Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.
Questions tagged [ac.commutative-algebra]
5322 questions
118
votes
5 answers
What do epimorphisms of (commutative) rings look like?
(Background: In any category, an epimorphism is a morphism $f:X\to Y$ which is "surjective" in the following sense: for any two morphisms $g,h:Y\to Z$, if $g\circ f=h\circ f$, then $g=h$. Roughly, "any two functions on $Y$ that agree on the image of…
Anton Geraschenko
- 23,718
53
votes
3 answers
Is it true that, as $\Bbb Z$-modules, the polynomial ring and the power series ring over integers are dual to each other?
Is it true that, in the category of $\mathbb{Z}$-modules, $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}[x],\mathbb{Z})\cong\mathbb{Z}[[x]]$ and $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}[[x]],\mathbb{Z})\cong\mathbb{Z}[x]$?
The first isomorphism is…
Maharana
- 1,742
47
votes
10 answers
Rings in which every non-unit is a zero divisor
Is there a special name for the class of (commutative) rings in which every non-unit is a zero divisor? The main example is $\mathbf{Z}/(n)$. Are there other natural or interesting examples?
lhf
- 2,942
42
votes
4 answers
Classification of finite commutative rings
Is there a classification of finite commutative rings available?
If not, what are the best structure theorem that are known at present?
All I know is a result that every finite commutative ring is a direct product of local commutative rings (this…
Puraṭci Vinnani
- 2,186
40
votes
1 answer
Is the radical of an irreducible ideal irreducible?
I originally posted this to math.stackexchange.com
here. I got a partial answer, but I now suspect that the complete answer is much harder than I thought, so I'm posting it here.
Fix a commutative ring $R$. Recall that an ideal $I$ of $R$ is…
Mary
- 401
36
votes
4 answers
Rings for which no polynomial induces the zero function
For any commutative ring $R$ let $R[x]$ denote the ring of polynomials with coefficients in $R$. Any polynomial $p \in R[x]$ naturally induces a function $\hat{p} :R \rightarrow R$. In some cases, a nonzero polynomial will induce the zero…
mweiss
- 525
33
votes
2 answers
Noetherian rings of infinite Krull dimension?
Since Noetherian rings satisfy the ascending chain condition, every such ring must contain infinitely many chains of prime ideals s.t. the heights of these chains are unbounded.
The only example I know of is the one due to Nagata [1962]: we take a…
moby
- 331
30
votes
5 answers
Atiyah-MacDonald, exercise 2.11
Let $A$ be a commutative ring with $1$ not equal to $0$. (The ring A is not necessarily a domain, and is not necessarily Noetherian.) Assume we have an injective map of free $A$-modules $A^m \to A^n$. Must we have $m \le n$?
I believe the answer…
CJD
- 1,058
29
votes
2 answers
Elementary proof of Nakayama's lemma?
Nakayama's lemma is as follows:
Let $A$ be a ring, and $\frak{a}$ an ideal such that $\frak{a}$ is contained in every maximal ideal. Let $M$ be a finitely generated $A$-module. Then if $\frak{a}$$M=M$, we have that $M = 0$.
Most proofs of this…
Phillip Williams
- 1,329
27
votes
5 answers
Why does the (S2) property of a ring correspond to the Hartogs phenomenon?
Hartogs Theorem says every function whose undefined locus is of codim 2 can be extend to the whole domain. I saw people saying this corresponds to the (S2) property of a ring. But I can't see why this is true. Can anybody explain this or give a…
Yuhao Huang
- 4,982
25
votes
1 answer
The Rabinowitz Trick
The recent question about problems which are solved by generalizations got me thinking about the Rabinowitz trick, which is used to prove a statement of Hilbert's Nullstellensatz, specifically, the inclusion of the ideal generated by an affine…
Grant Rotskoff
- 640
25
votes
4 answers
Is a domain all of whose localizations are noetherian itself noetherian ?
Is a domain $D$, all of whose localizations $D_P$ for $P \in Spec(D)$ are noetherian, itself noetherian ?
The question is motivated by proposition 11.5 of Neukirch's Algebraic Number Theory:
Let $\mathfrak{o}$ be a noetherian integral domain.…
KBuck
- 538
24
votes
2 answers
Discriminant and Different
First some context. In most algebraic number theory textbooks, the notion of
discriminant and different of an extension of number fields $L/K$, or rather, of the corresponding extension $B/A$ of their rings of algebraic integers is defined.
The…
Joël
- 25,755
23
votes
6 answers
Noether's normalization lemma over a ring A
Given a field $k$ and a finitely generated $k$-algebra $R$ without zero divisors, one knows that there exist $x_1, \ldots, x_n$ algebraically independent such that $R$ is integral over $k[x_1, \ldots, x_n]$.
Does one have a similar statement, under…
user2330
- 1,310
23
votes
5 answers
To prove the Nullstellensatz, how can the general case of an arbitrary algebraically closed field be reduced to the easily-proved case of an uncountable algebraically closed field?
In his answer to a question about simple proofs of the
Nullstellensatz
(Elementary / Interesting proofs of the Nullstellensatz),
Qiaochu Yuan referred to a really simple proof for the case of an
uncountable algebraically closed field.
Googling, I…
user2734
- 1,371