bio | website | TheBigQuestions.com/blog |
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location | ||
age | ||
visits | member for | 4 years |
seen | 43 mins ago | |
stats | profile views | 7,569 |
Nov 19 |
comment |
Robotics, Cryptography, and Genetics applications of Grothendieck's work?
I do not think that MO is well-suited to speculation about what might or might not have been in the mind of a reporter for the New York Times. |
Nov 19 |
comment |
Grothendieck -sad news
In view of the positive response to David Roberts's comment pointing to my first blog post, I am moved (though with some trepidation about self-promotion) to point to the far more ambitious blog post here: thebigquestions.com/2014/11/17/the-generalist |
Nov 17 |
answered | Probability spaces involved in using Bayesian Inference |
Nov 17 |
comment |
Consecutive primes all congruent to 4 mod 1
Vacuum my living room. |
Nov 16 |
comment |
maps $\mathbb{S}^{n} \to \mathbb{S}^{n}$
Please do not answer off-topic questions. |
Nov 12 |
comment |
Two (other) rings…are they isomorphic?
Your rings have different dimensions and hence are not isomorphic. Is there perhaps a typo in the question? |
Nov 5 |
awarded | Enlightened |
Nov 5 |
awarded | Nice Answer |
Nov 5 |
awarded | Nice Answer |
Nov 5 |
comment |
Is the unit tangent bundle of $S^{n}$ parallelizable?
@RyanBudney (and Peter Crooks): I don't think you are using language in a standard way. The statement that "$M$ is parallelizable" always (as far as I'm aware) means that the tangent bundle to $M$ is trivial. I've never (outside of your comments) seen "parallelizable" used to mean "trivial a as a vector bundle". |
Nov 2 |
awarded | Yearling |
Oct 30 |
awarded | Notable Question |
Oct 24 |
awarded | Nice Answer |
Oct 24 |
awarded | Necromancer |
Oct 24 |
revised |
What is the Beilinson regulator?
added 37 characters in body |
Oct 24 |
answered | What is the Beilinson regulator? |
Oct 20 |
comment |
Lifting a direct summand of a free module
After your edit, the answwer is trivially yes: Split the map $R_I\rightarrow S_I$ with a map $f_I:S_I\rightarrow R_I$. Lift $f$ arbitrarily to a map $f:S\rightarrow R$. Then the composition $R\rightarrow S\rightarrow R$ is the identity mod $I$, hence an isomorphism. |
Oct 14 |
comment |
Lifting a direct summand of a free module
Multiplication by $1+i$ becomes a split injection mod $I$. |
Oct 14 |
answered | Lifting a direct summand of a free module |
Oct 6 |
comment |
polynomials with roots on the unit circle
Even in the quadratic case, $P$ can be anything of the form $Ax^2+Bx+A$, with $-2A<B<2A$, which doesn't translate into any sort of congruence condition mod primes (other than the trivial condition that two coefficients that are equal remain equal after reduction). So I'm unclear on what you're looking for. |