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1:44 AM
@leslietownes I taught at UCLA for two years before being hired away by Apple.

traitor!

Not as a lawyer. I actually did math there (for graphics). I held some small classes for the other graphics programmers there.

i have sometimes worked on cases where our client is suing apple. small world

1:57 AM
tim cook nixed our using the apple logo for our little company.
despite the fact that we found lots of hardware bugs.
i am sure i out of nda by now. oops.

Huh?

:math.stackexchange.com/questions/4182396/… I've defined a negative dimension using the $\subset$ opposite inclusion of traditional Homology
That's right, I'm a badass

2:43 AM
I just need to prove that it's functorial
I'm guessing $A$-modules to $A$-modules will do

3:26 AM

you can put \ in front of ln and it looks better
trying to think of a simpler way to do it, this looks like it could get uggggly

Leslie: one question

we're even

Given two cyclic groups G and G', I can find no. of homomorphisms from G to G' by actually finding only where generator of G gets mapped in G'

@leslietownes it is ugly, I have had no simpler way come into mind :'-(
I edited the \ln though

3:29 AM
Now, if $G'$ is cyclic but nothing is given above G (except its order), what can we say?
About normal subgroups of G?

so you want to evaluate Hom(G,C_n), where G is a finite group and C_n is the cyclic group of order n?

Every subgroup is normal?

Hi Ted.

is G abelian?

both cyclic, Leslie.
hi, Koro

3:31 AM
No. G is not given to be abelian

well he then switched horses midstream and said what if "nothing is given above [sic] G"

excuse me — you said both cyclic?

Homomorphic image of G is given to be cyclic

ugh.

@TedShifrin : That's the case which I understand how to handle (in terms of finding no. of homomorphisms from G to G') but my next sentence asks a different case

3:33 AM
he switched streams midhorse

Leslie: order of G is given and its homomorphic image is given to be cyclic

so G is a finite group and n is a positive integer and what are the homomorphisms from G to C_n

Do examples.

Please let me know if the question doesn't make sense or I have used any terminology wrong
Let's do one example then: Suppose that I want to find no. of homomorphisms from 1) $Z_{10}$ to $Z_8$ 2) from $Z_{10}$ onto $Z_8$

my cat curled up on my daughter's bed on top of the sheet and blanket, so when we put our daughter to bed we didn't cover her with anything. it would have disturbed the cat

3:35 AM
not cyclic.

Let me build up the background just in case my question does not make sense.
I just want to say that the above example problem can be handled by me: For 1): I'll use first isomorphism theorem to conclude 0 and for 2) I'll see where $1\in Z_{10}$ will get mapped to in $Z_8$ and since mapping is onto I would wish that image to be generator of $Z_8$ so we have 4 possibilities (1,3,5,7) and hence 4 no. of such homomorphisms.

so G is cyclic.

This was well-handled. Now, let's go to a different example: Let G be a group of order 60 and its homomorphic image is given to be cyclic and is of order 12. Then we are interested in finding order of all normal subgroups that G has. My idea is to somehow utilize the above idea (that was used in example)
I don't understand how to do so. I only know from first isomorphism theorem that G indeed will have a normal subgroup of order 5 (that is kernel of the homomorphism)
But apart from that, I don't know how to get the order of other normal subgroups of G.
Ted, I asked this question yesterday and I was suggested a solution also but I feel that to be complicated for a group theory beginner such as myself.

if you have a homomorphism from G to C_n and a homomorphism from C_n to C_m you can compose them to get a homomorphism from G to C_m which might have a larger kernel. that's one thought.
it might be better to work with examples.

So I am just trying to work out an understandable way with the existing tools.
@leslietownes This infact answers my question indeed.
For example: I map $x$ in G to $x^4$ (say) and then $x^4$ to $x^3$ and then composition of homomorphism is also a homomorphism.

3:48 AM
@Koro mapping is onto?
impossible.

Yes Ted, I think so because its homomorphic image is given $\phi (G)$ (given to be cyclic and is of order 12)

Does 8 divide 10?

of course no. why 8?
Have I messed up again by causing some confusion?

totally wrong

What is wrong in that Ted? For the first part, I did say 0 (that is no homomorphism, which you would agree with)
which agrees with "8 does not divide 10"

3:54 AM
What is the possible image of a homomorphism from $\Bbb Z_{10}$ to \Bbb Z_8$? It can be of order 1,2, 4, 8. (that is {8},$\langle 4 \rangle, \langle 2 \rangle, \langle 1 \rangle$) right? I'm confused now. Absolutely wrong. I need to revise things more calmly. The order of the image must divide 10. yes, Ted you are absolutely right. I remember this result 4:01 AM Just the FHT. it should divide both - 10 and 8. Yup. Thanks a lot, Ted. Talking to you feels like I am getting sharper and my concepts are being refined. Don’t forget fundamentals! And do concrete examples. Yes Ted right :-) I'll try the original problem again (order 60 normal subgroup one), I hope to see some light this time. Many thanks. 4:06 AM Sure. Do smaller examples first. sure. Thanks a lot Professor Ted :-) @napstablook I posted my answer 4:24 AM @robjohn that sure was..... complex @napstablook It was pretty straightforward, but specifying the contour takes a bit. This particular integral allows a cute substitution trick. 3$\forall$has$\exists$as a "left adjoint" in Topos Theory. Let$M \xrightarrow{f} M' \xrightarrow{g} M''$be sequence of$A$-modules and$A$-module homomorphisms. Now suppose we don't have$gf = 0$or in other words no ability to compute traditional Homology at$M'$, but instead we have sort... Might be a new concept, but might be covered by "resolutions" as well, which I have not studied yet My goal was a grid of the form$\Bbb{Z}/a^n \to \Bbb{Z}/a^{n-1}$that contained every natural$a$So it's not just a double complex, nor a triple complex, but a$\Bbb{P}$-complex one dimension in the complex grid for each prime. So, those natural modulo maps are not chain maps, but they are "reverse" chain maps... Each dimension chain in the grid terminates at the$0$module so$0$is the "origin" You of course have commuting, squares, cubes, hypercubes, etc. But interestingly, there is a reverse homology concept instead of$\ker f \subset \text{im} g$, it's$\ker g \subset \text{im} f$! Take that as the definition of negatively dimensioned in the case of modules that have dimension defined. In this reverse homology exactness is exactly the same principal as traditional exactness. That's when the "dimension in homology" is zero. or reverse homology That's like saying$-0 = 0$. @TedShifrin I require that the chain maps all be surjective, but perhaps there is a way around that I mean it's$\ker g \subset \text{im f}$.* Wait, I said that. But the first one was the "traditonal homology" module and it's$\text{im} f \subset \ker g$. I'm just reversing things 1 hour later… 6:07 AM Odd functions are symmetric about what? The origin??? what does that mean? If i said Odd functions are symmetric about y=-x would i be wrong? @AdilMohammed rotational symmetry 2 hours later… 7:52 AM Hi. To write 'for all', we use inverted A, how to write this in Latex? @aarbee i dont use latex much but heres a link i found tex.stackexchange.com/questions/316575/… @AdilMohammed got it, thanks. it's simply \forall No problem @hyper-neutrino Thanks i have a little more doubt on the analysis of graphs. If -f(x) means the graph is a mirror of f(x), then what does f(-x) mean? Does it mean for all x, we go to -x, take the value of graph of y=f(x) there and put that value for f(-x) at x?? I don't understand your question. Are you asking about even or odd functions or neither? oh so its different in all cases? 8:01 AM I don't know what "If -f(x) means the graph is a mirror of f(x)" means "-f(x)" does not mean anything i was playing around with desmos and found that -Cos x was the mirror of Cos x on x-axis so i thought it would be like that in all cases no that is a property of even functions Formally, an even function is a function such that$f(x)=f(-x)$and an odd function is one that$f(x)=-f(-x)$or more commonly written,$f(-x)=-f(x)$9:06 AM I was actually trying to figure out how f(x), -f(x) and f(-x) are related generally. more specifically i wanted to know if there were graphically some way to plot -f(x) if i know f(x) @hyper-neutrino like here, if you look the cos and -cos are mirrored to each other along x axis but if you look at sin and -sin x i thought they too were mirrored to each other along x axis... are you telling they are mirrored along y axis? 9:57 AM Without using polylogarithmic functions, how can I evaluate$\sum_{k=1}^\infty \frac{\sin(k\pi/2)}{k^3}$? Via, polylogs, I can show that this is equal to$\Im(\operatorname{Li_3}(i))=\beta(3)=\frac{\pi^3}{32}$. But can this be evaluated without Polylog? what can I say about the partial derivatives of a lipshitz vector field$K: \mathbb{R}^n\to \mathbb{R}^n$@AdilMohammed if a function$f$is such that$f(x)=f(-x)$then it will be mirrored in the y axis ( we call such a function an even function ) I thought that$1/3k^3= \int_0^k \frac{1}{t^4}$. But then I cannot interchange the sum and integral since the upper limit of the integral is dependent on the lower limit of the sum How can I do it? Or can you'll just give a hint?$\sin(k\pi/2) \in \{-1,0,1\}$huh, it's mildly interesting that this series returns anything nice at all Hey @Astyx, perhaps you know this: if I have a Dedekind domain$A$and$0\neq f\in A$and localize at$\{{1,f,f^2,\dots\}$,$A_f$is Dedekind and I have an induced map$Cl(A)\rightarrow Cl(A_f)$between the class groups. I have calculated that the kernel is generated by the residue classes of the prime factors of$(f)$. Do you know if this is correct? I'm just trying to confirm, but haven't been able to find this in the literature (it surely is there, but I don't know where). 10:13 AM Sounds ok For any normal Noetherian separated scheme X, there is a short exact sequence 0 -> Z -> Cl(X) -> Cl(X \ Y) for any irreducible codimension 1 closed subscheme Y of X This is a well-known fact, see Hartshorne. Yours is a special case Y = Z(f)$4 > p^2$implies$2 > p \lor -2 < p$, which makes sense intuitively, but what would be the rigorous justification for getting$-2 < p$from$\pm 2 > p$? @BalarkaSen I hate you He dropped the H bomb thanks for the confirmation 10:59 AM I'm terrible at inequalities and it's killing me when I try to do analysis would it make sense to get a book on olympiad inequalities and practice that way? neat! thanks! given two probability densities$\mu,\nu$can I apply Jensen to the integral$\int f(x) (\mu(x)-\nu(x)) dx$to say :$ \big \int f(x) (\mu(x)-\nu(x)) dx \big)^2 \leq ...$2 hours later… 12:42 PM Is there a way to upload images here? there's a little upload button besides the chatbox Unable to find any, i am on phone. you need a bit of reputation to be able to 1:01 PM @AdilMohammed In general, there is no relation. Some functions have$f(x)=f(-x)$and we call them even. Some functions have$f(-x)=-f(x)$and we call them odd. Others, like$y=e^x$have neither. 3 hours later… 3:43 PM Are there any references that I can read on secondary characteristic classes? i looked at the one book on my desk that talks about characteristic classes, and it doesn't have them in the index. ted may know. he is contractually obligated to know this stuff as a chern student Yeah I know, I didn't find anything about them in the usual literature. I was told to look them up when I wanted to show a relationship between higher chern classes and the curvature forms on vector bundles 3:59 PM @SayanChattopadhyay This is the wrong name. If you want to understand the relationship between Chern classes and curvature forms, this is called Chern-Weil theory. It is explained in many references but I am fond of Morita's "Geometry of differential forms". Secondary characteristic classes (like secondary cohomology operations) are defined in a setting when the characteristic classes you already know vanish. The standard example is the Chern-Simons invariant of a flat connection. The simplest case is really quite simple, so I'll explain it. Suppose one has a *flat complex line bundle*, so Secondary characteristic classes are more complicated and not seen as often. You probably do not actually want to look into those 4:19 PM is the following true?$4 - p^2 > 04 > p^2\pm \sqrt{4} > p$I ask because I have no clue how to deal with the$\pm$here$4 > p^2$means$2 > |p|$now write out the cases and trying going from there ah that makes more sense thanks! is there some reason why root taking doesn't work like with equalities? negative things can square to positive things? i dunno if there is a deeper meaning to it because$4=p^2$means$2=|p|$not$\pm 2=p$ah 4:24 PM although the latter two happen to be equivalent, I consider it more accurate to say the first is the direct equality, and the latter is derived i see no difference between them, but if i had to take a position i would probably agree you just gotta be careful what the inequality is actually saying as you manipulate it. whether it's squaring or anything else. any kind of mathematical operation preserves equality in a tautological way this is, in a sense, what equality means in the first place otoh, there is no reason for anything to work well with inequalities that's comforting the world is dark and mean so we aren't actually taking a square root to go from$4 > p^2$to$2 = |p|$, we're finding an alternate meaning? sort of? of course we know$\sqrt{x^2} = |x|$4:28 PM you can think of it as taking a square root. if a and b are nonnegative numbers and a < b then sqrt(a) < sqrt(b). that's true and you can think of that as 'square rooting an inequality.' but yeah, you run immediately into what sqrt(p^2) is and that's where the thinking cap has to be on. okay here's another way you could do it$4-p^2>0(2+p)(2-p)>0$so the product of two things being positive means they have the same sign Yes, that's the correct way to do it if you want to avoid absolute value. and then if$p\geq 0$then$2+p>2-p$so you need$2-p>0$i.e.$0\leq p<2$that's neat thanks a lot! and smth similar for$-2<p\leq 0$idk exactly what, sorry my brain is not working atm 4:37 PM Well, be careful. You need the conjunction ("and"). So$2-p>0$(i.e.,$p<2$) AND$2+p>0$(i.e.,$p>-2$) .... or$2-p<0$and$2+p<0$. The latter says$p>2$and$p<-2$, which of course cannot occur. ah right, thanks i was thinking of a different approach for a different but similar thing i believe, lol Even for such simple math, the structure is complicated. Either P and Q or R and S. wait you need the four terms? I thought that if we have$2-p >0$and$2+p>0$it guarantees$(2+p)(2-p) > 0$it does, but what about$2-p<0$and$2+p<0$? But if we're trying to solve an inequality we must consider all possibilities. To solve means you have an if and only if situation. 4:42 PM you have to consider the other cases to get the full solution set (even though in this case the second set is empty) Understanding the logic is very important. ohhh right, if both are negative it works too So solving for$p$, we'd have to write: [$2>p$and$2>-p$] or [$2<p$and$2<-p$] wait the right side is a contradiction ah! but it's an OR so this just means that$-2<p<2$is the only solution set 4:54 PM right! nice! thanks, everything clicked algebra has been saved @robjohn But, I have eyes at the back of the room. =D Actually, I saw your message by pure luck. No ping or person informed me. When someone has not been active in the room for a while, it takes a superping to ping them. A simple @-ping won't do, I believe. superping? meaning the bat signal? @robjohn you can also reply to a message they've sent @leslietownes yeah. It's the bloodhound ping of the moderators. It will ping even people who have never been in the room. 5:02 PM not only that, it can ping people who don't even exist in chat if you superping someone by their main site profile and they don't have a chat profile it will forcefully create one IIRC pinging bernoulli himself to weigh in on this question about bernoulli numbers. arise, johann. the moderators command it @hyper-neutrino true, if you can find an old comment of theirs. the power of the diamond compels you @robjohn You're right, and I knew that. I also knew the risk of getting irrelevant pings after popping up here. Now you're at the mercy of the math chatroom ;-) 5:07 PM @hyper-neutrino If that's true, does it also grant permission to speak for low-rep users? @user21820 no, we have to add that manually per room @hyper-neutrino Oh, so moderators can abuse low-rep users by pinging them with taunts in chat that they can't reply to? =P well if a low rep / suspended user joins a room, anyone can lol @hyper-neutrino Yea, but you can forcefully pull them into a room they didn't even join. =P true :P 5:08 PM i don't know what my computer is for, if not abusing and taunting people who have lower than my amount of reputation on an SE site. @leslietownes It's for abusing and taunting you. It does that every day, even. my therapist raised this possibility. @leslietownes Aha, we're on to something. petition to call it "abusing, taunting, and trolling" so we can acronymize it to "AT&T" @hyper-neutrino Lol. 5:09 PM haha Also totally unrelated to mathematics, here's a video I just watched: what is the symbol § used for in math books (like §5) ? 5:25 PM @Prithubiswas Section. You can also search Google for it:-) 167 in ascii. love the interweb @leslietownes I've never understood why ASCII fonts make ∪ and ∩ look SO ugly. 5:44 PM it was not intended to render those symbols. @copper.hat Well, even the unicode fonts here render those same symbols ugly. Maybe I'm remembering wrongly (I distinctly recall that those symbols were available in some ASCII version), but anyway even the current unicode font here makes those two dual operations look terrible. MathJax:$\unicode{x2229}\unicode{x222A}$Unicode: ∩∪ @robjohn Indeed I know MathJax renders those symbols nicely. It's just crazy that the unicode font designers didn't make dual operations dual. What were they thinking (or not)? there. @robjohn Don't they look awful to you? On my system, ∩ is as tall as O, but ∪ is as short as o. 5:57 PM yes. I would think they'd be the other way around: ∪∩icode @robjohn Hahahaha... Maybe they were preempting lAtEx. ∪∩iCoDe ∪∩ιℂ°▷∈ I think that was graffiti on a wall outside CalTech once ;-) @robjohn Really? I was never there, in case you're wondering. =P 6:02 PM I am just kidding. I have no idea, but it would be a nerdy graffiti You can try it out at your university. Just kidding. i'll spray paint it on the wall of my house ourdoors, so the HOA gets on me about it Maybe they would be mesmerized instead. Though the first thing that came to mind was Higher-Order Arithmetic. @leslietownes speaking of HOAs, I wonder if those condo owners in Florida get their HOA fees refunded? 6:09 PM i'm very interested to know the cause of that. i don't know how florida works. the stereotype is shady developers putting stuff up overnight and lax building codes and minimal licensing requirements for contractors. Ack... the death toll is up to 4. 159 still missing. texas seems to do OK with that, however. all things considered. i'd need 20 stages of review to put a table on my balcony but if it keeps the house from falling down, i'm for it. @leslietownes Isn't there simply a loading limit per unit area?? The color has to be approved. A list of what will be eaten there should be submitted. If it will be used for business. HOAs are fun. Was it purchased at an approved Ikea? user, i dunno. my most recent experience with building codes was a biotech startup. they were prototyping instruments and i asked, uh, are you plugging these into the wall? and they were, which was in violation of about a billion local ordinances. they thought it was OK because the product wasn't on the market. strangely, a lot of regulation of what you can and can't plug in is decided on a city-by-city or county-by-county level. most refer to UL standards and such so it's almost the same thing as you go from place to place, but sometimes not. i told them to put it on a battery at least when the inspector showed up if you can't be good, be careful 6:18 PM @leslietownes Lol. @leslietownes: my wife is an attorney and the firm she is working for specializes in construction defect. She was very interested in the paper this morning. it's not my area but i'd love to know what went wrong here. something went horribly wrong. if it is regulatory failure there will be no recovery against the state but i would not want to have built that building right now. a math lab in a lower floor unit disintegrated. there was a case a few years ago, people were partying on a balcony in berkeley and the balcony fell off. shambolic construction. i think it settled. @leslietownes hopefully it was a low second story 6:27 PM it killed six. some from dublin. irishtimes.com/news/ireland/irish-news/… fourth floor :( i forget the details, somethign to do with wood supports and a design that guaranteed they would rot away. @mme Oh Mike!!! Thanks! Its been a long time yeah, that was big news in the intersection of my communities my high school senior class loaded a waterslide improperly at a water park, causing it to collapse and kill somebody. there was a game where you'd try to see how many people you could put on the waterslide at once. @leslietownes Wow; guaranteed expiry. dumb, but a kind of dumb approved of by the school and park authorities. until the disaster. 6:38 PM people only pay attention when someone gets hurt. that has been my broad life experience. are you listening? jk @copper.hat I'm paying full attention. is anyone hurt? then no i often needed to get the attention of my younger siblings... 7:04 PM I don’t understand this: “if G is a finite group and$\phi$is a homomorphism from G onto$Z_{10}$. Then,$\phi^{-1}\langle 2\rangle$is normal and of index 5.” Why? There’s absolutely no reason to believe that. All we should be able to say is$\phi^{-1}\langle 2\rangle$is normal in G. Period. Please let me know what I am missing. Is it that under homomorphisms indices of pullback of subgroups are preserved? i was going to say that's odd... @Koro false, take the trivial homomorphism in fact preimages of normal subgroups are normal (iirc) but i think you can only claim things about indices when you have some kind of combinatorial relation between them maybe surjectivity, injectivity, who knows I have surjectivity here @LucasHenrique. Now can we say something ? The statement that I find confusing above, is true or not? What are your views on this, Lucas? i think surjectivity is sufficient i.e. the statement is true Also, please note that$\phi$above is from G onto$Z_{10}$. 7:19 PM i'll sketch a proof here's the general fact: if$f\colon G\rightarrow H$is a surjective group homomorphism and$N\triangleleft H$is normal, then$f^{-1}(N)\triangleleft G$is normal and$(G\colon f^{-1}(N))=(H\colon N)$here's the general fact: if$f : X \to Y$is a coherent map between$(\infty, k)$-topoi... Hmm that’s what I’ve been missing all these days that said, I'm very confused by what you quote, because the subgroup generated by$2$in$\mathbb{Z}/10\mathbb{Z}$has index 2 and not 5 @BalarkaSen here comes Balarka and must use some terminology that is beyond my understanding right now 🥲 Coherent ^ @Thorgott: you’re right. Noted. That was a mistake there ☹️ @Thorgott and this should follow from FHT 7:25 PM given$f: G\to H$surjective and$N \unlhd H$normal, then$M := f^{-1}(N) \unlhd G$(prove). now take$\pi\colon H \to H/N$the canonical projection. prove that the map$\phi: G/M \to H/N$given by$\phi(gM) = \pi(f(g))$is well defined and a homomorphism. find its kernel (hint: trivial kernel!) and use the surjectivity of$f$to conclude that$\phi$is an isomorphism @Koro he's messing around.$\phi$is an isomorphism$\implies |G/M| = |H:N| \implies (G:f^{-1}(N)) = (H:N)$fwiw, it's also true for a non-normal subgroup$K\le H$that$(H\colon K)=(G\colon f^{-1}(K))$@LucasHenrique nice. Thanks a lot :-) I see some light now. I was stuck at a similar question for last 2 days @Thorgott computing a explicit bijection between the tranversal sets? in fact,$G/f^{-1}(K)\rightarrow H/K,gf^{-1}(K)\mapsto f(g)K$is an isomorphism of$G$-sets (the latter being viewed as such via pullback) lol, yeah 7:43 PM on this topic, here's a good (and, surprisingly, useful, yet not often mentioned) exercise, which I recommend: for a subgroup$H$of a group$G$, compute the$G$-automorphisms of the$G$-set$G/H$7:58 PM Thanks a lot @LucasHenrique. I got it now :-) 8:13 PM If$X\subset\mathbb P^n$is an irreducible curve, can it then always locally be described by$n-1$equations? (so$X=Z(F_1,\dots,F_m)$for homogeneous polynomials$F_i$, and is irreducible and of dimension$1$) And if so, would it then always be possible to choose these locally definining$n-1$polynomials from the fixed$F_1,\dots,F_m$? I'm wondering these things, because I've seen several definitions of algebraic curves now, and I'm trying to show some equivalences 8:27 PM @ShaVuklia Look at the twisted cubic curve in$\Bbb{P}^3$, these are given be the vanishing of three$F_1,F_2,F_3$hypersurfaces. Such things are called incomplete intersections Yes, so locally they're given by two polynomials, which is exactly my case My book calls them local complete intersections By local I am guessing you mean in the local ring of a point right/ Eh, I was thinking topologically (but probably what you say holds too) I think local complete intersections is the most general one can get if you deal with algebraic sets$X\subset\mathbb P^n$(smooth of dimension 1), but I can't prove it Hmm okay so if I am thinking this carefully, locally complete intersections are those subvarieties which at each local ring are generated by the right number of guys. The question we want to ask is if every incomplete intersection a locally complete intersection 8:56 PM Take$(t^2, t^3, t^5)$in$\Bbb P^3$. I think this cannot be described even locally by 2 equations. 1 The following is an Exercise 1.1.11 of Hartshorne's Algebraic Geometry. Let$Y\subset \mathbb{A}^3$be the curve given parametrically by$x=t^3, y=t^4, z=t^5$. Show that$I(Y)$is a prime ideal of height$2$in$k[x,y,z]$which can not be generated by$2$elements. We say$Y$is "not a local com... Lol, Balarka beat you to it:P Another H bomb Yeah I think$(t^p, t^q, t^r)$is a local complete intersection iff$p, q, r$solve the postage stamp problem IIRC you prove this by looking at$I/I^2$I worked this out once but I don't remember anything now. Huh, thats neat 9:08 PM OK I remembered the right condition. Does$(t^2, t^3, t^5)$work? Yes. The condition is that the least number which is not expressible as an$\Bbb N$-linear combination of$p, q, r$is odd.$2, 3, 5$works.$3, 4, 5$fails least number bigger than$p,q,r$? Largest number I meant of course. what why would there be a largest one 9:23 PM Because they are coprime Aye When you say 3,4,5 fail, you mean$(t^3, t^4, t^5)$is NOT a local complete intersection right? Yes I think so Ok ye$2\$ is the largest guy not expressible in terms of those. Even!!
Anyway I forgot how this works. Look at Kunze's commutative algebra text, it's somewhere there

9:56 PM
If, to each element of G, there is at least one element of F, and to each of F, also there is at least one G, is it enough to consider that both groups are homomorphic?

@BalarkaSen This is so obscure to me
But it seems to work

It's a pure manipulation proof
Afair

I just don't see how to link parity to things that matter

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