basis $1,z$. Thus $k[u,v]$ is a polynomial algebra and $T$ is integral over it.

Height measures prime-chain length within Krull dimension theory.

The height of an ideal $I$ is the infimum of the heights of prime ideals containing it:The height of a prime is the supremum of the lengths of strict prime chains ending at that prime.

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 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.

If an ideal $I$ is contained in a finite union of prime ideals ${\mathfrak{p}}_1\cup \cdots \cup {\mathfrak{p}}_r$, then $I$ is contained in one of the ${\mathfrak{p}}_i$. Equivalently, if none of the primes contains $I$, one can choose an element of $I$ outside all of them.

In a Noetherian ring, every prime ideal minimal over a proper principal ideal has height at most one.

For a proof, suppose instead that ${\mathfrak{p}}_0⊊\mathfrak{q}⊊\mathfrak{p}$ and that $\mathfrak{p}$ is minimal over $(a)$. Quotient by ${\mathfrak{p}}_0$ and localize at $\mathfrak{p}$; one may assume that $R$ is a Noetherian local domain, $0⊊\mathfrak{q}⊊\mathfrak{p}$ with $\mathfrak{p}$ maximal, and $\mathfrak{p}$ is the only prime containing $(a)$. For the symbolic powerseach ${\mathfrak{q}}^{(n)}$ is $\mathfrak{q}$-primary. The quotient $R/(a)$ is a zero-dimensional Noetherian ring and hence Artinian, so the descending chain $({\mathfrak{q}}^{(n)}+(a))/(a)$ stabilizes. For large $n$, writing an element of ${\mathfrak{q}}^{(n)}$ modulo ${\mathfrak{q}}^{(n+1)}$ and using $a\notin \mathfrak{q}$ givesThe Nakayama lemma applied to ${\mathfrak{q}}^{(n)}/{\mathfrak{q}}^{(n+1)}$ gives ${\mathfrak{q}}^{(n)}={\mathfrak{q}}^{(n+1)}$. Localizing at $\mathfrak{q}$ would therefore make the powers of the nonzero maximal ideal $\mathfrak{q}{R}_{\mathfrak{q}}$ stabilize at a nonzero ideal, contradicting the Krull intersection theorem for the Noetherian local domain $R_{\mathfrak{q}}$.

If a principal prime ideal $(x)$ in a Noetherian local ring has positive height, then the ring is a domain. Localizing at $(x)$ shows that every annihilator of a power of $x$ lies in $(x)$; stabilization of the annihilator chain then proves that $x$ is a nonzerodivisor and that zero is prime.

If an ideal of a Noetherian ring is generated by $n$ elements, every prime ideal minimal over it has height at most $n$. The proof inducts on $n$, using the Krull principal ideal theorem in a quotient and prime avoidance to choose a prime chain compatible with the induction.

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.

If $R$ is Noetherian and ${\mathfrak{p}}_0⊊{\mathfrak{p}}_1⊊{\mathfrak{p}}_2$ are prime ideals, there are infinitely many primes strictly between ${\mathfrak{p}}_0$ and ${\mathfrak{p}}_2$. After quotienting by ${\mathfrak{p}}_0$ and localizing at ${\mathfrak{p}}_2$, this reduces to a local Noetherian domain of dimension at least two. If it had only finitely many height-one primes, prime avoidance would give a nonzero element of the maximal ideal outside all of them, contradicting the Krull principal ideal theorem.

If a Noetherian ring has finite Krull dimension $d$, then it has infinitely many prime ideals of each height $n$ with $0<n<d$. In a prime chain of maximal length, apply prime ideals between a three-prime chain to the terms of heights $n-1$, $n$, and $n+1$.

For a Noetherian local ring $(A,\mathfrak{m})$, the Krull dimension of $A$, the Krull dimension of the associated graded ring ${\mathrm{gr}}_{\mathfrak{m}}(A)$, and the order of the pole at $t=1$ of its Hilbert series are equal.

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.

If $x$ is a non-zero-divisor in a finite-dimensional Noetherian ring $A$, thenEvery prime chain above $(x)$ can be extended downward by a minimal prime of $A$, because a non-zero-divisor belongs to no minimal prime.