Reputation
12,653
Next privilege 15,000 Rep.
Protect questions
Badges
1 14 38
Newest
 Nice Answer
Impact
~155k people reached

Jun
21
comment Soluble(solvable) and nilpotent groups
What have you tried? Question 1 is a pretty straightforward induction on $i$.
Jun
21
answered Extra-Special $p$ group and complement
Jun
21
comment “On Numbers and Games” or “Winning Ways for Your Mathematical Plays”?
That sounds like a reasonable thing to do.
Jun
21
revised “On Numbers and Games” or “Winning Ways for Your Mathematical Plays”?
added 25 characters in body
Jun
21
comment correspondence between prime ideals of coordinate ring and subvarieties
@Seven Yes, we are required to show that, but that is part of the proof of the general result. See e.g. www2.gsu.edu/~matfxe/commalglectures/lect6.pdf Proposition 1.10
Jun
21
answered “On Numbers and Games” or “Winning Ways for Your Mathematical Plays”?
Jun
21
comment correspondence between prime ideals of coordinate ring and subvarieties
@Seven This is a general fact about localizations of rings: If $R$ is a commutative ring and $S$ is a multiplicative subset of $R$, then the prime ideals of $R$ disjoint from $S$ are in 1-1 correspondence with prime ideals of the localization $S^{-1} R$. And the correpondence is the obvious one: from $R$ to $S^{-1} R$, send $P \subset R$ to $P R_S$, the ideal generated by $P$; from $S^{-1} R$ to $R$, send $Q \subset S^{-1}R$ to $Q \cap R$. In this case, $R = \Gamma(V)$, $S$ is the set of functions not vanishing at $P$, and $S^{-1}R = O_P(V)$.
Jun
16
answered Introductory Type Theory Questions
Jun
13
answered A peculiar decomposition of elements in a group
Jun
8
revised Continuity and uniform continuity of $f(x)$ over $\mathbb Q$
TeX'ed
Jun
8
answered Anomalous finitary objects
May
30
awarded  Nice Answer
May
15
awarded  Nice Answer
Apr
16
awarded  Enlightened
Apr
16
awarded  Nice Answer
Jan
30
comment Cosets in quotient rings over ideals of the form $(n,f(x))$
And even if you do have leading coefficient 1, it's still complicated. Suppose $p(x)=x^2+1$. Dividing by $p(x)$ gives a remainder $a+bx$. Now how do you divide this by $n$? The most straightforward thing to do is divide $a$ and $b$ separately by $n$ and replace them with their remainders. So $a$ and $b$ can be any integer from $0,1,\ldots,n-1$. This gives you the cardinality of the quotient ring ($n^2$), but it does not tell you the ring structure, which is a number theory question - it depends in a more complicated way on the prime factorization of $n$.
Jan
30
comment Cosets in quotient rings over ideals of the form $(n,f(x))$
If $p(X)$ does not having leading coefficient $\pm 1$, then it is not so clear what it means to divide a polynomial by $p(X)$ and take the remainder. (e.g. How do you divide $x^3+1$ by $2x^2$ in $\mathbb{Z}[x]$?)
Jan
30
answered Mutually exclusive AND independent event (help with examples)
Jan
30
answered Cosets in quotient rings over ideals of the form $(n,f(x))$
Jan
6
comment Implicit assumptions about axes
It doesn't matter. You're just rotating the entire diagram by 90 degrees but that won't change the volume.