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12:15 AM
Consider a set in the plane such that every intersection with a vertical line is at most countable. Can such a set have positive measure?
The set is not required to be closed

12:35 AM
@AkivaWeinberger do you have a proof if the set is closed?
porque podemos reducir a este caso gracias al hecho que la medida es regular

Remind me what a regular measure is
"every measurable set can be approximated from above by open measurable sets and from below by compact measurable sets" - Wiki
I do not know a proof where it's closed
or compact

es decir la medida de un conjunto es el supremo de los subconjuntos compactos (que es la condicion relevante a esta situacion)
(estaba buscando la palabra supremo jaja)
pienso que hay una clasficiacion para conjuntos cerrados de R, que no se si se aplique a R^2 tambien

DogAteMy: Sounds like Fubini's Thm, assuming the set is measurable.

es que cada conjunto cerrado es la union de un conjunto numerable y un conjunto perfecto
y cada conjunto perfecto es innumerable

Fubini does sound like the right tool to use

12:41 AM
hmm

I'll need to double-check how nice the sets need to be for that

it just needs to be R^2-measurable right

> "If you associate to each real x a countable set A(x) ⊆ ℝ then there must be nonequal reals x,y so that x isn't in A(y) and y isn't in A(x)." This is independent of ZFC.
Called "Freiling's axiom of symmetry" apparently (but I haven't looked that up to check)
graph A (i.e. $\{(x,y)|y\in A(x)\}$)
Flip $A$ across the diagonal, call that $A^\top$ I guess
$\{(y,x)\mid y\in A(x)\}$
Then this is saying $A\cup A^\top$ never equals $\Bbb R^2$
And I suppose that only can happen if $A$ is nonmeasurable, by Fubini
Oh, $A$ intersect any vertical line is countable
Wikipedia says this is equivalent to the negation of the Continuum Hypothesis (assuming the rest of ZFC)
So Freiling XOR CH
So I guess, metamathematically, CH is a bad axiom iff Freiling is a good one

1:31 AM
Ugh.

1:45 AM
@robjohn Thanks! I have found over the years that many of my attributions are incorrect, so appropriate, pithy quotes are now surrounded by legal disclaimers :-).

1 hour later…
2:58 AM
Henlo, anyone here?

3:11 AM
1

Let $F$ be continuous on $[a,b].$ let $f:[c,d]\to \mathbb R$ be differential function satisfy $f([c,d])\subseteq [a,b].$ Show that $G:[c,d]\to \mathbb R$ defined by $G(x)=\int_0^{f(x)}F(t)dt, x\in [c,d]$ is differentiable and $G'(x)=F\circ f(x) \cdot f'(x).$ My attempt $\lim_{h\to 0} \frac{ G(x... 3:33 AM why are you posting the question here? @robjohn is the mean green for Christmas? 3:49 AM Hello, can anyone take a look at my post regarding orthogonal probability measures? 0 Problem:$\alpha$and$\beta$are two mutually orthogonal probability measures in the sense that for some Borel subset$E\subset[0,1]$,$\alpha(E)=0$and$\beta(E)=1$. If$F(x)=\alpha\{[0,x]\}$and$G(x)=\beta\{[0,x]\}$for$0<x\leq 1$, show that for any$\varepsilon>0$, there is finite collectio... 4:26 AM Hi All. In group$G = \{ 1 , -1 , -\iota , -\iota \}O(1) = 1O(-1) = 2O(\iota) = 4O(-\iota) = 4$Question is that$n$is the order of$a \in G$where$n$is the least positive integer. How do we apply$a^k = e$Where$k$is some integer.$k = nq + r , 0 \le r < n$Where$q$&$r$integers. (1) What is$q$and$r$as in my example??? (2) Also what is the meaning of$0 \le r < n$. Why we take$r$value less than$n$and greater than equal to$0$??? Pls explain these questions with my example. 4:44 AM I know that if$R$is a PID, then any submodule of a finitely generated free R-module is free Can we drop the condition 'finitely generated?' 4:59 AM beautiful does someone know anything about the number of ways of representing a number as sums of cubes? Hi @TedShifrin Good Morning. hello @TedShifrin hello @123 you new to this room? Hello @LeonhardEuler . Yes i am new in this room. okay which field of math you study? maybe algebra University Math. But now i am focusing on math more deeply rather than computation. 5:09 AM I study NT Analytic NT Want to know about the number of ways of representing a number as sums of squares @LeonhardEuler Great yes sure definitely. But I can't find much information they are not studied very much @LeonhardEuler Pls see my question. If you can answer. which question? Pls see few comments above. 5:13 AM @123 this one? Yes this one. okay let me read it Sure Thanks.$k=nq+r,\,0\le r<n$reminds me of division algorithm it's like we are dividing$k$by$n$you can think$q$to be the quotient and$r$to be the remainder this is what division algorithm says a very trivial answer is $$q=\lfloor\frac{k}{n}\rfloor$$ @LeonhardEuler Yes it is Euclid division algorithm. 5:22 AM I dunno much about group theory Yes but in group theory, to find an order of element what$q$and$r$in my example? @LeonhardEuler Okay.. no prob. please ping me if someone gives a proper answer @LeonhardEuler Sure. Yesterday i have discussed with @TedShifrin may today someone gives me answer. 5:37 AM @123 Your question is very unclear. 5:47 AM Hello @copper.hat I want to understand theorem for finding the order of element of group with example. In this theorem we used division algorithm. If$n$is order of element$^n = e$. Why we use$a^k = e$Where$k = nq + r$and$0 \le r < n$. What is$q$and$r$here as in my example for elements in group??? 6:12 AM @123 i don't know what theorem you are referring to or how the division theorem is used. @copper.hat I sahred yesterday. let me tag. ok, just fyi i'm leaving in a few mins. @TedShifrin Pls the theorem above this comment. @copper.hat See here and here. :-) @123 sry, how do i find the theorem you are talking to. i don't have the energy to trawl through the comments looking 6:23 AM what would be the most important consequences of a formula for primes? just wondering 36 I have heard that any submodule of a free module over a p.i.d. is free. I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so it does not apply to modules that are not finitely generated. Does the result still hold? What... ah found it: 10 What would be the immediate implications for Math (or sciences as a general) if someone developed a formula capable of generating every prime number progressively and perfectly, also able to prove (or disprove) the primality of every N-th number. I know this is a very large and subjective answer,... does someone want to see an exact fromula for the nth prime? crazy @Measureme Theorem is below this comment. 7:13 AM I have a nonsense question. When we say 'a contravariant functor F is left exact' we mean that given an ES A -> B -> C -> 0, then 0 -> F(C) -> F(B) -> F(A) is exact. Is 3 a special number here? Will this not imply A->B->C->D->0 is ES then 0 -> F(D) -> F(C) -> F(B) -> F(A) is ES. ES = exact sequence 4 hours later… 10:46 AM Hey all, I was wondering whether anyone took Apostol's calculus 11:05 AM @AlexD I have that book on my laptop but didn't read it Could you check whether he covers infinite series, please? If so, would you mind sharing on what chapter it is? Okay it would take some time lemme check Thank you Tom M. Apostol calculus volume 1 One-Variable Calculus, with an Introduction to Linear Algebra has a chapter on infinite series Chapter 10 @AlexD Thank you 11:16 AM You're welcome Hello Henlo Yeah I greet everyone I like that :-) I'm trying to solve a problem on implicit differentiation using partial derivatives but I keep getting the same wrong answer, obviously something escapes me Can't I use the simple division rule we use in single variable calculus with partial derivatives? What is the problem you are trying to solve? I'll Latex it. It'll take me a few minutes Suppose$f: \mathbb{R}^3 \to \mathbb{R}$is$\mathcal{C}^2$and$\frac{\partial f}{\partial z} (\Bbb{a}) \neq 0$, so that$f=0$defines$z$implicitly as a$\mathcal{C}^2$function$\phi$of$x$and$y$near$\Bbb{\bar{a}}$.$\Bbb{\bar{a}}$is$\Bbb{a}$without the third component. Show that: 11:39 AM How about that for a fractal? $$\frac{\partial^2 \phi }{\partial {x}^2 } (\Bbb{\bar{a}}) = - \frac{\frac{\partial^2 f}{\partial z^2} \left(\frac{\partial f}{\partial x}\right)^2 - 2 \frac{\partial^2 f}{\partial x\partial z} \frac{\partial f}{\partial x} \frac{\partial f}{\partial z} +\frac{\partial^2 f}{\partial x^2} \left(\frac{\partial f}{\partial z}\right)^2}{\left(\frac{\partial f}{\partial z}\right)^3}$$ I can't believe I did it (more or less XD) What I did is use implicit differentiation with partial derivatives on$\phi$to get: Oh the partial derivatives on the right are all evaluated at$\Bbb{a}$with implicit differentiation I get: $$\frac{\partial \phi}{\partial x} = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}$$ But if I just differentiate a second time using the rule for fractions I get the wrong result. so that's clearly the wrong way to go about it. what is it that escapes me? 12:06 PM If I have a linear operator$T : Z^r \to Z^r$such that$T^3 = Id$and 1 is not an eigenvalue of T. How do I show 3 divides the index of the image of I-T ? So far I've shown that 3 divides the norm of all the eigenvalues of I-T and that I-T is diagonalisable if we extend the map to$\Bbb Q[w]^r$(where w is a primitive third root of unity). Also it's clear that the eigenvalues come in pairs: the dimension of the eigenspace for w is the same as the one for w^2 @123 @123 I don't really see a Theorem. 12:24 PM @Astyx the index of the image of$I-T$is precisely$|\det(I-T)|$as long as that is nonzero (which is the case since$1$is not an eigenvalue of$T$) Oh yeah Thanks! 12:35 PM @LeonhardEuler That looks to be based on $$\pi(n)=\sum_{k=2}^n\left\lfloor\cos^2\left(\pi\frac{(k-1)!+1}k\right)\right\rfloor$$ 12:45 PM @EdwardEvans Re this: I'm not using the fact that B is a DVR am I? I'm just thinking about how it works with the valuation err Maybe it's a superfluous hypothesis Yeah I don't think you need that B is a DVR, it's just any old ring between A and Frac A I think It's just that$Frac(A) = A[p^{-1}]$where$v(p) =1$1:02 PM what about smth stupid like Frac(A) = A[X]/(pX - 1) lmao I gotta go listen to someone talk about L-functions now what's stupid about that, that's how you localize at an element smh No Frac(A) is formed by writing A x 1/A even for 0 @EdwardEvans direct product, right 3 hours later… 4:35 PM @LeonhardEuler giphy.com/gifs/cauliflower-xasZeFOhWS5cQ this is smth for you uh How do I show that there are an inifnite number of primes of Z that totally split in$Q[\sqrt{d}]$? I've shown that for every nonconstant polynomial P, there's an infinite number of primes p such that there exists an n such that p|P(n) uhh I can't remember In other words I want to show that the decomposition group is trivial @Simone I haven't read everything, but since you already did some work yourself, this might be a good question for MSE. I did precious little I'm afraid I'm only confused as to how I should go about to do a second order partial derivative of phi again Why can't I use the calculus 1 division rule (converted to partial derivatives)? Even if I use the more rigorous approach of the Jacobian matrix and the differentiation rules on those I get the same (wrong) result. I think I haven't yet digested partial differentiation very well 4:48 PM If you include your attempt at calculating the second derivative, then it might be a good question on the site, just saying. I just tried the division rule, getting: $$\frac{\partial^2 \phi}{\partial x^2} = (-1) \left( \frac{\frac{\partial f}{\partial z} \frac{\partial^2 f}{\partial x^2} - \frac{\partial f}{\partial x} \frac{\partial^2 f}{\partial x \partial z}}{ \left( \frac{\partial f}{\partial z} \right)^2} \right)$$ Which is wrong. thanks for the consideration btw, you guys always help. I also tried some reverse engineering on the solution, and I saw that what I got is a part of the solution, but that didn't give me an epiphany. 5:07 PM @Simone I didn't do anything to help you :D then at least give some moral support damn you!! XD jk What about amoral support? Fine I'll have a look at it sure, or immoral support... if you meet me at the back of the alley... @supinf thank you. @Simone What you're missing is the points at which you actually take those derivatives. The complete formula reads$\frac{\partial\phi}{\partial x}(x,y)=-\frac{\frac{\partial f}{\partial x}(x,y,\phi(x,y))}{\frac{\partial f}{\partial z}(x,y,\phi(x,y))}$, so if you differentiate again, you don't only have to use the quotient rule, you also have to use the chain rule and the Jacobian of$(x,y)\mapsto(x,y,\phi(x,y))$will come into play 5:15 PM riiiiiiight. I didn't think of that that's why I didn't ask this on MSE, I knew I was being stupid Thank you Thorgott. ok someone was faster than me :) np 1 hour later… 6:48 PM Hello @MikeMiller How's it going? btw @SayanChattopadhyay we will introduce the poisson bracket and liouville operator so I guess symplectic stuff isn't that far after all Oo awesome! What was the course specifically on again?$M$is a semisimple means it is a direct sum of simple submodules. I am trying to prove that the direct sum decomposition is unique up to isomorphism and permutation. That is, if$N_1 \oplus ... \oplus N_{t} = M = M_1 \oplus ... \oplus M_{s}$, I want to show that$t=s$and there exists a permutation$\tau \in S_t$such that$N_i \cong M_{\tau (i)}$. I'm not sure where to begin. @SayanChattopadhyay stochastic processes in fluids so the liouville operator will determine the time-evolution of a stochastic process, if I understand it correctly Oh nice, maybe a naive question. But in such cases is the formalism more Lagrangian dependent or Hamiltonian? I have a feeling Lagrangian because there's more of a hands on approach, dealing with E-L equations and so on? 7:03 PM hamiltonian it is I think @user193319 try induction We're always dealing with the hamiltonian, at least @Thorgott On either$s$or$t$? ye we'll then be interested in "slow" observables, which apparently correspond to conservation laws, and "fast" observables, which are hard to measure and classified as noise 7:05 PM @user2103480 Yeah you would get$iL = \{\cdot,H\}$, where the right hand guy is the hamiltonian vector field @user2103480 Oh I have never heard of this so we calculate around with correlation functions, determine eigenvalues and resolvent of the liouville operator and then we go on to this stuff: The Mori-Zwanzig formalism, named after the physicists Hajime Mori and Robert Zwanzig, is a method of Statistical Physics. It allows to split the dynamics of a system in a relevant and an irrelevant part using projection operators, which allows to find closed equations of motion for the relevant part. It is used e.g. in fluid mechanics or condensed matter physics. == Idea == Macroscopic systems with a large number of microscopic degrees of freedom are usually well described by a small number of relevant variables, e.g. the magnetization in a system of spins. The Mori-Zwanzig formalism allows to... "Mori-Zwanzig projection operator formalism" Dude this stuff is so fascinating, it's like you're doing some kind of decomposition on the algebra of observables Yeah I hope I will be able to appreciate that and not get lost in the details at quite a few points I get lost since I don't know the underlying physics, but atm it's all about correlation functions so that's OK So is like stat mech is a prerequisite to this course? Yes and No I was told it's fine if I either took advanced stat mech or a course on stochastic analysis (technically I took neither but I'm fine on the stochastic analysis since we don't need anything like girsanov) But the course is, at the moment, still based on a previous run of a similar course Meaning that the sheets and the lecture notes are identical 7:15 PM I see And this time we're only mathematicians. He mentioned that it's unfortunate, since in the last run, there were many physicists, and the exercises/material/notation are accordingly more physics-ish and with more assumptions implicit than I can follow Oh that's bad. I always wanted to take a course on stat mech, but the major system in my university puts so much workload on you, I could never sit do the prerequisites and go do a course in stat mech from the physics department And we do not have anyone who takes a "mathematical" course on stat mech so that it can be offered as an elective or something I still hope that the physics leaves a trace of intuition for what's happening in these and those dynamical systems at the end of the semester. I've noticed that uni and studying for exams, then forgetting it all, made me compartmentalize a bit too much of the knowledge And that needs to be broken up; I think "applied" courses can really help there @SayanChattopadhyay And the intersection with other courses isn't large enough to learn stat mech while studying something else? @SayanChattopadhyay How would a mathematical stat mech course look like? @user2103480 Yeah for instance let me tell you what I have to take this upcoming semester. I have Enumerative Geometry, Riemannian Geometry, Geometric Group theory, Lie Groups and its representations, Knots and Braids, Homology and Cohomology, and maybe Algebraic Groups. With all this and a master thesis to think of and write (which is going to be something "applied" as of now, I think), I do not know where I can find the time to do some physics wtf that's too much to learn properly in one semester for a normal person 7:30 PM Exactly, I am thinking I will drop one course, that is the max that I can drop, or they change something in the next semester because of all these online shenanigans ?????? smartest to drop would probably be either enumerative geometry or riemannian geometry? I'd think these are the least interconnected with the rest Nah those are the ones I badly want to take, the enumerative course is very rarely offered, and what I maybe want to do for my thesis uses some sort of enumerative geometry, I believe. I am thinking of dropping either Algebraic groups, or GGT. sad @AlessandroCodenotti noises anyways that's a horrifying workload I am expecting the Lie groups course to be an easy one, Homology Cohomology should be manageable since I have seen some of it before. I have no clue what Knots and Braids is going to be like, and the Enumerative Geometry course has a weird ass course structure 7:36 PM I'm at my limit with 3 courses, each 1 lecture per week, and one 2-lecture course although 2 of the courses with 1 lecture move at a pretty fast pace @SayanChattopadhyay Don't drop algebraic groups. I never get to talk about algebraic groups in here otherwise :) If someone can tell me what exactly is happening with that course, it would be great. Here is what they plan to do: Examples of counting problems in geometry: Grassmanians and intersections. Chasles problems on conics. Fixed points of transformations. Bezout’s theorem and 9-point circles. How to “count properly”: Transversal intersections. Fundamental theorem of algebra and multiplicity. Grassmanians and Projective space: Cell decompositions. Schubert cells. Schubert calculus. Moving and blowing-up: Resultants and plane curve intersections. Singular intersections by moving. Singular interse ouch @TobiasKildetoft Lol, well I am pissing someone off in the chat anyway, its either Balarka, Alessandro or you :p ah yes, the formal grothendieck-riemann-roch theorem smells like dried tears in a corner 7:40 PM @Thorgott lmao yeh lmao The "references" are, Milnor's Characteristic Classes, and Fulton's Intersection Theory, with some random K-theory book. These are hardly references, they are damn mountains Thank god I don't do all that stuff. Yesterday I needed to show convergence of an infinite product by first completing squares, then taking the logarithm and doing the full log(1-x) taylor expansion, and afterwards bounding by a geometric series. Felt nice not to do some abstract shite Probably a course on algebraic groups will never get to the stuff I am mainly interested in anyway (i.e. their representation theory in positive characteristic) @Sayan Beautiful course! Classic stuff motivating modern technology. @Ted, help me understand, what does the course plan to do? I couldn't find any direct references and things seem very scattered 7:44 PM A friend of mine is interested in algebraic combinatorics and schubert calculus. She doesn't like topology and geometry, but her advisor makes her do equivariant cohomology of the grassmannian and k-theory You can go into the dried-tears-corner with her Lol, after this course, I hope to read Sheldon Katz's Enumerative Geometry and String Theory, maybe I can already die. Schubert calculus: arxiv.org/abs/1309.5055 @TobiasKildetoft About this, yeah I do not think there is any representation theory. The course ends at Borel's fixed point theorem Yeah, that is pretty inevitable. It takes a lot of prerequisites to really get anywhere My favorite exercise for you is: How many lines meet four lines in general position in$\Bbb P^3$? 7:50 PM By general position, do you mean no three of them meet at the same point? You can do this completely classically, so that a high school student could follow, or you can learn Schubert calculus and tge cohomology ring of the Grassmannian. No, pairwise skew. Oh okay Side exercise: Classify the (smooth] doubly-ruled surfaces in$\Bbb P^3$. Nice, let me give them a shot @SayanChattopadhyay but GGT is so good 7:57 PM You guys are going to mess my entire thing up lol, maybe now I will drop Lie groups and take GGT. I went to two different seminars today and they were both about hyperbolic group even though one was supposed to be a model theory seminar (I'm not complaining though) @Thorgott For the inductive step, assuming that I have$N_1 \oplus ... \oplus N_{t-1} \oplus N_t = M_1 \oplus ... \oplus M_s$, should I mod out by$N_t$or something like that? I'm not sure to use the inductive hypothesis. @SayanChattopadhyay take a course on spde as well, they're useful in quantum field theory! take an entire master's workload in one semester Yup, anyway the world's going to shit, so why not finish a master's in a semester 8:15 PM yeah, I'd try something like that Anyone want to think about some atrocious interesting point set topology? Hmm...The LHS will be isomorphic to$N_1 \oplus ... \oplus N_{t-1}$, but I don't know what the RHS is... The$N_t$can't be fully contained in any of the$M_i$'s, since they are simple modules... @AlessandroCodenotti I'll say yes and you better make me regret it Do you know what the Vietoris topology on the space of closed sets of a topological space is? no, but I'm pleasantly surprised to hear that there is something else named after Vietoris 8:23 PM lol So you take some space$X$and$F(X)$is the set of closed subsets of$X$. We put a topology on$F(X)$with basis$\langle U_1,\ldots,U_n\rangle=\{F\in F(X)\mid F\subseteq\bigcup U_i\land F\cap U_j\neq\varnothing \text{ for all $j$ }\}$where$U_i$range over open sets in$X$did you know that Vietoris was one of, if not the, oldest mathematician ever yep, oldest Austrian ever Almost 111 years @Thorgott Is it possible that there is some$i$and$j$such that$N_i$and$M_j$are equal? Then could I mod out by this common factor? @AlessandroCodenotti eleventy one* ^ @EdwardEvans or one-hundred-onety-one 8:28 PM onety onety one ah yeah$\ell\ell\ell$Juan Jandrad y Juanti Juan wait idk i'm bored isnt it onety-onety onety-one 8:29 PM It depends @AlessandroCodenotti Ok, if we have two collections$\{U_i\}$and$\{V_j\}$, then every closed set in$\langle U_i\rangle\cap\langle V_j\rangle$also belongs to$\langle U_i\cap V_j\rangle$for some subcollection of indices (maybe omitting index range was a bad notational choice), so they form a basis, alright yeah So this thing I'm reading claims that if you take a chain$Y$of closed subsets of$X$, meaning that for all$E,F\in Y$either$E\subseteq F$or$F\subseteq E$, then the closure of$Y$in$F(X)$is still a chain Which seems very reasonable, but I'm not seeing a formal argument Oh here$X$is even compact hausdorff just to be safe random general facts:$K(X)$, which is the subspace of$F(X)$of compact sets, tends to inherit many properties of$X$, metrizability (if$X$is compact metrizable then the Hausdorff metric induces the vietoris topology), connectness, being Polish, zero dimensional, separation axioms etc. 8:52 PM Easier way: Jordan-Holder theorem. Question: What does it mean to say$k^n$is the unique simple module of$M_n(k)$? 9:03 PM @AlessandroCodenotti I think every ascending sequence of closed subsets converges to its union. Without further contemplation, I'd like to claim the same thing holds for nets. Adding unions of elements of a chain to it doesn't make it not a chain, clearly. If$C_n$is a decreasing sequence of closed sets and$\bigcap C_n\subseteq U$, is$C_n\subseteq U$for large enough$n$? smells like some compact Hausdorff separation trick @AlessandroCodenotti what is$F(X)$? closed subsets of$X$@Thorgott But the union is not necessarily closed thanks Since$X$is compact Hausdorff convergence of sequences here is the same as convergence in the topological liminf=topological limsup sense (points whose neighbourhoods meet cofinitely many/infinitely many points of the sequence) 9:11 PM oops, that's a fair point that makes it more complicated @AlessandroCodenotti are those closed sets necessarily? @user193319 it just means what is say:$k^n$is a simple module over$M_n(k)$and it's the only one but more is actually true:$M_n(k)$is semisimple, so because$k^n$is the unique simple module, every module over$M_n(k)$is just$\bigoplus_{I} k^n$for some index set$I$ok, then all left to do is convince yourself that adding those to the chain doesn't destroy chain-ness, no? (and pray that replacing "sequence" with "net" doesn't change anything) yeah I can trust that nets won't mess this up :P Being in the closure of$Y$, for a closed$F$, should mean something like "if you cover$F$with finitely many open sets, then you necessarily also cover an element of$Y$" (this is not accurate, but that's my possibly misleading intuition) why is the topological limsup=liminf of an ascending sequence of closed sets not just the closure of their union? meeting infinitely many of them <=> meeting cofinitely many of them <=> meeting any of them <=> meeting their union and clearly adding closures of union would keep it being a chain @user193319 That works too, of course. For the record, what I had in mind was observe that the$N_t$on the LHS corresponds to some submodule on the RHS, which necessarily has to look like (up to isomorphism) some partial diagonal. Then quotienting also drops only one factor on the RHS and induction does the rest. 9:31 PM Ah, I see. That's what I was thinking would happen! Thanks for that! 9:48 PM Eyy @lukas what's up Hey @Amin Howdy, Demonark. How does one show that any simple module over the nxn upper triangular matrices (over a field) is one dimensional? Hey Ted! How are you guys doing? Hassling with semi-incompetence which may or may have nothing to do with the covid mess. 9:50 PM I don't know much about the nxn upper triangular matrices. Jacobson radical (all strictly upper triangular matrices), I know it has a nice direct sum decomposition...hmmm...I think that's about it. @LukasHeger I'm guessing 1 dimensional as a vector space. oops I didn't read the upper triangular thing That's okay. It happens (to me all the time). Rip, that's no good Ted @user193319 for any ring$A$, simple modules over$A$are the same as simple modules over$A/J(A)$, where$J(A)$is the Jacobson radical Wait, is$A/J(A)$always guaranteed to be a simple module? 9:53 PM no$A/J(A)$is another ring But it has an$A$-module structure. I'm saying that simple modules over$A$are the same as simple modules over$A/J(A)$Oh, I see. Hmm...I wonder if I have that theorem available. Let me check my notes. Also that's a nice starboard we've got going for$A$the ring of$n \times n$upper tringular matrices, then$A/J(A) \cong k^n$(as a ring) for$k^n$it's easy to show that all simple modules are one-dimensional 9:55 PM @Ted I think you might appreciate this, in my most recent meeting with my advisor some hyperbolic stuff came up that I wasn't totally sure about, and he's like yeah your hyperbolic geometry is rusty @user193319 it just says that$J(A)$acts trivially on each simple module And then I'm like :0 that's giving a bit too much credit to me lol Also AMS had a sale a couple days ago and I bought Helgason! Helgason is rather dense, but well-written I no longer possess him, of course @LukasHeger When you say simple modules over$A$are the same as simple modules over$A/J(A)\$, what exactly do you mean by "same as"?

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