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Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.
21
votes
5
answers
1k
views
Computation of fraction field of formal series over the integers
What is the fraction field $K$ of the domain $\mathbb Z[[X]]$?
It is strictly smaller than the field of Laurent series $L=\operatorname {Frac}\mathbb Q[[X]]$, since $\sum_{i\geq 0}\frac {X^i}{i!}\in …
2
votes
When is the radical of the extension of a prime ideal prime?
If you look at the situation geometrically, you have a morphism of affine schemes $Spec(S)\to Spec(R)$ and you are asking if the inverse image of the integral subscheme
$V(\mathcal P) \subset Spec(R) …
4
votes
Accepted
Different notions of associated prime (in the non Noetherian case)
Dear Andrea, let $A=K[X_1,X_2,\ldots ,X_n,\ldots]$, the polynomial ring in countably many variables and $I$ be the ideal $I=(X_1^2, X_2^2,\ldots )\subset A $. Then
$$\mathcal M=(X_1,X_2,\ldots)\in Be …
10
votes
What is the "strongest" non-local property of a ring/module that is true of all localization...
A simple example is obtained by taking $P$ to mean "has positive dimension".
Every local domain of positive dimension $(A,\mathfrak m)$ has $P$ at all maximal ideals (i.e. just at $ \mathfrak m$ !) …
8
votes
Is being finitely generated a local property?
Dear Andrea, the fact you mention is indeed true. Consider the associated morphism of affine schemes $Spec (\phi) = f : X=Spec B \to Y=Spec A$. Your hypothesis on unit ideal generation translates in …
34
votes
Atiyah-MacDonald, exercise 2.11
Dear CJD,
if you are still interested in your problem, already solved three weeks ago by Anton, here is another point of view.
Let $M:A^m \to A^n$ be injective. Let $B=\mathbb Z [\ldots,m_{ij},\ldots …
3
votes
1
answer
200
views
Does a regular pair of elements in a noetherian domain remain regular if their order is swit...
Recall that in a commutative ring $A$ an ordered pair of elements (a,b) is said to form a regular sequence if the ideal $\langle a,b\rangle $ is strictly included in $A$ ,if $a$ is not a zero-divisor …
21
votes
2
answers
2k
views
What is the dimension of the product ring $\prod \mathbb Z/2^n\mathbb Z$ ?
In an anwswer to a question on our sister site here I mentioned that a reduced commutative ring $R$ has zero Krull dimension if and only if it is von Neumann regular i.e. if and only if for any $r …
4
votes
Accepted
Atiyah-MacDonald: exercise 5.29 - "local ring of a valuation ring"
Let B be the subring of K containing A. I am going to prove that B is the localization of A at a prime ideal $P\subset A$, which seems a reasonable interpretation of the statement that B is a local ri …
13
votes
Accepted
Reference for tensor products of fields
Dear anon, the most complete reference might be Bourbaki's Algèbre, Chapter V.
For question 1, I suggest Bourbaki's Algèbre, Chapter V, §10, 4. Descente galoisienne, Corollaire . There the Master pr …
18
votes
Accepted
Is the tensor product of a power series ring and a field noetherian?
The answer is no: for example, $k[[x]]\otimes_k k((x))$ is not noetherian.
Indeed, if it were, so would be $ k((x))\otimes_k k((x))$.
But this would contradict the following interesting general theore …
23
votes
Classification of finite commutative rings
This is a very interesting question related to the Hilbert scheme $Hilb^n(\mathbb A^d)$ classifying $n$ points in affine space $\mathbb A^d$.
I don't think there is a classification but there is an es …
12
votes
Non finitely-generated subalgebra of a finitely-generated algebra
Dear Nicojo, since you now have many counter-examples, let me give you a situation where $B$ is finitely generated, in line with your question 2). I am going to adopt your notations with the important …
3
votes
Exotic principal ideal domains
Dear Qiaochu, if $A$ is a discrete valuation ring and if $B$ is an étale algebra over $A$, then $B$ is a discrete valuation ring. In a related vein, the henselization of a discrete valuation ring $A$ …
5
votes
Accepted
Comparation of dimensions of rings
Dear Li, first of all I think that when you write "... such that $q \cap B=p$, and $q$ is the minimal such ideal in the sense of inclusion", you mean "... and $q$ is a minimal ideal...".
The answer t …