groups-rings-fields

[barren sierra]

given a set of 2 elements, how do I find the number of closed associative binary operations on the set

I can't think of a way other than exhaustion of all cases

16 cases.

Which isn't that much but still sucks

I was able to figure out how many operations are commutative and how many operations have identity

but I am stuck on associativity

[next obsidian]

Idk lol

[south storm]

https://math.stackexchange.com/questions/105438/how-many-associative-binary-operations-there-are-on-a-finite-set

[next obsidian]

I tried doing some work based on what x^2 and y^2 are

[barren sierra]

bruh so I gotta go through the cases?

smh that blows

I mean this isn't due till next wednesday so I guess I'll just go to OH by skipping a class

and ask him what the intended solution is

[next obsidian]

I mean you can try to classify it in terms of functions

Idk how that would be easier but ¯_(ツ)_/¯

[urban acorn]

given the monoid of endomorphisms of some group, can we tell whether it's abelian?

[urban acorn]

I'm currently trying to see whether the following is a counterexample: Q with + vs. the semi-direct product of Q by Z where the 1 of Z multiplies the elements of Q by 2

the former is abelian, the latter is non-abelian, and I'm not sure whether their endomorphism monoids are different. the former has Q (including 0) with multiplication as the endomorphism monoid

I'm not sure about the latter...

I think there's an embedding Endo(Q) -> Endo(Q semiproduct Z), but not sure if it's surjective or not

[lethal cipher]

Alright, I tried to prove that the residue classes of integer modulo n is associative under multiplication.

This is my first time working with residue classes, so I want y'all to read my proof, rip me a new one, and then help me figure out how to improve it.

Most of my wording is probably off, so I could certainly use some help refining it.

I think I realized near the end that since equivalence classes are sets, we can just show that each set includes the other to show they are associative.

Here is a comparison between my first and second attempt. I feel a lot more confident in attempt 2.

[next obsidian]

This kind of thing is where I feel like just developing the idea of quotients for rings and showing that this lines up with Z/(n) is the cleanest way to do it lmao

[lethal cipher]

Well, we are in the preliminaries of abstract Algebra. We just started the class. Although you may be right, I can't say it is within my toolbox at this point.

[oblique river]

have you proven that multiplication of residue classes is well-defined?

like, (a+nZ)(b+nZ) = (ab + nZ) doesnt depend on the choice of a and b?

[next obsidian]

Sorry that wasn’t a suggestion on how to do it, more like this is annoying and it ends up being easier to do once you know more

[oblique river]

(for the record I think your second proof is good, but there's no reason to limit a, b, and c to be less than n.)

[next obsidian]

It’s like proving associativity for the multiplication on elliptic curves. You can do it with elementary methods but there’s a higher tech way to do it that’s so much cleaner I feel like it’s better to just take on faith it’s associative then once you know enough to do the nice proof just do it that way

[chilly ocean]

When I'm asked to prove in intro algebra that a binary operation is associative, I like to just pretend to do it

[next obsidian]

Lmfao

[oblique river]

I just wanna say that this is so much better than what ive seen before

when i TAed abstract algebra, there was a student who had to prove that addition in Z/4Z was associative

[next obsidian]

Oh no

Pls no

Stop

Don’t do this to me Buncho

[oblique river]

so he wrote out all 64 triples of the form [(a + 4Z) + (b + 4Z)] + (c + 4Z) = (a + 4Z) + [(b + 4Z) + (c + 4Z)]

[next obsidian]

:(

[oblique river]

no other commentary

he just wrote them all out, one on each line

[next obsidian]

Was this on paper?

[oblique river]

yeah

took over two pages

[next obsidian]

Jesus Christ

You didn’t read that shit did you?

[oblique river]

no

[next obsidian]

Just say “okay”

Assume it’s right

[oblique river]

i saw what he was doing and how many pages it was and just wrote a little note

I think I gave him like 6/10 or 7/10

maybe 7/10

[next obsidian]

Was this the entire problem?

[oblique river]

oh hmm good point, maybe the problem was to prove Z/4Z was a group under addition

[next obsidian]

I feel like I’d give 10/10 because I mean… idk it is a perfectly rigorous proof of the stated fact

[oblique river]

so he probably got like 8/10 or something

[terse crystal]

No. Choosing an element of End(Q) is equivalent to choosing an element a of Q such that the image of 1 is a . Choosing an element of End(semi direct product of Q and Z) is equivalent to choosing r,s from Q such that the image of (1,0) and (0,1) are (r,0) and (s,1) respectively.

[next obsidian]

Just a god-awful proof

[oblique river]

i mean, the point of math homework isn't to produce technically correct statements

it's to practice and apply techniques youve learned in class

[next obsidian]

Idk. I think if you get the job done using stuff up to what you’ve learned you did it

But maybe that says more about me than anything else

[oblique river]

another example from later in the class was during galois theory

they were asked to find all subfields of Q(sqrt(2), sqrt(3))

and the student did it by hand like

"let me calculate what Q(a + bsqrt(2) + csqrt(3) + dsqrt(6)) is for a, b, c, d in Q"

using really bad linear algebra

and he lost points

[next obsidian]

I guess for me I just feel like if you don’t have a clever idea and you just go to town and do what you can to make it work then you did it

If you specified to do it using some specific method then sure

[oblique river]

if youre supposed to be practicing using the fundamental theorem of galois theory

and you dont do that

then you havent met the requirements

[next obsidian]

I mean yeah that’s bad

Idk

¯_(ツ)_/¯

I haven’t graded a class yet so

[oblique river]

if this is an exam i would be more lenient

[next obsidian]

Who knows

I guess the thing for me is that on an exam you don’t have time for that

So I’d write down like “okay you did it, but you want to be doing ___ instead”

Because you won’t be able to do this for everything

But I guess I’ve never really thought much about what purpose HW serves lol

For me it’s just been problems to do and prove lmao

[oblique river]

i mean, you can tell me if this isn't accurate, but i would imagine that most students (including you) would find it odd/confusing if you had a homework assignment in a group theory class that asked you to prove the heine borel theorem and youd probably complain about it

"why is my algebra class making me prove analysis theorems"

[next obsidian]

Yeah that’s true

[oblique river]

so clearly homework should have some relevance to the class and isn't just a list of random true statements

[next obsidian]

Yeah that’s true

Like

I think the argument for scoring less than 100% for a technically correct solution that missed the point

[oblique river]

i'm going farther than just saying the statements of the problems should be relevant to the class, and im saying the proofs should be too

[next obsidian]

Is that you want to make them use the things they learnt to like get practice with them

But I don’t see how scoring less encourages them more than say, giving 10 and writing a note down saying to avoid excessive force and to try to use relevant theorems and stuff like that

And then maybe down the line if they keep doing it to like tske points off

If nothing else can stop them

I just as a student feel slighted if I get marked down on a perfectly valid solution

[oblique river]

people dont read comments if they get a 10/10

[next obsidian]

Like I had to prove the theorem about flatness being able to be checked on ideals

Oh I do

xD

I like seeing comments on what I wrote

[oblique river]

this isn't about the exact number but more just the existence of a number

[next obsidian]

And I had a long complicated proof because we couldn’t use Tor and i tried my hardest and this was super hard and I finally got a proof

[oblique river]

but in studies about grading work it's been found that whenever a piece of work has a numerical grade on it, students overwhelmingly dont read comments, or if they do, they dont internalize them

[next obsidian]

And I got 7/10 and the TA said “it would take a lot of time for me to verify all this”

And I was like bruh

[next obsidian]

Huh

I guess I can only speak for myself

[oblique river]

i mean, yeah, like I kinda agree with the TA here? i would at least skim it and see if it looked reasonable

but I would take off at least a point

if you gave some nasty proof which missed the point

[next obsidian]

I mean I literally don’t know how you could make a good proof without Tor

It wasn’t nasty it just is a hard problem lol

I don’t know I haven’t had to grade a bunch of proofs in a week as a TA

So maybe I’d be more inclined to agree after I have

But I was annoyed because I spent like 15 hours making that proof and checking all the details

And it was like “sorry too long”

[oblique river]

i mean i would try to leave something more detailed than that

honestly like I think the most optimal thing would be to just return it with a hint for the shorter proof and say "try it again"

[lethal cipher]

To be fair, excessive force is a terrible approach to proving something...

[next obsidian]

I don’t think there is a shorter proof though

Or at least anytbing significantly shorter

[oblique river]

did the TA give everyone a 7/10 with that same comment?

if so then you have a point and the TA just sounds bad

[next obsidian]

Idk

I think some people literally just used Tor

Even though we didn’t cover it

[lethal cipher]

Ouch, that's rough

[next obsidian]

Like I’ve thought about this problem a lot and have tried to shorten it and I can’t come up with anything

I can reduce to fg easier now because of direct limits but I didn’t want to prove that commutes without tensor

[oblique river]

i mean, going back to your point from way earlier, sometimes it's nicer to use more advanced stuff to prove something simple. i dont mind if students do that as long as what they write is understandable to another student in the class and doesnt completely circumvent the problem

[next obsidian]

then you have to reduce to a principal ideal by doing a bunch of quotients

[oblique river]

i get that that doesnt fly in all classes

cuz different instructors have different feelings about it

[next obsidian]

Yeah I mean that original comment wasn’t about like doing it for hw

I just was expressing I think these things don’t really teach much like

Proving associatibity of modular arithmetic doesn’t feel like you learn anything lol

So I’m annoyed stuff like this always get assigned as hw

[oblique river]

you learn that proving things are associative is really shitty

[next obsidian]

When you can do it way cleaner

I guess lol

[oblique river]

but yeah @lethal cipher i think your proof is fine

like i said i dont think you need to restrict to a, b, c < n

i think it kinda obfuscates your point

and makes me (the reader) ask "wait why are they doing this? is this necessary?"

oh @next obsidian maybe this will make you unhappy too but starting like halfway through the semester

i started taking off 1 point whenever someone proved something by unnecessary contradiction

like their proof of A implies B was

[next obsidian]

Bruh I would be mad

[oblique river]

assume A is true and B is false
[insert direct proof of A implies B]
therefore B and not B, contradiction
therefore B is true

[next obsidian]

Wait

Wtf

Okay well that’s really stupid lmfao

[oblique river]

I guess there's also another version which is

[next obsidian]

Like sometimes I write a contradiction proof because it gets me started

Then I realize I have an idea that can work for a direct proof

But I’ve already written down a proof that works

So I just say whatever

[oblique river]

assume A is true and B is false
[insert direct proof of the contrapositive not B implies not A]
therefore A and not A, contradiction
therefore B is true

again I think HW is more than just "find a logically correct proof of this"

but part of it is learning how to write math

[next obsidian]

I mean sure but like

I have finite time

And I don’t want to spend it rewriting a proof that’s totally valid to a proof that’s the same thing but not contradiction

[oblique river]

i guess i think like, HW should be 80% logic and 20% composition

[next obsidian]

Like the above is kind of stupid

[oblique river]

chm doesnt use tex moment

[next obsidian]

I do

I literally Tex everything

I mean if it’s as bad as your examples

Then like I will rewrite it since it’s so easy to

But like sometimes I find that like doing a minimal counterexample works easier

Like assume that x1,…,xn are such that X, then show that x1 isn’t X

And the idea could be used to directly prove it

[oblique river]

i'm not saying that writing down counterexamples to statements is bad lol

i'm literally talking about the two forms which i described above

[next obsidian]

I mean yeah

That ones really dumb and I agree there

But sometimes people extend past that and say “this can be proven directly so proof is bad”

[oblique river]

but i dont really buy the "i dont have time" argument cuz like, that's literally how any other class which requires writing works. if you write an english paper and you have a sound thesis but you cant communicate it cuz your writing is shit and full of nonsense and unrelated facts youre not going to get an A

learning how to write math well is an important skill

and I personally think HW is the best place to reinforce that

and as long as the instructor is very clear about the standards ahead of time

I think there's really no excuse to being lazy and writing bad math

[next obsidian]

I really don’t like stylistic stuff being done as a grade, I think math is pretty much about getting a correct proof and having it readable.

I’ve read plenty of weird proofs with a random unnecessary assumption in textbooks I feel like holding the students to higher standards than their textbook is silly

This holds for even things where a TA says not to say a statement is obvious and you can point to your textbook doing it over and over

[oblique river]

i mean i think the abc conjecture is pretty good evidence that most mathematicians dont agree. even before SS published what they thought the error was, large parts of the community had basically already written off the proof, and mostly because mochizuki's writing is impenetrable and he went through no effort to try to explain it to people

[next obsidian]

I mean that’s at a point it wasn’t readable. I meant understandable by readable in that context

Or meant readable to mean understandable?

[oblique river]

i mean i guess this is just getting down to some kind of philosophy of math stuff but like, i dont think that mathematics is just about finding facts. there is a community aspect to math which i dont think you can ignore

and part of that is communication and learning to write and talk in a way which other mathematicians can understand clearly

[next obsidian]

Oh well I disagree haha, like there’s a community around it but to me doing math and the actual content is just proving stuff

[oblique river]

but proving something is useless if you cant (or dont) communicate it to others

[next obsidian]

Yeah that’s true, but I don’t think an unnecessary proof by contradiction inhibits that

[oblique river]

do you watch sports at all

[next obsidian]

Yeah

[oblique river]

basketball? football? baseball?

[next obsidian]

I’ve seen all of them

Most recently I watched the NBA finals

[oblique river]

so like, in basketball, you can get penalized for taking an action which could have inhibited another player, even if it didn't actually inhibit them

if i foul you when you shoot a basket, you get free throws even if you make it

[next obsidian]

Yeah

[oblique river]

I think there's room to correct bad behaviors even if no actual abuse occurred

[next obsidian]

Yeah but as a grade? Maybe the idea of grade is different in our head but to me a grade is something like insanely valuable

Like I stress about grades more than anything in my life

[oblique river]

i mean... that's not good...

[next obsidian]

So being penalized for that when I feel like you’re doing something that is harmless

Is incredibly infuriating

Like it’s just some random stylistic subjective choice the person decided on

[oblique river]

i mean, yes. losing points because you used the wrong width of lined paper is infuriating

or because you wrote your name in print instead of cursive

or because you printed on yellow paper

but i'm saying that i think that "doing math" is so much more than just "find a logically correct statement"

[next obsidian]

Like I got marked down for not putting an equation on a separate line and I can point out dozens and dozens of published things that have the same length equation in-line

Like “put equations on their own line” and it’s a string involving two equals signs

[oblique river]

whether or not my students like it, i am very clear about what my standards are, and i also dont take points off on the first offense, and i take time in class to talk about what good proofs look like and why it's important to write succinctly

[next obsidian]

Okay well

Here’s the thing okay

[oblique river]

i make it a part of the class, not just some extra thing which gets tacked on at the end like

[next obsidian]

I’m way more okay with that now

[oblique river]

"oh btw sry you forgot a period -5 pointslmao"

[next obsidian]

Like if you’re clear about it

And gave warnings

Like that seems fair to me

I can swallow that a lot more

My experiences have been like

“You will be graded on style”

In the second quarter (not in the first)

First hw I lost like 5 points because I didn’t put equations on different lines

And they didn’t like that I have long sentences

And like the first hw “wasn’t graded on style so you can understand what the TA wants”

And we turned in the second hw before the first was graded

[oblique river]

so your issue is with bad course management then lol

[next obsidian]

So I literally had no idea what the TA wanted and then got assblasted for it

I guess haha

[oblique river]

lost 5 points
got assblasted

[next obsidian]

It felt like it

I mean okay that’s definitely hyperbolic lol

But like I said my grade is what I’m worrying about the most

I mean I won’t give a duck in grad school

[oblique river]

well learning and growing is what i care about the most

[next obsidian]

But thinking about not getting into a school because some TA thought my sentences were too long and marked me down

Pisses me off

[oblique river]

sorry but like if you actually think that losing a few HW points is going to make or break your grad school admission you have been really really misguided

[next obsidian]

I mean it probably won’t

But I lived my life until uni getting a 4.0 so I could not worry about it

Like I worked my ass off to be perfect so I couldn’t have that be a liability

Like none of this is logical I guess, but this is what I’m dealing with

[oblique river]

maybe you have done this but like, if single points give you that much anxiety, i think you should talk to a therapist about it

[next obsidian]

I mean it’s not like I get anxious I just get really annoyed

If I feel like I fucked up

Then that’s on me and I should try harder

But if it feels like it wasn’t a good call / justified then it really really annoys me

[oblique river]

but saying "i dont value what my instructors value and therefore i get annoyed when they tell me i dont meet their standards" sounds more like a you-problem than a them-problem

[next obsidian]

I guess but to me it feels like your English teacher marked you down because you didn’t write on the topic they wanted

To me it seems completely irrelevant to what the course should be graded on

[oblique river]

like, what youre saying isnt this egregious, but ive had a student tell me before "why should my grade be penalized for not doing any homework"

[next obsidian]

Lmao

[oblique river]

like, i'm sorry that you dont value homework, but I do value it and I clearly communicated that

as long as the instructor is transparent, and as long as the work aligns with the course goals, and is assessed consistently, I dont see a problem

[next obsidian]

Yeah I mean I have to agree there

[oblique river]

and on the issue of style, I do think it's actually a relevant skill for being a mathematician. (many) journals care about this stuff and (many) referees will ask you to do revisions if your proofs are a mess

[next obsidian]

I just haven’t had it be outlined before that these are things you’re graded on

It’s been randomly just thrown in

[oblique river]

yeah i mean i think that's just bad course management

[next obsidian]

Besides one time I guess and I dropped the course for other reasons

All stemming from the TA’s grading but whatever

[oblique river]

like, for a similar but maybe more cut-and-dry example, randomly taking off points because students dont use tex is not fine (imo), but saying "becoming proficient in tex is an important skill for a mathematician, it's something we're going to work on and i'll give you guidance in, and by week 8 all homework must be submitted in tex" is fine (imo)

[next obsidian]

Yeah I agree

[oblique river]

(as long as that's appropriate for the course, if you did that in calc 1 that wouldnt be appropriate hahaha)

[next obsidian]

I man if you’re outlining it in the course syllabus and clear

You can say that your students need to do your hw upside down and I think that’s in your right but I’ll think it’s fucking ridiculous and drop the course and complain

Okay maybe not that bad

But like if you’re clear what the expectations are and they fall into a reasonable territory I’ll say it’s within your right

[oblique river]

hahaha i mean yes, course goals should be relevant and meaningful

it's not just about being transparent

[next obsidian]

Okay well

I guess the final thought on this I can say is

I’ve seen people been overly anal about this sort of thing and mark down small points on hw and you can argue there’s room for improvement so no 10/10

But if your course is graded so you need like constant 9 and 10s out of 10 to get a 4.0 this seems trash

If you’re lenient on how the course is finally graded it’s a lot easier to swallow that pill for me personally

I think I’m just coming from the wrong place on all of this at this point because I care too much about the grade at the end lol

[lethal cipher]

Yeah you're right. I was initially planning on doing something more cumbersome which would want that requirement, but this ended up a lot cleaner and that restriction is not a necessary inclusion.

[oblique river]

i think it's always good once you finish writing a proof to kinda go back and "clean it up" a bit

because sometimes this happens when you write a proof -- you'll start out thinking you're going to take one path but then end up taking another

if you leave some of the relics of the original approach in there, it can be confusing to the reader

I think subconsciously a lot of people reading proofs sort of assume that every step they read is going to be important, "because otherwise why would it be there" and so if you accidentally leave some extraneous step in, then the reader can be like "wait, is this necessary? what goes wrong if I take this step out? everything seems ok even if I remove this... wait am I going crazy?"

@lethal cipher

[novel parrot]

hi

so a is not in Pi for i <= n-1

so there exists an xi in a not in pi

what do they mean by when j != i

[gritty sparrow]

They are saying that by the induction hypothesis, they can pick xi in a such that xi is not in p_j for all j, except when j=i

[novel parrot]

i dont understand

a is not completely in Pi for i <= n-1

so some xi in a not in pi

[gritty sparrow]

Yes that is also true, but that is not what he is saying

Ok let me be a little concrete, take i=1 and n=3

So he is saying there is x in a such that x is not in p2 and not in p3

And you are saying that there is an x in a such that x is not in p1

[novel parrot]

also not in p1 then?

[gritty sparrow]

Different x’s

Just because there exists an element of a that is not in p1 and there exists some element that is not in p2 and p3 doesn’t mean that they are the same element

[novel parrot]

no

by induction though there is atleast 1 element of a not in any p

[gritty sparrow]

No

By induction there is an element of a not in p2 and not in p3

[dusty river]

apply induction hypothesis on this first step itself

a is therefore not in the union of any n-1 of the P_i

[novel parrot]

can we n=3 and prove for n=4

[dusty river]

so you can pick x_i which is in a but not in the union of everything except P_i whbich is what saketh is saying

[novel parrot]

a is not in p1,p2,p3 so not in union of p1p2p3

now a is not in p4

so a is not in union of p1p2p3p4

[dusty river]

yeah and by similar reasoning, a is not in the union of p1p3p4, of p2p3p4 and of p1p2p4

[dusty river]

That reasoning doesn't work, {1,2} is not in {1}, not in {2}, but is in the union

[novel parrot]

ok

[novel parrot]

i dont understand this

[dusty river]

is this clear?

[novel parrot]

No

[dusty river]

So your assumptions are:
you have some n prime ideals P_i
a is not in any P_i
whenever a is not in any n-1 subcollection, it is not in the union of that subcollection

The last line is the induction hypothesis and how you phrase it is important because notice that we are assuming this for every n-1 subcollection

not just P_1,...,P_n-1

[novel parrot]

subcollections of n-1 things huh

[dusty river]

Yeah

[novel parrot]

hmm

ok

[dusty river]

So that induction hypothesis together with a not being in any P_i tells you that a is not in any of the unions (except possibly the union of everything, which is what we have to prove now)

[dusty river]

this should be clear now

[novel parrot]

what about p1p2p3

that is also true by induction?

[dusty river]

Yes, you already mentioned it so I didn't

[novel parrot]

ok lol

wouldnt the sum be inside pi

[dusty river]

inside which P_i? all of them?

[novel parrot]

p1 atleast

[dusty river]

It is not in P_i because n-1 of its summands are in P_i and the last one isn't

the ith one isn't in P_i

[novel parrot]

if we are doing x1 + x1x2+ ...

all terms are in p1

[dusty river]

oh no

we aren't doing that

we're doing x1x2x3 + x1x2x4 + x1x3x4 + x2x3x4

in the case n=4

omitting each x_i exactly once

[dusty river]

Then you should be able to see that the ith summand here is not in P_i but is in the rest

[novel parrot]

i see

[dusty river]

So the whole sum can't be in P_i for any i

[novel parrot]

so we assume that a is not contained in any p_i for i <= n

by induction we know that a is not contained in any n-1 subcollection

[dusty river]

yes

the union of any n-1 subcollection

[novel parrot]

for the n=3 prove for n=4 case

a is not completely contained in P1,P2,P3,P4

are we calling x4 the element not in P1 cup P2 cup P3

[dusty river]

yep

[novel parrot]

and if x4 was not in P4 we would be done

or any index

[dusty river]

yeah, you can say the same for x1 x2 and x3

[novel parrot]

yeah

why do we require prime ideals then?

[dusty river]

to say that x1x2x3 is not in P4

because x1, x2, x3 aren't

[novel parrot]

alrighty

it makes sense now

thank you very much

[dusty river]

np

[novel parrot]

ive looked at a variation of this proof

and it said we dont need all primes

2 ideals can be non prime

so how would that work

[dusty river]

Ye you can do something like that, its a simple modification of this argument

[novel parrot]

ok

[dusty river]

but I will have to recall how it went

[novel parrot]

is just a modification of the sum

at the end

i think ..

[dusty river]

ye

[gritty sparrow]

No modification mecessary

*necessary

Oops, I got it wrong, modification is necessary

[novel parrot]

it feels weird using a lower case a for an ideal

[dusty river]

not so simple

can't recall

[trim grove]

while finding units of a eculidain domain does norm is nescessey, and if we change norm does it effect the units?

[golden pasture]

nop

you could have different euclidean functions

and it turns out that it is 1 precisely when you input a unit

<try proving it, isnt too hard!>

[trim grove]

ok i will try

[honest magnet]

Hi, i am going over some abstract algebra notions for my research project, and I came across this one: https://encyclopediaofmath.org/wiki/Transformation_group. Specifically, it is written that: "if M is a vector space over a skew-field, then groups preserving this structure are called linear groups (cf. Linear group)". Why does it specifically give as an example skew-fields? what about fields? are there transformation groups that preserve a field structure, which are not linear?

[novel parrot]

fields are skew fields

[trim grove]

i have to show that $F$ is isomorphic to $Z_{p}$ for some prime $p$, where $F$ is a field and there is a ring homomorphism form $Z$ onto $F$, I think here mod P will work because by isomorphism theorm we will have $\frac{Z}{<pZ>} \cong F$ , but the problem is iam not getting how to prove it is onto ( mod p). Please help

[cloud walrus]

Algebra

[novel parrot]

is that isomorphism given?

[trim grove]

no , homomorphism is given

[dusty river]

You can't prove that it is onto, because it need not be

[novel parrot]

u can show pZ is prime

[dusty river]

There is a homomorphism from Z to Q

but this is not onto

[trim grove]

thats great idea

[dusty river]

And I mean there are fields which are not isomorphic to some Zp

[trim grove]

is't it should be maximal?

[dusty river]

oh nvm the homomorphism is given to be onto, yes it should be maximal

[novel parrot]

oops

yeah i meant maximal

[trim grove]

wait but how can this help me to show Z/<pz> iso to F?

[dusty river]

first isomorphism theorem

[trim grove]

so map will be mod p, and image will be F. and kernal will be <pZ>?

[dusty river]

Yeah you just have to prove the last part that kernel is <pZ> for some prime p

[trim grove]

but one more thing how to show mod p is onto, (wapis wahi a gaya)

[dusty river]

forget mod p for now

You are given an f, onto, from Z to F

first isomorphism theorem says that F is isomorphic to Z/ker f

Compare this to what you have to prove and you see that it is enough to show that ker f is pZ for some prime p

[prime sparrow]

Excuse me for butting in, as I wait to ask my own question, but why even formulate a problem like that. They basically want you to show that the factorgroup by a ideal made by a prime is a field, which is just proving few properties. Either showing that you can "solve" in the field or you have no zero divisors with some other properties I cannot remember.

[trim grove]

got it

[dusty river]

example application of the first isomorphism theorem

[dusty river]

.

[prime sparrow]

I guess, I saw that exact statement you are trying to prove in the book I read now, sorry for being this abrupt.

I read in Russian, so I am not sure I even translated correctly 😢

[dusty river]

No problem

[prime sparrow]

can I do mine question?

[dusty river]

sure

[prime sparrow]

I am having a trouble understanding a bit more advanced problem. I am reading der Warden's Algebra to get to know modern abstract Algebra.

In a paragraph about determinants it is defined elegantly as a polylineal antisymmetric form of vector space over a given field o. And one of the intermediate equalities puzzles me, specifically that it always equals 1 for the vectors of the basis, i.e. $D(x_i) = 1$. However I am struggling to prove this without using the representation as list of elements of o, in which case proof is trivial as a sum $\sum \pm x^i y^j z^k\cdot \ldots$ has only a single non-zero term. Is there a way to prove this without going to this representation? I have troubles with this because it gives a circular dependency since you assume that determinant will be the same for in any transformed basis, but you use this property (det=1) to prove basis invariance. Thank you!

[cloud walrus]

Artej

[prime sparrow]

+I wonder if circular proofs A->B & B->A are that bad

[dusty river]

yes if you accept circular proofs then you can prove anything 0=1 → 0=1 → 0=1

[prime sparrow]

point taken

[dusty river]

By D(x_i) do you mean the determinant of the matrix which is x_i in the ith column?

[prime sparrow]

yes, $x_i$ are vectors, so really it is $D(\mathbf{x}_i)$

I guess no amsfonts 😢

[dusty river]

But then that is not true because if you take your basis to be (2,0) and (0,2) for R^2 then the determinant of the matrix with these is 4

[dusty river]

you didn't put $ signs around the x_i

[prime sparrow]

I did it around the whole expression

[dusty river]

the first x_i

[prime sparrow]

oh!

[cloud walrus]

Artej

[prime sparrow]

ic! works 🙂

well, in 'own' representation basis vectors $\mathbf{p}_k$ are just $\mathbf{e}_k={0,\ldots, 1, \ldots}$ at k'th position

[dusty river]

ah

[cloud walrus]

Artej

[dusty river]

So you're talking about the identity transformation in every case

because in any basis, the identity matrix represents the identity transformation

[prime sparrow]

so $D(\mathbf{p}_k)$ is 1 in that basis, and by theorem it is invariant and should be 1 everywhere else, here D is a form and not an operation on matrix

[cloud walrus]

Artej

[prime sparrow]

I wondered if you could prove it in arbitrary basis by the symmetries of D

[dusty river]

Yeah, its matrix with respect to any basis looks the same

The identity matrix doesn't change if you use other bases to represent it

Do you know any change of basis formula for matrices?

[prime sparrow]

oh, because $A^{-1}IA$=$I$

[dusty river]

Yep

[prime sparrow]

f

[dusty river]

I commutes with anything

[cloud walrus]

Artej

[prime sparrow]

Thanks!

now it is not a circle :3

[dusty river]

😌

[prime sparrow]

I should just English der Warden somewhere to just check over English terminology. I learned non-abstract algebra before, just my pet project to understand stuff better.

[prime sparrow]

basically now you use the derivation of matrix transform to prove invariance for special case of det over basis, which lets you extend it over any group of vectors. Neat!

They should have included it like that in book originally 😦

[dusty river]

Yeah it's really nice

[strong valve]

hello ❄️ E

[rustic crown]

[strong valve]

hey @rustic crown long time

[rustic crown]

uwu

[strong valve]

what are you doing these days?

[rustic crown]

grading 😦

and reading category theory

Moldi ❤️

[hidden haven]

[strong valve]

what is a group abelian?

asking doubts to keep discussion going on

did i do it right?

[hidden haven]

You're gonna be told to move to chill

[rustic crown]

you can fail enough so that they demote you to second sem?

[strong valve]

Some one should try

[rustic crown]

#groups-rings-fields is the best place to chill

[hidden haven]

I'm TAing alg1 and 2 review sessions for first years

[strong valve]

can you guys move to #chill?

[chilly ocean]

det

u move to chill

[rustic crown]

[chilly ocean]

det owns this channel

[strong valve]

I wish I had TAs as based as you

[rustic crown]

Moldi and Saketh took it from me 😛

[strong valve]

guess who owns det

[chilly ocean]

[rustic crown]

[strong valve]

isomorphism theorem

[rustic crown]

lol you're still trying

[strong valve]

keeping the discussion pertinent

i don't any alg beyond this

[hidden haven]

You took 4 semesters of algebra

[rustic crown]

bro we did galois theory together

[chilly ocean]

ok guys lissen I improved the notation

no one thought of it before

how do u denote trans degree?

[strong valve]

2 of them online 🌻

[rustic crown]

trans.deg()?

[chilly ocean]

yeah

but imo better

is

|E/F|

[strong valve]

you just T pose to show that you have transcended

[chilly ocean]

absolute value

so perfect

[rustic crown]

yea pretty nice

[strong valve]

are you grading python?

[chilly ocean]

@hidden haven dont thonk homie

[rustic crown]

[E:F]_t should also be good? like some people use it for separable degree

[rustic crown]

i should right now, but

[chilly ocean]

[] is for vector

[strong valve]

@hidden haven what are you doing nowadays? (except for reacting to msgs, ofc)

[chilly ocean]

t!catting

[hidden haven]

Teaching det cat thy

And learning it too 😌

[rustic crown]

:catLove:

why

[strong valve]

awesome

[hidden haven]

emote artist asked for it to be removed

[strong valve]

#chill

[rustic crown]

[next obsidian]

tr. deg_k(L) is what I’ve seen for the transcendence degree of L over k

If that was what ppl were talking about earlier

[chilly ocean]

ye

but |L/k| better

[next obsidian]

That looks like you’re talking about the degree of the extension to me

[chilly ocean]

wdym degree of the extension

[L:k]?

thats very distinguishable

[next obsidian]

Yeah I feel like that’s what you’d mean if you wrote |L/k|

[chilly ocean]

no

ofc if you dont define it then ye ure not gonna guess, if we popularized it then it would be better

[chilly ocean]

Could we just like use ordinals and combine normal degree and transcendence degree ?

Or something like that (I don't think ordinals work exactly)

We need a new set of numbers that describes how big an extension is

[chilly ocean]

wdym

trans degree kinda describes how big it is

[fossil shuttle]

you usually use cardinals for transcendence degree

i guess you could do it w ordinals idk. but a transcendence base should not really be ordered.

[chilly ocean]

Like how would you describe how big Q(pi, sqrt(2)) is over Q? Something like "omega times 2" ? (This is not right as an ordinal because order of pi and sqrt(2) don't really matter I think, but also not right as a cardinal since we then the degree would be the same as if we included just pi)

(I don't really know any transcendental field theory, maybe I am saying dumb things)

[gritty sparrow]

I feel like another argument for using cardinals rather than ordinals is the fact that if two purely transcendental / algebraically closed extensions of a field have the same cardinality of transcendence basis, then you can find an isomorphism. So having an ordinal structure will probably differentiate too much

[next obsidian]

I feel like you shouldn’t really be using ordinals in most situations

[chilly ocean]

Yes, then we should make a new number system

[fickle pewter]

hei, can I ask a question?? if I have 4 statement that have to be equivalent, how many statement should I prove?

so it looks like this.
The following statements is equivalent :
a
b
c
d

should I prove all of them? or just some of them?

[chilly ocean]

what do you mean "prove all of them"?

[fickle pewter]

a iff b, a iff c, a iff d. isn't it enough?

[next obsidian]

Yeah

[chilly ocean]

yes, that is enough

[next obsidian]

You can also prove a => b => c => d => a

[fickle pewter]

ohh I see

so I have to prove it like that. Okai. Thank you.

[zealous mortar]

Hi, can anyone tell me how to simplify a ring like $\mathbb{Z}[x]/(4x-1, x+1)$

[cloud walrus]

jm

[next obsidian]

you’re dictating that x = -1 by wuotienting by x + 1

And then 4x-1 turns into -5 so then you’re quotienting Z by (5) which turns into F_5

You can prove this directly now by taking a map Z[x] -> Z by evaluating a polynomial at -1, and then by quotienting by (5)

You should find that the kernel is (4x - 1, x + 1) I think

I think

[zealous mortar]

Right. That makes sense. So, for $(2x-1, 4x+3)$, then $2x = 1$ and then = $(4x-1) = (5)$

[cloud walrus]

jm

[next obsidian]

Don’t think so, 4x-1 turns into (3)

Err

It should become (1) no?

[zealous mortar]

oh.. wait.. (1)

agree with 1.

[next obsidian]

Yeah

[zealous mortar]

Perfect.. Thanks so much!

[next obsidian]

And this is easy to see like… with your hands

2(2x - 1) + 4x + 3 = 1

And that’s in the ideal

[proud bear]

are Z[x] and Z[x,y] free Z-modules?

[oblique river]

yes

[proud bear]

wait so are Z[x] and Z[x,y] isomorphic as Z-modules

[oblique river]

yep

[proud bear]

bruhhhh

[oblique river]

do you believe that there's a bijection between N and N x N?

[proud bear]

yeah

[oblique river]

because that fact is essentially exactly the same as this one

take your favorite bijection f: N --> N x N; write f(i) = (a_i, b_i). Now define a map Z[x] --> Z[x,y] by x^i --> x^(a_i)y^(b_i)

there's your isomorphism

[proud bear]

i see

[oblique river]

a weird somewhat related fact is that the power series ring Z[[x]] is not a free Z-module

[proud bear]

is that because you can't write every element of Z[[x]] as some finite Z-linear combination of elements or something?

[oblique river]

it means that there's not a Z-basis, yes

there isn't a set of elements s.t. every element of Z[[x]] can be written uniquely as a finite Z-linear combination of the elements of that set

[proud bear]

ahh ok. bery cool

[oblique river]

it's not too hard to see (imo) why {1, x, x^2, x^3, ...} is not a Z-basis for that ring

[proud bear]

yeah

[oblique river]

but it was surprising to me that in fact no Z-basis exists

[proud bear]

hmm

[strong valve]

hi

[urban acorn]

yeah, you might sort of expect that because we fail to create a basis in the obvious way for F[[x]] for exactly the same reason that we fail to do so for Z[[x]], but F[[x]] still does have a basis, just weirder.

Also it's kind of immoral to talk of R[x], R[x,y], R[[x]] or such when you only care about the R-module structure

these things inherently have more structure

[gritty sparrow]

Yo

[novel parrot]

$a \in R - N$ so $1-ay = (1-ay)^2 \rightarrow x(-3y+xy^2) = 0 \in N$ x is not in N so $y(-3 + xy^2)$ y is not in N so $-3 + xy^2 \in N \rightarrow x\in N$

[cloud walrus]

ActiveChapter

[novel parrot]

is this ok?

i can say that y was not in nilradical because y could be anything so choose something outside of nilradical

[urban acorn]

commutative algebra by atiyah-macdonald

based

[novel parrot]

indeed

[hidden haven]

I don't understand your argument at all

[urban acorn]

okay, suppose the jacobson radical was not equal to the nilradical, then it has to properly contain it. therefore, by hypothesis, we have idempotent non-zero x in the jacobson radical. So 1-x is a unit. But then x(1-x) = x - x^2 = 0, so 1-x is a zero divisor, a contradiction.

[urban acorn]

I couldn't follow it either with all the algebraic expressions with numbers

[novel parrot]

-3 = -1-1-1

so we can say not in nilradical

and y isnt

so x must be

was my thought

[urban acorn]

-3 = 1 + 1 + 1 in characteristic 2 or 3 😎

[novel parrot]

char 0

[hidden haven]

-3 will be in the nilradical in Z/9Z

[novel parrot]

i was kinda assuming char0

[urban acorn]

char0 is based

[novel parrot]

this is better though

[next obsidian]

THAT CHAR 0 IS DEF A CHMONKEY MOMENT

[novel parrot]

😆

[urban acorn]

write perfect algebra textbook
write "characteristic infinity" instead of 0
no one will read it

[novel parrot]

which book is that

characteristic infinity

[next obsidian]

Doesn’t exist yet

Maybe you’ll write it

[novel parrot]

lol

[next obsidian]

You know

It wasn’t until I was taking grad algebra I found out why we call it char 0 lmao

It’s because of the map Z -> R

[urban acorn]

yep

[next obsidian]

Which just happens to agree with the order of 1 for all but char 0

Lmfao

I mean “just happens to”

Isn’t quite fair

It does by like construction

[chilly ocean]

i don't follow

[next obsidian]

Wdym?

[chilly ocean]

the map agrees with the order of 1 (order of 1 is the characteristc?), but what does this mean?

[urban acorn]

the logic for characteristic 0? it's the (non-negative) generator of the kernel of the map Z -> R given by sending 1 in Z to the identity of R

[next obsidian]

The order of 1 as an abelian group

For char n

Is n

Also I mean that characteristic agrees with the order of 1

When the characteristic isn’t 0

But to understand why we call it char 0 and not char infinity is because we actually define it as the generator of the kernel of Z -> R

[chilly ocean]

oh

[next obsidian]

Not as the order of 1 as an abelian group

[chilly ocean]

i'm still not sure i like the name characteristic 0, despite that there may be a rationale behind it

[next obsidian]

Honestly

I prefer it at this point because 0 is shorter than infinity

Lmao

[urban acorn]

yeah lol

[next obsidian]

And it’s easy to tex unlike \infty

I need to go to sleep lmfao

[urban acorn]

and also "characteristic infinity" feels a bit unnecessarily esoteric

[next obsidian]

I think that’s because it’s not common

[urban acorn]

yeah

[next obsidian]

If we called it that then I think char 0 would feel esoteric

[chilly ocean]

sounds cooler imo than characteristic 0

[urban acorn]

that's how conventions work

also there's something less formal about it, 1 + 1 + ... never becomes 0 after finitely many sums so we say it takes "infinity" for it to become 0?

[next obsidian]

Well here’s a property that we have

For there to be a map R -> S it’s a requirement that char S divides char R

Actually

Nvm

If you say everything divides infinity

It still holds

[urban acorn]

but you have to make this non-obvious choice for it

[chilly ocean]

it feels sort of more analytic minded, like inf{n: 1 times n = 0}

[urban acorn]

with 0 you get it as a consequence of ordinary arithmetic

[next obsidian]

Yeah

[novel parrot]

if jacobson radical = nilradical then all primes are maximal right?

[next obsidian]

No

[novel parrot]

oh

😦

[next obsidian]

Any finite type integral domain k-algebra has this property

Almost more generally a ring like this is called a Jacobson ring

[novel parrot]

i see

[next obsidian]

But actually that requires that sqrt(I) is the intersection of all maximal ideals containing I

[urban acorn]

Z[x] is like that

[next obsidian]

So it’s like that property about nilradical = Jacobson but for all R/I

[urban acorn]

nilradical and jacobson radical are 0, but (x) is a non-maximal prime ideal

[next obsidian]

Jacobson rings are nice though, I think maps between them with some slight assumptions maybe added on top have the property that the inverse image of a maximal ideal is maximal

[urban acorn]

inverse image of maximal ideal is maximal
yo that's actually really awesome

[next obsidian]

If you’re curious the Stacks Project has a page on it

[urban acorn]

I've actually been interested in functorial max spectra

Consider a compact Hausdorff space X, then consider the ring of continuous maps X -> R with pointwise operations, I believe the maximal ideals correspond to points of X

which gives us a contravariant functor

[next obsidian]

https://stacks.math.columbia.edu/tag/00GB

[urban acorn]

so I've been interested in finding a functor going to the other way

[next obsidian]

this says if R is Jacobson and R -> S is finite type

Then S is Jacobson and closed points map to closed points on Spec

Which says maximal pulls back to maximal

[urban acorn]

it's "almost represented" by R, kind of like how Z "almost represents" the identity functor on Ab

[wooden ember]

shouldnt this be $Z_i(D_{2^n})=D_{2^n}^{n-i}$ where here $Z_i(D_{2^n})$ is the ith component of the upper central series, with $Z_0(D_{2^n})=1$ and $D_{2^n}^i$ is the ith component of the lower central series with $D_{2^n}^0=D_{2^n}$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[urban acorn]

@wooden emberwhat does ${D_n}^x$ denote? the x-fold direct product of $D_n$?

[cloud walrus]

All groups are abelian

[wooden ember]

as i said it is the ith component of the lower central series

defined inductively by $D_{2^n}^0=D_{2^n}, D_{2^n}^1=[D_{2^n},D_{2^n}], D_{2^n}^{i+1}=[D_{2^n},D_{2^n}^i]$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[urban acorn]

oh, okay

this isn't the kind of thing I typically think of, but I think you could know whether $D_{2^n}^{n-i}$ or $D_{2^n}^{n-1-i}$ are the right choice by looking at the nilpotency class

[cloud walrus]

All groups are abelian

[wooden ember]

yeah exactly my point

$D_{2^n}^{n-n}=Z_0(D_{2^n})$ since $D_{2^n}$ has nilpotency class of $n$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

and the general result follows easily from $\left|Z_{i+1}(D_{2^n}) : Z_{i}(D_{2^n}))\right|=\left|D_{2^n}^{i} : D_{2^n}^{i+1}\right|=2$

[urban acorn]

I think you're right, but I'm gonna wait for someone else to look at it

[wooden ember]

aight, thanks though

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

ohhh i see what i did wrong

the nilpotency class is n-1 because D_4 is abelian

[novel parrot]

whats going on here

after seeing det(matrix)(xi) = 0 for all i

$\phi$ is a root of $det(xI - (a_i_j))$ ?

[cloud walrus]

ActiveChapter
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hidden haven]

Yeah

[novel parrot]

but my next question is ..

[hidden haven]

Like where are you stuck exactly

[novel parrot]

what does that det expand to

[chilly ocean]

expands to eevee

sry continue

[hidden haven]

Some polynomial with coefficients in a, doesn't matter what polynomial

Monic polynomial

[novel parrot]

sorry i like pikachu better

[novel parrot]

ok

hmmmmmmmmmm

why did they say that det was the zero endomorphism

[hidden haven]

Because it's an endomorphism of M that equals 0

This is happening over the ring of endomorphisms of M

That's why you can take phi as an entry in the matrix

constants of A embed into this ring because they act by multiplication

[novel parrot]

ok mmmmmmmmmmmmmmmm

[hidden haven]

So when you expand out the det it is an element of this ring

ie an endomorphism of M

[novel parrot]

ok

i think it makes sense

what about this?

why is xM = 0

x = 1 = 0?

[hidden haven]

Putting phi = identity in the previous expression you get exactly the expression he writes for x

[novel parrot]

ye

[hidden haven]

And you know that that expression is 0

As an endomorphism

So applying that endomorphism (which is just multiplication by that element) makes everything 0

[novel parrot]

but multiplying by 1 also sends everything to 0

?

[hidden haven]

No

Multiplying by 1 is the identity endomorphism

[novel parrot]

ok

[hidden haven]

@novel parrot this is actually wrong, this won't work in that ring because it is non commutative (multiplication is endomorphism composition). This is in the subring of End(M) that contains all elements of A and phi, and that makes a commutative subring so you can work with determinants and adjoints

There might be a simpler way to resolve this actually, I didn't think about this properly earlier, maybe it's just something obvious that I'm missing

[vast quiver]

I think maybe the way to look at it is that we’re upgrading M from an R-mod to an R[t]-mod where t acts by the specified endomorphism phi. (this is same sorta thing that’s used to get jordan/rational canonical form from fundamental theorem of finitely generated modules over PIDs)

Gathmann’s notes on commutative algebra might have a tiny bit better explanations?
You can check out the module chapter here (check out the “cayley-hamilton” prop in particular for this) https://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c3.pdf

Here’s a link to the whole thing

https://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013.pdf

[hidden haven]

Thanks! I was doubting it because AM doesn't say anything about it

[vast quiver]

yeah no problem!

[novel parrot]

ok

ill look back at it

[urban acorn]

I like the fact that every non-zero element of a finite field solves a polynomial of the form x^n - 1

with this really nice one-line proof

let n be the order of the permutation x -> ax

[hidden haven]

order of that permutation - 1 I think

Otherwise take x^n - x and remove non zero as a requirement

[urban acorn]

no, pretty sure it's not -1

try an example, i.e. 2 in F_3

[hidden haven]

Wait what is a

The proof of this that I've seen is to take n = |F|-1 because Lagrange

Are you allowing n to depend on element chosen?

In that case you can also use pigeonhole and then take LCMs to get a uniform n but you won't get a bound

[urban acorn]

yes, this works as well, and is actually more elegant

[urban acorn]

are non-commutative rings really that interesting? hearing about stuff like the difference between right and left ideals is just yucky

but I guess in theory I see why one should care about them

they're the objects that act on abelian groups

which is really neat

[oblique river]

My unpopular opinion is that “all rings are commutative” and that “what people call non-commutative rings are really just algebras over commutative rings”, i.e., algebras dont have to be commutative

[thorn delta]

The only part of studying noncommutative rings I cared for were free modules over said rings. I thought the weirdness of no well defined rank was pretty cool

[next obsidian]

God

Cursed

[south patrol]

The best rings don't have a (multiplicative) identity either

[next obsidian]

[oblique river]

http://www-math.mit.edu/~poonen/papers/ring.pdf

that's why all rings have an identity

[south patrol]

😔

[uncut girder]

Poonen

Poonen embodies

He is pure energy

[barren sierra]

I don't get the hint

shouldn't it say that h * g = g => h = e?

why do we care about g * g specifically?

[kind temple]

this should be the right hint i think

[unreal portal]

yeah that looks like a typo to me

[wooden ember]

why may we assume $G''' = 1$? Ive figured out the rest under this assumption (after resolving the fact that $G/G''$ was probably a typo and should have been $G/C_G(G'')$) but I cant figure out why we may assume $G'''=1$. The structure of the question screams 3rd isomorphism theorem but I cant figure it out

[cloud walrus]

𝓛ittle ℕarwhal ✓

[hidden haven]

Is G > G' > G" > G''' with everything normal?

[wooden ember]

yeah since $G^{(i)}$ is characteristic in $G$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

also the DF errata says it should conclude $G''=G'''$ not $G''=1$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

and again ive managed to prove it for the case G'''=1 but i cant see how to reduce to that case the rest of the time

[hidden haven]

Yeah then hint makes sense

Quotient all numerators and denominators by G'''

Nothing changes because isomorphism theorem but now your new groups are G/G''' > G'/G''' > G''/G''' > G'''/G''' = 1

[wooden ember]

oh right

yeah that makes sense

thanks a bunch

[hidden haven]

[wooden ember]

and then G''/G'''=1 gives us G''=G'''

[hidden haven]

Yep

[wooden ember]

coolio

danke

[urban acorn]

so you'd just call a non-commutative ring a Z-algebra

TIL there are two interesting important groups which we don't know whether they are isomorphic

the absolute Galois group of Q, and the Grothendieck-Teichmuller group.

I don't understand the details, but a rough sketch is this:
The absolute Galois group is the group of field automorphisms of the algebraic closure of Q that fix Q itself.
The absolute Galois group is given in a certain way, and there are particular relations that we know must be satisfied, but it is unknown whether they generate all relations.
The Grothendieck-Teichmuller group is given in that way, but only with these relations. If these relations do indeed generate all relations for the absolute Galois group, then they must be isomorphic.

[hidden haven]

You don't need to say "that fix Q", all field automorphisms fix the prime subfield (subfield generated by 1) because they fix 1

In fact all field homomorphisms fix the prime subfield for the same reason

[urban acorn]

oh, true

[oblique river]

Yes, but they could have more structure than just a Z-algebra. For example, the matrix “ring” M_n(R) is an R-algebra

[urban acorn]

Sure.

It's like how abelian groups are Z-modules, but you could also have abelian groups that are R-modules for a more particular ring

[oblique river]

Yeah

[wooden ember]

getting nowhere with part a of this one, any help would be appreciated. I only showed $\overline{N_G(P)}\subset N_{\overline{G}}(\overline{P})$ since that's pretty easy but im really struggling with the other inclusion, and especially where frattini's argument comes in

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

from what im seeing online I need to work with PN and ive tried doing so but cant figure out what to do

okay i think ive got it

[chilly radish]

can you classify the subgroups of Z^n? It feels like there should be some that are not free abelian but I can't seem to come up with a simple counterexample

nvm turns out it's always free abelian lol

[urban acorn]

I can think of a proof for that on top off my head

you just need to show that they're all finitely generated, then this follows from the classification of finitely generated abelian groups and the fact that Z^n is torsion free

nevermind though, I lost my train of thought for showing that they are finitely generated

[chilly radish]

you show that they're finitely generated with SNF

it's part of the usual proof for the fundamental theorem of finitely generated abelian groups

[next obsidian]

Let $(A,m)$ be a regular local ring. Let $\nu$ be the valuation obtained by letting $\nu(a)$ be the maximal $n$ such that $a \in m^n$. If $f \in m$ is non-zero, how do I show that $\nu(f)$ is equal to the multiplicity of $A/(f)$?

[cloud walrus]

Chmonkey

[uncut girder]

what is multiplicity

@next obsidian

[next obsidian]

I got a reference for the result, but there’s two ways to define it

For a finite module M, if dim A = d then you can define it as $\lim_{n\to\infty} d!/n^d l(M/m^nM)$ where l is the length of the module

[cloud walrus]

Chmonkey

[next obsidian]

The other way is, if you know what the Samuel function is, it just is l(M/m^nM), for n >> 0 this is a polynomial which you can write as
e/d! n^d + [lower degree shit], then the multiplication of m on M is that number e

The other definition is basically just a formula you get from this definition here, the other is probably more useful but it’s helpful to know both

[untold cloud]

Hi, guys, why it says R is an integral domain here? 0=deg(1)=deg(fg)=deg(f)+deg(g), why R needs to be an integral domain?

[dusty river]

2x*3x = 0 in Z/6Z[x]

[untold cloud]

Thanks, and sorry maybe i didn't say it clearly, i mean how R is an integral domain implies 0 = deg(1) = deg f + deg g, i didn't see any relation here

[dusty river]

Since the leading terms can't cancel when you are over an integral domain, you get deg(fg) = deg(f) + deg(g)

The leading term of fg is just the product of leading terms of f and g

which gives you the equation

[untold cloud]

Oh, thanks

[lunar spruce]

If f is a group homomorphism f(0)=0. I know why. But what about f(1)=1. Why is this not possible

[sharp sonnet]

what's 1?

or 0 for that matter

[lunar spruce]

neutral element addition, neutral multiplication

like why do they say in fields f(1)=1 but not in groups

[sharp sonnet]

a group just has one neutral element

[hot lake]

because groups are only required to have one composition law

for which the neutral element may be called 0,1,e, or any other letter you fancy

[lunar spruce]

Ok. What if I have a ring?

There I can have two right

[hot lake]

if you have rings with unity, then f(1)=1 should be in the properties of ring morphisms

usually

[lunar spruce]

If I have a ringhomomorphism from a ring to a field, it doesnt havr to be injective compared field to ring. Is it because 1=0 is allowed in a ring? And in field its not.

[hot lake]

that field morphisms are injective is a theorem, and it has a proof using the field axioms.

and it uses some axioms or properties that rings don't have so you can't turn it into a proof that all ring morphisms are injective

also there are lots of examples of ring morphisms that are not injective

and I'm not sure how 1=0 being allowed or not has anything to do with any of that

[hidden haven]

And morphisms from rings to fields need not be injective if that's what you're asking

Because Z → Z/2Z is not injective from a ring to a field

[lunar spruce]

Yes. The proof I read used the fact that 1 is not equal 0 in fields: let f(z) =0, 1=f(1)=f(z^-1z)=f(z^-1)f(z)=0 so in fields the contradiction works.

But not in rings is that the intuition.

[hot lake]

most rings have 0 different from 1

[hidden haven]

You need 1≠0 because if you don't then the 0 field becomes a thing and then yeah the proof doesn't work

But 1≠0 isn't the only reason that field morphisms are injective

[hidden haven]

This example holds even though 1≠0 in both rings

[hot lake]

the difference between rings and fields isn't that 0 has to be different from 1

it's that in fields, all nonzero elements are invertible

[hidden haven]

Like multiple field axioms are being used to prove invertibility, and one of them is also the existence of inverses

[lunar spruce]

The proof fails for rings maybe because I can do f(z^-1)=f(z)^-1 which doesnt have to exist?

Or is 1<>0 in rings is enough to be injective

[hidden haven]

Both

[hot lake]

the proof fails for rings because z^-1 often doesn't exist

[lunar spruce]

ok. Even there it fails already.

[terse crystal]

If D is an integrally closed domain with field of fractions F,is any irreducible polynomial from D[x] still irreducible in F[x]?

[mild laurel]

I want to say that you can't. This is known as Gauss's Lemma for polynomials and is known to hold for UFDs. In fact, if Gauss's lemma holds, then your ring must be at least a GCD domain. Thus if you take D to be an integrally closed domain that is not a GCD domain, then this fails

@terse crystal

[next obsidian]

This is true if your polynomial is monic

See exercise 9.6 of Matsumura Commutative Ring Theory

[mild laurel]

[next obsidian]

moment indeed

[mild laurel]

Thats neat though

[next obsidian]

Basically you just assume it factors as f = gh in the field. Roots of g are roots of f, and thus integral over D, then you can express the coefficients of g as a product of those so the coefficients are integral over D. Do the same for h, now all the coefficients of both are in D

Technically there’s some “extend to a splitting field of g and h” snuck in there but w/e

You use that thing about like “if D is an integrally closed domain, for an algebraic extension of its FOF an element is integral iff its min poly has coefficients in D”

IIRC

[mild laurel]

yeah that makes sense

wait so where is monic used

oh I guess min polys are monic

[next obsidian]

Yeah

[terse crystal]

But those roots belong to an algebraic extension of F not F itself

The integral closure of D in this extension field is not necessarily D

[next obsidian]

There’s a theorem that gets you past that like I said

[next obsidian]

Namely this

[terse crystal]

Thanks. Got it.

[untold cloud]

Hi, guys, is this I ideal? I think it is not even a subgroup of Z cross Z star. The zero is (0,1). But (0,2) has no additive inverse.

[urban acorn]

Huh, yeah, this addition is really weird. It's not at all immediately obvious that this forms a ring.

where is this from?

[untold cloud]

it is from my lecture note...

[wooden ember]

am i correct in saying that all of these are trivial (Phi is the frattini subgroup)

[terse crystal]

You can induce a ring here. As a set it’s the quotient of ZZ# over the equivalence relation ~ where (a,b)~(c,d) iff there exists (e,f) from ZZ# and non-zero integers u,v such that (a,b)=(eu,fu) and (c,d)=(ev,fv)

It’s easy to prove that this is an equivalence relation and the addition and multiplication induced from it are well defined, independent with the choice of equivalence classes.

The zero in this ring is the equivalence class of (0,1) and the additive inverse of equivalence class of (a,b) is the equivalence class of (a,-b)

[terse crystal]

Yes all of them are {e} by considering F_i in S_n and intersection of F_i and A_n in A_n where F_i is the subgroup of S_n containing permutations that fix i.

[wooden ember]

Yeah thought so, thanks for checking 🙏

[lunar spruce]

I have two definitions of Ideals of rings: One is: I subset of R is additiv subgroup of R and RIR is subset of I. and the other one just requires second part ri is element of I.

are these definitions equivalent?

for me RIR element of I is kinda a stronger definition.

And the other one is just looking at one side...

[hidden haven]

Yeah the first one defines a 2 sided ideal, the second one only a left ideal. They are the same when you have a commutative ring

[chilly ocean]

yo if B1 is trans basis for E1/K and B2 for E2/K then why can we find set S \subset B1 \cup B2 such that S is a basis for E1E2/K?

[hidden haven]

The union of the bases should have a maximal algebraically independent subset, and that has to generate everything in the union, else the things that aren't generated could be added to it

[chilly ocean]

'should have' is some zorn argument?

[hidden haven]

Yes

[lunar spruce]

if R/M is a field, R a commutative ring and M an ideal of R, why 1 cannot be element of M? the book says 0!=1 in fields, but that doesnt help me.

[oblique river]

If M contains 1 then in the quotient you will have 1 = 0

[hollow imp]

If M has 1 in it, then M = R.

[lunar spruce]

where is the problem if M = R? What quotient?

we defined additive classes like R/I = {r+I|r element of R}

with addition (r+I)+(s+I)=r+s+I, and multiplication (r+I)(s+I)=rs+I ;_; I just dont see a problem with R/R being a field rn.

[hollow imp]

because M = R means that R/M has one element

which means 0 = 1

[lunar spruce]

oh true r+R would not make disjunct classes. So there is only one class?

I got it thxx

[potent briar]

Under what conditions on $\xi$ are the vectors $(\xi, 1, 0)$, $(1, \xi, 1)$, and $(0, 1, \xi)$ in $R^3$ linearly dependent?

[cloud walrus]

vik

[potent briar]

i get a system

$\alpha\xi+\beta=0$

$1+\beta\xi+\gamma=0$

$\beta+\gamma\xi=0$

[cloud walrus]

vik

[potent briar]

beta = -gama*xi

so alpha*xi - gamma*xi = 0

so (alpha - gamma) * xi = 0

if xi = 0

does the system have infinite solutions?

[next obsidian]

I’m pretty sure you can throw the three vectors into a matrix, then they’re linearly independent if that has non-zero determinant

The determinant should be a cubic in \xi

[potent briar]

i havent gotten to that part of the book yet

[next obsidian]

Well rip

[potent briar]

matrices and determinants dont exist in my brain

[chilly radish]

If xi=0 then two of the vectors are identical and so in particular linearly dependent

If $\xi \neq 0$ then $\alpha = \gamma$, subbing in you get
$$1-\alpha \xi^2 + \alpha=0$$
So you can solve this polynomial in terms of $\xi$

[cloud walrus]

ShiN

[chilly radish]

Or actually you wanna solve for alpha probably, and see when it's not 0 depending on xi (Cuz you only have 1 degree of freedom since beta is determined by alpha, so if alpha is 0 all the coefficients are 0 and therefore the vectors are independent)

[inner acorn]

Anyone got any favourite books on group theory topics like the nielsen schreier theorem (subgroups of free groups are free) or presentations of abelian groups (the smith normal form stuff)?

[novel parrot]

if overline(v) injects, why is v surjective

i dont understand the maps

[next obsidian]

Do you not get the definition?

It’s taking a map M -> N say? And then composing with v to get the composition M -> N -> N’’

So let’s assume that v-bar is injective, this means that different maps M -> N give different maps M -> N -> N’’

Oh rip I was describing what happens in (5)

For (4) you’re precomposing sorry so a map M’’ -> N gives you a map like M -v> M’’ -> N

[novel parrot]

ok

[next obsidian]

So really what’s happening here is this is saying that surjevtive <==> epimorphism in R-mod

[novel parrot]

and we could precompose becuz m to m'' surjective

[next obsidian]

But I really don’t think that’s gonna make you go aha!

So let’s just prove what’s stated there

[next obsidian]

No you aren’t assuming that. you can precompose because these are functions

So here’s how to go

Suppose v wasn’t surjevtive, then let L < M’’ be the image

Consider N = M’’/L and then the projection map pi:M’’-> N

Now note that v-bar(pi) = pi•v = 0, because if you do v(x) for x in M, then this lands in L, and then pi will kill it because you reduce mod L

But wait!!

V-bar is a map between modules because Hom-sets of R-modules are actually R-modules

And pi is a non-zero map, but v-bar sent it to 0

So v-bar is not injective

If you don’t want to appeal to linearity of v-bar, note that both the 0 map and pi get sent to 0 by v-bar

[novel parrot]

oh alright

i see