@PedroTamaroff i wish i could write that video on their papers

@PedroTamaroff you know that's a pretty good idea

@AlexanderGruber I want copyright.

@AlexanderGruber Was this one really found on a paper by a student? It looks like one of the greatest horrible (horror) mistakes I've ever seen. :-)))

@Chris'ssis it was found on several students papers

about 3 weeks ago

@AlexanderGruber Geezzz

i wish i could say it's the worst i've ever seen but i'm sure i could come up with more

@AlexanderGruber lol :-)))

i had a big problem earlier this semester with students distributing exponents across addition (i.e. $(a+b)^2=a^2+b^2$) to the point where i had to issue an outright ban on it, zero points on the entire problem even if everything else is right

@AlexanderGruber These errors are really famous! There should be a hall of fame containing all of them.

@AlexanderGruber These things happened in an university? Or is it a high school?

@AlexanderGruber WHOA MAN.

this week i had this on several quizzes $$\frac{1}{\cos(x)-1}=\frac{1}{\cos}-\frac{1}{1}=\sec(x)-1=\tan(x)$$

WHOA.

@Chris'ssis yes, these are college students

@AlexanderGruber I would give them one week off at Mordor.

it's a good university too, hard to get into. i don't know what to say sometimes, it baffles me.

It's funny how they recall that $\sec(x)-1=\tan(x)$ but make that stupid mistake before

@PedroTamaroff my plan for my next class is to give papers like that back to their owners and make them correct it

I know no trig identities besides the ones you can derive from $e^{it}$

@FernandoMartin even then it should be $\tan^2+1=\sec^2$

Haha ok I have no clue about trig identities

you'd be surprised what you can get from $\sin^2+\cos^2=1$. i haven't found any that can't be reduced to that yet (or some other pythagorean theorem).

@AlexanderGruber My favorite way to show $s^2+c^2=1$ is to use $\cos(x-y)=\cos x\cos y+\sin x\sin y$ and $\cos 0=1$.

the only trig ids I know are that one and the $\sin(a+b)$, $\cos(a+b)$ ones

i just draw the unit circle

those follow easily from $e^{i(a+b)}$

You complexifiers.

Shun you all.

@FernandoMartin you can get double and triple angle formulas from that too, by simplifying e.g. $e^{i2\theta}=e^{i\theta} e^{i\theta}$

@Pedro: How do you even define $\cos$ without complex analysis?

@FernandoMartin ¬¬

adjacent over hypotenuse

power series is cheating

@FernandoMartin Using integrals pal.

How?

can't you just define it as the solution to $y^{\prime\prime}+y=0$?

You use $\sqrt{1-t^2}$-

like defining sine as the inverse of the arcsine?

Aha.

That's absolutely contrived

How so?

Nah man.

It's geometric.

@AlexanderGruber Interesting. My professor would have cut my head off for those mistakes. :-)

@Chris'ssis a dark cloud rose over the classroom that day, believe me.

it hadn't been that bad since logs.

@AlexanderGruber hehe. Sometimes a dark cloud is need.

@FernandoMartin The area of a semicircular sector given by an angle of $\theta$ radians is $x=\cos \theta$, $$\frac{x\sqrt{1-x^2}}2+\int_x^1\sqrt{1-t^2}dt$$

Well, I guess it's a matter of taste/opinion @Pedro

So you use that.

It works for $|x|\leqslant 1$.

And then you use that.

@FernandoMartin Suppose $A\oplus A\simeq B\oplus B$, $A,B$ finite abelian.

Then $A\simeq B$.

Thinks counterexamples for infinite groups.

It works for f.g. abelian groups as well, I think

Yes, of course you can reduce that to the finite case because free part is easy. =D

Right

@FernandoMartin "Our favorite PIDs are $\Bbb Z$ and $k[X]$."

Rotman dixit.

$\Bbb Z$ and $k[X]$ are my only PIDs

I couldn't come up with an infinite counterexample

$\otimes_{\mathbb{Z}} \mathbb{Z}_2$

What does that mean?

like, a countably indexed product of $\mathbb{Z}_2$s

and what's your other group?

@AlexanderGruber Yes, me thought the same, but counterexample to what...?

@FernandoMartin $\mathbb{Z}_2$

@AlexanderGruber But it is $A\oplus A=B\oplus B$.

But that's not a counterexample

I think Alex wasn't reading what we were discussing.

oh, well use $\oplus \mathbb{Z}_2$ then it's the same

@Pedro: Tensoring with $\mathbb{Q}$ one gets that the "free part" of $A$ and $B$ coincide

@FernandoMartin Yas.

so I think that a counterexample involves torsion subgroups necessarily

Do you know if it's false?

@FernandoMartin No idea.

Sneaky Peter is sneaky.

i think it works as long as there is ACC and DCC

okay here is where you can find the answer: Corner, A. L. S. On endomorphism rings of primary abelian groups. Quart. J. Math. Oxford Ser. (2) 20 1969 277–296

@AlexanderGruber Ah.

Then you Krull the f*** out of them.

@PedroTamaroff yes

Cool.

without that you can find a group isomorphic to its triple product but not its double product

Ah, yes, Rotman says that. Let me find his reference.

and then you have $G\times \left(G\times G\times G\right) \cong \left(G\times G\right)\times\left(G\times G\right)$ but $G\not \cong G\times G$

@FernandoMartin You heardz of Krull's theorem?

Nope

(i was misreading your question before)

What does it say? @Pedro

@FernandoMartin It is actually Krull-Schmidt. It says that a group that if both Noetherian and Artinian can be decomposed into a product finitely many idecomposable groups.

And this product is unique up to permutations.

In fact you can glue the decompositions up.

So it is a bit stronger, I guess?

I mean.

Yup

If you have $G=G_1\times G_2\times\cdots\times G_s=H_1\times\cdots\times H_t$, all factors indecomposable, we can reindexing so that $G_i\simeq H_i$ we can write $G=G_1\times \cdots \times G_l\times H_{l+1}\times \cdots H_t$.

Still, if the group has infinite rank then it isn't noetherian

anyone know books that cover flows, lie derivatives, differential forms, lie groups, and lie algebras?

@Alex

there is the Peter Olver's

have you seen this?

No I haven't, I am currently using Geometry, Topology, and Physics by Nakahara

amazon.com/…

Thank you for link

I've seen [ just seen! ] Nakahara...I could not learn by using nakahara

but you can also see Arnold's on ODE...

amazon.com/…

Going to see if I can get copies of these from SpringerLink.

i'd like to study some of that

finally worked

=_= School's VPN keeps disconnecting me....

I came up with a new concept maybe: math.stackexchange.com/questions/761144/…

None of that notation is standard. @EnjoysMath It is incredibly unclear what you are trying to say.

@ThomasAndrews help me make it more clear please

PLEASE!!!

oh math god

What is the first stuck spot that you encounter?

@ThomasAndrews I keep asking for comments on related material and no one seems to get it. I've formalized and formalized and that just makes things worse. I don't know how to make it clearer if the community doesn't speak.

@Mike @AlexanderGruber I declare Rotman the supreme master of groups.

Untill I read Isaacs.

I can change my mind.

And now the downvote -_-

No, that's me.

@PedroTamaroff best be ready to ;)

you ever have one of those teachers where you're listening, understanding everything, and thinking "when is he going to get to the hard part?"

and then the lecture ends and you realized you just learned some really hard stuff without barely any effort

isaacs is one of two people i've ever gotten that with

it's like reverse mind reading or something.

You aren't owed answers, @EnjoysMath. Don't whine.

Okay, I should just ask questions that have already been asked

@ThomasAndrews what's not standard about my notation?

And that is why you get no help. Classy, @EnjoysMath.

Yeah, because we're all keeping tabs on each other. BS

@PedroTa

I help people on MSE all the time, even with poor posts. Elitists... all you

@PedroTamaroff... never heard about Isaacs, just rotman...

I've helped you three times today through two badly written questions. You might consider gratitude rather than swearing at me. @EnjoysMath

@EnjoysMath... what is going on?!

He's being persecuted! He's halfway on the way to being a triangulator, I suspect.

anyway @Enjoys, I can send you some tips on how to make your questions more well-received if you like. people can be pretentious and impatient here, but there are tricks you can use to make your questions more likely to be answered.

@EnjoysMath Dude, you need to be clearer.

Give your ideas some time.

Before posting them here.

@JoshPetrie "down votes".

That's a paddlin'.

@Pedro: Do you have your copy of AM near you?

@FernandoMartin Yes.

Do you understand what $M$ is on exercise 5 from chapter 2?

$M$ is any $A$-module?

OK, I'm stupid

Yes, it is - nevermind

@FernandoMartin HOLY BONKERS.

Your exercise 5 is my exercise 6.

I'm worried now.

It actually is exercise 6 here as well

DAMN YOU.

I don't know how to type

It says "For any $A$-module, let $M[X]$..."

Fernando goes so fast he gets confused on what number he's on.

It should say for any $A$-module $M$

but I'll let it pass

It's clear what it means bro.

One whamburguer for Fernando.

And a serving of french cries.

It has to be a whaaaamburger.

Whamburger sounds like you're hitting him.

Whineburguer then?

The point is to pollute the original word the least possible.

I suspect the downvoters (I am not one of them) are doing so because your question is a definition (a very unclear one, from my reading,) and then you ask, "Is this new? Is this useful?" Those kinds of questions are generally discouraged. We can define lots of things in mathematics, so if you have no motivation for defining that thing, then it's unclear why we'd care. It's more about, "Have I invented something new?" @EnjoysMath

<3 Julie Bowen.

I like whining

it's fun

Which was also your announcement when you came here, @EnjoysMath - "I came up with a new concept, maybe: ..."

I like sandwiches, @FernandoMartin

I'm particularly fond of Pastrami

Hi all

can u guys plz don't post

WAT

?

what

"can u guys plz don't post"

I don't even.

Downvote?. Is something wrong here?

Maybe they're afraid it's unjustified $-\frac 1{12}$ stuff.

It's not, but it looks like it at a glance.

Maybe they think you're smelly and are downvoting you out of spite.

I've been getting a lot of downvotes on old questions in the last few days. One question got downvoted twice, so it wasn't just one person, unless he was really determined...

@PedroTamaroff Somebody downvoted all the 1 hour or older answers to that question, except for one. I think we have a suspect.

@ThomasAndrews SIGH

@user3222184 maybe

I apologize for my disrespect and outburst

@EnjoysMath I accept your apologization.

And your outburstation is pardoned.

@Pedro: apology :)

@EnjoysMath already forgotten, my friend

hi @Mike @Pedro @AlexG

@EnjoysMath so anyhow, are you considering rings defined conventionally, where addition is commutative?

@TedShifrin Well I'm not Sirius.

You mean Siri?

I'm considering mostly polynomial expressions in commutative rings, since my goal is having to do with computation and our everyday numbers commute

evening @TedShifrin

@EnjoysMath so then any permutation of the monomials is the identity

Hi @Ted

My every night numbers are quaternions ...

Hi @Fernando

@AlexanderGruber I don't understand

I've decided what to buy for my birthday

Oh yeah, I have to find @Pedro a birfday present

*numbers used for computation on a modern computer

@EnjoysMath $abc + def + ghi = ghi + abc + def$ because addition is commutative in any ring

(unless i'm misunderstanding your question)

Oh I deleted it, but it's $s^+ (abc + def) = def + abc$ the operator on the left is part of a group acting on all related expressions. Where related expression means it can be gotten to by using your ring axioms.

@TedShifrin I think we might be able to meet after all.

One confusion for me was the question said $abc+bde + xyz$. The repeated $b$ seemed a problem, because you couldn't then explain the question as a permutation of variables. @EnjoysMath @AlexanderGruber

Cool, @Pedro. So, a real birfday present!

@TedShifrin YAS. Remember, I arrive July 28th.

Any suggested day on your side?

I'll take almost any day.

It's not a permutation of variables, as $xy - zw \not \equiv zw - xy$ as maps from $R^4 \to R$.

@EnjoysMath what i'm saying, though, is that $def + abc = abc + def$, so $s^+$ acts by the identity on any monomial expression ($s^+(abc+def)=abc+def$)

You arrive in NJ on 7/28?

@TedShifrin Correct.

Well, it wasn't just permutations of variables.

And stay up to 21st of August.

No, $def + abc \equiv abc + def$ as polynomial maps but not as expressions

@EnjoysMath what's the difference?

Does the string $aabb$ = $bbaa$ ??

not usually

if $a$ and $b$ are members of a commutative ring, yes

We start classes 8/21. Do you want to visit here on your way back to the southern hemisphere?

Two expressions can be equivalent on their polynomial maps, but if you try to compute using one over the other, one could lead to more computations, if by "computing using" you take to mean reading the expression directly and performing all seen operations.

@TedShifrin I can buy a plane ticket inside the US, right? From a state to another?

@EnjoysMath okay i see what you're trying to do

Sure. But it might be cheaper to do it en route. I dunno.

@AlexanderGruber $a + b$ as an expression, there's a map from expressions to their underlying map. I just use the same notation because that's what we're used to working with. What else would I use to represent a polynomial expression other than the expression itself?

@TedShifrin Right, I didn't do that. =/

Hadn't enough information.

Yeah, that was the word that was heavily non-standard. It still strikes me as deeply uninteresting - the actions that you listed never reduced computations. @EnjoysMath

@EnjoysMath well, you have to stipulate first where your expression lives

All math is expressions on paper any way, the other stuff is chaotic electrical clouds

i had assumed that you meant $abc+def$ as a member of $R[a,b,c,d,e,f]$, which is a commutative ring

Well, @Pedro, I gave you dates months ago! :)

@TedShifrin I am a bad planner!

Growl.

That's why I will never wed anyone.

Not "expressions" in this sense, no. @EnjoysMath. Again, no reason to be insulting.

$abc + def$ as an expression of a member of that ring

LOL, don't be silly.

No, it is not an expression of a member of the ring, it is just an expression.

what you mean is that you want to look at members of the coset of $abc+def$ in the free ring and see which have the smallest length

@ThomasAndrews the expression operators do reduce computations. As you can get to expression $(x + 1)(y+1)$ from $x^2 + y^2 + xy + 1$

@Pedro: Generally things are cheaper when one plans ahead. There is also Amtrak.

where length is going to be some function we have to define

@TedShifrin I CAN TAKE A TRAIN ALL THE WAY DOWN SOUTH?

YAY.

I was talking about the group actions you lsited.

I know Amtrak.

Yeah

I could explicitly list the operations to reduce that expression, if you want

Let me look at that.

ok ... It stops in Gainesville, GA, or ATL.

Well, $(1+x)(1+y)$ is not $1+xy+x^2+y^2$, except on $\mathbb Z_2$ :).

@TedShifrin Gainesville huh.

ah, you got me, I meant $xy + x + y + 1$ = 4 ops vs 3

@TedShifrin How long should the ride take?

not yours, @alexG

@EnjoysMath anyhow what you're asking is something i think has been studied

5 or 6 days:)

That's why I can't find any info on it?

it has a name let me see if i can remember it

That would be great

@TedShifrin What...?

Probably 12-15 hours ...

I mean comp. complexity has, and there's algebraic complexity theory, but I've never seen this specifically

@TedShifrin Your scared me.

I like scaring you :)

From NY more than 15 I bet

@TedShifrin Heh, OK.

It's one way, though, right?

This will hurt my pockits.

I dunno how much cheaper than airfare

@EnjoysMath okay i can't figure out what it's called for rings

but in group theory, it's called a word metric

I am not sure if plane gives me a two way ride for 300 or something.

It's smallest grammar for strings, which is somewhat related

Or only a one way ride.

I wouldn't hope so.

The set of "expressions" is just a set of binary trees with some labeling - leaves are either constants from $R$ or variables, and the other nodes are selected from the binary operations, $+$ and $\cdot$ here. It's still not clear what set the group action acts on.

The group acts on the set of all expressions obtainable from the ring axioms, from the expression.

(the relevant axioms)

e.g. distributive, associative, commutative

A lot of things can be written as a tree, but that doesn't necc. help study them.

Except you haven't actually defined that action. For example, if $g$ sends $1+x+y+xy$ to $1+x+y+yx$, what does $g$ send $(1+x)(1+y)$ to? You have to define the action on all the elements of your set.

for whatever purpose

You're acting on the last term or something?

I was using the closed term on the left

@Pedro : Plane roundtrip $323+tax

i'll check Amtrak

within the topmost scope, so (((a+b))c)) a+b means the leftmost additive term is (((a+b))c))

And there you seem to be misunderstanding what a group action is. @EnjoysMath

what's important isn't the formalism

I'm not making much use of the group yet besides the fact that you can undo operations

@TedShifrin Oh, that sounds delightful.

what he means is, take any expression. there are many expressions which it is equivalent to. which one requires the fewest operations?

Well, it was in the title of your question, so I thought it was important. @EnjoysMath

If I can make each of those operators well-defined and into a group, then we have a group action, simple. As long as (g(h(s)) = (gh)s.

Just because operations can be reversed doesn't mean that it is a group action.

Yes, that's true

Amtrak cheaper, but 16 hours each way. $170 or $270, not sure what the diff is.

I don't know what to say except I'm pretty confident that it's a group action

I think it has to do with the way I defined the operators using the underlying expression.

Okay, well, you haven't defined a group action on the set, nor have you defined a group, and I'm not sure you have given me any reason to trust your gut.

You certainly have an equivalence relation on expression trees.

@EnjoysMath

Anything I prove too formally automatically becomes uninteresting it seems

That's the risk. Did you really think you'd change the world?

I was young once. I tried to prove Fermat's Last. The question is, can you have the humility to recognize that a lot of very very very smart people have come before you and banged their heads on problems to no avail. It doesn't mean you shouldn't try, but in trying, keep some perspective.

One problem I see is that $(x_1 + 1)\cdots(x_k + 1) + x_1 x_2 x_3$ could be a smallest expression for that polynomial but unreachable unlesss I include the $a + (-a) = 0$ axiom.

dang

I certainly don't see anything that will result from this formalization which is gonna help you solve $P=NP$ or similar complexity questions.

hello @Ted

Well, when they invented group theory, people probably had the same doubts.

lel

nando, irony?

I'm a bit busy

I'll join in a while

Perhaps, but not everybody "invents group theory." That's the thing, you have delusions of grandeur well before you have actually accomplished anything. Galois invented groups and solved a problem with them. There was motivation. Here, you are talking about a formal manipulation that is trivially part of the question of evaluating polynomials, and tons of people have investigated those operations.

The formalization you are talking about is essentially trivial. It's certainly possible that something could pop out of it, but it seems highly unlikely. Galois was not formalizing something trivial when he invented groups - he found groups, and then formalized.

I have a tensor product question I'd like to run through someone. Anyone up for it?

i can try

it's boring

still up for it?

hmmm.

I have to prove that if $B$ is a flat $A$-algebra and $N$ is a flat $B$-module, then $N$ is flat as an $A$-module

@FernandoMartin nah

lel

I think that what I did is ok but it's kind of weird

Nevermind, Pedro will have to solve it sooner or later :)

What is a flat A-algebra? Does that just mean an A-algebra that is flat as an $A$-module?

yes @Thomas

It all works because $M\otimes_A N \simeq M\otimes_A (B\otimes_B N) \simeq (M\otimes_A B)\otimes_B N$

but one needs to be careful with what the isomorphisms actually are

@EnjoysMath anyway if you want an answer to your question, that's it. it has been studied, and that's where you can read about it.

@What, where?

@AlexanderGruber

@AlexanderGruber Ic, thanks

Actually I mainly want to compute the distance between a center of a circular set and the furthest edge of a bigger set that contains the first one

@Bel the first sentence is very very long

@AlexanderGruber ok I will edit that

@Bel yeah, and add motivation too

i've read through it like 3 or 4 times now and i still can't really tell what it's about

there's just too many symbols and definitions all crammed together, people aren't answering it because by the time they get halfway through reading it they decide it isn't worth the effort to sort out

put some story behind it - why is this particular problem interesting, where does it come from, which parts are which

ok I think I'm on chat now...

Is anyone else here from the meta.math.stackexchange.com/questions/13448/… reopening process meta post?

@NotNotLogical You're in the right place. ml should follow if he wants to continue the discussion.

@AlexanderGruber "Form follows function" :) We were just using edits because it was convenient (and actually, someone told me to in the comments)

@NotNotLogical it spams the front page of meta when posts are repeatedly edited, which is disruptive to other users.

if ml doesn't come, it means he doesn't want to continue the discussion, or that he has left.

@AlexanderGruber Well at least I've drawn attention to the issue. Could you perhaps unlock the post tomorrow in case anyone else wanted to answer? I will not use it for discussion, I will use the chat room now that I know it exists.

@NotNotLogical this lock only lasts for an hour

oh ok?

i mean oh ok.