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5 votes
0 answers
128 views

What are the conditions for the dual of the exterior algebra to be isomorphic to the exterior algebra of the dual?

The exterior algebra $\Lambda^*_kM$ can be defined for a $k$-module $M$, where $k$ is a commutative ring. A number of sources mention, without condition or proof, a (canonical) isomorphism $$(\Lambda^*...
Cameron's user avatar
  • 171
4 votes
0 answers
128 views

Length of dual module

It is well known that, given a commutative ring $R$ and an $R$-module $M$, the dual module $M^\vee = \operatorname{Hom}_R(M, R)$ does not always satisfy $M^\vee \cong M \ (1)$, and not even $M^{\vee \...
JBuck's user avatar
  • 223
5 votes
0 answers
138 views

Rings over which free modules of a certain rank are reflexive (satisfy Specker's theorem)

Following this question about the case of $\mathbb{Z}_{(p)}$, I've got to ask what is known more generally about rings and dimensions for which Specker's theorem holds. Let me make the following ...
Gro-Tsen's user avatar
  • 32.5k
7 votes
2 answers
270 views

Double dual of free $\mathbb{Z}_{(p)}$-modules

For an abelian group $A$, put $DA=\text{Hom}(A,\mathbb{Z})$ and $D_{(p)}A=\text{Hom}(A,\mathbb{Z}_{(p)})$. It is a theorem of Specker that when $A$ is free abelian of countable rank, the natural map $...
Neil Strickland's user avatar
10 votes
1 answer
603 views

Isbell Duality and Dualizing Scheme Objects

I'm not sure if this question is too elementary for MO; nevertheless, I have seen many helpful discussions surrounding this topic here. I'm interested in studying the adjunction $\operatorname{Spec}\...
LiminalSpace's user avatar
5 votes
1 answer
349 views

A generalization of integral Poincaré duality

In this paper, Felix, Halperin and Thomas define the notion of a Gorenstein space over a field $\mathbb{k}$: An augmented differential graded algebra $R$ over $\mathbb{k}$ is Gorenstein if $\text{Ext}...
Matt's user avatar
  • 208
2 votes
0 answers
166 views

Do Poincaré duality algebras need to be defined over a field?

I asked the below question here on MSE, but after some time and a bounty offering I have not received an answer. A graded commutative, connected $\mathbb{k}$-algebra $A$ is called a Poincaré duality ...
Matt's user avatar
  • 208
16 votes
1 answer
1k views

What is a module over a Boolean ring?

Recall that a (unital) Boolean ring is a (unital) commutative ring $A$ where every element is idempotent; it follows that $A$ is of characteristic 2. There is an equivalence of categories between ...
Tim Campion's user avatar
2 votes
0 answers
66 views

Reference request: Matroid cryptomorphisms for arbitrary monomial ideals

For a matroid $M$ let $C$ be the circuit ideal of $M$, that is, the Stanley-Reisner ideal of independence complex of $M$. Then there are simple ideal-theoretic operations that take $C$ to the facet ...
Aaron Dall's user avatar
9 votes
2 answers
513 views

Divided power algebra is artinian as a module over the polynomial ring

I already asked this on math.stackexchange.com, but did not receive much responses. I hope this is also appropriate for mathoverflow. In the paper Homological algebra on a complete intersection, with ...
red_trumpet's user avatar
  • 1,286
6 votes
1 answer
123 views

Reflection-invariant monomial ideals and Alexander duality

First we give some definitions from Section 3 of the paper Monomials, Binomials, and Riemann-Roch by Manjunath and Sturmfels and then we restate a claim from that paper offered without proof. Finally ...
Aaron Dall's user avatar
3 votes
1 answer
247 views

Support of cohomology of a dualizing complex

Let $A$ be a commutative noetherian local ring, and let $D$ be a dualizing complex over $A$. Let $i$ be the minimal integer such that $H^i(D) \ne 0$ (I am assuming cohomological grading, so the ...
Dualizing's user avatar
7 votes
2 answers
847 views

Criterion for being reflexive via Ext

In this question it was claimed that if a module $M$ over a noetherian domain $R$ satisfies $\rm{Ext}^i(M,R)=0$ for $i=1,2$, then $M$ is reflexive. Is this true? Does someone know a reference or a ...
Hans's user avatar
  • 3,031
49 votes
4 answers
4k views

Why is there a duality between spaces and commutative algebras?

1) The category of affine varieties over $\mathbb{C}$ is equivalent to the opposite category of finitely generated reduced algebras over $\mathbb{C}$. The equivalence associates to an affine variety ...
Yonatan Harpaz's user avatar
19 votes
3 answers
2k views

Classification of rings satisfying $a^4=a$

We have the famous classification of rings satisfying $a^2=a$ (for each element $a$) in terms of Stone spaces, via $X \mapsto C(X,\mathbb{F}_2)$. Similarly, rings satisfying $a^3=a$ are classified by ...
Martin Brandenburg's user avatar
1 vote
1 answer
218 views

Does projective duality preserve arithmetic-Cohen-Macaulay-ness?

Let $V$ be a vector space over $\mathbb{C}$. Suppose $X\subset \mathbb{P} V$ is an algebraic variety, and consider its projective dual variety $X^\vee \subset \mathbb{P} V^*$. If the coordinate ring $\...
Luke Oeding's user avatar
5 votes
1 answer
926 views

Matlis' dual of injective modules

Let $(R, \mathfrak{m})$ be a commutative Noetherian complete local rings ($R$ can be regular, if you need). Let $E(R/\mathfrak{p})$ be injective hull of $R/\mathfrak{p}$, if $\mathfrak{p}= \mathfrak{m}...
Pham Hung Quy's user avatar
6 votes
1 answer
670 views

Base change of trace for Gorenstein or Cohen-Macaulay morphisms

This is basically a question of functoriality for base change of CM morphisms. EDIT: $\text{ }$ Brian Conrad sent me an email explaining the that this is indeed true, and follows from his book. I'...
16 votes
1 answer
2k views

Questions about spectra of rings of continuous functions

I have been thinking a bit about rings of continuous functions of various kinds -- how they motivate the more modern notion of the Zariski topology on the prime spectrum as well as how they fit into a ...
Pete L. Clark's user avatar
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. ...