User fred rohrer - MathOverflow

Answer by Fred Rohrer for Surjectivity of the natural map of injective module to its localization

2013-05-31T14:25:17Z

A ring $A$ is said to have the ITI property with respect to an ideal $\mathfrak{a}$ if $\mathfrak{a}$-torsion submodules of injective $A$-modules are injective.

Noetherian rings have ITI with respect to every ideal, but a ring with ITI with respect to every ideal is not necessarily noetherian (e.g. absolutely flat rings have ITI). However, there are rings without ITI, even with respect to some principal ideals.

If $A$ has the ITI property with respect to (the ideal generated by) $f$ then the canonical morphism $I\rightarrow I_f$ is an epimorphism. Hence, noetheriannes is not necessary for this property to hold. But it is unknown to me (and I think also in general unknown) whether noetheriannes can be omitted.

Answer by Fred Rohrer for Why did Bourbaki ignore the theory of categories?

2013-05-24T21:20:24Z

It might be interesting to look at the Appendix to Exposé I of SGA 4. In a footnote this is described as follows:

Nous reproduisons ici, avec son accord, des papiers secrets de N. BOURBAKI.

While the appendix treats mainly the theory of universes, it makes use of the language of categories. Moreover, some internal references hint to the existence of some more "papiers secrets" containing a draft of a chapter about categories.

Answer by Fred Rohrer for Examples of polynomial rings $A[x]$ with relatively large Krull dimension

2013-05-16T05:35:42Z

Your question is essentially completely answered in the paper The dimension sequence of a commutative ring by Arnold and Gilmer (Amer. J. Math. 96 (1974), 385--408).

EDIT: The aforementioned paper by Arnold and Gilmer treats the case of an arbitrary finite number of variables. Since you are interested only in the case of one variable, the general but still rather concrete construction of rings with the desired properties given in Bourbaki's Algèbre commutative VIII.2 Exercice 7 should be sufficient.

Answer by Fred Rohrer for "Axiom of global choice"

2013-04-21T09:49:45Z

Dear Sergei, you might be interested in first reading Bourbaki's Théorie des ensembles (at least chapters I--III) and then have a look at section 0 and the appendix of SGA 4.I. This gives a slightly different approach using Hilbert's almighty symbol $\tau$.

Answer by Fred Rohrer for How to cite math journals?

2013-04-08T13:46:49Z

I agree with the other answers suggesting to use the standard abbreviations. The question is of course, Which are the standard abbreviations? Just recently (while arguing with the Springer Correction Team about, well, abbreviations of journal titles) I learned that for example Springer uses the abbreviations as given by the ISSN database. To me this seems indeed somewhat more standard and moreover more universal than any other list. Unfortunaetly, this list seems not accessible to mere mortals. But said page can, with some care, at least be partially useful, as it provides this accessible list of standard abbreviations of words that may occur in titles of serials.

Answer by Fred Rohrer for How to call covers not covering anything else?

2013-03-05T21:07:38Z

The construction "covering of $A$ by [something]" seems handy. For (1), one can use "covering of $A$ by subsets of $A$", and for (2) one can use "covering of $A$ by nonempty sets".

Answer by Fred Rohrer for Subgroup of lattice-ordered group

2013-02-19T06:04:26Z

No. A counterexample (essentially from Bourbaki's Algèbre VI.1 Exercice 12 a)) is the following.

We furnish $\mathbb{Z}$ with its usual structure of ordered group and consider the product of ordered groups $G=\mathbb{Z}^3$. This is a lattice, and for $(x,y,z),(u,v,w)\in G$ we have $$\textstyle\sup_G((x,y,z),(u,v,w))=(\sup(x,u),\sup(y,v),\sup(z,w)).$$ Now we consider the subgroup $H=\{(x,y,z)\in G\mid z=x+y\}$ of $G$, furnished with its induced structure of ordered group. This is also a lattice, as one readily checks that for $(x,y,x+y),(u,v,u+v)\in H$ we have $$\textstyle\sup_H((x,y,x+y),(u,v,u+v))=(\sup(x,u),\sup(y,v),\sup(x,u)+\sup(y,v)).$$ However, since $$\textstyle\sup_G((0,1,1),(1,0,1))=(1,1,1)\neq(1,1,2)=\sup_H((0,1,1),(1,0,1))$$ we see that $H$ is not a sublattice of $G$.

Answer by Fred Rohrer for What kind of subset is Spec(R_P) in Spec(R)?

2013-02-09T13:12:05Z

Take $R=\mathbb{Z}$ and $P=0$. Then, ${\rm Spec}(R_P)$, considered canonically as a subset of ${\rm Spec}(R)$, consists precisely of the point $0$. In particular, it is not open in ${\rm Spec}(R)$. Hence, it is not a neighbourhood in ${\rm Spec}(R)$ of $0$, despite $0$ being the generic point of ${\rm Spec}(R_P)$ and ${\rm Spec}(R_P)$ being an open neighbourhood of its generic point in itself.

You might want to have a look at Section I.2.5, especially Proposition I.2.5.2, of EGA [1970 edition] (or Section I.2.4 in the first edition) to get some more general information about your situation.

Answer by Fred Rohrer for Is the empty graph a tree?

2013-02-01T21:30:54Z

In Bourbaki's terminology, the empty graph is a tree - cf. LIE.IV.Annex.3.

On the definition of delta-functors

2011-11-07T08:24:15Z

A $\delta$-functor is usually (e.g. in Grothendieck's Tohoku paper or in Cartan-Eilenberg's Homological Algebra) defined to be a pair $$((T^i)_{i\in\mathbb{Z}},((\delta^i_{\mathbb{S}})_{i\in\mathbb{Z}})_{\mathbb{S}\text{ short exact sequence}})$$ consisting of a sequence of functors $(T^i)_{i\in\mathbb{Z}}$ (with common Abelian source $\mathscr{C}$ and common additive target $\mathscr{D}$) and a family $((\delta^i_{\mathbb{S}})_{i\in\mathbb{Z}})_{\mathbb{S}\text{ short exact sequence}}$ , indexed by short exact sequences in $\mathscr{D}$, of sequences of morphisms in $\mathscr{D}$ - the so-called connecting morphisms - subject to some conditions.

However, sometimes (e.g. in Rotman's Introduction to Homological Algebra (First Edition)) a $\delta$-functor is defined to be a sequence $(T^i)_{i\in\mathbb{Z}}$ of functors as above such that there exists a family $((\delta^i_{\mathbb{S}})_{i\in\mathbb{Z}})_{\mathbb{S}\text{ short exact sequence}}$ as above, subject to the same conditions as above.

One might wonder if the omission of the additional data of the connecting morphisms is done on purpose. Or, more precisely:

Are the families of connecting morphisms of a $\delta$-functor uniquely determined by its sequence of functors?

(I guess that the answer is no, and I guess it is still no if we consider only exact or even universal $\delta$-functors.)

EDIT: As Martin's example shows, the answer to the original question is no for trivial reasons. But is there an example with essentially different connecting morphisms (whatever that means, but it should rule out Martin's example, compositions with automorphisms, etc.)?

More general: Can we say that the connecting morphisms are unique up to something, and even say what something is?

Finally, concerning Martin's remark about the definition of universality: We can of course define this by using all connecting morphisms instead of a given one. Then we may wonder whether the usual notion of universality implies this stronger one...

Answer by Fred Rohrer for Does every regular Noetherian domain have finite Krull dimension?

2013-01-23T08:04:20Z

No. An example is given in K. Fujita, Infinite dimensional Noetherian Hilbert domains, Hiroshima Math. J. 5 (1975), 181-185.

Answer by Fred Rohrer for Laurent Polynomials

2013-01-04T06:33:25Z

Invertible elements of Laurent algebras, and more generally of algebras of torsionfree, cancellable commutative monoids, are characterised in Theorem 11.3 and Corollary 11.4 of Gilmer's Commutative Semigroup Rings (Chicago Lectures in Mathematics, 1984). The proofs given there are quite accessible.

Answer by Fred Rohrer for Lattice of Prime ideals

2012-12-30T08:47:51Z

First, construct a totally ordered group $G$ of height $n$ by iterating the construction in Bourbaki's Algèbre commutative, VI.4.2 Exemple 1. Second, construct a valuation ring $A$ with group of values $G$ following the recipe in loc.cit., VI.3.4 Exemple 6. Then, the spectrum of $A$ has cardinality $n$ and is totally ordered by inclusion.

Answer by Fred Rohrer for Direct product of rings

2012-12-29T21:30:55Z

Products of fields are semihereditary. This follows from the facts that products of fields are von Neumann regular and that von Neumann regular rings are semihereditary. A proof can be found (as suggested by David White) in Lam's Lectures on modules and rings, Example 2.32 d).

(In the above, fields and rings are not necessarily commutative.)

Dimension of polynomial algebras

2012-12-04T15:01:30Z

Let $R$ be a commutative ring of Krull dimension $d$, let $n\in\mathbb{N}$, and let $R[X_1,\ldots,X_n]$ denote the polynomial algebra in $n$ indeterminates over $R$. One can show that then we have $\dim(R)+n\leq\dim(R[X_1,\ldots,X_n])$. So, it is natural to wonder about the class of rings $R$ for which this inequality is an equality.

I know of only two subclasses of this class: Namely, the class of noetherian rings (Krull 1951) and the class of Prüfer rings (Seidenberg 1954).

Are there other interesting classes of rings with the property that the Krull dimensions of their polynomial algebras are minimal in the above sense?

Answer by Fred Rohrer for Question on localization technique

2012-12-04T13:35:02Z

First, if $h:A\rightarrow B$ is a morphism of rings, $M$ is a $B$-module, and $N\subseteq M$ is a sub-$B$-module, then it holds $M=N$ if and only if the underlying sets of $M$ and $N$ are equal, hence if and only if the $A$-modules obtained from $M$ and $N$ by scalar restriction along of $h$ are equal.

Second, if $A$ is a ring, $M$ is an $A$-module, and $N\subseteq M$ is a sub-$A$-module, then it holds $N=M$ if and only if $N_{\mathfrak{p}}=M_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$ of $A$ (cf. [Bourbaki, Algèbre commutative, II.3.3 Théorème 1]).

Third, putting the above together answers what seems to be your first question.

Finally, it is not completely clear to me, what you want as an answer to your second question. Let me just point out that in the last paragraph of the proof one applies Lemma 15.1.4 (ii) and thus needs the base ring to be local with infinite residue field, and one moreover uses the facts that $R$ is *local and thus needs the base ring to be local.

Answer by Fred Rohrer for Question on local cohomology

2012-12-04T09:19:16Z

First, claim 1 as stated is wrong - consider the zero module.

Second, I guess that you talk about a step in the proof of Theorem 15.3.1 in Brodmann-Sharp. If so, then you have more hypotheses than you mentioned. Beside others, $R$ is noetherian and - most important - $M$ is $0$-dimensional. (And $M$ is not "positively graded", a notion that seems not reasonable for graded modules.)

So, what you want to show is that under these hypotheses, $M$ is $R_+$-torsion. For this it suffices to show that $M$ is $\mathfrak{m}_0+R_+$ -torsion. More general, it suffices to show that a $0$-dimensional finitely generated graded module $M$ over a *local graded ring with *maximal ideal $\mathfrak{m}$ is $\mathfrak{m}$-torsion. And this is indeed the case. Namely, $0$-dimensionality means that the graded ring $R/(0:_RM)$ is $0$-dimensional. Now, $\sqrt{(0:_RM)}$ is the intersection of the graded primes containing $(0:_RM)$. But as $R/(0:_RM)$ is $0$-dimensional, $\mathfrak{m}$ is the only such prime, implying $\sqrt{(0:_RM)}=\mathfrak{m}$. Since $\mathfrak{m}$ is finitely generated (as $R$ is supposed to be noetherian), there exists $t\in\mathbb{N}$ with $\mathfrak{m}^t\subseteq(0:_RM)$, and this implies that $M$ is an $\mathfrak{m}$-torsion module as desired.

Answer by Fred Rohrer for Are grothendieck universes enough for the foundations of category theory?

2012-12-01T20:28:57Z

I suggest you have a look at Grothendieck's SGA 4, Exposé I, where a lot of category theory is developed (and applied) on the basis of Bourbaki set theory plus Grothendieck's axioms UA and UB about universes.

Answer by Fred Rohrer for Length of a resolution

2012-11-20T14:46:13Z

This is answered in my answer to a related question of yours.

Answer by Fred Rohrer for Projective dimension of zero module

2012-11-20T07:34:53Z

Let me explain a definition of projective dimension which gives the same result as the one given by Sándor Kovács, but without any restriction on the ring or the module we are talking about. This is, by the way, the one chosen by Bourbaki (A.X.8.1).

Let $A$ be a ring.

0) We write $\overline{\mathbb{Z}}=\mathbb{Z}\cup\{-\infty,\infty\}$ and furnish $\overline{\mathbb{Z}}$ with the ordering that extends the canonical ordering on $\mathbb{Z}$ and has $\infty$ as greatest and $-\infty$ as smallest element. We convene that suprema and infima of subsets of subsets of $\overline{\mathbb{Z}}$ are always understood to be taken in $\overline{\mathbb{Z}}$.

1) If $C$ is a complex of $A$-modules and $C_n$ denotes its component of degree $n\in\mathbb{Z}$, then we set $$b_d(C)=\inf\{n\in\mathbb{Z}\mid C_n\neq 0\}$$ and $$b_g(C)=\sup\{n\in\mathbb{Z}\mid C_n\neq 0\},$$ and we call $$l(C)=b_g(C)-b_d(C)$$ the length of $C$. Note that if $C$ is the zero complex then we have $b_d(C)=\infty$ and $b_g(C)=-\infty$, hence $l(C)=-\infty$.

2) If $M$ is an $A$-module and $(P,p)$ is a left resolution of $M$, then the length $l(P)$ of the complex $P$ is called the length of $(P,p)$. Note that if $P$ is the zero complex (which may be the case if and only if $M=0$) then the length of $(P,p)$ is $-\infty$.

3) If $M$ be an $A$-module, then the infimum of the lengths of all projective resolutions of $M$ is called the projective dimension of $M$. Hence, if $M=0$ then we have a projective resolution of length $-\infty$, and thus the projective dimension of $M$ is also $-\infty$. Conversely, if $M$ has projective dimension $-\infty$ then - since every $A$-module has a projective resolution - it necessarily has a projective resolution of length $-\infty$, and thus it follows $M=0$.

Note: This clearly makes sense in every abelian category with enough projectives, and there are obvious variants of the above that yield analogous definitions of injective or flat dimensions.

Answer by Fred Rohrer for Finitely generated resolutions

2012-11-16T11:07:06Z

The last sentence in the answer by Simone Virili can easily be generalised as follows:

If $R$ is left coherent, then a left $R$-module has a projective resolution whose components are of finite type if and only if it is of finite presentation.

Answer by Fred Rohrer for On some finiteness properties for schemes

2012-11-07T07:15:42Z

An example of a noetherian ring of infinite dimension can be found in Nagata's Local Rings, Appendix A1, Example 1.

Edit: An interesting generalisation of Nagata's construction yielding noetherian rings of infinite dimension whose maximal ideals have prescribed heights was given by Fujita in his article Infinite dimensional Noetherian Hilbert domains, Hiroshima Math. J. 5 (1975), 181–185.

Linear algebra over principal rings

2012-10-29T14:09:12Z

Consider an extension $R\subseteq S$ of commutative rings, and suppose that $R$ is principal (i.e., $0$ is the only zero-divisor of $R$ and every ideal of $R$ has a generating set of cardinality $1$). By means of scalar restriction we consider $S$ as an $R$-module. Let $M$ be a sub-$R$-module of finite type of $S$ containing $R$.

In this situation, Gilmer and Heinzer claim (in Remark 2 in their article "On the complete integral closure of an integral domain") that there exists an $R$-module $N$ containing $M$ such that $R$ is a direct summand of $N$.

Their argument is just the remark that $M$ "has a linearly independent module basis containing the identity of $R$".

Unfortunately, I cannot follow this argument, nor can I prove the claim in a different way. Even worse, I meanwhile have the feeling that the claim is not true.

Does someone know either a proof of this claim or a counterexample?

Answer by Fred Rohrer for A construction of tensor product (Coutinho's book: D-module)

2012-10-15T06:16:28Z

I suggest to also have a look at Bourbaki's Algebra (edition from 1970 or later), Chapter II, Section 3, especially Paragraph 4 (about multimodule structures on tensor products).

When is a torsionfree subgroup contained in a torsionfree direct summand?

2012-09-28T15:34:22Z

Let $F$ be a torsionfree subgroup of a commutative group $G$. Are there nontrivial conditions known under which there exists a torsionfree direct summand of $G$ containing $F$?

I would already be happy with such a condition in case $F$ is free or $G$ is of finite type.

Motivation:

We consider a property $\mathcal{P}$ of graded rings and its behaviour under coarsening. Given an epimorphism of commutative groups $\psi:G\rightarrow H$ and a $G$-graded ring $R$ with $G$-graduation $(R_g)_{g\in G}$ , the $\psi$-coarsening $R_{[\psi]}$ of $R$ is the $H$-graded ring with underlying ring the ring underlying $R$ and with $H$-graduation given by $(R_{[\psi]})_h=\bigoplus_{g\in\psi^{-1}(h)}R_g$ for $h\in H$.

Suppose we know that $\mathcal{P}$ is reflected by $\psi$-coarsening for every $\psi$, i.e., if $R_{[\psi]}$ has $\mathcal{P}$ then so does $R$, and that if $\ker(\psi)$ is a torsionfree direct summand of $G$ then $\mathcal{P}$ is respected by $\psi$-coarsening, i.e., if $R$ has $\mathcal{P}$ then so does $R_{[\psi]}$.

Then, we want to know whether $\mathcal{P}$ is also respected by $\psi$-coarsening if $\ker(\psi)$ is torsionfree but not necessarily a direct summand of $G$. And this would follow immediately if $\ker(\psi)$ is contained in a torsionfree direct summand of $G$.

Answer by Fred Rohrer for Using Function Terminology for Functors?

2012-09-19T14:14:05Z

As proven by Jon's comment, the statement is not clear in meaning. A precise statement is the following, but it is of course unclear whether it is equivalent to the statement you had in mind:

$F$ induces by restriction and coastriction a functor from $B$ to $D$.

(The very useful word "coastriction" seems to be not so well-known, but it is explained in this answer.)

Answer by Fred Rohrer for union of infinitely many prime ideals

2012-08-06T06:39:05Z

As an addition to the answers already given, let me mention the interesting paper Baire's category theorem and prime avoidance in complete local rings by Sharp and Vamos (Arch. Math. 44 (1985), 243-248). Beside other things, it contains the following:

A neat proof of Burch's Lemma (cf. the answer by Neil Epstein) on use of Baire's category theorem.
An example showing that in Burch's Lemma the hypothesis of completeness cannot be omitted; this is essentially the same as the example given by Mahdi Majidi-Zolbanin.
A noetherian local ring with uncountable residue field has countable avoidance (i.e., if an ideal is contained in the union of a countable family of ideals (not necessarily prime!) then it is contained in one of these ideals); this can be viewed as a generalisation of a special case of the Hochster-Huneke result mentioned by Neil.

Answer by Fred Rohrer for Why should morphisms between two graded vector spaces preserve grading?

2012-07-11T13:02:38Z

You might want to have a look at Section 3 of Chapter 2 of the Introduction of Saul Lubkin's Cohomology of Completions (North-Holland Mathematics Studies 42, 1980), where the differences between the two possibilities of defining a category of graded vector spaces suggested by you (with the obvious correction suggested by Mark) are discussed in detail and in a very general setting.

Also related, less abstract, but still very interesting are the following two papers (for whose authors names I am not capable of producing here the necessary diacritics):

J. L. Gomez-Pardo, C. Nastasescu, Topological aspects of graded rings, Comm. Algebra 21 (1993), 4481-4493;

J. L. Gomez-Pardo, G. Militaru, C. Nastasescu, When is HOM$_R(M,-)$ equal to Hom$_R(M,-)$ in the category $R-gr$? Comm. Algebra 22 (1994), 3171-3181.

Answer by Fred Rohrer for Blackbox Theorems

2012-06-14T06:20:07Z

Embedding theorems for abelian categories (Freyd, Mitchell, Lubkin, ...) seem to qualify.

A reducible connected scheme with pairwise disjoint irreducible components

2012-05-31T14:55:42Z

Without finiteness assumptions, the irreducible and the connected components of a scheme may behave in strange ways. More precisely, let us consider a scheme $X$ and the following properties:

(1) $X$ is the sum of its irreducible components;

(2) The irreducible and the connected components of $X$ coincide;

(3) The irreducible components of $X$ are pairwise disjoint.

It is clear that (1) implies (2), that (2) implies (3), and that if the set of irreducible components of $X$ is locally finite then all three statements are equivalent (see [EGA 0.2.1.6]). However, (2) does not necessarily imply (1) in general: An affine counterexample is given by the spectrum of the product of infinitely many fields (which is non-discrete and totally disconnected). So, out of pure curiosity we may ask the following:

Is there an (affine) scheme fulfilling (3) but not (2)?

One can note that this is equivalent to the following:

Is there a nonempty, reducible, connected (affine) scheme whose irreducible components are pairwise disjoint?

(One can also note that for topological spaces that are not necessarily underlying spaces of schemes it is easy to find an example that fulfils (3) but not (2) - every connected, separated space with at least two points does so.)

Comment by Fred Rohrer

2013-06-17T04:26:51Z

@rghthndsd: I would not say it is a good discussion, but mathoverflow.net/questions/120536 contains what I was referring to.

Comment by Fred Rohrer

2013-06-16T20:35:39Z

Wait... we will again have to discuss whether or not the empty scheme is connected.

Comment by Fred Rohrer

2013-06-04T21:08:31Z

Well, "usually" seems to be a slight exaggeration...

Comment by Fred Rohrer

2013-05-31T13:02:07Z

Ah, thank you for the explanation. So your answer does not answer the question.

Comment by Fred Rohrer

2013-05-31T12:48:49Z

Without understanding everything you wrote, I do not think it is correct. It seems to imply that - without any condition on a ring $R$ - torsion submodules of injective $R$-modules are injective. This is false, cf. mathoverflow.net/questions/47043.

Comment by Fred Rohrer

2013-05-24T22:11:10Z

@Ryan: I do not know if good old Nick is writing right now, but his newest book - the new edition of Algèbre VIII - appeared only last year.

Comment by Fred Rohrer

2013-05-24T21:08:20Z

Oops, I will delete my wrong answer. Thanks to Fred.Fred and Torsten!

Comment by Fred Rohrer

2013-04-21T10:42:59Z

Dear Sergei, I refer to SGA 4, Expose I, Appendix ("Univers" by N. Bourbaki). I do not know of any translation of SGA, so you have to go with the french original.

Comment by Fred Rohrer

2013-02-20T22:59:06Z

First, $\sup(x,u)+\sup(y,v)$ is greater than $x+y$ and than $u+v$. Second, if $(a,b,a+b)\in H$ is greater than $(x,y,x+y)$ and $(u,v,u+v)$, then $\sup(x,u)$ is smaller than $a$ and $\sup(y,v)$ is smaller than $b$. Hence, $\sup(x,u)+\sup(y,v)$ is smaller than $a+b$. This yields the claim. (Note that the third component needs to be the sum of the first and the second in order for the triple to be an element of $H$.)

Comment by Fred Rohrer

2013-02-20T21:41:23Z

Dear Rajnish, I do not understand your question. Please clarify.

Comment by Fred Rohrer

2013-02-18T06:39:43Z

Dear @Qing, thank you for your explanation.

Comment by Fred Rohrer

2013-02-17T07:16:27Z

@Liran: The $I$-torsion functor, considered as taking values in the category of $I$-torsion $R$-modules, is right adjoint to the inclusion of the category of $I$-torsion $R$-modules in the category of $R$-modules. But considered as taking values in the category of $R$-modules, you are of course right.

Comment by Fred Rohrer

2013-02-16T08:18:12Z

Is it known that projective modules of finite type over polynomial algebras in countably many indeterminates over fields are free?

Comment by Fred Rohrer

2013-02-15T13:07:00Z

@Martin: Why does Tom's comment answer the question?

Comment by Fred Rohrer

2013-02-11T06:27:11Z

It is not true that Bourbaki's set theory "does not include the axiom [scheme] of replacement". Replacement (or some version thereof) is included in axiom scheme S8 (E.II.1.6). (At least in the 1970 version; I do not know about older versions.)