MathOverflow
user26857's user avatar
user26857's user avatar
user26857's user avatar
user26857
  • Member for 11 years, 4 months
  • Last seen more than a week ago
Profile Activity
Stats
1,259
reputation
44k
reached
17
answers
8
questions
Communities
View all
Badges
View all badges
This user doesn’t have any gold badges yet.
13
silver badges
27
bronze badges
Top tags
View all tags
52
Score
23
Posts
92
Posts %
9
Score
1
Posts
4
Posts %
8
Score
2
Posts
8
Posts %
6
Score
4
Posts
16
Posts %
4
Score
2
Posts
8
Posts %
4
Score
1
Posts
4
Posts %
Top posts
View all questions and answers
All Questions Answers
Score Newest
answer
26
Nontrivial theorems with trivial proofs
Nov 30, 2015
answer
13
Is a domain all of whose localizations are noetherian itself noetherian ?
Nov 2, 2019
question
11
Are there irreducible ideals that are not primary in $K[X_1,\dots,X_n,\dots]$?
Oct 29, 2014
question
9
Equation with $q$-binomial coefficients
May 3, 2015
answer
9
Hilbert's Nullstellensatz on polynomials with integer coefficients
Feb 2, 2013
answer
8
Radical generation of ideals in Noetherian rings
Nov 13, 2012
question
5
Macaulay's example of prime ideals in $\mathbb C[X_1,X_2,X_3]$ having large number of generators
May 18, 2015
question
4
Example of fractional ideal whose inverse does not commute with localization
Feb 5, 2015
question
4
When does a faithful module have an element with zero annihilator?
Mar 15, 2015
answer
4
Strongly Noetherian property. When is the tensor $A\otimes_{k}B$ Noetherian for Noetherian rings $A$ and $B$?
Nov 12, 2012
Top network posts
View all network posts
59
Is Pythagoras the only relation to hold between $\cos$ and $\sin$?
32
Is quotient of a ring by a power of a maximal ideal local?
21
Integral domain with a finitely generated non-zero injective module is a field
21
The number of ring homomorphisms from $\mathbb{Z}_m$ to $\mathbb{Z}_n$
20
Let $n$ be an integer greater than $1$. In a ring in which $x^n = x$ for all $x$, show that $ab=0$ implies $ba = 0$.

Your privacy

By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.