groups-rings-fields

[coral spindle]

Man, I feel like there's no standard notation

[rotund aurora]

Im not sure there's a notation everyone agrees with tho. From my experience that was the case at least (that people didn't agree)

[silk bramble]

yeah I am surprised

[coral spindle]

I've also seen $\mathcal M^A_B(T)$ and things like this

[cloud walrus]

Boytjie

[silk bramble]

rotman uses this

a pain to write

[rotund aurora]

I have also seen $\text{Rep}_{A,B}(T)$, which I guess is pretty explicit

[cloud walrus]

Croqueta

[cloud walrus]

det

[rustic crown]

me like this

[next obsidian]

me like u

[rustic crown]

me like you too uwu

[flint crater]

What's a good abstract algebra book?

[rustic crown]

a book that works for you

check #book-recommendations for popular books

[flint crater]

Awesome thank you

[long geyser]

bet none of u seen this beautiful notation

[delicate orchid]

if ur mathematics isn't basis independent u suck ngl

[chilly radish]

I wish I could go back to a time where I didn't know this was a thing

truly the evil of mankind is unending

[long geyser]

lol it actually grew on me, definitely looks hideous at first though

[silk bramble]

I now think this is the best for me

[signal fractal]

which book is this? terrible

[silk bramble]

base Delta

[long geyser]

knapp - basic algebra

and as I said I personally like it, makes some important results look simpler imo

[rustic crown]

why do you need notation for that >.<

[long geyser]

you don't need notation to express matrices w.r.t. bases?

[rustic crown]

nope

[silk bramble]

because base changing is a common thing in Linear algebra

[rustic crown]

you can use words

[silk bramble]

good luck with that

[long geyser]

^

[rustic crown]

you can do a lot of stuff basis free

so no need to worry about base change

a matrix is just a linear map k^n --> k^m with standard basis

[silk bramble]

good luck teaching linear algebra without it. just try.

[rustic crown]

if you think about it like that, then no probs uwu

.<

[long geyser]

I think applied ppl would literally punch you in the face rn if they could

[elder wave]

LOL

[rustic crown]

oh i didn't read applied

oopsie

[elder wave]

I saw that

[rustic crown]

.<

[elder wave]

I wouldn't tho

[rustic crown]

[elder wave]

or would I

[rustic crown]

arigato

[elder wave]

[rustic crown]

F

[rustic crown]

anyway, i was trying to say to think more diagramatically instead of notationally

for example if you have a matrix A : k^n --> k^m and you decide to change basis, on both left and right via the invertible matrices P : k^n --> k^n and Q : k^m --> k^m, then the new matrix would be Q^-1 A P, and you don't need to remember where that inverse will go you if you have the diagram in your head

[silk bramble]

everyone knows in linear algebra and diff geo you eventually transition to basis-free and category theory for the theoretical stuff
but that means nothing when teaching someone before that

and when you need to actually calculate things you have to specify a basis

[long geyser]

saying just have a diagram in your head is unhelpful when visualizing that diagram is like one of the major roadblocks to someone learning this stuff

I struggled a lot with change of basis stuff when I was first learning linalg and using better notation unironically helped me a lot

[silk bramble]

yep, good notation ensures correctness and tells you how to calculate correctly for students

Tv_k for k-th column, that kind of thing

[rustic crown]

i got overpowerd >.<

[elder wave]

can't be having that, shoot back

[rustic crown]

okie lemme say my perspective a little differently... this "notational calculus" can be useful to some, but i've always sort of forgotten the details whenever i worked with weird notations. what you call [v]_V, i would just think of as the preimage of v under the map k^n --> your vector space with which sends the standard basis to V. ofc that notation is good, but just the notation without a diagram is far less useful

the equation you wrote would be replaced by this diagram

[cloud walrus]

det

[silk bramble]

as long as you instinctively know what I mean without explanation that's good enough

[wicked patio]

does the second isomorphism theorem for groups allow for some kind of "unique factorization"

like you can chop off quotients in any order, or something

[next obsidian]

Maybe try to more precisely say what you mean?

I think you use the 2nd isomorphism theorem to prove Jordan Holder which is maybe in some really really creative way a statement about “unique factorization”

But it’s hard for me to see what you really mean by that

[barren sierra]

isn't the whole "chopping off quotients" thing the 4th isomorphism theorem?

https://en.wikipedia.org/wiki/Correspondence_theorem
this thing

[somber sleet]

If we have a ring R, when is a principal ideal (a) for a in R, the whole ring R ((a)=R) ?

[coral shale]

recall the defn of (a)

[somber sleet]

Is it only then the case, when a = 1?

[coral shale]

no

[somber sleet]

(a) = {ax | x in R} right?

[coral shale]

yes

so when is this set equal to R

try to have a go using that defn

[somber sleet]

when ax= x for all x in R?

[coral shale]

that is one solution

How do you show 2 sets are equal?

[somber sleet]

(a) in R and R in (a), showing both inclusions

[coral shale]

yes

im thinking

===
So we already know (a) is a subset of R

thats a given

[somber sleet]

ah wait, if a is a unite, then it should also hold orght?

[coral shale]

so we need necessary and sufficient conditions for R subset (a)

[coral shale]

thats better

[coral shale]

so try to prove a is a unit is equivalent to this

I was thinking in terms of a mapping, you want to justify an injection from R to (a)

since u already know (a) subset R

[somber sleet]

let x be in R, than ax in (a), for a unity, this means that there exists an a' s.t. a'a = 1, so we get that x = a'ax which is contained in (a), because a' is in R and ax is in (a). So it just follows by definition of ideal

[coral shale]

ah right thats better

all u need to do is show 1 is in (a)

👌

[somber sleet]

I don't know if it's right, but I'd do it like this

[coral shale]

u show this

a is a unit <=> 1 in (a) <=> (a) = R

[somber sleet]

that makes a lot of sense now, thanks a lot

[somber sleet]

is Z2 an integral domain?

[chilly ocean]

how do I show that f is a bijection, does it suffice to show it is an injection and surjection?

i.e. $f(x) = f(y) \implies \frac{1}{x} = \frac{1}{y}$ and if we multiply by $xy$ we get $x = y$

[cloud walrus]

isomorphism

[tender wharf]

err we can't use that notation

but notice it is injective due to closure and existence of inverses

[chilly ocean]

wdym?

[tender wharf]

you know how bijectivity works yeah

[chilly ocean]

injectivity and surjectivity

yeah

[tender wharf]

ok then injectivity follows from a property of inverses

[chilly ocean]

what about surjectivity

[tender wharf]

which one?

[chilly ocean]

o

hm

[tender wharf]

surjectivity is even easier

[chilly ocean]

a property of inverses huh

hm

[tender wharf]

yup

[chilly ocean]

(xy)^-1 = y^-1 x^-1?

[tender wharf]

not that

can we have 2 inverses?

[chilly ocean]

uniqueness of inverses

[tender wharf]

yep

[chilly ocean]

but how does injectivity follow from that

[tender wharf]

you know how to prove injectivity?

[chilly ocean]

not sure

[tender wharf]

I recommend reviewing the definition

[chilly ocean]

f(x) = f(y) implies x = y

[tender wharf]

yup

[chilly ocean]

so

$x^{-1} = y^{-1}$

[cloud walrus]

isomorphism

[chilly ocean]

can I multiply the equation by $x$ on the right and then $y$ on the left

[cloud walrus]

isomorphism

[tender wharf]

think about it carefully

[chilly ocean]

the inverse of x is x

and the inverse of y is y

hence x = y

right?

[tender wharf]

your left and right multiplication also works was what I meant but yeah

[chilly ocean]

gotcah

what about surjectivity

[tender wharf]

I'm going to leave that one to you since I am walking rn

[chilly ocean]

ok tyvm

[coral shale]

dont use quotient notation for inverse in algebra

unless its actual numbers

[chilly ocean]

why not

[coral shale]

because its just not good practice

[formal ermine]

yes

[tender wharf]

.

[coral shale]

stick to one notation which is ^-1

[formal ermine]

Z/kZ is a field/integral domain iff k is prime

[coral shale]

to denote inverse

[formal ermine]

-x

[coral shale]

1/x is ex^-1

where e is identity

there is no / operation basically

in a group there is only one operation defined

[formal ermine]

why was this sullied

-x is the common notation for inverses in abelian groups

[chilly ocean]

$x^{-1} x = y^{-1} x \implies 1 = y^{-1} x$

[cloud walrus]

isomorphism

[chilly ocean]

then $y1 = y y^{-1} x$

[cloud walrus]

isomorphism

[chilly ocean]

right? @tender wharf

[coral shale]

this is the approach yes
xx^-1y = xy^-1y

[lethal dune]

f ∘ f= id

[nocturne bone]

I’m trying to do this question, but I’m confused on how to define an action of D_12 on the units mod53

[delicate orchid]

is D_12 order 12 or 24 here

[hot lake]

You don't need to actually know the action to answer the question

But yeah "the action" is bad writing if you don't say what it is

[delicate orchid]

actually yeah, it doesn't matter, lol

[nocturne bone]

It has order 24

[nocturne bone]

Hmm okay. Do you have a hint for how to approach this question then?

[lethal dune]

orbit-stabiliser

[blazing viper]

h

[foggy merlin]

does anyone know when singular value decompositions of matrices over noncommutative rings exist?

[lethal dune]

[foggy merlin]

I can also require skew fields instead of rings if that helps

[bleak abyss]

ALright lez do this

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Also elements of C are called chambers

As an example

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

That's an example

Rank of the system is the size of I, a gallery is a sequence of adjacent chambers (no consecutive repetition)

Subgallery is a subsequence of consecutive elements

The gallery has type i_1,...,i_n if c_{k-1} is i_k-adjacent to c_k

And if each i_k \in J, we say it's a J-gallery

Finally, we say C is J-connected if any two chambers can be joined by a J-gallery

J-residues are connected components

Panels are rank 1 residues

Meaning we're taking a bunch of chambers which are i-connected for some i (and are the i-connected component)

So let's look at our example for a sec

[cloud walrus]

Sloth King Daminark

[hot lake]

I thought chambers were the elements of C = G

[bleak abyss]

C = G/B in this case

[hot lake]

ah silly me

[bleak abyss]

All good

[bleak abyss]

So, thing to note about this example is

If gB is i-adjacent to hB is i-adjacent to kB

Well gB is i-adjacent to kB

So if we can find an i-gallery starting from gB and ending at kB, then they're just i-adjacent

[hot lake]

isn't that always the case since each i determines a partition of C

[bleak abyss]

Oh wait tru

So yeah in that case in general, i-panels are just parts of the partition

Good catch

[hot lake]

and you can talk about panels for any subset of I ?

[bleak abyss]

Well those are called resides

Also my internet on my computer died sadge

Back

So i-panels in the G/B example are what? gB and hB are equivalent if gP_i = hP_i. So we index equivalence classes by cosets of P_i

Now let's consider higher rank

What's an {i,j}-residue?

Yeah okay so wait I think the same logic from earlier should be able to say that if x and y are connected by a type J gallery, they are connected by a type J gallery of length |J|?

At least let's see for |J| = 2

So let's say we have a gallery c_0 ... c_n

Actually yeah this logic doesn't work

All I can really say is that if I have a path of shit that's all i-adjacent I can compress it

Good to know

Dogshit internet

I'm actually gonna take potato and throw him at the WiFi if they don't get this shit taken care of

[atomic mesa]

any hints for this?

"order" makes me think that Z_4 x Z_8 mod <(0 2)> is a group

idk if the notation is universal, but i think its the set of all left cosets of <(0,2)> by elements in z4 x z8

[coral spindle]

Yes, it is a group. Is it unclear that this is the case?

Are you in the context of modules? Every module is, in particular, an Abelian group.

[atomic mesa]

no clue what a module is

we've just recently been introduced to left cosets

[coral spindle]

Right. Well, it does indeed turn out that when you mod by certain subgroups, you get groups.

If you have not seen quotient groups yet, this question really shouldn't have been posed to you

[atomic mesa]

oh we did touch on quotient groups actually

[coral spindle]

So in fact it should be clear that this is a group.

[atomic mesa]

oh i see

does it make sense to talk about order of an element if the set isnt a group?

[coral spindle]

No.

Not typically.

[atomic mesa]

got it

now, doesnt (3 1) + <(02)> give multiple elements

[coral spindle]

No.

That is one element of the quotient group.

I strongly suggest you go back to your notes and read about quotient groups.

[atomic mesa]

yh i will, our lecturer told us we didnt need anything to do with normal groups so thats why i thought i could do it

[coral spindle]

There is a good reason that you don't need to worry about normal subgroups. Maybe you should think about why.

Also, there is no such thing as a normal group – only a normal subgroup of a group.

[atomic mesa]

ahhh ok i need ot look back through my notes

slightly unrelated question, if i want to partition a group G into left cosets of H (supbg of G)

say i manage to do it with k left cosets

must k divide |G| ?

[coral spindle]

Do you recall the statement of the Theorem of Lagrange?

[atomic mesa]

yeahh i was playing around with the idea of cosets. lagrange says |H| divides |G|

[coral spindle]

OK

This is not the full statement of lagrange.

[atomic mesa]

oh?

[coral spindle]

Lagrange in full states that $|H|\cdot[G \colon H] = |G|$, where $[G \colon H]$ is the number of cosets of $H$ in $G$, aka the index of $H$ in $G$.

[cloud walrus]

Boytjie

[coral spindle]

QED.

I hope it's clear why I say QED :P

[atomic mesa]

we didnt do this but i think i get why

[coral spindle]

It should be clear if you inspect the proof.

[atomic mesa]

yeah if we partition G into a union of k left cosets (they are distinct because they are each an equivalence class) ?

then look at |G|, and get in the end |G| = k|H|

and observe k is [G:H]

[pliant forge]

I am a bit confused on how to show $A \cong \oplus_{i=1}^n S_i$ as a left $A$-module, where A is a ring of $n\times n$ matrices over a field $k$, and $S_i$ is the i-th column vector of A.

[cloud walrus]

Eso

[pliant forge]

mainly how the module homomorphism would be

(left module)

[rich patrol]

Try 2x2 case

[pliant forge]

how would one think of the vectors as being direct

[rustic crown]

the notation is kinda weird... A is a ring and not a matrix

S_i are just k^n as vector spaces over k

[rich patrol]

Being direct? Wdym

[rustic crown]

ig i see the problem now

[pliant forge]

rf(x) = f(rx)

r here is a matrix right

from A

[rustic crown]

r * [s1 ... sn] = [r*s1 ... r*sn] which is why it's direct

[pliant forge]

you can write the elements of the direct sum as tuples with the column vectors i suppose?

[rustic crown]

yep, more "internally", given a matrix M, you can write it as a sum of matrices M_i where the i-th cloumn of M_i is same as M, but rest other columns are 0

so different i's don't interact with each other

[pliant forge]

yeah

that makes sense

the separation of these isomorphic vectors were bugging me

[white oxide]

why do they need to introduce the part in blue? i don't see how it plays in the proof of case II

so in words they're saying that a^some integer s is equal to a^the remainder of s when it is divided by n

im still confused about how that applies

also why is a^0 = e, a, a^2, a^3? is it becuase they're all less than n so they have to equal a^0? but that doesn't make any sense, because how can, say, a^2 and a^3 be distinct if they both equal a^0

a^2 = a^0 = a^3

therefore a^2 and a^3 cannot be distinct

[formal ermine]

this means

(a^0 = e), a, a^2, a^3

so like

we have a^0, a^1, a^2, a^3

but we know that a^0 = e

this doesn't say anything about a^1, a^2, a^3, ..., etc tho

[white oxide]

wait how does it not say anything about them?

because if we have a^0 = e, a, a^2 ... a^n-1

then if a^0 = e

we have a^0 = a^2 = e

because a^0 = e

[formal ermine]

why?

a^2 = a*a

a = a^1 not a^0

[white oxide]

ohhhhh that's what you mean

the notation they used was kind of confusing

okay that makes sense

[formal ermine]

this is common notation

[white oxide]

but why did they introduce the part in blue?

[white oxide]

nah

never seen it before

[formal ermine]

no idea lol (no idea as in I cba reading all of it)

this proof looks more complicated than it actually is

[white oxide]

yeah this textbook kinda mid ngl

but thanks anyways

imma just ask my prof

[jagged ocean]

Hi any math wizards here

[formal ermine]

what's up

[jagged ocean]

So I thought I chose a Basic math class. I was wrong. Now i'm stuck

[formal ermine]

are you talking about abstract algebra or hs algebra

[white oxide]

bro thought abstract algebra was algebra

[jagged ocean]

Idk the difference

To be honest

[formal ermine]

well what topic are you having difficulties with

or rather what is your question about?

[jagged ocean]

Erm equations and relation with x and y

[formal ermine]

that's #prealg-and-algebra

this is for uni level algebra (groups, ring, modules, fields) lol

[jagged ocean]

Oh

It's in my uni course

[formal ermine]

huh

do you major in mathematics?

[jagged ocean]

Yea i had to choose another class in computer science and i chose math

No computer science

[formal ermine]

can you like state your exact question and I can tell you where it belongs

[jagged ocean]

okay

Consider the equation x^2+(y-2)^2=1 and the relation “(x, y) R (0, 2)”, where R is read as “has distance 1 of”.

For example, “(0, 3) R (0, 2)”, that is, “(0, 3) has distance 1 of (0, 2)”. This relation can also be read as “the point (x, y) is on the circle of radius 1 with center (0, 2)”. In other words: “(x, y) satisfies this equation x^2+(y-2)^2=1 , if and only if, (x, y) R (0, 2)”.

Does this equation determine a relation between x and y? Can the variable x can be seen as a function of y, like x=g(y)? Can the variable y be expressed as a function of x, like y= h(x)? If these are possible, then what will be the domains for these two functions? What are the graphs of these two functions?

Are there points of the coordinate axes that relate to (0, 2) by means of R?

[formal ermine]

lol I thought you meant like "solve for y in y - 2 = x"

I think this belongs in #proofs-and-logic

[jagged ocean]

oh okay

[pliant forge]

it kinda reminds me of discrete math

[jagged ocean]

Do you get it?

[long geyser]

#proofs-and-logic and #discrete-math share a lot of topics, so it can go in either

[pliant forge]

i believe its just showing that for any integer you choose for the exponent of a, it will always lie between 0 and n-1, because one can just apply the division algorithm to this number, s, finding the multiples, say m, of the divisor n and a remainder r, which lies between 0 and n-1. Then since s can be written as mn+r the mn part vanishes in the exponent (its an exponent of the identity).

[glossy crag]

How "big" should an algebraically closed field be in order to accommodate transcendental extensions, i.e. if you have a field K and an algebraically closed field C containing it, are there any conditions on C that would imply it contains all extensions of K? E.g. as far as I can tell, all extensions of Q (not just algebraic ones) embed into C.

[rustic crown]

it can't

you can make a field as large as you want

for example take X to be a set of variables such that |X| = |P(C)| the cardinality of power set of C. then K(X) has cardinality larger than that of C, so no way you could embed it into C

[indigo ridge]

I need to find a
partition of Z into 10 infinite sets, and for each partition, what is the
corresponding equivalence relation?

so I know there's a theorem that states let P be a partition of a nonempty set A. Then there exists an equivalence relation R on A such that P is the set of equivalence classes determined by R

My guess is, make the partition of the set that is 10k and another set 10k+1,...,10k+9 for an integer k.

how can I find this corresponding relation?

[rustic crown]

when are two integers going to be in the same partition?

[coral spindle]

If you have a partition P of a set A, you therefore have a function f : A -> P sending each element to its part. Then this partition is defined uniquely by the equivalence relation a ~ b iff f(a) = f(b)

This is half a joke. If you want to 'find the equivalence relation,' you're going to have to be more specific. If you just want a nice description of it, you're going to have to be creative.

Anyway, I am going to become the joker.

[elder wave]

[indigo ridge]

if those integers are either 10k, 10k+1 ... or 10k+9 for integer k

[coral spindle]

See, I was right

[indigo ridge]

hm?

[coral spindle]

Because I'm annoyed that this isn't algebra-y enough, I'll mention that this is basically the first isomorphism theorem for sets. If you substitute sets for algebras with a particular signature, this (with some interpretation) is exactly the 1st iso

[indigo ridge]

interesting, but I don't know what any of that means at this point ;P

unfortunately.

I vaguely know what an isomorphism is tho

[coral spindle]

I am saying this for other people

What I am replying to is relevant to your question

[indigo ridge]

so my understanding of what you said is given a partition of a set A there exists a mapping from A to P (partition) and it is uniquely defined by some equivalence relation aRb iff f(a)=f(b)

I didn't get the very last part

aRb iff f(a)=f(b)

[coral spindle]

If you have a more specific question than 'I don't get it' I might be able to help

[indigo ridge]

basically what @rustic crown said?

[coral spindle]

If not I suggest you try and prove what I say here

[indigo ridge]

why does the equivalence relation have to be f(a)=f(b)

do you know of any elementary examples so i can imagine it better?

[coral spindle]

An equivalence relation is a set.

[indigo ridge]

Yeah

[coral spindle]

I can possibly describe it in several different ways, but as a set, these ways will all be 'equal' in that sense.

So indeed, the equvalence relation 'has to be f(a) = (b)'

[indigo ridge]

Ok

[rustic crown]

what did det say

[coral spindle]

But as I said here...

[rustic crown]

i said nothing

.<

[indigo ridge]

you said this

[coral spindle]

So true det

[indigo ridge]

Ok gotcha

[rustic crown]

oh okie

[indigo ridge]

@coral spindle so is there not a way where we can define a partition on a nonempty set A and from then on deduce a unqiue relation R on a nonempy set A?

[coral spindle]

That's what I've been saying

I told you what R was

[rustic crown]

so like boytjie says, you ofc told me the equivalence relation, but you have to be a little creative to write it "nicely"

[coral spindle]

to be very specific, it is the set R = {(a,b) in A^2 | f(a) = f(b)}

So you are asking "what is the equivalence relation" but what you really mean to ask is "what is a nice description of this equivalence relation"

And I am replying

"be creative"

[indigo ridge]

Im trying lol 💀

let me just ponder over what you said

i need some time to think.

[rustic crown]

oh ferb

can you tell me a non-trivial polynomial in that set?

[indigo ridge]

Ok I get it now ;P

[rustic crown]

that's not a polynomial >.<

[coral spindle]

Doesn't look like a polynomial to me...

[indigo ridge]

silly ferb

[coral spindle]

That's better

Hey here's a funny hint

We call such functions 'even'; those functions such that f(x) = f(-x)

[rustic crown]

okie, and if it was an ideal then x^2 * g would also be in the set for any g in Z[x]

[coral spindle]

Nvm det's idea is more straightforward

[indigo ridge]

whats so funny about this hint, seems like your jokes are worst than your explanations

[rustic crown]

nuu

-x^2 is also even

[coral spindle]

It's not funny haha, it's funny hmmm

[rustic crown]

yee

[next obsidian]

It’s a subgroup tho 🗿

[rustic crown]

you could show abstractly that every ideal in Z[i] is principal, and then it would be just a consequence of that

or you can compute the generator

gcd

if (1+5i, 5+i) = (d) then d divides both 1+5i and 5+i

[coral spindle]

Euclid's algorithm, what a banger

[rustic crown]

so d divides gcd(1+5i, 5+i)

have you seen how to calculate gcd's so far?

[delicate bloom]

that works

[rustic crown]

yee looks like it

[coral spindle]

There is no algorithm in general that will give you an example given a ring

You're going to have to be creative!

[rustic crown]

lol

[coral spindle]

Hint: x and y don't mix nicely

[coral spindle]

I'm a stuck record the joker

[rustic crown]

.<

too hard to read

[glossy crag]

Shit, I knew this would be the answer even as I typed it out...

Thanks

[rustic crown]

okie so you're trying to give me an ideal which isn't generated by a single element

so the description of it should involve at least 2 generators

else it's principal already

what are the two nicest elements in R[x, y] which might give you a nice ideal

but that's still a single polynomial

you need to give to polynomials that don't mix properly

F

it's already 1:41 am

how time so quick

[next obsidian]

Show that nonzero primes in a PID are maximal

Then exhibit a nonzero non-maximal prime ideal

[rustic crown]

that's so big brain

okie i'll spoil

show the ideal ||(x, y)|| is not principal

[next obsidian]

What does PID stand for

[rustic crown]

because this would give an example of an ideal whcih isn't principal >.<

[chilly ocean]

[rustic crown]

yee

but why make life harder

(x, y) is a simpler ideal to work with

[lavish nexus]

assume it is principal

[rustic crown]

say (x, y) = (d)

get a contra

[lavish nexus]

show something goes wrong

what can d be

[rustic crown]

show math breaks

because you assumed something you weren't supposed to

[tawdry crystal]

What does it mean for two lie algebra to be orthogonal?

[rustic crown]

(idk, maybe [u, v] = 0 for u in one and v in other? this assumes both are sub-lie-algs of a single thing)

idk, i shouldn't >.<

[lavish nexus]

what can d be

can it be x

can it be y

what can it be

in order for b_1 and b_2 to exist

xyb_1=x

What is b_1

There’s no y^(-1) in R[x,y]

[rustic crown]

it not polynomial

[lavish nexus]

Positive powers only

[rustic crown]

it a rational function

[rustic crown]

(non-negative )

[lavish nexus]

ummm

what is b_1

then

[rustic crown]

good luck iteribus

i'll go

.<

[lavish nexus]

you have the same problem

[rustic crown]

sowwy

okie maybe i'll give one hint and leave

show that if f divides g then x-degree of f is smaller or equal to that of g

and same for y-degree

[lavish nexus]

no it must contain x and y

[rustic crown]

x-degree of f := largest power of x appearing in f

[lavish nexus]

ok before anything

do you see any problem with the ideal (5)

[rustic crown]

good luck uwu

is sum of two integers an integer?

🙈

well think about what we need to show

.<

[lavish nexus]

no

[woeful sand]

i don't have a solid background on category.
Is it semantly wrong to call the field of rationals initial?
i have seen a demonstration that it's initial in the fields of characteristic 0.

[lavish nexus]

you don’t know I is an ideal yet

[rustic crown]

yea, it's not initial in the category Fld, but it is initial in the subcategory Fld_0

[lavish nexus]

nvm I misread

[woeful sand]

thank you

[rustic crown]

(if Q was initial, then you would need to have a map Q --> F_2)

[woeful sand]

thats what i thought

[rustic crown]

the category of fields is weird, you never have maps between two fields of different chars

[woeful sand]

why it's so boring to revisit analysis after you get in love with algebra 👀

[rustic crown]

hehe

yep, that's because J was a subset and I is an ideal, so closed under addition, and that sum of integers is an integer

[tulip cedar]

you could look into more algbraic versions of it also

[rustic crown]

it can be (3, x)

but there are many choices for I

[woeful sand]

last time i asked to recomendations for books here people were kinda rude with me
if it doesn't bother you can you pls send me a dm any names of good books like that?

[rustic crown]

which give the same J

[tulip cedar]

i dont know if I know any books specifically for that, except lawvere's which I havent read

but ive been looking stuff up online recently about smooth topoi and the like

[woeful sand]

lawvere is on the recomendations of a national book im reading

sorry for interrupting you guys, have fun.

[rustic crown]

okie i think you should spend more time with getting comfortable with the definitions... think about these for a few days >.<

i shouldn't answer to you more >.<

[tulip cedar]

he wrote a lot of stuff about trying to make categorical style physics, it seems

[tulip cedar]

wait what

[woeful sand]

altho i don't agree it's necessary, the book im reading tries to contextualize polynomials and completeness (of real line) as a motivation to dedekind results

[tulip cedar]

couldnt that just mean they need to pick a better form of arrow?

[rustic crown]

yee

[woeful sand]

i feel more confortable that way

[rustic crown]

that won't be called the category of fields then... we want it as the full subcategory of CRing

(det is tired, det will go, bai bai)

[chilly ocean]

CRing(e)

[rustic crown]

that's me

[vagrant zinc]

Guys, could someone guide me in this exercise?

[vivid tiger]

Hint 0: ||Try to find it for n=3,4,5, and notice a pattern||

Hint 1: ||try to prove n-1 is such a k||

Step 1: ||(n-1)^2 = n^2 - 2n + 1 = 1 mod n. So if n-1 is a unit, we are done.||

Step 2: ||n-1 is coprime to n. so it's invertible|| (do you know why?)

so we are done.

[vagrant zinc]

the pattern with respect to U(n)?

[vivid tiger]

a pattern of elements that have order 2 (you just need to find one of them for each Z_n)

then, if you picked the right ones with the right pattern, you have units and are done

(don't overthink it - brute forcing the n=3,4,5 examples should give you a somewhat obvious way to find the k)

[vagrant zinc]

I was looking at U(n) and I should have looked at the Z first ok I got it I found the pattern

[vagrant zinc]

answering your question in step 2 no, I don't know.

[vivid tiger]

ahh, ok

so, do you see why if they're not coprime, it's not a unit (as in, the converse)?

[vagrant zinc]

nop

[vagrant zinc]

❤️

[vivid tiger]

let's try some examples

first, the converse is easier

namely, show that not coprime -> not invertible

example: 3 in Z_6

since 3*2=0, 3 cannot be invertible (why?)

(that last why is a ring property)

[vagrant zinc]

Ok, Ok I understand

[vivid tiger]

if k is not coprime with n, you can find an m ≠ 1 so that n=ma and k=mb for some a, b. then ka=bn=0 mod n. but then a is strictly less than n and so you can find it in the ring Z_n and so k*a=0 so k is not invertible

now, let's go backwards. try to reverse that argument

[vagrant zinc]

okay okay I understand, but I don't know much about abstract algebra I'm just getting started.

[haughty hollow]

How exactly do I do this? Find all the elements of the subgroup ⟨(1, 2)(3, 4)(5, 6), (1, 3)(2, 5)(4, 6)⟩ of Sym(6)

[vivid tiger]

if it's not invertible, then there exists a in Z_n such that ka = 0 mod n so there exists a b such that ka=bn. so, since both k and a are strictly less than n, ka < n*n so b is less than n. so then n's factors are either in k or in a. To not be coprime, the non-1 factors must all be in a, but that implies a=n which contradicts, so n shares a non-1 factor with k and we have that they are not coprime

so, we proved that in Z_n, invertible aka being a unit is equivalent to being coprime with n

oh, wait

[vagrant zinc]

How do I show that n-1 such k?
I can think of something about U(n) over the mcd(n-1,n).

[vivid tiger]

n-1 is coprime with n and squaring gets only a term of 1 that's not a multiple of n

[vagrant zinc]

my head is going to explode from so much I have to study, I understand what you are going to I remember my previous course of modulo number theory and it makes me think a little about what you are explaining but it is difficult I have just been studying analysis in several variables and I have a lot to digest, I will tell you that I understand but not at that level, I can only say that I know the basics.

now well I am studying algebra to change the subject hehe

I don't want to disappoint you because you have explained me very well 😄

[vivid tiger]

lemme quickly come up with why there must be an a such that ka=0 mod n.

[vivid tiger]

at what level?

i haven't looked all that much into algebra

yes, lots to digest. best thing to do is to somehow find the time to digest

[vagrant zinc]

well you know you have to be studying, finding out a lot and simplifying your doubts and problems until you get it, maybe I still have a long way to go, but I will remember every word you tell me because everything is useful to me.

[celest furnace]

Awuita i would say another tip would be to look for it in other domains. What about Z? Eventually you might stumble upon that (-1) satisfies this, so maybe it also will in Z_n

[vivid tiger]

that's equivalent to looking at n-1

[celest furnace]

yep

[vivid tiger]

injectivity of the map that sends x to kx implies 0k != ky for all nonzero y. then k must be a zero divisor. likewise if k is not a zero divisor then it must be injective as otherwise kx=ky without x=y implies k(x-y)=0 but then k is a zero divisor which contradicts. so injectivity of this map is equivalent to being a non-zero divisor

suppose k is a zero divisor, then kx = 0 for a non-zero x, but if k has an inverse then x=0 so k is not invertible

now, suppose k is not invertible, then the map that sends x to kx is not surjective nor injective, but then we have zero divisor

[vagrant zinc]

[vivid tiger]

sorry

now, we do know invertible <-> not zero divisor

so we know is a unit <-> coprime

[vivid tiger]

so we know the first line of this is true, and the rest follows

sorry for the ring around the rosie

i appear to have forgotten about subtraction, which is really why i took so long

[vagrant zinc]

thanks bro ❤️

[pliant forge]

Let $\Gamma$ be a quiver, $k$ a field, and a two-sided ideal $I \subseteq k\Gamma$ such that $\exists m \geq 2 \colon R_{\Gamma}^m \subseteq I \subseteq R_{\Gamma}^2$ where $R_{\Gamma}$ is the arrow ideal of $k\Gamma$. Let M be a $k\Gamma/I$-module and assume it's simple.
Choosing a maximal path $p$ such that $pm \neq 0$ for some $m \in M$, and a vertex $i \in \Gamma_0$ such that $e_im \neq 0$, where $m_i := e_im$ I get stuck with trying to show that the subset ${am_i | a \in k}$ of $M$ is a $k\Gamma/I$-module.

[cloud walrus]

Eso

[pliant forge]

(to show submodule^ of M) And thus this would imply the dimension would have to be 1 as its non-empty and a submodule of a simple.

[woeful sand]

my head almost exploded reading this

[white oxide]

a bit of a silly question, but say if we have a bijective function f from A to B. are we allowed to write f(A) = B?

[chilly ocean]

what do you think?

[white oxide]

yes

[chilly ocean]

yes. why?

[white oxide]

because the whole set A gets mapped to the whole set B, or in particular each element in A is mapped to a unique element in B

i'm not so sure of the reason for the notation behind it though

it makes sense

i can see sets mapping to sets

the reason i ask is because

well let me post it

[chilly ocean]

f(A) means the set of all things in B that get mapped to by f. bijections are, in particular, surjections, so everything in B is mapped to by something in A. so B = f(A)

also, this has nothing to do with abstract algebra

[white oxide]

right

let me get to the reason

so im trying to prove 42, and my whole idea was since phi is an isomorphism it is clearly bijective

so we have f(G) = (G')

since G is cyclic there exists a in G such that <a> = G

or in other words, for an integer n, a^n = G

[chilly ocean]

no

[white oxide]

no?

[chilly ocean]

that's not what <a> = G means

[white oxide]

well

for every integer n

[chilly ocean]

no

for every g in G, there is an integer n such that a^n = g

[white oxide]

really?

because my book uses this notation

[chilly ocean]

think about how that is the same thing as what i wrote

[white oxide]

oh wait ur right

ok so that method doesnt work

maybe i could take some arbitrary g in G

and since it is equal to a^n for some n

do smoething with that

the only issue is how i would express something of the sort like

a^-n

because my whole idea was take some arbitrary g in G

[chilly ocean]

you assume that G is cyclic and that you have an isomorphism of G with another group G'. then you need to show that G' is cyclic. i think you need to carefully review all of the definitions

[white oxide]

yea i know my g

i'm trying to figure out how yknow?

[bleak abyss]

Alright let's make some more progress

I did figure out the whole thing about J-residues of G/B

So let's continue

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

So let me give one more example which I thought about on my own a bit but didn't quite explain here

The chambers/singletons here are the faces of the octahedron

I'll draw it why not

1-panels are {a,b}, {c,d}, {e,f}, {g,h}
2-panels are {a,d}, {b,c}, {e,h}, {g,f}
3-panels are {a,e}, {b,f}, {c,g}, {d,h}
{1,2}-residues are {a,b,c,d} and {e,f,g,h}
{1,3}-residues are {a,b,e,f} and {d,h,c,g}
{2,3}-residues are {a,d,e,h} and {b,c,g,f}
And the whole thing is connected

Now the thing is

The panels just correspond to edges. 1-panels are the edges labeled with a 1

The {i,j}-residues are vertices

So using the face terminology

This makes sense

That cleared shit up for me a good bit

New lemma just dropped

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Okay what does that mean?

So R is some residue of cotype J. So yeah starting from c in R, R is just anything you can get to be a sequence where everything is I\J-adjacent

Well, relax the condition to I\K

You'll still get a residue, it'll be of cotype I\K

And obv shit being i-adjacent for i in I\K is less restrictive than for i in I\J since K subset J

So this includes R

Keep getting distracted....

But yeah uniqueness is because we're fixing a residue and we have established which elements it contains. Well in particular, if S_1 and S_2 are faces of R of cotypes K_1 and K_2

Then S_1 and S_2 have the same face of cotype K_1 cap K_2

This is just thinking about uniqueness a bit

[rotund aurora]

What is this all about?

[bleak abyss]

There's only one face of R of cotype K_1 cap K_2. Well, S_1 has a (unique) face of that cotype, so that'll be it. So does S_2. gg

Croqueta: chamber systems and buildings, eventually those built from p-adic groups

https://people.maths.ox.ac.uk/demartino/LectureNotes.pdf

[bleak abyss]

So let's look at the octahedron for a sec

Let's say R = {a,b}, which is of cotype {2,3}. Now let's say K = {3}. Look at faces of cotype {3}, meaning type {1,2}. They're {a,b,c,d} and {e,f,g,h}. The one which contains R is {a,b,c,d}

Reset notation, let R = {b}, which is an empty-residue (aka a chamber). S_1 = {a,b}, cotype {2,3}. S_2 = {b,c}, cotype {1,3}. They have the same face of cotype {3}, namely {a,b,c,d}

Last thing I'll do for now before going to sleep

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Intuitively a simplicial complex is, you have a bunch of vertices, you put a bunch together as simplices

Vertices are simplices, and if you give me a simplex, every subset of the vertices is another simplex

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Which sends the right vertices to the right places

And yeah I'm out of steam good night

[somber sleet]

Can someone explain to me, why there exist only p-1 automotphism on Zp?

as an additive group

[rustic crown]

because any map of abelian groups Z/pZ --> Z/pZ is given by multiplication by some element in Z/pZ

now when is this going to be invertible?

[somber sleet]

when x^-1 is in Zp I guess?

[rustic crown]

yep

[somber sleet]

but this should be the multiplicative inverse right?

If I define f(a)= x*a for a in Zp arbitrary

[rustic crown]

right, so which elements in Z/p have a multiplicative inverse?

[somber sleet]

all except 0 I guess(?)

[rustic crown]

yeap :3

and so p-1

[somber sleet]

okay that makes sense

but why are the only possible automorphisms defined by multiplication with an element?

couldn't there be something more

[rustic crown]

all homomorphisms are like that

if f : Z/n --> Z/n is a map of abelian groups, then it's given by multiplication with f(1)

because f(m) = f(1 + 1 + ... + 1) = f(1) + ... + f(1) = m * f(1)

[somber sleet]

allright, that makes much more sense, I get it, thanks det

sorry, was kinda easy as a question hahaha

[rustic crown]

the same reasoning shows that Aut(Z/nZ) = U(n) the group of units in Z/nZ

[rustic crown]

issokie uwu

[pliant forge]

im still wondering about this

[rustic crown]

what's e_i?

[pliant forge]

e_i is the trivial path on vertex i. So multiplying by e_i from the left on any arrow/path which ends at vertex i will be the same path

[rustic crown]

oh okie

[rustic crown]

okie i have a question

[pliant forge]

sup

[rustic crown]

@pliant forge so you know that the arrow ideal in kGamma/I is R_Gamma/I which is nilpotent by assumption, so it has to lie in the jacobson radical, so if M is a simple kGamma/I module then it's also a simple (kGamma/I)/(R_Gamma/I) module

but this is just a direct sum of k for each vertex in Gamma

[pliant forge]

i have not been introduced to rad

[rustic crown]

does this not imply that simples are one-dimensional

[rustic crown]

oh i just mean that the arrow ideal will kill the simple module

because it is nilpotent

say J is that ideal, then JM = 0 otherwise by simplicity JM= M and so by induction 0 = 0M = J^mM = M

[pliant forge]

wait

what if its infinite

[rustic crown]

you told me that I contains R_Gamma^m for some large enough m

[pliant forge]

right

nvm

[rustic crown]

so what i'm saying is that M = km = km_i

which is ofc a module over kGamma/I

does that make sense?

but the main idea is that the nilpotent thing will kill M

because of simplicity

[pliant forge]

wait im trynna see why that implies its 1 dim

[rustic crown]

oh i can say that directly

[pliant forge]

ok sure

[rustic crown]

so look at that m

multiplying on the left by any arrow will kill it

[pliant forge]

ohhh yeah

because its the arrow ideal

[rustic crown]

so only things that can act non-trivially are e_j

[pliant forge]

yes yes yes

thansk det ❤️

[rustic crown]

but e_j *(m_i) = 0 unless j = i

uwu

[sweet echo]

I can show the inequality using some rank arguments but I'm not sure how to prove we can always achieve the lower bound dimh = 2n - rank(A)

also confused that they say they usually define $\mathfrak{h} := \bigoplus_{i=1}^{n} \mathbb{C}\alpha_{i}^{\vee}$ as surely then $\mathfrak{h}$ would just always have dimension $n$

[cloud walrus]

ΣAC

[sweet echo]

(i know that eventually when looking at cartan matrices they have max rank and so the dimension of the cartan subalgebra will be n but clearly this is not true for a generic matrix?)

[somber sleet]

does someone have a nice intutition of free abelian groups and free groups?

I can't really visualize it

[void cosmos]

ig think of the universal property

[analog zephyr]

Haven't heard about free group for long time

I think that I forget what was it about

The finite products of the elements of a group or something like that

[rotund aurora]

Diagram

Analogy with vector spaces, which you are probably familiar with already

[analog zephyr]

U mean on Cayley Diagraph?

[void cosmos]

no

[rotund aurora]

(All vector spaces are free)

[analog zephyr]

Just diagram

[rotund aurora]

The arrows thing

Im lying on my bed rn so lazy to provide the diagram myself

Maybe someone else will, or maybe try googling

[void cosmos]

think of being faithful

to something

to a gf

[rotund aurora]

Ok its this

Lul

[void cosmos]

yea

think of like

F is faithful to G such as in like

if X tries to hook up

he rides the function to come beat her up

or what

shee*

wtf

[analog zephyr]

Arrows where arrow mean generator

---->, ------, <------->

For example

[rotund aurora]

Mmh I think Im not following you

[analog zephyr]

Wait a minute

u want to see a diagram for free group, no?

[analog zephyr]

Sorry, I'm no the job right now, so I get busy sometimes

Fk it was a misunderstanding XD I was talking about "Free torsion and torsion groups"

But for i see, free group is very similar to what I was thinking

are very similar*

Basically u have a group G, then a set X Wich is a subset of G where the finite products of the elements of X can generate G(I'm not sure about it) or some group A. And I think A will be the smallest group containing X

Yeah, like a basis for a vector space

Then X is a generating set

[somber sleet]

how do you find the elements of this group, and how do you knon when it's finite?

[chilly ocean]

in general proving a group is finite given its represented this way is a hard problem

here the condition yxyx^5 = 1 should remind you of a semi-direct product

indeed, this group should be a semi-direct product of Z_8 and Z_2

its elements should be e, x, ..., x^7, y, yx, ..., yx^7

how to multiply two elements of this group should be given by the condition xy = yx^3

[chilly ocean]

well.... I wouldn't really say it explains what the free group is

it's you that the free group with generators X is a group with X as their elements and all the expression that can be formed from them, so that no relations are imposed on it

[somber sleet]

my problem is that I don't even know how to approach this problem, like I found the identity and everything, but I can never come to a conclusion

[chilly ocean]

what's the problem though

to find its elements?

[chilly ocean]

Like, to understand what the free group is just try to look at what the free group with 2 generators x, y is

you have expressions of the form $$x^{n_1}\cdot y^{m_1}\cdot ...\cdot x^{n_k}\cdot y^{m_k}$$ there

[cloud walrus]

Blitz

[chilly ocean]

where n_i, m_i are integers

and multiplication is a bit like concatenation of strings

just that the terms when combined with each other can disappear like x^-1 multipled with x

[somber sleet]

I came to the conclusion that I have to generator, but afterwards they prove somthing, I don't even know how, by using the normalizer

and I don't understand why it's necessary to do so

you don't understand german right?

[chilly ocean]

wait, what are we talking about

[chilly ocean]

lol, no

[somber sleet]

this is the solution of the exercise

and at one point they prove that <x> is normal in G

[chilly ocean]

I don't think they are saying much there

[formal ermine]

damn gurl

[chilly ocean]

just analyzing the group G

it probably helps to construct this group at the end of this

to show that it really does have 16 elements

for me though, it follows from construction of semi-direct product

x^8 = 1 is basically Z/8Z

y^2 = 1 is Z/2Z

and f(y)(x) = yxy = x^3 is the automorphism in the definition of semi-direct product

defined on the generators

https://en.wikipedia.org/wiki/Semidirect_product#Inner_and_outer_semidirect_products

like in the definition of outer semi-direct product

[toxic zephyr]

this is equivalent to det(A) not congruent to zero mod p right?

[chilly ocean]

yes

A is invertible iff det(A) =/= 0

works for any field

(weird notation btw, why mention mod p when you already write elements of A are from F_p?)

[toxic zephyr]

awesome tysm

[toxic zephyr]

i think because A is an integer matrix, not Fp.

[somber sleet]

guys, how would you write Z[1/2] down as a set?

[chilly ocean]

{n/2^k : n, k integers}

[formal ermine]

Fp is iso to Zp

[toxic zephyr]

true but i mean $A\in\bZ^{n\times n}$ rather than $\bZ_p^{n\times n}$

[cloud walrus]

nilpotent nix

[formal ermine]

ah

[chilly ocean]

Yeah it makes sense then

Operation "mod p" is a ring homomorphism though so that det(A mod p) = det(A) mod p

Since det is defined using addition and multiplication

So no issue here

[barren sierra]

dumb question / sanity check, making sure I remember my field extension stuff

[cloud walrus]

Spamakin🎷

[somber sleet]

How do you get it? And does it always work if we have fractions?

[next obsidian]

I cannot help at all, but what in the fuck is an algebraic circuit over a finite field lol

[toxic zephyr]

the automorphisms on $\bZ^n$ are given by matrices $A\in\bZ^{n\times n}$ with $\det(A)=\pm1$ right?

[cloud walrus]

nilpotent nix

[next obsidian]

What sort of automorpuisms

If you’re looking for as a group, then yes

[toxic zephyr]

yeah. okay cool thanks

[chilly ocean]

Z[1/2] is the set of polynomials in 1/2 with integer coefficients, so it's the set of numbers a_1 + a_2/2 + ... + a_n/2^(n-1)

where a_i are integers

but number like that is already of the form k/2^n

all you need to see is that it equals (2^(n-1)a_1 + 2^(n-2)a_2 + ... + a_n)/2^(n-1)

nominator being an integer

[barren sierra]

My question is only about the hint which is all algebra

But a algebraic circuit is a circuit whose inputs are field constants and variables and whose gatss are addition and multiplication (typically) and the output is the computation of a polynomial

[somber sleet]

Why do have then hat Z[i] ={a+ib | a, b in Z}

[barren sierra]

Super cool stuff

[next obsidian]

Da fuq is a circuit ☠️

[somber sleet]

Is it because i to a n power is just +-1 or +-i

[next obsidian]

Like

[chilly ocean]

because powers of i are 1, i, -1, -i, 1, i, ...

[next obsidian]

A circuit that you’d draw in some physics class?

[somber sleet]

You opened me a world

[barren sierra]

Pretty much

I can explain more after my class but can you confirm my sanity check about the hint plz

[chilly ocean]

so like if you have a polynomial in i then the sum of even powers time integer will be again an integer, and sum of odd powers time integer will be an integer times i

so it's of the form a+ib for some a, b integers

[coral spindle]

We usually just call this GL_n(Z), by the way

[formal ermine]

yes it's in Fp

[somber sleet]

This is so easy, why the freak does my prof not explain anything

[formal ermine]

obligatory I think

[chilly ocean]

idk

[coral spindle]

To elaborate: since for square matrices any commutative ring you can always compute the adjugate matrix, and A adj(A) = det(A) I, it's clear that an inverse matrix exists iff det(A) is a unit in R. An indeed, the only units in R = Z are \pm 1.

[formal ermine]

I have the urge to say skill issue but this is #groups-rings-fields...

[coral spindle]

Well, I should say that one direction of this iff is clear; the other isn't.

(the adjugate matrix is exactly what gives us the other direction)

[chilly radish]

is this a numerics thing? or a cryptography thing?

[chilly radish]

about the hint - yes

it's basically saying that as a field extension, a finite extension of a finite field is generated by a single element

i.e. F_p^k = F_p(\gamma)

[barren sierra]

Dope

I remember field theory 🤩

[barren sierra]

Algebraic Complexity Theory

[chilly radish]

ah

[barren sierra]

Will be happy to explain stuff later after class

Prof talks at the speed of light

[chilly radish]

no worries

[rotund aurora]

Let M be a module over some ring and consider the set S_n={N | N is a submodule of M with |M/N|=n}. Is |S_n| always finite for all naturals n ?

I guess it is, but its not obvious to me

Never have thought about this

[next obsidian]

hell no

Consider the module Prod F_2

infinite

then take the submodule which is 0 in one factor and F_2 everywhere else

each has quotient F_2

[rotund aurora]

Thanks

[coral spindle]

Of course, Prod F_2 is a non-noetherian module

Could there be a noetherian module for which S_n is still finite? Maybe you can produce a counterexample to this :)

In fact in some ways the generic example of a module over a ring gives you a counterexample... hint hint

Oh wait mb, the example that I was thinking of was incorrect!

[rotund aurora]

I think you can do exactly what Chmonkey did but over Z

[coral spindle]

Not quite, because that's only for all n but not one specific n, right

I made this mistake

[rotund aurora]

take infinite product of Z, and quotient out by infinite product of Z, except at the i-th summand you take nZ

[coral spindle]

Right and this is still not a Noetherian module

[rotund aurora]

uh right

not finitely generated

[coral spindle]

I will go back to thinking about whether or not there is a Noetherian module for which S_n isn't finite

[rotund aurora]

this is actually so fun

I think these two things gives that S_n is finite for finitely generated free modules over Z

not entirely sure, Im tired rn

or like R PID instead of Z I guess

[coral spindle]

I agree, I'm trying to come up with a general argument though

I strongly suspect it will have to be finite if the module is noetherian

[rotund aurora]

it would be something nice to figure out

I was also thinking about imposing random condiitons on the numbers S_n, like what if S_n=1 for all n. This looks very much like Z

[coral spindle]

That's a cool idea

I think I have the beginning of an argument

Let's say we have two submodules N and N' of M, such that |M/N| = |M/N'| = n.

Now I want to inspect N + N'.

If N + N' = M, then since M / N = (N + N') / N is isomorphic to N' / (N \cap N'), well, we must conclude that |N'/N \cap N'| = n

This is a bit bothersome.

Well, I'll give it some more thought later

[barren sierra]

Ok I am presume you know what a graph is, and specifically a directed acyclic graph (it's what it sounds like)

So an algebraic circuit is a DAG with a single sink (the output), sinks labelled with field constants or variables, and all the intermediate nodes are operation gates (typically addition and multiplication) where the incoming edges are the inputs and the outgoing edges are the result of that gate. Since it's a DAG you can reuse computations as needed which is cool. If you want to remove this, then your graph is actually a tree and we call this a formula.

This model of computation leads to alot of super interesting questions. For example studying why computing the permanent is hard versus why computing the determinant is easy.

This sits in a field of study called algebraic complexity theory

[gilded osprey]

I am given the set ${\epsilon, \zeta, \tau, \mu }\subseteq S_4$ with composition, where $\epsilon$ is the identity and
$$\zeta = \left(
\begin{array}{cccc}
1 & 2 & 3 & 4\
2 & 1 & 3 & 4
\end{array}
\right),
\tau = \left(
\begin{array}{cccc}
1 & 2 & 3 & 4\
3 & 2 & 1 & 4
\end{array}
\right),
\mu = \left(
\begin{array}{cccc}
1 & 2 & 3 & 4\
4 & 2 & 3 & 1
\end{array}
\right)
$$
and asked to determine if it is a group. Am I correct in saying it is not since it is not closed under composition. For example $\zeta\tau$ is not in the set

[cloud walrus]

JacobHofer

[formal ermine]

you should write this as cycle notation

but yeah you're correct

we have zeta = (12) and tau = (13) so zeta o tau = (213)

which is none of those

[gilded osprey]

Okay, that's what I figured. I am not familiar with cycle notation, this was just the notation we have been given

[formal ermine]

cycle notation is (abcdefgh...)

e.g. here zeta = (12)

2 gets send to 1 and 1 gets send to 2

(213) means 2 gets send to 1, 1 gets send to 3, 3 gets send to 2

[gilded osprey]

Oh interesting

So is zeta * zeta = (121)?

[coral spindle]

No! Elements may not repeat in cycle notation

zeta * zeta is the identity element. Usually people just write this as 1, or e, in cycle notation

[gilded osprey]

Okay, gotcha. Again, I've never seen that style of notation before

So why would one write zeta * tau = (213) and not just (23)?

[coral spindle]

For the same reason one would write 1 + 2 = 3 and not 1 + 2 = 4... those are completely different things

(213) is not equal to (23)

[gilded osprey]

Oh wait, yeah I just can't read. (213) implies 2 goes to 1 and 1 goes to 3, (23) implies 2 just goes immediately to 3

[next obsidian]

[barren sierra]

It's super cool stuff

Huge fan

[delicate orchid]

that shit does sound cool

[barren sierra]

Heavily debating studying it for grad school

[delicate orchid]

what do you mean by "reuse computations" though?

[barren sierra]

1 sec

Left is a circuit, right is a formula

Note that in the circuit, I reuse for computation of xy to compute xy^2

But in the formula I have to compute xy twice

[delicate orchid]

I see

I think

yes, yes I do

[pliant forge]

🎄 🎅

[chilly ocean]

Christmas

[bleak abyss]

Alright mine again

So

[cloud walrus]

Sloth King Daminark

[bleak abyss]

So in my octahedron example from yesterday, ${a,b,c,d}$ and ${e,f,g,h}$ have cotype ${3}$, while ${a,b,e,f}$ and ${d,h,c,g}$ have cotype ${2}$. Well, the residues of cotype ${2,3}$ are ${a,b}$, ${c,d}$, ${e,f}$, and ${g,h}$. So ${a,b}$ is gonna be an edge connecting the vertex of cotype ${3}$ containing ${a,b}$, namely ${a,b,c,d}$, with the vertex of cotype ${2}$ containing it, namely ${a,b,e,f}$. Which makes sense by the picture here

[cloud walrus]

Sloth King Daminark

[bleak abyss]

More generally

If you give me a residue R of cotype {i_1,...,i_r}

That'll be an (r-1)-simplex

[delicate orchid]

and here we cope

[bleak abyss]

no u

But yeah so, that residue has a unique face of cotype {i_1,...,i_{r-1}}

That face has an associated (r-2)-simplex already built

[delicate orchid]

https://media.discordapp.net/attachments/832423825076912128/1060257838821298306/feel_that.gif

[bleak abyss]

So yeah we just identify it all

[vagrant zinc]

someone could give me a hint

[rustic crown]

if a and b are elements of G, then so is ab

(what does the condition on G say now?)

[vagrant zinc]

that aa=e then if a=a^{-1} belongs to the group so that if
aa=(aa)^{-1}
aa=a^{-1}a^{-1}
aa=aa
therefore it is an abelian group

[rustic crown]

it would be nice if you called the other a a b

because powers of a would commute in any group

a^m * a^n = a^(m+n) = a^(n+m) = a^n * a^m

[bleak abyss]

Continuing

They give an example where the geometric realization of a chamber system isn't that of a simplicial complex

Let's work things out

The way they organized it we know the panels

But then the {i,j}-residues are just the whole thing

Ex 14 is confusing

[bleak abyss]

Now the A_n(k) building

[sacred pier]

why we need to mention orthogonal matrices (is in general linear group)?

thank you

[chilly ocean]

are you asking why orthogonal matrices are important?

[lethal dune]

[rustic crown]

ryuwu

[sacred pier]

So the thing inside the bracket {Q in GL(R) |Q^T = Q^{-1}} Is trying to define orthogonal matrices. Thank you