Articles of krull dimension

An equivalent condition for zero dimensional Noetherian local rings

Let $(A,m)$ be a Noetherian local ring. Why $A$ is zero dimensional if and only if a power of $m$ is $\{0\}$ ?

Krull dimension and transcendence degree

What is the simplest proof of the fact that an integral algebra $R$ over a field $k$ has the same Krull dimension as transcendence degree $\operatorname{trdeg}_k R$? Is it possible to use only Noether normalization theorem?

An example of a (necessarily non-Noetherian) ring $R$ such that $\dim R>\dim R+1$

What is an example of a non-Noetherian ring $R$ such that the Krull dimension of $R[T]$ is greater than dim$R+1$?

Do there exist semi-local Noetherian rings with infinite Krull dimension?

Do there exist semi-local Noetherian rings with infinite Krull dimension? As far as I know, Nagata’s counterexample to the finite dimensionality for general Noetherian rings is not semi-local.

$A$ is an affine $K$-algebra and $f$ a non-zero divisor of $A$. Can one say that $\dim A=\dim A_f$?

Let $A$ be an affine $K$-algebra and $f$ be a non-zero divisor of $A$ then can one say that $\dim A=\dim A_f$ ? What I proved that if $A$ is an affine domain and $f$ is a non-zero element in $A$ then $\dim A=\dim A_f.$The sketch of the proof goes as following: Since $A_f \cong […]

Noetherian ring with finitely many height $n$ primes

If $R$ is a Noetherian commutative ring with unity having finitely many height one prime ideals, one could derive from the “Principal Ideal Theorem”, due to Krull, that $R$ has finitely many prime ideals (all of height less than or equal to $1$). It may comes to mind that, in general, if $R$ is a […]

When will $A$ satisfy the dimension formula?

What property should $A$ satisfy so that $A[x_1, \ldots, x_n]$ satisfies the dimension formula, $$\mathrm{dim}(A[x_1, \ldots, x_n]) = \mathrm{dim}(A[x_1, \ldots, x_n]/\mathfrak{p}) + \mathrm{ht}(\mathfrak{p}),$$ for any prime ideal $\mathfrak{p}$ in $A[x_1, \ldots, x_n]$? For instance, this property holds when $A$ is a field. Is there a general property that ensures this formula is satisfied?

The Krull dimension of a module

Let $R$ be a ring, $M$ is a $R$-module. Then the Krull dimension of $M$ is defined by $\dim (R/\operatorname{Ann}M)$. I can understand the definition of an algebra in a intuitive way, since the definition by chain of prime ideals agrees with the transcendental degree. So, why dimension of module $M$ should be $\dim (R/\operatorname{Ann}M)$? […]

Examples of rings whose polynomial rings have large dimension

If $A$ is a commutative ring with unity, then a fact proved in most commutative algebra textbooks is: $$\dim A + 1\leq\dim A[X] \leq 2\dim A + 1$$ Idea of proof: each prime of $A$ in a chain can arise from at most two prime ideals of $A[X]$. The left equality holds when $A$ is […]

A chain ring with Krull dimension greater than one

Recall that a commutative ring $R$ with identity is a chain ring if the set of ideals of $R$ is linearly ordered under inclusion. I want to know if there a chain ring with Krull dimension greater than one. Or does every chain ring have Krull dimension at most one?