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Mar
9
comment How to prove that any perfect complex on an affine scheme is strictly perfect?
I might be mistaken, but I'd have thought that the usual definition of "strictly perfect" (as found in, say, SGA6) would replace your "complex of finite rank free sheaves" with a complex of finite rank locally free sheaves. This would rule out Fernando's counterexample.
Mar
9
answered What is $K_2(\mathbb{Z}[x,x^{-1}])$?
Mar
4
comment Communal problem books
Perhaps at least marginally relevant: thebigquestions.com/papers/jimmys.pdf
Mar
3
awarded  Popular Question
Mar
3
awarded  Nice Question
Mar
3
asked Do the real numbers “know” that they are countable in a larger model?
Feb
5
comment map in K-theory
At the level of rings, this is the map $K_n(R)\mapsto K_n(R/I)$ given by taking the class of a projective $R$-module $[P]$ to the class of the (finite projective dimension) $R/I$-module $P/IP$. (Take the alternating sum of the classes of the projective modules occurring in a projective resolution of $P/IP$.) The reason you need smoothness is to insure that $P/IP$ has finite projective dimension so that this class is well defined.
Jan
24
comment Why are they called 'pernicious' numbers?
It looks like Michael Stoll posted while I was typing.
Jan
24
comment Why are they called 'pernicious' numbers?
Let $t(n)$ be the digit sum of the binary representation of $n$. Then Google will tell you that $n$ is called odious if $t(n)$ is odd and evil if $t(n)$ is even. Thus every number is either odious or evil, and therefore the words "odious" and "evil" cannot be pejorative in this context. It seems very likely that "pernicious" was chosen in the same spirit because it starts with the letter "p".
Jan
23
comment Is it possible to create an infinite sequence in which no subsequence is repeated 3 times in a row?
Aside from other issues, the Threefold Repetition Rule described in this post bears very little resemblance to the Threefold Repetition Rule from the actual game of chess.
Jan
8
comment Nepero game (by Yacov Perelman)
continued --- so you might expect to find rare counterexamples when $n/e$ is just slightly less than a half-integer.
Jan
8
comment Nepero game (by Yacov Perelman)
Suppose just for a moment that instead of taking $n$ to be an integer, you take $n=ke/2$, where $k$ is an odd integer. Then $n/e$ is a half-integer, so the two choices for $m$ are equally close, but the higher of the two choices always gives a (slightly) better payoff --- in fact the ratio of the two payoffs goes rapidly to $1$ from above. This suggests that when your $n/e$ is very close to a half-integer, there should be a slight preference for the larger of the two $m$'s, which is almost always outweighed by the strong preference for the closer one.....continued
Dec
30
comment Probability, Topology, functional analysis problem
Why do you "have to prove" this?
Dec
29
accepted Stable Household Formation
Dec
29
asked Stable Household Formation
Dec
28
comment Ext functor for more than two modules?
What exactly are you claiming is functorial in what?
Dec
15
awarded  Necromancer
Dec
15
comment Why do we teach calculus students the derivative as a limit?
@quid: Well, of course. The Greeks were very primitive and had not yet discovered the letter $h$.
Dec
14
answered Why do we teach calculus students the derivative as a limit?
Dec
9
comment Upper bound on number of cells created by varieties of co-dimension 1
Are these polynomials supposed to take their values in an ordered field? Otherwise what's a "region"?