15
votes
Krull dimension of a local ring and completion
Let $\Gamma$ be the group $\mathbb{Z}\times\mathbb{Z}$ with the lexicographic ordering. Then any valuation ring $A$ with value group $\Gamma$ is a counterexample.
Indeed, the maximal ideal of $A$ is ...
- 2,928
15
votes
"Noetherian" and "finitely generated" for polynomial algebras
The following example of Eakin [Eak72] says that $n = 2$ already suffices. I have tried to fill in some details to (hopefully) make the example independent from the rest of Eakin's paper.
Example [...
- 1,813
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
5
votes
Accepted
Algebras whose subalgebras are finitely generated
Let $k$ be a commutative ring. Is there a name for those commutative $k$-algebras with the property that every subalgebra is finitely generated? $\ldots$ Where can I read more about them?
The paper
...
- 14.6k
4
votes
Accepted
Non-noetherian cohomology and base change
The answer to the question about flatness is no and here is a counterexample. Let $S_n = \mathbf{A}^1$ with coordinate $t$ for $n\geq 1$ (over some base field). Let
$$
\mathcal{A}_n = \mathcal{O}_{\...
- 16.1k
4
votes
Algebras whose subalgebras are finitely generated
I'll assume that $k$ is noetherian. I'll just write $k$-ACC, or ACC if no ambiguity, to mean the ascending chain condition on $k$-subalgebra.
(For $k$ arbitrary, $A=k$ is the only $k$-subalgebra of ...
- 63.9k
4
votes
Accepted
Completion of a local ring is noetherian (under some hypothesis)
We can just follow the proof from the stacks tag linked from the one suggested by Keerthis Madapusi but under your hypothesis.
Let $a_1,\dots,a_r$ in $R$ generate $\mathfrak m/\mathfrak m^2$. Then the ...
- 148k
4
votes
Accepted
Is a universally closed monomorphism a closed immersion?
There is a non-surjective epimorphism $B\to C$ where $B$ and $C$ are zero-dimensional local rings (D. Lazard, see http://www.numdam.org/item/SAC_1967-1968__2__A8_0/). Then $\mathrm{Spec}\,(C)\to \...
- 19.6k
4
votes
Accepted
Cohomology and base change without Noetherian assumption
There is a fairly general version of base change for schemes in Lipman's "yellow book":
Lipman, Joseph; Hashimoto, Mitsuyasu: Foundations of Grothendieck duality for diagrams of schemes. ...
- 9,229
3
votes
Krull dimension of a local ring and completion
There exist examples of commutative local rings of finite dimension whose completions have infinite dimension (all dimensions are Krull, of course). Take $A$ to be a complete valuation ring, whose ...
- 131
3
votes
Algebras such that the tensor product with any Noetherian algebra is Noetherian
A (not necessarily commutative) algebra $A$ over a commutative noetherian ring $R$ is called strongly noetherian if for every noetherian $R$-algebra $R'$ the extension $A \otimes_R R'$ is noetherian. ...
- 5,407
3
votes
Higher direct images along proper morphisms in the non-Noetherian setting
According to Kiehl, R. in
Ein ``Descente''-Lemma und Grothendiecks Projektionssatz für
nichtnoethersche Schemata (Math. Ann. 198 (1972), 287–316.)
If $f \colon X \to Y$ is a proper finitely ...
- 9,229
3
votes
Moduli space of almost complex structures as an algebro-geometric object
Let $M$ be a closed $2n$-dimensional smooth orientable manifold, and let $TM$ be its tangent bundle.
Let $\mathcal{A}(M):=\{J\in C^\infty(M,\mathrm{End}(TM))\ |\ J_x^2=-1,\ \forall x\in M\}$. This ...
- 8,529
3
votes
Tensor products of two domains
First of all, as abx says, let us assume that both $S$ and $T$ are contained in some larger field $L$, in order to make the notation $k(S)\cap k(T)$ well defined.
In the above answer by Jason Starr, ...
- 331
3
votes
Tensor products of two domains
I made a mistake in the comment above. The comment is valid if both $R$ and $S$ are regular. However, as stated, there are counterexamples.
For instance, let $R$ be $k[x,y^{-1},x,y^{-1}]_{\mathfrak{...
3
votes
Accepted
Torsion submodules of non-noetherian modules
Take the $k[x,y]$-module with generators $a_n, n\in \mathbb N$ and relations
$$x a_1= ya_1=0$$ $$x a_{2n}+ ya_{2n+1} =a_n$$
Let $\mathfrak a=(x,y)$, then clearly $\mathfrak a M = M$. The $\mathfrak ...
- 148k
3
votes
Proof of the coherence of ${\Bbb F}_q[[X_1,\dots,X_{\infty}]]$
I do not see where the inequality
$$
|(A_{n,m})^∗|^{δ_{n,m}}≤|M_{n,m}|
$$
comes from. Different choices of $δ_{n,m}$-tuples of elements of $(A_{n,m})^∗$ could give rise to the same element of $M_{n,m}$...
- 331
2
votes
Relative Bertini Theorem
The answer to the question Q is "no". Take $n=d=2$. Let $\frak P$ be the ideal generated by $X_1^2+X_2^2+S_1$ and $X_1^3+X_2^3+S_2$. Substituting any two complex numbers $s_1$ and $s_2$ for $S_1$ and $...
- 331
2
votes
Accepted
On the relation between two definitions of torsion functors
The conjecture is not true.
By Quý's comment, the conjecture implies that $\Gamma_{\mathfrak{m}}$ is not a radical if $R$ is a $0$-dimensional local ring whose maximal ideal $\mathfrak{m}$ is ...
- 6,700
2
votes
Accepted
Is $Hom_R(S_X^{-1}R, E)$ the minimal injective cogenerator of $S_X^{-1}R$?
I give a counter-example. Let $R$ be a semi-local domain with two maximal ideals $\mathfrak{m}$ and $\mathfrak{n}$. So the minimal injective generator module is $E = E(R/\mathfrak{m}) \oplus E(R/\...
- 2,773
2
votes
Accepted
Integral domain satisfying a.c.c. on radical ideals and with algebraically closed fraction field
As requested, I am making my comment an answer. For every integer $n\geq 1$, denote by $R_n$ the power series ring, $$R_n := \mathbb{C}[[z_n]],$$ where $z_n$ is a variable. For every pair of ...
2
votes
Removing Noetherian condition from cohomology and base change
Since the question is about non-noetherian base change, I have just released a preprint with some collaborators that has a result in this direction (apologies for the self-promotion). The paper is ...
- 9,229
1
vote
Removing Noetherian condition from cohomology and base change
I'll fill in the gaps in your proof. The following is the proof I found while going through Vakil's book.
The first issue is using the given surjectivity of $\phi_y$ to show the surjectivity of $\phi_{...
- 11
1
vote
An example of a special $1$-dimensional non-Noetherian valuation domain
This isn't possible in any $1$-dimensional quasi-local domain $D$. If $a,b\in D$ and $b$ is not a unit then consider the multiplicative set $S$ generated by $b$. Clearly $S$ is not disjoint from any ...
- 938
1
vote
Power series ring $\Theta[[X_1,\ldots,X_d]]$ and prime ideals
The answer is ``yes'' if $\Theta$ is noetherian.
Indeed, order the monomials $X_1^{\alpha_1}X_2^{\alpha_2}\dots X_d^{\alpha_d}$ by the lexicographical ordering of the $(d+1)$-tuples $\left(|\alpha|,\...
- 331
1
vote
Power series ring and monomials
No. For instance, take $n=\epsilon=1$, $p=2$. Then the only $\epsilon$-monomials are 1 and $X_1$ and the $2\epsilon$-monomials are 1, $X_1$, $X_1^2$ and $X_1^3$. Take $\alpha=\beta=X_1$, $x=y=X_1^2-...
1
vote
Heights of contracted ideals
The answer is no in general. Actually, the opposite inequality is true!
The inequality fails when all the primes of minimal height containing $I$ are destroyed by the localization. Here is an example:...
- 2,851
1
vote
Prime ideal of $A[X_1,...,X_d]$
If $f$ is not required to be a morphism of $A$-algebras, a stupid counteraxmple exists. For instance, $A=k[x_1,x_2,\dots]$(infinitely many variables), $d=0$ and $$f:A\to A/(x_1,x_2,\dots)\cong k\...
- 7,377
1
vote
Structure theorem for non-Noetherian local rings
Here is a paper by Nagata that provides an answer for rings satisfying Krull's intersection theorem.
Only top scored, non community-wiki answers of a minimum length are eligible