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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.