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Questions tagged [ac.commutative-algebra]

Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.

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Is completeness of a field an algebraic property?

Pretty straitforward: If a field has a metric in which it is complete can it have a metric in which it is not complete? By metric I mean field norm of course
Adam Gal's user avatar
  • 700
8 votes
1 answer
371 views

How to construct a ring with global dimension m and weak dimension n?

Given two integers $m,n$ such that $n < m$, it is easy to construct a ring with global dimension $m$ or weak dimension $n$. But I wonder whether there exists a ring satisfying both the conditions?
TmobiusX's user avatar
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3 votes
2 answers
376 views

How to compute the ring of invariants of SO_3(k) acting on a polynomial ring

Let $k$ be a field and let $A$ be the polynomial ring over $k$ in $3n$ variables: $A = k[X_{ij} \vert i=1,2,3 \quad j=1,2,\cdots,n]$. ${\rm SO}_3(k)$ acts on $A$ in the following way: Given $g \in {\...
tkf's user avatar
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16 votes
3 answers
3k views

Is being torsion a local property of module elements?

Say $R$ is a ring, not necessarily a domain, and $M$ is an $R$-module. All rings are commutative with 1. An element $m\in M$ is called torsion if $r.m=0$ for some regular element (non-zerodivisor) $r\...
Andrew Critch's user avatar
0 votes
0 answers
165 views

Support sets along a ring homomorphism.

Let $(R,m)$ and $(S,n)$ be commutative noetherian local rings, and $f: R\rightarrow S$ be a local homomorphism (i.e., $f(m) \subseteq n$) with $S$ flat as $R$-module. If $M$ is a finite generated $R$-...
TmobiusX's user avatar
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9 votes
3 answers
3k views

Why are divisible abelian groups important?

I just quote wikipedia: "Divisible groups are important in understanding the structure of abelian groups, especially because they are the injective abelian groups." I am asking for detail ...
2 votes
2 answers
866 views

Modules over a Gorenstein ring

$A$ a Gorenstein ring, $M\neq 0$ a finite $A$-module with finite injective dimension. According to Bruns, this implies that $M$ has finite projective dimension. How do I see that?
ashpool's user avatar
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3 votes
3 answers
670 views

Algebraic, analytic, formal modules

Consider torsion free modules over the germ of a fixed isolated algebraic hypersurface singularity {$f=0$}$\subset\mathbb{C}^n$. There are natural functors (using categories of finitely generated ...
Dmitry Kerner's user avatar
0 votes
1 answer
502 views

Finiteness of injective hull of residue field for Artin local ring

$(A,\mathfrak{m})$ an Artin local ring, $E(A/\mathfrak{m})$ the injective hull of $A/\mathfrak{m}$. How do I see that $E(A/\mathfrak{m})$ is a finite $A$-module?
ashpool's user avatar
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-1 votes
2 answers
671 views

Do Gorenstein rings necessarily have a finite projective dimension (as a module over itself)? [closed]

Do Gorenstein rings necessarily have finite projective dimensions?
ashpool's user avatar
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9 votes
2 answers
3k views

Projective & injective dimensions

$A$ a Noetherian local ring, $M\neq 0$ a finite $A$-module. I'm not quite sure about the relation between finiteness of projective and injective dimensions of $M$. Does the finiteness (or infiniteness)...
ashpool's user avatar
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3 votes
2 answers
1k views

Depth and dimension

$A$ a Noetherian local ring, $M\neq 0$ a finite $A$-module. Then is it true that $\mbox{depth }M\le\mbox{depth }A$ just like $\mbox{dim }M\le\mbox{dim }A$? I don't see any relation between an $M$-...
ashpool's user avatar
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5 votes
2 answers
731 views

What is the completion at a family of ideals?

Let $A$ be a (commutative with unit) noetherian ring. If $I$ is an ideal of $A$, the $I$-adic completion of $A$ is by definition $$ \widehat{A} := \underset{\leftarrow}\lim A/I^n. $$ This operation is ...
Ricky's user avatar
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13 votes
4 answers
4k views

Size of a Groebner basis

I've spent some time recently looking at some Groebner bases for some specific ideals coming from problems in computer vision. The generators are not sparse, and they all have the same degree (...
Chris Aholt's user avatar
-3 votes
1 answer
1k views

An elementary question about the Krull dimension of modules [closed]

Let $R$ be a commutative ring. If $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$ is an exact sequence of modules, we have that $\operatorname{Supp}M=\operatorname{Supp}M'\cup \operatorname{...
TmobiusX's user avatar
  • 1,207
1 vote
2 answers
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Real solutions to underdetermined system of polynomial equations

I have the system of multi-variable polynomial (quadratic) equations with real coefficients. The number of equations is given scales as $K$ and the number of unknowns goes as $K^2$. So for for large $...
Raisa's user avatar
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14 votes
4 answers
6k views

When is an algebra of commuting matrices (contained in one) generated by a single matrix?

Let C be an nxn matrix, then the polynomials in C (with appropriate coefficients) form an algebra of commuting matrices. I feel that I should know if the converse is true but I do not. So my first ...
Aaron Meyerowitz's user avatar
2 votes
1 answer
980 views

Spectral sequence for Ext

Why is $H^p(X, \mathcal{E}xt^q(F, I))=0$ for $p>0$ and $I$ an injective sheaf and $F$ an arbitrary sheaf? I'm trying to check the hypothesis in the grothendieck spectral sequence applied to the ...
J Delzant's user avatar
7 votes
2 answers
542 views

Does every nontrivial sheaf of rings have a maximal ideal?

Let $R$ be a sheaf of rings on a topological space $X$. Assume $R \neq 0$. Does then $R$ have a maximal ideal? So this is a spacified analogon of the theorem, that every nontrivial ring has a maximal ...
Martin Brandenburg's user avatar
5 votes
2 answers
689 views

Irreducible components of quotients of Cohen-Macaulay rings of the "correct" dimension

Suppose $R$ is a Cohen-Macaulay ring. It is well known that if $I$ is an ideal of $R$ generated by $n$ elements, and $I$ has codimension $n$, then $R/I$ is also Cohen-Macaulay. Now suppose that $I$ ...
Alexander Woo's user avatar
43 votes
5 answers
3k views

Explicit elements of $K((x))((y)) \setminus K((x,y))$

In an answer to the popular question on common false beliefs in mathematics Examples of common false beliefs in mathematics I mentioned that many people conflate the two different kinds of formal ...
Pete L. Clark's user avatar
9 votes
1 answer
792 views

Reconstructing a polynomial from resultants

I am trying to compute a monic polynomial $f(x)$ with integer coefficients and known degree $d$. I am given $n$ pairwise coprime polynomials $g_1(x),\ldots,g_n(x)$, also with integer coefficients, ...
David Mandell Freeman's user avatar
3 votes
2 answers
2k views

Extension problem

As I understand, if $0\rightarrow A\rightarrow X\rightarrow B\rightarrow 0$ is a short exact sequence of abelian groups, $\mbox{Ext }_{\mathbb{Z}}^{1}(B,A)$ gives all the isomorphism classes of what ...
ashpool's user avatar
  • 2,857
2 votes
2 answers
420 views

Homological dimensions of module

$(A,\mathfrak{m})$ a Noetherian local ring, $M\neq 0$ a finitely generated $A$-module. As I understand, $\mbox{Ext }^{j}(A/\mathfrak{m}, M) = 0$ for $j<\mbox{depth }(M)$ and for $j>\mbox{inj. ...
ashpool's user avatar
  • 2,857
7 votes
2 answers
2k views

Ideals in the ring of single-variable Laurent polynomials with integer coefficients

I'm writing some software to automatically compute things like Alexander modules, Alexander ideals, Milnor signatures and Farber-Levine pairings for 1 and 2-knot complements. So among other things I'...
Ryan Budney's user avatar
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14 votes
3 answers
2k views

Intuition for Model Theoretic Proof of the Nullstellensatz

I recently read the model-theoretic proof of the Nullstellensatz using quantifier elimination (see www.msri.org/publications/books/Book39/files/marker.pdf). I'm convinced that the Nullstellensatz is ...
David Corwin's user avatar
  • 15.4k
4 votes
2 answers
2k views

Existence of a minimal generating set of a module

Does a module (over a commutative ring) always possess a minimal generating set? When the module is not finitely generated, the typical Zorn's lemma type argument doesn't seem to work. More precisely, ...
ashpool's user avatar
  • 2,857
36 votes
4 answers
12k views

Flatness and local freeness

The following statement is well-known: Let $A$ be a commutative Noetherian ring and $M$ a finitely generated $A$-module. Then $M$ is flat if and only if $M_{\mathfrak{p}}$ is a free $A_{\mathfrak{p}}$-...
ashpool's user avatar
  • 2,857
4 votes
2 answers
2k views

Non-finite version of Nakayama's lemma?

Let $A$ be a local ring with nilpotent maximal ideal $\mathfrak{m}$ (i.e., some power of $\mathfrak{m}$ vanishes), and $M$ an $A$-module (not necessarily finitely generated). Let $\bar{S}\subset M/\...
ashpool's user avatar
  • 2,857
12 votes
2 answers
3k views

Is tensor product of local algebras local?

In general, the tensor product of two local rings is not local. For example, $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}\ $ is not a local ring. Let $\mathbb{F}_{p}$ denote the finite field with $p$ ...
N. Kumar's user avatar
  • 592
13 votes
3 answers
3k views

Zero divisor conjecture and idempotent conjecture

Let $G$ be a torsion-free group and $C$ the ring of complex numbers. The zero divisor (idempotent, resp.) conjecture is that there is no nontrivial zerodivisor (idempotent, resp.) in $CG$. The wiki ...
yeshengkui's user avatar
  • 1,373
16 votes
4 answers
3k views

Cardinal of maximal linearly independent subsets of a free module

Is it true that the cardinality of every maximal linearly independent subset of a finitely generated free module $A^{n}$ is equal to $n$ (not just at most $n$, but in fact $n$)? Here $A$ is a nonzero ...
ashpool's user avatar
  • 2,857
2 votes
0 answers
797 views

How can I prove R[x] is integrally closed iff R is integrally closed ? (R: integral domain) [closed]

$R$ is a integral domain. How can I prove $R[x]$ is integrally closed iff $R$ is integrally closed?
Qijun Yan's user avatar
6 votes
3 answers
3k views

Tor and projective dimension

Is it possible that $\mbox{Tor }^{r+1}(M,N)=0 \ \ \forall N$ yet $\mbox{proj. dim }M>r$? What I do know is that if $(A,\mathfrak{m})$ is Noetherian local and $M$ is finitely generated over $A$ ...
ashpool's user avatar
  • 2,857
1 vote
4 answers
2k views

Reduced rings, idempotents and their prime spectrum

Let $B$ be a commutative unitary reduced ring and let $A$ be a subring of it. Let $e$ be an idempotent of $B$. Then we have a natural surjective ring homomorphism $A\rightarrow Ae$ defined by $a\...
Jose Capco's user avatar
  • 2,275
4 votes
0 answers
179 views

Global dimensions of orders over non-Gorenstein centers

This question concerns the following Lemma 4.2 in this paper by Van den Bergh: Lemma: Let $R$ be local, normal Gorenstein ring of dimension $d$. Suppose $M$ is a reflexive $R$ module such that $A=\...
Hailong Dao's user avatar
  • 30.5k
15 votes
1 answer
1k views

Countable Hom/Ext implies finitely generated

Today I learned this interesting fact from Jerry Kaminker: If $A$ is an abelian group such that $\mathrm{Hom}(A,\mathbb{Z})$ and $\mathrm{Ext}(A,\mathbb{Z})$ are both countably generated, then in fact ...
Greg Kuperberg's user avatar
5 votes
2 answers
2k views

Is there a non-projective flat module over a local ring?

Is there a non-projective flat module over a local ring? Here I assume the ring is commutative with unit.
ashpool's user avatar
  • 2,857
14 votes
0 answers
899 views

Frobenius upper shriek/flat of a dualizing complex

Let $X$ be a separated connected scheme of characteristic $p > 0$. I am going to assume that $F : X \to X$ (the absolute Frobenius) is a finite map. This condition is called being $F$-finite. ...
48 votes
4 answers
4k views

Are there more Nullstellensätze?

Over which fields $k$ is there a reasonable analogue of Hilbert's Nullstellensatz? Here is a more precise formulation: let $k$ be an arbitrary field, $n$ a positive integer, and $R = k[t_1,..,t_n]$. ...
Pete L. Clark's user avatar
9 votes
3 answers
1k views

Vector spaces with natural bases

Sergeib's question asks about vector spaces without a natural basis. Actually, I would claim (apparently in accord with many comments and answers to Sergeib's question ) that this is the default ...
4 votes
2 answers
883 views

Group & modules of arbitrary cardinality [closed]

How do I see that there is a group of an arbitrary cardinality? Is this also true for abelian groups? Also, given a commutative ring $R\neq 0$ how do I see that there is an $R$-module of arbitrary ...
ashpool's user avatar
  • 2,857
13 votes
0 answers
496 views

Are the supports of $Ext^i(M,N)$ eventually periodic?

Let $R$ be a Noetherian, commutative ring and $M,N$ be finitely generated $R$-modules. Question: Do the sets of minimal primes of $\text{Ext}^i_R(M,N)$ (for a fixed pair of $M,N$) become periodic ...
Hailong Dao's user avatar
  • 30.5k
2 votes
1 answer
918 views

Is a formally smooth morphism a filtered inductive limit of smooth algebras?

Given a unital commutative ring $A$ (not necessarily noetherian) and a formally smooth morphism of rings $f:A \to B$, where $B$ is not necessarily noetherian, is (or when is) $B$ a filtered inductive ...
Lutz Geissler's user avatar
2 votes
1 answer
1k views

For which rings does there exist an invertible Vandermonde matrix?

Suppose $R$ is a commutative ring, and $S \subset R^{n\times n}$ is an $R$-module. We are given $H_0,\dots,H_n \in R^{n\times n}$, and we know that for all $r \in R$, $$H_0 + r H_1 + \dots r^n H_n \in ...
Laurent Lessard's user avatar
1 vote
0 answers
534 views

Integral element in the quotient of a polynomial ring

Hello, I'm writting a "report" (to learn) on algebraic geometry, and was looking to write a proof for the following statement : Theorem : Let $K$ be an algebraically closed field and $C_1, C_2 \...
warsomekey's user avatar
11 votes
1 answer
1k views

The associated prime ideals of $Ext^i_R(M,N)$

If R is a commutative noetherian ring, M and N are modules with M finite. It is well known in commutative algebra that $AssHom_R(M,N)=Supp(M)\cap Ass(N)$. But I want to know whether there is a ...
TmobiusX's user avatar
  • 1,207
28 votes
6 answers
5k views

Expressing $-\operatorname{adj}(A)$ as a polynomial in $A$?

Suppose $A\in R^{n\times n}$, where $R$ is a commutative ring. Let $p_i \in R$ be the coefficients of the characteristic polynomial of $A$: $\operatorname{det}(A-xI) = p_0 + p_1x + \dots + p_n x^n$. I ...
Laurent Lessard's user avatar
20 votes
1 answer
3k views

On a theorem of Jacobson

In a comment to an answer to a MO question, in which Bill Dubuque mentioned Jacobson's theorem stating that a ring in which $X^n=X$ is an identity is commutative (theorem which has shown up on MO ...
Mariano Suárez-Álvarez's user avatar
4 votes
1 answer
914 views

Elements of trace zero in a field extension

Let $K=F_q$ and $F=F_{q^3}$, define the set A={$x \in F$ : $Tr_{F/K} (x)=0$}. Is it true that for every $x \in A$ there are $y,z \in A$ such that $x=yz$?
Izhar Oppenheim's user avatar

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