Fred Rohrer
|
Registered User
|
|
|
May 16 |
revised |
Examples of polynomial rings $A[x]$ with relatively large Krull dimension added 348 characters in body |
|
May 16 |
answered | Examples of polynomial rings $A[x]$ with relatively large Krull dimension |
|
Apr 21 |
comment |
“Axiom of global choice” 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. |
|
Apr 21 |
answered | “Axiom of global choice” |
|
Apr 8 |
answered | How to cite math journals? |
|
Mar 12 |
awarded | ● Necromancer |
|
Mar 5 |
answered | How to call covers not covering anything else? |
|
Mar 5 |
awarded | ● Critic |
|
Feb 20 |
comment |
Subgroup of lattice-ordered group 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 $H$.) |
|
Feb 20 |
comment |
Subgroup of lattice-ordered group Dear Rajnish, I do not understand your question. Please clarify. |
|
Feb 20 |
accepted | Subgroup of lattice-ordered group |
|
Feb 20 |
revised |
Subgroup of lattice-ordered group added 725 characters in body |
|
Feb 19 |
answered | Subgroup of lattice-ordered group |
|
Feb 18 |
comment |
Are f.g. projective modules free over total quotient ring of a reduced non-noetherian commutative ring Dear @Qing, thank you for your explanation. |
|
Feb 17 |
comment |
Left/right exact functor “in nature” which is not a right/left adjoint @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. |
|
Feb 16 |
comment |
Are f.g. projective modules free over total quotient ring of a reduced non-noetherian commutative ring Is it known that projective modules of finite type over polynomial algebras in countably many indeterminates over fields are free? |
|
Feb 15 |
comment |
Are f.g. projective modules free over total quotient ring of a reduced non-noetherian commutative ring @Martin: Why does Tom's comment answer the question? |
|
Feb 11 |
comment |
Where in ordinary math do we need unbounded separation and replacement? 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.) |
|
Feb 9 |
comment |
What kind of subset is Spec(R_P) in Spec(R)? "...does occur in some valuation rings": This is in fact the case in every valuation ring. |
|
Feb 9 |
revised |
What kind of subset is Spec(R_P) in Spec(R)? added 506 characters in body |
|
Feb 9 |
comment |
What kind of subset is Spec(R_P) in Spec(R)? @Georg: You are of course right, since spectra are not necessarily totally ordered by inclusion. |
|
Feb 9 |
answered | What kind of subset is Spec(R_P) in Spec(R)? |
|
Feb 5 |
comment |
Is the empty graph a tree? @ACL: As far as I know, Bourbaki does not use "nonpositive". |
|
Feb 2 |
comment |
Is the empty graph a tree? Ah, I see. There seems to be some confusion between "connected spaces/graphs/things..." and "connected components of spaces/graphs/things...". If the empty thing is one of the former but none of the latter, then the problems about unique decompositions disappear. |
|
Feb 1 |
comment |
Is the empty graph a tree? @Angelo: I do not understand your argument. May I ask you to explain? |
|
Feb 1 |
answered | Is the empty graph a tree? |
|
Jan 26 |
revised |
On the definition of delta-functors added 650 characters in body |
|
Jan 23 |
comment |
Does every regular Noetherian domain have finite Krull dimension? Nagata's Lemma E1.1 (that yields noetherianness of the example in question) is similarly elementary as Matsumura's proof (or the original proof) of Cohen's Theorem, but it finishes the whole thing. |
|
Jan 23 |
accepted | Does every regular Noetherian domain have finite Krull dimension? |
|
Jan 23 |
comment |
Does every regular Noetherian domain have finite Krull dimension? @nosr: cf. Neil's answer and my comment on it. |
|
Jan 23 |
comment |
Does every regular Noetherian domain have finite Krull dimension? This is indeed (a special case of) the original Example 1 in the Appendix of Nagata's Local rings. Your last sentence might be confusing, as it remains true without the word "regular". |
|
Jan 23 |
answered | Does every regular Noetherian domain have finite Krull dimension? |
|
Jan 4 |
answered | Laurent Polynomials |
|
Dec 30 |
accepted | Lattice of Prime ideals |
|
Dec 30 |
answered | Lattice of Prime ideals |
|
Dec 29 |
revised |
Direct product of rings corrected typo |
|
Dec 29 |
revised |
Direct product of rings deleted 131 characters in body |
|
Dec 29 |
comment |
Direct product of rings Oh dear, let me edit my silly mistake. |
|
Dec 29 |
revised |
Direct product of rings deleted 3 characters in body |
|
Dec 29 |
answered | Direct product of rings |
|
Dec 5 |
comment |
Dimension of polynomial algebras The converse of the implication you mention in the second paragraph does not hold; see p. 406 in the paper by Arnold and Gilmer mentioned by J.C. Ottem in his answer. |
|
Dec 5 |
comment |
Dimension of polynomial algebras Dear @François, this is a very interesting reference! May I ask you to add it as a proper answer? |
|
Dec 4 |
revised |
Question on localization technique edited body |
|
Dec 4 |
accepted | Question on localization technique |
|
Dec 4 |
comment |
Dimension of polynomial algebras This - of course very interesting paper - contains a lot of information that is related to my question. But does it indeed concretely describe classes of rings with the desired properties? |
|
Dec 4 |
comment |
Question on localization technique ... If the result holds in case we have a local base ring then it holds in this situation, and thus the claim will be proven once we prove it for the case of a local base ring. |
|
Dec 4 |
comment |
Question on localization technique Dear @Axy, concerning scalar restriction you should take a look at Bourbaki's Algèbre II.1.13. Concerning the proof, I explained that if $M_{\mathfrak{p}}=N_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$ in $R_0$, then $M=N$. So, if we want to show that $M=N$, then we have to show that $M_{\mathfrak{p}}=N_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$ in $R_0$. To do this, we take such a prime ideal and localise everything at it, obtaining a similar situation as before but this time over a local base ring... |
|
Dec 4 |
asked | Dimension of polynomial algebras |
|
Dec 4 |
comment |
Question on localization technique For clarity's sake, the book under discussion is the one by Brodmann and Sharp. |
|
Dec 4 |
answered | Question on localization technique |
