groups-rings-fields

[woven delta]

It is

Prove it

[ivory dust]

Wait that is the structure?

[woven delta]

If it's easy

[ivory dust]

Gimme a sec I'm in the middle of something

Just look at the generators though

[woven delta]

😮

[ivory dust]

Like it follows from the free abelian group

[woven delta]

What does this mean

[ivory dust]

There's a homomorphism from the free abelian group over n generators to any abelian group with a minimal generating set of cardinality n.

(I could write this out more formally but I'm walking to dinner right now)

I think that works though. I'm pretty sure it's proven in section 1.11 of Jacobson

[woven delta]

I mean literally that's what a generating set is

So sure

This is not a novel statement

But go on

[ivory dust]

So each element of the group is a_1^k_1 a_2^k_2 ...

Then that's literally just a direct sum

[woven delta]

No

You're wrong

[ivory dust]

What am I missing

[woven delta]

It's not unique

You haven't proven anything

[ivory dust]

Oh not neccesarily unique

Like it is unique but I haven't proven it

[woven delta]

So prove it

Then you can say it's easy

Until then don't

[ivory dust]

Ok fine

Like I said I'm at dinner

[woven delta]

So prove it later

But actually the proof is a bit hard

So I doubt you'll be able to do it

[ivory dust]

Oh then I think I never saw the proof

But I didn't need it for this problem anyways right?

Or is my proof wrong

[woven delta]

Idk

The point is don't say shit is easy

Until you've proven it

[ivory dust]

I mean I thought I had proven it

So that was my mistake

I'll try to prove it later though

And probably fail

[woven delta]

I'm not sure what your proof idea even is tbh

[ivory dust]

I don't have one

I thought something was a proof when it only proved existence and not uniqueness

[woven delta]

So you're saying there is a map from G to the product of some cyclic groups

So why is that map well defined

Thats where uniqueness comes in

[ivory dust]

I'm saying there's a map from the product of some cyclic groups to G

[woven delta]

Sure, Z^n

But that map is not an isomorphism

[ivory dust]

Well it is right? I just didn't prove it

[woven delta]

The existence of the map from Z^n is trivial

But it doesn't do anything for you yet

[ivory dust]

Well it's true that there is an isomorphism between any finitely genrated abelian group and a direct product of cyclic groups right? The proof is just much higher level that I thought I was

[woven delta]

Yeah that's the statement of the theorem

But it's not so easy

[ivory dust]

Right that makes sense

[merry pollen]

Let f be a monic integer polynomial, let g be another integer polynomial. If f divides g in Q[x], then f divides g in Z[x]

I don't understand this proof

"since f is monic we can do division with remainder in Z[x]: g = fq+r, and this equation remains true in the ring Q[x] and gives the same result. In Q[x] f divides q. Therefore r=0 and f divides g in Z[x]

[mild laurel]

What exactly do you not understand?

[merry pollen]

why does the fact that f divides q in Q[x] imply that f divides g in Z[x]

and why is monic important

[mild laurel]

monic is important because otherwise division won't work correctly, i.e., dividing x^2 by 2x would be x^2 = (x/2)(2x) + 0 but that doesn't work in Z[x]

[merry pollen]

sorry why wouldn't that work in Z[x]>

[mild laurel]

That's a typo, it should say "In Q[x] f divides g"

because (x/2) is not in Z[x]

[merry pollen]

oh right thanks

but what about the first question

[mild laurel]

That's a typo, it should say "In Q[x] f divides g"

[merry pollen]

whats a typo?

[mild laurel]

"In Q[x] f divides q."

[merry pollen]

uh isnt that what I wrote

[mild laurel]

And I'm saying that's a typo

[merry pollen]

oh the q and g yeah

I think I see now thanks

[lavish gale]

cyclic group makes thing hard 😱

🤕

[chilly ocean]

why, aren't cyclic groups 'easier' because of their structure?

[lavish gale]

well they r easier but i dont uderstand well enough to solve problem 😅

[mild laurel]

If you don't understand cyclic groups I'm not sure you understand any groups

[lavish gale]

😂

[solemn rain]

xddd same curiosity zero

[upper pivot]

cyclic groups are... what do you mean you dont understand them??

[simple agate]

if I have a field K, and f,g non-zero are in K[X] and h = gcd(f,g) isn't this question just bezout's identity? it seems trivial lol

[mild laurel]

Yes it's just bezouts

[simple agate]

my lecturer listed this under the intermediate questions had me questioning myself

thanks

[mild laurel]

But that doesn't mean it's easy

I mean the proof of bezouts for integers isn't all too easy, you have to invoke the well ordering principle

[simple agate]

we wouldn't be expected to prove that in this class lol

[mild laurel]

I mean sure

But I'm just saying that proving bezouts for polynomials might not be too easy

Since proving bezouts for integers is not easy

[chilly ocean]

isnt bezouts like one line proof? f = g(X-a) + c and its clear that f(a) = 0 iff c =0 which proves that for a in K f(a) = 0 iff (x-a) | f .

assuming K field

[solemn rain]

@upper pivot idk im.trash

[chilly ocean]

I guess bezouts also said that number of roots of f is <= deg f but it follows with induction iirc

[mild laurel]

wait what

how do you get that f = g(x-a) + c?

[chilly ocean]

K[x] is euclidean

where c is polynomial of degree 0

[mild laurel]

okay you're using g for something else here

how does this show bezout's then

[chilly ocean]

f(a) = 0 iff (x-a) divides f right

thats what we want to show

[mild laurel]

thats not bezouts no

[chilly ocean]

what is bezouts then?

[mild laurel]

bezout's identity

or his picture

[chilly ocean]

wow I didnt know THATS called bezouts identity. Sorry, I didn't know that has a bezout name, the only big bezout theorem I ever heard was the polynomial one

[mild laurel]

I've never heard that called bezout's

[chilly ocean]

LOL really?

[mild laurel]

nope

[chilly ocean]

What I meant by Bezouts theorem is known as polynomial remainder theorem

and that bezouts identity I always identified with euclidean algorithm, never heard the actual name

[woven delta]

Why does bezout for integers require well ordering theorem?

That doesn't seem right

Bezout is gcd(a, b) = na+mb for some n, m right?

[chilly ocean]

yes

yeah I dont think it requires it

[woven delta]

So you can prove that for integers easily

[chilly ocean]

couldn't you just like do the euclidean algorithm and go back?

[woven delta]

Wlog assume gcd(a, b)=1, then use euler's theorem

Which is just Lagrange's theorem for finite cyclic groups

That definitely doesn't require well ordering theorem

Also Euclidean algorithm works

@mild laurel what do you mean?

[mild laurel]

Can you prove that euclidean algorithm or more precisely, that Z is euclidean without well-ordering?

[tepid citrus]

do the euclidean algorithm

and go backwards

do you mean prove that it works?

that's what being an euclidean domain means

[mild laurel]

you can't show that Z is euclidean without well-ordering I think?

[chilly ocean]

The only proof I found uses the fact that integers form an ordered integral domain

[woven delta]

@mild laurel you can use eulers theorem

Like I said

And that's just Lagrange's theorem for finite cyclic groups

So no well ordering problems

[mild laurel]

Sure but you said Euclidean algorithm also works and I was disputing that claim

But I also don't see how to use Euler's theorem here

[woven delta]

So you assume wlog that a and b are coprime

Otherwise divide the gcd out

Then by eulers theorem you know a^phi(b) cong 1 mod b

So a(a^(phi(b)-1)) cong 1 mod b

So there exists some n so that a(a^(phi(b)-1))+nb =1

And you're done

[tepid citrus]

all you need for euclidean algorithm is the fact that you can divide

[mild laurel]

Yes and I think division with remainder only works because of well-ordering

[woven delta]

Anyway zoph do you see now?

[tepid citrus]

there's a finite amount of naturals x such that ax <= b for fixed a,b

you don't need well ordering

unless you mean the fact that it terminates

I mean there's an order going on

and "well ordering" is equivalent to elementary properties of natural numbers

I don't know what you mean by proving well ordering

it's like saying proving induction

[woven delta]

Yeah for the naturals this is a weird hill to die on

[mild laurel]

Yeah liquid that works

[tepid citrus]

for polynomials k[x] it's even easier

degree goes down in each step

so it terminates

[mild laurel]

I'm not sure where I talked about proving well-ordering

[tepid citrus]

the point is you misled Godel

and what he said immediately gives the identity

[mild laurel]

I'm not sure how I misled him, he was trying to prove something that was not Bezout's, which he admitted, even if it does immediately lead to Bezout's

Also this is probably a dumb question, but how do you show that there are a finite number of naturals such that ax <= b, because the only way I can think of uses well-ordering

[bleak abyss]

I don't understand why the use of well-ordering is particularly relevant

[tepid citrus]

ab > b

[mild laurel]

Liquid claimed that you don't need it to prove bezouts

[bleak abyss]

Scissors: to be fair you'd probably induct at that point, again uses well-ordering

[tepid citrus]

what do you mean

[bleak abyss]

I mean induction and well-ordering are the exact same thing

And you'd say something like

[tepid citrus]

induct where

to prove what

[bleak abyss]

ab > b, and if x > b then ax > b. How? By induction

[mild laurel]

@woven delta even using Z/bZ implicitly assumes that division works I think, which I'm arguing that requires well ordering

[bleak abyss]

I'm guessing that's what you're trying to do

[tepid citrus]

I mean I was just saying ab>b and there's finitely many numbers less than b

but if you wanna prove arithmetic of naturals then you need its properties

sure

[bleak abyss]

"I mean the proof of bezouts for integers isn't all too easy, you have to invoke the well ordering principle" seems to be the first instance of well ordering being mentioned in this convo

And like

[mild laurel]

Yeah, and I was just bringing that up to show that proving bezouts might not be super easy, not anything about well ordering in particular

[bleak abyss]

I mean well-ordering isn't really not easy tho

[mild laurel]

I was thinking that the idea of trying to use well ordering to prove bezouts isn't obvious

[woven delta]

But the actual proofs of bezouts are very easy

[bleak abyss]

I mean practically anything you wanna prove about integers that isn't like

Oh this follows from the fact that addition is exists

Will use either induction or well-ordering

[tepid citrus]

addition uses induction too

it's such a weird thing to say

[bleak abyss]

Or I mean successor functions

[tepid citrus]

and it's led to such a weird conversation

[woven delta]

Yeah anyway this sort of foundational thing is just not worth thinking about

And let's all forget we had this discussion

[mild laurel]

Yeah okay true, I guess the only proof I knew used well ordering so that's why I thought bezout for integers might be hard to prove

[tepid citrus]

yeah i'll drink some bleach

[mild laurel]

Yeah someone teach me algebraic geometry now pls

[tepid citrus]

affine schemes just means opposite category of rings

[bleak abyss]

In the beginning you have solutions to polynomials

[tepid citrus]

you glue them to make schemes

[bleak abyss]

And that's mspec

But you remember that if you wanna think about non-algebraically closed fields, which you do because of number theory, then mspec doesn't cut it

Lo and behold there's Spec

[tepid citrus]

and you pick other scary words to rename things that are easy

like "quasicoherent sheaf" just means module

[bleak abyss]

And now let's manifold some Specs together because idk may as well copy differential topology

And you have a scheme

qed

[chilly ocean]

□

[mild laurel]

Wait I have an actual question though

If you take the sheaf O_X(1), it seems like this is supposed to be an O_X module but I'm not sure how this works out

[tepid citrus]

it's locally O_X

and the transitions play nice

[mild laurel]

Oh so it's just O_X being an O_X module locally?

[tepid citrus]

yes

[mild laurel]

Ah that makes sense

This works for all n too

[chilly ocean]

just do 4/cosx2

[willow garden]

could someone explain to me the difference between an automorphism and a permutation?

I've got that an automorphism is an isomorphism from G -> G, and a permutation is a bijection from a set to itself.

it seems to me that the only difference is that a permutation is when it's a set, but an automorphism is when it's a group?

[mild laurel]

vaguely

automorphisms must still be group homomorphisms

You could still consider permutations of a group, like how right multiplication is a permutation on a group but not an automorphism

[willow garden]

I'm trying to wrap my head around that

if it's a permutation, it's a bijection. and doesn't right multiplication define a homomorphism?

(I could be wrong about that)

[mild laurel]

Well maybe I should be clearer

Take some nonidentity element g \in G

[willow garden]

taken.

[mild laurel]

Consider the function f: G \to G, f(h) = hg

This is what I mean by right multiplication

[willow garden]

ok

yeah, I'm with you

ah! gotcha

ok

[mild laurel]

This isn't a homomorphism

[willow garden]

yeah, I can see how it wouldn't be.

all right, that helps make things a lot clearer

[mild laurel]

But it is a bijection(permutation)

[willow garden]

and this is what D&F means when they say that Aut(G) ≤ S_G

it's a subgroup for sure, but not necessarily the same as the whole group

because there (may) exist permutations that don't represent homomorphisms

that really helps clear things up for me. in particular, I was trying to get what the book's saying about a group acting by conjugation on a normal subgroup, and how it's "structure preserving"

[mild laurel]

Yes that's true

[willow garden]

I wasn't really clear on why that's true. but given your explanation, it's a lot better now, thanks

[mild laurel]

I'm not sure what they mean by that last bit either but maybe context would help

Usually structure preserving just means its a homomorphism

[willow garden]

well like, elements of order n map to other elements of order n, and so forth

true for an automorphism, not necessarily true for a permutation

[mild laurel]

Sure

[solemn hollow]

But isn't the determinant of a strictly triangular matrix 0?

[mild laurel]

I think they want the subgroup with 1's on diagonal but yes you're right

[solemn hollow]

ah that makes sense cool thx

[fading wagon]

@solemn hollow It's just that all the entries on the diagonal are nonzero

it's not just the subgroup with just 1s

[bitter mauve]

can i get help pls

[mild laurel]

What have you tried

[bitter mauve]

hmm... so i looked at the examples lmao

to see how being a subset of nil(A) matters

didnt really get how it works

[potent birch]

what would a unit of the quotient ring be in this case?

[bitter mauve]

1+I

obviously

i looked at Z/8Z and mod it by {0,4}

which is a subset of {0,2,4,6}

works as said

dunno what else to do

[mild laurel]

Haven't tried it, but you might be able to do this one directly

[bitter mauve]

checked that if a maps to a unit in A/I then there exists a b s.t. ab=1 mod I

[mild laurel]

i.e., if a + I is a unit in R/I, then there exists some b such that ab - 1 \in I

And then show that a times b to some power maybe is just 1

[bitter mauve]

where did we use the nilpotent thing here

[mild laurel]

idk I should probably check this actually works

Yeah

Use the fact that ab - 1 is nilpotent

[bitter mauve]

oooo

thanks a lot

[waxen iron]

Not a question but I'd like to thank everyone in this server that helped me during this last semester. With your help (and a lots of youtube lel)I was able to not only pass the class but more importantly feel like I reasonably understand the basics of Group theory!`
Again, many thanks!

[willow garden]

👍

[outer estuary]

How do I calculate the cokernal of an mxn matrix?

[mild laurel]

What exactly are you confused about?

[outer estuary]

Just the general algorithm @mild laurel

[mild laurel]

Well do you know how to calculate the image of a matrix

[outer estuary]

Row reduction?

[toxic scaffold]

I don't think so

[mild laurel]

It works, but you have to be careful

[toxic scaffold]

isn't the image of a matrix just the span of the columns?

[mild laurel]

The idea is that you row reduce and see which columns have pivots

Sure, but that doesn't give you a basis

[outer estuary]

But reduced it gives kernel as well

[mild laurel]

No reason to think it'll work the same for the image

Idk, you should just google how to find the image of a matrix

[unique prism]

Yooooo, so I'm taking the first of two classes in abstract algebra at my college. My professor is total garbo and I've gotten to the point where I have to basically read the textbook to teach myself all of the content but the first test is on tuesday on all of group theory (or at least everything we have done so far). This test includes the basics such as the Euclidian algorithm as well as more complex stuff like dihedral groups, cyclic groups, and so many proofs I want to cry. I hear from other students that if you can understand the basics you can start to understand everything else soooo,

my question for <@&286206848099549185> is do you have any tips of understanding these basics so I can do well and ace this test? Much thanks and many love to anybody who can answer my question

*tips for understanding the basics

[mild laurel]

Okay first read #❓how-to-get-help

I mean of course you want to understand the basics before you move on to more complicated things

But its pretty hard to give advice on this general of a question, the advice is going to be the basic advice

of reading the textbook and other sources, doing exercises, etc

[unique prism]

Ok, then to get more specific, i also have a quiz on disjointed cycle decompositions tomorrow.

lets say, for example, a=(1 4 5) and b=(2 5 1)

if i wanted to find ab:
then my current understanding of it is i start with the first element of the second cycle, see what that maps to, (in this case 2->5) then i would look to the first cycle and see if that will map to anything in that cycle (which it would, to 1). and I repeat this process until i get
ab=(2514)

is this correct?

(also sorry for pinging helpers when i shouldnt have)

[mild laurel]

The example you wrote out is correct, but the rest of the numbers aren't correct

You have to repeat that same process with every number

So 5 -> 1 under b, then 1->4 under a

[unique prism]

i figured that the process would be repeated until completion no matter the size of the given cycles

[mild laurel]

I'm not sure what you mean

[unique prism]

wait, i see what you mean

i thought the process was, if a=(a1 a2 a3 ...) and b=(b1 b2 b3 ...)
b1->b2 then if b2 is in a, b2 to what b2 maps to in a
now jump back to b
b2->b3 then if b3 is in a, b3 to what b3 maps to in a

Instead what I'm getting from you is
b1->b2 then if b2 is in a, b2 to what b2 maps to in a (say an)
an->an+1 then if an+1 is in b, an+1 to whatever an+1 maps to in b.
eventually,
ab=(b1 b2 an an+1 bn bn+1 ...)
or, in the case that b2 is not in a, then...
ab=(b1 b2 b3 [check if b3 maps to anything in a and continue from there] ...)

does that make sense? its hard to do not on paper

[mild laurel]

Uh no, I'm trying to say the first one

Maybe check your work, what happens if you apply the first idea starting with 5

Really, you should be thinking of a and b as functions from {1,2,3,4,5} to {1,2,3,4,5}

And ab is just the composition of these two functions

[unique prism]

wait, i see what you mean

[mild laurel]

yeah idk, those lines of code are complicating it

[unique prism]

my prof complicated it with arrows all over the freaking board, man

Is that where its derived from?

so... a(b(2))=a(5)=1 so 2->5->1
and a(b(5))=a(1)=4 so 5->1->4

[mild laurel]

Both the calculations you wrote down are correct yes

But what about those question marks

Go read the definition of cycle notation again

[unique prism]

in this case im saying that from the information given we dont know what things like a(5) maps to, assuming that all that is given are the two cycles a and b

if we were given an actual permutation then i know it would be a different case

[mild laurel]

But you are given an actual permutation

Like I said, go review the definition of cycle notation

[unique prism]

i did, this is just a random example i made with random numbers

unless you are indicating that since they arent indicated in the cycle notation that they map to themselves?

so a(5) = 5?

which, in hindsight, would make sense

sorry, wait, bad example

a(2) = 2

i misread the picture

[mild laurel]

That is how cycle notation is defined yes

[unique prism]

BIG BRAIN

Thank you so much omg

Last question to bother you with. for the quiz, the prof sent an email saying what we would be tested on and in addition to this, he said, "Also compute the order of an element in the integers modulo n with addition and the order of an element in a symmetric group." and im not sure what he means by that

[mild laurel]

What exactly don't you understand

[unique prism]

i know that the order of, say, 9-bar under integers mod 20. its 20/gcd(9, 20) or, in general:
the order of m-bar under integers mod n is "n/gcd(m,n)"

im not sure how to compute the order of an element in a symmetric group

and flipping through my notes i cant find an example of it

[mild laurel]

Play around with it

What's the order of (1 2)

[unique prism]

2? since 1->2 and 2->1?

[mild laurel]

Yep

What about (1 2 3)

Idk, you should go play around with it

see if you can come up with an answer

[unique prism]

it must be the number of elements factorial, so for (1 2 3 4 5 6 7 8 9) its 9! ?

[mild laurel]

What's your reasoning?

[unique prism]

each element has to relate to each other element so its a permutation

i also did sets with 2-4 elements to find the pattern

[mild laurel]

Is this what the definition of the order of an element in a group is?

[unique prism]

no... the order of x in a group, G, is the smallest positive, non-zero integer, n, such that x^n=e is the definition

[mild laurel]

so think again

always check that you actually know what the words mean before you try to do something with them

[unique prism]

if this is the case, then im not sure how (1 2)(1 2) results in the identity

since we already established that (1 2) ^2 = e

[fickle brook]

you've already established that (1 2)^2 = e but aren't sure how (1 2)^2 = e?

[mild laurel]

He's saying that I confirmed that it was the case, but now he's not seeing why it is the case

[unique prism]

lol, yes, because rather than composing theme together we counted the number of relations in the element

that too

[fickle brook]

(1 2) is the permutation that swaps points 1 and 2

what happens when you swap them and then swap them again

[unique prism]

OH, DUH

[fickle brook]

that's right you go back to where you started

[unique prism]

so in the case of (1 2)^2, 1->2 (from a) and 2-> 1 (from b)

so that implies that (1 2) is the identity?

or am i just being more stupid

[fickle brook]

you are being more stupid

[unique prism]

NO! 😭

[fickle brook]

(1 2) itself isn't the identity

(1 2)^2 is

[unique prism]

bit (1 2)^2 = (1 2), right?

*but

so rather than looking to make a cycle turn into an identity, the identity should be the starting cycle?

this is probably easier done with a cycle of size 3: (1 2 3)^6 =(1 2 3)^0

once you have made a complete cycle

then you have reached the identity

shit, its 4 am

i gtg, but thanks for the help up to this point, youve done more than my prof has done in 3 weeks, lol

[fickle brook]

bit (1 2)^2 = (1 2), right?

no! (1 2)^2 = e!

[leaden finch]

can someone help me with my problem, i still dont get it

[mild laurel]

did you think about what I said at all

[leaden finch]

i did but i still dont it

do we have to do this by a proof by contradiction?

[mild laurel]

you can

[chilly ocean]

You need to use the notion of ideals and the theorem that says they are closed under operation modulo

You can derive a contradiction with the set aZ + bZ

[woven obsidian]

this is the proof of the fundamental theorem for finite cyclic groups

I'm unsure why the last row implies m_i | n

I understand that by definition $a^n = a^k$ where m_i | k

[cloud walrus]

AoiKunie:
Compile Error! Click the reaction for details. (You may edit your message)

[mild laurel]

okay so since a^n = a^k, is there some relationship between n and k?

[woven obsidian]

Well k is a multiple of n

n is the order of G btw, it was stated in the beginning of the thm

But in this case k=0 works since everything divides 0. I'm not sure how that gives me information about dividing n

[mild laurel]

I mean

All the elements of H_1 are a^(k m_1) for some integer k, and only those elements

[woven obsidian]

What I have a problem with is different representations of an element

b is in H_1 if b=a^(k m_1) for some k. But does that imply that if b = a^q, then m_1 | q

In the proof we know e is in H_1 because a^(0 m_1) = e. But I'm unsure how that implies m_1 | n just because a^n=e

[mild laurel]

Yes, if b = a^q, then m_1 | q

Because an element is in H_1 if and only if it's of the form a^(k m_1)

By the definition of the subgroup generated by an element

[woven obsidian]

By the definition doesn't it suffice that it can be written as a^(k m_1) for some k

Not that every representation of it as a power of a is of that form

[mild laurel]

Yeah no you're right

[woven obsidian]

So any idea how they drew that conclusion? :/

[prime jackal]

What should I take after AP abstract algebra?

I know it is crazily hard but I can stand something further.

[fading wagon]

what's available?

[prime jackal]

I'm doing it online.

So anything.

@fading wagon

[fading wagon]

hmm, maybe algebraic NT?

if you haven't yet

[prime jackal]

Ok.

It's available.

How complicated would I be looking at?

[solemn rain]

wtf

AP abstract algbera??

@prime jackal ?

[chilly ocean]

Why does linear independence imply a^2=-1 here?

[fading wagon]

@chilly ocean because the components split

so in particular, 2ab=0 and b^2=0

[chilly ocean]

what does that mean?

for components to splkit

[fading wagon]

Suppose there's more than one way for x+y(-2)^(1/4)+z(-2)^(1/2)=-1

then subtract the 2 ways from each other and you have a rational coefficient relation between 1, (-2)^(1/4), (-2)^(1/2), contradiction because they are linearly independent

@chilly ocean

Your two relations are the one by a^2+2ab..., and -1+0+0

[chilly ocean]

oh wait ok yeah of course

I get it

bruh thx I dont know why Im so slow

probably the weed

[prime jackal]

@solemn rain yes.

[solemn rain]

can u send any resources for it?just curious

[prime jackal]

Have you even done AP linear algebra? @solemn rain

https://cty.jhu.edu/online/courses/mathematics/abstract_math.html

Look through requirements. It requires linear algebra which requires calc 3 which requires calc 2 which requires precalc which requires algebra 2.

[solemn rain]

okay lmao

tysm

[prime jackal]

Yw.

[toxic scaffold]

linear algebra doesn't really require calculus IMO

[uncut girder]

It doesn't

It's stupid to make calculus a prereq

[golden pasture]

lol what calc

[prime jackal]

IKR.

[delicate bloom]

you can get around requirements usually by just asking permission anyways

or just sign up for it and no one will check if it's online and the system doesn't deny you

[mild laurel]

@prime jackal what does ap abstract algebra cover?

[prime jackal]

https://cty.jhu.edu/online/courses/mathematics/abstract_math.html

Topics include:

logical reasoning

set theory

properties of functions

binary operations and relations

properties of the Integers

countable and uncountable sets

real and complex numbers

unique factorization of polynomials

[mild laurel]

There's no abstract algebra here

You should learn abstract algebra next

[chilly ocean]

lol

[leaden finch]

can someone help me with part b

[solemn rain]

u=v=1

a=2 b = 3

@leaden finch

[oblique river]

did you just click on the first channel under "advanced mathematics"? I don't think this belongs here.

[chilly ocean]

Idk exactly where to put it

[bitter mauve]

😮 @oblique river

u here

anyway, guys i need help

what is this theorem really saying

like the stuffs on the left is a 0 endomorphism?

I^i doesn't make sense to me tho

[mild laurel]

I is an ideal, and there's a notion of ideal multiplication

Yes, the thing on the left is an endomorphism, and it is equal to the 0 endomorphism

[bitter mauve]

oh, thats what it is

hate this

weird theorem appears out of nowhere

[mild laurel]

This is a very, very, very, very important theorem lmao

You've heard of Cayley Hamilton? For matrices and linear algebra?

[bitter mauve]

yes

[mild laurel]

This is a generalization of cayley hamilton to modules

[bitter mauve]

yea, i read the proof to see if it'll help me understand what the statement is about

[mild laurel]

helps you prove a ton of important theorems

[bitter mauve]

but sth doesnt fit in my head

[mild laurel]

It's really just Cayley hamilton, matrices are endomorphisms of vector spaces

[bitter mauve]

i think ill just have to read this like 10 more times and ill get it

[woven delta]

This is not abstract algebra

[tardy marlin]

where should i post it then?

[delicate bloom]

#elementary-number-theory

[tardy marlin]

danke

[wanton current]

I have a question regarding first order logic. I have a language L which is true in a structure M iff the domain has exactly 5 elements. I have to create a sentence which is satisfied in this condition.
My attempt is that, "there exist 5 distinct elements such that every element is equal to one of these 5 distinct elements."
Is this attempt correct and is there an easier statement I could come up with?

[hot lake]

so you have a language L and you have to build a sentence which is true in a structure M if and only if M has exactly 5 elements ?

[wanton current]

Yea

[hot lake]

then yeah your sentence sounds good

[wanton current]

I am wondering if a simpler inference can be made

[hot lake]

not really

the actual sentence is boring to write lol

all those non-equalities

[wanton current]

Lol, that's why I was looking for a more straight forward one. Still thanks

Got a follow up question

If the problem statement changes from exactly 5 elements to at least 5 elements, how would my solution change?

[hot lake]

actually this is more #foundations than #groups-rings-fields

then you would remove the second part

[wanton current]

that's all I have to do?

[hot lake]

think about it

[wanton current]

It makes sense, I am just trying to think counterexample

[static seal]

in response to some of the "this is not abstract algebra" comments , sometimes abstract algebra and number theory overlap (regarding the module question, not the logic question).

[scarlet estuary]

Liquids comment wasn't referring to the module question

It was some now-deleted #elementary-number-theory question

Elementary number theory would not talk about modules lmao

[solemn rain]

can some1 show me that the 2 definitions

of group actions are equivalent

1st definition :

a group action is a map GxX ---> X : (g,x)--->g*x

such that (gh) *x = g(h *x)

and (e,x) ---> x for all x in X

where e is the identitiy

now the second definition

a group action is a group homomorphism G ---> Aut(G)

if im sure

basically some1 define for me group actions pelase lmao

we say a group acts on X if there exists a homomorphism G--->Sym(X)

where Sym(X) is the set of all bijections on X

[woven delta]

Given a group homomorphism h:G\to Sym(X), define a map (g, x)\mapsto h(g)(x)

[solemn rain]

whats h(g)(x)?

what does this mean

[woven delta]

Given such a map f:GxX \to X and an element g of G define h(g)(x)=f(g, x)

[solemn rain]

okay

g*x basically

[woven delta]

h(g) is a permutation of X

So h(g)(x) is h(g) evaluated at x

Now you have to show that these maps I just defined are homomorphisms with the desired properties

[solemn rain]

if i show h(g)(x)=f(g, x)

is a map with the properties i hav e

im done right?

and the otheer way around

show that the map with such properties

is a homomorph

?

[woven delta]

Yeah

[solemn rain]

got it

ty

[robust coyote]

Is there a good way to solve this problem that doesn't just involve guessing random values?

[hot lake]

euclidean algorithm ?

you can also look for k(1),k(2) and k'(2) and see what you can make of it

[delicate bloom]

this is a classic euclidean algorithm type problem

[robust coyote]

Hm, okay, thanks! I'll look into that algorithm then.

[pulsar briar]

you can check that degree only goes down in the euclidean algorithm at each step

[robust coyote]

I moreso just encountered this problem when solving a different problem in linear algebra. Wondered if there was a better way to solve it than guessing.

[pulsar briar]

so you may assume the k_i are degree just enough to make both sides the same

then k1 is a constant and k2 = aT + b

[mild laurel]

Yeah, and from there you can plug in values for T like 1 and 2 to easily find the coefficients

[robust coyote]

Great! Thanks.

[delicate bloom]

I guess first step if you were to do euclidean algo is $(T-2)^2 = (T-1 -1)^2 = (T-1)^2 -2 (T-1) + 1 = (T-1)(T-3)+1$

[cloud walrus]

Merosity:

[delicate bloom]

and hey, that's also the last step, just subtract now

$1 = 1*(T-2)^2 -(T-3)(T-1)$

[cloud walrus]

Merosity:

[delicate bloom]

done

[subtle granite]

Hi, can I ask a question? Or is the channel busy?

[mild laurel]

think we're done, so go ahead

[subtle granite]

Thank you.

So I have a group G, of order |G| = 15, with subgroups A, B of order |A| = 5, |B| = 3.

I need to show that the intersection A∩B = {1} and that the smallest subgroup containing both A and B is G itself.

I thought of approaching it with Lagrange's Theorem

since |A| and |B| are primes, then elements in A can only have orders of 1 or 5, and similarly for B, they can have orders of 1 or 3.

since no element can have two different orders

can I just follow on to say that A∩B = {1}?

[solemn rain]

try to generlize

[subtle granite]

(I am denoting the group identity to be 1)

[solemn rain]

suppose |H| = m

|K| = n

and gcd(m,n) = 1

show that |H intersects K| = 1

[mild laurel]

You have to be a bit more specific here, basically, you have to A intersect B can only have elements of order 1

[subtle granite]

yes

[mild laurel]

Then you should say that thats only the identity

[subtle granite]

and the only element of order 1 must be the identity

right

ok

it seems awfully wordy

maybe I should get used to that

@solemn rain that's a nice approach too, I like that

[solemn rain]

why

[mild laurel]

Yeah I mean when you're just starting out, its good to write every detail

[subtle granite]

Ahah this is my 2nd course in abstract algebra, did my first course over a year and half ago so I'm super rusty at basic problems

[solemn rain]

woo

cool af

mind after you finish your problem

share any resources?

xd

[subtle granite]

uhh sure lemme see if I can just send my lecture notes lol

[solemn rain]

after ur problems

ofc

[subtle granite]

here

It starts off assuming that people haven't done the intro course

so they go over the basics real quick (Ch1 and 2)

[solemn rain]

yea

thats like similiar to hernsteein

idk

anyways gl

[subtle granite]

ty

[solemn rain]

thats like really similar to ictp

[subtle granite]

what's that sorry?

[solemn rain]

https://www.youtube.com/playlist?list=PLLq_gUfXAnknLXjNSnKKLT4LI1AfTy9PS

same content

but covers field and galois theory

but same for gt

[subtle granite]

yeah this is only a 2nd year course in the UK

[solemn rain]

yeea

gl

[robust coyote]

Oh

So my polynomial question is like solving a linear diophantine equation but using polynomials instead of integers

Extended Euclidean Algorithm to solve the coefficients

Interesting. I vaguely remember doing this during a math summer camp in high school.

[leaden finch]

can someone help me with this one

i did this

[mild laurel]

You really should stop posting these here and post them in #elementary-number-theory

[leaden finch]

this is abstract algebra btw

[mild laurel]

It fits #elementary-number-theory better

[leaden finch]

can you tag me where the section is

i cant find it

[mild laurel]

Just click on #elementary-number-theory

[leaden finch]

nvm got it

lol

[subtle granite]

Hey I need some help. I have to prove that the groups

$$ \mathbb{Z}_p \times \mathbb{Z}p $$ and $$ \mathbb{Z}{p^2} $$

[cloud walrus]

Timo:

[subtle granite]

Are not isomorphic

Where p is a prime number

I’ve not done a proof where I have to prove groups are NOT isomorphic

[south coral]

I think isomorphism should preserve some nice properties of a group such as being abelian, or cyclic

@subtle granite if you could show that one of your groups has some property which isomorphism doesn't change, but the other group doesn't have it, then they can't be isomorphic

[golden pasture]

the order of an element under a isomorphism has to remain the sme

subgroups under isomorphisms are subgroups too

[subtle granite]

@golden pasture @south coral thank you!! I showed that one of the groups can have an element of order p^2 but the other cannot, and since isomorphisms preserve order, I arrive at the contradiction that they cannot be isomorphic

[south coral]

Nice! At this point I'm just thinking aloud and you probably already realized this, but one way you could think of doing this is: you know that $\bZ_{p^2}$ is cyclic, and all cyclic groups of the same order are isomorphic, so if you want to prove that it's not the same as $\bZ_p\times\bZ_p$, the latter can't be cyclic too, and proving this boils down to what you said

[cloud walrus]

Astianthus:

[south coral]

Ugh texit doesn't work nicely with discord markdown of course

[subtle granite]

I want to show that given an abelian group G, of order p^2, where p is prime, that G is isomorphic to either Z_p^2 or Z_p X Z_p.

I have already shown that if G is cyclic then it must be isomorphic to Z_p^2. I am not sure how to to show that if G is not cyclic then it must be isomorphic to Z_p X Z_p.

sorry for so many Qs

[pulsar briar]

1. find an element x of order p, so (x) = Z/p
2. show that G/(x) = Z/p
3. conclude that G = Z/p x Z/p

[subtle granite]

I get 1,2

not sure about 3

[upper pivot]

no that doesnt work.

[subtle granite]

oh

[pulsar briar]

3 follows from abstract nonsense, basically you have shown

0 -> Z/p -> G -> Z/p -> 0

maybe you aren't familiar with this notation

but the idea is, you can show that G is generated by two elements, (x) and a representative of a generator of G/(x)

call it y

[mild laurel]

Uh, the cyclic group of order p^2 has elements of order p, and their quotient is Z/p

[upper pivot]

^

in general

$G \neq G/N \times N$

[mild laurel]

Idk why you're using short exact sequences to try to justify your claim

[cloud walrus]

JohnDoeSmith:

[subtle granite]

I am still very confused. I thought about applying this theorem given to us, but I still cannot make the connection of how to bring in the direct product

[upper pivot]

the trik here is to notice something

[pulsar briar]

modules over a field are projective

that's why it immediately follows

[upper pivot]

what can you say about Z?p^2 and Z/p x Z/p

[pulsar briar]

i'm trying not to use this language because he's obviously not familiar with it

[upper pivot]

what property do they both share

[mild laurel]

This is a module over Z?

Z isn't a field?

[pulsar briar]

over Z/p

[subtle granite]

@upper pivot I'm not sure what you're looking for sorry :((

[mild laurel]

Like you can do your first two steps with Z/p^2Z

In exactly the same way

[pulsar briar]

and not the last

[upper pivot]

well just think about properties of both of those groups

[pulsar briar]

this is probably more confusing for the poor guy

[mild laurel]

Then you need to use some property about Z/pZ x Z/pZ in the third step that isn't true about Z/p^2

[pulsar briar]

the property is that it's a Z/p module

if you don't understand a solution from someone helping you should just ask for clarification or back off

instead of confusing the person

[subtle granite]

oof

[mild laurel]

Why do you think bringing up short exact sequences to someone just starting to learn algebra is even helpful at all

[upper pivot]

he is obv intro to group theory

[pulsar briar]

I wasn't, you forced me to bring it up this way when you claimed the solution was wrong

you can do this in an elementary way

[upper pivot]

using module argument is not gonna help him lmao

[pulsar briar]

this is such a clusterfuck

anyway as I was saying, to do part 3, take x and y a coset representative of a generator of G/(x). note that y in G must have order p, and then show that G = (x)*(y)

[subtle granite]

The link I'm not able to make is that final step

[pulsar briar]

okay so what you have is

two elements x, y of order p

i.e. (x) = Z/p and (y) = Z/p

which are disjoint in G

define a map

(x) * (y) -> G

[subtle granite]

ah

okay

[pulsar briar]

and show the kernel is 0

[subtle granite]

then

by this

we're good

?

[pulsar briar]

yes

[subtle granite]

ok

I got it

thank you

thanks all of you who helped, got a bit mixed up in between but thanks for taking the time :)

[cerulean siren]

Does someone mind checking one of my homework problems? Sorry for the potato quality

(that's a } ., not a 3.)

[golden pasture]

type it out

my eyes

[upper pivot]

hmm this isnt quite right

are all the residue classes you described distinct?

whats the residue class of 4 for example @cerulean siren

[cerulean siren]

not sure what you mean by residue class, do you mean coset?

[upper pivot]

yeah

[cerulean siren]

4 would be in the 4 + I coset

[upper pivot]

but is 4+I different than the other cosets before it?

||hint, 4=(2+2i)(1-i)||

[cerulean siren]

yeah, because the ones before it will have + i by the time it gets to 4

ohhh I always forget about subtraction

let me revisit this haha

[bold furnace]

I know i have to prove transitive, reflexive, and anti-symmetric properties, but i can’t figure out how to prove it

[subtle granite]

Hi, I have a simple Q about normal subgroups. If H is a normal subgroup of G, it means that for any g in G, gH = Hg. Is it then true that g^-1 Hg = H?

[solemn rain]

yees

@subtle granite

[subtle granite]

ah ok

cheers

[solemn rain]

cheers

[solemn rain]

let H = {(1),(12)}

is H normal in S_3

is the only way i can do this is just

write out S_3

and like bruteforceE?

[bitter mauve]

(1 2 3)(1 2)(3 2 1) is not in H

anyway, @mild laurel

[solemn rain]

ty

[mild laurel]

I feel like I'm missing something here, I is an ideal right

[bitter mauve]

just a subset of A

[mild laurel]

Ah that makes more sense

What does it mean to add subsets of A then?

[bitter mauve]

oh damn wait

i mean

let me make it more sensible

$\sum{a_i\omega_i}=0\implies a_i\in \mathfrak{m}$ for all $i$

[cloud walrus]

nnutannep:

[bitter mauve]

is this obvious?

adding sets of A would be all sums, i believe

the book i used didnt say anything beforehand but just used this fact

[mild laurel]

so the \omega's are in M?

[bitter mauve]

yes

they are minimal basis of M

[mild laurel]

Oh right

In that case, you should remember everything not in the maximal ideal m is a unit

If one of the coefficients is a unit, you can rewrite the equation and have like \omega_1 = stuff

which would violate the minimal basis

[bitter mauve]

damn

so it WAS obvious

[mild laurel]

its kinda your typical linear algebra proof

[bitter mauve]

indeed

[mild laurel]

like how you show that if your coefficients aren't all 0, you can rewrite one in terms of the rest

I was thinking about Nakayama's lemma for a while cause this vaguely looked like that

[bitter mauve]

yea im trying to read an application of NAK

Let $(A,\mathfrak{m})$ be a local ring ; then a projective $A$-module is free.

[cloud walrus]

nnutannep:

[bitter mauve]

the author proves for the finite case first. but then there are these horrible lemmas that allows for proof in general case

[mild laurel]

What are you reading?

[bitter mauve]

matsumura

[mild laurel]

Ah

I've never read it

[bitter mauve]

goes on for few more lines

[mild laurel]

lmao nice

[inner holly]

Hello, for this question am I allowed to divide this into cases where I assume 1) x = 0, 2) y = 0?

[bitter mauve]

no

[inner holly]

oof

Oh wait

I can just multiplicative inverse on xy = 0

and show that either of them can equal 0

[bitter mauve]

nope

[inner holly]

now that's allowed right?

[bitter mauve]

what is F

[inner holly]

oh

a field

[bitter mauve]

use that

just assume the opposite

say x not 0

then u can multiply by inverse of x

qed

[inner holly]

Okey dokey, thanks 😄

[cerulean siren]

but is 4+I different than the other cosets before it?
hint, 4=(2+2i)(1-i)
Wait no I'm still confused

what does multiplication have to do with cosets?

I though cosets were formed by the addition group operation

[mild laurel]

Not exactly?

Look at the ideal axioms

[cerulean siren]

what do ideals have to do with cosets?

[mild laurel]

everything?

I mean I is an ideal here

your cosets are based off of I

[cerulean siren]

right

so the cosets of a ring are r + I for r in the ring

using the addition operation I thought

and where I is an ideal

[mild laurel]

R/I is a ring, not just a group

You have a notion of multiplication of cosets too

[cerulean siren]

yes ... but aren't the cosets themselves are formed by adding to elements of R?

[mild laurel]

You can think about it like that sure

Going back to what john said

4=(2+2i)(1-i)

(2 + 2i) is in your ideal I

[cerulean siren]

I'm going to do some easier problems first I think

[pulsar briar]

the problem is what the "..." means in your list

you are only thinking about adding elements of the form k(2+2i)

for k an integer

but elements of I are of the form r(2+2i)

for r = a+bi

I isn't the group generated by 2+2i under addition

[cerulean siren]

ohh I see, I didn't have every element in I

this is a very large ideal

(or I'm missing some pattern)

so far I have { 0, 2+2i, 4+4i, -2-2i, ..., (2+2i)^2 = 8i, 2+8i, -2+6i, ..., (2+2i)(8i) = -16+16i, -14 + 18i, ... } and I'm sure there's a lot more

hmm they're all even

[mild laurel]

how did you get 2 + 8i?

[cerulean siren]

err that should be 2 + 10i

let me redo it lmao

[bitter mauve]

noooooo

[mild laurel]

Maybe as hint

4 + 4i = 2(2 + 2i)

-2 - 2i = -1(2 + 2i)

Also wait, you've calculated (2 + 2i)^2 wrong

[cerulean siren]

damn it

isn't it 4 + 8i - 4?

[mild laurel]

oh wiat no I'm wrong you're right

so 2 + 10i = (3 + 2i)(2 + 2i)

[cerulean siren]

I hate complex numbers

they're so hard to reason about

I got 2+10i by adding 8i to 2+2i

[mild laurel]

Yep

Anyways, look at the things I noted, and maybe you'll see a pattern

[cerulean siren]

you always get 2y + (8+2z)i for y, z integers?

or am I missing a multiplication again

[mild laurel]

Does -2 - 2i follow that?

[cerulean siren]

yes, 2*-1 + (8+(-5*2))i

[mild laurel]

Oh I misread sorry

8 + 2z for integers z is just the same as 2z for integers z

[cerulean siren]

yeah you're right

[mild laurel]

You're right that they are all of that form, but you don't get everything of that form. Like you haven't gotten 2 yet

Maybe let me write out a couple more for you

-2 + 6i = (1 + 2i)(2+2i)

-14 + 18i = (1 + 8i)(2 + 2i)

2 + 10i = (3 + 2i)(2 + 2i)

[cerulean siren]

how did you get 1+2i in I?

[mild laurel]

It's not

[cerulean siren]

oh right it's an ideal

[mild laurel]

Look at the second factor in all the examples I've given

[cerulean siren]

it's always 2+2i, because this is the principle ideal of 2+2i

the real and complex coefficient are always a multiple of 8 apart?

[mild laurel]

That's what I was trying to get at

[cerulean siren]

(for 2+2i, that multiple is 0)

[mild laurel]

Every element you have written down is a multiple of 2 + 2i

And all multiples of 2 + 2i are in the ideal, by the ideal properties

[cerulean siren]

ok, so maybe coefficients that are a multiple of 7 apart won't be in the ideal

I don't know how to show that though

I might do some other homework for a while, I appreciate you being so patient though

[mild laurel]

I don't see what you're getting at? We've completely described the elements in the ideal

[cerulean siren]

well but how do we know that's everything? maybe we're missing something

like I was missing 8i before

I'm just going to say the answer is 8 lol

[mild laurel]

Yeah, I mean I only showed examples

But you have to prove that there are only the multiples of 2 + 2i

[cerulean siren]

that's how we defined the ideal though

there's nothing to prove

the original definition is the ideal generated by 2+2i

[mild laurel]

Ah you defined the ideal like that okay, in that case yeah

That's how we know that the multiples of 2 + 2i are exactly the elements in the ideal

But yeah, you have to go back and think about why 4 = (1 - i)(2 + 2i) matters now

[cerulean siren]

the coefficients aren't a multiple of 8 apart ahhh

so we were missing something

[mild laurel]

I still don't know why you're thinking about those coefficients

We've described the ideal in a very nice way completely

[cerulean siren]

the original question is what is the size of Z[i] / <2+2i>

and I'm trying to find some element that's not in <2+2i> so I can get an intuition

but every one I find ends up being in the ideal haha

[bitter mauve]

just see <2+2i> has elements of form p + (4q+p)i for every p and q in Z

[cerulean siren]

(Z[i] being the gaussian integers, { a + bi | a, b integers })

[mild laurel]

1 is not in your ideal

[cerulean siren]

@bitter mauve same question, how do you know 2 isn't in there?

[mild laurel]

since it is not a multiple of (2 + 2i)

[cerulean siren]

yeah, so I know there's at least 2 cosets, but I don't know if there's more

[mild laurel]

When are two cosets equal?

[cerulean siren]

when they have the same elements

regardless of the names for the cosets

[bitter mauve]

scratch that I meant even p

[mild laurel]

One way to think about that is that a + I = b + I if and only if a - b is in I

[cerulean siren]

ok, that makes sense

so then (2+2i) - 1 = 1+2i is not in I

[bitter mauve]

yes

[cerulean siren]

but my question is how do I know if that's in the same coset as 1 + (2+2i) = 3 + 2i or not?

or better phrased, how do I know if 2 is in I?

(wolfram alpha is failing me btw https://www.wolframalpha.com/input/?i=(2%2B2i)*x+%3D+2)

[bitter mauve]

yea the quotient u got is not a gaussian integer tho

[cerulean siren]

yeah, so it's not conclusive

[bitter mauve]

wut

[cerulean siren]

isn't division a multifunction on the complex numbers?

[bitter mauve]

anyway <2+2i> has elements of form 2p+(4q+2p)i for every integer p and q

[cerulean siren]

2/(2+2i) could have many possible answers

[bitter mauve]

no

it has one answer

[cerulean siren]

maybe I'm thinking of something else

I'm terrible with complex numbers lol

[mild laurel]

Division algorithm might help here

[bitter mauve]

is my argument correct

[cerulean siren]

I'm just going to go with I, 1+I, 2+I, and 3+I, I don't have any more time to spend on this

[mild laurel]

i + I is a coset

[cerulean siren]

so it is

ok, then I, 1+I, 2+I, 3+I, i+I, (1+i)+I, (2+i)+I, (3+i)+I

[bitter mauve]

wait

whats the actual problem

[cerulean siren]

and if there's any others I'll just be wrongn

[bitter mauve]

number of elements in Z[i]/I?

[cerulean siren]

@bitter mauve find the size of Z[i] / <2+2i>

yes

lmao so 8 was right all along

[bitter mauve]

hmm.... i was wrong

[bitter mauve]

I could understand the proof here but the claim that the author makes about the diagram being commutative is new to me. I know that he tells me to read appendix B but its huge. So if anyone could help me understand why that'd be great

i know that a:L_1 --> N exists and phi*a=f because L1 is free and hence projective. but i dont get why the left arrow makes the diagram commutative.

[cinder mountain]

Suppose ϕ : R → S is a Ring homomorphism. If ϕ is onto and R has a unity element 1_R, then S has a unity element ϕ(1_R). If, additionally, ϕ is an isomorphism and S has a unity element 1_S, then ϕ(1_R)=1_S```
Hey, I'm a bit confused by this proposition. I don't really understand the difference in the "additionally" part. if ϕ(1_R) is a unity element in S, won't it already be equal to 1_S? how could 1_s not exist if ϕ(1_R) is a unity element?

also I should disclaim that I am translating the prop to English too, so any awkward wording could be attributed to that

[latent anvil]

Yeah you're right, the first bit+uniqueness of 1 gives you the second immediately

Unless I'm also missing something

If x and y are both multiplicative identities in S, then x = xy = y

[cinder mountain]

I thought of that, but wouldn't ϕ(1_R)=1_S still hold even without that assumption?

[latent anvil]

Which assumption? That φ is an iso?

[cinder mountain]

I was responding to someone that deleted their message

[latent anvil]

φ(1_R) can fail to be 1_S if φ is not surjective

[cinder mountain]

derp

[latent anvil]

Oh sorry

[cinder mountain]

but yeah, thanks

can't help with your problem, sorry 😅

[latent anvil]

Np, good luck with ring theory!

[cinder mountain]

thanks!

[warped bay]

but yeah, i think it's actually saying that the pre-image of 1_S is unique

if under a ring iso

[cinder mountain]

I think that's the only thing it would make sense for it to say, but as it's written, it doesn't seem to be saying that

[latent anvil]

Reposting because it got buried:
Let G = GL(n, C) and let H be the subgroup generated by matrices in rational canonical form. Is G = H? Same question for H generated by Jordan canonical form matrices and C = k a field (algebraically closed if we're thinking about Jordan canonical form)

[wind steeple]

what's rational canonical form ?

[latent anvil]

https://www.math3ma.com/blog/rational-canonical-form-a-summary

Every matrix is similar to a block matrix of the form indicated here

[latent anvil]

Someone on reddit pointed out that Jordan canonical form matrices are upper triangular, so they don't generate everything

Is there an elementary way to show everything in C is similar to a matrix in upper triangular form? I.e. without invoking Jordan canonical form

[mild laurel]

This is Schur's decomposition

[latent anvil]

Cool

And it's very explicit in the 2×2 case

My algebra students have a hw problem this week to show that if G is a finite group and H a proper subgroup of G, then G is not the union of conjugates of H

Then to show this for possibly infinite G but finite index H

Then to find a counterexample in general

They wouldn't let me take the algebra course last year

So I started a study group, which got accredited as an independent study and a senior undergrad helped out with it

It went so well that I'm running it again this year as a mentor

[mild laurel]

lmao wtf

[latent anvil]

It's pretty great

Anyways, G = GL(2, C) and H = upper triangular is a counterexample

H is a proper subgroup of G, but everything in G is conjugate to something in H

The example I came up was G = mobius transforms acting on the riemann sphere. Then every element of G has a fixed point, so the union of stabilizers is all of G, and they're all conjugate, but any given stabilizer is a proper subgroup

But the students won't do complex analysis until spring so I can't use it

[mild laurel]

Those are both nice

[latent anvil]

Yeah. I thought brouwer fixed point gave an example but the action of homomorphisms of the disk on the disk isn't transitive because of invariance of boundary

[mild laurel]

Right

[latent anvil]

I think they might secretly be the same?

A 2×2 matrix is upper triangular iff it has (1, 0) as an eigenvalue

Equivalently, it fixes span (1,0) under the action of GL(C, 2) on the space of lines in C^2

Which means we're really looking at stabilizers of an action of GL(C, 2) on P^1

But P^1 is the Riemann sphere!

Holy shit it is!

The action of a matrix M = [[a b] [c d]] on [z : 1] (considered as a line in C^2) is [az+b : cz + d]

And the action of M on z/1 as a mobius transform is (az+b)/(cz+d)

And M [1 : 0] = [a : c] is the mobius transform f(z) = (az + b)/(cz + d) on the point at infinity

[mild laurel]

Oh yeah shit, that works

[latent anvil]

That's really cool

I came to the same answer in two completely different ways

Trying to find a "geometrical" transitive group action where everything has a fixed point vs thinking about matrix canonical forms

[mild laurel]

math is pretty interconnected

[solemn rain]

math is pretty cool

[latent anvil]

Math is pretty cool

[golden pasture]

cool is pretty math

[solemn rain]

Math is very nice.

[bitter mauve]

Is math related to science?