groups-rings-fields

[kind temple]

told you the universal algebra arc was in the works

lol

[novel star]

what is the argument that hes talking about?

i dont have access to paper and i have no mental scratch space

[tame sierra]

what book is that?

[novel star]

daddy lang

chapter 5 section 9

[tame sierra]

huhh weird my copy's chapter 5 only has 6 sections

oh it's chapter 6 isnt it

[wraith cargo]

It's identical to the one above
If it were a p-th power you'd have that a is a p-th power

[novel star]

ohh ty

idk why i didnt see that

[novel star]

yes my bad

for some reason the word "galois" is associated to the number five in my head

probaably because french

[round anvil]

how to prove that a 3-cicle is equal to other 3-cicle conjugate?

[elfin wraith]

Could you post the specific problem?

3 cycles being conjugate depends on the group, in A_4 there’s two conjugacy classes of 3 cycles

[round anvil]

n>=5, sorry

[round anvil]

i got it

[glossy crag]

I assume I must be mistaken given the amount of upvotes, but how is this MO answer https://mathoverflow.net/a/13372/137264 correct, specifically the justification for dim_hW*<=dim_kV*? They say to "take elements of W dual to the functionals", but the whole point is W* is not isomorphic to W, so there aren't dual vectors to an arbitrary basis of W*.

[fleet cairn]

can there be a morphism that behaves just like the identity morphism but it doesnt come from an end set?

[next obsidian]

What

[fleet cairn]

in some cateogry where a morphism is defined in such a way, is it possible that there exist some morphism g: Y->X where X and Y are distinguishable as objects (X != Y)
but for all f in Hom(X, Y), g o f = f and f o g = f (acts like identity morphism)

[next obsidian]

This doesn’t even make sense

How can g•f = f, the source and target of f can’t exist to make this work

Similar for f•g = f

[fleet cairn]

why cant f exist, how come

[next obsidian]

Stare at the equations you wrote down

And realize it doesn’t even make sense to say they are equal

[tribal moss]

More explicitly, writing "g o f = f" compares morphisms from different homsets, which is considered meaningless.

[fleet cairn]

ohhh, i was thinking that if you defined morphism in a way where the soruce and target arent meaningfully used that a g could exist that acts like ID

[velvet hull]

well, you can always formally define morphisms to behave however you want

in particular you can just create a new morphism that behaves like the identity with respect to composition with every non-identity morphism

[next obsidian]

[fossil spade]

this was given to me as a practice problem for my upcoming midterm in my abstract algebra class and I feel like its a bit out of place. I was trying to do it but honestly I have no clue how to begin, does this have to do with the order of an element or maybe inverses?

[velvet hull]

the integers under multiplication mod p form a group

[fossil spade]

something that i do think maybe is similar is that if G is a group and |g| = n then g divides n, where g \in G

[fossil spade]

yeah i was thinking of that

seems like a version of fermats little theorem

from what ive seen

[velvet hull]

yeah, you're close

[fossil spade]

i see

so from what i can think It might have something to do with having inverses for elements and if a^k is congruent to a mod p (where k is an integer and p is a prime) then that guarantees that every element has a inverse therefore integers under multiplication mod p form a group?

but im not sure how that works maybe i should read through the proof...

[velvet hull]

can you rephrase the question in terms of group theory?

that might help

turns out the multiplication isn't that important here

[fossil spade]

oh so the binary operation doesnt matter?

[velvet hull]

well it does actually

but i think it will be informative try and answer the more general group theory question first

and then specialise to the multiplicative group

[fossil spade]

alright...

i need time to think

[tribal moss]

I think you also need to know that the multiplicative group mod p (or of a finite field more generally) is cyclic.

[kind temple]

its easy to see that V_d(a ^ d) is less than a ^ VD, not quite sure how to show the other direction.

[kind temple]

okay, this was a case of me not being familiar enough with definitions. i think i have it now.

let x <= a ^ VD be compact. since x <= VD, there is a finite subset D’ of D such that x <= VD’.
since D is directed, there is some d_0 in D such that d’ <= d_0 for all d’ in D’.
now x <= a ^ d_0 <= V_d (a ^ d)

since a ^ VD is the join of all of the compact elements below it by algebraicness, then we win.

these two problems really helped me with the intuition here. each object is the join of elements below it, but in an algebraic lattice, we can restrict our attention to only the compact elements below it.

they act as probes almost

this reminds me of compactly generated spaces

and compact test functions

[tall igloo]

damn that's crazy, it's almost like these lattices are compactly generated as categories

[novel star]

how do you go about this?

i guess it should be Z/nZ

but my intuition feels shaky

ig im not entirely sure that all the roots are related by nth roots of unity

oh wait nevermind

because if a and b are solutions

then 1=t/t=(a^n)/(b^n)=(a/b)^n so a/b must be an nth root of unity

[kind temple]

sorry, what is the question?

[novel star]

oh find the galois group

whoops

does the method of reducing mod primes generalize to more general fields to calculate galois groups

like C(t) in the message above?

[chilly radish]

You have a more general version for dedekind domains, but here you can just find all of the automorphisms by exhaustion (since you know there should be exactly n and you can find one of order n pretty easily)

[swift root]

this is what makes them "algebraic", as algebraic operations are finitary!

so the philosophy is that you only have to check on finite subsets (i.e. finitely generated or compact closed sets)

[vivid kestrel]

in an integral domain R with a prime ideal p, can the intersection over all powers p^k be a non trivial ideal

[quiet pelican]

Consider the ring R = k[x, y, x1, x2, …]/(y = x x1, y = x^2 x2, …, y = x^n xn, …)
By the map R -> k[x1, …] sending x, y to 0 and everything else to itself, we see that (x) is prime
Hence if there is a zero divisor, x is a zero divisor (if ab = 0, then WLOG a \in (x), so a = xc, so either x is a zero divisor or bc = 0, but then we’re done by induction on deg a + deg b)
But the map R -> k[x] by sending every except x to 0 is well defined, and maps x to a non-zero divisor, so x is not a zero divisor and we’re done

Then the intersection of all ideals (x)^k contains (y)

The way I thought of this is essentially if any ring has this property, then this ring must be non-zero (universal property nonsense)

Then construct enough nice quotients to show that this ring works

[vivid kestrel]

dont fully get it rn but thanks a lot still!!

[rocky cloak]

Consider
R = k[x^2^-n : n in N]
i.e. the polynomial ring where you adjoin square roots of square roots of x.

Then the ideal generated by all powers of x is maximal and has p^k = p for all k>=1

It cannot happen if R is Noetherian though

[vivid kestrel]

thanks thats insightful

why does R noetherian prevent it though

[rocky cloak]

So if I is the intersection of the p^k, then the Artin Rees lemma tells you that pI = I.

Then by Nakayamas lemma there is some x in p with xy = y for all y in I.

So if y is nonzero in a domain that would mean x=1

[vivid kestrel]

thanks!

[chilly radish]

I don't see why the end holds? Image of a zero divisor isn't necessarily a zero divisor

[next obsidian]

This theorem is called Krull’s intersection theorem

[restive sparrow]

completely lost on this one, does anyone mind helping me out?

eta(R) is the nilradical

should be frak(N)(R) but typo ig

2->3 is doable, 3->1 is something i almost have

but 1->2 is just ??

[next obsidian]

Prove that the nilradical is the intersection of all primes

And 1->2 becomes trivial

This also handles 3->1

I mean actually all of these follow from this fact

[wraith cargo]

use a zorn's lemma argument

it's a bit nontrivial but the idea is that like

suppose you have an element that's neither nilpotent nor a unit

then consider that set of all ideals that don't contain a power of this element

and use Zorn's lemma to prove the maximal element is prime

[restive sparrow]

i have that the nilradical is contained in the intersection of all primes

[next obsidian]

It is equal

[wraith cargo]

the other direction is a lot harder and you need to use Zorn's lemma to prove it

[restive sparrow]

i know, but idk how to get it the other way 😔

ah okay

[next obsidian]

This is a fundamental basic fact of commutative algebra

[restive sparrow]

i am reading dummit and foote as my first exposure to this field oof

[next obsidian]

I would try to prove that because it is fundamental for further stuff

[wraith cargo]

we had this problem on an exam but we wen're allowed to use this fact we had to prove that all the elements are either nilpotent or a unit from first principles

[restive sparrow]

ouch

[wraith cargo]

which isn't super hard if you're at all familiar with Zorn's lemma type arguments

[rocky cloak]

It can be proven in a pretty slick way using localization.

Do you know the relationship between prime ideals and localization?

[next obsidian]

Did they also give you a gun to kill yourself with?

[wraith cargo]

no the last problem on the algebra exam is always some Zorn's lemma bullshit

[elfin wraith]

I mean this is a good technique to know in general lol

[wraith cargo]

this year we had to prove that there's a prime ideal contained in the set of all zero divisors of a commutative ring

[restive sparrow]

i do not know what localization is

[elfin wraith]

You use the same kinda idea to show like normal is a local property

[restive sparrow]

am i too dumb for this channel

[rocky cloak]

Then I guess don't worry about it for now

[tribal moss]

Most people manage fine without knowing any techniques for killing themselves.

[wraith cargo]

no don't worry you're still learning

[next obsidian]

No

[next obsidian]

This is interesting

I wonder what the simplest lowest tech way to approach it is

[wraith cargo]

Have you ever seen a Zorn's lemma argument @restive sparrow

[delicate orchid]

CRAZY message to open this channel to btw

[wraith cargo]

I did it using Zorn's lemma but idk if there's a simpler way

[restive sparrow]

yeah

[wraith cargo]

tho the class is kinda hell for most students

this is an intro algebra class for undergrads

and they use the noncommutative ring definition for prime ideals

[restive sparrow]

if every chain contains an upper bound then the set contains at least one maximal element

er, poset

[wraith cargo]

yesss

and usually the arguments are framed in such a way that the maximal ideal in this poset is prime

[elfin wraith]

Based

[rocky cloak]

3 -> 2 is not that hard to prove though, so you could go through that

[elfin wraith]

I was going to um actually chmonkey about the nilradical only being the intersection of primes for commutative or Noetherian rings but I saw the question does specify commutative 🌾

[restive sparrow]

those are certainly words

[wraith cargo]

so what you should do is try to show using Zorn's lemma that if there exists an element that's neither a unit nor nilpotent, then it's not contained in R

a hint would be to consider the set of all ideals that don't contain a power of this element

now you just have to prove that this set is nonempty and that every chain has an upper bound

[restive sparrow]

nonempty coz of the zero ideal

these ideals are partially ordered by inclusion

and so naturally contain an upper bound? do i have to justify that?

oh ya i do

[wraith cargo]

ya you should even if it's obvious on some level better to write it out

the hardest part will be showing that this maximal ideal is prime

it'll take some thinkin'

[restive sparrow]

every maximal ideal is a prime ideal because R/M is a field and thus an integral domain makiing M prime

[next obsidian]

I got it

[wraith cargo]

well it's not maximal in the entire ring

[restive sparrow]

bruh

[wraith cargo]

it's maximal in this set of ideals

and it might not be a maximal ideal in the whole ring

[restive sparrow]

oh ya

[next obsidian]

Let S be the set of non-zero divisors, this is multiplicativelt closed and consider the map R -> Q(R) = S^-1R. This is an injection by knowing what the kernel of a localization map is.

Pullback a prime ideal of Q(R) and since all non-zero divisors got inverted there this has to lie inside of the zero divisors of R

[restive sparrow]

true

that is... more high tech 😭

[wraith cargo]

lol that's really elegant actually

[next obsidian]

I was thinking things over and really wanted to work in the total ring of fractions and realized I can

In fact the primes of the total ring are exactly those contained in the set of zero divisors

At first I wanted to prove every minimal prime consists of zero divisors which I can do assuming you know nilradical stuff by just localizing at the minimal prime, but this lets you get part of the way there without any Zorn’s

I came up with this in my very short shower lol

[novel star]

can you use eisensteins criterion on polynomials over C(t) with t transcendental?

its the field of fractions of C[t] right?

[delicate orchid]

exercise: prove that it is

[novel star]

it contains C[t], and any field containing C[t] clearly must contain C(t)

[delicate orchid]

I'm convinced!

[south patrol]

What do you mean by this? I assume you mean for polynomials over C[t]. And then yes, in fact you can generalise the statement to prime ideals in any integral domain

(The proof is the exact same)

One subtlety is that you may want to go between irreducibility over an integral domain A and its field of fractions. But you will be fine here e.g. by a form of Gauß' lemma applied to the UFD C[t]

[novel star]

thats comforting

ty!

[south patrol]

Np

[novel star]

is it known if the field obtained by adjoining all roots of unity to the rationals is quasialgebraically closed?

lang claims it is an open problem but that was a while ago

[spring pilot]

I’m getting confused with the wording of question 5. It refers to a ‘cyclic group’ generated by a, and then asks to show that this is a subgroup of G. What is there to show if we already assumed that a generates a group?

[kind temple]

i'm not entirely sure what you are asking.

problem 5 is saying to show that (i) for a in G, the set <a> = {a^n : n is an integer} is a group, and (ii) this group is a subgroup of G

[spring pilot]

I’m confused because it refers to the set as cyclic. So it just seems like there’s nothing to actually do.

[verbal valley]

i mean there isn't much to do, you basically just have to say that for y in {a^n : n in Z}, y^-1 is also in that set and also in G.

and that the product of two such things is in that set and also in G

actually since G is finite you don't even need to show the inverses lol

[kind temple]

for (i), you do what Dirichlet says, just check the group axioms.
(ii) is trivial

the steps for (i) that i would do are:

• show that the identity is in <a>
• show that for a^n in <a>, (a^n)^{-1} is in <a>
  once you have done this, (ii) follows immediately

[spring pilot]

So ‘cyclic’ is just doing all the heavy lifting then?

Closure: a^m o a^n = a^ m + n.
Assoc: Inherited from G.
Identity: Since it’s cyclic there in some integer (other than 1) k such that a^k = a. Then a^k-1 is the identity.
Inverse: a^m o a^k-1-m = a^k-1.

[kind temple]

well, a^0 = e is the identity too

[kind temple]

but this works because G is finite

[spring pilot]

Okay. This book only allows positive powers, but I see your point.

Thanks!

[kind temple]

oh? that is kind of odd i guess?

[spring pilot]

Yeah this is like a general math book, so I guess they just wanted to make things simpler.

[south patrol]

It's probably just cause it's a finite group and so it doesn't matter

[kind temple]

does it talk about the subgroup generated by a subset at any point?

[south patrol]

(But this is a bad definition of cyclic because e.g. Z is a cyclic group generated by 1)

[rocky cloak]

I feel like that makes it more complicated in the relevant cases and just wrong in general

[south patrol]

Wdym

Lol

[spring pilot]

No.

[south patrol]

Ah phew

lmao

[spring pilot]

My b

[south patrol]

Why did it say a has to not be the identity anyway tbh lol

[kind temple]

i think they wanted the divisibility criterion in part (b)

[spring pilot]

Because that would only generate a singleton set containing the identity?

[tribal moss]

But that's a perfectly good subgroup, and the order of e (that is, 1) certainly divides |G|.

[kind temple]

what is the order of e? is it 0 or 1? i forgot

[spring pilot]

Yeah I just mean it’s a trivial case.

[spring pilot]

Idk though, I didn’t write the book lol.

[tribal moss]

The order of an element g is the smallest positive integer n such that g^n=e.

Alternatively you could say that { n in Z | g^n = e } is an ideal in Z and the order of g is the generator of that ideal ... except that would lead you to say that a non-torsion element has order 0 instead of infinite order. (But on the other other hand, that would make the characteristic of a ring always agree with the order of 1 in its additive group).

[latent solstice]

Does someone wanna explain the N/C theorem to me?

I kinda get it

But not really

[knotty badger]

what's the N/C theorem?

[rocky cloak]

That the normalizer modulo centralizer is a subgroup of the automorphism group?

[latent solstice]

Yeag

The proof used a mapping f(a)(h) and im not really sure what that means

[rocky cloak]

Well you have a map
N(H) -> Aut(H)
By sending g to conjugation by g

And the kernel is exactly the centralizer

[rocky cloak]

Well presumably a is an element of N(H), so then f(a) should be an element of Aut(H), i.e. an automorphism

So then f(a)(h) is the image of h by this automorphism

[knotty badger]

this is typical for mappings that output functions

for each element $a \in N(H)$, you get a function $f(a) : H \to H$ defined by $h \mapsto a h a^{-1}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

the "name" of this function is f(a)

and the way you evaluate a function named, say, $\text{bob}$, would be $\text{bob}(h)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

thus we use the notation $f(a)(h)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

if you prefer, you can think of this as a two-argument function $g(a, h) = a h a^{-1}$, so a function $N(H) \times H \to H$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

and indeed in general you can always convert between "function that outputs functions" and "two-argument function"

you do this already for group actions, for example

where you can either view it as a function $\rho : G \times X \to X$ satisfying some properties, or a group homomorphism $\phi : G \to \text{Sym}(X)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[elfin wraith]

I also dislike this notation personally, I tend to think currying the variable looks better and shoving it down as a subscript but alas it’s what people seem to do

[knotty badger]

wait wdym by "currying the variable looks better"

[swift root]

f(a, h)

[knotty badger]

uh, i thought that was uncurrying

[swift root]

currying = applying the fact that Set is closed monoidal

[swift root]

oh could be

I refer to both as currying more to speak of the general concept of hom(A x B, X) = hom(A, hom(B, X))

[knotty badger]

yeah but my impression was that $\text{Hom}(A \times B, X) \overset{\text{currying}}{\longrightarrow} \text{Hom}(A, \text{Hom}(B, X))$

[swift root]

wikipedia seems to agree with you

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

that looks nicer

[knotty badger]

i usually call this tensor-hom

[swift root]

that's specifically for R-mod

[knotty badger]

yeah Set is F1-mod

[swift root]

brudging sounds of annoyed agreement

fine but only because I recently saw the q-analogue of the binomial coefficient counting the number of k-dimensional subspaces of (F_q)^n

[south patrol]

Currying.

[swift root]

internal hom

[south patrol]

Yes

[long nebula]

The real ones call it Schönfinkelization

[elfin wraith]

Oh yeah I have it backwards apparently, oops

Or honestly defining like g_a(b)=f(a)(b), I personally really don’t like the curried (as I’ve just learned ) notation, I just find it cumbersome and confusing

It’s common enough that I’ve like accepted it, but I try to avoid it if I can and basically never use it in my own notes, though if I’m writing something others will see I’ll usually just do whatever’s standard

[tardy hedge]

i like how straightforward this proof is

[velvet hull]

-# hehe, ass

[somber goblet]

ass ass ass ass ass

[tardy hedge]

p in Ass

[tulip otter]

I want to find Gal(L/Q) where L is the splitting field of x^3-2. So |Gal(L/Q)|=[L:Q]=6 and its easy to write all elements of Gal(L/Q) by looking at how each element of the group acts on a=cbrt(2) and one of the complex roots, call it w. But then what do i do?

do i start looking at powers of each element to see their orders and try composing them with each other

doing that randomly would take forever tho lol

[novel star]

S_3 is generated by a swap and a cycle

the swap can be given by complex conj

the cycle can be obtained by guessing

[cursive spindle]

Proof: a bunch of asses

[tulip otter]

and the 3-cycle is given by the map σ which sends a to aw and fixes w? because then σ(a)=aw, σ(aw)=a\bar w and finally σ(a\bar w)=a which forms a 3-cycle?

[novel star]

yes

[tulip otter]

I see

[novel star]

or you can notice that the only group of order six that acts on 3 elements is S_3 ig

[cursive spindle]

tfw u need to know enough about the standard groups to determine the Galois group

[tulip otter]

real

[tulip otter]

but in general it wouldnt be too efficient to write out all elements of the galois group and then look for relations lol

I mean there are too many things to try even for relatively small orders no?

[novel star]

you can also reduce mod primes to find cycle types

lang speaks of this

[cursive spindle]

You essentially learn a bunch of tricks to do this in general

Of course there is no "one in all" solution

[tulip otter]

yea thats clear

tho rn I am only allowed to write out the elements of the galois group and try to find a presentation that tells me what group this is

[novel star]

rip

[cursive spindle]

Working mod p is really nice

[tulip otter]

we havent developed any other tools in class yet so rip

[novel star]

or you prove dedekind theorem on test

[tulip otter]

yea it seems that there are nice results about this

I have to lock in tho, otherwise i wont know anything

[cursive spindle]

"Let's work modulo p like the computer scientists"

I think nothing pissed me off these days more than this quote

[novel star]

you have to cover it with fancy words to hide it from the cs people

[south patrol]

Lmao

Yeah that would ragebait me a lot

[velvet hull]

any and all radical ideals will be annihilated. especially the left ones

[cursive spindle]

Wait till you know it was Ravi Vakil

[azure cairn]

sorry for necroposting but i saw this in aluffi!!

and (more necroposting) does orthogonality have anything to do with the morphisms being orthogonal lol

there's a categorification of orthogonality/independence called naturality
perhaps more elaboration on this if you can?

[tardy hedge]

That was good

[velvet hull]

hehe ass

[knotty badger]

Wdym

[azure cairn]

The homomorphism is orthogonal to the group operation in the diagram

Although i guess that's stylistic choice

[vocal pebble]

looking at part b) here:
Let zeta_m denote a primitive mth root of unity. Then it suffices to show that [Q(zeta_m) : Q] = |Z/mZ^*| goes to infinity as m goes to infinity. I think that this is true, since all primes in between m and m/2 are coprime to m, and by the prime number theorem, there will be approximately m/ln(m) - m/2ln(m/2) many such primes, and this quantity goes to infinity as m goes to infinity.

but i feel like this is not required and some other counting argument will suffice but i cannot see what needs to be done

[knotty badger]

Yes that’s true

[vocal pebble]

oh hmm i guess i can just use the formula for phi(m)

[quiet pelican]

I claim phi(m) >= m/2^{#primes dividing m}
The result is immediate for powers of 2
phi(m) = (product over primes dividing m of (1 - 1/p))*m
And if p^r divides m, where p is the smallest odd prime that divides m and p^{r+1} doesn’t divide m, then phi(m) = (p-1)p^{r-1} * phi(m/p^r) > (p^r)/2 phi(m/p^r) >= m/2^{#primes dividing m} (by induction)

And #primes dividing m <= log_3(m), as long as at least 3 different primes divide m, so the original result follows after changing logarithm basis on the first inequality (do the cases of 1 and 2 primes dividing m separately, they’re easy)

[vocal pebble]

oh, so i was looking at this instead: [see below] and it seems like phi(n) >= sqrt(n). where n = product p_i^{k_i}

[quiet pelican]

Modulo some problems for p_1 = 2, yeah that works

(But you can easily deal with those problems separately)

[spark veldt]

Heya, so I'm currently trying to learn how to compute Galois groups of polynomials. In some of the solved problems I've seen, through some deduction (that I can still follow), they come to the conclusion that the Galois group of some polynomial is a subgroup of some well known group like Sn. Then they proceed to show that it's C2 x C2 or D4 or whatever though a series of random facts (like... transitive groups of order X are Y, or that subgroups of groups of order Z are A and B). At some point it seems like to find these Galois groups you need to have encyclopedic knowledge of these groups and subgroups, and I'm wondering if it's normal to have trouble with these?

Like.. is there a 'megalist' of these group properties?

[quiet pelican]

For Galois groups, it’s a good idea to know what subgroups of S_3, S_4, S_5 look like

Because you’ll rarely be asked to compute larger

[spark veldt]

ah I see, are there anything else worth spending time digging other than Sn for n=3,4,5?

[quiet pelican]

Not really
Like knowing how to compute the relevant info is useful
But at some point it comes down to brute force

You know your group is a subgroup of S_n for some specific, known value of n, and that it has xyz properties
Then list out all the subgroups of S_n with those properties and done

[spark veldt]

ah i see

say, if you're an algebraist working with larger groups say S_50, is there like a software that gives you all its subgroups?

[quiet pelican]

Yup
GAP is the usual one
Maybe not as large as n = 50 (that’s a very big group)
But certainly up to like 10

[spark veldt]

ooh nice i'll check that out sometime

thank you micoi!

[small yacht]

i understand that the first 3 diagrams represent associativity in groups and the identity. But what do the last 2 represent?

[quiet pelican]

The other axiom (inverse)

[small yacht]

Hm I guess i don't see well how they do it

[swift root]

writing all the maps down explicitly helps

[small yacht]

They're there i just didn't capture htem in the screenshot

[quiet pelican]

So tracing around the left diagram in the 3rd pair we get
(Going from top left to top right to bottom right): g -> (g, g) -> (g, g^{-1}) -> gg^{-1}
(Going around the other way): g -> 1 -> e
So putting it together, the commutativity says gg^{-1} = e

And the right diagram gives g^{-1}g = e

[swift root]

that's what I meant

[small yacht]

Ohhhhhh

thxx i get it now

[balmy python]

is there a way to prove the size of D_2n is 2n using its presentation?

[long nebula]

Which presentation

[quiet pelican]

You can prove from the presentation that every element can be uniquely expressed as s^k r^j
The hardest part is proving that s =/= r^{n/2} for n even
Which I’d do by just exhibiting a homomorphism where their images are not equal

[quiet pelican]

The usual one I assume

[long nebula]

Yea, establish a canonical form to write the elements in

[rocky cloak]

For presentation
<r, s | r^n = s^2 = 1, srs = r^-1 >

I'd go (r) is normal of order n.
D2n/(r) = <r, s | r = s^2 = 1 > = C2

[zealous sandal]

Can somebody help me with part a of this problem? I think I understand the idea, but I'm struggle to grasp what elements of R/M look like.

this is what I have so far

and my thought is that since f(x) + M and g(x) + M are inverses, I might be able to pull them up into R and if they are inverses in all of R, then we would have a problem since f(x) isn't invertible at c

[tough raven]

I can't formulate a formal consequence of the following stronger than what you have said, but I think it is in a very similar spirit! I attended a talk yesterday where the assignment of ultraproducts to families of models (and an ultrafilter) was interpreted as a "compact Hausdorff topology" on the category of models, by analogy to the fact that the category of compact Hausdorff spaces is monadic over Set and we can take (ultra)limits of elements of a compact Hausdorff space indexed by a set S along an ultrafilter on S (= extension of f: S → X to the Stone-Cech compactification S).

[swift root]

huh, interesting

i cant pretend to understand this well but it is cool

oh nvm i do

didnt read properly

[swift root]

the most important part for my purposes is that closure under direct limits and products implies closure under ultraproducts

this means that quasivarieties are equivalently prevarieties closed under direct products

so when you have some property that holds in a prevariety and you want to show it holds in a quasivariety you just have to show it is closed under direct limits

which is nice cuz imo they are far nicer to work with directly

+work nice with the functors i consider

[velvet hull]

actually no; on second thought I don't think this approach is going to be fruitful.
think about what kind of consequences the zeros of a function will have for that function and M

[rocky cloak]

Are you familiar with partition of unity, and compactness of [0, 1] ?

[vocal pebble]

In general you can show that if Z(M)={x in [0,1] | f(x) =0 for all f in M}, then Z is an inclusion reversing map (try to show it afterwards) ignore this message

[rocky cloak]

Is there a nice way to show that Z reverses inclusion without just showing the original problem?

[vocal pebble]

Okay I guess the argument reduces down to one point anyways when trying to show that Z reverses inclusions

[rocky cloak]

Also, it doesn't reverse inclusion for all ideals right?

Like something like
f s.t. lim[x->c] f(x)/(x-c) = 0
Should be an ideal that only vanishes at c, but is not M_c

[vocal pebble]

Oh yeah lol that's also related to part (c) of the same problem

I remember writing similar flawed arguments for a question in a test and getting full credit for them.. it seems like I managed to scam the graders

Ok but I guess you can use some sort of similar argument without mentioning Z.. ||if an ideal vanishes nowhere, then by compactness you can find finitely many functions whose sum is positive everywhere and so generate the whole ring. If an ideal I vanishes at more than one point, let c be some such point. Then I subset M_c|| which i guess is what you had in mind

[balmy python]

how do you determine that quotient group?

how do you know it has that representation

[rocky cloak]

To get a presentation for a quotient group you just add all the elements of the normal group as relations

I guess you still need to convince yourself that (r) has n elements though

That is less obvious

[balmy python]

is that not trivial from the definition lol

r^n = 1

i guess n may not be the order of r

[rocky cloak]

So it has at most n elements, that's clear

[balmy python]

order of r must divide n, so if n is prime then it's also clear

[rocky cloak]

Could still have order 1

[balmy python]

oh yeah true

ah this is so confusing

[balmy python]

would this n not be maximal by definition?

since if there was a smaller order m, you would just have a smaller D_2m

[rocky cloak]

Well you would need to prove that D2n and D2m are not the same

[twilit wraith]

Alright somethings bugging me

Let L be the Galois closure of Q(a) and let p be a prime dividing the order of G = Gal(L/Q). Show that there is a subfield F of L such that [L:F] = p and L = F(a).

Im struggling to prove that a is not in F

Somehow I have to show that, if H is the subgroup fixing F, then every automorphism sigma fixing a somehow gives a problem

[south patrol]

Kinda need more context lol like it depends how you define F

[twilit wraith]

Wdym

[south patrol]

What tools do u have access to

[twilit wraith]

All of galois theory

[south patrol]

I mean like they asked you construct an F and you're struggling to prove smth about your F

[rocky cloak]

I would say think about the conjugates of Gal(L, Q(a))

[twilit wraith]

If H is the subgroup fixing F, and a were in F, does that imply that H fixes the conjugates of Gal(L, Q(a))?

[warped pilot]

What is F.

[twilit wraith]

F is a subfield of L

[warped pilot]

Are you saying suppose you had such an F what properties would it have to have.

[twilit wraith]

Well there is certainly such an F

[warped pilot]

Ohhh

[twilit wraith]

Since theres a subgroup of G of order p

[warped pilot]

Right sorry yeah. Wasn't clear from how you worded it.

[twilit wraith]

Apologies, dummit and foote can be kinda unclear sometimes

[rocky cloak]

It implies H is a subgroup of Gal(L, Q(a))

[chilly radish]

g Gal(L/Q(a)) g^-1 = Gal(L/Q(g(a))

So if H fixes a it normalises this subgroup

[chilly radish]

Oh yea I guess that's more trivially true lol

[rocky cloak]

But yeah moving to the field side: what happens if F contains g(a) for all g?

[twilit wraith]

Then H fixes L clearly

Which is a problem

[rocky cloak]

So there must be some g(a) not contained in your F.

Then the next step would be to modify F so that it's a it doesn't contain instead of g(a)

[twilit wraith]

Would this be conjugation by g inverse

I see

The rest is straightforward

I was unaware that conjugation by g changes the indeterminate in Q in the gal group though

[rocky cloak]

I don't understand that that means

What are you saying?

[twilit wraith]

I did not know that g Gal(L/Q(a)) g^-1 = Gal(L/Q(g(a)))

[rocky cloak]

Ah I see.

Yeah that's just a general group action thing

[twilit wraith]

Im a bit rusty on groups is probably why

[rocky cloak]

If H fixes x then
gHg^-1 fixes g(x)

[twilit wraith]

Ah nvm that makes complete sense

[rocky cloak]

Because
( g h g^-1 ) (g(x)) = g h(x) = g(x)

[twilit wraith]

And im allowed to do this because group order is preserved under conjugation

Alr this is making total sense now

[tender linden]

i've done all of these and i don't really have a question about that but is this ring isomorphic to $(\mathbb{Z} /2\mathbb{Z})^{\lvert S \rvert}$ perchance

[cloud walrus]

artemetra

[tender linden]

or something like that

where |S| is cardinality and we have a direct product of all of the Z/2Zs

[quiet pelican]

Yes it is

[tender linden]

ayy nice

pretty cool

[knotty frigate]

Does anyone have any recomendations on how to tackle this?
My only idea works for solvable finitely generated abelian groups

but idk what to do if its infinitely generated or non-abelian

we havent gotten to sylows theorems and the previous chapters just covered the isomorphism theorems and normal groups

[knotty frigate]

the <= implication feels rather easy. if the order of P_{i+1}/P_i is prime then its Z_p so we can start working using that i think

[rocky cloak]

It's only true for finite groups as written

[knotty frigate]

oh

[knotty frigate]

What about finitely generated groups?

wouldnt they work?

[rocky cloak]

The order of G will be the product of
|Pi+1/Pi|

So the product of finitely many primes, so finite

[knotty frigate]

oh right lmao

so ig i only need to worry about the non-abelian case

which we really have little familiarity with. I was told smth similar to the fundemental theorem of finite abelian groups would be sylow theorems

[knotty frigate]

but can i write the group using a direct product the factor groups. idk why but smth in me wants to try smth like that or smth related to that

[rocky cloak]

So I'm assuming the definition of solvable you have is that G has such a series with Pi+1/Pi abelian.

In which case it's just about refining the series

[knotty frigate]

but then (G/N)xN isnt isomorphic to G

[knotty frigate]

yeah thats the definition

we ve been given

[rocky cloak]

So if you know how to do it for abelian groups and you know the correspondence theorem then youre done

[knotty frigate]

Okay ill try to see if i can make it work.

Thanks

If not ill come back tmrw

[chilly radish]

Well they're contained in one another so it's definitely not going to be a direct product

[hardy swan]

does anything interesting happen if you combine complex numbers and dual numbers?

[next obsidian]

You cause global unrest in the economy

[frail shoal]

something like C[ε] ?

that is, we have the complex numbers, and also we have ε such that ε² = 0

off the top of my head, the main application i can think of would be in AG because this encodes what's known as a "fuzzy point"

[wicked patio]

Maybe something with complex analysis and infinitesimals

Like, use it for calculus

[storm tiger]

How do I start it?

[rocky cloak]

The first thing to do is probably to identify some relations between those statements (cyclic groups are abelian for example).

Next you might want to try to use some of these statements or their complements to prove something about the order of gh

Or think about what having an order m element says about the order of the group

[storm tiger]

It divides the order if group

if we take klien group 4

and elements order is 2

[rocky cloak]

Right, so if every positive integer divides the order of the group what does that tell you?

[storm tiger]

D option is wrong

[rocky cloak]

Why do you say that?

[storm tiger]

K4 is abelian

[rocky cloak]

And how are you relating that to the problem?

[storm tiger]

order of its elements are integer(2)

[rocky cloak]

Okay, but 3 is also an integer.

Does it have an element of order 3?

[storm tiger]

It does not have any element

Of order 3

[rocky cloak]

So then it doesn't satisfy the property in the image

Because given m=3 you are supposed to find a g of order m (which is 3)

Perhaps a tangent / spoiler, but ||I would think whoever made the problem meant m, n and r to be strictly larger than 1.||

||It doesn't change the answer, but right now the problem is a little silly, since there clearly aren't any examples given m=n=1||

[storm tiger]

Actually I didn't understand the properly or its poorly formed

[rocky cloak]

Well, the group G is supposed to have that property that for any triplet of positive integers m, n and r you're supposed to be able to find g and h in G such that g has order m, h has order n and gh has order r.

So for m=2, n=2, r=1 we should be able to find such g and h, but also for m=3, n=7, r=46, etc

In particular if we just ignore n and r, then for every m we're supposed to find a g of order m

[storm tiger]

The problem is slightly confusing...can you elaborate it more
..last line

[rocky cloak]

Well for example: given m=4, n=something, r=something
Then your group is supposed to have g and h with
g order m (so g order 4) and then the other properties (h order n, gh order r)

So in particular the group has a g of order 4

(and nothing special about 4 of course)

[storm tiger]

So we need a group which has all types of elements?

Z2?

0,1 ...and Order 1 and 2 available

[rocky cloak]

It needs to have an element of order m for every positive integer m yes

[storm tiger]

Option D discard

Option A and C

Thank you

Next doubt

It satisfies dihedral group property

So D3=6 order of group

But doubt is

Elements of order >=3 always exists in pair and x,xy are self inverse then no. Of elements will not odd ?

[rocky cloak]

I cannot parse what you're trying to say here

6 is not an odd number... So it's correct that the number of elements isn't odd...

[tribal moss]

I think he's saying that the elements of order 2 are exactly y and xy, and a group with an even number of order-2 elements ought to have odd order.
But the presentation doesn't imply that y and xy are the only elements that square to 1 ... and without doing further investigation, it's not even certain that y and xy are different elements, or that one or both won't be the identity.

[rocky cloak]

y is the element with order 2, but I guess typo(?)

[tribal moss]

Whoops, fixed.

[rocky cloak]

Well undercops typo, not (solely) yours

[azure cairn]

this is a lot funnier with the new channel name

[toxic shoal]

,w cosets

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by yeetusdeletus5

[delicate orchid]

why is blud reading posts from 400 years ago

[azure cairn]

i like backreading stuff in large servers

[true bolt]

Well if ur going to comment on something can it be on topic at least

[azure cairn]

lmfao ok

could someone check this proof then

[wraith cargo]

Correctamundo

[azure cairn]

ty

this problem was funny because it's always the topic of those brainrot math videos from 2019

in similar disposition to "WHAT IS $\sqrt{2\sqrt{2\sqrt{2\sqrt{\dots}}}}$?"

[wraith cargo]

Kinds freaky that the answer is 2 ngl

Makes you think

Clearly 2 is some kind of fundamental constant in nature

[azure cairn]

true i just verified

take limits of both sides of x_n = 2sqrt(x_{n + 1})

[delicate orchid]

or just x = sqrt(2x) => x^2 = 2x => x = 2 (or the spooky answer x = 0)

[tall igloo]

this only works if you assume it has a limit from the start!

[vocal pebble]

I think you can just say that it's increasing and bounded from above.. √(2(√2...√2) ) n times = 2^{1/2+1/4+1/8+...+1/2^n} <= 2

[tall igloo]

yup exactly my thoughts

and once you have a limit you can do the algebra

[azure cairn]

Mct

I'm aware I did this in ch2 of my analysis book 🥸

[azure cairn]

[verbal valley]

the fundamental constant of ruining my theorems 😤

all my homies hate 2

[azure cairn]

LOL

Saved

[south patrol]

Lol

I have worked on two projects in my phd and in both characteristic 2 is the issue

[atomic python]

I'm a fake mathematician so I have no experiences myself.
But I have a friend who's earning her PhD and I've heard on two occasions her swearing about it.

Now that I think about it, the fact that it's been two occasions is rather apt.

[south patrol]

Lmao

[elfin wraith]

Its such a nepo baby prime anyway, hardly counts

[dire wren]

I think char 2 is nice

Like everything is self- inverse, is just so nice.

[tough raven]

when 0 ⨯ 0 = 1

[tribal moss]

Same as when 1 × 1 = 2

[dire wren]

Self additive* inverse

Is there anything nice done on complete(as in no gaps and no jumps) totally ordered non negative abelian monoids?

I am asking because I was thinking about metrics, but mapped to things other than R

I didn't say group, cuz that would just reduce back to R

[tall igloo]

monotone convergence theorem?

is that what this is called lol

i use that name exclusively for the thing in measure theory. i dont have a name for this thing, it is just trivial given completeness + the definition of limsup 💀

[azure cairn]

Yes

[azure cairn]

Yes it asserts bounded monotonically increasing (resp. Decreasing) sequences converge to their supremum (resp. Infimum,l

[tall igloo]

ye

[azure cairn]

You haven't heard of monotone convergence thm before? That's odd actually lol it deserves a name

[tall igloo]

i mean i know this statement i just dont think ive been calling it that

[azure cairn]

Yeah ofc

[kind temple]

is the mct from measure theory a generalization?

[azure cairn]

But it deserves its name because it's in a list of things that are equivalent to completeness of R

I.e. from mct you can derive completeness iirc

And it'd just very useful I think I've used it at least once every section of my ra book

[tall igloo]

yeah i was wondering if this analysis thing is that applied to the stupid measure space (a point)

i dont think so? integration there is just evaluation but MCT doesnt tell you the things actually exist

just that if they both either do and are equal or both dont

[kind temple]

it tells you something is measurable too, right?

[tall igloo]

ohh true

but MCT doesnt even assume boundedness

maybe it follows from DCT? lol

[tall igloo]

true

[kind temple]

i know almsot zero measure theory lol

measure(measure theory) = 0

[tall igloo]

shoot, at some point presumably soon my advisor will ask me to choose whether i will work at even or odd characteristic for the rest of my life

[south patrol]

You can do it for sequences of reals allowing convergence to oo

Indeed you are allowing for oo in the usual (= measure theoretic) MCT anyway

[tall igloo]

@south patrol what is the derived MCT

[south patrol]

Uhh

"Filtered colimits commute with everything"

Never fails

[tall igloo]

any functor worth its weight in gold

[south patrol]

I remember having to put my hand up and ask like lol are you assuming all functors preserve sifted colimits in a talk

[tall igloo]

anyways we should probably stop yapping in groups rings fields lol

[south patrol]

Which was correct fortunately lol like it is common in Goodwillie calculus

Yeah go tp #1203471755449073774

[kind temple]

omg i thought we were in #real-complex-analysis wth

[azure cairn]

No, this is the correct shitpost channel, it's not useful anyways

[tall igloo]

where is groupoids E_oo-algebras and uhhh (the theory is still being developed)

F_1...

[maiden crater]

Kind of stupid question

But here can't I say for the polynomial to be irreducible, (x-1)(x-2)..(x-n) would have to 1 for some n

which isn't possible

so we have a contraidction

Oh, that's false is it noy

[wraith cargo]

assume you can factorize your polynomial as f(x)g(x)

then what can you say about the factors if you evaluate them at 1,2,...,n

[maiden crater]

it will come out to be -1

[wraith cargo]

what does that imply about f(x) and g(x)?

[maiden crater]

f(n) and g(n) are inverses( effectively)

which is a contradiction?

[wraith cargo]

why is it a contradiction?

think about the degree of f(x) and g(x) respectively

[maiden crater]

their degrees has to be 0?

[wraith cargo]

yes, but why?

you're missing a key step in concluding that

[tribal moss]

Inverses is not really the right word here. Can you be more specific?

(The way I can think of concluding this with does not pass through "their degrees must be 0", by the way).

[maiden crater]

I mean |f(n)||g(n)|=1?

[maiden crater]

hmm?

[tribal moss]

And what are the possible actual values of f(x) and g(x) that satisfy that condition?

[maiden crater]

1

±1

[tribal moss]

Hmm, actually that's not specific enough.

You need to say f(x)g(x)=-1 without the absolute-value signs, and then consider what you can say about f(x) and g(x) as a pair, together.

[maiden crater]

lemme think a bit I guess

I'll come back to this in a bit?

[tribal moss]

Actually let's just reveal this step.

The level of specificity you need here is:

For each x in {1,2,...,n}, either f(x)=1 and g(x)=-1, or f(x)=-1 and g(x)=1.

Then continue with: "In each of these cases ..."

[maiden crater]

in each of theses cases, the OG polynomial has a root at each 1,2,....,n

[tribal moss]

What? No.

The original polynomial has value -1 as each of 1,2,3....,n.

And irrespective of that, that statement does not say anything about f or g at all.

[maiden crater]

well if I consider f+g, I find each n is a root?

[tribal moss]

So it would be weird to conclude "no matter whether f(x)=1 and g(x)=-1 or f(x)=-1 and g(x)=1, the original polynomial has such and such property"

[tribal moss]

That is progress.

Now what is the degree of f+g?

[maiden crater]

less than n

[tribal moss]

And thus you conclude ...

[maiden crater]

the polynomial f+g is 0

or f(x)=-g(x)

[tribal moss]

Yes.

[elfin wraith]

I missed the -1 at the end of this for so long and thought everyone here was going insane

[tribal moss]

My strategy was to keep my mouth shut until I (eventually!) noticed the -1.

[elfin wraith]

It took me a solid 5 re-reads and even grabbed a piece of paper to make sure I wasnt being crazy lol

[tribal moss]

Speak up if you give up on the final step from here.

[maiden crater]

From here we conclude that we have a contradiction as if f and g have opposite signs in the leading coefficient ,we won't get the deised polynomial with 1 as a leading coeff

[tribal moss]

Yep.

[maiden crater]

Thanks!

[rapid cave]

How to see that the obvious map $R \otimes_\bZ \bZ[X] \rightarrow R[X]$ is injective? This map is given by $r \otimes f \mapsto rf$

[cloud walrus]

ExpertEsquieESQUIE

[quiet pelican]

Show that it’s injective as maps of Z modules, using that Z[X] is the countable direct sum of Z (as a Z module), and R \otimes_Z Z = R as Z modules

[rapid cave]

I see

[tardy hedge]

what was the context for this map?

[rapid cave]

showing the tensor product is R[x]

its shown in a chapter about fibered products of affine schemes

[tardy hedge]

isnt it easier to define an inverse R[x] -> R (x)Z Z[x]

[rapid cave]

what inverse

[wraith cargo]

assume that r(x)f maps to 0
then you can write f as sum a_nx^n where a_n are in Z
Since rf is zero we have that r is annihilated by every element a_n
Hence we can write r(x)f = r(x) sum a_nx^n = \sum r(x)a_n x^n = \sum ra_n (x) x^n = \sum 0 (x) x^n = 0

[gusty thistle]

I would use that tensor product commutes with direct sums

[tardy hedge]

send every term rx^n to r(x) x^n?

[wraith cargo]

this works but it might be a bit of a pain to prove is like well defined

tho ig it shouldn't really be that hard

[rapid cave]

what about a sum of pure tensors that is sent to 0?

[wraith cargo]

well you can write every element as r(x)x^n

[gusty thistle]

When trying to prove things about the tensor product it’s usually a good idea to avoid talking about explicit elements. Maybe that’s taste.

[wraith cargo]

so you're reduced to the single tensor case

by linearity

since the coefficients are elements of Z so they can jump across the tensor product

[wraith cargo]

how would you prove this isomorphism without looking at elements?

[gusty thistle]

.

[wraith cargo]

ig that's fair lol

[rapid cave]

do you prove this using universal properties?

[rapid cave]

.

[tardy hedge]

You can define maps out of the tensor product by using its universal property

I think what he is saying is just R (x) Z[x] = countable direct sum R (x) Z = countable direct sum of R = R[x]

[rapid cave]

I understood this

[rapid cave]

I will prove this then

[wraith cargo]

yeah you can probably

definitely*

[gusty thistle]

Ideally yes

Like using the tensor hom adjunction

It’s just writing out the formula

But maybe that is not the most responsible (c) way to do it if you see it first time

But I would do it using tensor hom adjunction

Probably the responsible things is to do a proof by writing out elements first so you know how to do it. Maybe I was giving bad advice

[south patrol]

R[x] corepresents the forgetful functor from commutative R-algebras to sets. So does R (x)_Z Z[x] by adjunction

You can unwind this and use that the iso Hom_R(R[x], B) -> B is induced by evaluation yo check that this iso from above is indeed the usual map you'd write down

[pliant forge]

Suppose M is a R-S bimodule for (possibly noncommutative) unital rings R,S.

I am wondering in which instances ModS(M, -): ModS -> ModR has a right adjoint. I was hinted to consider the instance when M is finitely generated projective S-module, and that the right adjoint would be -(x)_S ModS(M, S), but that doesnt really add up as far as im concerned.

well firstly because the right S-action wouldnt make sense here

[rocky cloak]

So for it to have a right adjoint it would need to preserve colimits.

It preserves cokernels iff M is projective, and preserves filtered colimits iff M is finitely presented. So fg projective is correct.

In this case you can show that
Hom(M, X) = X (x)_S Hom(M, S), so the right adjoint would be
Hom_R(Hom(M, S), - )

In general a functor ModS -> ModR has a right adjoint iff it's naturally isomorphic to tensoring with a bimodule. In which case the adjoint is Hom as per the hom tensor adjunction

[pliant forge]

I was reading sources talking about the isomorphism, but somehow reading this kinda cleared up my confusion. thanks

[rapid cave]

the forgetful functor is covariant, so how can R[x] can corepresent it?

[rocky cloak]

Hom(R[x], -) is also covariant

[rapid cave]

corepresenting is not Hom(-,R[x])?

[tall igloo]

we call this represneting for some reason

[south patrol]

That's what corepresent means

[rocky cloak]

I'm assuming both conventions exist, but for me it makes sense that the representable functors are those in the image of the yoneda embedding and not the dual notion

[south patrol]

I would say this is what is usually done but sometimes people use representable to mean either

[rapid cave]

so this aligns with what is in the picture?

[rocky cloak]

No

[south patrol]

No

[rocky cloak]

Opposite of picture

[south patrol]

Usually representable means Hom(-,A)

[rocky cloak]

Or dual anyway

[south patrol]

Or maybe means both lol

[rapid cave]

lmao

ok, so its just a matter of context

[south patrol]

Well idk if that is how I'd put it

Matter of author

[rapid cave]

yes

[tall igloo]

i think topologists like represents for Hom(-, A) due to brown representability

[rocky cloak]

Algebraists work with compactly generated categories as well

[marble hinge]

Definition: Take the largest class of objects that seem to act the way we want.

Example: They sometimes still don't act the way we want.

Definition: An object is called "normal" if it really does act the way we want.

that's a funny thread on Reddit that I found :)

https://www.reddit.com/r/math/comments/dpbkwl/graduate_algebra_classes_in_a_nutshell/

[verbal valley]

see also:

Defintion: A fneeb is a polysemantic foo satisfying the...

Student: Don't we need the axiom o-

Lecturer: Henceforth all fneebs are finite dimensional

[swift root]

infinite dimensional case will never pop up anyways...
(famous last words)

[quiet pelican]

I mean I’m literally thinking of rewriting my paper to be like
Definition: We call X a foo if it is nice
Theorem: X is a foo if and only if [horrible technical conditions that are relatively easy to check]
Theorem: X is a foo if and only if [nice theoretical thing that we care about that’s very hard to check]

[tribal moss]

Hide the two theorems in a hideous 20-part "the following conditions are equivalent", and then prove them as a network of implications that make a non-planar graph.

[tardy hedge]

wait that reminds me when i did comp sci classes in undergrad they LOVED foo man

in all the toy exercises they loved foo

WHY

[novel star]

because bar

[tardy hedge]

wat dat mean

i remember bar too

it was foo and bar

if foo = 5 and foo = foo + 1 what is print foo

type shi

[novel star]

i love strange conventions

but i dont trust computer scientists

[tardy hedge]

ya they weird

(and stinky)

[novel star]

exactly

all the good ones r just mathematicians

[tardy hedge]

most interesting comp sci class i did was learning assembly

[novel star]

that was the point when i fully switched to math

scary

[novel star]

but then again i cant really say ive swirched from or to anything

i used to be somewhat interested in computer architecture

but then i realized i dont have an engineer brain

modern architecture is magic

truke

[tardy hedge]

Lol

[kind temple]

is there no constructive proof that ab = 0 —> a = 0 v b = 0, say, in the integers?

[long nebula]

Assuming you can't use the law of excluded middle, I really can't see how you would prove that

[swift root]

wtf

thats crazy
so without LEM unique prime decomposition doesnt necessarily exist either?

without LEM, Z can just fail to be a domain

[velvet hull]

not even prime decomposition, just any proof by contradiction

[swift root]

yes i know

[long nebula]

How does prime decomposition rely on LEM

[swift root]

i guess i just never realized

[long nebula]

That seems entirely constructive: given a number, I can tell you an algorithm to compute its factors

[swift root]

uniqueness?

[long nebula]

Wait you can still cancel stuff on both sides though? Like ab = ac implies that b = c for a ≠ 0

I think?

[swift root]

ab = 0 = a0 => b = 0 for a ≠ 0

0 = a ⋅ 0 is by definition of multiplication

[long nebula]

Hmm I'll have to think about it more

[long nebula]

What axioms are you starting from

Peano arithmetic?

[kind temple]

sure

[kind temple]

how do you prove this constructively?

[swift root]

i was just showing that right cancellation implies ab = 0 => a = 0 or b = 0

[kind temple]

ah

[swift root]

i.e., you cannot assume right cancellation

and uniqueness most likely breaks down

So suppose a, b are nonzero but ab is. Then a = a(1 + b). Both a and 1 + b have prime decompositions, and multiplying them gives a new, definitely different prime decomposition for a

so uniqueness must fail

[long nebula]

I think you're assuming LEM here?

You need a=0 or a≠0 for this to work

[swift root]

how does one prove X or Y then?

like genuine question

[long nebula]

You have to either prove X or prove Y (for intuitionistic logic)

[swift root]

doesnt there always need to be some casework though

can you show me an LEM free proof of a statement of the form P -> X or Y?

[long nebula]

Actually hm "a=0 or there exists b such that a=Sb" is provable by induction for natural numbers a

[tribal moss]

If I recall correctly, you can build up to "x=y or x!=y" on naturals by (double) induction too.

And then you have a fair amount of LEM that you can make use of.

[somber goblet]

fr

its all Z/2Z-algebras

everywhere

[tribal moss]

Always has been.

[kind temple]

yea, been playing around with a lean proof.

[long nebula]

Thinking about it more, this actually is straightforwardly provable for natural numbers:

Induction on a.
Base case: a=0, obviously proves a=0 v b=0
Inductive case: a=Sc. Then, 0 = ab = (Sc)b = cb + b. So b=0, proves a=0 v b=0.

[kind temple]

i think i have one using double induction/cases

[tribal moss]

If there's any justice, you should be able to prove trichotomy too.

[long nebula]

(Lemma I think should be easy to prove but not sure: adding two natural numbers and getting 0 means they're both 0.)

[kind temple]

0 >= a + b >= a,b >= 0

[long nebula]

Yea this is easy with induction

[kind temple]

as long as you know that a + b >= a,b, then you are good

[long nebula]

I found these nice notes on Heyting arithmetic https://www.cs.cmu.edu/~crary/317-f22/lectures/08-heyting.pdf

[tough raven]

In the integers equality is decidable, so I don't think constructive makes any difference.

More explicitly, say you use a definition of the integers like "natural or -natural except 0 = -0", with naturals defined by (intuitionistic) Peano arithmetic. Then reason by 4 cases if a, b are naturals or -naturals. In each case, if either natural is 0 we are done, otherwise we have (Sx)(Sy) = (Sx)y + Sx = S(...) ≠ 0 (i.e. you use a principle of explosion here) for the naturals and the product of the integers is this or -this.

[novel star]

i was trying to show that for any prime ideal P of k[X_1,...X_n], ht P + coht P = n, but i couldnt do it for some reason. could i have a small hint please?

maybe i just need to know more about transc extensions

[next obsidian]

This shit is lowkey annoying as fuck man

Imma be honest I think it’s worth it to just read a proof

Unless it’s hw

In which case you gotta say more about what you can use and what you’ve done

I lowkey forget the best way to do it

[crystal vale]

If A is ring and S is a multiplicative set, then S^-1A \otimes_A S^-1A is isomorphic to S^-1A as A module , right?

[next obsidian]

Yes

Prove that tensoring with S^-1A is the same as localizing at S

And it just becomes the statement that S^-1(S^-1A) = S^-1A

[crystal vale]

Yes i shown that

[crystal vale]

So this is isomorphic as S^-1A module

If M -> N is isomorphic as B module then is that isomorphic as A module if A -> B ring homomorphism exists

Yes it will be

[next obsidian]

Yeah

[rocky cloak]

Consider
0 < P1 < P2 < ... < P
Giving the height of P.

Then P1 is generated by some irreducible polynomial, and let's say wlog the polynomial contains X1. Then changing variables Xi |-> Xi + X1^m we may assume it's monic in X1.

Calling the polynomial f we see that f, X2, X3, ..., Xn is a trancendence basis, so generates a polynomial subring, let's call it S.

Then by the going down theorem PnS has the same height as P, and by the going up it has the same coheight.

But as PnS contains f, we can reduce to S/(f) which is the polynomial ring in X2, ..., Xn. So by induction the theorem holds

[granite hull]

Hello! Why do we study objects such as fields, groups, etc?

[next obsidian]

This is an incredibly broad question

A lot of stuff that we care about have this structure on them

And so by learning how to study that we learn more things

Maybe one example I can say is that some guy… I wanna say… Dedekind??

That doesn’t quite sound right, but he “proved” Fermat’s Last Theorem a long time ago

But implicit in his “proof” was an assumption which we would now state as being that all rings of integers of number fields are UFDs

These are basically taking the integers and also allowing you to use some extra complex number, so like take numbers of the form a + bsqrt(-5)

Where a,b are integers

This forms a ring, and it turns out there isn’t “unique factorization into primes” like you have for the integers because 6 = 2•3 = (1+sqrt(-5))(1-sqrt(-5))

This failure meant that the proof of FLT was actually invalid, and we even have a group called the ideal class group for rings of this form which tells us exactly how far the ring is from having unique factorization

This group is just the identity precisely when it’s a UFD, so in some sense the large the ideal class group is the “farther” away from being a UFD you are

All this stuff requires the machinery of groups, rings, fields in order to state, but it all ends up relating to something very concrete like solutions to the equation x^3 + y^3 = z^3

[elfin wraith]

There’s a million possible answers to that.

One might be that groups encode symmetries of an object and this can be useful to exploit in physics. Or that the geometry of curves, surfaces etc can be explained by polynomial rings. Maybe you like number theory where there are just endless connections. Maybe you like topology where many of your most powerful invariants are algebraic.

Slightly more philosophically, and I guess my motivation in a sense, they’re a very natural feeling thing to study. In primary school you start learning how to add positive numbers. Eventually you get good at this and start doing subtraction. Eventually you start to look at multiplication and then division. In highschool you start to study polynomials and factorisation.

All of these things behave quite differently, and learning how they behave is algebra. Here you’ve passed from a monoid, to a group, to a ring, to a field, to a polynomial ring, you’ve been studying a small aspect of these all your life. It makes sense then to pick out exactly what makes those structures tick and study them abstractly

[next obsidian]

Ur answer isn’t as good as mine haha

[elfin wraith]

The first example I can think of where that would be useful would be the end of highschool when you start to do linear algebra and learn about matrices. You now have objects which you can add and multiply and subtract, but in a fundamentally different way. Now your multiplication isn’t commutative anymore, and you can’t always “divide” matrices. This is new behaviour, but still really very similar to what you already know

[elfin wraith]

https://tenor.com/view/blows-you-up-with-mind-gif-2633489430227586185

[next obsidian]

End of highschool
Linear algebra

Bro nope your European is showing

[elfin wraith]

Idk NA chud skill issue

[void marsh]

more philosophical
Yesh! I just wanted to say... number systems are constructed very naturally, almost generatively. The fact that you go from free monoids, to the group of integers and so on until algebraic completion, it feels very transcendental... learning about abstract algebra is like learning the foundation to the universe ^_^

[granite hull]

Sorry, I was offline, let me read all of your answers

[granite hull]

So, they act like backbones, in a way?

[next obsidian]

Read mine first it’s the best

[granite hull]

Or, categorizing species into families?

[next obsidian]

It’s more like, if you had a thing and you cared about understanding it, and you found out it’s a bird, wouldn’t it be good to study birds?

[void marsh]

cute analogy

[elfin wraith]

Similar to chmonkeys example though is the classic fact that there’s no formula for the roots of a degree 5 polynomial. You definitely learned how to solve the roots of ax^2+bx+c in highschool using the quadratic formula.

You may not know, but there does exist similar formulas for degree 3 and 4 polynomials (they’re just very huge and horrible lol), but you can show that there’s in fact no formula for any higher degree polynomial and this is a profoundly algebraic problem. In fact this is in some sense what started algebra as a field of study. This fact falls out of a certain correspondence between groups and fields (the Galois correspondence)

[next obsidian]

Somehow incredibly similar to the fact Nope just said, you can disprove various compass and straightedge constructions which are problems dating back to the Greeks

Such as squaring the circle, and trisecting an angle

This uses Galois theory, which finds a remarkable relationship between fields and groups, and the structure of the subfields and subgroups

[elfin wraith]

This fact is still a mystery to me because I never cared to read that section of my Galois notes

[granite hull]

So, from what I understand from this,

A lot of different things, have similarities between them? So we sort them into uh.. little categories and study their common features?

[void marsh]

Idk lots of galois theory but its so cool... remember trying to understand the proof for this like 2 years ago,,

[granite hull]

Sorry, I’m very very new to this, what is galios theory?

[next obsidian]

Tbh I just took as granted that the actions they described corresponded to the algebraic operations.

Like I had to learn right then what the compass and straightedge constructions are so it was kinda like ok

[elfin wraith]

Yeah that’s how I think about. Then maybe you come across some object you’re interested in for whatever reason, and if you realise that “oh hey this is a ring!” You now have access to a whole toolbox that applies to any ring

[next obsidian]

Given a field F < L, you can find a relationship between intermediary K that fit between them, and subgroups of a certain group

That group is the automorphisms of L which fix F

[elfin wraith]

And sometimes things have a bunch of different levels of structure you could look at, and it’s often useful to look at them at these different levels

[granite hull]

Wait, I just learnt the definition of a field today please have mercy

🥀💔

[next obsidian]

So subgroups of that correspond 1-to-1 to those intermediary fields

[verbal valley]

because they are interesting :)

[next obsidian]

Yeah I mean I don’t expect you to understand the Galois theory aspect of it

[quiet pelican]

Groups are the cutest mathematical object :3

[next obsidian]

Our point is just that we have classical problems which ae couldn’t solve for ages

And this purely algebraic gadget called Galois theory somehow cracked the problem

So that’s why we needed to study groups and fields

[elfin wraith]

You’ll get there don’t worry, it’s just all about maps between fields, and studying how different fields are contained within another.

Think about how x^2-2 has no solutions over Q, but it does over R, or how x^2+1 has no solutions over R, but does over C

Galois theory studies this sort of phenomenon

[next obsidian]

In a sense yeah

You took calculus right?

You have some function you wanted to understand

And you’re like damn I wanna know how fast this grows

Then you realize there’s something called continuous functions and you realize cool things about them, like the intermediate value theorem

[quiet pelican]

It’s essentially the observation “ruler and compass constructions correspond to repeated quadratic extensions of the coordinate field, and so produce extensions of degree 2^n, but [coordinate needed for XYZ construction] is not contained in any such extension”

[next obsidian]

Then you realize a lot of those can be differentiated so you develop that theory

And now you can understand your cool function way more

[granite hull]

I, hate to disappoint you all, can I explain my situation

[next obsidian]

No.

[elfin wraith]

Yeah I can see that, makes sense. I just never cared to read it because like, what year is it 126? I don’t care about ruler and straight edge constructions

[next obsidian]

I actually gotta get ready for bed

Nope is gonna take over and do an okay job at taking my place

:^)

[void marsh]

i wanna know ur situation

[elfin wraith]

Aww thanks chmonkey

[quiet pelican]

Yeah the interesting thing to me is that it basically requires no theory except some very basic linesr algebra

[verbal valley]

bro has never been threatened at gunpoint to trisect an angle \silly

[granite hull]

So, technically, while I did take calculus, I only learnt the most basic of the concepts properly. Like, power rules, chain rules, and maybe a bit of trig, like, sin x = opp/hyp. Other things, I’ve been just, self studying, from random textbooks and YouTube videos, so my knowledge, is full of holes

[granite hull]

My country, is ass.

[quiet pelican]

Give me a protractor :3

[elfin wraith]

That does sound like a normal first introduction to calculus. Depending on where you are in the world the more developed “mathematical” might be called analysis. Are you at uni right now?

[granite hull]

I,

[elfin wraith]

Not appropriate

[void marsh]

why u learning math

[void marsh]

SORRY i just wanted to say

[granite hull]

I am starting uni in a few months, actually. Maybe in like, 2

[elfin wraith]

That’s reasonable then, don’t worry

[granite hull]

I love it, I don’t think there’s any other reason

[verbal valley]

you can have a tomahawk but no protractors :3

[elfin wraith]

A lot of people, especially in America, go to uni knowing no calculus and certainly no algebra

[void marsh]

why do u love it

[elfin wraith]

You’ll get there in time, it’s a very very long ultra marathon, not a sprint

[void marsh]

what are the memories that come to mind that make you love math

i remember learning about the mandelbrot set in highschool... or about neural nets and backpropagation recreatively back then... I got lots into logic from compsci and learning formal logic also helped a lot...

[swift root]

youre not even expected to know pure math of any kind at all, really

because that's (unfortunately) just not taught in standard curricula

[granite hull]

Thank you so much,

[granite hull]

For this question,

I don’t have any particular incident that made me love maths, I think

Hm actually I do

[void marsh]

@granite hull me an artist. You can apply abstract algebra directly to many seemingly unrelated things. Lots of formal poetry applies group theory, I myself have superficially researched formal poetry. Same for music, music is deeply mathematical. My whole life I applied math concepts to art and I think math is a beautiful muse

so yeah thats also another cool reason to learn abstract algebra me thinks

[void marsh]

i wanna knowww

[granite hull]

For some reason, I bought a maths textbook a few years ago, maybe 6

It was like, an introductory textbook? I think? It had things like, rational expressions,

I just did the textbook, and idk, I think I fell in love with it

Anyways, thank you for the help, yall. I really appreciate your answers, they did help me gain clarity

[marsh fulcrum]

SPURS SCORED A GOAL

GLORY GLORY TOTTENHAM HOTSPUR

[elfin wraith]

#discussion please

But also no, it would be incredibly funny to see spurs relegated

[next obsidian]

The spurs are in danger of being relegated? Lol

[long nebula]

Idk, seems pretty clearly aboit fields to me........

[vestal jay]

Can I just say that gN = Ng implies gNg^-1 = N through left multiplying both sides by g^-1? Or am I butchering some rigour?

[quiet pelican]

Yes you can say that

[rocky cloak]

That's fine yes

[vestal jay]

Thanks

[rocky cloak]

Ng is by definition the set
{ng : n in N}
So
Ngg^- is
{n g g^-1 : n in N} = N

You can always apply the same operation to two equal things and get something equal

[tardy hedge]

is the justification for this some universal algebra thing

[rocky cloak]

I wouldn't say so.

It's just the definition of a function / well definedness.

[quiet pelican]

It’s the definition of being well defined

[rocky cloak]

If x = y then f(x) = f(y)

[swift root]

what jagr and micoi said

formally you can see that G acts left-and right on its subsets

and this is a biaction in the sense that g(Sh) = (gS)h for all subsets S and g, h ∈ G

this all follows from the associativity of G

[kind temple]

can’t quite see why there is this bound on |Sg(X)|

[swift root]

Do you understand the bound |E^n(X)| <= |X| + |F| + omega?

3.2 says that Sg(X) is the union of all E^n(X), so the result follows from the above

[kind temple]

yea, i can’t see how the proof of 3.2 is being used

[rich granite]

Did enpeace recommend you that UA book 😭

[kind temple]

like sum |E^n(X)| <= w(|X| + |F| + w)

[kind temple]

i think once before but i’m studying it on my own lmao

[rich granite]

Ah okay lol

[kind temple]

is it just this but fiddling with the cardinal arithmetic?

[rich granite]

Yes that seems to be it. Because unioning two countable sets leaves it countable

[rich granite]

As is any countable union of countable sets

[swift root]

Sg(X) is the union of all E^n(X), then |Sg(X)| is the sup of all |E^n(X)|. Since |E^n(X)| \leq |X| + |F| + omega, we have that the sup of |E^n(X)| which is |Sg(X)| is \leq |X| + |F| + omega

[kind temple]

ah, that’s much more clean thank you

[swift root]

yeah its kinda fiddling with cardinals but its a good result

so in particular if |F| is countable infinite then Sg(X) \leq |F| for all X with |X| < |F|

[kind temple]

when is this result useful?

[swift root]

I don't quite recall exactly where but I do remember it is used somewhere

hold on

nvm cant find anything 😔

I do know i've used it before though

[kind temple]

lol alr all good

is the irredundant basis theorem worth learning? seems kind of random

[swift root]

eh

Its a cool result but i dont care abt it

oo now you're gonna get to congruences

[kind temple]

yess

i skimmed this and it looked very fun

i was also thinking about the section on term algebras

sections 10 and 11 of this chapter

those are the ones im looking forward to the most

[verbal kraken]

bro finkes hes 1/mu e

can I be the epsilon to your c? together we will mu

one of me, two of you, multiply together and we mu

well actually one of you cus your already c^2

but that doesnt rhyme

[kind temple]

https://tenor.com/view/say-whaaat-gif-18261283115155956121

[verbal kraken]

whats your elo? I bet Im below

c? yo?

Maybe theres no chemistry between us but Im atleast feeling an electrical tension

[tardy hedge]

wat u doin pal

[verbal kraken]

rizzing

[tardy hedge]

NOT. ALLOWED.

[verbal kraken]

I was about to rizz you but I fucked up so now idk what Im gonna do

[tardy hedge]

🥀

[kind temple]

im unrizzable

[verbal kraken]

mb it was a sign that I should only rizz c^2

because he is the one*9x10^16 for me

[tardy hedge]

Ts is poetic

[verbal kraken]

It's

I bet its kinda toothy. Cus by the communitive property of multiplication you are kinda23

[verbal kraken]

technically its 1/mu but sometimes math isnt on our side

I bet theres more rizz hiding here in the abs alg server but I kinda only know hs algebra...

and calculus but thats not relvant

[verbal kraken]

is that why you named yourself kiand123. Did I find the secret puzzle?

how do I apply for moderator?

[swift root]

for me my favorite chapter was Malcev conditions i think

the most frustrating too

cuz they are hard

[swift root]

poetic? this looks more like a psychotic episode to me

[kind temple]

what did you enjoy about them?

[swift root]

the idea that you can make terms magically appear out of thin air by just requiring that the variety satisfies a property that on the surface has nothing to do with terms

where tf did ideals come from

[swift root]

take for example the variety of semigroups. We know that sets can be interpreted as trivial semigroups, so it follows that e.g. semigroups cannot have "special properties" that hold for all semigroups

because else a Malcev condition would make nontrivial terms appear, which sets cannot satisfy

[frail shoal]

whao

that's beautiful

[kind temple]

wait that is pretty cool

[frail shoal]

this reminds me of a useless brainrot fact

(i collect a lot of these)

the singleton set has 2 distinct cosemigroup structures and they correspond to the two "natural" ways to equip a set with a semigroup structure

in particular, ab := a and ab := b

[tough raven]

What was the original exampla again?