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1
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2answers
691 views

Rational power series

If we let $R=\mathbb{Z}[x]$ and $D=\mathbb{Z}[[x]]$. We say that $z\in D$ is rational if there is $g\in R$, $g\ne 0$ such that $zg\in R$. Let $S$ be the set of all rational elements in $D$. Then $S$ ...
1
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2answers
161 views

Is there a relationship between the right global dimensions of R and R[1/v]?

A few days ago I asked a similar question about Krull dimension and got fantastic answers. Unfortunately, for the application I have in mind (a question on ring spectra), Krull dimension doesn't ...
30
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3answers
3k views

What the heck is the Continuum Hypothesis doing in Weibel's Homological Algebra?

On page 98 of Weibel's An Introduction to Homological Algebra he mentions that the ring $R = \prod_{i=1}^\infty \mathbb{C}$ has global dimension $\geq 2$ with equality iff the continuum hypothesis ...
14
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4answers
2k views

Is there a Galois correspondence for ring extensions?

Given an ring extension of a (commutative with unit) ring, Is it possible to give a "good" notion of "degree of the extension"?. By "good", I am thinking in a degree which allow us, for instance, to ...
6
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2answers
819 views

Is every poset the poset of prime ideals of a ring?

The answer to this question, as it is, is trivially false, for one necessary condition is the existence of maximal element(s), i.e., maximal ideals exist and are prime. My question was inspired from ...
9
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6answers
2k views

Applications of commutative algebra

Hi. I'm preparing a thesis in commutative algebra, and when I say this to my friends they always ask me what are the applications to "real-world", and I don't know what to answer. This let me think ...
1
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1answer
283 views

A problem on Moebius transformations

We have the following result: Let $R=\mathbb{C}[t]_f$, with $f=(t-a_1)(t-a_2)\cdots (t-a_n)$. Then the automorphism group of $R$ is isomorphic to the group of all Moebius transformations which fix ...
2
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2answers
608 views

A problem for finite dimensional commutative algebra

Let $(A,m)$ be a local commutative associative algebra over the field of complex numbers, $m^n\ne 0$, $m^{n+1}=0$ for some $n>0$, and (1) $A$ is finite dimensional as vector space (2) for any ...
10
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4answers
866 views

Are units of rings of functions on algebraic varieties finitely generated (mod. constants)?

Hello, Consider the following question. Let $A$ be a finitely generated reduced algebra over an algebraically closed field $k$. Consider the group of units of $A$, modulo the group $k^*$. Is this ...
20
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3answers
1k views

Invariance of $Z[x]$ under a self-equivalence of the category of commutative rings with 1.

Let $\mbox{Rings}$ be the category of commutative rings with $1$. Is there an equivalence of categories $F: \mbox{Rings} \to \mbox{Rings}$ s.t. $$F(\mathbb{Z}[x])\not\cong \mathbb{Z}[x]?$$
4
votes
1answer
581 views

module of differentials of formal power series ring and of its field of quotiens

For any $A$-algebra $B$ ( commutative ring with 1 ), we have the existence of $\Omega_{B/A}$, the module of relative differentials of $B$ over $A$, which can be defined by an universal property. In ...
1
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1answer
370 views

Unimodular column property

Hi, I know that if $R$ is a ring such that every projective $R$-module finitely generated is free then $R$ has the unimodular column property. I would like to know if there is a ring $R$ that doesn't ...
7
votes
1answer
691 views

Direct sum of injective modules over non-Noetherian rings

Hi. I know, by the Bass-Papp theorem, that if every direct sum of injective $R$-modules is injective then $R$ is Noetherian. I would like to know if there exists a direct sum of injective $R$-modules ...
1
vote
3answers
521 views

Stably free module not finitely generated is free

Hi. I have read that stably free modules not finitely generated are free; this is proved in M.R. Gabel, stably free projectives over commutative rings, Thesis, Brandeis Univ., Waltham, MA 1972. But ...
12
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2answers
2k views

Generalized Euler phi function

Let $n$ be an integer, there is a well-known formula for $\varphi(n)$ where $\varphi$ is the Euler phi function. Essentially, $\varphi(n)$ gives the number of invertible elements in ...
2
votes
2answers
357 views

Related to fractional ideals

$K$ a field, $A\subset K$ a subring, $M\subset K$ an $A$-submodule. Define $$(A:_{K}M):= \lbrace s\in K|sM\subset A\rbrace$$ Then it is easy to see that $$M\subset A\Longleftrightarrow A\subset ...
5
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2answers
880 views

Is being finitely generated a local property?

I am trying ot figure out a proof of the following fact, that I believe is true, but it seems to me that something is lacking. Suppose we have commutative, unitary rings $A,B$ and a (unit preserving) ...
4
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0answers
670 views

Commutative ring Notes by M. Artin

In 1966, Professor Michael Artin gave a course for first-year graduate students at MIT on commutative algebra. In that course he covered many classical topics, (the Spectrum of a commutative ring, ...
2
votes
0answers
477 views

Decomposition group vs Galois group of completed extension for height > 1 primes

Assume Let $R$ be a Noetherian normal excellent domain, $F$ its field of fractions. Let $S$ be a finite $R$-algebra, $L$ its field of fractions. $L/F$ a (finite) Galois extension $S$ normal in $L$ ...
32
votes
7answers
3k views

Does $SL_3(R)$ embed in $SL_2(R)$?

Is there any non-trivial ring such that $SL_{3}(R)$ is isomorphic to a subgroup of $SL_{2}(R)$? $SL_{3}(\mathbb{Z})$ is not an amalgam, and has the wrong number of order $2$ elements to be a subgroup ...
1
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0answers
313 views

Functoriality of a standard integral domain construction.

The evident forgetful functor from fields to integral domains has a left adjoint, namely the construction of the quotient field for a given integral domain. Another standard construction is taking the ...
5
votes
0answers
886 views

Are local, Noetherian rings with principal maximal ideal PIR?

A question asked by a friend. I believe it's false, but lack a decisive counterexample. This question shows that it is true for valuation rings, but I know too little about them. In the wider ...
3
votes
0answers
143 views

Generic Rank of R^{1/p}

Suppose $R$ is a local Noetherian domain of dimension $d$ in characteristic $p>0$. Suppose $R^{1/p}$ is a finitely generated $R$-module, and suppose $k$ is the residue field of $R$. Is the ...
1
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0answers
487 views

nilpotent matrices over polynomial rings

I am looking for an analogue of the Jordan normal form for nilpotent matrices over the polynomial ring ${\mathbb Z}[x_1, \dots, x_n]$. More precisely, is there a description for the orbits of action ...
4
votes
0answers
216 views

When is the ring of invariants of a finite group generated by symplectic reflections a complete intersection ring?

Let V be a finite dimensional symplectic vector space over $\mathbb{C}$. Let $G$ be a finite subgroup of the symplectic group $Sp(V),$ which is generated by symplectic reflections, i.e. by elements ...
3
votes
0answers
561 views

Basic commutative algebra question.

Suppose that A is a local ring (commutative with unit), finite over a field k. Let L be the residue field A / m where m is the unique maximal ideal of A. Does the dimension of L (as a k-vector space) ...
2
votes
2answers
346 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. ...
1
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1answer
621 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, ...
17
votes
4answers
3k views

Flatness and local freeness

The following statement is well-known: $A$ a commutative Noetherian ring, $M$ a finitely generated $A$-module. Than $M$ is flat if and only if $M_{\mathfrak{p}}$ is free for all $\mathfrak{p}$. My ...
2
votes
2answers
657 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 ...
1
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4answers
848 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 ...
17
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1answer
2k 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 ...
6
votes
1answer
1k views

Can an infinite commutative ring have a finite (but nonzero) number of non-nilpotent zero-divisors?

By a theorem of Ganesan, if a commutative ring not a domain has only finitely many zero-divisors, then the ring must be finite. (There are analogous results for non-commutative rings.) There are ...
12
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2answers
1k views

What are the units in the ring of Laurent polynomials?

What are the units in $R[X,X^{-1}]$, where $R$ is a commutative ring with $1$? I know that the question for polynomial rings is a standard textbook exercise. However, I couldn't find a reference for ...
43
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1answer
2k views

A condition that implies commutativity

Let $R$ be a ring. A notable theorem of N. Jacobson states that if the identity $x^{n}=x$ holds for every $x \in R$ and a fixed $n \geq 2$ then $R$ is a commutative ring. The proof of the result for ...
12
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4answers
762 views

What does the semiring of ideals of a ring R tell us about R?

Here is something I've wondered about since I was an undergraduate. Let $R$ be a ring (commutative, let's say, although the generalization to noncommutative rings is obvious). Ideals of $R$ can be ...
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5answers
2k views

Axiomatic definition of integers

The real numbers can be axiomatically defined (up to isomorphism) as a Dedekind-complete ordered field. What is a similar standard axiomatic definition of the integer numbers? A commutative ordered ...