11
votes
Accepted
An infinite dimensional local domain whose chains of primes are finite
Choose a field $k$ and a Noetherian $k$-algebra $R$ of infinite dimension. (I know you know such a thing exists.) Let $R' \subset R[x]$ be the set of polynomials $f = \sum a_i x^i$ whose constant term ...
- 126
7
votes
Accepted
Local ring of infinite dimension
Let $k$ be a field. Let $x_{i, j}$, $1 \leq j \leq i \in \mathbf{N}$ be variables. Consider the ring
$$
R = k[x_{i, j}]/(x_{i, j} x_{i', j'}, i \not = i')
$$
Let $\mathfrak m$ be the maximal ideal ...
- 86
4
votes
Accepted
Krull dimension of the smooth locus
$\DeclareMathOperator\Spec{Spec}$
Maybe I'm misreading this, but I don't see why you need dimension $\geq 4$, normal, domain, etc.
EDIT: I originally wrote this requiring R1 but I don't think we need ...
- 20.5k
4
votes
Accepted
Is every universally catenary ring a going-between ring?
OK, let $R \subset S$ be an integral ring extension with $R$ universally catenary. Let $\mathfrak q \subset \mathfrak q'$ be primes in $S$ such that there is no prime strictly in between them. We have ...
- 56
4
votes
Accepted
Decomposing Noetherian hereditary rings of Krull dimension $1$ into product of hereditary domains (i.e. Dedekind domains)
Certainly finite products of Dedekind rings with fields work as well. These are the only ones, even without the assumption on the Krull dimension:
Lemma. Let $R$ be a Noetherian (commutative) ...
- 28.7k
4
votes
Should Krull dimension be a cardinal?
I realize this question was asked some years ago, but some coauthors and I recently published a paper on this topic in J. of Algebra: "An infinite cardinal-valued Krull dimension for rings".
...
- 41
4
votes
The Krull dimension of the tensor product of rings
You are looking at infinite tensor products so we really shouldn't expect the infinite product to have finite dimension in most cases. Your second example was simply an infinite tensor product of $\...
- 542
4
votes
Accepted
Is the projective dimension of finite torsion-free modules over regular ring of dimension $n$ smaller that $n$?
Yes. If $\operatorname{pd}_A(M)=n $, there exists a maximal ideal $\mathfrak{m}$ of $A$ such that $\operatorname{pd}_{A_{\mathfrak{m}}}(M_{\mathfrak{m}})=n $. By the Auslander-Buchsbaum theorem, this ...
- 38k
3
votes
Finiteness of Krull dimension of commutative Noetherian ring for which maximal length of regular sequence in maximal ideals have a uniform upper bound
I believe we can just modify Nagata's counterexample (see for example Exercise 9.6 in Eisenbud's book) as follows. Let $k$ be a field, and $R_n$ be the ring $k[x_{n,1},\dots,x_{n,r_n}]/I_n$ where $I_n$...
- 16.2k
3
votes
On analytic transcendence degree and Krull dimension for homomorphic images of power series rings
Clearly $R_P=(R/P)_P$, and so we may replace $R$ by $R/P$. Since its dimension is $s$, one can find a system of parameters $y_1,y_2,\ldots,y_s$ in the maximal ideal of $R/P$. Then the inclusion $k[[...
- 6,262
3
votes
Accepted
When is the Tor-dimension of $R/(r)$ strictly smaller than that of $R$?
There is no difference between flat (Tor) dimension and global dimension for Noetherian rings. Now, it is well known that for an arbitrary local $R$ and a nonzerodivisor $f\in\mathfrak{m}$ either $\...
- 1,810
3
votes
Algebraic characterization of commutative rings of Krull dimension 1,2, or 3
A remark about the one-dimensional case. By the characterization of Krull dimension given by T. Coquand and H. Lombardi, dim($R$)$\,\leq 1$ iff $\forall_{x,y\in R}\exists_{a,b\in R,m,n\in\mathbb{N}}\,...
- 1,388
2
votes
Valuation ring whose maximal ideal and every ideal of finite height are principal
This is false. For example, there can be no nonzero ideals of finite height.
Remark. Note that for every totally ordered abelian group $\Gamma$, there exists a valuation ring with value group $\Gamma$...
- 28.7k
2
votes
Accepted
Integral domain satisfying a.c.c. on radical ideals and with algebraically closed fraction field
As requested, I am making my comment an answer. For every integer $n\geq 1$, denote by $R_n$ the power series ring, $$R_n := \mathbb{C}[[z_n]],$$ where $z_n$ is a variable. For every pair of ...
2
votes
Accepted
Are integral extensions of a catenary ring still catenary?
No. Nagata's famous family of examples of non-catenary rings yields a non-catenary finite extension of a catenary noetherian local domain.
Reference: M. Nagata, On the chain problem of prime ideals, ...
- 6,700
2
votes
commutative ring satisfying descending chain condition on radical ideals
Answer (2) here Valuation ring satisfying either a.c.c. or d.c.c. on prime ideals gives a counterexample to the first 2 questions (a radical ideal in a valuation ring is prime).
- 2,665
2
votes
Accepted
Valuation ring satisfying either a.c.c. or d.c.c. on prime ideals
For any totally ordered set $W, R=k[x_i/x_j^n: i,j \in W, i \lt j, n \ge 0]$ is a valuation ring with value group $\oplus_{i \in W} \mathbb{Z}$ and Spec($R$) = {initial segments of $W$}. Thus $W$=...
- 2,665
2
votes
Accepted
Krull dimension and elimination theory over the integers
This should be a comment, but my reputation is low, so I have to post this.
I assume that you want the solutions of $p_1=\dots=p_r=0$ to be in $\mathbb{Z}^n$. Then the answer is no. Take, for example, ...
- 228
2
votes
Accepted
A proof of $\dim(R)+1 \leq \dim(R[T]) \leq 2 \dim(R)+1$ with the Coquand-Lombardi characterization of Krull dimension
(As requested by OP, I convert my comment into an answer. However, I am not technically equipped on constructive mathematics, thus further edits are welcome, and I make it a community wiki.)
It turns ...
2
votes
A proof of $\dim(R)+1 \leq \dim(R[T]) \leq 2 \dim(R)+1$ with the Coquand-Lombardi characterization of Krull dimension
If the Krull dimension of $R$ is $k+1\ge 1$, it is not $\le k$. Hence there exists an $(x_0,\cdots,x_k)\in R^{k+1}$ such that $y:=x_0^{m_0} (\cdots ( x_k^{m_k} (1+a_k x_k)+\cdots)+a_0 x_0)\ne 0$ for ...
- 1,388
2
votes
Accepted
On "minimal presentation" of local rings essentially of finite type over a field
If $R$ is the local ring of the point $P$ on a smooth $n$-dimensional $k$-variety $V$ then any such homomorphism would be an isomorphism. So it cannot exist if $V$ is irrational. For example, take $V$ ...
- 3,137
2
votes
Accepted
How bad does a ring have to be for a failure of "going-in-between"?
This is a partial answer, giving a two dimensional Noetherian counterexample. It starts from your observation that one has to consider the case where $A/\mathfrak{p}_0$ is not integrally closed. ...
- 5,339
1
vote
Accepted
Transcendence degree and Krull dimension for homomorphic images of power series rings
Let $k$ be countable and let $I=(0)$. Then $R_P$ is the field $k((x_1,\dots,x_n))$, which is uncountable provided $n>0$, and therefore of uncountable transcendence degree over $k$, since an ...
- 16.2k
1
vote
Dimension of the associated graded module at an ideal
$\DeclareMathOperator{\gr}{gr}$This follows from Theorem 4.5.6 in Bruns and Herzog's Cohen-Macaulay Rings. In particular, let $R$ be a filtered ring with Noetherian filtration $F=\{I_i\}_{i\geq 0}$ (e....
- 135
1
vote
Dimension of the associated graded module at an ideal
In Algèbre locale, multiplicités, J.P. Serre shows that the dimension is equal to the degree of the Hilbert--Samuel polynomial. In turn, this polynomial is the same for the module and its associated ...
1
vote
Change chain of prime ideals so that $a \in P_1$
A proof is given in Bourbaki, Algèbre commutative, VIII.3.1 Lemme 1.
- 6,700
1
vote
On GCD and LCM of elements in integral domain with Krull-dimension 1
This is only a comment about classes of rings in which no counter-examples can be found.
Since a one-dimensional GCD domain is a Bézout domain [1, Corollary 3.9], GCD domains will not provide any ...
- 7,893
1
vote
torsion free modules $M$ over Noetherian domain of dimension $1$ for which $l(M/aM) \le (\dim_K K \otimes_R M) \cdot l(R/aR), \forall 0 \ne a \in R$
$l(M/aM) \le r \cdot l(R/aR), \forall 0 \ne a \in R$ should be true for all torsion free modules $M$. Every module is the direct limit of its finitely generated submodules, and the functor $ M \mapsto ...
- 401
Only top scored, non community-wiki answers of a minimum length are eligible