groups-rings-fields

[cloud walrus]

Shuri2060

[coral shale]

Hence...

I would try to write these proofs as minimally as possible without omitting detail

[pastel cliff]

oh that's even better actually

[chilly ocean]

Basically what was already said but my first instinct would be to just left multiply the last line by ab and you no longer need to justify the inverses thing, you just have ab=ba

[lapis trail]

So many domains

A is a Sussy Baka Domain

[pastel cliff]

is this also valid feels super short

oh maybe this only works for positive exponent

but expanding to negative wouldnt be much work

[lavish nexus]

if gh=hg
1 = h^(-1)g^(-1)gh = h^(-1)g^(-1)hg

[coarse storm]

Corollary: Odd numbers are also even!

[chilly ocean]

curious question

equivalence classes come from equivalence relations

are there other relations that have classes with more interesting properties?

[pastel cliff]

whatd i do lol

[chilly ocean]

not equivalence

[chilly ocean]

General argument makes sense but the way you paired everything in twos doesn't check out lol
I think you meant to just yeet every g into one big thing of parens and the hs into another big thing of parens?

[pastel cliff]

yeah pretty much

this class is teaching me more about proof writing than algebra kek

ohhh i see i thhink

then in the second line can i just jump to a full parens of g and full parens of h

[chilly ocean]

Yeee cause if the length of the product of (gg)s is k, then we have g^{2k} which doesn't make sense for odd n

[pastel cliff]

like $(gg \cdots g)(hh \cdots h) = g^n h^n$

[cloud walrus]

μ₂

[chilly ocean]

Yeah that makes the implied induction more clear

[pastel cliff]

and i dont need to do anything to generalize it to negative powers

[chilly ocean]

You def gotta do that case but it's just a light tweak

You got dis

[pastel cliff]

ok i thought so

merci

[barren sierra]

This reminds me alot of Euclid's Lemma

is that why they're called Prime ideals?

[tribal moss]

Exactly, yes.

[barren sierra]

neat

also

is there a clean way to do the second part of this?

other than "let x = a + bi + cj + dk and y = w + xi + yj + zk" and going into expansion hell?

cause I can do the expansion hell but I'd really rather not ya know?

[next obsidian]

Just do the computation

[barren sierra]

pain

ok

[thorn delta]

How do you define prime ideals for an arbitrary ring?

because this happens to be how they are defined in hungerford

[barren sierra]

oh the definition assumes R is commutative with 1

[thorn delta]

ah okay

[barren sierra]

even tho in general rings don't have to have either

[next obsidian]

I think you have to define left prime and right prime in general rings

[barren sierra]

I think so yea

my notes says that formulation is a bit more complicated

[thorn delta]

why, P is prime if IJ \subset P implies I subset P or J subset P seems completely symmetric to me

[tribal moss]

But "ideal" itself isn't symmetric in non-commutative rings.

[barren sierra]

it seems like

commutativity you can say ab in P => a in P or b in P
noncommutativity you can say AB subset of P => A subset of P or B subset of P

an example where the latter works but the former doesn't work I think is the 0 ideal in the ring of n x n matricies?

that's what wikipedia says but I don't quite follow lol

I probably misread that

[chilly ocean]

I looked up a solution to that problem I was stuck on after trying something related to the hint Mat gave me

It was a little complicated but I digested it into a couple parts: first recognizing that the set of remainders, if they are all constant, of division of elements of an ideal by a fixed polynomial, is an ideal in the ring of constants. This is very useful when quickly switching views between a polynomial ring with several variables and the naturally isomorphic one in any one of those variables

The rest of it is not hard to show with the goal in mind of generating everything with just two elements

Basically having a 1st degree polynomial show up is nice in polynomial rings and can often tell you about structure just by playing with division since you're guaranteed to get constants

Definitely didn't expect that one of these problems was gonna include an entire hidden "hey to really obviously crack this you need to prove some set that's not necessarily obvious is an ideal" so it was definitely a harder one

In hindsight it kinda makes sense from a problem solving perspective, we want to show anything in some ideal, containing this other ideal generated by k, can be written (p, q) for polynomials p and q. Just say p=k and now you have, for f in the ideal, f = ak+bq. Compare with the division f = gk+r. If we can find b so that r=bq then we're done, so you might think we need to show the remainders are an ideal, or at the very least an ideal in the constants

[nocturne cobalt]

hey im a bit confused on how there is a homomorphism between Z/2Z and S_3 the symmetric group, can anybody help?

Im guessing Z/2Z is defined as the elements {0,1}

so how do we recover the group of S3 from this?

[tribal moss]

You don't need to recover all of S3.

Remember that a "homomorphism" is simply a map whose values lie somewhere in S3 (and which preserves the group operation).

At the very least there's always the homomorphism that maps both elements of Z/2Z to the identity permutation in S3.

(There are exactly three other homomorphisms in addition to that).

[nocturne cobalt]

ah right that makes more sense

[nocturne cobalt]

what does this look like explicitly?

so if we had phi(1)

[tribal moss]

phi(0) = e and phi(1) = e?

(Such a zero morphism exists between every pair of groups).

[nocturne cobalt]

mhm im still confused 😅

i just cant see how its mapped

how does phi(0)=123-->123

[tribal moss]

I'm confused by your confusion. Do you agree that the identity map is an element of S3?

[nocturne cobalt]

yes

so how do we have two identity maps?

[tribal moss]

We have only one identity map.

phi(0) and phi(1) are both the same element of S3.

[nocturne cobalt]

why is that?

[tribal moss]

Because that's how I defined the phi we're talking about.

[nocturne cobalt]

oh right

so i could pick any phi

for any z/2z element

i.e.

phi(1)=(12)

[tribal moss]

Yes, as long as your choices satisfy the homomorphism condition: phi(a) o phi(b) = phi(ab) for all choices of a and b.

[nocturne cobalt]

so the operation on LHS is modulo

what is the RHS operation?

permutations?

[tribal moss]

The group operation in S3 is function composition.

[nocturne cobalt]

oh yeah my bad 😅

so if I wanted to prove that Im (phi) is not a normal subgroup in this case

how would i go about doing that?

I have to show conjugation is not invariant right?

[tribal moss]

That would be one way. (Notice that the image of the zero morphism I showed is a normal subgroup, though).

"conjugation is not invariant right" should be "that the subgroup is not invariant under conjugation".

[nocturne cobalt]

so with the phi i picked

can we say let a be an element of S3

then we do something like phi(1 o a o 1)=phi(1) o phi(a) o phi(1)

so we have (12) o phi(a) o (12)

[tribal moss]

Since a lives in S3, you cannot take phi of it.

[nocturne cobalt]

oh yeah

oof

[tribal moss]

"Invariant under conjugation" would mean that whenever you pick a in Z/2Z and b in S3, the value of b^-1 o phi(a) o b always lies in the image of phi.

[nocturne cobalt]

yeah i understand

im trying to show that it isnt a normal subgroup

[tribal moss]

Right. So you're looking for a counterexample of this property.

[nocturne cobalt]

yah :X

[tribal moss]

I.e. a concrete a and b such that b^-1 o phi(a) o b isn't in the image.

[nocturne cobalt]

correct

the inverse of a permutation say 123 is 321 right

[tribal moss]

The parentheses are required in cycle notation, but the inverse of (123) is (321), yes

[nocturne cobalt]

and phi(1)=(12)(3) right? for my defined phi

[tribal moss]

Yes.

[nocturne cobalt]

so now i just have to compose say (123) o (12)(3) o (321)?

[tribal moss]

Yes.

[nocturne cobalt]

and show that it isnt in S3?

[tribal moss]

Oh, it will automatically be in S3 -- it's a permutation of {1,2,3} after all, and S3 is the set of all such permutations.
You need to show that it isn't in the image of your phi.

[nocturne cobalt]

isnt the image of my phi S3 tho?

[tribal moss]

No, the image consists only of the two elements phi(0) and phi(1), whereas there are six elements in S3.

The "codomain" if your phi is S3, but the "image" is only the elements that are actually hit.

[nocturne cobalt]

oh yeahhh

so (123) o (12)(3) o (321) gives (1)(32) which is in S3 but isn't in the image of phi since phi(o)=e and phi(1)=(12)(3)

is that right? :X

[proud bear]

yes

[nocturne cobalt]

thank you so much :))

[nocturne cobalt]

Do inner automorphisms form a normal subgroup ?

[proud bear]

yes, they are a normal subgroup of the group of automorphisms

[nocturne cobalt]

and the group of automorphisms is a group of isomorphisms that map back to the same codomain?

[proud bear]

yeah, for a group G the elements of the automorphism group of G are isomorphisms from G->G

[nocturne cobalt]

so just to be clear

G is >= Aut(G) >= Inn(G)

[next obsidian]

G does not contain Aut(G)

[nocturne cobalt]

i mean in terms of

elements/dimension

[next obsidian]

No

[coral shale]

sounds wrong

[nocturne cobalt]

G is a bigger group than Aut(G)

[next obsidian]

No

[nocturne cobalt]

kk

[next obsidian]

Let me try to produce an example

[coral shale]

Thinking of one too lel

[next obsidian]

I believe the Klein 4 group has automorpjism group S_3

[nocturne cobalt]

klein 4 is Z2 x Z2 right?

trying to remember

[next obsidian]

The reason is the Klein 4 group has 1 identity, and 3 elements a,b,c

Yeah

And a,b,c are totally symmetric

They satisfy like

[coral shale]

In particular, note all automorphisms map generators to generators of the same order when trying to construct them (the converse is not necessarily true, though - this process doesn't always produce an automorphism, I believe)

[next obsidian]

The square of any of a,b,c is just e

And like the product of any two is the other

So we can permute a,b,c in any way we want

This is the like qualitative reason as for why it’s automorpjism group is S_3

To actually prove it, you can just prove it haha

Fixing e and swapping a,b,c is a bijection

And showing it’s a group homomorphism is real easy

[nocturne cobalt]

hmmm right

what would be a general characteristation of the ker of this mapping from klein 4 to S3?

[next obsidian]

What do you mean

The kernel of the map G -> Inn(G) is always the center

So in this case it’s all of the Klein 4 group

[nocturne cobalt]

yes exactly what I was asking

right right

ty

[next obsidian]

[coral shale]

hmm 🤔

[next obsidian]

You send g to “conjugate by g”

[coral shale]

Inn(G) are all the auto which act by conjugation

[next obsidian]

And this has kernel the center

[coral shale]

ah ok yh

gxg^-1 = x <=> gx = xg
being slow 💤

[next obsidian]

[nocturne cobalt]

yes :)) ty guys <33

so in general when proving a map is an automorphism do we prove it has a group homomorphism then prove it is injective/bijective etc?

[next obsidian]

Yup

Or well

A map doesn’t have an automorphism

It is an automorphism

[nocturne cobalt]

yes mb

[next obsidian]

And yeah you just prove its bijective

Either by showing injective and surjective manually or like finding an inverse

[nocturne cobalt]

kk great ty

and also 1 more thing

just to wrap my head around this

G--> Aut(G)

The image of this map is the Inn(G)?

[sturdy marsh]

yup

[nocturne cobalt]

ty

[pastel cliff]

does this look like a valid proof

idk if i should be explicit about the fact that inverses commute

or if it's obv enough

[spice whale]

are automorphisms interesting?

[coral shale]

Not just interesting. Important.

[spice whale]

hm

[next obsidian]

I mean, I guess, but why even use inverses?

I think inducting is the easiest

[chilly ocean]

Can someone help me approach this problem?

[pastel cliff]

induction would have to go both ways still tho

[coral shale]

you can substitute to show the inverse statement

[spice whale]

why are automorphisms important

[coral shale]

What have you done in algebra so far

[next obsidian]

(gh)^n-1(gh) = g^{n-1}h^{n-1}gh then you can just shove g through each of the h's

[spice whale]

really only basic-intermediate group theory

so like

[coral shale]

just the latest topics u have seen

[spice whale]

kernels and stuff

[next obsidian]

and if you want to show that h^{n-1}g = gh^{n-1} super explicitly

you can do that by induction too

[coral shale]

Have you seen isomorphism theorems?

[spice whale]

just getting there

[next obsidian]

you go to hgh^{n-2} = gh^{n-1} by g and h commuting

[pastel cliff]

unrelated but chmonkey can we assume u got into chmolubia - if so congrats

[spice whale]

this right

[coral shale]

yes

Not that it is necessarily directly related to what we're discussing

[next obsidian]

:)

thanks

[coral shale]

but this result gives intuition for a bunch of stuff (makes things make more sense)

[spice whale]

so

[coral shale]

I'm not sure I can make an example that would be enlightening 🤷‍♂️

[spice whale]

I'll encounter them properly soon i imagine

[next obsidian]

Here's a really simple example of automorphisms being useful

[coral shale]

I think group actions are probably very relevant when talking about automorphisms? If you haven't seen those yet, it is harder to motivate

[next obsidian]

if you want to prove some statement about like, something in R^n

like n-dimensional euclidean space

[spice whale]

hm

[next obsidian]

and it's something like "on an open ball near x..."

[spice whale]

hm

[next obsidian]

the map which just shifts the entire domain is an automorphism

as in y -> y - x

[spice whale]

ohh

[next obsidian]

and often times the thing you want to prove is preserved under isomorphisms

[spice whale]

so you can move it to the origin?

[next obsidian]

so you can assume x = 0

yeah

this is just one example of them being useful

[spice whale]

yeah

[next obsidian]

in other times automorphisms are bad

[spice whale]

oh

[next obsidian]

as in you might know that two things agree up to an automorphism

as in like f = g\circh

for some automorphism h

[spice whale]

oh

[next obsidian]

now for technical reasons you might really want f = g

[chilly ocean]

Feeling humbled by this one person's solutions to Artin, did not at all realize during the exercise showing formal power series are rings that if it's over a field you get lots and lots of units
I had some handwavey justification that only the constants could be units but goddamn practically everything is a unit! So used to the nice degree properties of usual polynomial rings

[next obsidian]

and if you know the only automorphism is the identity then you can conclude f = g

[spice whale]

but

if there are others

[next obsidian]

yup

they might not be the same "on the nose"

and you can sometimes work around this

[spice whale]

like if the group isn't abelian

[next obsidian]

well no an abelian group can have automorphisms

like x -> -x in Z

[spice whale]

oh wait

sorry

it's inner automorphisms

[next obsidian]

right

anyway, this issue with automorphisms

[spice whale]

other than id

[next obsidian]

causes sometimes bad things to happen for technical reasons

[spice whale]

hm

[next obsidian]

like in algebraic geometry there's the notion of a stack which, I wikll not even begin to describe

but they are messy, technical, blech

but their existence is basically necessitated because the presence of automorphisms screws things up

[spice whale]

ooh

[coral shale]

have u seen normal subgroup?

[spice whale]

yes
H subgroup of G such that forall g in G, gH = Hg

[coral shale]

$$(\forall g\in G, n\in N)(gng^{-1}\in N)$$

[spice whale]

i think

[cloud walrus]

Shuri2060

[coral shale]

Alternatively

N being normal in G

Now normal subgroups come up everywhere

Because you can only quotient with normal subgroups

G/N

[spice whale]

is left and right cosets being equal an equivalent condition

[coral shale]

Yes

[spice whale]

ah

[coral shale]

gN = Ng is equivalent to that statement because

gNg^-1 = N

is exactly that statement I wrote

[spice whale]

oh

yeah

[coral shale]

Ok, so gng^-1

Is otherwise known as 'conjugating n by g'

Have you heard of it?

[spice whale]

that looks like an inner automorphism

[coral shale]

It comes up quite a lot

yes Inn(G)

[spice whale]

i have heard of conjugacy classes

[coral shale]

Is the automorphisms which are conjugations

[spice whale]

yes

[coral shale]

so back to here

we can see conjugation by g is an automorphism on N

[spice whale]

yes

[coral shale]

Because it maps elements in N to another

[spice whale]

yeah

[coral shale]

This idea is important when considering various stuffs 👀

[spice whale]

interesting

[coral shale]

If u want a harder example

I am currently doing Galois Theory where automorphisms are important

[spice whale]

yeah

like

extension fields mapping to groups i think

[coral shale]

So the field C extends R

[spice whale]

yep

[coral shale]

The galois group is

the automorphisms which 'fix' R

ie. f(x) = x for anything in R

[spice whale]

oh

[coral shale]

It turns out there are only 2 possible automorphisms

Sorry I should be more clear

We want automorphisms of C

which fix all elements in R

f : C -> C

[spice whale]

so galois groups are groups of field automorphisms?

[coral shale]

yes, a specific type of auto

Basically

L : K,

L extends K

We want automorphisms L -> L that fix all elements of K

[spice whale]

hm

[coral shale]

You know the theorem which says there is no quintic formula?

[spice whale]

yep

[coral shale]

In particular, some quintic solutions cannot be written in terms of +-*/ and rooting

same for higher degree

this result comes from galois theory, basically

[spice whale]

isn't it to do with the galois group having a non-cyclic group as a factor

[coral shale]

That, I wouldn't know lul

[spice whale]

oh lol

[coral shale]

I am still near the starting line for Galois

[spice whale]

i think it has to do with A5 being a simple group

[prisma ibex]

It has to do with the fact that A5 is not solvable

[coral shale]

wish my lecturer would stick the full notes up

rather than chapter by chapter

[spice whale]

I'm trying to speedrun through the start of godement

because it's all quite simple stuff

[prisma ibex]

Yes

[coral shale]

Homomorphisms must be bijective to have inverses

ie. isomorphisms

[full panther]

i know that part

[coral shale]

You can verify the inverse of an isomorphism is also an homomorphism 👀

and hence iso

proving what you want

[spice whale]

i thought godement would never cover cat theory
but it's in an exercise

[open pilot]

It's somewhat tangential but you can have f^(-1)(b) = {a in A: f(a) = b} for a non-bijective homomorphism

[south patrol]

Isn't this just the definition of inverse image

I am confusion

[next obsidian]

I think they're saying not to confuse f^-1(b) actually saying f has an inverse

like the fiber f^-1(b) [which really is f^-1({b})] could be confused with evaluating b at the inverse of f

[chilly ocean]

Is this valid

My first attempt

[lavish nexus]

yeah

this is 3rd iso thm

[chilly ocean]

yeah but i didnt say anything about the correspondence theorem

[lavish nexus]

it's your 3rd line

[chilly ocean]

my friend said this

but idk what that means

i think my proof is valid

[lavish nexus]

that is fancy H = H/N

which is your 3rd line

[chilly ocean]

okay

im good then right?

[lavish nexus]

put|| on fancy G and H when appropriate

[chilly ocean]

wym

[chilly ocean]

The person gave a back-of-the-envelope proof with the original intuition wiped clean away, and after mulling it over for a while I figured out a very pretty argument for existence (and hence iff) of inverses of polys in F[[t]] with unit constant term:

View it as trying to find an inverse "mod" x^k for every positive k. Guaranteed since if the constant term is a unit, gcd with x^k is 1, and extended euclidean algo gives us our inverse. Repeat ad infinitum

[coral shale]

I think of it as division alg

but yes

[lavish nexus]

|fancy G|

[chilly ocean]

where did i miss that ? @iterbus

[chilly ocean]

Oo how do you mean?

[grizzled elm]

I have a question about this proof of K splitting field ⟺ any irreducible polynomial with a root in K splits completely over K

[chilly ocean]

@lavish nexus

[coral shale]

probably what you said more or less

Polynomial long division except it goes on forever

You want to keep killing terms

[lavish nexus]

look at your first line

[chilly ocean]

i dont put a a order there

do i?

[lavish nexus]

well you're counting the order of fancy G

[chilly ocean]

oh wait i do

[grizzled elm]

Why must xi be in K' ? so like why isnt this just changing xi to another root in the isomorphism

[chilly ocean]

Like writing 1 as 1+0x+0x^2+... Right

because this right ? @lavish nexus '

[coral shale]

yes @ zd

[lavish nexus]

what is [|G|:|H|]

[chilly ocean]

the right side

[lavish nexus]

last line you're counting cosets why would you put ||

[chilly ocean]

im so confused

Pog

thats the only way i would do it

[lavish nexus]

that's it

[chilly ocean]

i did do that

now its good

good?

[lavish nexus]

yeah

[grizzled elm]

I think this is only true if we assume \alpha is a root of g(x), does this break the proof?

[dire summit]

i have a question on the language of this homework problem:

When you “roll” an octahedron, each of the eight faces is equally likely to be facing up. Ordinary octahedral dice have the numbers 1 to 8 on the faces. Weird pairs of octahedral dice have certain positive integers on the faces (not simply the numbers 1 to 8 and not necessarily the same for the two octahedra in the pair) but have the amazing property that when you roll the two of them, you get the same probabilities for the sum as you do for ordinary octahedral dice. Find all weird pairs of octahedral dice.

we had a problem before where we had to answer the question of "can you create a pair of weighted dice such that each of the 11 outcomes have equal probability" and i know this is a similar/related problem, but i'm having a rough time understanding what specifically is being asked of me

[dire summit]

this is just the start of whatever the solution will end up being, but i would like a second opinion that this is indeed what the problem is asking (this was the manner in which the professor handled the earlier version of the problem i mentioned)

[molten viper]

So, I need to show the other direction of this subset relation at the bottom

[next obsidian]

Oh my god

I’m sorry, I don’t know anything about Grobner bases

[molten viper]

o p e

[next obsidian]

I don’t even know their definition

[molten viper]

fair enough

Well how much do you know about ideals

[next obsidian]

I just know it’s a computational tool that helps you compute ideals

A lot

[molten viper]

ah well you may still be able to help me then

I just need to show that bottom subset relation is equality given that division property

[full panther]

my course covers that soon should i be worried

[next obsidian]

Wait so

[molten viper]

If you go to lecture (unlike me) you'll do great

[next obsidian]

What are you asking?

I don’t think the divisibility conditions tells you about >

[molten viper]

I need to show two monomial ideals are equal

I'm not doing anything with >

[next obsidian]

The divisibility condition gives you <

[molten viper]

oh

[next obsidian]

Oh are you asking why?

Why < holds?

[molten viper]

Well no, I'm asking how to show >

[next obsidian]

Yeah I don’t think that follows from the divisibility

The divisibility conditions only can guarantee you <

[molten viper]

Well it's true, but idk why

[next obsidian]

Afaict

[molten viper]

Ok well, how might I show 2 monomial ideals are equal

[next obsidian]

Idk what a monomial ideal is

Also from the setup of the problem it looks like you’re assuming they’re equal

Like is this the entire problem?

It looks like this is only part of the question

[molten viper]

here's the problem

It's a biconditional

[next obsidian]

I mean

Idk what LT(g) is but

It looks like it should be like

[molten viper]

Leading Term of the polynomial g

[next obsidian]

And what’s a Grobner basis

[molten viper]

oh fuck I've seriously misunderstood this problem

[next obsidian]

Yeah it looked like it to me

Like under any reasonable interpretation of what it’s asking

I feel like the > containment ought to be trivial

But idk what any of this avtually is so idk

[molten viper]

The > containment is trivial

I just misunderstood why it's trivial lol

[next obsidian]

[chilly ocean]

wtf LT(g)?

leading term of g in I?

[molten viper]

It's the leading term of g

[chilly ocean]

I dont know definition of grobner basis

I should know though

apparently it is useful in fracking

[molten viper]

It seems a niche topic, apparently

[chilly ocean]

Nah

its useful apparently

[molten viper]

It's just so confusing

all this notation

Time for some good old contradiction maybe?

CONTRAPOSITIVE MWAHAHAHAH

eh nevemind

I choose to be satisfied with what I have

[rapid slate]

I could do the (i) part in the following way:

I tried defining an element y by conjugating it by an element of k but couldn't really proceed

Could someone give any hints?

[viscid pewter]

clearly order of xK divides n, as (x^n)K = 1K

no wait you finished (i), ok

[fresh vessel]

Any ides for how to show x^6 + 3x^4 + 4x^2 + 1 is irreducible over Q? A couple grad students and I have been talking about it for a few hours, but so far we have made next to no progress. Wolfram says it is irreducible, but all the standard tricks dont seem to work.

[rapid slate]

Yeah...but what's your argument for the second one?
I don't know how to use the normal subgroup condition

[viscid pewter]

i think the normal subgroup condition is being used by the fact that G/K even exists

and xK is a nice enough thing to talk about

[rapid slate]

Oh yeah

Other than that it's not really useful

[viscid pewter]

eh, it's foundational

but it could be useful with the conjugation thing idk

i remember doing this q but i don't remember how it goes, lemme see...

[rapid slate]

Yeah but i couldn't decide an element

To be conjugated

Such y has order n

[viscid pewter]

lemme try something out

[rapid slate]

Cool... let me know
Thanks!

[pastel cliff]

i talked to wew lads about this before but am still stuck

the thing on the right, that elements of SO2(R) are rotations

im having trouble showing that Mx dot My = x dot y using the properties of SO

[wise igloo]

that is a nice large whiteboard

[pastel cliff]

indeed

@hidden haven 👉👈

[hidden haven]

[pastel cliff]

sorry for ping, a friend of mine solved it using elements of SO in terms of sin and cos but that is lame and i really wanna get this way

[hidden haven]

Gonna meet det in like 30 mins I'll ask him lmao

[pastel cliff]

yall go to the same uni 👀

[hidden haven]

I don't wanna think rn after sitting in car for 2 hours

Yes

[pastel cliff]

moldi lore

[hidden haven]

[pastel cliff]

do ping if you or they have any opinions tho pls

[rapid slate]

I tried this

But doesn't work until it commutes

[pastel cliff]

im down cataclysmically atm and am stubborn

[hidden haven]

Ye sure

[next obsidian]

Also Saketh

Who rn is chmonkeynumberonefan

[pastel cliff]

[pastel cliff]

does it look at all reasonable

[next obsidian]

Kinda funny 3 ppl at the same university in India are all active users

[chilly ocean]

I'm one of your twitter followers does that count

[wise igloo]

who

[pastel cliff]

who’s the third

[pastel cliff]

same here kek

[next obsidian]

Det, Saketh, and Moldi

They all go the same school

[rapid slate]

Me too from India

[pastel cliff]

ahh

[chilly ocean]

Can someone double check my proof for me? I'll be presenting it in class, so i wanna make sure that I'm not making any egregious blunders—the problem seemed pretty trivial to me.

[viscid pewter]

nah

'let g1, g2 both be the identity element'

so you're assuming implicitly that there is only one identity element

it doesn't affect the rest of your proof tho

however the notion of the order of a group also sorta assumes that there is only one element? try and do it with even more basic methods

we're talking like the four group axioms, basic

[pastel cliff]

kai any thoughts on my thing up there

[viscid pewter]

god

i'm trying to do like 3 different things rn, i'll maybe get to your thing later lol

[pastel cliff]

all g

im switching to another question for now

[hidden haven]

There are a lot more inactive people from my school in this server

[strong valve]

Hi!

[paper flint]

When you're proving that a certain structure has a unique element satisfying a property, assume that there exist two elements that have the said property (no need to assume they're distinct because that's what you want to prove anyway), and show that the property will force both elements to be equal.

[lethal dune]

det is inactive now?

[paper flint]

No

There's a reaction literally 3 messages above

That's the det activity indicator

[lethal dune]

oof lmao

[pastel cliff]

hi det

[rustic crown]

Hewwo

[pastel cliff]

can i bother u to look at this pls

[rustic crown]

So like Mx • My = x^t (M^t M) y = x^t y = x • y

[pastel cliff]

ok ig that's just not an equality ive seen before

i think wew lads used it as well

[rustic crown]

Oh, so how are you defining the dot product?

[pastel cliff]

no different than normal for R2 i suppose

[rustic crown]

that's the usual definition x • y = x^t y

[pastel cliff]

wait

wow

i am dense

ok well in general tho is that line of reasoning enough of a proof though

bc the claim is this

so it's implying that we should use cos and sin shit

but me like visual stuff

[hidden haven]

What's the definition of SO though

That's just a property

[pastel cliff]

detA = 1 and AA^t = I

or like

matrices with those properties

[hidden haven]

Ye so that's good enough

[pastel cliff]

let's fucking go

my group didnt believe me lmao

[hidden haven]

[pastel cliff]

good thing im the only one who knows latex so im the one typing up the hw

[hidden haven]

Group hw?

[pastel cliff]

yes

prof imposed it

[hidden haven]

That's the stupidest shit I've ever heard

[pastel cliff]

and when i paired up with someone and we agreed privately to do it independently he slapped another person on us

[dire summit]

hey y'all, i'm once again having a case of "i don't understand what this homework problem is asking of me", if anyone here can make sense of what i'm supposed to do i'd appreciate it

[hidden haven]

[dire summit]

"Eisenstein’s criterion says that for any prime number p, a certain conclusion follow. Show that for any number p that is not prime, the same conclusion does not follow."

[hidden haven]

You are supposed to find a counterexample to the statement you get when you replace "prime" with "not prime" in the statement of eisenstein

[dire summit]

correct me if i'm wrong but eisenstein's criterion says that if there exists a *specific* prime number p that satisfy certain conditions, then the conclusion follows

i'm not sure where "for any prime number p" comes from

[hidden haven]

Eisenstein is for all primes

It's just that the polynomial in question will not satisfy the hypothesis for all primes

[dire summit]

ah ok that makes more sense

bleh this'll be annoying to write up

[hidden haven]

[warm holly]

Yo I have one question. In my linear algebra course we define a root of polynomial over a field. Then we proved the theorem : b is a root of f in K[x] iff (x-b)| f. But why do we define roots over a field? The theorem which I stated might work for rings in general and we can speak about roots in rings as well no?

[dreamy jewel]

You are right that this holds for commutative rings in general, however the following is true in an integral domain R but may not be outside of it

For n > 1, if a_1, ..., a_n are distinct roots of f(x) in R[x], then f(x) = (x-a_1)...(x-a_n)g(x)

But we usually want our ring to be a field because a polynomial ring R[x] is a PID if and only if R is a field

And PID's are usually very nice to work with

For instance, non-zero prime ideals in a PID are maximal

[pastel cliff]

see i figure it's cuz like

"math should be collaborative"

but

[warm holly]

Alright, thank you very much!

[dreamy jewel]

No problem!

[wooden ember]

Does this work for all commutative rings or just UfDs?

[lethal dune]

we were given once, a single guy did the whole thing, I didn't even know what the assignment was

[wooden ember]

I thought it was only guaranteed to be true of UFDs

[hidden haven]

[pastel cliff]

i mean we did the hw "together"

[hidden haven]

You just divide by x-a with remainder. The remainder is then a constant, so if you evaluate the polynomial at a you get that constant which must be 0

[pastel cliff]

but im redoing most of it lol

[wooden ember]

You don’t have Euclidean division for general rings though do you

[hidden haven]

You can always divide by monic polynomials

[pastel cliff]

experiment time

which proof is more readable

[hidden haven]

It won't be Euclidean

In the sense that you won't get a Euclidean domain

[wooden ember]

But you still have the division I see

[hidden haven]

Yes

Only if leading coefficient of divisor is unit

If it were possible for all, then it would be ED

[wooden ember]

Yeah makes sense

So what exactly is the subtlety with poly rings over ufds then

[hidden haven]

They are also UFDs

[dreamy jewel]

I guess you can do GCDs in UFDs

[hidden haven]

But not EDs

[dreamy jewel]

Maybe that’s helpful

[hidden haven]

Because you still can only divide by monic

[wooden ember]

Right but they’re all factorable if they have n roots though right

(n is degree)

[dire summit]

coming back to say thanks for the clarification, the proof turned out to be easier than expected thanks to you :)

[dreamy jewel]

Yeah that holds in general for all integral domains

[wooden ember]

Why integral if we can do what moldi just said

[dreamy jewel]

For n distinct roots

[wooden ember]

Oh cause the degrees won’t add up right

Okay thanks for the answers

[hidden haven]

Not sure how degree not being additive is relevant

It's a direct from definition of domain argument for distinct roots

[dreamy jewel]

Try inducting on n

[dire summit]

if i have room to slide in here and ask one more question i'd really appreciate it

my last hw problem is a beefy paragraph and i'm not quite sure what it is they're asking me to prove

here let me copy paste

When you “roll” an octahedron, each of the eight faces is equally likely to be facing up. Ordinary octahedral dice have the numbers 1 to 8 on the faces. Weird pairs of octahedral dice have certain positive integers on the faces (not simply the numbers 1 to 8 and not necessarily the same for the two octahedra in the pair) but have the amazing property that when you roll the two of them, you get the same probabilities for the sum as you do for ordinary octahedral dice. Find all weird pairs of octahedral dice.

[dreamy jewel]

Bruh that is indeed a handful

[hidden haven]

🥴

[dire summit]

so for a weird pair, are they supposed to have the same probability together as a single die? as a pair of normal dice?

[dreamy jewel]

Imagine having a homework where the prof actually asks you creative and thoughtful questions that are worded in a “fun” way

Couldn’t be me

[dire summit]

i have issues with the american education system and you summed most of them up in that one joke

[dreamy jewel]

I, too, share your issues with the American education system

[dire summit]

...i'm having the creeping realization that this is probably a global thing isn't it

[dreamy jewel]

More or less lol

[dire summit]

damn

[wooden ember]

Well it should be right? I mean if it’s not an integral domain you can only guarantee f(x)=(x-r_1)…(x-r_n)g(x)

[lapis trail]

You might check out Sichermen dice because that's what this sounds like

[wooden ember]

With g non constant possibly

[dreamy jewel]

Ah try x^2 - 1 in Z/8Z

[wooden ember]

Since the degrees of the linear factors don’t necessarily add up but in integral they do and so g has to be constant

[wooden ember]

That’s not integral though 🤔

[dreamy jewel]

It has 4 roots in Z/8Z but you can’t factor 4 roots out

Oh sorry I thought you were saying

[dire summit]

you're absolutely right yeah

[dreamy jewel]

You could still factor out n roots in a non-integral domain

[dire summit]

ok glad to know there's actually a frame of reference for this problem

i can work with this, thanks

[wooden ember]

I mean if each linear factor divides out x^2-1 can’t you do it but then you multiply by a « correction » polynomial g(x)that gets rid of two of the factors

[dreamy jewel]

Can you elaborate on what you just said here

[wooden ember]

Write it as in here in such a way that two of the factors cancel out with g

Or is that some of the linear factors divide each other in Z/8Z then 🤔

[dreamy jewel]

Can you construct such g(x)?

[hidden haven]

I don't see how you guarantee that

[wooden ember]

Hmm no you’re right I can’t

But so I can’t think of a good reason for why it doesn’t factor, does one of the factors fail to be irreducible or smth

[hidden haven]

Try proving by induction on number of roots

[wooden ember]

Aight I’ll do that when I get to a desk

Ah but wait x^2-1 has degree 2 and 4 roots

I was talking deg roots

Anyways I’ll talk about this later I have class now

[dire summit]

have fun :)

[dreamy jewel]

It’s amazing that degrees aren’t additive in a general commutative ring

So I know that a polynomial is invertible if and only if its constant term is a unit, and all of its other coefficients are nilpotent

Is there some general conclusion of this but more degree n polynomials

IE like if f has degree r and g has degree s, r + s > n, but fg has degree n

What can i conclude about their coefficients

[wooden ember]

Okay I tried it out and see the issue I think

Linear factors aren’t necessarily prime in a non integral domain

[pastel cliff]

this claim is just wrong right

[wooden ember]

So the induction can’t be completed

[pastel cliff]

dont they have to be integers

sorry for interruption, but if it's wrong then proof is a quick fix

[wooden ember]

Rational works

If you write a = p/q, take A^q to get the identity

Integer a only gives the identity

[pastel cliff]

ohhhhhhh ok

ty zote

[dire summit]

what do you guys think of this proof so far, i'm starting to lose confidence in it by the second

[pastel cliff]

group member wrote this and im having a small stroke

[dire summit]

weather update: bleh

[pastel cliff]

mood

,ti

[cloud walrus]

The current time for nitezba is 02:54 AM (EST) on Thu, 24/02/2022.

[dire summit]

ayy same

,ti

[cloud walrus]

You haven't set your timezone! Set it using the interactive timezone picker with ,ti --set.

[dire summit]

,ti --set est

[cloud walrus]

Your timezone has been set to EST!
Your current time is 02:55 AM (EST) on Thu, 24/02/2022.

[dire summit]

east coasters be we

[pastel cliff]

pain

but i dont wanna ping moldi again

[viscid pewter]

,ti --set gmt

[cloud walrus]

Your timezone has been set to Etc/GMT!
Your current time is 07:59 AM (GMT) on Thu, 24/02/2022.

[pastel cliff]

this isnt finished but it's wrong right

i think im using the claim in the proof after the second equal sign

but idk how else

you still busy kai

[next obsidian]

Bruhhhhh why do you have to prove this

[pastel cliff]

dumb bad homework

i can bitch and moan all day about this class but i wanna finish

ive been polishing this hw for literal hours bc i dont trust my group

theyre smart but im not risking stupid point deductions again

but yeah this class is emphasizing some of the less interesting stuff it seems

[dire summit]

yeah i feel like in group theory you kinda have to slog through some boring proofs unfortunately :/

[dire summit]

on the bright side, my huge paragraph up there turned out to be an alright proof to work through, i'm glad i'm done with that

so hopefully it gets better for your work too

anyway

now i am become sleepy, the goer to bed

[pastel cliff]

is this wrong

i still think im using the claim in the proof but

,ti

[cloud walrus]

The current time for nitezba is 04:05 AM (EST) on Thu, 24/02/2022.

[rapid slate]

Any ideas?

[lapis trail]

Hint: rational roots theorem

[chilly ocean]

Hint: 1st isomorphism theorem

[rapid slate]

[prisma shuttle]

i heard that $(\mathbb{Z}/p\mathbb{Z})^{\times}$ can be finitely generated using $-1$ and $5$ for any prime $p$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

and that $-1$ has order $2$ and $5$ has order $2^{n-2}$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

can someone epxlain why $5$ has order $2^{n-2}$ cuz id see why

[cloud walrus]

JustKeepRunning

[sharp sonnet]

what is n

i dont think this statement is correct either, have you tried any examples?

5 has order 2^{n-2} in (Z/2^{n}Z)^\times though

a proof can be found in e.g. ireland rosen

[prisma shuttle]

mod 2^n sorry

p=2

[prisma shuttle]

yes this is what ima trying to prove

[sharp sonnet]

it's in chapter 4 of ireland rosen

dont think there is an intuitive way to "just see" it

[tender mist]

Show that $n$ coprime with $|K|<\infty$ implies that
$$\begin{aligned}
K& \rightarrow K\ k & \mapsto k^n
\end{aligned}$$
is injective, hence surjective. So, there is some $\bar{k} \in K$ such that $\bar{k}^{n} = (x^n)^{-1}$. Take then $y=x\bar{k}$ and show why $n$ is the smallest positive integer such that $y^n = 1$, keeping in mind that $n$ is the smallest positive integer such that $x^n \in K$.

[cloud walrus]

Mat

[rapid slate]

But for that shouldn't $\bar{k}$ commute with $x$?

[rapid slate]

I did this

[cloud walrus]

ℝ𝕂

[chilly ocean]

You could probably do it just considering 2-adic valuation, i.e. lifting the exponent lemma

Actually yeah it's exactly that, basically a two line proof that way lol

Letting k be the order of 5 mod 2^n since 4 divides 5-1 the lemma apllies. Hence v(k) = v(5^k - 1) - v(5-1) = v(m2^n) - 2 >= n-2

And the smallest such k is a multiple of the order so it is the order

That being 2^{n-2}

[fresh vessel]

Wouldnt that only show it has no linear factors? What about quadratic and cubic polynomials?

[tender mist]

Oh good call, that didn't cross my mind, I'm sorry. I hadn't seen you already solved for commutative case as well
I thought about it a bit but seems hard, maybe I'll try again some other time, keep me updated if you're able!

[nocturne cobalt]

So im a bit stuck on trying to construct a diffeomorphism between SU(2) --> S^3, how do i map the matrices to points on the 3-sphere?

i can see its possible since C^2 is isomorphic to R^4, just confused how to construct it :x

oh mhm

if we say (a,b) is in C^2

and just use the sphere formula we get |a|^2 + |b|^2 = 1

so would our mapping be A=[matrix of SU(2)] |----> (a,b)?

[dull root]

Can anyone help me clarify if I am understanding this right: Lets say I am working over $\C$ and have $a$ which has minimal polynomial over $Q$; $x^3 + 2$. This means that in the extension $Q(a)$, $1,a,a^2$ is a Q-basis. Does this mean the degree of the extension is 3, but $Q(a)$ is still a simple extension?

[cloud walrus]

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

[thorn delta]

yes, in general if a is algebraic over K, then the simple extension K(a)/K has degree equal to the degree of the minimal polynomial of a over K

[dull root]

is there a trick to see how to write for example u^3 in the case as a Q-linear combination of basis: 1,u.u^2 if I know the minimal poly

[thorn delta]

yes. in Q(a), a^3 + 2 = 0 so a^3 = -2 for example

[dull root]

yea that was the easy case, but something like a^4

[thorn delta]

-2a

[dull root]

how do you see that?

[thorn delta]

multiply both sides by a

[dull root]

ah, so this works in geenral for any a^n

[thorn delta]

ye, basically the image of any polynomial from Q[x] in Q(a) can be reduced in this way to get some linear combo of 1, a, and a^2

[runic hemlock]

Given an injective $\mbb{Z}$-module $Q$, and a (commutative, unitial) ring $A$, I'm trying to prove that
[\text{Hom}_{\mbb{Z}} (A, Q) ]
is an injective $A$-module. This is an exercise a book I'm reading. Any idea how to go about this?

[cloud walrus]

Porphyrion

[runic hemlock]

Probably useful: Just before this it was proved that for an A-module M to be injective it suffices that for any ideal I of A, and any A-linear map I->M, it factors through some map A->M.

[dull root]

If I have fields $K \subset L$, where $L$ as an $K$-vector space has dimension 2. I can take a K-basis $1,a$ and it is clear to me then $L=K(a)$ is a simple extension of K. This should mean that a has minimal poly of degree 2 over K. How do I see that $a^2$ is in K? or is this not true in general

[cloud walrus]

MasakaBakana

[waxen hedge]

Try to take an explicit minimal polynomial for $a$.
And write what does it mean that $a$ is a root of this polynomial

[cloud walrus]

Adrien

[dull root]

I see it now from the quadratic formula 🙂

[lapis trail]

You are correct, I made a mistake

[spice whale]

i just did a proof that left divisibility implies right divisibility in a monoid (group because it has division but whatever) and I'm happy

first step i proved that for all x,y there exists z such that xy = yz

then proved that left inverses are also right inverses

then that y\x = (y^(-1))x

and finally concluded

[potent briar]

hello I'm learning more about vector spaces

how can I do the following exercise

Consider the quotient spaces obtained by reducing the space $\mathscr{P}$ of polynomials modulo various subspaces. If $\mathscr{M} = \mathscr{P}_n$, is $\mathscr{P} / \mathscr{M}$ finite dimensional? What if $\mathscr{M}$ is the subspace consisting of all even polynomials? What if $\mathscr{M}$ is the subspace consisting of all polynomials divisible by $x_n$ (where $x_n(t) = t^n$)?

[cloud walrus]

vik

[thorn delta]

did you figure this out?

[chilly radish]

Ok this is really weird. The book i'm reading is defining the grading by degree of the free algebra F[x_1,...x_n] over F, but they insists the grading starts from degree 1. it's weird to me that they're ignoring constant polynomials. In fact in the entire book so far they're ignoring the constant polynomials, but not putting in the constant polynomials makes F[X] not an algebra, on the other hand, the subject of the book is polynomial identities, and you can't really talk about a homomorphism from the free algebra to a nonunital algebra. I think it may be implicit that all algebras are unital associative but i'm not sure. Even if all algebras are unital associative, it still doesn't explain why they start the grading from 1.

idk if there's something i'm missing or if this is just inconsistent

[delicate orchid]

I hate simple looking statements that are deceptively hard to prove

[delicate orchid]

proofed it (I think )

[next obsidian]

You can actually do this in a very neat way using adjoints. A right adjoint to an exact functor will send injectives to injectives

[barren sierra]

Abstract Algebra: "Oh yea the stuff we're covering is pretty heady at times. I don't even know how to explain it"

Also Abstract Algebra: "Today we are talking about how to evaluate polynomials"

[delicate orchid]

omgggg homomorphisms from R[x] to R

[barren sierra]

ye lol

we covered these in my undergrad course also so this is review

it's just funny

I get why we have to do it

[delicate orchid]

evaluating polynomials as functions isn't really something that's built into the polynomial ring so you have to do a slight modicum of trolling

[waxen hedge]

Try first to prove the following:
Let E be a vector space, and F,G be subspaces of E such that E is the direct sum of F and G
Take (f_i) a basis of F and (g_j) a basis of G
Then the images of the (g_j) through the projection E -> E/F form a basis of E/F

[barren sierra]

ye

[waxen hedge]

Considering the number of elements in these basis, you can deduce the dimension of the quotient

[barren sierra]

actually I do have a notation question. Everytime I've seen this taught I see people say "Oh write ev_a: R[x] -> R as the evaluation map of f to f(a)" but this notation is only used when introducing this topic. I've never seen anyone actually use such notation. Does anyone?

[delicate orchid]

which part of the notation

the ev_a part?

[spice whale]

is the set of homomorphisms from R[X] to R the same cardinality as R

[delicate orchid]

jesus that's a hell of a question

ok slightly less so (unless R is infinite )

[spice whale]

by R i mean the reals

[delicate orchid]

ohhhhh lord

ohhhh heavens

[spice whale]

start with finite fields then

[delicate orchid]

wait no hold on

is this a first iso moment

[spice whale]

for finite fields it's no i think

but idk

[delicate orchid]

I'd be inclined to agree

that was my first instinct

[spice whale]

hold on

let's do F2

it's not

because F2[X] is countable

wait

no

it's not

[delicate orchid]

we don't care about the ring itself

[chilly radish]

There are only 2 homomorphisms from F2[x] to F2

Homomorphisms.from R[x] to R are uniquely decided by the image of x so this is true

[spice whale]

oh of course

[delicate orchid]

it is

[spice whale]

ohh so I wasn't capping

[delicate orchid]

oh yeahhh the generator images thing

[chilly radish]

We can think of R[x] as the free algebra of rank 1 over R then this is obvious by the universal property

[delicate orchid]

if it was groups I would've remembered that

[potent briar]

@waxen hedge i havent yet seen general morphisms between spaces so i dont know what a projection is

[spice whale]

so every point of evaluation for a polynomial ring is a homomorphism to the base ring and nothing else is

[delicate orchid]

I considered thinking about R and R[x] as R-modules but started getting flashbacks so I stopped

[potent briar]

this book studies spaces before transformations

[chilly radish]

Yes

[delicate orchid]

the technical definition is that a transformation T is a projection if and only if T = T^2 but that definition smells

you can think of it instead as mapping one basis vector to 0 and keeping all the others the same

[potent briar]

im more worried with quotient spaces rn

im trying to follow the book

[chilly radish]

I think u need the ring to be commutative and unital for this tho alison

[spice whale]

hm

wait

non-unital ring??

I thought all rings had to have monoidal multiplication

[delicate orchid]

they do

[chilly radish]

Depends on your definition of ring

[next obsidian]

This is not true

Those are for F_2-algebra maps

[spice whale]

I've heard the term rng for these

[next obsidian]

Okay wel let me amend

[chilly radish]

No one uses the term rng

[next obsidian]

In this case you’re right

There are 2

[delicate orchid]

I swear I've seen tropo use rng

[next obsidian]

But this point

This is only for R-algebra maps

[chilly radish]

Yea you're right

I was thinking of Algebras

[next obsidian]

It ends up for F_2[x] it doesn’t matter

Because of how it ends up working out lol

[delicate orchid]

ok, good

[spice whale]

R-algebra map?

[chilly radish]

It did seem sus to me

[next obsidian]

Namely for Z or any Z/nZ

Every map from Z/nZ[X] is an algebra map

Because all integers are forced to map to themself

Just by 1 mapping to 1 and additivity

[delicate orchid]

because the trivial ring homomorphism isn't an evaluation map

[chilly radish]

True

My bad then

[spice whale]

what's the inverse image?

oh wait

nvm

I'm stupid

[chilly radish]

Do u have an example of a non trivial ring homo from poly ring to base ring that is not an R algebra map @next obsidian

[nocturne cobalt]

does the kernel give a tangent space on a group in general?

[chilly radish]

What is a tangent space on a group without any extra structure?

[delicate orchid]

something something direct sum

[next obsidian]

If you have any endomorpjism

Given a map f:R -> S

You can just make a map R[x] -> S

[nocturne cobalt]

idk, what do you mean without any extra structure?

[next obsidian]

By sending x to anything in S, and mapping the scales via f

[chilly radish]

True that's fair

[coral shale]

Catching up on Algebraic Number Theory. I am going through the definitions of trace, norm, discriminant. Trace and norm I mostly get, discriminant not so much (I am looking for some intuition/interpretation more than anything else) @ if you reply pl 🙂

[chilly radish]

What's your definition of a tangent space

[nocturne cobalt]

vector space at some point p on our manifold whose elements are the tangent vectors on the curve passing through the point p

[chilly radish]

So what do you mean in general in this case, a general lie group?

[next obsidian]

So actually what I did there was just write S as an R-algebra in multiple ways

Each of those f’s turns S into a new R-algebra

But if you strip the algebra structure and consider S just as a ring this gives lots of different maps

[nocturne cobalt]

yeah i guess so

[chilly radish]

This only works if R is commutative unital right

[next obsidian]

Don’t even ask me about non-unital stuff

But I think so

[chilly radish]

Like otherwise you can't give an Algebra structure I think

[spice whale]

how useful is cat theory for intuition in algebra

[next obsidian]

But also like

[next obsidian]

Literally backwards

IMO

[spice whale]

hm

[next obsidian]

Besides some sorts of heuristics

Like “oh mapping out properties are usually colimits, and those commute. So localization and quotients maybe commute”

But in general you’re going the other direction

[spice whale]

ah

[languid meteor]

can anyone tell what the author is trying to get at here?

I asked my lecturer and he said he's trying to say we're assuming we can compute these complex m'th roots, that if we want to solve them in general you have polynomials in cosines and sines and it's not obvious we can solve them

though I didnt totally understand what he was getting at

[delicate orchid]

that's pretty much it tbh

[languid meteor]

ok cool so we're basically saying assume 'm'th (complex) roots exist'

and we know they do by the complex exponential formula?

[runic hemlock]

If I understand correctly, your meaning is to use that the functor F: Ab-> Mod_A given by N->Hom(A,N) is adjoint to the forgetful functor Mod_A->Ab?

[next obsidian]

Right

Or is that quite right?

It should be adjoijt to restriction of scalars

Which is right adjoint to extension of scalars

So I think explicitly

Hom_A(M, Hom_Z(A,Q)) = Hom_Z(M (x)_A A, Q) = Hom_Z(M,Q)

And right, so all you need to note is that

If M^• is an exact sequence of A-modules

Then considering it as Z-modules it’s still exact

So I guess it is just the forgetful functor, yeah

[runic hemlock]

Oh right, tensor and Hom are adjoints 😀

[next obsidian]

Yup

I wrote it this way because I’ve seen this trick used before

In particular, I saw it used to show that O_X-mod has enough injectives

Taking a stalk is like f^* from a point or something IIRC

And it has an adjoint and blah blah blah

Or something like that

[runic hemlock]

Actually this exercise is part of the proof that Mod_A has enough injectives...

[next obsidian]

Right

I kind of figured

I just remember this idea popping up again sometime

[runic hemlock]

Yeah that's a nice trick, thanks

[next obsidian]

Obviously there’s a dual notion

For projective

[runic hemlock]

Like if P is projective then Hom(P, A) is?

[next obsidian]

Right it’s like

A left adjoint to an exact functor sends projectives to projectives

Just because Hom_D(fP, C^•) = Hom_C(P, gC^•)

So if g is exact then gC^• is still exact

And then the latter complex is exact because P is projective

[runic hemlock]

Right

Yeah cool

[next obsidian]

[tender mist]

@rapid slate I looked at it again and noticed that argument can be adjusted to general case, like so (the normality of K is needed for the steps)

[cloud walrus]

Mat

[tender mist]

Does it sound good in your opinion?

[white jackal]

How do you do part b?

[chilly ocean]

hmmm'

[barren sierra]

I guess first step is

do you know what bijection they are talking about

[white jackal]

honestly idk bro

[barren sierra]

do you know what the correspondence theorem is?

[white jackal]

yea if H is a normal subgroup of G then there exists a bijection of G mod H

[dull root]

If I have a,b are algebraic over Q. I see that [Q:Q(a)], [Q:Q(b)] is finite, but why is [Q(a,b) : Q(a)] finite. Is this dimension trivally bounded?

[next obsidian]

b is still algebraic over Q(a)

And Q(a,b) = Q(a)(b)

[dull root]

since b is algebraic over Q, it must be trivally algebraic over Q(a)

[next obsidian]

Yeah

Just take a polynomial it satisfies over Q

That has coefficients in Q(a) too

Cuz they’re in Q

[dull root]

yeah i'm dumb lol, that is indeed obvious

[next obsidian]

It’s okay, you’re just manifesting your pfp for a moment

[spice whale]

would you say autoisomorphism or isoautomorphism

[next obsidian]

What

[chilly ocean]

automorphism

[spice whale]

oh

[dull root]

autobijectivehomormophism

[thorn delta]

Homoautoisoendomorphsim

[spice whale]

wait do automorphisms have to be bijective

[chilly ocean]

yes

[dull root]

they are isomorphism