groups-rings-fields

[wraith cargo]

But I'm asking you why (r)/(p) is principal

(if you wanna argue this tell me what element it'd be generated by)

[maiden crater]

gcd(a_1,\dots,a_n)

[wraith cargo]

How do you know that gcd's exist in R/(p)

Also it's much simpler than that

||show that (r)/(p) is generated by the image of r in R/(p)||

[maiden crater]

idt I will get it

[wraith cargo]

It's important to understand every part of your proof

You shouldn't be making jumps in logic of you can't argue it in detail

Still, good to learn this lesson :)

[maiden crater]

I'll work on this tomorrow maybe, I'm getting a bit bogged down( brain fog)

or maybe in a hour

thanks

[errant wedge]

If you have fields $E\subseteq F\subseteq K$ such that $F/E$ is a finite extension and $K$ is a finite-dimensional $F$-vector space, is $K$ a finitely-generated $E$-algebra?

[cloud walrus]

Sara

[wraith cargo]

Doesn't this imply that K is a f.g. E-modules and hence a f.g. E-algebra since the algebra structure is compatible with the module structure

Like just the fact that K is a f.g. F-vector space is the same as saying that K/F is a finite extension...

[errant wedge]

oh yeah whoops

[reef rivet]

This was something that popped up in my lecture notes. While I do somewhat intuitively see why this would be the case, the proof does not make any sense to me. Can someone help walk me through what this proof is saying?

[velvet hull]

case 1 or 2? or both?

[long nebula]

Are there any specific lines that you're stuck on

[reef rivet]

Both

[long nebula]

what's the definition of "prime subfield"?

[reef rivet]

Its the intersection of all of the subfields of F, right?

[long nebula]

right, so we need to prove two things

1. F_p is contained in all the subfields of F
2. F_p is itself a field

do you see how those two things imply that F_p is the prime subfield?

[reef rivet]

Yes... and p has to be prime, right?

[long nebula]

yes

was that proved earlier in your lecture notes?

[reef rivet]

Yes

[long nebula]

okay great

[reef rivet]

Otherwise, we would have zero divisors

[long nebula]

exactly

[long nebula]

so now let's try to tackle each of these

first, why is 1) true?

[reef rivet]

I am struggling to figure out why that is true

[long nebula]

no problem

let's think about it sort of case by case

F_p is just {1, 2, 3, ..., p-1}

first off, is 1 contained in every subfield?

[reef rivet]

Yes, the multiplicative identity should be in every subfield

[long nebula]

yup

what about 2

do you think 2 is contained in every subfield?

[reef rivet]

It should be as well, right? Since 1 + 1 = 2?

[long nebula]

yup

since subfields are closed under addition

what about 3, do you think 3 is contained in every subfield?

[reef rivet]

Yes. Because its closed under addition by the same logic

[long nebula]

yea

so do you think you can prove that F_p is contained in every subfield?

[reef rivet]

Every subfield, must have 1,1 + 1, 1 + 1 + 1... up to p - 1(1). Since that in itself is also a field, then it must be the prime subfield, since its the smallest subfield that contains elements that are in every other subfield

[long nebula]

yup exactly

that proves case 1

the logic for case 2 is really similar

try to reason about it intuitively

[reef rivet]

Intuitively, it makes sense. I need 1, 2, 3.... which would lead to me the naturals, but, since a field also requires multiplicative inverses, then I would also need 1/2, 1/3, etc... which will get me to the rationals, which is Q. But I have no clue where or what the universal property of localization means or how it is used here.

The only reason case 1 did not need the fractions, was because the inverses were also other integers mod p

[reef rivet]

Or why I need to define a homomorphism

[long nebula]

Can you state the universal property of localization? Do the lecture notes talk about it?

[long nebula]

You also used a homomorphism in case 1, it was just implicitly

Like the field that you find inside F is not literally Z/pZ

Rather you have an isomorphism mapping Z/pZ into F such that n in Z/pZ goes to 1_F added n times in F

[reef rivet]

Not in this lecture. He mentioned it somewhere in a previous lecture, but I genuinely understood nothing from that lecture.

He did spend a lecture talking about localizations

[long nebula]

Okay well the idea is pretty simple actually, despite the fancy words

So you saw that if we send 1 in Z/pZ to 1_F in F, then all the other things logically follow, right? Like 2 in Z/pZ has to go to 1_F + 1_F in F, and so on, for it to be a homomorphism of fields

So just by saying that 1 in Z/pZ goes to 1_F in F, you can tell me where all of the rest of the elements in Z/pZ go in F

Are you following me there?

[reef rivet]

Yes

[long nebula]

yeah, so this is nice because it both shows existence (there is a homomorphism that extends your original one) and uniqueness (there's exactly one choice for that)

the universal property of localization is just the same thing, it's saying (in more precise terms) that if you know how to map 1 in Z to 1 in F (and you can invert every natural number in F, since it's characteristic zero), then you extend the whole thing to get a map phi from Q to F

and therefore every subfield of F contains phi(Q)

and then obviously phi(Q) is a field since Q is a field and phi is a field homomorphism

[reef rivet]

How did we get from 1 in Z -> 1 in F and extending the map phi from Q to F?

[long nebula]

By exactly your logic here

You know where 1 goes, so you also know where 2, 3, ... go, and then by taking inverses you know where 1/2, 1/3, ... go

So really, by continuing this process, you know where any rational goes

The universal property says that 1) you can reach every rational by doing this process of adding and taking inverses and 2) no matter which path of additions and inverses you use to get there, you'll always get the same answer for what the map should be

i.e. existence and uniqueness

[reef rivet]

Oh I see, ok

Thank you so much for clarifying

[crystal vale]

Let F is the field of char p, and K is algebraic extension of F, then it is not necessarily that K is inseparable extension, right?

I mean yeah, I am just thinking of an example

Let me think

[rocky cloak]

Probably harder to come up with inseperable examples than seperable ones

[tough raven]

K = F

But I'm sure you meant something more interesting, so that this is not a spoiler.

[crystal vale]

I thought so but yeah I need a different one

[tough raven]

Right so figure out what more it needs to do other than this

Usually helps guide the thought process

[crystal vale]

I think let F = F_2, and K = F_2[y] where y is the solution of equation x^2 + x + 1 = 0 over F_2.

So this minimal polynomial has distinct roots

So I think one of the equivalent condition to be separable is the number of morphism K -> K fixes F should be equal to degree of extension, which holds here

[rocky cloak]

It's the number of morphisms from K to the algebraic closure of F.

What you're describing is equivalent to being galois (seperable and normal)

[crystal vale]

Oh thank you for pointing out

[crystal vale]

But i think in this case still it is 2, because y has only two choice

[rocky cloak]

Yes, this is a galois extension. So it's in particular seperable

[crystal vale]

Oh

[long nebula]

you're welcome!

[fleet cairn]

What are the names of the categories C_{A, B} and C^{A, B}

the first one are objects (Z, A, B) such that varying Z; Z -> A and Z->B

the second one have the arrows reversed

[primal junco]

category of spans?

[primal junco]

usually the opposite category is prefixed by co-, so category of cospans

[reef rivet]

Suppose F:K is a finite normal extension and F:L:K. Then L:K is normal if and only if L is preserved by Gal(F:K).

Can someone help guide me through why?

[rocky cloak]

What do you have so far?

[reef rivet]

Like, I know that in general F:K is a normal extension, and F:L:K means that F:L is also normal. I just do not see... how the Gal(F:K) condition matters at all

And why we can assume anything about the relation between L:K other than the fact that K is a subfield of L

[rocky cloak]

Well, what is the definition of normal

[rocky cloak]

The easy direction should be ||if L is normal, then the Galois group preserves L (use that automorphisms permutes roots of polynomials)||

Conversely ||if L is not normal you'll have an irreducible polynomial with a root in L and another root not in L.||
||If you let a be the root in L you get a map K(a) -> F with image not in L, by sending a to the other root.||
||Now use that F is normal to inductively extend this map to an automorphism F -> F.||

[reef rivet]

Normal is like... all of the polynomials K[x] will all have their roots in F.

If L is normal, then Galois group preserves L makes sense.

The converse makes a lot less sense to me. You will have an irreducible polynomial with a root in L and another not in L, but why do we need the map? I am not sure what sending a to the other root in the map does, or how F is normal comes into play here.

[rocky cloak]

An automorphism that takes a to a root not in L will not preserve L

So that's the point of the map

The normality of F is used to construct the automorphism

[reef rivet]

So you are saying if L is normal, then the group preserves L. But if L is not normal, then the automorphism will not preserve L, and that completes the iff statement?

why is the normality of F needed to construct this automorphism?

[rocky cloak]

Well, try to construct it and see what you use

[reef rivet]

Ah, okay. There was a theorem in a previous lecture that allows you to add in one element at a time, and make a new homomorphism. And the condition for that Theorem holds if F:K is normal

Is that what you meant by Inductively extending the map to F -> F

[restive idol]

What would be the universal property of localisation be for modules?

I'm thinking of a setup like this. Let $f: M \to N$ be an $R$-module homomorphism where $M$ is an $R$-module and $N$ is an $S^{-1}R$-module (we can think of it as an $R$-module by the localisation map). Then I want to have an $S^{-1}R$-module homomorphism $$\widetilde{f}: S^{-1}M \to N$$which an appropriate diagram commutes

[cloud walrus]

Lucas

[next obsidian]

The universal property comes from the fact that there’s an adjunction coming from push-pull

Specifically, maps from M into an S^-1A module N are exactly the maps from S^-1M into N

I think you can describe this more explicitly by something like you want the image of f: M -> N to be such that for all s in S, the multiplication by s map on the image is bijective

[restive idol]

Aww, sounds really good! I don't think I know this adjunction. What is push-pull??

[next obsidian]

It’s one of the names for the restriction / extension of scalars adjunction

Between B-mod and A-mod for an A-algebra B

Normally it involves the tensor product but when B = S^-1A this is just localization

[restive idol]

Thank you!! Also, is this normally what is called the universal property of localisation for modules?? I just made it up

I haven't seen it being mentioned in commutative algebra texts

[tribal moss]

In addition to this, you need to require that \tilde{f} is unique -- otherwise S^-1M (+) P would also qualify for every P.

(This is generally the case for universal properties.)

[restive idol]

Yes yes, thank you!

[waxen plover]

Is there any postgrad here who finished Herstein? I would like to talk in private regarding something.

[tribal moss]

Most likely you'll have better chances if you just start your conversation here.

[swift root]

why does it have to be a postgrad lol

[foggy tartan]

picky eater

[swift root]

qwq

[fleet cairn]

allufi is trying to tell me that a group is a groupoid with a single object. He deduce that it being a single object means it has one hom set, and then give the homset a name G. and then proceeds to say if operation on elements of G satisifes these axioms (that looks like 90% of category axioms) then it is a group. am i misunderstanding or is this completely overengineered?

[woeful sage]

they're gonna eat the postgrad!?

[rocky cloak]

I'm not sure what you mean by overengenierd.

The point is that if you have only one object then the axioms for a group and the axioms for a groupoid are the same

[glass sinew]

isn't that literally Joke 1.1

but yeah what jagr said, they are the same. the groupoid definition is just kinda perverse

[fleet cairn]

like why is the groups axioms needed if thats just what a category is anyways

idnetity

composition

inverse just comes from it being a groupoid

[tribal moss]

"Needed" for which purpose?

[quiet pelican]

Because this is how sane people define it, and “a group is a groupoid with one element” is usually used to motivate groupoids rather than the other way around

[woeful sage]

This is like "Why are integers needed if they're just rational numbers with denominator 1"

[dense walrus]

are extensions of $\mathbb Q$ of the form $\mathbb Q(a^{{\frac{1}{n}}}, b^{\frac{1}{m}}})$ same as $\mathbb Q(a^{{\frac{1}{n}}} + b^{{\frac{1}{m}}})$? feels like you should be able to separate a and b with suitable polynomials

[cloud walrus]

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

[dense walrus]

latex seems fine this time around

[quiet pelican]

No because you can do stupid things like a = 4, b = -2, m = 3 and n = 6
Where the latter field is just Q

Usually this should work though I think

[dense walrus]

ahh i see, ty
i was working with the particular case of 2^(1/2) and 4^(1/3) and wondered if this worked in general

[quiet pelican]

Namely, the proof of the primitive element theorem gives you that a^{1/m} + c*b^{1/n} should work for all but finitely many choices of c

Where “finitely many” has some explicit bound which I think looks like nm or something

[tribal moss]

Because it is useful to have a word for "endomorphism of an initial commutative ring as a module over itself", of course.

[dense walrus]

will look into that

[waxen plover]

ummm, not a question to be honest. I want a teacher. Never mind guys. Sorry.

[waxen plover]

🤣

[waxen plover]

thats the direction of motivation I guess? The whole point of generalisation of anything is to search for something non trivial. If you just define a group to be set of actions which fix something in a certain set, then these are the axioms you can abstract out. Then you ask the question the other way around, "does every group have to be a set of actions which fix something. The answer happens to be YES." (Sorry if that is not what you are looking for)

[azure cairn]

You can treat a group like a groupoid with 1 element if you like

It says the same thing as the traditional associative binary operation with certain properties

I don't think it's a bad definition to internalize what a group is /shrug

[fleet cairn]

wait my whole confusion was the associative binary operation with certain properties were built directly on the group is a gorupoid with a single element def

[azure cairn]

Yes, the point is that most intro algebra books don't define groupoids before groups

So to be compatible with other books and their pedagogy, he explicitly writes out what it means to be a groupoid with 1 element

Which is a good idea anyways, since you should be able to make the link between the two defs (groupoid with 1 element, structuee equipped with associative binary op with certain axioms)

[azure cairn]

He notes right after Joke 1.1 that other people would (rightfully) criticize how he defined a group if he didn't elaborate, so he does

[quiet pelican]

(And also, no one actually uses the groupoid definition
Well, except maybe some people with terminal category brainrot)

[south patrol]

To be fair this perspective is helpful for various things and can be helpful in that you can consider group(oid) objects in other cats

[quiet pelican]

Isn’t the group objects perspective generally more along the lines of the morphism based defn of groups?

Like, a one object subgroupoid of a category is just a subgroup of an automorphism group
That’s more like, representations than group objects

[south patrol]

Yes typically but i mean more to emphasis groupoids and then the fact i can view group objects as a special kind

Amusingly e.g. this is a way group objects are defined in oo land

But also like in AG often groupoids appear and then groups are a special kind

[reef rivet]

Can someone explain what a dual representation is, and how it is any different from just a normal representation

[south patrol]

Well a dual representation is a type of representation. It's entirely analogous (and a generalisation of) how you have a "dual vector space" V* associated to a vector space V

[dim widget]

as was said above it’s more a relationship between two representations than it is an absolute descriptor. If V is a representation and V^* is the dual vector space, then G acts on it by sending a functional f to the functional (gf)(v) = f(g^-1v)

and that is the dual representation of the original V

[torpid grotto]

I have a problem that I'm not even sure how to approach. I know somehow I need a mapping to turn odd cycles into even cycles, and I know we have to do this by extending an odd cycle by one. For example, if we were given (1, 2) we might multiply it by (2,3) to make it an even cycle. I'm just not sure how to write this in notation. Does anyone have a pointer as for how to write this mapping?

Actually I might be silly. I think I can simply write something like sigma(n+1 n+2)

lol nevermind I figured it out

[sweet spindle]

Does anyone have good bibliography for supersingular elliptic curves/isogeny based cryptography? I can't find it anywhere

[fleet cairn]

does the set of all continuous functions and + forms an abelian group

(cuz off the bat 0 is identity, commutativity is true, and closure is ensured)

[elfin wraith]

What’s the last thing to check

[fleet cairn]

but im not sure what we should do abt its inverses

but it feels like it should work

oh nvm

(continuous function) + -(the same continuous function)

[verbal valley]

continuous functions on what, to what?

this is an important part of the question

[fleet cairn]

i dont get it

its just Set morphism of R->R satisfying continuity

then composition = addition in terms of picking the next morphism

is that what u mean

[verbal valley]

if X is just some topological space, and Y is just some topological space, how do you add two continuous functions X -> Y

[elfin wraith]

Similar question, what about under multiplication/composition?

[verbal valley]

R -> R works fine, as you've seen, but you need to be specific when you're doing math :)

[fleet cairn]

oh i just meant continuous functions in calculus 🙂

[fleet cairn]

closure for sure (simple fact in calc)
inverse is its recipricol
commutative is obviously true

and lastly identity

its one sided

so not a group?

[karmic moat]

If f : R -> R is continuous, is 1/f?

[fleet cairn]

oh i misunderstood

no

because f can be 0

f being 0 on its own is defined and continuous, but as a recipricol its not

[elfin wraith]

If we’re thinking about multiplication, what can you tell me about say 1/sin(x)?

If we’re thinking about composition, does every continuous function on R have a continuous inverse on R?

[elfin wraith]

Yep

[fleet cairn]

inverse is not satisifed

and for identity

its multiplication so to preserve continuity

*1 and 1*

[elfin wraith]

Yeah similar issue, take like x^2, which is continuous but doesn’t have an inverse on all of R

If you require a slightly* stronger property than just being continuous you do actually get a ring of such functions

[karmic moat]

slightly stronger as in all functions must be of the form f(x) = ax for some constant a?

[elfin wraith]

No nvm I was being stupid lol

[karmic moat]

i don't think you're stupid

#WNope

[shut rose]

For pointwise addition, you would need Y to have an addition operation, so like a TAG or somehing.

[novel star]

is there anything valuable to be gained by viewing the algebraic closures of a field as a groupoid

bc i feel like the fact that distinct algebraic closures are not uniquely isomorphic must be suggestive of something

[quiet pelican]

except for the entirety of Galois theory and half of number theory? /hj

[novel star]

wait is that actually true

i remember someone saying that number theory was the study of the absolute galois group of Q

oh ig this would give u an absolute galois groupoid

but i dont know what a galois groupoid is

[quiet pelican]

The “half of number theory” may not quite be true, but a lot of AlgNT uses Galois theory
And Galois theory is pretty much predicated on “fields have non-trivial automorphism groups, and this gives us a very good description of subfields”

[novel star]

yes i understand that part

im just bothered by the non-uniqueness of algebraic closure

there must be some way to make those extra isomorphisms disappear

[verbal valley]

yea the fact that algebraic closure fails to be universal is kinda the whole of galois theory

it is a good fact rather than a bad fact

it makes things more interesting

[novel star]

oh

how can you use this tho

oh well yeah ig it mustnt be universal otherwise absolute galois group would be pointless

[swift root]

yes there would be no absolute galois group

like, the fundamental group of the groupoid of algebraic closures is exactly the absolute galois group

[novel star]

yes

but what does the groupoid mean

im tempted to grasp at some geometric interpretation

[swift root]

well you'd be grasping at something good, because schemes have a notion of fundamental that, in the case of Spec(k) is the absolute Galois group

in general the whole idea of some group action determining certain "extensions" always feels geometric

i admit i dont know much more about this

[novel star]

oh is this related to the etale fundamental group thing ngroupoid told me about yesterday

[swift root]

yes

[twilit wraith]

today i had my first day for an REU im in, and ive been tasked with finding a basis for the lie algebra sl(n,C)

i think ive done it but i just wanna check that im not missing anything

so since this lie algebra is defined as the matrices with trace 0, it just suffices to inspect the diagonals since any matrix in this space with zeroes on its diagonal is just generated by the set of matrices with 1 in a nondiagonal entry and zeroes elsewhere

so that means that the lie algebra is at least n^2 - n dimensional

and is at most n^2 dimensional since its a subspace of gl(n,C) as a vector space

because the trace has to be zero, the diagonal of any matrix in this lie algebra is determined by n-1 elements on the diagonal

since the last remaining element is just the negation of the sum of the rest

[south patrol]

Ye

[twilit wraith]

so i suspect this lie algebra is n^2 -1 dimensional

[south patrol]

I suspect you are overcomplicating it

[south patrol]

Like this conditions is equivalent to being in sl(n,C)

[twilit wraith]

and a basis would be the union of the set of matrices with 1s in a nondiagonal entry with 0s elsewhere and the set of diagonal matrices with a 1 in the top left corner and a -1 somewhere else in the diagonal, rest being 0s

[twilit wraith]

yes

but i feel like my train of thought is fairly straightforward

maybe theres a simpler way of thinking about it but this seemed the most apparent to me

[south patrol]

Yee this works nice

[south patrol]

Hint is that it is defined as the kernel of a homomorphism

[rocky cloak]

gln is n^2 dim. Trace is a map gln -> C
C is 1 dim, so sln is n^2 - 1 dim

[south patrol]

Gotem

But ye also can like write down a general element

[twilit wraith]

oh thats smart

trace is a linear transformation true

i did have to come up with a basis constructively

but having n^2 - 1 as the dimension before doing so makes it way easier

bc i can just find a linearly independent set of size n^2 - 1, which the set i provided earlier is clearly that

[swift root]

whats an REU?

[karmic moat]

research exprience for undergraduates

it's a usa thing

[twilit wraith]

what anamono said

[swift root]

thats cool

[twilit wraith]

im specifically doing something on applications of lie algebras to a topic of modern physics called celestial holography

[swift root]

wish we had that here

[karmic moat]

national science foundation gives you money and you go to a school and you do a research project over the summer

[twilit wraith]

definitely makes prospects of grad school a tad less stressful since grade-wise ive been barely less than id like

[swift root]

i guess we've got honours program here? but thats uninteresting to me tbh, its very applied focused

[swift root]

thats great!

[karmic moat]

actually i think the nsf gives the money to the school and the school gives the money to the student

[twilit wraith]

yeah this

my own university has one so it makes it to where i can do an REU every summer which is very nice

[karmic moat]

damn that is nice

[novel star]

to what extent can a ring be identified with its category of modules

[wraith cargo]

Morita equivalence is a thing that seeks to measure this

[novel star]

ty! this seems interesting

[next obsidian]

If the ring is commutative then it is the same

But eg, a ring and a matrix ring over it are morita equivalent so it’s weaker for noncomm stuff

[novel star]

thats reassuring

[rocky cloak]

More specifically the center of a ring is determined by its category of modules (it is the endomorphism ring of the identity functor)

[novel star]

i think this makes sens

[reef rivet]

What is Schur's Lemma? I lowkey cannot understand it for the life of me

[verbal valley]

Let (\phi : M \to N) is a homomorphism of modules (eg (R)-modules, (G)-modules, etc) where (M) and (N) are simple modules. Then (\ker \phi) is a submodule of (M), and so by simplicity of (M) is either 0 or (M), and similarly (\operatorname{im}(\phi)) is a submodule of (N) and for the same reason is either 0 or (N). Moreover, if (\ker \phi = M) then (\operatorname{im} \phi = 0), and if (\ker \phi = 0) then (\operatorname{im} \phi) is nonzero and hence is (N). Therefore the (\phi) is either 0, or it is an isomorphism.

[cloud walrus]

Dirichlet

[reef rivet]

I'm... not sure what this means?

Ok. If phi: M -> N is a homomorphism of modules, then phi must be 0 or an isomorphism.

[verbal valley]

no

[reef rivet]

And there's like another part about phi = lambda (Identity Matrix) over C where lambda is an eigenvalue?

[verbal valley]

they have to be simple modules

[reef rivet]

What are simple modules?

[verbal valley]

It's a module M where the only sub-modules are 0 or M.

an irreducible representation of a group G is the same thing as a simple G-module for example

[reef rivet]

Ah Ok, this was the definition he used since this came before the lecture on characters and rings

Schur's was used to prove this, but I lowkey do not understand this solution at all. How would Schur's apply to this?

[verbal valley]

The key here is the step where he shows that rho(a) and rho(g) commute. This is checking that "composing with rho(a)" is a homomorphism of representations from (V,rho) -> (V,rho).

once you have that, and given that (V,rho) is irreducible, you can apply Schur's lemma

(and the fact that the only isomorphisms (V,rho) -> (V,rho) are scalar multiples of identity, which i consider to be a seperate fact but which many authors bundle into the statement of Schur's lemma)

[swift root]

yeah i mean its an easy corollary and needs the fact that C is alg closed

but it does hold for arbitrary simple findim representations of C-algebras

[reef rivet]

So there are two representations? or only 1?

[verbal valley]

only one

[verbal valley]

my main reasoning for this is that schur is true for wayyyyyy more than just f.d. reps of groups over C lol

[reef rivet]

Imma munch on this for a bit

[swift root]

sure

but its true for all fd reps of k-algebras for k algebraically closed thats good enough for me 🔥

[elfin wraith]

Its not so hard, but is also harder than actually proving Schurs itself lol

[swift root]

ok true

schur is like one sentence that is like 4

[quiet pelican]

Also Schur needs 0 ideas, and that needs 1 idea

[reef rivet]

Is being Abelian the same as being G equivariant?

[swift root]

abelian is a property of the group

[verbal valley]

No. Abelian is a fact about the group. G-equivariance is a fact about linear maps

[reef rivet]

🥀

[swift root]

but left multiplication is a morphism of the regular representation iff it is in the center

[verbal valley]

the point is that when G is abelian, the linear map "left multiply by rho(a)" is G-equivariant

[reef rivet]

Ok. I will remember that too

[verbal valley]

(and more generally this yes)

[reef rivet]

For my final tmrw

[verbal valley]

ah gl gl

[swift root]

🙏 you got this

[reef rivet]

My previous exam scores says otherwise 💀

[swift root]

something something its about the journey

[fleet cairn]

in a group G, does the set of all powers of g which has finite order. Form a group if you inherit G's morphism. It looks like it, is this result any important?

[rocky cloak]

Are you asking if the set of elements in G of finite order forms a subgroup?

Then no, this set isn't in general closed under the group operation.

[fleet cairn]

no just a group

[rocky cloak]

Then you'll need to elaborate more what you mean

Like (basically) any set can be given a group structure, by just defining some arbitrary group operation on it

But that's sort of meaningless

[fleet cairn]

ok so an element g has finite order

so then consider the set of its power

g1 g2 g3 gn

and inherit the same mapping from G

[rocky cloak]

So you're asking if the cyclic subgroup generated by g is a subgroup?

Yes

[swift root]

a subset can only inherit the group operation if it is closed under said operation, which isnt in general true for the torsion (finite order) elements

if youre asking about { 1, g, g^2, ... } then this does form a subgroup if g has finite order

[rocky cloak]

That is true

[swift root]

in the case that g has infinite order you need to explicitly include g^-1 and all its powers too

this is usually denoted <g>

[fleet cairn]

so is a subgroup a group but some extra conditions

then it does form a group

it should have closure

[swift root]

if G is a group, then a subgroup of G is a subset containing the identity, closed under products and taking inverses

this subset will then inherit the group structure of G and be a group itself

[tardy hedge]

hello there

[swift root]

haia

[sturdy rock]

all finite abelian groups are isomorphic to an external direct product of cyclic groups

[delicate orchid]

"external" 💔 also you can easily extend this to finitely generated abelian groups if u include the infinite cyclic group

the real question is what did you stand to gain by posting ts msg

[sturdy rock]

"ts" is not "this" unc

[tardy hedge]

yes it is lil boy

[sturdy rock]

all uncs dont gang up on me pls

[delicate orchid]

people say ts like I give one

[karmic moat]

wew is calling ur message shit

u just gonna take that?

[delicate orchid]

now answer the god damn question brainiac

[sturdy rock]

well i just made a statement...

[delicate orchid]

are we here to discuss the details of outdated internet slang or the smith normal form

[sturdy rock]

god forbid a man recite a line of poetry

[delicate orchid]

the question is, can you prove it? and you can't use the calc

[sturdy rock]

i cannot

[delicate orchid]

that's ur homework twin

[sturdy rock]

yea but the course is over

[delicate orchid]

try proving it for "internal" direct products first

[sturdy rock]

and the proof was bit lengthy so i just skipped it

[swift root]

(calc is slang for calculator btw)

[ornate torrent]

Doing group theory for my 2nd year of university and idk why but it just makes me feel warm and fuzzy inside

Its one of the few things in my maths classes that ive just been like "yeah i get that!" For most of it

I suspect it might not stay this way when i go into 3rd year but im okay with that

[quiet pelican]

Group theory is best :3

[south patrol]

Lol

[fleet cairn]

does this theorem or whatever have a name, is it only for groups?

[quiet pelican]

Yeah this only holds for groups

[fleet cairn]

in group theory, order is used in two ways, the order of element (smallest n such that g^n = e) and the order of a group (its cardinality). Is this a coincidence in notation, or is there some weird connection between them (dont just tell me its cuz they are both in the contexts are groups)

[quiet pelican]

The order of an element always divides the order of the group (by Lagrange’s theorem)
And if the order of an element equals the order of the group, the group is cyclic

[somber goblet]

the order of $g \in G$ is equal to the order of the cyclic subgroup $\langle g \rangle \leq G$ generated by $g$.

[cloud walrus]

lexi

[swift root]

quasigroups*

:3c

[tall igloo]

i guess to answer the original question, ive never heard this thing referred to with a specific name

[swift root]

i usually would call it the Latin square property

[torpid radish]

It is really famous, and its (almost direct) application is the Cayley's thm.

[torpid radish]

Every quasigroup has the same property, although it's not generally a group.

[torpid radish]

In fact, such a 'multiplication table' is sometimes called a 'Cayley table', named after him. (I wouldn't say often though...)

[torpid radish]

In simple terms, Cayley's theorem states that every left and right multiplication is a permutation, and the composition of such permutations form a (sub) group (of some symmetric group). The 'latin square' you are seeing now is a collection of permutations, for each row (=left multiplication) and column (=right multiplication). Once you realize this, the rest can be shown almost trivially.

[past fjord]

in a 2x2 matrix it isnt a group under multiplication, i used the fact that AB != BA why is this not sufficient enough to prove its not a group? Doesn't AB!= BA prove multiplication doesn't commute?

[azure cairn]

groups dont need to be commutative wrt their operation

groups whose operations are commutative are special and called abelian groups

note that, however, all elements must have an inverse under the group operation for it to be a group

[past fjord]

ahh forgot about abelian

ty

does a ring need to be communative to be a field? or does this go into the division ring definition

[azure cairn]

yes

[past fjord]

ty

i have my abstract algebra exam in like 3 hrs im like stressing rn

[mint seal]

a field is a commutative division ring

[verbal valley]

in modern terminology yes -- though some authors, especially historically, do use "field" to include some noncommutative division rings

[elfin wraith]

The french and their consequences

[maiden crater]

Let $r \in \mathbb{Q}$ and $r \geq 0$. Then $\mathbb{R}/(x^2-r) \cong \mathbb{Q} ( \sqrt{r})$
\
Let $\theta = \bar{x}$ then $\bar{x}^2 = r$
\
Define $\phi: \mathbb{Q} [x] \to \mathbb{Q} ( \sqrt{r} )$
with $x \to \sqrt{r}$
\
$ker \phi = (x^2-r)$
\
Then $\mathbb{Q} / \ (x^2-r) \cong Q \left ( \sqrt{r} \right)$

How does this look?

[velvet hull]

do you mean Q/(x^2-r)?

[maiden crater]

yes

I originally wrote it for R, which I supposw is right too?

[velvet hull]

no it's not, R[x]/(x^2-r) is just R for r>0

your statement also is not true when r=0

[cloud walrus]

what a wonderful world(wai)

[maiden crater]

oops, but yea, it's fine overall I guess?

[velvet hull]

I don't really see a good justification for why the kernel is x^2-r exactly but the details are not hard to work out

[maiden crater]

I can do that

one min

well, (x^2-r) goes to 0 , obviously

so any multiple of it does too

[maiden crater]

@desert verge what's wrong

[swift root]

x^2 - r may not be irreducible

[velvet hull]

oh yeah there's also that

[swift root]

i.e., if r has a rational square root then ℚ[√r] = ℚ

[velvet hull]

you explain it to them not to me

[swift root]

the kernel of ϕ certainly contains (x^2 - r), but it may not be (x^2 - r). In fact, kerϕ = (x^2 - r) if and only if x^2 - r is an irreducible polynomial over ℚ, which is if and only if r = a/b for some perfect squares a, b ∈ ℤ.

in the other case, ℚ(√r) = ℚ, so √r ∈ ℚ and ker ϕ = (x - √r)

[maiden crater]

oh right

I meant it to be for irreducible

my bad

[vivid kestrel]

for a field K of characteristic not equal to 2, I know that every quadratic extension is of the form K(sqrt(a)) for some a in K, how do i show that if two quadratic exteions K(sqrt(a)) and K(sqrt(b)) are isomorphic, that then a = c^2 b

[quiet pelican]

We know sqrt(a) \in K(sqrt(b))
Write sqrt(a) = c + d sqrt(b) and expand out

[vivid kestrel]

so is the above only true if we assume that the isomorphism of K(sqrt(a)) and K(sqrt(b)) restricts to the identity on K? because if not, how would we get that sqrt(a) in K(sqrt(b))? since we wouldnt know that (let f be the isomorphism) f(sqrt(a))^2 ) = a?

[quiet pelican]

Isomorphic as extensions includes the requirement that it’s the identity on K

[vivid kestrel]

ah okay i didnt read it that way but it makes sense

so there are examples where two quadratic extensiosn are isomorphic as fields but not in a way that restricts to the identity on K?

[quiet pelican]

Q(x)(sqrt(x)) and Q(x)(sqrt(x + 1)) over Q(x) are not isomorphic
But over Q they’re both isomorphic to Q(x) (as they are Q(sqrt(x)) and Q(sqrt(x + 1)) respectively)

[vivid kestrel]

thanks !!

[maiden crater]

stupid notation question

what does F(x)[y] mean

ah

😭

[crude kiln]

it's always vaguely bothered me that you use round parentheses for field extensions and square brackets for ring extensions when rings are round and fields are square

[next obsidian]

Idk wtf ur cooking but I love it

[crude kiln]

this isn't like a "math is red english is blue" type thing where i'm associating mathematical objects with shapes based on vibes i mean the things that the words "ring" and "field" refer to outside of math

[rotund dragon]

like an actual ring (one on finger) and field (field of grass or whatever)

[crude kiln]

yeah

[elfin wraith]

Does a field need to be square?

[crude kiln]

it would be kind of weird if one was round

i'm picturing a field of crops specifically

[rotund dragon]

how else are we gonna make "compute the area problems" for people to post in #algebraic-geometry

[elfin wraith]

I usually think of a field as just any kinda large enough patch of grassy land, maybe thats just me, but yeah I guess farmed fields tend to be roughly rectangular

Further fuelling the agricultural geometry memes

[quiet pelican]

Agrarian groups

[tardy hedge]

im just trying to understand the setup of this formula, the mN in N/mN there really means the extension of the image of m times N right

And do we have phi(m) subset n?

ty

[tough raven]

They are the same.

That is, mN for the R-module N and (mS)N for the S-module N.

[tardy hedge]

Oh i didnt know you were raghuram

Didnt realize the name change. Thought simplicial bicycle was someone new lol

[tawdry coral]

hi

[karmic moat]

https://en.wikipedia.org/wiki/Center-pivot_irrigation

it's pretty fun to look out while flying and see these circular fields

it's like a big quilt

[crude kiln]

woah

[karmic moat]

on the other hand, a circle inside a rectangular field is what those aliens from mars do or something

[tardy hedge]

Ok yea idk why this confused me for a sec but ye

[fickle dirge]

Stupid question.. but what is this "simple computation"

Like does the book mean that a simple computation can be used to get a(x) and b(x) or just that its easy to show that these work

[quiet pelican]

The simple computation is just carrying out Euclid’s algorithm and back substitution as they described

[fickle dirge]

ah

[quiet pelican]

Like, they say “by the Euclidean algorithm”, and that’s a procedure that’ll give you what a and b are explicitly

[fickle dirge]

Yes got it, I just thought that they only used the fact that a(x)(1+x) + b(x)(x^3-2) = 1

thanks

[vapid vale]

I feel similarly but not for rings/fields but rather multiplicative inverses feels more rigid and akin to square brackets

[hybrid oriole]

is this dummit and foote

[fleet cairn]

For the connection between groups and groupoid with a single object. It's only sort of meaningful if defining group <- groupoids and vice versa is defined this way.

Given an arbitrary group (G, ), you can construct a groupoid with a single objects:
-Let the single object be a placeholder {x}
-Define Hom(x, x) = G
-Composition of morphism, f;g := fg

and

Given an arbitrary groupoid A with a single object, you can construct a classical group:
-Let G be a set such that G = Automorphism(A)
-Define the operation/function GxG -> G, such that f*g -> h := f;g -> h

the main idea im conveying is that for example when defining the definition of morphisms or operation, it has to "mirror" the referenced. because otherwise you can always define any weird looking 1groupoid from a group and vice versa though it wouldnt be very useful and satisifying result. is there a more formal and precise idea to explain that this particular definition is more special? i hope this makes sense.

[tardy hedge]

yes

[crude kiln]

the thing that makes this special is that if you take a group, convert it to a groupoid, and then convert it back into a group, you end up with the same group, and vice versa (up to the dummy variable {x} possibly being different)

you can freely convert between the two objects without losing information

[fleet cairn]

i feel like i expect some more idk but ur right

[crude kiln]

i guess that would be naturality?

or no it would be functoriality in this case

basically just meaning it plays nice with morphisms

[fleet cairn]

all quasi group's multiplication table is a latin square, but how about the converse

[swift root]

for finite magmas, one is a quasigroup iff the multiplication table is a latin square

[fickle dirge]

Any hint with this? I think I can do it by just brute force taking theta = a + bi , ab in Q then subbing in x^3 -2. But could there be another way? Or is that just the standard way

[dim widget]

I think that's a pretty normal way, but if you are learning field theory/Galois theory there is a better way

[fickle dirge]

yeah this is field theory, I was looking for a better way

this is from dummit

[dim widget]

Namely: if L/K/F are fields then [L:F] = [L:K][K:F]

if one of those polynomials was not irreducible then (since it is cubic) it would have a root, so there would be a map Q(\cuberoot{n}) \to Q(i)

since that is a map of fields it would be injective, so there would be a subfield Q(i)/Q(cubetoot{n})/Q

[next obsidian]

That’s a cool way to do it, but I think the easiest is that you just know what the roots of this polynomial are by using sqrt(2)e^2ipi/3 (or with sqrt(3)) and then you can just see these aren’t in Q(i)

[maiden crater]

einestein criteria mayhaps

[next obsidian]

You don’t need to show there is not root in all of Q(i) by starting with generic elements, you know all the roots already

[next obsidian]

You’d need to know the primes of Q[i] tho which idk you might or might not

[maiden crater]

true, I think this is in exercise of D&F after gusaain integers

unless I'm rememebr wrong

[dim widget]

if you know a little bit you only really need to factor 2 and 3

[next obsidian]

Yeah

[dim widget]

and that could be done by hand

[next obsidian]

I was gonna say you can maybe do that with just the norm

[dim widget]

yeah

[next obsidian]

But you also need to know Eisenstein lol

[dim widget]

well I think a lot of students know that

[maiden crater]

that is done well before fields in D&F which is where OP says this is from

[next obsidian]

Oh I see

[dim widget]

since it's the only systematic way to show a polynomial is irreducible it's pretty often taught in ring/filed theory

[maiden crater]

so yea, Eisenstein should work

[next obsidian]

Me when I make a theorem about how to file my taxes

[maiden crater]

unrelated minirant ||I for some reason on my test today, thought algebric fields are algebrically complete, lost 5 marks in a snap, but I digress||

anyways, sorry star

what were you saying

[fickle dirge]

So using this injective map I get a contradiction right [Q(i) : Q] = 2 = [Q(i) : Q(cbrt(n)] [Q(cbrt(n) : Q] = x * 3 right does that work?

[next obsidian]

What’s an algebraic field

[dim widget]

Yeah

[maiden crater]

oops, I meant algebric extension

[next obsidian]

Wait what???

[fickle dirge]

thanks!

[next obsidian]

No no no

That presupposes the polynomial is irreducible

Oh wait

Over Q

[dim widget]

the basic point is that a field cannot contain a subfield of larger degree

[next obsidian]

An algebraic extension of what lol

[maiden crater]

or of deggree that doesn't divide it no?

[dim widget]

or that

[swift root]

https://tenor.com/view/no-no-no-wait-wait-wait-gif-23893070

[fickle dirge]

I thought about that but I wasn't sure let me try to do it that way

[fickle dirge]

yeah

[next obsidian]

Okay well how do you know [Q(cbrt(n)):Q] = 3 haha

[maiden crater]

unless you've done field extensions in your course, you should yeah

[next obsidian]

I guess it’s easier to show that?

[dim widget]

use einestein’s criterion

[next obsidian]

You can even use real analysis

[maiden crater]

That's not too hard is it

[dim widget]

I think you can show that with einestein’s criterion

[next obsidian]

I feel like it’s like

[fickle dirge]

yes i can show it with eisenstein

[next obsidian]

Virtually the same as showing it over Q(i) tho haha

[dim widget]

you mean einestein

[next obsidian]

It’s like ever so slightly easier

[dim widget]

You can also show it with the rational root theorem which is much easier

[next obsidian]

I feel like what’s more swag is you conclude either Q(i) = Q(cbrt(n)) or Q(cbrt(n)) = Q

[fickle dirge]

is it a typo in dummit?

[next obsidian]

And then you see that’s impossible

[dim widget]

no it actually is Eisenstein it was a typo above

I am just joking around

[maiden crater]

that would require you to know what gcd means in Q(i), but being an ED that's obvious I think?

[fickle dirge]

oh okay haha

[maiden crater]

Sorry if I'm yapping too much

[next obsidian]

Wait lmfao TTEG you’ve been carrying that from this one lol

I didn’t even notice

[dim widget]

No we’re talking about showing degree of Q(cuberoot(2)) is 3

which comes down to showing that poly is irreducible over Q

[maiden crater]

||oh, that can also be shown by showing the minimal poly is of degree 3,no?||

[dim widget]

so you can use the usual rational roots theorem

[next obsidian]

That’s literally what we’re doing

[maiden crater]

oops

sorry :(

[dim widget]

yeah that is equivalent

[fickle dirge]

yes i think just eisenstein works right..? I'm not sure how to apply it to Q(i), I think the contradiction argument above was clean enough

[maiden crater]

you'll need primes in Q(i)

[next obsidian]

Q(i) = Frac(Z[i])

And Z[i] is an ED is a PID is a UFD so you can use Eisenstein

So all u need to do is know 2 and 3 are prime in Z[i]

[fickle dirge]

right

that works too thanks everyone:D

[next obsidian]

U should write up the one that u think is most swag

That’s what math is about

[next obsidian]

Nothing

[maiden crater]

?

oh right

no primes in Q(i)

[fickle dirge]

Q(i) is a field haha

[maiden crater]

yes

[next obsidian]

[maiden crater]

I wanted some advice

[dim widget]

(0)

[next obsidian]

Does that mean 0 is prime?

I honestly forget if 0 is excluded by convention

[dim widget]

(0) is always prime on a domain

[next obsidian]

Buy a magic 8 ball and shake it

[maiden crater]

so for someone who has done basic group, ring and field theory what's the enxt step

[next obsidian]

The ideal yeah but what about the element

I guess 0 is prime iff domain

[maiden crater]

modules, gradute group/ring theory, or gallois or something like that

[dim widget]

do advanced group theory ring theory and field theory

if you haven’t learned Galois theory definitely learn that

[maiden crater]

which book would you suggest

[next obsidian]

You should purchase this book called Commutative Ring Theory by Hideyuki Matsumura and read it cover to cover

[maiden crater]

D&F would work for this?

ooh, I forget

I have a commutative alg book

[dim widget]

yeah summit and Foote has a surprisingly deep selection of topic

[maiden crater]

the classic don't recall its name

mcdonald I think?

[dim widget]

It even gets into quite advanced commutative and homological algebra

[next obsidian]

Is it called Commutative Algebra by Sir Michael Atiyah and someone McDonald

[maiden crater]

yup, that

[next obsidian]

Fuck that book

It’s trash

Jk

[maiden crater]

😭

[next obsidian]

But the book is the exercises lol

[dim widget]

It’s a pretty good book

A bit dense though

[maiden crater]

eh, I can do it over my third year, doing mostly analysis courses next year

will do algebra stuff in year 4( cat theory,homological, alg, commutative alg)

[fickle dirge]

I'm taking it next semester u scared me

[dim widget]

I learned commutative algebra from atiyah and Macdonald and it was pretty good

[next obsidian]

It’s legit probably like the best “here’s what you need to be able to get through introductory AG without being really confused” really fast

[next obsidian]

Like from the two of them directly? That’s cool

[dim widget]

I think atiyah had already gone crazy when I was a college student so that wouldn’t have been possible

[next obsidian]

If you’re hardcore though you’ll purchase a book called Commutative Algebra by this guy named Bourbaki

[maiden crater]

Also, uh I'll probably get biased answers here, but. graph or metric spaces would be better next sem.

Intersted in both, but as I'll do measure can only do one

[next obsidian]

Just be a HSer?

You’re doing measure theory?

But haven’t seen metric spaces before?

[maiden crater]

yes

[fickle dirge]

thats weird

[next obsidian]

Do you know topology?

[maiden crater]

on R yes

but not in general

[next obsidian]

That’s interesting…

[maiden crater]

the uni wants to offer toplogy the sem after the next

[fickle dirge]

its basically the same

[maiden crater]

and that's after hearing our interests

[next obsidian]

I feel like if you’re gonna do measure theory you should just say fuck it and learn metric space shit as a byproduct of knowing topology

Graph theory is p cool

If it was a gen top class I’d say do that over graph theory

[swift root]

yes, famously every topology appearing in math behaves like the real numbers

[maiden crater]

||just never wanna see stats again||

[next obsidian]

But idk to do a whole class on metric spaces it’s like whatever

[maiden crater]

thats not possible this coming sem :(

[next obsidian]

That’s why I said if

[fickle dirge]

they did say metric spaces not topology in general

[swift root]

right

still

[next obsidian]

Like honestly unless you’re doing really weird shit I don’t know why you need to know a lot of stuff about metric spaces

[fickle dirge]

yeah i was joking anyway

[next obsidian]

And if they show up in some odd situation, your intuition from real analysis mostly works

[swift root]

yeah usually the topology is the most interesting part

unless you do like ggt idk

[next obsidian]

Either that or it’s an ultra metric inequality and then you need to rebuild intuition

[swift root]

ask micoi

[next obsidian]

But like

[maiden crater]

we'd be doing baire category in this

[next obsidian]

Unless you’re doing the fucked analysis on general metric spaces or whatever the hell the Dieudonné book does

The baire category theorem is literally easy

[maiden crater]

though that's normal I think

[next obsidian]

It’s in the start of Munkres

https://stacks.math.columbia.edu/tag/0CQN

Like it’s literally also on the stacks project

Lol

[swift root]

Bredon spends like a couple pages cooking up some fucked up manifold counter example to the baire category theorem

I believe

[next obsidian]

Counterexample with what hypothesis lacking?

[next obsidian]

Wait I think this is a different one to what you’d see in the class lol

[swift root]

idr

[next obsidian]

You are not gonna be a locally (quasi)-compact space lol

[swift root]

hold on I think they've got bredon here in the uni library

[next obsidian]

Maybe the metrizable thing

But idk, locally you are so

[next obsidian]

Wait does this not require the neighborhoods to be open?

Wait lmfao a compact neighborhood is defined in a fucked way

Nvm lol

[swift root]

okay nvm its not an explicit counterexample?

its like a fucked up "manifold" (not required to be second countable) with a countable dense subset, and an uncountable discrete subset

[acoustic igloo]

These are subgroups of achiral tetrahedral symmetry

Are they grouped by conjugacy class?

https://en.wikipedia.org/wiki/Tetrahedral_symmetry

I think so. The group is isomorphic to S4, which has 11 conjugacy classes of subgroups according to https://groupprops.subwiki.org/wiki/Subgroup_structure_of_symmetric_group:S4

[fickle dirge]

How can I show that x^3 -pi^3 is irreducible over Q(pi^3)

do i just get all the roots in C and show none of t hem are in Q(pi^3)?

[acoustic igloo]

Huh, achiral tetrahedral symmetry is not a subgroup of full icosahedral symmetry?

[tribal moss]

You shouldn't need to do it separately for each root.
(The two complex roots of x³-pi³ are obviously not in Q(pi³), but the proof that the real one is not there ought to work for every root without changes anyway).

[fickle dirge]

yeah, I guess I was wondering if there would be some other way.. I feel like in field theory there is so many ways to do things so I'm trying to get some deep understanding

[tribal moss]

What I would do is assume there is a root in Q(pi³); then it is f(pi³)/g(pi³) for some rational polynomials f and g. Now x³-pi³=0 expands to a rational polynomial that has pi as a root; if you can show that this cannot be the zero polynomial, you have a contradiction with the known fact that pi is transcendental.

[fickle dirge]

ah i see I will try to write this one out, thanks

[tepid light]

I already got the set of subgroups from partitioning S10 into two groups of 5 numbers being permuted over, what are some other ones?

[south patrol]

This is an easy problem because you can just say how many you found and the answer will be correct

[tepid light]

yea

my algebra class was less than successful this semester though so idk what else there is (we didn't cover much content)

I crave knowledge

[swift root]

what did you cover

[tepid light]

Symmetric groups, dihedral, whatever groups are notated A_n, signs, homo/iso/automorphisms, some graph theory and graphs being automorphic, and lagranges theorem/consequences of it

That's pretty much it

[swift root]

not even the isomorphism theorems?

[tepid light]

Nope :/

[swift root]

tf

[tepid light]

Also did cartesian products, centers, conjugation, counting transpositions, odd/even, Cayley tables, thats about it

looked through my notes real quick

But no major theorems excepting lagrange's theorem and some consequences of it.
The prof honestly had an alergic reaction to proofs, he taught almost entirely through examples. "Here's a dihedral group, which is actually the symmetry group of a hexagon, and isomorphic to this graph btw, which can actually be represetented as a subgroup of X, ... etc"

kinda all over the place

[quiet pelican]

I think I can prove that these are all of the possibilities but it’s bleh

[tepid light]

really, that's it?

[quiet pelican]

I believe it should be

[tall igloo]

I gave this some thought and would also be surprised if there's more

[tepid light]

interesting

Why is that? I would've thought there was some weird way to label things to get other mappings

[tall igloo]

yeah I like this phrasing. I also think I can prove this however I don't actually want to lol

[tepid light]

proof by trust me

This is the actual question statement

[quiet pelican]

Essentially it’s a very similar proof to the fact that Aut(S_5) = S_5

[tall igloo]

I would think about the generators of S_5 and note that the generators of each S_5 you have must not be able to talk to each other

[tepid light]

Made me think there must've been way more than the 126 I found

[quiet pelican]

I was thinking about it by considering the 5 cycles
Because having two order 5 elements in S_{10} which are not part of the same cyclic subgroup and which commute is very restricting

[quiet pelican]

Actually it’s proof by

[swift root]

proof by eepy

thats so felt

[tall igloo]

this also seems like it'll work ye

[ivory anvil]

Can infinitely generated groups be isomorphic to finitely generated ones?

[velvet hull]

depends on what you mean by infinitely generated

[ivory anvil]

I thought of an example so instead is there a group with countably infinite generators that isn’t isomorphic to some finitely generated group?

[ivory anvil]

In what way?

[velvet hull]

you can take any infinite but finitely generated group G. then {G} is an infinite generating set for G

[ivory anvil]

What’s the term for the smallest possible generating set of a group?

[velvet hull]

not sure if there's a word for that, but then if you use that as your definition of finite and infinitely generated group then the answer is no, by definition

[karmic moat]

Minimal generating set?

[velvet hull]

that doesn't work because two minimal generating sets of a group is not guaranteed to have the same size

there are special cases in which case this is true (p groups come to mind)

[karmic moat]

I mean minimal wrt inclusion

But yeah I should’ve specified

[velvet hull]

that doesn't fix the problem

since I assume OP wants minimal in size, not inclusion

[karmic moat]

Hmmm fair enough

[velvet hull]

really, I think the only meaningful phrasing of the question is this:

Can every infinite generating set for a finitely generated group always be reduced into a finite generating subset?
Let me think try and think of an example, but I suspect the answer is no nevermind its true straightforwadly

[maiden crater]

(1) under addition is also generated by the set {-1,0,1,2...} where all positives are primes

[wraith cargo]

That isn't a minimal generating set tho

[maiden crater]

oops, right

[maiden crater]

isn't this the questiom?

[wraith cargo]

It is

[maiden crater]

The question doesn't then ask for a minimal generating subset does it

[wraith cargo]

Well the question is asking

Are there infinitely large minimal generating sets of finite groups

[maiden crater]

o

[long nebula]

Is there an example you're thinking of that prompted this question? Sounds very interesting

[ivory anvil]

No just popped up

[hybrid oriole]

im reading subsets on groups with presentations nad relations, idk how to find the minimum size of the groups

[tardy hedge]

How would you intuitively describe what a flat module is?

[hybrid oriole]

i dont know what is a flat module, but im going to see what it is

[tardy hedge]

That wasn’t a response to your question it was a different question lol

[hybrid oriole]

oh ok

[tardy hedge]

How

[velvet hull]

It’s called flat because when you localize the module at the prime ideal, its dimension over the residue field as a vector space is the same across all prime ideals

[hybrid oriole]

i dont think every set can be reduced

[tardy hedge]

How do we know its a module over the residue field

[velvet hull]

You have to take quotients

Residue module over residue field

[tardy hedge]

Oh ok

I had no idea about that characterization of flatness

I’m assuming that has some geometry interpretation but i havent learned much alg geo yet

[velvet hull]

Yes it’s ag

[velvet hull]

Il have to double check the exact statement when I get back home tonight

[tardy hedge]

Not many people talk about depth and cohen macaulay stuff here, is it kind of niche or something?

[karmic moat]

Maybe it's niche in the community in this server but it's definitely not niche in commutative algebra

[next obsidian]

Bro thinks cohen Macaulay rings are niche, vro everything is CM when you go to a comm alg conference

[swift root]

elementary lattice theory :>

[tardy hedge]

Lol

I hope to more often

[untold garnet]

~~How will they ever get to hard problems if all their rings are CM? ~~

[tough raven]

Alternatively assume everything is smooth

[dim widget]

ChMonkey

[dim widget]

I think just not so many young people are interested in more advanced commutative algebra

[wraith cargo]

These damn youngins not wanting to learn PROPER commutative algebra! They're satisfied with reading that dribble ATIYAH-MACDONALD instead of a proper text like ZARISKI-SAMUEL. This generation is DONE FOR!!!

[untold garnet]

My impression is that advanced commutative algebra is not fancy and quite difficult to motivate in a vacuum.

[verbal valley]

i mean it's pretty easy to motivate once you have enough AG

but

yea

i guess that's not in a vacuum

[chilly radish]

AG is just motivation for doing more commutative algebra anyways

[untold garnet]

I prefer Eisenstein descent or R=T theorems as motivation for some advanced commutative algebra, but explaining how this comes into play in the argument seems usually hard to explain.

This is a bit like localization: schemes are a great motivation, but… go explain that to a student taking their first class in commutative algebra.

[verbal valley]

some day i want to write an AG book that is jsut a collection of a bunch of practical examples worked through scheme theoretically, to both motivate the scheme theoretic pov and to illustrate how it's not all just a bunch of abstract nonsense

[tardy hedge]

in R->R[x,x^-1] grade of an ideal of I extended in R[x,x^-1] can't get smaller than grade(I,R) right? Is it just the same grade

[fickle dirge]

Any hint with this?

[quiet pelican]

So take F = Q, and a, b \in R such that F(a) \cap F(b) is larger than Q

[velvet hull]

morally, you are looking for a and b that are distinct but partially related

[fickle dirge]

but still F(a) ≠ F(b)

[quiet pelican]

Yeah

[maiden crater]

||Ideally I'd say you want a to belong in F and b to belong in the extension and not in F?||

This is wrong, my bad

[fickle dirge]

Oh I can just take something like Q(sqrt2 + sqrt3), Q(sqrt2+sqrt5).
Then the degree of the extension Q(a,b) is the degree of Q(sqrt2,sqrt3,sqrt5) = 8. But degree of Q(sqrt2,sqrt3) is 4 and same for Q(sqrt2,sqrt5)

does that work?

8 < 16

[maiden crater]

I think it does?

[fickle dirge]

ok yay

[balmy python]

when someone says how many different ways are there for a group to act on a set, are they asking the number of different mappings G -> S(X)?

[tribal edge]

up to conjugates i think ? cause two conjugate actions act the same way

[noble nexus]

yeah up to symmetries

though it kinda depends what kind of symmetries you want to consider

you could say up to isomorphisms of G-sets, which basically means up to permutations on X

but you can also account for the automorphisms of G

[balmy python]

wait wdym?

[tribal edge]

two actions are considered the same if one can be obtained from the other by conjugating with a permutation of X

[balmy python]

is this to do with the fact that the stabiliser of a set element is invariant under conjugations?

[uneven bobcat]

Personally I think the wording is a little vague and I would ask them to clarify if they meant to count group actions, ie homomorphisms G --> S(X), or if they meant isomorphism classes of actions.

Since, it would seem from the discussion that you haven't been told what an isomorphism of group actions is yet, I would actually assume they just mean homs G --> S(X), because if you're taking a course/reading a book, one would think you would've come across the definition of isomorphism before being asked this question, if they meant iso classes.

[uneven bobcat]

I might be misunderstanding what you're saying, but in general the stabiliser need not be normal, which is to say, invariant under conjugation.

Where it "comes from" is just people asking themselves what a reasonable notion of isomorphism/equivalence of group actions ought to be. In general, two objects should be equivalent if there are morphisms f:A --> B and g:B --> A such that fg and gf are both the identity. So the question becomes, what is a reasonable notion for a map between two G-actions. I would encourgae you to spend a little time meditating on what you think a good definition would be, and then looking up what other people use.

[uneven bobcat]

I doubt this is what is meant, in my experience it's pretty uncommon for people to consider this. Although, I do think it's interesting to think about how Aut(G) acts on G-set, or say the irreps of G.

[noble nexus]

Yeah I agree, usually people don't (though perhaps they should more?)

I remember when doing finite group theory this coming up for classifying semi-direct products

because you need to account for the symmetries of the group itself and the other group it's acting on if you want to isolate the truly "unique" actions

[karmic moat]

Yes, assuming R is Noetherian. If R is Noetherian, R -> S is faithfully flat, and I is an ideal of R such that IR ≠ R, then grade(I, R) = grade(IS, S)

I’m pretty sure this statement is true

[karmic moat]

Proof:

Actually this has got me thinking

Do people study the grade of non-finitely generated modules over a Noetherian ring? or finitely generated modules over a non-Noetherian ring? or non-finitely generated modules over non-Noetherian rings?

I assume one would lose the nice properties of grade, namely, the theorem of Rees saying that all maximal sequences have the same length given by non-vanishing of Ext

But maybe one could just define grade by non-vanishing of Ext

Actually now I have another question. If R is a Noetherian ring, is there an example of a non-finitely generated R-module M and and an ideal I such that IM \ne M, but there are two maximal M-sequences in I with different length?

[karmic moat]

Looks like this is the generalization to non-Noetherian rings (Northcott “Finite Free Resolutions”)

[balmy python]

if the characteristic of a field K is 0, why is the smallest subfield isomorphic to Q?

[next obsidian]

Z embeds into it cuz char 0 and then from there you get Q

[lusty marlin]

Maybe they're asking why there can't be a subfield smaller than ℚ.

[next obsidian]

But this holds for any char 0 field so Q would be in any such subfield

Unless you want me to note that there can’t be a positive char subfield in a field of char 0 haha

[lusty marlin]

Yeah that's what I had meant. I remember being confused about this exact thing back when I first learned algebra

[maiden crater]

it has 1, and then it follows you can construct something that looks like the rationals

it as 1 ,so it follows n*1_F exists

so you have something like N

now by adding 0 and their inverses you must have something that looks like Z

now you're missing inveses

so you add 1/n

but the the rationals arise at once

[maiden crater]

that is once you have 1, all of this naturally arises

it's trivial I suppose to see this is a subfield?

[balmy python]

ah yeah i get it now 😭

the homomorphism must also abide by the inverses

[maiden crater]

ye!

[thin stream]

hi :) my course notes have a proof for this lemma, but it's very direct-proof and doesn't feel enlightening, so i've been trying to prove it a different way with a bit more algebraic flair?

here's my attempt:

(the blue arrow at the top is the free group functor)

im not sure how i can show it commutes?

(and im not even sure if my proof is correct, esp the last step)

could somebody help me clean up my idea?

[swift root]

the last step feels very convoluted, showing all those isomorphisms

do you know the homomorphism theorem?
If f : G → H is a morphism then this factors through a unique morphism f* : G/N → H if and only if N ⊆ ker f

Then you can replace your last steps after <<R>> ⊆ ker φ with a reference to this

[thin stream]

ok lowkey i forgot that was a theorem

and was accidentally trying to prove it in my answer

😭

ok so my idea is correct... yay!

[swift root]

its probably one of the most most useful theorems in elementary group theory

i mean, the fact you were reproving it here just goes to show it lol

[next obsidian]

Elementary algebra of all forms period

[swift root]

yes its the standard way of proving it

[swift root]

sure

[thin stream]

thanks enpeace! see yall

[swift root]

i can stand by that

[next obsidian]

At some point this statement will become so embedded into your blood you will believe it to be an axiom

[swift root]

the universal algebra formulation is particularly enlightening because it's about relations

[next obsidian]

Universal algebra ppl like vegans, always gotta be mentioning it any chance they can

[swift root]

i will not stand for vegan slander

[elfin wraith]

Bros living in 2015 🌾

[swift root]

top 10 cringiest vegan feminist compilation

[next obsidian]

Is the smugcatto not dripping in enough irony for yall

[swift root]

[karmic moat]

Girl who got a septum piercing: didnt even hurt btw

[south patrol]

Lol

Difference is that veganism is a moral imperative whilst universal algebra is the opposite

[next obsidian]

Moral imperative my ass

[swift root]

whats the opposite of a moral imperative even

warcrime?

[south patrol]

I mean any moral prohibition

[tough raven]

Morality prohibits universal algebra?

[karmic moat]

when you make mapo tofu is the substitute for the meat just more tofu

[long nebula]

shiitake mushrooms

[karmic moat]

could be the move

[long nebula]

plus peas and diced carrots

[karmic moat]

hmm yeah that sounds good

[long nebula]

[karmic moat]

i wish i could find chinese rapeseed oil in the states

菜籽油

i cant find it in the asian markets near me

[south patrol]

Mapo tofu is also just often served without meat right

Or am I having fake mapo

[karmic moat]

usually it's got ground beef or pork

[maiden crater]

mapou tofu is group theory now

TIL

/j

I mean whatever sustains group theorists Algebrists

[next obsidian]

I ain’t never had no mapo tofu without meat

[quiet pelican]

Everyone should be group theorist :3

[swift root]

does quantum groups count

[quiet pelican]

We will convert you in time :3

[swift root]

uhuh

quantum groups are already groups but better

[next obsidian]

What about abelian groups which also have an additional commutative, associative, unitary binary operation which distributes over the group operation?

What about that theory

[quiet pelican]

Commutativity doesn’t exist

[swift root]

well that theory is obviously trivial because it commutes

[karmic moat]

Interesting…

This is worth looking into, seems promising

[chilly radish]

https://themalamarket.com/products/caizi-you-sichuan-roasted-rapeseed-oil?srsltid=AfmBOorwA7iriQ9AWqAU-ErOeCqe0-kaHxNzncAtL8GLrgwT3TO_wuXI

Mala market delivers

[white oxide]

Can I have a hint for 3c please, I’m proceeding by contradiction

[next obsidian]

Show that dim Ann(W) = dim V - dim W

Then show that if W1 ≠ W2 then dim Ann(W1) [or W2] is too small

[novel star]

what is it called if you have 2 groups and you interleave their elements

say G and H

[sly crescent]

Direct product?

[novel star]

and you have strings of elements of G and H

[verbal valley]

free product?

[novel star]

yes i think its this

ty!

[karmic moat]

👀👀👀

[delicate orchid]

Groups rings fields btw ✌️

[karmic moat]

Roasted rapeseed oil btw

[swift root]

well there certainly are groups rings and fields there

[somber goblet]

if $E/F$ is a field extension where $[E : F] = mn$ for $m, n \in \bN_{>1}$, is there an extension $K/F$ such that $E/K = m$ and $E/F = n$?

[cloud walrus]

lexi

[somber goblet]

i.e. does any composite-degree field extension factor into sub-extensions

[rocky cloak]

Not necessarily, no.

[tough raven]

Just going to sketch the following but this is easy for purely inseparable extensions so it reduces to separable extensions. For those you can embed in the Galois closure and it becomes the following group theory problem: given a finite group G and subgroup H (such that the intersection of all conjugates of H is the trivial subgroup), is there a chain of subgroups H ≤ ... ≤ G with prime indices?

[swift root]

if the extension is Galois then this is related to asking if the Galois group has a subgroup for each divisior

[somber goblet]

oh true

[rocky cloak]

I guess the smallest example would be that A4 doesn't have any subgroup of order 6

[balmy python]

what is good about the CRT for rings?

[cursive spindle]

One of the many ways to decompose your ring into simple ones

[balmy python]

whats useful about this though?

[cursive spindle]

???

Don't you see the usefulness in this?

[next obsidian]

Don’t worry about it and you’ll see it get used at some point

But as an example say you want to count how many numbers are invertible mod n

Write n = p_1^n_1 • …•p_k^n_k and then CRT gives you that
Z/nZ ≈ Z/p_1^n_1 x … x Z/p_k^n_k

And so
(Z/nZ)* ≈ (Z/p_1^n_1)* x … x (Z/p_k^n_k)*

[long nebula]

The CRT is great because it lets you solve math contest problems :^)

[next obsidian]

So the number of units mod n is
(Number of units mod p_1^n_1)•(number of units mod p_2^n_2)•…•(number of units mod p_k^n_k)

So all you need to do is compute the number of units mod a prime power

And this is very simple

[long nebula]

Chill man

[verbal valley]

me when rational root theorem

[cursive spindle]

Didn't mean it in a bad way. I was curious what's not useful about this

[cursive spindle]

Here are another two examples to keep in mind on top of what Chmonkey said.
If you consider the ring Z[i], one thing you can ask is how does this ring behave relative to the usual ring Z we all know and love. For example it turns out that not all primes p in Z stay prime in Z[i]. One example of such is p=5. CRT gives you a way to see this. Take Z[i]/(5) = F_5[x]/(x^2+1) but then x^2+1=0 mod 5 has 2 distinct solutions i.e x=2,3 so then we have F_5[x]/((x-3)(x-2)). Notice that the factors we have there are comaximal so CRT applies here and we get Z[i]/(5) = F_5[x]/(x-3) x F_5[x]/(x-2) = F_5 x F_5. Now what this is saying is that 5 splits into two distinct prime ideals in Z[i] and sure enough you can check 5 = (2-i)(2+i). As a good exercise I recommend doing this in general for arbitrary p and see what you notice.

The other one is more geometrical. (well really the former one is the same sort of behavior but whatever). Geometrically CRT tells you if "geometric sets" are disjoint. For example take A= R[x,y]/(y(y-1)). Again (y) and (y-1) are comaximal so CRT applies here and you get that this ring is isomorphic to R[x,y]/(y) x Rx,y. Simplifying futher you get A = k[x] x k[x]. This is saying that the ring A is describing 2 disjoint geometric sets i.e we have the two lines y=0 and y=1. (Btw if you're curious this sort of ring is called the coordinate ring)

[tardy hedge]

For grading polynomial rings (in Z^n lets say), are we allowed to set deg(xi) to anything we want, and it still creates a valid grading?

[elfin wraith]

Yeah, this happens a lot in AG contexts, for reasons I should really be able to explain better

(Weighted projective space is useful in studying 4-folds and I forget exactly why)

[tardy hedge]

the grading stuff seems weirdly powerful

I guess it makes sense cause we use degrees to argue stuff about polynomials all the time

[swift root]

yee

[tough raven]

Grading = group action so yes

[next obsidian]

You can use weird gradings so that ideals become homogeneous under that grading and then apply graded algebra methods

[somber goblet]

oooh that's cool

[tardy hedge]

Can I show that if M and N are S-1A-modules then M(x)N over S-1A and M(x)N over A are isomorphic by showing every bilinear map M x N -> L factors through M x N -> M(x)N over S-1A?

[somber goblet]

wait do graded algebras induce a group algebra structure

[swift root]

no

[somber goblet]

does a ring graded by monoid $M$ have a natural embedding into a $M$-ring $R[M]$

[cloud walrus]

lexi

[swift root]

no

[somber goblet]

[rocky cloak]

Sure

[swift root]

the homogeneous components may be more than 1-dimensional

every ring is trivially graded by 1 for example

[somber goblet]

every ring R is a monoid ring for the trivial monoid over R

i don’t mean over k i mean over some ring

[swift root]

still, if the component at the identity is very small compared to the other components then im not sure you can embed it into R[M] for some R

[velvet hull]

there is a natural projection onto the components

but its not necessarily an embedding because of what enpeace said

[somber goblet]

ohh

[swift root]

not a ring map

[velvet hull]

yes thats why i said projection

[swift root]

oh lol mb

[swift root]

why do you trust HChan not me 💔

[swift root]

maybe you can find a large enough R but at that point you loose any structure of your original graded ring so its kinda useless

[somber goblet]

oh yeah hmm

[swift root]

point is you have the grading structure because its much more flexible and general than monoid algebras

[somber goblet]

yeah ig

[swift root]

its like going from monoids to algebras

[somber goblet]

oh but every monoid algebra is graded true

isnt’t it

[rocky cloak]

I mean if R is your graded ring and you consider R[M] (with a grading from M, not related to the grading of R).

Then you should be able to send
r to r * deg(r)

[somber goblet]

wait this is kinda interesting ngl

it’s a free way to get M-graded algebras

[swift root]

well, its the trivial case

boring

[somber goblet]

[swift root]

oh true

[somber goblet]

ℝ-graded algebra

[rocky cloak]

So they're more cofree than free I guess

[chilly radish]

Yea if you show they satisfy the same universal property then you win