It's nice that you're enthusiastic about mathematics, @epic_math, but these are all lame jokes that everyone has heard a million times and it's kinda tiring
@trivialmathisdifficult you can ask what you want here, just if it's relentless spam then you might get ignored. Genuine questions will get answered if someone decides to answer them.
@mathguy I strongly recommend looking at Abstract Algebra: Theory and Applications. It is free online, and is updated every year: abstract.ups.edu/download.html
@Clarinetist That looks like an interesting reference, though it seems really weird to me that it has linear algebra so late, introducing vector spaces after both groups and rings
It has been a long term fascination of mine that if you sample $n$ values $X_1, \cdots, X_n$ from the standard 1D normal distribution, look at the sample mean $(X_1 + \cdots + X_n)/n$ and the sample variance $1/n \cdot \sum_{i = 1}^n (X_i - \overline{X})$ (some scale by $1/(n-1)$, which is more appropriate), then they are independent.
Right. There's a slick proof that I think I briefly saw in Erik's thesis, but I never really liked that proof as much. There's a "pure geometry" reason that takes some work to set up but is fascinating.
I think he'll know it, but I have never seen that stuff phrased geometrically. I have said this here before but I feel it has to do with having a foliation, with a family of measures parametrized by $\Bbb R$ on the ambient manifold, such that along the leaves the measures don't change, but transversally they do
Yes, it's a good exercise in working with equivalence classes and forcing students to confront correct notation. This was something I worked extremely hard at when I taught my algebra course.
Oh, I see. So you're not even using normality of $N$. I did in my proof, using the group structure on $G/N$.
So that makes it a stooopid question for a stooopider reason.
Yeah, I'm actually giving a talk at the math club this Friday about the Monster Group. I want to try and explain that a group can be realized geometrically
A geometric interpretation of the Monster would be ideal, but I don't know if that's a thing that can be managed. Been looking for a while and it's too complex for me to get a grip on anything
For me, so much of education is about interactions — and most of my students spent a lot of time in office hours talking with me and working with their fellow students. I would have to learn to be a totally different teacher.
I'm attending Alg Top I, Adv Geom I, Topological Manfifolds, Intro to Surgery Theory and Algebraic Geom I for the moment but I might not write the exam for all
Quite happy to now be in a line of work where I am only minimally affected by Covid (mainly the lack of social interaction in person with my coworkers that I miss)
Well the thesis+thesis seminar is 36 credits so if you can do around 30 credits per semester in the first three semesters you don't need to do more classes in the last term
Yeah we had an information talk where they went over all that with us
For the lectures this semester, I'll actually just be happy if I can absorb like 70% of the material, my background is admittedly quite lacking in some areas
Anyway, it's been nice chatting to you all once again, I gotta head out now :)
The pro trick everyone does for the 3 areas requirement is to take Global Analysis I because it's an intro to smooth manifolds course but it gives analysis credits
That's a sneaky way to make everyone learn what a manifold is.
Sort of how I included differential geometry of curves and surfaces as a way to fulfill the undergraduate analysis requirement (real analysis — hard, complex analysis — easy, diff geo — medium), arguing that it would make people understand multivariable calculus better (like the so-called advanced calculus courses in the US).
Here's a question: So, zero divisors can be nilpotent, idempotent, and stuff like $\bar{x},\bar{y}\in R[x,y]/(xy)$ (I don't have a name for that one). Is this an exhaustive list of type of zero divisors or are there others that fit into none of the above?
I was thinking that in some sense the third form was universal, but I would probably have needed to be working in a $k$-algebra and map to $k[x,y]/(xy)$. Trying to think it through gave me a headache.
$\overline{x} \in \Bbb Z[x,y]/(xy)$ is a versal (=universal -uni(que)) zero divisor in the sense that for every zero divisor $r \in R$ there's a ring map $\Bbb Z[x,y]/(xy) \to R$ which maps $\overline{x}$ to $r$
but the ring map is not unique because one might do different stuff with $y$
however $(\overline{x},\overline{y}) \in \Bbb Z[x,y]/(xy)$ is the universal pair of elements that multiply to zero
Choose any infinite sequence of elements $z_1, z_2, \dots \in U$ such that $\{z_i \mid i \in \Bbb N\}$ is discrete. Consider the following chain of ideals $I_n:\{f \in \mathcal O(U) \mid \forall i\geq n:f(z_i)=0 \}$. Then $I_1 \subsetneq I_2 \subsetneq \dots$ doesn't stabilize
you can use the Weierstraß product thoerem to construct $f \in \mathcal{O}(U)$ such that $f_i(z)\neq 0$ for $i<n$ and $f_i(z)=0$ for $i \geq n$
Weierstraß products can also be used to show that $\mathcal{O}(U)$ is a gcd domain when $U$ is a domain
@TobiasKildetoft the idea is if we take some discrete infinite set $S$, then the ideal of functions which vanishes for almost all elements in $S$ is not f.g.
here in the chat if I type " is just get that, but in a program like LibreOffice Writer if everything is set properly, it automaticall replaces " with „ and “, respectively, depending on the context
@LukasHeger Ah, nice. So, for every $n,m\in\mathbb{N}$, we can find a ring with a zero divisor $x$ such that $x$ is not idempotent, not nilpotent, and $x^n=x^m$
So, technically speaking, there are two classes of zero divisors. Ones such that there exist nonzero $n\neq m\in\mathbb{N}$ where $x^n=x^m$ and ones where if $x^n=x^m$ then $n=m$. Idempotents and nilpotents are just a special case of the first
