groups-rings-fields

[slim kayak]

without 1

[final dragon]

oh yea

[chilly radish]

I think they mean modular arithmetic

[delicate orchid]

barely

[final dragon]

our professor always says when we do rings we have a ring containing elements of blue giraffes bro thinks he funny

[chilly ocean]

@formal ermine maybe ur teacher can help here

😭

[final dragon]

we did some quanternions or what are they called

i didnt understand nothing

[fossil hound]

[slim kayak]

I have a certain feeling they arent spending much time on modular arithmetic

[final dragon]

yea

we just skipped through that and got to rings

[chilly radish]

It's a weird pedagogical decision but I don't think vector spaces are per se easier than rings to introduce

Both are easy to motivate

[formal ermine]

real

[slim kayak]

Yeah idk this might be a professor taking a very liberal approach that vaguely fits into the criteria of the course

[slim kayak]

but non-commutative and non-unital rings?

[final dragon]

yea so

[chilly radish]

Commutativity is a prison of our own creation

[final dragon]

you guys what does it mean when there is a ring R and there is something like this R/I

[chilly ocean]

anything non-commutative and without one ie like putting pineapple on pizza

you can do it but you're crazy

[south patrol]

Quotient rings (quotients by ideals)

If you forget the multiplication everywhere then it is just the normal quotient group

Then you put a multipliction on it in the only way that makes sense

[chilly radish]

Non commutativity is everywhere in nature

[south patrol]

Fr

As soon as you try to consider endomorphisms for example

[chilly ocean]

pineapple as well

[final dragon]

yo can sombody help me with this problem i dont even know what im supposed to do here

[delicate orchid]

non-unital on the other hand.....

[final dragon]

I and J are ideal comutative ring R and f:R->R/I x R/J deffined as f(r)=(r+I,r+J),r from R proov that f is homomorfism of a ring wich its core is I intersection J

[formal ermine]

homomorfism

[slim kayak]

chase the diagram definitions

[final dragon]

yea we say it like that here

homomorphism

[slim kayak]

the translation is kernel

[final dragon]

yea kernel

right

ker

[slim kayak]

das me

[south patrol]

oh is this a german moment

no

[final dragon]

we say it kernel here also but i think english its core

[slim kayak]

Bild(f) is worse than Im(f) imo

[south patrol]

no it's kernel in English

[slim kayak]

Although only slightly, but aesthetically ker(f)>>>kern(f)

[final dragon]

oh

[south patrol]

das Image, right

🔥

or is it der

[slim kayak]

die image, natürlich

[final dragon]

is ker where all elements are paste into 0

[south patrol]

well i found this here

[slim kayak]

who decides this

[formal ermine]

yes this is correct

[south patrol]

https://www.duden.de/rechtschreibung/Image

[formal ermine]

r u a native german

[south patrol]

I just went off vibes for das tbf

it would be die if it weren't pronounced french ig lmao

lol imagine if it were like

der Image, den Imagen

[slim kayak]

mmhm

💀

[formal ermine]

so in german the rule is gender(english word) = gender(translation into german) in 99.9% of the cases

[slim kayak]

i dont want to imagen that

[south patrol]

but image is french right lol

but yeah i guess still holds

e.g. Portemonnaie

[formal ermine]

portemonnaie is a german word if u ask me

[south patrol]

ye fair

or is das Image a "new" word for german imported from english

[formal ermine]

nah

well

yes

[south patrol]

i assumed it's an older french one

[formal ermine]

on second thought

its like public image

[south patrol]

ah sure

[formal ermine]

schlechtes image

[south patrol]

thank

[slim kayak]

btw do you also find most german versions of math terms being way less cool than the english version

[formal ermine]

No

[south patrol]

personally i find the german names cooler lol

[final dragon]

yo did you guys start with groups or rings

[formal ermine]

groooops

[slim kayak]

garben, halme and keime make me think I am about to buy 40 eggs and a sheep

[south patrol]

Hauptidealsatz, Hauptvermutung, Nullstellensatz, usw.

[formal ermine]

thats too real

[south patrol]

Like nice snappy names

[formal ermine]

Aaron Pfister Hauptsatz

[south patrol]

Lol

[formal ermine]

say that twenty times in a row while doing a handstand

[south patrol]

Die, die die, die die deutsche Wörter präferieren, hassen, sind kringelig

sag das 20mal

[formal ermine]

took me a solid 30 seconds to understand the dies

[slim kayak]

i am allowed to count with my fingers

[formal ermine]

zehn zahme ziegen ziehen zehn zentner zucker um zehn nach zehn am zehnten zehnten

[south patrol]

this has made me now wonder the difference between bevorzugen and vorziehen

cause we were taught bevorzugen but i'd instinctively say vorziehen more commonly lol

[formal ermine]

those words couldnt be more apart

they mean completely different things

[slim kayak]

eh

ich bevorzuge X vs ich würde X vorziehen

[formal ermine]

bevorzugen = to give someone an advantage
vorziehen = to put someone before someone else (e.g. in a queue)

[formal ermine]

yes thats the same

but jemanden

[south patrol]

i mean in the sense of preferring something

[formal ermine]

bevorzugen vs jemanden vorziehen is different

[south patrol]

but sure yes makes sense

i wasn't actually aware of that 2nd meaning of vorziehen

though it is very literal i guess lmao

[slim kayak]

yeah

[south patrol]

thank u illum TIL

by "give someone an advantage" what do you mean

[slim kayak]

i cant remember having heard vorziehen used in this literal sense in forever

[south patrol]

lol

[slim kayak]

if you mean this use case, then they have the same content, altho slightly different connotations

the former expresses a general preference, whereas the latter id interpret as a weaker preference or a preference that emerges from a provided choice

maybe we are algechillng a bit too hard hm...

[formal ermine]

wir mussten den patienten vorziehen da sein zustand sich stark verschlechtert hat

[slim kayak]

a bit fringe but yeah...

einen termin vorziehen

thats pretty standard

[formal ermine]

i can only think of discriminatory examples for this rn lol

der schüler wurde aufgrund seiner herkunft bevorzugt

=> er hat bessere noten bekommen

[glossy crag]

That's just because you haven't "lived" with the equivalent English terms, I'm sure Americans/Brits feel the same way about sheaves and stalks.

[slim kayak]

yeah probably

[formal ermine]

prägarben 💀 💀 💀

[south patrol]

ah okay so like advantages as a result sure

[slim kayak]

prä goes kinda hard dont lie

[formal ermine]

prä ist ein schlechter präfix

[south patrol]

I think sheaves and stalks are used/mentioned so infrequently in everyday life (at least for me) that they sound cool to me lol

(as a Brit)

[formal ermine]

my condolences

[south patrol]

Maybe farmers think about it differently lol

[slim kayak]

at least sheaves are sometimes used in other concept, like a sheaf of papers or clothes lol

[south patrol]

Uh I have never heard anyone talk about sheaves of papers or clothes

Lol

Literally only example is a sheaf of wheat to my mind

[glossy crag]

IDK, can't speak much for the English terms, but in Russian they sound just as agricultural.

[slim kayak]

maybe i was thinking of a heap

[glossy crag]

Dunno about clothes, but "sheaf of papers" is a perfectly mundane expression I think.

[slim kayak]

the founders of AG were too hard into cottagecore aesthetics it seems

[glossy crag]

Mwah, the French~~

[chilly ocean]

germans must feel scammed

grothendieck being a native german but saying fuck you after ww2 and just using french

[formal ermine]

fr

bro imagine EdAG being in german

[chilly ocean]

that would be much easier than french

agreed

did you know you can actually visit his house in Berlin 🤣

[delicate orchid]

did you know you can visit my house too

[chilly ocean]

like not go in but there is like a sign and all

[chilly ocean]

do you have pizza

[delicate orchid]

I live right next door to a pizza place

[formal ermine]

ive never been to berlin

[chilly ocean]

well your house is superior to grothendieck

[chilly ocean]

its not worth the visit dw

i only went there for the plot

and it was disappointing

idk what i even expected

[slim kayak]

the plot?

[chilly ocean]

it came up and i was like

"watch me ill go there"

wasted afternoon

[slim kayak]

was it even sunny that day

for the mediocre sightseeing

[chilly ocean]

barely

[formal ermine]

Yeah my french is terrible

https://www.instagram.com/reel/C035SogIbVT/?igsh=MTl6enlmaGR1cjBsMw== this is my level of french

panchbantz

Thank god I’m bilingual

Likes

9251

Instagram

[chilly ocean]

lol

🤣

[wooden ember]

just to be sure im not being stupid, if I have a quotient of G by a compact subgroup H, then the preimage of any compact subset of G/H is compact right?

just cause itll look like KH which is a product of compact sets and thus compact?

[rustic crown]

why would it look like KH?

for locally compact hausdorff spaces, a map f : X --> Y is proper iff f is closed and each fiber is compact iff inverse image of each compact is compact.

[keen loom]

Is your G compact?

[wooden ember]

Right I see my mistake

[wooden ember]

Okay I can still work with this I didn’t know that criterion

[wooden ember]

What’s wrong with this argument though? Take an open cover of p^-1(K), quotient it via p which is an open map to get an open cover of K, extract a finite cover and take preimages for a finite cover of p^-1(K)

It’s naive and must be wrong im just not seeing it

Nvm I see

Stupid argument

The point is I wanted to show PSL_2R -> PSL_2R/PSO_2R is proper and thought I could get away without really using too much about those specific groups

But I guess I’ve got to work for closedness of the map

[keen loom]

https://mathoverflow.net/questions/186764/the-properness-of-a-submersion

And your fibers should look like the compact subgroup so should be compact

[wooden ember]

Oh yeah it is a submersion

Fair enough then closedness is trivial

Perf thanks a bunch

[stark helm]

Can someone provide me with some ideas or hints for this one?

[dull marsh]

Well, do you think the statement is true or false?

(Either way, try plugging in some small values of n and see what happens)

[carmine mantle]

yeh!
do some small computations and determine some pattern, that will give u a intuition. That's the best idea for any elementary number theory problems, but may not always work in advanced number theory (like algebraic or analytic).

and they try to prove that the pattern holds everyone, which shouldn't be that bad

that is only if you are sure that it is true

[severe linden]

Is it correct that this is the case iff a) 0 < k < n and b) k > 0?

[rustic crown]

you could also use tube lemma to get closedness, if K is compact and X is closed in G, then need to show G \ KX is open, pick g in G \ KX and notice K^-1g ∩ X is empty. So under the map G x G --> G given by (a, b) --> a^-1b, K x {g} maps inside X^c. so by tube lemma we can find U containing g such that K x U also maps inside X^c, i.e. K^-1U ∩ X is empty, so U is in G \ KX and contains g.

[keen loom]

We would need to know what the action is from the "previous 2 exercises"

[delicate orchid]

I imagine it would just be permuting the subsets in the obvious way

[severe linden]

Exactly, it's just the trivial action

[delicate orchid]

Well no it’s not the trivial action unless n = k lol

[severe linden]

σ * {a_1, ..., a_k} = {σ(a_1), ..., σ(a_k)} and same for tuples

[severe linden]

Oh yeah sorry I meant "the action obvious from the context", my bad

[delicate orchid]

No worries I was being nitpicky

[keen loom]

Oh I thought it was the sphere lol

I should go to bed

[severe linden]

I guess that establishes that at least it's not faithful when k = n ;)

For a)

[delicate orchid]

I think both your answers are correct though

[severe linden]

My reasoning was that for a), if 0 < k < n you can always write some {i} as the finite intersection of k-element subsets, so if σ(A) = τ(A) for all k-element subsets A then σ(i) = τ(i) for all i, and in b) well you can just look at the first index so the permutations have to be the same

[final dragon]

guys can sombody help me with this proove that there exists an non abelian group G s.t. 2 consecutive natural numbers i,i+1 holds that for every element a,b from G , (a*b)^i=a^i * b^i, (ab)^(i+1) = a^(i+1) * b^(i+1)

[slim kayak]

What's your go to example for a nonabelian group

[final dragon]

the permutations group

idk what to call it in englishj

[slim kayak]

Yeah those work if I understood the question correctly

[final dragon]

but listen you can do this by hand go trhoug all the groups maybe to the 10 like power idk what to call it

and i did it like that by hand

but i want algebraic

[slim kayak]

When is (ab)^n=a^n b^n true, generally?

[final dragon]

10 cyclic

[south patrol]

I think every finite group has this property

Lol

[final dragon]

what

[slim kayak]

"there exists"

So you need to provide one example

[final dragon]

oh

yea mb

[south patrol]

This doesn't hold for the free group on the generators for example

[final dragon]

yea its no problem to go through group of permuations and finde example

but can this be done like algebraic

[south patrol]

Yes

[slim kayak]

Maybe you should consider reading a bit of a proof book on the side, you seem a bit rusty

[final dragon]

who me

[slim kayak]

Yeah

[south patrol]

No me

[slim kayak]

Yeah potato

Go to trig and geometry now potat

[south patrol]

Tbf I probs suck at trig now lol

[slim kayak]

Whenever some acquaintance asks me to help with proof stuff I am happy to help

Unless it's geometry class

[final dragon]

so is there a way without going trough and checking manually

[south patrol]

Yes

[final dragon]

so how

[south patrol]

Not sure what hint to give hm

[slim kayak]

This has an easy and stupid solution, but I think it's better if you try to find a more interesting example

[south patrol]

I suggest you consider any theorems in group theory u know

As there is one that makes this 1 line

I assume Kerr has the exact same idea

[slim kayak]

So like, start playing around. Again, what is a condition that gurantees (ab)^n=a^n b^n generally?

[stark helm]

i actually think it might not be true since I have tried it, like n=4, n=6, but at least I can not factorize it into reducible over Z pattern?

[slim kayak]

The finite group nuke argument

[dull marsh]

n = 4 is a counterexample, yeah

Not sure what you meant by the end though

[stark helm]

I mean can you factorize this polynomial, I fail

[stark helm]

fail to factorize it into product of Z[x]

[slim kayak]

But I dont like your solution potato tbh

[dull marsh]

x^3 + x^2 + x + 1 = x^2(x + 1) + (x + 1) = (x^2 + 1)(x + 1)

[slim kayak]

its very olympiad-y, one trick that you either see or dont

[final dragon]

wait you guys did it

[slim kayak]

There are several ways to do it, altho more important is that you find something that gives you practice with problem solving here

ok enough yappin sry

[slim kayak]

what have you tried so far

[final dragon]

we only did lagranges theorem

[south patrol]

Bingo

[final dragon]

what does that have to do with this doe

[slim kayak]

well, when is (ab)^n = a^n b^n no matter what? Which one n value

[final dragon]

when its 1

idk

[slim kayak]

and for finite group there are more n's

[copper nacelle]

hi

[copper nacelle]

kerr

[slim kayak]

me

[final dragon]

why does that hold

[slim kayak]

remember the argument you used yesterday?

[final dragon]

yea

[slim kayak]

or a slightly different hint, what are the "simplest" (sub)groups?

[final dragon]

how do you mean simplest

[slim kayak]

the way you'd use it

[cloud walrus]

Spamakin🎷

[narrow wagon]

When solving a system of linear equations we can eliminate the variables by adding or subtracting

is it something that’s is only possible if the algebraic structure is a field

[formal ermine]

why should it

[narrow wagon]

This is my guess

say 2x+ 3y=10

-2x + 7 y=3

eliminating a variable is same as adding the same thing both sides

2x -2x +3y+7y= 13

3y+7y=13

[formal ermine]

the only issue arises when you wanna divide

[narrow wagon]

y(10)=13

I have used every axiom of a ring

till now

[narrow wagon]

It should be atleast an integral domain?

[formal ermine]

every non zero element is invertible => ur working in a field

[narrow wagon]

yes

this must be the case for me to have a solution for linear system of equation right?

[formal ermine]

not necessarily

[stark helm]

I am wondering why we can not have unique remainders in Eulicidean domain, since I don't see any real difference between this one and a field with division algorithm?

[celest furnace]

There’s no “canonical” choice

If you think of Z as a Euclidean domain then negative numbers can also be valid remainders

If you restrict to only positive then it’s unique

[final dragon]

how can i proove that a factor ring is isomorfic to another ring

[coral spindle]

First iso theorem

[final dragon]

yea ty

if I and J are ideals of a ring is I U J still ideal of R

[rocky cloak]

Only if one of them contains the other

[final dragon]

why is that

[coral spindle]

I think you should try proving it, it's a simple enough exercise and you will gain some intuition.

[final dragon]

ty guys i wouldve just said this holds but it doesnt

[celest furnace]

This is also false at a higher level of generality—for subgroups N,H of G N cup H is a subgroup iff one contains the other

So you can just apply this to these ideals since they r abelian groups under addition

[final dragon]

but we do rings before groups

[prime quail]

I am looking for a calculator.
Input is 2 permutations.
Output is number of elements in the group generated by the input.

example
input 1=(1,2)
input 2=(2,3)
output=6

[prisma ibex]

use sage

http://sporadic.stanford.edu/reference/groups/sage/groups/perm_gps/permgroup_element.html

[cobalt heath]

Mathematica would also work well (& easier to use) if you have the money.

[barren sierra]

sage 🤩

[summer path]

Sage fun

[deft pecan]

baby question about group generators: I'm trying to prove that the closed interval A := [-2, -1] generates R^*, i.e. the real multiplicative group. Intuitively I can see that A isn't closed under multiplication, and given some x in R\{0} we can factorize it in terms of some elements from A -- but I'm not entirely sure how to show this, or if this is what I need to show here. Also: I'm guessing there's nothing "special" about this particular closed interval, and that any closed interval in R\{0} would do?

The definition of generator I'm working with is that, given a group G and a subgroup X, Gen(X) is the smallest subgroup of G that contains X, i.e. it's the subgroup K such that X is a subset of K, and if Y is any subgroup containing X then K is a subset of Y. The only other fact I've shown already is that Gen(X) = Gen(X^{-1}) where X^{-1} is the set of inverses of elements of X

[crystal forge]

so the goal would be to show that for any x in R^*, as you said, there’s a “factorisation” in terms of elements in [-1, -2] - perhaps try some concrete examples with x and then working from there? (say, 2, 3, 4)

[formal ermine]

if you have a raspberry pi you can get it for free i believe

[celest furnace]

You can show that you can get all of [1, 2] then all of [1/2, 1] then all of [1/4, 1/2] and so on until you get all of (0, 1]

Then think about how x -> 1/x works and see if you can get everything in R_>0

[celest furnace]

Also you are likely right about any interval using the same tricks—dividing by the upper limit first then continuing and getting all of (0,1]

[rapid junco]

For a communitave Ring R that is non unital if we introduce a multiplicative identity 1. Then define 1 + 1 = 0 and define a + 1 just to be a new element does this give rise to a new ring?

In the homework I was doing apparently the trick was to look at Z x R but are these similar?

[rocky cloak]

Remember that multiplication should distribute over addition.

So think about what (1+1)*a would equal in this case

[delicate orchid]

unit quaternions, I think you mean

unitary groups are implied to be matrix groups over the complex numbers, usually

[rocky cloak]

I guess unitary matrices over quaternions aren't much studied, so they don't get their own special notation.

[delicate orchid]

well I mean we can look at them from either perspective

well yes that's what an isomorphism is? I'm confused now

the group of unit quaternions is denoted H^1, which is isomorphic to SU(2)

as sets they're different

I'd write it as C^1 but S^1 is more common (cause they're a circleeee)

sorry let me correct myself lol

[rocky cloak]

Guess there's two different definitions of unit floating around here

[delicate orchid]

yeah I realise that now which is probably my source of confusion

yeah, just extend the domain of the isomorphism to GL(2, C) or GL(1, C) by composing with the inclusions SO(2) -> GL(2, C), U(1) -> GL(1, C) if you want to make it explicit

[uneven bobcat]

Careful, the word "representation" has a formal meaning in group theory, usually meaning a vector space with a group action.

[uneven bobcat]

In that case, I don't see how it could be true. Like I said, SO(2) isn't a vector space.

They're isomorphic as Lie groups though.

So, you can get representations of one from representations of the other.

Definition: Let G be a group. A representation of G is a pair (V, p) where V is a vector space over some field k, and p: G --> GL(V) is a group homomorphism where GL(V) is the group of invertible k-linear transformations of V.

So, yes, part of the definition involves a homomorphism from you group, into the group of linear transformations on a vector space.

If the above definition is too abstract, you here's the following simplification which is a special case of the above, and the case one usually studies first:

Let G be a group. A finite-dimensional complex representation of G is a homomorphism G --> GL_n(C) where GL_n(C) is the group of n-by-n invertible complex matrices.

[rocky cloak]

R^2 (with the usual action) and R with trivial action would be two (real) representations

The isomorphism SO(2) = U(1) also gives a 1D complex representation.

Which is really just the same as the R^2 one in disguise

[daring burrow]

Can you suggest me lecture on yt about splitting field

[uneven bobcat]

Given that this conversation spawned from them being confused about this isomorphism is and the definition of a representation, you should parse what you're saying for them.

[rocky cloak]

They didn't seem confused to me

[uneven bobcat]

Check out Matt Salamone's channel.

[rocky cloak]

Seems they understood it pretty well

Trivial meaning acting by doing nothing (every element acts like the identity)

The usual is how 2x2 matrices usually act on 2d vectors, matrix vector multiplication

[uneven bobcat]

So to conform with the definition I gave you above, for the usual action let V = R^2 realized as column vectors. Then, I define the homomorphsim
p: SO(2) --> GL(V) by sending the matrix X in SO(2) to the linear transformation

v --> Xv for all v in V = R^2.

For the trivial representation, let V = R. and define p: SO(2) --> GL(V) to be the map sending everything to the identity in GL(V).

[stark helm]

if there is an ideal (d), and a is in (d), can I say (a) is the subset of (d). I am considering in this way: we can have a=dp where all p is in ring R, for all t from R, we have at=(dp)*t, therefore all elements from (a) is in (d)?

[south patrol]

Yes

Also like (a) is the smallest ideal containing a

So since (d) contains a it contains (a)

[old vapor]

can you provide your thoughts on combintronics..i have doubts if i understood it correctly

or what would be the appropriate channel to discuss about it

can you provide your time for discussion about this #groups-rings-fields message

[barren sierra]

https://media.discordapp.net/attachments/496784958430380033/1197961460073111604/image.png?ex=65bd2b8d&is=65aab68d&hm=76111ced93b071d24d97e1758126c5024710af2d2c7fd88e4e95e10fa68451de&=&format=webp&quality=lossless&width=1921&height=225

[cloud walrus]

Spamakin🎷

[barren sierra]

so those computations seem to imply the claim in the last sentence of the question here ^ is wrong
so I must be misinterpreting the question but I'm not sure

[rapid junco]

Ah so a + a should be zero, but then nessexwriky means we could be introducing new in verses.

[keen loom]

There is a different combinatorics channel https://discord.com/channels/268882317391429632/576511723574525962

[final dragon]

what is the factor ring R/I where I is the ideal of R and R is the set of matrices (a b 0 c) where a,b,c are real numbers , and I is matricies (0 b 0 0) , b is from real numbers

how do cosets look

[keen loom]

It's often the case that the first isomorphism theorem helps. If you can think of a surjective homomorphism to another ring that has your ideal as the kernel, then the factor ring is isomorphic to the codomain.

To think about what the codomain could be, you can get some intuition by thinking of all the things in the ideal as "collapsed" to zero, as if you've imposed a set of relations on your ring.

[uneven bobcat]

Short answer: yes.

Long answer: Technically you need to first define what you mean by "unitary matrix over the quaternions". Recall that over C, unitary matrices are defined to be those that satisfy XX^ = I, where the ^ means conjugate-transpose. Of course in the quaternions we have a notion of conjugation, but I highlight this because for matrix groups over other fields (or division algebras, meaning they have multiplicative inverses but need not be commutative, like the quaternions) we might not have such a notion.

However, if your field F has a quadratic extension, there is a way to make sense of U(n, F). So yes, U(1, H) is isomorphic to the group of unit quaternions H^* which is isomorphic to SU(2, C).

[stark helm]

I am wondering why we can have R[x]/(x) is isomorphic to the integral domain R?

[coral spindle]

Like most times we show that a factor ring is isomorphic to something, we use the first isomorphism theorem. Try it

[south patrol]

Also like think about what elements of R[x]/x look like

[stark helm]

can we just say that (x) is the kernel when we construct a ring homomorphism between R[x] and R?

[stark helm]

since it seems that R[x]/x is of the form r(constant coefficient), when all r is in R?

[stark helm]

it should be of the form r(constant coefficient) when r is in R, and is it convincing to say that (x) is kernel when we construct ring homomorphism between R[x] and R?

[slim kayak]

Well, what is the kernel of the R[x]->R sending x to zero?

[stark helm]

only pick 0 from R[x]? or picking a constant coefficient a so that a is in R(since R can only contain elements of the form a?)

[coral spindle]

You can certainly say it, but it's up to you to construct such a homomorphism and prove it has kernel (x).

[stark helm]

Maybe I need to ask a similar question, if there is a mapping from Z[x] to Z, then how will you state that (x) is the kernel of this ring homomorphism?

[round hull]

it's best illustrated in two steps. -1 via multiplication represents getting the additive inverse, and taking additive inverses twice corresponds to doing nothing

the latter part is really a statement about abelian groups

hence (-1)(-1) = -(-1) = 1

[coral spindle]

I think you should try discover the answer to this yourself. Write down a few mappings from Z[x] to Z, and see if the kernel is (x). A good way to begin to check if the kernel is (x) is to see if the kernel even contains x.

[round hull]

account deleted

[summer path]

That was a while ago

[slim kayak]

another victim to discord message wormholes

[keen loom]

what does that even mean..."makes sense outside of university math"...like, applicable to the real world?

[slim kayak]

People who have had no experience with math at an university level

Clearly an impossible task. How could anyone understand even + odd = odd without years of study?

[keen loom]

lol

I would go rolle's thm/MVT

[slim kayak]

those are nice ones

[chilly ocean]

Things like bezouts theorem can be understoof pretty easily

Even tho they require a lot of machinery to prove

[celest furnace]

2+1 = 3 which is odd, 4+3 = 7 which is odd… yep we’re good, it’s true

[keen loom]

I cannot tell you how many papers I grade look like this

[celest furnace]

Division algorithm too

[celest furnace]

What class?

[keen loom]

intro to discrete math, we have compsci students take it

[celest furnace]

Lmaoooo

2 of my friends r CS students and they TA the intro to proofs class for CS and they say the same lol

[keen loom]

it's so bad...it's one of the few times when I'm just like "it is what it is"

when it comes to teaching

[celest furnace]

My friend showed me one combinatorics example that would be good to keep in mind

The pattern of something was like powers of 2 up to like n = 4 then the next one is just 31 outta nowhere

Something related to diagonals of an n gon

[keen loom]

Dihedral group order? (Shot in the dark)

Naw that's 2n

[cobalt heath]

2 + 1 is odd.
2 + 3 is odd.
3 + 1 is odd.
The rest of the proof is left to the reader as an exercise.

[celest furnace]

https://oeis.org/search?q=1%2C2%2C4%2C8%2C16%2C31&language=english&go=Search

Looks like it might be the composition one

[keen loom]

Hahah that's nice

[celest furnace]

Yep when in doubt look it up lol

[golden turtle]

spotted

[celest furnace]

!status

[flat tree]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[golden turtle]

7

if you want a challenging problem from a non-mickey mouse 335 tho I can show u one from basedrian

[celest furnace]

This is more appropraite for #multivariable-calculus

[slim kayak]

Hm, maybe it only works for small numbers?

[golden turtle]

you wouldn't know about that

[slim kayak]

1001+2=1003
1005+8=1013
Hm..

[golden turtle]

maybe it only works for numbers involving a 1 or a 3?

[cobalt heath]

I mean, it illustrates how to show it

You can get all the combinations by adding 2 on each entry

[slim kayak]

Good point

[ivory trail]

maximum number of parts you can divide a disk into with n chords

even a 3b1b video on it https://www.youtube.com/watch?v=YtkIWDE36qU

[celest furnace]

Awesome

[cloud walrus]

Riku

[boreal inlet]

If I do not assume this map is a ring homomorphism, can I even continue to extend this map to the whole ring?

I am asking this because we have been asked to prove this as an isomorphism (automorphism, to be precise.)

K is a field here.

[keen loom]

What makes you say it's not necessarily a homomorphism?

[boreal inlet]

Considering it to be a set map, we have no useful relations to define the map for other polynomials at all

[keen loom]

Sure you do: x is a generator for the whole thing. You only need to know it for x.

[boreal inlet]

That is true

[keen loom]

Polynomial rings always work that way: choose where to send x, and it can be extended to the whole ring

[boreal inlet]

Ok so, x is a generator, so x^n can be determined. How exactly we do it?

x^2 maps to the square of the image of x right?

and after we determine each x^n in the format, we can scale it using any scalar from the field, and also we can add the images.

All of these are properties of homomorphisms

My question is, if we do not start with these assumptions, how can we arrive that this is naturally true for this map?

[keen loom]

This is a good question

[boreal inlet]

Exactly.

[keen loom]

Polynomials are the "free" rings. They have the special property that if you send their generating set somewhere (anywhere at all, it can be an arbitrary set map), you can uniquely extend that map to a ring homomorphism.

[boreal inlet]

Wait, actually?

[keen loom]

This is like in linear algebra: if you choose to send your basis elements somewhere, then you can automatically extend that to a linear map

You just construct the map according to the rules of homomorphism

[boreal inlet]

This answers my question.

[boreal inlet]

I have to show {(ax + b)^n | n \in {1, ..... N} } is linearly independent in K[x] over K.

a and b are fixed elements of K, a not zero.

I thought of doing induction, but now I am stuck at the inductive step

Showed the case for n = 1 and 2

If K was R I'd use the continuity of polynomials but we don't know that for any arbitrary field

Never mind, I did

[boreal inlet]

Okay so, I used the linear indepedence to show this map is injective.

Now I have to show this is surjective

This seems to be difficult though.

Because the map x |-> (x - b)/a doesn't seem to work as an inverse

This gives back x when applied to ax + b

But fails for (ax + b)^2

(ax + b)^2 = ax^2 + 2abx + b^2, and now if you put (x - b)/a in place of x, it doesn't give back x^2

Oh my god.

Ignore me

I saw the mistake

Also, now I realize this map itself is the inverse and it would have been much easier.

[celest furnace]

U can do it more directly too

If they were linearly dependent a finite sum of them would be 0 with not all coefficients 0

Let a_m be the coefficient of the highest power of (ax+b)^n. But how can you kill the x^n when everything else has smaller power?

[boreal inlet]

That's the exact inductive argument I applied

Assuming upto m -1 is L.I, then showing the coefficient for x^m has to be also zero, is sufficient

[celest furnace]

Ah yep that works

Was wondering if you could use that f(x) = ax+b is affine so an isomorphism somehow

Like is the image of the basis elements under an isomorphism still a basis probably I think this is what u we’re getting at

[boreal inlet]

Yes, if I was able to use it I'd Just write this

[celest furnace]

Or rather the linear transformation fixing K and sending x to ax+b

Ahhh so u found an inverse it all makes sense now sorry

[boreal inlet]

Nah it's okay, this is also a viable idea

Okay just confirming, if I have a group homomorphism f : G -> K and I have found a set mapping g : K -> G such that

f • g = Id(G) and g • f = Id(K). Notice the fact that g is a set mapping.

Is this sufficient to say f is an group isomorphism?

[celest furnace]

Yes. Why?

[boreal inlet]

Or do I have to prove g is also a group homomorphism

[celest furnace]

No its true in general that if f: G -> H is a group homomorphism and bijective then f^{-1} is also a group homomorphism

I recommend you prove this fact if you dont believe it

[boreal inlet]

No no, I believe it, I just was asking if it was needed

Thanks

[celest furnace]

Ah yeah something like that is assumed

[boreal inlet]

This reduces my workload just a bit

[regal star]

Could someone tell me to follow the book "a first course in representation theory" by Fulton Harris, what are the prerequisites required?

I have gone through course in group theory although a basic one and courses in real analysis and linear algebra by axler

[dim widget]

probably the only requirements are a first course in group theory and a good understanding of linear algebra

the book is pretty self contained

[regal star]

Is multilinear algebra required?

[summer path]

There's an appendix section on multilinear algebra in F&H, I believe

You'll probably want to be familiar with it one way or another

[formal ermine]

whats multilinear algebra

[lavish sigil]

If we have a real vector space with inner product, $(V,g)$, and if we let $(V_\mathbb{C},g_\mathbb{C})$ be its complexification, where the inner product has been extended by complex bilinearity (i.e.\ \textbf{not} to a Hermitian inner product), then the Clifford algebra $\mathrm{Cl}(V_\mathbb{C},g_\mathbb{C})$ is canonically isomorphic to the complexification $\mathrm{Cl}(V,g)\otimes_\mathbb{R}\mathbb{C}$. The complex conjugation on the latter thus induces a complex antilinear map $c:\mathrm{Cl}(V_\mathbb{C},g_\mathbb{C})\to\mathrm{Cl}(V_\mathbb{C},g_\mathbb{C})$ such that $c^2=\mathrm{id}$. If we assume that $w_1,\dots,w_k\in V_\mathbb{C}$, so that the product $w_1\cdots w_k$ is an element of $\mathrm{Cl}(V_\mathbb{C},g_\mathbb{C})$, is then $c(w_1\cdots w_k)=\overline{w_1}\cdots\overline{w_k}$, where the conjugation on the right is the standard conjugation on the complexification of a real vector space?

[cloud walrus]

gustavn64

[dim widget]

multilinear algebra is part of linear algebra

who learns about linear algebra without bilinear forms and determinants?

I guess maybe from Axler

[rapid junco]

in a general ring, can there be an element a such that a^2 = -1?

[dull marsh]

Well, the ring of complex numbers is such an example

[south patrol]

take any ring R and consider R[x]/(x^2 +1)

you can force it to be the case

There's also the funn case of the trivial ring

[rapid junco]

what am i missing then to show this:

i have deduced that N(x) \in {+- 1} if x is a unit

a^2 + b^2 = 1, where a and b are in Z

without using order

[dull marsh]

Pretty sure that N can't be negative

[rapid junco]

Yeah but how do you show this without using the order on Z?

Or is this literally the way.

[dull marsh]

Yeah

[long obsidian]

Can someone help me think about direct limits of groups?

For instance if I consider a group A with the identity map on itself can we form a direct limit? A->A->A->A...?

I feel like this should be A where we just use the identity to get the commutative diagram and in fact A->A->A with the identity maps will commute with A->A with an identity map

Is this correct?

[rocky cloak]

Yes

[long obsidian]

Do you always use this universal property? I saw something that made me think it was more natural to think of these as sequences (a potential example being the p adics).

[mighty kiln]

A filtered colimit of sets / groups / modules etc can be explicitly described by taking the union and quotienting by the relation "a ∼ b iff a is eventually equal to b"

[rocky cloak]

Sometimes it's easier to work with the universal property, sometimes it's easier to work with an explicit construction.

[mighty kiln]

Where "eventually equal" means they get mapped into the same element in some group

[long obsidian]

Ohh okay I didn't know about the eventually language that is helpful. I also forgot it was an equivalence relation to the union thank you.

[formal ermine]

that's mutlilinear algebra? i thought its just linear algebra lol

[slim kayak]

You usually can do it with any colimit, the filtering condition just means the quotient is significantly easier. Also, being able to consider cofinal families makes them much more friendly objects

[mighty kiln]

I think in general you take coproduct instead of union?

[slim kayak]

I thought that was what you meant with union?

[mighty kiln]

No

Like disjoint union

[slim kayak]

Setwise that's what they are

I am not sure why you would want to pass to the union description instead of working with the coproduct itself?

It's usually the nicer object

[mighty kiln]

Cuz it be useful

Say stalk

[slim kayak]

Yeah?

[mighty kiln]

You can do much arguments by taking representatives

Cuz is quotient of disjoint union

[slim kayak]

I never had a situation where there being a disjoint union helped with the stalk arguments

You mean that you can take sections of some neighbourhood in the pre-image of a germ?

Or wait, do you mean that you can talk about germs in general?

Due to the coproduct/coequalizer construction?

[mighty kiln]

Well like whenever you take a local section as a representative

You're using the fact that the elements of the stalk are represented by local sections

[slim kayak]

Yeah I get what you mean now

I suppose it clearer if you go the setwise colimit -> equip it with extra structure route for stuff like rings/modules/abelian groups that any germ always corresponds to some local section so you can pass over to them, like when making - say - the argument that being integral passes over to staks and vice versa

[mighty kiln]

Yea

Thus filtered be pog

[slim kayak]

Cofinality is a savior

Means that your stalks are determined up to iso by your base

[tough raven]

That fact should be true in any “reasonable” “algebraic” category and is helpful for explicit computations when coproducts are not so simple (e.g. in the category of groups or rings).

Like, words are fine, but union is better.
I think.

[dim widget]

for groups it is not the disjoint union

Products and limits in algebraic categories are the same as those in set, but coproducts and colimits are usually unrelated

[slim kayak]

Yeah I was thinking abelian groups

[dim widget]

even in abelian groups the coproduct is not the same as disjoint union

It’s the same as the product in set in some nice cases but that is basically an accident

[slim kayak]

Hm... so the constructiong does require the filtered assumption somewhere in order to construct the colimit step by step as first a disjoint union with an equivalence relation and then equipping it with the right structure?

[south patrol]

In which content

[slim kayak]

I am refering to exercise 1.4E (page 43/44) in this here
https://math.stanford.edu/~vakil/216blog/FOAGjul3123public.pdf

[south patrol]

Do you mean coproduct

Oh okay filtered case

[slim kayak]

Coproducts wouldnt work since they arent filtered

Yeah, sloppy language on my end

[south patrol]

ye the forgteful functors here preserve filtered colimits

which is nice

[slim kayak]

does "here" extend further than modules and maybe rings?

[south patrol]

Apparently it should be for most like equational algebraic things but not sure

[slim kayak]

non-abelian group feels like they would be too cursed for this

[dim widget]

The “forgetful functor” of any monadic adjunction preserves all sifted colimits

[south patrol]

Sure

Hot

(thank)

[dim widget]

and sifted colimits are basically all filtered colimits and mutually split coequalizers, and the colimits they generate

[dim widget]

non abelian groups would also work

basically any algebraic structure built out of a set which admits a “free” functor would work

[viral mountain]

Consider a permutation $\sigma \in S_n$ written in disjoint cycle notation as $(1, 3, 6)(2, 4)$

If we consider $\tau = (1, 3, 6)$ and $\phi = (2, 4)$ as two separate permutations, is it a coincidence that $$\sigma = \tau \phi$$ or is there a reason for this?

[cloud walrus]

BlazingSaber

[barren sierra]

0 coincidence

That's the definition of the operation of S_n, which is composition

Remember these are just bijections [n] ↔ [n]

And what you've written there is how that operation is reflected when you express these as cycles

[viral mountain]

Having a tough time wrapping my head around this, maybe because the book I'm following taught me how to write in disjoint cycle notation given the two-row notation, as an algorithmic procedure

I understand that the definition for how these cycles are written is motivated by function composition, but trying to see how/why exactly

Okay, the fact that these cycles are "disjoint" helps make it a lot clearer
Given that $\tau$ and $\phi$ are disjoint, they share no common elements and so you can first apply the action of $\phi$ and then $\tau$, which can essentially be done as a single permutation $\phi$

[cloud walrus]

BlazingSaber

[cloud solar]

Let I be an open interval in R and (I,•) commutative group with the property that for every a,b in I with a<b and for every x in I we have x•a<x•b. a) Show that e is between b and b^-1 for every b in I-{e}. b) If H≠{e} is a subgroup of (I,•) with the property that for every sequence (x_n)n>=1 with numbers from H and for every b in I with b>e, the set {x_n | n>=1 } intersected with (b^-1,b) is a set with no elements or a set which has a maximum, show H is cyclic.

a) is trivial

b) is interesting

[cloud solar]

if I look at e<h with h in H-{e} we get e<h<h^2<...<h^n

and take the sequence x_n ={h,h^2,...,h^n}

and take b>h^n. If b^-1<h^n we get S=(b^-1,b) intersected with {x_n} has elements so S has a maximum but thats false because of the open interval. So b^-1 > h^n. But h^n>...>h>e. And because b^-1<e<b we get something false

Can someone explain where is the problem

[viral mountain]

Is this result incorrectly stated? Did they mean $ab^{-1} \in H$ instead $G$?

I tried proving the theorem with the current hypothesis without any luck, so I'm wondering if they meant H instead

[barren sierra]

yes should be in H

[cloud walrus]

BlazingSaber

[barren sierra]

also "for all a, b in G"

should be "for all a, b in H"

we don't care about elements outside of H

[viral mountain]

okayy cool

thanks!

[boreal inlet]

I have a field K, where K(t) = Frac(K[t]). Let u \in K(t)\K, such that u = f(t)/g(t) for non zero f, g \in K[t] and gcd(f,g) = 1.

Gcd is defined as K[t] is an Euclidean domain

I have to prove that h(x) = f(x) - ug(x) is irreducible in K(u)[x], and t \in K(t) is a root of h(x).

The latter part is very obvious

But how do I show irreducibility

[celest furnace]

Is K(u)/(u) iso to K? Maybe Eisenstein?

[slim kayak]

Are you familiar with localizations? Maybe how they commute with quotients?

Ofc if K didn't contain u in the first place.

The straightforward path is checking the map sending u to 0 and making sure the image is exactly K

[south patrol]

I mean, u is a unit

[celest furnace]

Ah shit ur right

[south patrol]

if it's K[u]/(u) then agreed

[lime junco]

im just confused by this definition

what is F? (wasnt mentioned before)

for context, E is a field extension of k

[boreal inlet]

u can't be in K by the question. u is taken from K(t) - K.

[boreal inlet]

F is a subfield of E, containing the field k and the elements in S. If you gather all such F and take an intersection, you get k(S), which is the smallest of all such fields

[lime junco]

ohhhhh

[boreal inlet]

This notion is very common

[lime junco]

its all F such that F contains k and S, i see

[lime junco]

ill keep that in mind lol

[boreal inlet]

Sadly I'm not. I was thinking of reading about what localizations are in my spare time but we formally haven't had any course on Commutative Algebra yet.

[alpine island]

yeah that first bit is kinda weirdly phrased

[lime junco]

follow up question

lets say I have it set up as k subset E where k is my field and E is my extension

is k(S) as denoted above

equal to all linear combinations of elementsof S with coefficients in k?

analogous to a vector span

[boreal inlet]

It doesn't have to be I think

[lime junco]

oh, unfortunate

[boreal inlet]

It definitely contains the linear combinations yes

But it could be bigger

[lime junco]

is there a nice way to like

explicitly describe that object in general

or no

[boreal inlet]

I'm new to field theory as well and up until now this is how we define it

I think for particular fields and particular subsets you can calculate that to isomorphism

[lime junco]

ah fair enough

I suppose then theres probably no real general case

but in "nice" fields one can compute it

[crystal turtle]

In general, you'll need to consider linear combinations divided by nonzero linear combinations to also have inverses

Finite extensions (deg(E) < ∞) you have a basis however, so you can always write things in terms of that

(or uhh deg(k(S)) < ∞ here)

[lime junco]

Ah i see

that makes sense

[boreal inlet]

Oh right, yes, my bad

This you can do

[boreal inlet]

. Btw, may someone help me here? I have been stuck for hours 😭

This is the original text if anyone needs

t is a root that is very easy to see, you just plug t in there

But that kind of also means t isn't in K(u), because otherwise h(x) won't be irreducible

I'm also confused on how different f(x) and f(t) is. At start it is also an indeterminate variable

[cloud walrus]

Riku

[boreal inlet]

Does that also say we are welcome to try contradiction, that is, assume beta is a root of h(x) in K(u). This extension (K(u) over K) is finite and has the degree as the degree of some minimal polynomial having u as root (for the time being I don't know what it is).

From there can we reach somewhere

Oh wait this assumption is also only true if u is algebraic over K

[marsh quarry]

we can show without much work that matrix multiplication is associative. but if we instead show that matrix multiplication is the same as composition of their corresponding linear maps, then does associativity come for free since function composition is associative?

if so, can we generally show some operation is associative by treating its inputs like functions and the operation as composition?

[alpine island]

Yeah you can do that, but finding a set of functions to work with is usually harder than just proving associativity directly

[boreal inlet]

This is true because of the obvious isomorphism between the groups of matrices over a field K under multiplication and group of linear maps over a field K under composition

But yeah in general the corresponding functions would be hard to find unless it's something like this

[marsh quarry]

do we know any other examples of proving associativity by treating objects like functions

[alpine island]

Sometimes the definition of your object just happens to be closely tied to functions

and it's easy to do this

[marsh quarry]

ya ok that makes sense

[boreal inlet]

😭

[alpine island]

Unfortunately no, I cannot

[boreal inlet]

From here, if I assume beta is a root of h(x) that is in K(u), I can write u = f(beta)/g(beta), provided both f(beta) and g(beta) is non zero

And we already know u to be f(t)/g(t) with gcd 1 between f and g

Both f(x) and g(x) are in K(u)[x] so evaluated at beta f(beta) and g(beta) both should be in K(u)

[boreal inlet]

From here I am unable to proceed anywhere

[boreal inlet]

Ok anyways this doesn't work for degree more than 3 so this shouldn't work in general

I think the problem boils down to the fact that f(x)g(t) - f(t)g(x) is irreducible in K[t, x]

Which would be true iff the same polynomial is irreducible in K(t)[x] and coefficients have to have gcd 1, Gauss's Lemma

[boreal inlet]

Now how to show this 😭

[boreal inlet]

https://math.stackexchange.com/questions/4849137/regarding-the-isomorphism-pgl2-k-cong-gl2-k-scal2-k-and-a-degree-of-a

I have given up and raised a stackexchange post.

[cloud solar]

Let I be an open interval in R and (I,•) commutative group with the property that for every a,b in I with a<b and for every x in I we have x•a<x•b. a) Show that e is between b and b^-1 for every b in I-{e}. b) If H≠{e} is a subgroup of (I,•) with the property that for every sequence (x_n)n>=1 with numbers from H and for every b in I with b>e, the set {x_n | n>=1 } intersected with (b^-1,b) is a set with no elements or a set which has a maximum, show H is cyclic.

if I look at e<h with h in H-{e} we get e<h<h^2<...<h^n

Can someone help me

[boreal inlet]

I'm actually not sure what is the operation supposed to do.

It certainly doesn't seem a subgroup of R

Or is it the multiplication

If it is, the open interval cannot contain zero

[cloud solar]

It is just some operation

And it is a subgroup of I

[mighty kiln]

This group is isomorphic to (R, +)

[cloud solar]

Ok I understand

So the problem is wrong?

[mighty kiln]

?

[cloud solar]

@boreal inlet said H doesnt look like a subgroup of R

[mighty kiln]

(I, ·) is not a subgroup of (R, +) with the standard inclusion I ⊂ R

But (I, ·) is isomorphic to (R, +)

[cloud solar]

Ok so how I can show H is cyclic

Im so confused

[mighty kiln]

Well any subgroup of (R, +) s.t. all bounded sequences have maximum is cyclic

Or perhaps (?) the same argument for proving this can work directly on I

[mighty kiln]

The easiest way to prove this is probably: Find all the integers, find all the rationals, find all the reals

[mighty kiln]

Tho I think this works actually

[cloud solar]

Where can I find the proof of this

[mighty kiln]

Consider the infimum of elements in H larger than the identity

[viral mountain]

I'm trying to prove that the subgroup $H = {a^n : n \in \mathbb{Z}}$ of a group $G$, for some $a \in G$ is the smallest such subgroup containing $a$

Is this a valid argument by contradiction to show the same?

[cloud walrus]

BlazingSaber

[viral mountain]

(sorry for the crappy handwriting, wrote this kind of informally)

[boreal inlet]

Wait, so the dot is multiplication?

[mighty kiln]

no

It's a group operation

[boreal inlet]

(\psi: (I, \cdot) \to (\mathbb{R}, +)) s.t.

[
\psi(a \cdot b) = \psi(a) + \psi(b)
] is an isomorphism?

[cloud walrus]

Riku

[slim kayak]

Not enough information

[rocky cloak]

For the first part, compare e and b, then multiply by b^-1.

For the second, call an element positive if it's bigger than e. Then you want to show that H has a minimal element and is generated by its minimal element.

Assume H doesn't have a minimal positive element, then let xn be a decreasing sequence, such that any positive element is eventually larger than xn (we know such exist because every well ordered subset of R is countable). Then consider b(xn)^-1 in (b^-1, b)

[cloud solar]

The sequence xn has all the terms positive?

[rocky cloak]

Yeah, xn positive

[cloud solar]

Can you explain more the part with the existence of the sequence and how this proves H is cyclic please?

[viral mountain]

In the proof of this theorem, it isn't clear how they arrived at the highlighted conclusion.
Is it because in any nontrivial subgroup of G, we must have an inverse for every element, so for every positive power of a, we must also have a negative power of a, and vice versa?

So we can be sure that there exists a power of a that's present in H if its nontrivial

[rocky cloak]

What it's proving is that H has a minimal positive element. After that you should try to prove that H is generated by this element.

Remember your hypothesis says that b(xn)^-1 has a maximal element inside (b^-1, b), how can you use this to get a contradiction

[rocky cloak]

That's right, if a^n is in H then also a^-n will be there

[cloud solar]

If A finite ring isomorphic with A1×A2×...×Ak and the number of nilpotent elements of A is 2^k then the number of nilpotents of Ai is 2?

[cloud solar]

If each Ai has 2 idempotents

And A has 2^k idempotents

If some ring Ai has only one nilpotent then there must be a ring Aj which has 2^r nilpotents with r>=2

[slim kayak]

The image of each projection is nilpotent (why?), go from there

[cloud solar]

Ik that but then?

[slim kayak]

Think through it

An element is nilpotent if all it's components are

[cloud solar]

Yeah this is a property

I used this to show A has 2^k elements because each Ai has 2 idempotents

[slim kayak]

Right

Now see what happens if some A_i doesn't have exactly 2 nilpotents

[cloud solar]

So if some A_i has only one nilpotent then there must be some A_j which has more than 2 nilpotents (2^p where p>=2 because the product it must be a power of 2)

In that ring A_i because it is finite there is some n s.t. a^=a^(2n) for a in A_i so a_i idempotent so every element must be invertible or nilpotent

And because A_i has only one nilpotent that means A_i is a field with only 2 idempotents

[slim kayak]

Excuse you?

A field?

[cloud solar]

It is commutative

A_i can be iso to Z/2Z for example

So can we have an A_i that has only one nilpotent?

@slim kayak so each A_i must have only 2 nilpotents or not?

[slim kayak]

Fields have no zero divisors, so the only nilpotent is 0

[cloud solar]

Yes. A_i has only one nilpotent

This is what we supposed

Oh we cant have

That

In the problem we also have that A has 2^k invertible elements

Or can we have such an A_i? We will get that A_i must be iso to Z/2Z

[viral mountain]

How do we justify that the only subgroups of $\mathbb{Z}$ under addition are $n \mathbb{Z}$ $\forall n \in \mathbb{Z}$?

[cloud walrus]

BlazingSaber

[slim kayak]

You can find a smallest element in it

[viral mountain]

My guess is that we know that the subgroups of any cyclic group are also cyclic

Since $n \mathbb{Z}$ are the only cyclic subgroups of $\mathbb{Z}$ under addition, they're also the only subgroups

[cloud walrus]

BlazingSaber

[slim kayak]

And given any larger element, this smallest thing must divide it.

[viral mountain]

Smallest element in which set?

[slim kayak]

Subgroup ofc

[viral mountain]

which subgroup tho, I mentioned all the subgroups nZ of Z

[slim kayak]

Any

[cloud solar]

@slim kayak I am trying to figure out if there is such an A_i with only one nilpotent. If it is true, then there must be A_j with 2^p nilpotents with p>=2 (so at least 4 nilpotents). Because A has 2^k invertible elements we know that A_j has 2^q invertible elements with q>=0. We also know A_j has only 2 idempotents (0 and 1) and the fact that A_j is finite and commutative.

Another thing I see is that we cant have q=0 because of the function from the nilpotents to the invertible elements f(x)=x-1 (picking a and b distinct both nilpotents)

So A_j has at least 2 invertible elements

Another thing: because A_j has only 2 idempotents it means that every element in A_j is nilpotent/invertible

[viral mountain]

I understand this proof, but I don't understand where the corollary below comes from

We'd seen in a previous section that the subgroups $n \mathbb{Z}$ of $\mathbb{Z}$ are cyclic. But how does that imply that all the subgroups of $\mathbb{Z}$ are of the form $n \mathbb{Z}$? What if hypothetically we're not considering subgroups of $\mathbb{Z}$ which are not of the form $n \mathbb{Z}$ but are still cyclic (as Theorem 6.6 shows)?

[cloud walrus]

BlazingSaber

[viral mountain]

I understood, never mind
Every cyclic subgroup of Z can be expressed as nZ
And since every subgroup of Z is a cyclic subgroup, it follows that every subgroup of Z can be expressed as nZ

[cloud solar]

Another thing: A_j has 2^q invertible elements and q>=p>=2 because the function f(x)=x-1 is injective

So this means that there must be A_t ring s.t. A_t has only one invertible element

[cloud solar]

This means that char(A_k)=2 because -1=1

So that means A_k has 2^m elements. But A_k also has 2 idempotent elements so every element is nilpotent/invertible. But A_k must have 2^s nilpotent elements with s>=0. so 2^s+1=2^m and this is only possible if s=0 so A_k has only one nilpotent, only one invertible element and exactly 2 idempotents so A_k is isomorphic with Z/2Z

[slim kayak]

Can you show that a ring that isn't the product of more rings can only have 2 or odd numbers of nilpotents?

[cloud solar]

Commutative ring?

[slim kayak]

If that's what you can assume here, sure

[cloud solar]

And how this helps me with my problem

Oh I see it

We will apply this for every ring A_i right?

[slim kayak]

The idea is that the number of nilpotents in A is just the product of all elements you can choose

[cloud solar]

Yes I know that

[slim kayak]

So if we can limit the prime factors of the nilpotents of any ring which can't be expressed as a non-trivial product of rings, then it follows

[cloud solar]

What do you mean

How do we know all the A_i's cant be iso to some non trivial product of rings

[slim kayak]

By finiteness

Like, make them sufficiently small

[cloud solar]

I know but I want to keep the property that all the A_i's have exaclty 2 idempotents

Oh

The property remains

[slim kayak]

Right, so split up the A_i's further into something that has nilpotents and something that doesn't, then move those without them to the side

[cloud solar]

Because 2 prime

[slim kayak]

Or whatever. If each A_i has r_i nilpotents, then the number of nilpotents of A are 2^k=r_1...r_k

So clearly each A_i has some power of 2 many nilpotents. Maybe you can show that the only possible values are 1 and 2

[cloud solar]

But if we șplit some A_i into B1× B2 the number of idempotents of B1 × the number idempotents of B2=2 but every ring with unity has at least 2 idempotents so we cant split A_i into a non trivial product of rings

So that means all the original A_i's cant be split

[slim kayak]

yeah, go this route

no splitting there

[cloud solar]

If these are the only possible values, if we have some A_i with only 1 nilpotent then there must be A_j with at least 4 nilpotents and that is false. So the only possible value would be 2

[slim kayak]

exactly

so show why 4,8, ... cant occur

[cloud solar]

So if all A_i's have more than 4 nilpotent elements the product will be striclty bigger than the number of nilpotents in A

[cloud solar]

To show why we cant have some A_i that has more than 4 nilpotents or all A_i's?

[slim kayak]

We we cant have any A_i with more than 2 nilpotents while still being a power of 2

[cloud solar]

How

[viral mountain]

I don't understand. How can we "choose" a generator to be positive?
Isn't what it means to be a group to be cyclic that it must contain an element whose powers represent every element in the set, but we need not know anything of the nature of the generator until we actually find what it must be?

[slim kayak]

You can if your group is ordered

If a is a generator, so is a^-1 or -a (depending on notation)

[viral mountain]

In other words, I know that a group $G$ is cyclic if there exists $a \in G$ s.t. $G = {a^n : n \in \mathbb{Z}}$
But we're only claiming that $a$ exists, we don't know the nature of $a$ until we find it, correct?

[cloud walrus]

BlazingSaber

[viral mountain]

hmm, I see
Wish the book would've mentioned this

[slim kayak]

Not to be rude, but I think thats reasonable to expect to figure it out

Like (a^-1)^-1=a

[viral mountain]

Not sure if it's supposed to be obvious, but I don't know anything about ordering

[slim kayak]

Ah, yeah the ordering is inherited from Z itself

Without it you cant really talk about "positive" elements

[viral mountain]

No, I meant literally I don't know what ordering means

[thin bough]

I don't understand the closure axiom for a group.

Let the symbol * be a binary operation on a set G. A binary operation means the calculation that combines two elements to produce another element. For example, addition 1 + 2 = 3 is a binary operation because it combines two elements, 1 and 2, to produce another element, 3. And the symbol * can be any binary operations including subtraction(−), multiplication(×), etc.

what does the symbol * mean ?

[keen loom]

Sure ya do, it's the thing the inequality symbol gives you

[slim kayak]

Yeah, thats as good as it gets without going into theory

[viral mountain]

Yeah I have a vague idea from analysis, but besdies the < symbol I literally don't know how to define an ordering

[slim kayak]

you have some method by which you can use < satisfying all the properties you know and love

[keen loom]

A total order is really the exact thing you already know

[slim kayak]

Usually you just go "a < b iff a and b fulfill P" where P is some property

[viral mountain]

a relation basically

[slim kayak]

and then you can check that it mets the axioms of what an ordering is (it acts like the < sign from the real numbers)

Its a relation fulfilling some rules, yeah.

[cloud solar]

How can I prove this? Can you explain please? I just dont understand.

[slim kayak]

Idk, I was suggesting some different thing you could try to approach

[cloud solar]

We can have one A_i with 1 nilpotent and other A_j with 4 nilpotents and the rest of them with only 2 nilpotents for example and the product of the numbers of nilpotents would be exactly 2^k.

I proved that if there is some A_i with only 1 nilpotent, then there is an A_t iso with Z/2Z

[slim kayak]

wait how

[cloud solar]

Above

[cloud solar]

.

[slim kayak]

its a function f(a,b) with the property that f(f(a,b),c) = f(a,f(b,c))

you can forget the second half since it just says * is associative.

[cloud solar]

So the question is all A_i's must have exactly 2 nilpotents or there is a chance that some A_i can have only 1 nilpotent?

[slim kayak]

do you have a picture of the exericse?

[cloud solar]

It is not an official problem. It is just some problem I was thinking about. If A is a finite commutative ring with # idempotents = # nilpotents = # invertible elements, who is A?

And the idea with the product of A_i's is from @rustic crown . I think he said each A_i has 2 nilpotents, 2 invertible elements and 2 idempotents, but how? Cant we have one A_i iso to Z/2Z. This is what I understood. Correct me if I am wrong.

[slim kayak]

Commutative rings with unity?

[cloud solar]

Yes

[slim kayak]

so the idempotents are fixed to be 0 and 1. The invertibles must include 1, so we have some self-inverse a. And now we also have a single nilpotent element b

besides 0

[cloud solar]

From where did you get that has 2 idempotents, 2 nilpotents and 2 invertible elements

[slim kayak]

just looking around

[cloud solar]

So we dont have an answer

@rocky cloak what do you think?

[rocky cloak]

Didn't you already ask this @cloud solar

[cloud solar]

Yeah but cant we have one A_i iso to Z/2Z, where A iso A_1×A_2×..×A_k ?

[rocky cloak]

Yes....

F2 is just a different name for Z/2Z here

[cloud solar]

So now the problem is not only about finding the finite commutative rings with 2 nilpotents, 2 idempotents and 2 invertible elements. If there is some A_i iso to F2 then there must be A_j in that product where A_j has at least 4 nilpotent elements

So idk if we can classify this type of rings

[rocky cloak]

Yeah, so what you do is choose A_i to be local rings with residue field F2. That will give you rings with the same number of nilpotents as units. But typically not that many idempotents, then you just tack on a few more F2 until you have enough idempotents

If you want to describe all local rings with residue F2, that will be a lot.

Things like F2[x, y]/(x, y)^2 and also Z/4Z, but also things like Z/4Z[2x] and a bunch of crazy stuff

[thin bough]

what does the . mean

[barren sierra]

the group operation

so when you define a group, it's a set + this operation satisfying certain conditions

so if G is the integers, then that dot could represent addition of two integers

or if G is a vector space, that dot could represent addition of two vectors

or if G the set of invertible 2 x 2 real matrices, that dot could represent multiplication of two matrices

etc etc

those are all examples of groups

[thin bough]

I see

Could you define your own operation ?

or does it only have to be addition and multiplication

[barren sierra]

well you'd have to show your operation satisfies the other conditions

but yes, in theory it could be anything

I just gave examples you are hopefully familiar with

[thin bough]

i see okay thanks

[coral spindle]

Note that this should be clear from the definition of a group!

You should take this opportunity to reread the definition.

[barren sierra]

(I think this is the definition lol)

[coral spindle]

There should be a line saying something like "a group is a set G with an operation . such that ..."

[barren sierra]

like I'd bet right above that screenshot is a sentence along the lines "A group is a set G with an operation * satisfying the following axioms"

lol

[coral spindle]

Is there grad school for psychics or sth

I should apply

[barren sierra]

I read that as physics

lol

[coral spindle]

ew no

[coral spindle]

Anyway, notice how the paragraph above your screenshot also answered your question.

[celest furnace]

So based

[thin bough]

I am trying to prove that the set of integers modulo 13 form a group with respect to addition

is there any association between prime numbers and it forming a group with respect to addition ?

[coral spindle]

With respect to addition?

No, the particular value of 13 does not matter.

[thin bough]

i see okay

Am I doing this aright?

Where M=the set of integers modulo 13

[barren sierra]

Looks good to me

[thin bough]

thank you

[cloud solar]

G is an abelian group

How to prove that for every n>=1 there is an unique subgroup H with n elements?

I was thinking using the fact that a matrix A is in subgroup of order n if A^n=I2

[formal ermine]

what is R

[cloud solar]

And then using the binomial expression

[cloud solar]

The set of real numbers

[viral mountain]

If $G$ is a cyclic subgroup, does it make sense to define the order for every one of its elements, or only for the element $a \in G$ s.t. $a$ is a generator of $G$?

[cloud walrus]

BlazingSaber

[celest furnace]

Each one

You define the order of an element a in an arbitrary group G as |<a>|

For example the order of 5 in Z/15 is |<5>| = |{0,5,10}| = 3

[viral mountain]

Right yeah fair

I forgot that <a> exists for every element of a group

I thought only those elements which generate the entire group have <a> defined for them

[cloud solar]

If H1, H2 two distinct subgroups of order n, because G abelian H1,H2 are normal so H1H2 subgroup of G and ord(H1H2)>n (idk if this helps)

[rotund aurora]

||I think G is C||?

Ye

[glossy crag]

@rotund aurora I see you're back to being a croq.

[south patrol]

A1 is also redundant lol

[tiny jolt]

bump

[long obsidian]

Hey I'm confused about the terminology of normal rings versus integrally closed rings. Are they exactly the same thing?

[slim kayak]

Depends on who defines it, normal rings on stacks are the same thing as integrally closed rings

What wikipedia defines as normal rings is that the localizations at all prime ideals (not just (0), if R is a domain) are integrally closed rings

[cobalt heath]

I know I am out of nowhere but; Thanks Kerr, now I do not have to look up wikipedia/hartshorne to learn what normal rings are

[formal fractal]

Let G,H be solvable groups. Show G x H is solvable

Can i do this constructively

<@&286206848099549185>

[barren sierra]

Do you know that G is solvable if and only if for all normal subgroups K, K and G / K are both solvable?

if not, that would be worth proving and using that

maybe there's a more direct way but idk

[static mauve]

I have the book "A First Course In Abstract Algebra" by Fraleigh. Im wondering what sections/chapters of this book might be found in your average introductory undergrad course. Im essentially looking for a knowledge base that goes deep enough to advance my studying in Topology and Diff Geo.

Here is a breakdown for the high chapters
-Groups and Subgroups
-Permutations, Cosets, and Direct Products
-Homomorphisms and Factor Groups
-Rings and Fields
-Ideals and Factor Rings
-Extension Fields
-Advanced Group Theory (maybe this ch name is a bit too vague)
-Groups in Topology
-Factorization
-Automorphisms and Galois Theory

[scarlet estuary]

undergrad courses tend to be paced fairly differently from school to school

i'd consider fraleigh to be roughly the contents of a two-semester introduction to abstract algebra

some schools might do it in one semester but that'd be fast paced

[barren sierra]

that'd be a fast one sem IMO

[static mauve]

I see. So maybe like the first half is roughly the "bare basics"?

[barren sierra]

yea probably

at minimum first 5 of what you listed

[scarlet estuary]

kinda? but like

[barren sierra]

although I'm curious what makes factorization come so late

[scarlet estuary]

i'd expect many schools to instead group the chapters by subject

[barren sierra]

and what is "advanced group theory" lol

[scarlet estuary]

i.e. there might be a group theory course that does the first 3 chapters and then the last 4, save galois theory

and then a ring/field theory course that does the rest

[static mauve]

I can send some pics to get into specifcs 😄

[barren sierra]

If my first abstract algebra course was just group theory

I would not have done a second

[cobalt heath]

Idk, maybe group homology stuffs?

[barren sierra]

no way homology is appearing that early

[scarlet estuary]

anyway

[scarlet estuary]

as for what you need for topology

you mostly just need the group stuff

to get started

ring/field theory are relevant later but initially you just need comfort working with groups

[static mauve]

@scarlet estuary cool! well that is definitely helpful!

I didnt want to spend TOO much time going off on a tangent with this book, but maybe starting with just groups to start is a good idea 🙂

[scarlet estuary]

yeah so like for context my undergrad did 1-3 and 7 in a dedicated group theory course

4-6 and 9-10 in a dedicated ring/field theory course

and then 8 in introductory algebraic topology

(actually i think at the end of general topology, but same diff)

[cobalt heath]

Idk, an algtop prof said that homology is the easiest thing possible in math

[static mauve]

I did plan on going through Lees Topological Manifolds book... so maybe that should cover chapter 8 of Raleigh 🤷

[scarlet estuary]

for general topology you dont really need anything besides basic familiarity with the properties of common fields (namely ℝ)

for algebraic topology, you need group stuff to start and further topics in algebra are gradually introduced as you develop the machinery to study them

[static mauve]

I have some familiarity with basic Topolgy via Apostols Real Analysis book

[scarlet estuary]

for differential geometry, you only need algebra insofar as you need comfort with vector spaces

so a bit of field theory is helpful but not super necessary

[static mauve]

Yeah, im going through an intro diff geo book (not Lees) and I havent really needed any algebra

Well thanks for the input! I do appreciate it

[sinful kite]

Fraleigh mentions in the preface how he covers the material across two courses and what he includes/excludes. And, there should be a chart of topics as well there.

Actually he tallks about only one intro course. In the 8th edition, Neal Brand talks about the two course split. If it helps you I can post the snippet here...

[static mauve]

Sure, that would be awesome! 😄

[sinful kite]

This is Neal Brand's take btw.. for Fraleigh's you can look at your edition's preface

oh wait the section numbers are going to be all different 😐

[static mauve]

Ah right. No worries, let me look a bit harder at the preface. I didnt initially see a breakdown for what Fraleigh thought was a good breakdown per semester

[sinful kite]

ok so the instructor's preface and the ToC are available on amazon.com as samples.

[formal fractal]

Can anyone else help

[rocky cloak]

Did you consider what spamakin said?

If N is a normal subgroup of G and both N and G/N are solvable, then so is G. This should actually be immediate from the definition, so try carefully writing down what the definition of solvable entails.

[meager fractal]

Z/6Z ~= Q / Z/2Z

can I conclude from this that Q is abelian?

[coral spindle]

I think Q is an odd choice to name a group, I thought for a moment you were talking about the rationals!

[meager fractal]

it is

call it M?

[coral spindle]

Why not G? :)

I'm thinking about the question still, I do think you should be able to conclude it is Abelian but I would like to double-check

I believe the dicyclic group of order 12 is a counterexample to this claim.

[meager fractal]

Z/6Z x Z/6Z?

[coral spindle]

No.

A presentation for the dicyclic group of order $2n$ is given by $\langle a, x \ |\ a^{2n} = e,\ x^2 = a^n,\ xa = a^{-1}x\rangle$.

[cloud walrus]

Boytjie

[coral spindle]

The subgroup $\set{e, x^2} = \set{e, a^n}$ is the centre, according to GroupProps.

[cloud walrus]

Boytjie

[coral spindle]

Then evidently the quotient is cyclic of order 6, and the dicyclic group is not Abelian.

[meager fractal]

hmm

[coral spindle]

Typo! This should be of order 4n not 2n.