12:17 AM
i seem to be able to prove that, the existence of a ring homomorphism $\phi:A \to B$ implies the characteristic of $B$ divides that of $A$, but Gallian says this is implied when the ring homomorphism is surjective
do we need surjectivity or not?
wikipedia also omits surjectivity

Check definitions of everything.

12:34 AM
oh, i think in one case we're allowing the zero homomorphism and in the other, we're not
seems like Gallian's homomorphisms don't require unity preserving, i think that explains that

Whence the trouble with surface-skimming internet learning.
But you can be unital and still be far from surjective.

yeah, i think in this case unital homomorphism guarantees that the existing homomorphism has enough structure to make the characteristics divide

12:53 AM
Ted, could you tell if my last proof is rigorous enough? Few messages above.

turns out $(\mathbb Z/4\mathbb Z)[1/2 + 4\mathbb Z]$ is the zero ring
i had many hopes for it, and they all came crumbling down
$(\mathbb Z/4\mathbb Z)[1/3 + 4\mathbb Z]$ however might be promising

What does that even mean? I’ve never seen anything like that.
@noballpoint A set does not take a maximum value. I don’t understand why you defined $s$ and $t$ as you did. Isn’t it just min and max of all four?

polynomial of $1/3 + 4\mathbb Z$ with coefficients in $\mathbb Z/4\mathbb Z$, with addition and multiplication defined as in normal quotient rings
we're making $3 + 4\mathbb Z$ into a unit
we gotta hope it is accidentally isomorphic to some usual quotient ring

This is stupid, with coefficients in mod 4, the $4\Bbb Z$ dies. So rewrite, please.
What does $1/3$ mean? Mod 4, $1/3=-1$. So … bye.

1:03 AM
You’re writing formal symbols without any understanding,

well, no: i'm trying to see if the normal rules are well-defined in this case

I guess it's localization

It’s nothing,

I second that what you're saying seems nonsensical fiddling.

do you know for a fact it's not well defined? @BalarkaSen

1:05 AM
I didn’t say that. Read carefully.

If you mean that I should've added all fours in the min\max functions, it is a waste of symbols, all $s$ are less than all $t$.
On the wording. Should I have said "The restriction of $f$ to $[s,t]$ has its max. min..."?

Or the set has a maximal element. Both are fine.

What is not well-defined? You're just mashing together a bunch of symbols, @shintuku.
It's less than ill-defined.

No, all $s$ less than all $t$ is false.

1:06 AM
@BalarkaSen do you know the definition of a natural quotient ring given a ring and its ideal?

Sure.

It’s defined. It’s a ton of symbols saying very little.

Give me a second... I think I didn't say the correct thing.

@BalarkaSen do you know whether a ring extension can seem to be a nonzero ring, but end up being a zero ring?

WTF

1:07 AM
I don't think you understand what a "ring extension" is.

Bingo.

a better phrasing would have been: it seems to be a ring extension, but ends up not being one

One $s$ must be leftmost, one $t$ must be rightmost. No?

Yes, that’s right.

Well...? What's a problem.

1:09 AM
But I think it makes it harder for the reader not to just say leftmost, rightmost. What did you gain?

Writing $R[S]$ for some ring $R$ and a random subset $S$ is nonsense notation, unless $R$ is already understood to be embedded in some larger ring $R'$, and $S$ is a subset of $R'$.

You cannot just invent notation

How would you define $s$ and $t$ so it is better for the reader?

Beautifully said @Balarka.

1:10 AM
no one is inventing notation, we're attempting to check whether the familiar notation is well-defined according to its exact same usual use
and this is definitely not a "new" usage of notation
you're being deliberately dense, nothing controversial is happening here

It is not incorrect, noball. I already told you my preference. This is why I don’t want to keep doing this. A comment or suggestion turns into a time-wasting battle.

checking well-definedness of an expression is perfectly normal mathematics
if you instead want to argue that it is not well-defined, sure, go ahead

Lol, whatever makes you happy.

@shin You’re being an ass again. I will quit interacting with you. I’m fed up.

if you don't have anything interesting to add, and if you don't recognize perfectly valid mathematical investigation for a learner, you have nothing of use or interest to add
that applies to both of you

1:13 AM
Bye!

grow up

You’re going to get kicked from the room in the future.
Impudent twit.

Well, then I apologize for interpreting suggestions as "this is incorrect" :)

this is a room associated with Math.SE officially, you're not free to make deliberate use of your moderation powers when you can't have the humility to accept you're wrong

Right, I just suggested an improvement.

1:14 AM
It's ok, we'll let shintuku do his thing. No big deal, I thought he actually wanted to learn that's why I commented.
Anyway, SE responded to the whole fiasco finally

there's an appropriate way to communicate with learners, balarka, and you're faling spectacularly

I’ve wasted so many hours on him and he has no respect.

Learners need to not think they know better.

I still wonder why shintuku called that user (don't remember the nickname exactly) a monkey.

1:15 AM
i don't. separate the morality from the mathematics, stick to the latter, please.

Because he thinks he’s a smart ass.

lol its a'ight leave it

@shintuku, just cooldown... it is you who will feel guilty for yourself making people around think bad of you.

oh hey freud, thanks for the perfectly useless intervention

Hi @Thorgott

1:23 AM
hey

Hi @Thor

A gentle notice: I am not engaging in any moderation activity right now. The reasons for this have been posted on the main meta, and on the math meta. However, this does not mean that you have free reign to behave badly in chat---when the strike ends, if I still have a diamond, it is likely that moderation action will be taken. Also note that @TedShifrin is a room owner here and, unless he, too, is striking, has the power to kick users from this room, or to put the room in timeout.

Hi @XanderHenderson
Long time

I’m not a mod, so the strike is beyond my purview. But I signed the petition and support the effort.
So many room stars who became mods have moved on, sadly. Daniel, Pedro, and a few others.
Alex Gruber, too.

Drawing some pictures for my thesis

1:30 AM
Where’s the woobly thing with no words?

I ended up putting it to word, finally.

I miss having my graphics program. I have no ability to draw easily.

Inkscape is no Adobe Illustrator, but it's free
You may like it

I’ve tried and gotten nowhere after a few hours.
Can you tutor me?

I only know the basics, which are good enough for my purposes now. What do you want to know?

1:34 AM
I can’t ask an intelligent question. But after 30 years of Illustrator, I knew what to do automatically. Now the simplest things don’t work.

Yeah, understandable.

looks like a 70s prog rock album cover. i like it

When I really need it, I’ll wrestle with it, or just quit entirely.

Does Illustrator also use cubic splines? That's the thing I really understand intuitively and works for me the best
Because I'm always drawing cusps $y^2 = x^3$ for contact topology

I wonder if @D.C.theIII is buried in orange smog. Or maybe it’s just NY.

1:36 AM
Oh yeah heard the news

@BalarkaSen Yes, if you click and drag.

Gotcha
Same thing in Inkscape. I put some random PL curve down by the Bezier tool, then edit it using the splines tool.
Double click to add nodes, options to delete them. Change node type using the options on the bar above, fiddle around with the slopes.
That's all I know
@leslietownes Lol, good point
@TedShifrin This is mostly because of the wildfires in Canada, right? Or has it come all the way into US as well?
I heard at the moment NY is more polluted than Delhi which is... quite an accomplishment

NYC is at the top of the list of all cities. Yes, from Canada.

Is it normal for Canada to have wildfires
Too cold usually

Climate change. Yes, way early, and going on for weeks.

1:43 AM
Terrifying.
Things changing way too rapidly.

It’s all a Democrat hoax. Don’t worry.

Lol
I'll tune in to the republican primary debate, should be entertaining
@leslietownes You may also like these:
We call these things "wrinkles", to no one's surprise.

Make them fatter and they look like me.

Haha.
I can parametrize these things algebraically, but the Mathematica plot looks worse than the hand-drawn ones.
Not worse, but less insightful.
It's supposed to be a movie of a pair of cusps in birth-death

You might need shading or specification of PlotPoints

1:50 AM
Ah, that's probably what I was missing. The thing is, the algebraic parametrizations aren't so neatly compactly supported.

No swallowtail?

Indeed, @Ted, the birth and death points (drawn in red above) are a kind of swallowtail points.

You can delimit the PlotRange, too.

We call them "unfurled" swallowtail. It's embedded, as opposed to immersed as in the usual swallowtail.
@TedShifrin Ah, fair.

I see blue and black, no red.

1:53 AM
Two red points, in the top picture. I should probably make them thicker

Oh, there they are

Balarka, you can post SVG here. You can't upload it to Imgur, but you can put it on Github. Eg,

Oh, nice!
That's a nice projection of the Borromean ring

You can even do animated SVG, as long as it doesn't use JavaScript. However, Xander is allergic to anims. ;)

I am not nearly as techy enough to make those, though.
techy? techie?

2:00 AM
Thanks! I can't remember what eccentricity I used, let me check...

How did you make it?
Looks extremely high-quality.
It's very funny, the perspective makes me think that the above picture should embed in a torus. But of course that's not true, because the Borromean link is a hyperbolic link -- these are never torus links.
I think the point is any two $(1, 1)$-curves on the standard torus in $\Bbb R^3$ actually link once in $\Bbb R^3$, i.e. they make a Hopf link
Whereas all pairs are unlinked for the Borromean

The semi-axes are $1+\sqrt3$ and $2\sqrt3$

If you switched some crossings it'll become toral

@BalarkaSen Handwritten SVG. Well, almost. I wrote some Python to write the SVG. So it doesn't have all the fluff you get from Inkscape, etc.

Oh my god.
Can't imagine handwriting SVGs

2:09 AM
AkIva posted this 6 months ago, so I figured it would be fun to do in SVG. And I did a few variations.
Nov 8, 2022 at 23:22, by Akiva Weinberger

Ah, of course.
Akiva is a master at these things.
I should learn how draw better pictures digitally someday. I have a cool, pictorial solution to a problem I'd like to send to AMM, but it'll take effort to render those pictures.

Here's a 3d one.

Gimbal lock.

Not SVG, though. :)
^ Interactive version

Neat!

2:14 AM
Animated version of the 1st one: raw.githubusercontent.com/gist/PM2Ring/…
I just wish there was a nicer way to make elliptical tori. Those ones look a bit distorted when the eccentricity is high.

You mean how one half of it looks a bit less convex than the other?
That should just be an optical issue, so maybe you mean something else.

They look a bit pinched near the vertices. They're just scaled normal tori.

Are you sure that's not an optical illusion because two other torii are intersecting it at exactly those points

It easy to make a torus with the small circles replaced by ellipses. But there isn't a sensible way to keep the small circles as circles while turning the big circles into ellipsea.

Oh I see

2:23 AM
@BalarkaSen Play with the interactive version. Set k to a small value, eg 1/5. You'll see what I mean by "pinched".

Yeah, I see it now.

so Maschke's theorem says that if the characteristic of a field $k$ does not divide $|G|$ then reducible module and decomposable module are the same in $kG$-module?

Yes, @onepotatotwopotato

thanks

1 hour later…
3:50 AM
$\mathbb Z/4\mathbb Z[1/3 + \mathbb Z]$ is isomorphic to $\mathbb Z/4\mathbb Z$
whoops, didn't write the coset properly above, i meant $1/3 + 4\mathbb Z$
$\mathbb Z/4\mathbb Z[1/2 +4\mathbb Z]$ is the zero ring
hypothesis: $\mathbb Z/n\mathbb Z[1/a + n\mathbb Z]$ is isomorphic to $(\mathbb Z/n\mathbb Z)[x]/\langle ax -1\rangle$

4:54 AM
$\mathbb Z/6\mathbb Z[1/2 + 6\mathbb Z]$ looks like it might be isomorphic to $\mathbb Z/3\mathbb Z$

5:16 AM
at this point, you might as well divide by zero

sure champ, turns out the structure is well defined

lmao
my comment wasnt rhetorical, you're trying to adjoin 1/2 to Z/6Z. 2 is a zero divisor, so you're pretty much trying to divide by 0
2*3 = 0 mod 6

exactly, now force a ring structure on the polynomial generated by $1/2 + 6\mathbb Z$
you'll have to say 1/2 = 1
3 = 0, 2 = 1/2
you'll end up with 0, 1, and 5 acting as the 2 in Z/3Z

5:34 AM
but uh, the notation isn't strictly correct since we're not actually adjoining the cosets mentioned
we're still doing some adjoining of something tho

I finally understand the idea behind the proof of why {(0,0)}\cup {(x,sin 1/x): 0<x<=1} is not path connected.

yes, the notation is wrong. just write what you want to write as a set. R = {values of polynomials with coefficients in Z/nZ at 1/a}. this is sensible iff a and n are coprime (so 6 and 2 is actually nonsensical, not even notation-wise). and in that case, R is indeed a ring and what that ring is is just Z/nZ

but we can make it path connected by attaching an arc to the origin at one end and joining its other end to {(x,sin 1/x): 0<x<=1}.

true

this will allow joining of the origin to the sine part by a path.

5:40 AM
oh, nice, it makes more sense to speak of localization

yeah, you can also note that this space is not locally connected.

@Koro if a connectng path existed it would have to be discontinuous therefore it could not exist
i'm pink therefore i'm spam

@copper.hat or if a path started at the origin 0, it has to be constant loop.

indeed. but even if you add $\{0\} \times [-1,1]$ it is still not path connected.

@shintuku Oops, you were talking about something else. 😅😅

5:43 AM
yeah eheh

@copper.hat yes, in this case any path starting at the origin will have its image contained in {0}\times [-1,1].
:)

@Koro this apparatus is known as the "closed topologists' sine curve"
heres an exercise (slightly surprising): the closed topologists' sine curve disconnects R^2

oh, I thought it was Warsaw circle.

yah i forgot thats also a name for it
so my exercise is R^2 \ W is disconnected, where W = Warsaw circle
(do you see why this is surprising?)

the Warsaw circle is path connected

5:47 AM
@BalarkaSen thinking
@copper.hat but not locally pc.
am I right?

yes

looks like a duck on its back eating crackers

vivid

@TedShifrin Thanks for your answer. I agree that I haven't talked about geodesic curvature there. But wasn't this the normal curvature I described there?

5:50 AM
@BalarkaSen I see why it is surprising (to me): there seems to be a 'gap' between y-axis and the 'sine' part.

yeah

So it is expected that R^2\W should be connected.

naively, yeah

no gap, the picture is bad

its like, R^2 \ (unit circle \ a point) is connected
so why shouldnt this be

5:53 AM
@BalarkaSen notice: 2 is not a zero divisor in the denominator since the only possible denominators are powers of 2

as copper said, this is an instance of picture being misleading
@shintuku bro. 1/2 = 3/2*3 = 3/0

but you don't have multiplication by 3 in the denominator!
so that won't happen
1/3 is not a member of the set

gn folks!

what does a fraction mean to you, pray tell

can I say that $W$ is homeom. to circle?

5:54 AM
its not, Koro. circle is locally connected

@copper.hat gn :)
@BalarkaSen oh right

at balarka: i'm pretty sure we're looking at localization of Z/6Z at (2), no?
so there's no zero division going on
no 3 accessible in the denominator

you're throwing words out with a near nil understanding of them. localization of a ring R at an element x doesn't say the only denominators are x^n. you allow fractions of the form a * b/(x^n * b), because rules of arithmetic has to be consistent for it to be a well-defined ring. which is exactly you cannot localize at zero divisors
well, you can, you just get the zero ring.
the definition of localization for zero divisors is slightly different to adjust for this

sure, but 1/3 is never in the set!
at no point do we add 1/3 to the set

i didn't say it is. if you allow (a * b)/(a * c) = b/c, you need to have (3 * 1)/(3 * 2) = 1/2

6:00 AM
we only allow the first operation if 1/c and 1/a are in the set
since we only have powers of 1/2, we won't get 1/3

which means you're changing what a fraction means
its not just polynomials in 1/2, which is indeed not meaningful
you have to tell me when a/b = c/d in your ring

it's not given in advance, that's why it's nice to be able to tell there exists an isomorphism with Z/3Z

if youre not given when a/b = c/d its not defined

all you know is that your set has z/2^n, with z in Z/6Z
if you suppose it is a ring, we're forced to make a few adjustments that make it well-defined, but these adjustments are not obvious at first glance

i dont know what that means buddy
if youre setting up an isomorphism of some object with Z/3Z you better know what that ring is. or even the set.

6:06 AM
yeah we know: draw a grid, the top row has 1 through 5 (coefficients in Z/6Z), and under each write multiplication by 1/2, 1/4, 1/8...

a set isnt a grid

these are the elements in your set (the grid is infinite, you write the first four columns)
now you wonder if there's a ring structure that will make arithmetic consistent
we start discarding all elements we discover to be superfluous
for example, we don't care about the columns 2 and 4, because division by 2 is already in another column
so those elements are all equivalent to something else
now you want modular arithmetic to be consistent, so you have 3*2, and since you also want 2*1/2 = 1, you must have 3 = 0

this is too far into the realm of speculation to be mathematics, really

it's perfectly legitimate number theory mate

youre not really the arbiter of what is legitimate, shin. but if you want to learn this, i recommend reading a book
localization, for zero divisors, are really pathological and fractions arent what they mean usually.
FWIW, localization of Z/6Z at (2) is indeed Z/3Z, but that really doesn't mean dividing by zero is legitimate, or what you think it means

6:11 AM
i don't think it takes any sort of arbiter to determine that testing whether the set z/2^n with z in Z/6Z s.t. arithmetic is consistent defines a ring
but uh, keep on rocking with your odd standards

the set is not {z/2^n : z in Z/6Z}
its that modded out by an equivalence relation
the premise is wrong here

yeah you're right about the set definition, but anyhow it's clearly a mathematical problem
especially in the grid form, where it's not evident at all you're gonna get a ring if you force arithmetic to be consistent

what is a mathematical problem? so far i do not see a problem.
i do not know what Z/6Z[1/2 + 6Z] means, nor what Z/6Z[1/2] means. (Z/6Z)_(2) is meaningful, but it is not what you say it is.

are you actually interested in the problem?

i would be if you could state it

6:17 AM
yesterday, by leslie townes
it certainly can get more confusing if there's no ring playing the role of Q here, i.e., there's no known ring that all of our constructions are playing out inside of

but alternatively, why not try and pick it up from a ring theory book?

i saw this all in my crystal ball

alright: we draw the grid, the top row has numbers 1 through 5, the subsequent rows divide by powers of two. hypothetically the grid goes on to infinity. now: suppose multiplication and addition are defined as usual but all results must be modded by 6, and from division the only thing we have is (1/n)n = 1
how many elements can we eliminate from the grid, because we discover they're equivalent to other elements of the grid?

i dont know what a "top column" is. do you mean "top row"?

6:19 AM
what is a fractional number mod 6?
i dont know what that means, first of all

we don't mod fractions that don't produce integers
but we can mod the numerator

i dont follow. a/b = c/d if a = c mod 6 or (a/b - c/d) is an integer divisible by 6?

I'm pretty sure you studied hyperbolic geometry once. What book or note did you read @BalarkaSen?

@onepotatotwopotato I don't know too much, I picked it up from people around me who do hyperbolic geometry
I don't really have a suggestion, sorry.

ah, ok

6:23 AM
at balarka: we're forced to have a/c = b/c iff a = b mod 6 and a/c, b/c aren't integers

@shintuku so 1/2 = 7/4 mod 6??

the issue is that we begin with n(1/n) = 1, and we're not sure what multiplication by (1/n) does, so we're not sure what fractions are
at balarka: it will turn out that the only way to make arithmetic consistent while keeping $1$ behaving as a multiplicative unity is to have 1 = 1/2

no hold on. you said a/c = b/c iff a = b mod 6. when is a/c = b/d mod 6?
give me a general definition
forget arithmetic, define what equality mod 6 means for now

we don't know from the start, that's the mathematical problem. can you come up with a ring structure knowing only n(1/n) = 1, and the only certitude being that you can mod n and not 1/n?

dude forget ring structure, define equality mod 6
i dont know what two fractions being equal mod 6 means
you realize not defining this means you do not even have a set, right?

6:29 AM
we don't know what fractions being equal mod 6 means, unless the fractions divide into an integer, like 4/2 = 2

what does "fractions divide into an integer" mean?
ok so you do not even have a set

4/2, 8/2, 6/2

and you want to put a ring structure on your nonexistent set?
im confused

we do have a set: the elements of the grid

but you want to pass to elements of the grid mod 6, and then the next problem is "define arithmetic"
i want to understand the first of these two steps

6:31 AM
right: for a/b = a(1/b) in the grid, all we know is we can simplify it by (a mod 6)*(1/b)

and im telling you, that's not a definition of what equality mod 6 means

we don't know what modding a fraction means, we only know what modding an integer means

so you do not know even the words that form the premise of your problem. how do you hope to solve the problem?
its like saying "let us consider the Rumpelstiltskin of the grid, and consider the ring structure of these Rumpelstiltskins"
this is not a math problem, its an undefined mess

that's not what I meant, i meant that the problem begins with the fact we don't know what modding fractions means, but we know we have a/b = (a mod 6)/b

Here's a definition. a/b = c/d iff 2*(ad - bc) = 0 (mod 6).
This is not how fractions work, but this is the only way to do what you want, if you even know what you want.

6:41 AM
i think we might have enough with only n(1/n) = 1

i dont know what that means

for instance, from $\frac{1}{2}2 = 1$ and $3 \cdot 2 = 0$, we have $3 \cdot 2 \cdot \frac{1}{2} = 0$, so $3$ must be zero
from which we eliminate the entire third column of 3 as an equivalent of zero

why can't i just define 1 = 0 and call it a day? all columns become 0, and its the zero ring
what's stopping me?

well you're trying to figure out if there's a nontrivial ring that's formed out of this minimal ruleset
i haven't had time to think of the truly minimal ruleset, but the problem consists of forming a nontrivial ring with a lacking ruleset

thats a new shift of goalpost
but ok
your "rule sets" are ill defined, but i can see one stumbling upon (Z/6Z)_(2) in this fashion, yes

6:49 AM
my bad i've just encountered it

this is nowhere close to a ring extension of Z/6Z though. Z/6Z[1/2] or Z/6Z[1/2+6Z] were both equally meaningless

right

thats not what you said the last time we were saying that!

well i thought i was saying something and the notation implied polynomial ring evaluation, but it's not really what it ended up being, my bad

maybe you should be less confrontational before you're wrong than being confrontational and apologizing afterwards for being wrong.
people tend to dislike you otherwise. math is, after all, a social activity - not a personal ivory tower of self-righteousness where you shower your disdain towards others from
2

6:57 AM