i liked chemistry a lot. i thought about majoring in it after my first college class, but it would have required changing colleges. which was about two more pieces of paperwork than i wanted to do that year.
no. good luck with that. side note, i didn't realize there was an oxfam america
speaking of diverse interests when i was in law school i was very interested in taxation for about a year. never did anything with it.
tax policy could be a very effective way of achieving a wide variety of goals less expensively or coercively than other ways. in the US, however, very difficult to get a lot of ideas off the ground.
A is a commutative ring with unity and X = Spec(A). Then denote $V(f) = {P\in X: f\in P}$. In this notation, for $f,g\in A$, if $rad((f)) = rad((g))$ then $V(f) = V(g)$ is obvious? here, rad(f) means radical ideal
This is a complete argument as far as I'm concerned. I can't judge your specific situation completely, but I would at the very least ask the TA to elaborate on what they think is insufficient and then maybe you can work from there.
Yes I mailed him. He replied : I should've written because the radical of prime ideal is itself (i.e. rad(P) = P), if a prime ideal P contains $f$, then it contains rad(f) = rad(g) so it contains $g$.
Where should I post a question asking what is the origin of the term 'b-thousand' that means 2^3? Actually what I really want to know is anything at all about this term, which doesn't yield anything on Google. There's also b-million, b-billion, and so on.
With the greatest respect, this conversation is a complete waste of time. Just type the name, it takes like 2 seconds and there's no need to find a shortcut. Also, do you not have a mouse? Why are you looking for a way to ping people without a mouse? I have no idea what's going on wtf
So.. why don't you just click on the relevant name? There has to be some additional information about you that means that you can't easily click or something, else this is just a waste of time
I did nothing wrong, because I pinged him as part of a conversation we were already having. So why did you say ' You are the one who used his name to practice pinging @MatthewChristopherBartsh'
Yea i know. This is a long shot, but perhaps we want to find $x\in \mathbb R^n$ that gives $f(x)\in \mathbb R^m$ such that $|f(x)|$ will be closest to zero as possible.
If $R$ is an integral domain and $S\subset R$ is a multiplicaitve set and $M$ is a $R$-module, then I already know $T(S^{-1}M) = S^{-1}T(M)$ where $T(M)$ denote the torsion submodule.
I want to show $M_m$ is torsion free for all maximal ideal $m\subset R$ then $M$ is a torsion free moduel
(T(M))_m = T(M_m) by the property I stated. and by assumption, (T(M))_m = 0 for all maximal ideal $m\subset R$. Now as $N_m = 0$ for all maximal ideal $m$ implies $N =0$ where $N$ is a $R$-module
Hi, if $X$ is a set for which there is a filter $\mathcal{F}(x)$ of sets containing $x \in X$ assigned to every point $x \in X$, and these filters are such that for all $U \in \mathcal{F}(x)$ there is some set $V \in \mathcal{F}(x)$ such that for all $y \in V$, $U \in \mathcal{F}(y)$, if we say a set $O$ is open if it is empty or such that for all $o \in O, O \in \mathcal{F}(o)$, does every member of $\mathcal{F}(x)$ contain an open set containing $x$?
it seems like intuitively this has to be true for us to call $\mathcal{F}(x)$ the neighbourhoods of $x$, but I can't see to prove it
I'm being a moron: given G-modules M, N and the set of group homs Hom(M,N) "is made into a G-module" via $(g\varphi)(m) := g\varphi(g^{-1}m)$. But it doesn't look like $(gg^\prime)\varphi = g(g^\prime \varphi)$ to me (which is why I'm being a moron)
Can we say the the volume of a cuboid is 0 when all the sides are coplanar . If I imagine it , it’s like a cuboid on the floor where I can’t store anything inside.
Since there is no height of the cuboid , volume is 0.
@Astyx Cool Brauer spam fact of the day: computing the Brauer group of a number field is "almost" equivalent to proving global Artin reciprocity (which establishes an isomorphism of a quotient of the idèle class group corresponding to a finite abelian extension of your number field with the Galois group of that extension)
In abstract algebra, an adelic algebraic group is a semitopological group defined by an algebraic group G over a number field K, and the adele ring A = A(K) of K. It consists of the points of G having values in A; the definition of the appropriate topology is straightforward only in case G is a linear algebraic group. In the case of G being an abelian variety, it presents a technical obstacle, though it is known that the concept is potentially useful in connection with Tamagawa numbers. Adelic algebraic groups are widely used in number theory, particularly for the theory of automorphic repr...
The adèle ring of a number field $K$ is the restricted product of $K_v$ over all places of $K$, the idèles are the invertible elements of this guy, and the idèle class group is idèles modulo principal idèles (which are the image of $K^\times$ in the idèles)
There are two ways I solved this Q. 8N + 4N = 12N . So , classify it in vectors form. So , I did found the resultant of 8i + 4i = 12i and 8i + 4i only. So , for 12i = sqrt of 12^2 = 12N But for the other one it is sqrt of 64+16+8*4cos theta where eta would be 0.(Angle between i and i is 0) . So , I get 112 as answer. Why are both the answers different
@TedShifrin if I have a sequence of nested simply connected sets $\Omega_n$, each with a biholomorphic $f_n\colon\Omega_n\to D$ with $f_n'(a) > 0$ and $f_n(a) = 0$ (i.e. uniquely determined by the Riemann mapping theorem), then the local behaviour of $f_n'$ at $a$ is always the same, right, even in the limit? I can't have a subsequence converge to $f$ such that $f'(a) = 0$?
I am having troubles wrapping my brain around what could change. If Riemann mapping theorem uniquely determines the $f_n$, and the sets are all nested and simply connected, then shouldn't f'_n(a) = f'_m(a) for all m and n?
@Shobhit no worries. In the future, try a few actual Wolfram language commands and see if they work. They often have implementations in wolframalpha.
How to prove that the series formed by the sequence $ 3^n sin(x / 4^n) $ converges absolutely and uniformly on $(a , \infty ), where $a>0$. I calculated the pointwise limit, which came out to be zero. Now to prove uniformness, i know, weierstrauss M-test, dirchlet test and abel's test. I am not able to prove its uniformness using any of the three ( i dont see how). How can i apply any of these tests ( if applicable, here)? If not, how to do this?
@RajorshiKoyal Please don't repost images repeatedly to chat. I see that you haven't asked any questions on main; that would be a better venue for some of these questions which tend to turn into long drawn out discussions about problems in a book about which we have no more information than you.
@Shobhit That does converge pointwise to $0$, but not uniformly.
@VVKK77 I know that some other Chrome users have had problems. I don't know if there is a fix that anyone has found, but I don't have a Chrome device, so it is hard for me to debug what's wrong.