groups-rings-fields

[next obsidian]

Yea

This is much stronger than just being closed under multiplication

[solar glacier]

so they sort of act like normal subgroups in a sense?

[next obsidian]

Because it absorbs stuff from outside of it

Well, they’re exactly the things you can quotient by so in that sense yes

[solar glacier]

cause u can mod out by them

^^

[next obsidian]

But this absorptive property is entirely unrelated

[solar glacier]

so question

does Z[i]/(i) equal the trivial ring

since (i) has a unit namely i itself as (i)(-i)=1

[rotund aurora]

(1) is the whole ring

so you are killing everything

[solar glacier]

but isnt i a unit as well

[next obsidian]

No

Err

Yes

Lmfao

[rotund aurora]

(i)=(1) I meant

[solar glacier]

so (i) IS the whole ring of Gaussian integers ? since it contains a unit

[next obsidian]

Yes

[rotund aurora]

btw, how do you guys look at ideals? I just think of them as a byproduct of wanting to add extra relations on rings (quotiening out on rings). Or perhaps as submodules of the R-module R, where R is your ring. Is this the right way to think about them?

[solar glacier]

gosh theres so many definiitions in ring theory lol

[next obsidian]

Idk they’re ideals

[chilly ocean]

i think of ideals as the things you can quotient by

[next obsidian]

They’re just things

You do stuff with them

[solar glacier]

^^

[solar glacier]

lol

[chilly ocean]

i'm stuff 😳

[next obsidian]

I mean in actuality I think of them most naturally as closed subspaces

Because I do AG

Lmao

[solar glacier]

alg geo?

[chilly ocean]

closed subspaces are exactly what you quotient by in FA...

hmm

makes u think

[rotund aurora]

FA?

[next obsidian]

Yeah so in AG it’s actually the quotient which is the closed supsacr

[south patrol]

do i ask multilinear algebra here or in lin alg lol

[next obsidian]

Spec A/I inside Spec A

Lmfao

[chilly ocean]

functional analysis

[next obsidian]

Here

[south patrol]

probably here

lol just wondering like is there a nice way to look at / describe the exterior algebra of a tensor product of (say, f.d.) spaces

with direct sums it behaves very nicely

but i'd not be surprised if it becomes a bit of a mess here

[solar glacier]

i havent seen much diff forms so

[chilly ocean]

you mean like \wedge (V \otimes W)?

or whatever the notation is

[south patrol]

Yup

[chilly ocean]

it's going to be a quotient of the tensor algebra of V \otimes W, right? can you write that out and simplify from there maybe?

seems messy but it's what i'd do first

[next obsidian]

It commutes

With tensor

[south patrol]

Hm it can't quite commute right

[next obsidian]

Well I guess it depends on what you mean by tensor product

[south patrol]

Tensor products of exterior power is exterior power of direct sum

[next obsidian]

Are you tensoring stuff over a fixed thing?

[south patrol]

OK so I mean vector spaces over some field F and tensoring over F

[next obsidian]

Let me say what I did mean

If you take exterior powers of M an A-module then base change to B by tensoring with B over A

This is the same as base changing to B then taking exterior powers

At least for free modules

I think in general?

Anyway like uhhh

[south patrol]

Oh okay hm

Okay yes sure that's cool but yeah I guess that's a diff situation

But good to know lol wasn't aware

Tbh if there is no nice formula or whatever it isn't the end of the world for the computation I'm doing

[pastel cliff]

what do you call a ring without eyes? a non-unital ring

proud of my math pun (please laugh)

[next obsidian]

I don’t get it

[south patrol]

I don't get it either

Oh like rng is ring with an i

what do you call the complex numbers without eyes

the complex numbers because numbers don't have vision

[chilly ocean]

more like without eye cuz rings are cyclopes

[knotty frigate]

bruh

[glossy crag]

Quick double-check, the division theorem works in R[x] (i.e. existence and uniqueness) for R an integral domain when dividing by a monic, just like for Z[x], right?

[rotund aurora]

yes, but idk what happens if R is noncommutative

it works whenever the leading coefficient is a unit

But polynomial division also works for non-integral domains right?

[lapis trail]

I got this question. What are the number of conjugacy classes of homomorphisms from Klein 4 to GL_2(C)?

[south patrol]

So like number of dimension 2 reps of K4 up to iso

Probably time for a character table right since dimension is only 2

[lapis trail]

Ok thank you

I didn't know where to start

[south patrol]

But then all irreps are 1 dim cause abelian

So it simplifies a lot

[sly crescent]

Commutator magmas

[solar glacier]

I have a math pick up line now that we’re in topic of@jokes

I wish I was a derivative so I could lay tangent to your curves lol

[chilly ocean]

Is compactness equivalent to sequential compactness to you? Because you look like an angel

[solar glacier]

I don’t get it lol

[chilly ocean]

Only some people will get it

[solar glacier]

Hmm 🤔 🧐 🤨

[chilly ocean]

and they'll probably be graduate students

[solar glacier]

Cause I’m metric spaces

They’re the same

In

But I’m a graduate student lol I should get this

The word angel has something to do w metric spaces I assume?

[chilly ocean]

I won't tell you. That's the point. Maybe I made it up

[pastel cliff]

👼

back to algebra - are there non noetherian euclidean domains?

me thinks not but perhaps google/my brain have missed something

[rotund aurora]

Euclidean domains are PIDs

So no

Euclidean domains are the nicest thing in the world

Literally Z

[ripe basalt]

I have a group |G| = 24 with a non-normal group |H| = 8 and i have to find: (1) how many of conjugates there are of H (2) homomorphism from G into S_3 and (3) Showing G is not simple

for (1) I just said there are 3, since there are 3 left cosets when G acts on H. eH, g_1 H, g_2 H and since H isn't normal then g_1 H gives g_1 H g_1^-1 as a conjugate of H

and so on

is that right?

i guess i could just use sylow's theorems..

but (2) sort of implies that this should be a group action view

oh yeah sylow's theorems give (1) and (3) actually

so could anyone help me with (2) and (3) if there's a way to do (3) with the class eqn?

[pastel cliff]

so what's a PID that isn't an ED

[rotund aurora]

There must be but I cant think of examples

[pastel cliff]

http://www.m-hikari.com/imf/imf-2013/29-32-2013/wongIMF29-32-2013.pdf

[rotund aurora]

There are probably examples coming from quadratic rings of Z

[pastel cliff]

this feels so contrived im gonna assume it's not terribly relevant at my level lol

[rotund aurora]

https://math.stackexchange.com/questions/857971/ring-of-integers-is-a-pid-but-not-a-euclidean-domain

I dont think this is pathological btw

But idk. PIDs are rare in number fields anyway I think

[ripe basalt]

is there a channel im supposed to ask my question

[rotund aurora]

This one is fine for algebra questions

[ripe basalt]

ok well i got (1) with Sylow's theorems. I just want help understanding (2) and if there's a class eqn solution to (3)

[rotund aurora]

I dont think I can help you. If you are patient, probably someone will help you

[ripe basalt]

ok

[pastel cliff]

ping me in like 2 hours if im awake i'll try

it'll help me study for my final

[ripe basalt]

hopefully ill have moved on from this problem in 2 hrs

i am also studying for a final

[lapis trail]

I got my algebra final tomorrow

[ripe basalt]

ok i got it

[pastel cliff]

mine on thursday

best of luck soldiers

[lapis trail]

😫😫😫

I would take it now if I could just so it wouldn't be looming

I think I've studied as much as I can

@ripe basalt do you need just one homomorphism, or do you need to count how many there are?

[ripe basalt]

I just need 1 but i figured it out that part

[lapis trail]

Was it the trivial one?

[ripe basalt]

what do you mean

[lapis trail]

Map everything in G to ()

[ripe basalt]

it's the one that sends g to gHg^-1

?

to ()?

[lapis trail]

Identity

[ripe basalt]

oh no

that wouldnt be useful

[lapis trail]

No

[ripe basalt]

it just sends g to its conjugacy class pretty much

im trying to figure out (3) actually

concluding that G can't be simple

and im trying to say the kernel is nontrivial pretty much

[next obsidian]

What’s the size of the image?

Or at least an upper bound?

Then you can see the map can’t have a trivial kernel

[ripe basalt]

the image of the action?

it's just 3

yeah that gives a nontrivial kernel by first iso

although the image should also be a group

and I guess it is, it just inherits the group law from G

[next obsidian]

This doesn’t make any sense

You act on a set of size 3 and this gives you a map into S_3

The group law is just the group law for S_3

[lapis trail]

I got a question becauseim learning this too. Can we use sylow? Because I think we have 4 sylow 3 subgroups and we only know they're normal if they're unique

[ripe basalt]

ah @next obsidianso the image of this homomorphism is non-trivial and it's upper bounded by 6 🙂

so the kernel is nontrivial

cool

the size of the image is upper bounded by 6*

[next obsidian]

This is a Sylow 2, the number of which has to be 1,3,5,7,9,11,13,15,17,19,21, or 23 by one of the Sylows

It also has to divide 24

The only option you have is 3

So there’s 3 conjugacy classes

[ripe basalt]

sylow counting for (3) is the worst way to do it IMO

this way is much better

but yes i used sylow counting for (1)

you couldve had 1 2-Sylow group, but H is non-normal by hypothesis

[south patrol]

G is not simple because only two primes divide its order

Jk

[next obsidian]

Very true

This post approved by Burnside

[south patrol]

Thank

[green locust]

anyone willing to give a hint as to how to start this? i saw an answer on stack exchange that used the universal mapping properties of product/coproduct but it didn't make much sense

[main needle]

Is this true? Any ideas how to prove it?

[pastel cliff]

that's a weird way to state that question

but cayley's theorem no?

could be wrong

[chilly ocean]

cayleys theorem just says that every group can be viewed as the subgroup of a symmetric group? i think this is more subtle

[toxic zephyr]

are dihedral groups solvable?

i've never really learned what a solvable group is

[next obsidian]

Dihedral groups are a semi direct product of C2 and Cn

So they have a three-term abelian series

So they’re solvable

I guess it’s actually two term

[main needle]

I feel like it's not correct but I have no counter example.

[toxic zephyr]

okay cool thank you. i don't really understand that but i sort of get the gist 😆

my professor also mentioned that jordan canonical form can be derived directly from modules. does anyone have some reading i could do on that?

[chilly ocean]

d&f goes into it quite a bit. i mentioned how it goes in a post somewhere, let me find it

[formal ermine]

I think if G acts transitively on {1, ..., n} then you can use orbit stabilizer

[next obsidian]

That says nothing about minimality

But actually this does give me an idea

[main needle]

thinking that If n is minimal then the action is transitive ?

[little root]

^^ that’s pretty smart

[main needle]

not sure it's true tho. But if it's true then we would be done by Orbit-Stabilizer indeed

[next obsidian]

At the very least you know it can’t fix anything

[little root]

Oh wait minimality might not nvm I was being dumb

[next obsidian]

Because if so you embed into a smaller S_n

Namely the subgroup which fixes any fixed number

So like, you get that it embeds into a product of S_ni where Sum ni = n and ni > 1

This is basically just looking at how it acts on the orbits and collecting these together

But idk if this gets you much

[main needle]

The statement might not be true to begin with but I couldn't find a counter example

[next obsidian]

Yeah I tried thinking of some possible examples, didn’t get that far lol

Wait there’s a super obvious example

Wait is there

Uhhhhhhhh

Yes there is

I don’t wanna spoil but it’s false

[main needle]

Lmao

[next obsidian]

Okay maybe not “super obvious”

But like, I was thinking of exactly this example

Then went “ah it doesn’t work”

[main needle]

I tried Z_p^m, Klein-4 didn't work

[next obsidian]

Then realized it does

[main needle]

is it quaternion ?

[next obsidian]

Nah

Is this hw?

[main needle]

Nah practicing for final

[next obsidian]

Ah

Consider Z_6

You want to find an order 6 element

It clearly doesn’t exist in S_3, so the next option is S_6

But you can find an order 6 element in S_5

Something like (123)(45)

[main needle]

Oh yeah I see

I was thinking of cyclic groups too but for some reason I was only thinking of prime power order

[next obsidian]

Yeah so

That’ll never work

By how divisibikity works

Like, if there’s an element of order p^n in S_m

Then p | m

Cuz necessarily then there’s an element of order p

[main needle]

Yup I see

[next obsidian]

Or we’ll

I’m not TOTALLY sure anymore

But it seems probably impossible

So I went to composite numbers which aren’t prime power

Which led me to 6

Then I was thinking about S_4 and realized it doesn’t work

So I abandoned the idea

Then I was like “wait bruh wtf I’m stupid”

[main needle]

Ah I should've seen that but it's already 12 Am and I want to sleep lmao

[pastel cliff]

,ti

[cloud walrus]

The current time for stμ₂dying is 02:04 AM (EST) on Wed, 07/12/2022.

[pastel cliff]

job isn't finished

[pastel cliff]

does eisenstein's criterion apply to Z[sqrt{-5}]

i want to show that (1 + \sqrt{-5}) is irreducible

[next obsidian]

Do this using the usual norm for complex numbers

And some basic estimates

[pastel cliff]

found this example but (1 + \sqrt{-5})(1 - \sqrt{-5}) = 6

[next obsidian]

What are you saying

The norm of 1 + sqrt(-5) is 6

And the norm takes integer values

This restricts things immensely then it’s just an easy analysis of possible norms of elements in the ring

[pastel cliff]

yeah this is my sleepy time cue then

,ti

[cloud walrus]

The current time for stμ₂dying is 03:10 AM (EST) on Wed, 07/12/2022.

[pastel cliff]

i dont think this is in any of my notes, what's the relationship between norm and irreducibility

[next obsidian]

If you can factor the thing you can factor the norm

[rotund aurora]

eisenstein works over any domain

[next obsidian]

Once you observe that N(x) = 1 implies x is a unit

You have to show that you can’t write 1 + sqrt(-5) as the product of things of norm 2 and 3

[shell agate]

Same argument as before btw

[pastel cliff]

:O forgor about that thank you

ok i actually understood that^ now

im obliged to still be awake so i'll ask for one more hint

just a pointer on this would be cool and good

[chilly ocean]

do it in Z and Q

always the example to have in mind!

[rotund aurora]

this is like the proof that sqrt 2 is irrational

[pastel cliff]

i feel like this is close to a contradiction

can i say that d = x/y contradicts the fact that it's in a UFD

ehhh probably not nvm bc im pretty sure you can arrive at x = 1/y

[next obsidian]

This is not true because x doesn’t live in D

[pastel cliff]

im seeing why this makes sense though

[chilly ocean]

Can anyone help?

[hot lake]

Z/5Z is a counter example

[shell agate]

no, there's 0 non-identity element in Z/5Z, and 0 is a multiple of 5

[warm wyvern]

?

The identity in a group is unique

[shell agate]

Sorry, I meant satisfying this equation

[warm wyvern]

Amd there are 5 elements in the group

[shell agate]

wait I mixed up

[warm wyvern]

They all do

[shell agate]

you're right

[warm wyvern]

[warm wyvern]

Maybe they meant a multiple of 4?

Coz i think that'd be true

[shell agate]

Isn't it also true that if you include the identity you get a multiple of 5?

(that's just Lagrange aint it)

[warm wyvern]

No coz it didn't specify abelian group

So the set all such elements don't necessarily make up a subgroup

[shell agate]

Do you have an example where it fails?

What are elements in S5 of order 5? Exactly the 5-cycles, right?

[warm wyvern]

I can't come up with an example tbh, sry

[shell agate]

me neither

[warm wyvern]

Prolly D_5 would work or smth

[shell agate]

D_5 has 4 elements of order 5 (so 5 with the identity)

[shell agate]

Where's this from ?

[untold basin]

Hello, I asked that question 2 days ago, but I realized now that I haven't used Galois Theory to show that X⁶+X³+1 is the minimal polynomial on zeta on Q
#groups-rings-fields message

I used the fact that
Q[X] -> Q[X] : P(X) |-> P(X+1) is an automorphism so P(X) is irreducible because P(X+1) is, but it is difficult in term of calculations

Does anybody know an argument using Galois Theory or am I searching for nothing ?

[shell agate]

is it easier to show that P(X+1) is irreducible than to directly show that P is?

And no I think you need to prove minimality before you can do any Galois theory

,wolf factor X^6+X^3+1

[cloud walrus]

Results provided by WolframAlpha

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

Requested by Blaxapate#3807

[untold basin]

Yes, with criteria of E

[shell agate]

Oh yeah, it's immediate

nice

[slender hamlet]

Not true. For G=Z/6Z, n=5

Ok I should have read the chats a bit, someone already posted that

[chilly ocean]

Gallian abstract algebra

[restive birch]

how would one go about showing that the preimage of a normal subgroup under a homomorphism is itself a normal subgroup?

[formal ermine]

show the following things: \

• $\varphi\inv(N) \neq \emptyset$ \
• for any $g_1, g_2 \in \varphi\inv(N): g_1g_2 \in \varphi\inv(N)$ \
• for any $g \in G, x \in \varphi\inv(N): gxg\inv \in \varphi\inv(N)$

[cloud walrus]

i believe in mathemagic

[warm wyvern]

Sounds like it's related to the 3rd iso theorem

Lemme think

@elder wave why sully? Lel

[chilly ocean]

You also have to show that if g is in the inverse image then so is g^-1

[shell agate]

It's indeed multiples of 4 in the 2021 edition of the book

[simple mulch]

I am currently having an issue with proving the commutator subgroup is normal in G

Let $g \in G$ and $[g_1,g_2] \in G^*$ (the commutator subgroup)

[cloud walrus]

mns

[simple mulch]

Then we have $gg_1^{-1}g_2^{-1}g_1g_2g^{-1}$

[cloud walrus]

mns

[simple mulch]

oh

I see what to do

[rustic crown]

also in general notice that this construction of commutator subgroup is natural!
If f : G --> H is a homomorphism of groups, then f([a, b]) = f(aba'b') = f(a)f(b)f(a)'f(b)' = [f(a), f(b)]
in other words, image of a commutator is again another commutator. and the restriction of f to G' = [G, G] defines a map f' : G' --> H'

you get what you need if you take f:G --> G, defined by f(a) = gag'

we would just get that the automorphism f restricts to an automorphism of G' = [G, G]

[rotund aurora]

So sure, the dimension is multiplicative over L and over \otimes, but I was wondering, given f in L(U\otimes V,U'\otimes V') how do we write it as a tensor (i.e., as a sum of pure tensors as described in the theorem)?

I cannot think of a simple way to do this, so its a bit confusing

wait I think I can just look at bases

I think I was just over complicating things

[nocturne bone]

I don't understand why this result shows that this is indeed the group action (for c)

[rustic crown]

yea you have to look at bases in such things as it being an isomorphism in general is not true

[rotund aurora]

the key is

[rustic crown]

another way to do this is by realizing that there is a nice isomorphism (under finiteness conditions) U* ⊗ V --> L(U, V) which sends f⊗v to the linear map ( u --> f(u) * v )

then reducing it to properties of tensor and duals

[rustic crown]

ig they're leaving the verification of it being a group action to you. what they did show is that the given action is transitive, so given two elements y = gx, z = hx in Gx you can find something in G such that (??) * y = z

[rotund aurora]

The key idea is that if you have two vector spaces $U$ and $V$ with bases $u_1,u_2,\dots$ and $v_1,v_2,\dots$ respectively, then the linear transformations $f_{ik}\colon U\to V$ defined by $f_{ik}(u_j)=\delta_{ij} v_k$ form a basis for $\mathcal L(U,V)$, where $\delta_{ij}$ is the Kronecker delta

no?

[cloud walrus]

Croqueta

[rustic crown]

yee, makes sense

[rotund aurora]

Then the vector space on the right is of this type (I mean L(something,something)), so when all the vector spaces are finite, you have a basis of that type (except you will have to keep track of many indices if you write it in terms of the basis u_i\otimes v_j lmao). Then each of these linear transformations of the basis come from tensoring two analogous linear transformations for bases of L(U,U') and L(V,V'), so you if you choose a basis, you can write all of this

[rustic crown]

and image of this basis would also be a nice basis on the other side

(i didn't read the indices btw :p)

usually i just think them as matrices in this case as one can concretely just say matrix with 1 only at the (i,j)th entry or something

[rustic crown]

oh i was thinking you're gonna take basis of both L(U, U') and L(V, V'), then the pair-wise tensors will give you a basis for their bigg tensor product

[lapis trail]

Is there a way to count the conjugacy classes of homomorphisms from Sn to An? I don't know where to begin

[hot lake]

think about their kernels I guess ?

ah wait idk what are the normal subgroups

is there anything else than An ?

[lapis trail]

Specifically this question is S5 to A5, S5 has only A5 as the proper normal subgroup.. and A5 is simple

[hot lake]

yeah

[lapis trail]

That's why it's weird. I thought it might be the number of conjugacy of A5

[hot lake]

so not many choices for the kernel

[lapis trail]

*conjugacy classes

[hot lake]

not exactly no

[south patrol]

Okay good it's A5 cause lol ouchies for bigger stuff ig

[lapis trail]

This prof loves "count the homomorphisms " question but this is the first time I've seen one like this

[hot lake]

I'm not sure what's troubling you

[lapis trail]

They're good questions I'm learning a lot

[south patrol]

Not too sure why they added in the conjugacy classes right

can't you just explicitly write down the number of maps

[hot lake]

it's a pain to count the exact number without the "up to conjugacy" lol

[south patrol]

Okay true yes here

I guess it boils down to finding the number of elements of A_n of order 2 but yes

[hot lake]

yeah

[lapis trail]

Oh I think I see why

[rotund aurora]

I think thats what I wanted to say

I think this is accurate, tho you dont have to read haha

[green locust]

Anyone?

[rotund aurora]

Where is this from?

[green locust]

Aluffi’s algebra chapter 0

Chapter VI

[next obsidian]

Why is it true when V = k

So why are maps k^2 -> k^2 just an element of k^4 interpreted in a suitable fashion

Think about what linear maps are for vector spaces, and how you arrive at this description

[rotund aurora]

If V and W are vector spaces, then in general dim(L(V,W))=dim(V)dim(W) does not (never?) hold when V and W are not finite right? So Im thinking, let V have countable dimension over R and consider L(V). Then L(V)=R^N, but R^N over R has uncountable dimension (N is the naturals)

[green locust]

I forgot R is not just an arbitrary ring, it is specifically End(V). Makes a lot more sense now, thanks!

[next obsidian]

Bruhhhhhhhhhhh

[south patrol]

Yeah as a special case the dimension of the dual of an infinite dimensional space is strictly larger than that of the original space

[next obsidian]

Your mother is a… with a bucket of… hippopotamus….

Damn are you even old enough to know what I’m talking about

[south patrol]

i haven't heard of it rip

[next obsidian]

The elder swear

[south patrol]

lol

nice

[rotund aurora]

Oh yeah right

[formal ermine]

why?

[rotund aurora]

[formal ermine]

ah wait I got it I think. we have deg(minimal polynomial) = [R[a] : R] so [R[a] : R] = 2 and because a is algebraic it's [R(a) : R] = 2 <=> R(a) = R x R = C

is that correct?

[coral shale]

construct the isomorphism explicitly and show it is one

[formal ermine]

wdym

[coral shale]

im assuming a is any root of a deg 2 minimal polynomial?

[formal ermine]

uhhh so K/R is a non-trivial finite field extension

[coral shale]

Then yh, you need to show R[a] === R(a) === C

[formal ermine]

a is in K but not in R

[coral shale]

the first step, what you've written, then the 2nd step, I would write down the explicit iso (sorry if I missed your Q)

[formal ermine]

we constructor its minimal polynomial

which is in R[x]

so it must be of degree 2

then $\operatorname{deg}(\text{minimal polynomial}) = [\bR[a] : \bR]$ so $[\bR[a] : \bR] = 2$ but because $K/\bR$ is algebraic we have $[\bR(a) : \bR] = 2 \iff \bR(a) \cong \bR \oplus \bR \cong \bC$

[cloud walrus]

i believe in mathemagic

[next obsidian]

C is not isomorphic do R^2 if you’re asking for this iso as a field

The latter isn’t even an integral domain

[coral shale]

well in particular, because a is algebraic, R[a] === R(a)

as for showing R(a) === C, yh i would write down the explicit isomorphism

[formal ermine]

I don't get what you mean

sorry

[coral shale]

so let f be the isomorphism say

f : R(a) -> C

ok?

I'm saying I'd write down explicitly what exactly f does

[formal ermine]

huh?

[coral shale]

to elements of R(a)

And yeah, R + R is a ring, not a field

[formal ermine]

I'm ultra mega confused

[next obsidian]

Multiplication in C is not the same as that in R^2

In R^2 you multiply component-wise

In C it’s twisted, you have (a,b)•(c,d) = (ac -bd,ad + bc)

[formal ermine]

in my lecture we said that if for a field extension $\bL/\bK$ we have $[\bL : \bK] = n < \infty$ then $\bL \cong \underbrace{\bK \oplus \ldots \oplus \bK}_{n ~\text{times}}$

[cloud walrus]

i believe in mathemagic

[next obsidian]

C is only isomorphic to R^2 as a vector space

[formal ermine]

yes that's what I meant

[next obsidian]

Then you can’t go through R^2

If you’re trying to show R[a] is isomorphic to C as a field

[formal ermine]

yes

wait

I'm really confused

wait what

[coral shale]

this is vector space isomorphism right? this statement

(been a while)

[formal ermine]

yes I believe so

idk linear algebra that well

[L : K] is the vector space dimension dim_K(L)

[next obsidian]

You can’t say that R(a) ≈ R^2 ≈ C and then conclude R(a) ≈ C as a field

[formal ermine]

then what did det mean yesterday

[next obsidian]

Because you are using an isomorphism to R^2 which only exists as a vector space

[formal ermine]

#groups-rings-fields message

[coral shale]

R(a) === C as fields

Is the claim

[formal ermine]

I was trying to understand that

[coral shale]

Your method of proof using vector space isomorphisms does not show this

i was suggesting to construct the field isomorphism explicitly and show it is one

im not sure how to expand on this hint since idk what part exactly u r confused about

[solar glacier]

I have some questions regarding the proof of P is a prime ideal in ring R iff R/P is an integral domain.

So for the backwards direction I assume P is not prime then there exists a,b such that ab \in P but a,b \not\in P

then computing (a+P)(b+P)=ab+P=0+P \in R/P thus a+P and b+P are zero divisors and thus R/P cannot be an integral domain

does this work for backwards

[shell agate]

yes

[solar glacier]

and going forward I assume R/P is not an integral domain then there exists zero divisors a+P,b+P \in R/P

does this imply a \not\in P and b \not\in P

[shell agate]

Well, a is in P iff a+P=0, by definition

so if you take non-zero zero-divisors...

[solar glacier]

so if its nonzero then a \not\in P

[shell agate]

yes

[solar glacier]

then I can compute (a+P)(b+P)=ab+P=0+P as a+P,b+P are zero divisors forcing ab \in P

but a,b \not \in P thus P cannot be prime

[shell agate]

exactly

[formal ermine]

why [K : R] = 2 immediately shows that K = C

[solar glacier]

thx!

[coral shale]

it does not immediately show this

?

[south patrol]

uh this is relatively deep right lol

well

[coral shale]

not sure what u mean by 'immediate', there is one more step

[formal ermine]

#groups-rings-fields message

[south patrol]

no i messed up smh

[coral shale]

But it is true all degree extensions of R are isomorphic to C

[formal ermine]

what did det mean here then

[coral shale]

he skipped that explicit step im suggesting

which is to show that isomorphism explicitly

[formal ermine]

that's what I'm trying to show in the first place

[coral shale]

R[a] === R(a) youre happy with?

thats a result

[formal ermine]

yes because a is algebraic

[coral shale]

ok

next R(a) === C

you need to show this

[south patrol]

oh i need to review the proof from galois theory that C is alg cloed lol

[rotund aurora]

wait

galois theory proves that?

[coral shale]

To show 2 fields are isomorphic, the most straightforward way is to construct the isomorphism and show it is one

(a bijective homomorphism)

[coral shale]

perhaps itll clear some confusion up if I point out this isnt an isomorphism. This is an actual equality

[formal ermine]

yes I know

[coral shale]

R[a], the ring, is a field if a is algebraic

[formal ermine]

I know

[coral shale]

ok

[burnt flower]

isn't that just the fundamental theorem of algebra

C is the algebraic closure of R

therefore C is algebraically closed

[next obsidian]

That’s… very circular

[burnt flower]

I might be wrong it's been a while

[woven obsidian]

You can prove it using galois theory and a minimal amount of analysis

[next obsidian]

Also that isn’t the Galois theory proof

[rotund aurora]

its correct lmao but thats not the difficult part

I have seen the typical proof of looking at p(z) with z small radius and z big radius, z running through the complex circle of a given radius

[woven obsidian]

That's the complex analysis proof

You can prove it using only the intermediate value theorem from analysis

And galois theory

[rotund aurora]

But people usually say that there's no "truly" algebraic proof that C is alg closed, because how you construct R is unnatural (from an algebraic point of view)

[woven obsidian]

Yeah you need IVT

[rotund aurora]

I c

I will check that out, thanks

[woven obsidian]

I find the complex analysis proof much nicer though, but I guess that's a matter of taste

[rotund aurora]

yeah, definitely its one of my favourite proofs

[molten viper]

Hey folks, I’ve got a group and a non-normal subgroup, I’ve got to show the quotient group isnt well defined.

So the way I think I have to do this is find 2 elements which are equal that produce different cosets yeah?

[woven obsidian]

The cosets are well-defined

You can define the collection of cosets for any subgroup

[molten viper]

Specifically, I’m doing D4 with the subgroup {e,s}

Ah so I show that G/H isn’t a group?

[woven obsidian]

The multiplication won't be well-defined

[molten viper]

I seeeee

So I have to find 2 cosets which don’t have a well defined product

[woven obsidian]

So find elements where a + H = c+H, b + H = d + H, where ac + H \neq bd + H

Yeah

[next obsidian]

Alternatively, and in my opinion easier, show if the operation is well-defined then the subgroup is normal

[molten viper]

Yeah but that’s not the assigned problem

We’ve actually proven that I believe

[next obsidian]

It’s the contrapositice of the assignment…

[molten viper]

It’s definitely not

[next obsidian]

Yes it is…

[woven obsidian]

It's the contrapositive of a generalization of the assignment

[next obsidian]

The assignment is to show
non-normal => operation isn’t well-defined. The contrapositive is that well-defined operation => normal

I don’t see how this is anything other than directly the contrapositive

[woven obsidian]

Apparently the assignment was only for the specific case of D_4

[next obsidian]

Oh, well I read the first part which set it up in full generality

[woven obsidian]

Yeah it's a bit of a silly exercise, because it's easier to do it in generality

[molten viper]

The full problem was “show this isn’t normal, then give 2 cosets that illustrate the operation of G/H is not well defined”

We’ve literally proven the general case

Anywyas, I believe I’ve found my cosets so thanks for the help

[rotund aurora]

You can take the vector space V to L(V), where L(V) is the vector space of linear operators on V. Is this L a functor? (I have no idea what functors are)

[molten viper]

Isn’t that category theory

[rotund aurora]

no

because I dont know category theory

[south patrol]

I don't think there's an obvious action on morphisms

[rotund aurora]

even if you could define such an action artificially, functors are only interesting when that action is something natural, right?

[south patrol]

Uh they can be interesting depends

[coral shale]

iirc V -> LV unnatural but V -> LLV natural right?

[rotund aurora]

well I guess the condition on respecting composition is pretty strong

[south patrol]

That is dual u mean slurp right this is L(V)

[rotund aurora]

also, the surjection onto the dual is only when the dimension is finite

[coral shale]

wait i thought L is the dual

[proud bear]

L(V) is Hom(V,V)

[rotund aurora]

^

Dual is L(V,F)

[coral shale]

ah ic

[rotund aurora]

we shorten L(V) for L(V,V) usually

[rotund aurora]

where F is the field of scalars

[solar glacier]

question if an ideal is given as

[cloud walrus]

MyMathYourMath

MyMathYourMath

[coral shale]

( 2.2, 2.(1+r5), (1+r5)(1+r5) )

[solar glacier]

namely (2 \cdot 2, 2 \cdot 1+\sqrt{-5},(1+\sqrt{-5})^2)

ok thats what i thought thanks!

[tribal drum]

how to show ((Z/7Z), multiplication modulo 7) is a group?

[coral shale]

the axioms?

[tribal drum]

i trialed an errored to 3, found that 3^n generates the elements but is there another way other than trial error

[coral shale]

huh?

[solar glacier]

show associativity, existnece of inverses and identity

[tribal drum]

oh cyclic group

not just group

[coral shale]

first show it is a group

then find a generator

[tribal drum]

yh the generator bit is there some method to finding it

[coral shale]

whats wrong with trial and error when there are so few to try

[tribal drum]

i suppose there isnt

[coral shale]

you can try convincing yourself/proving

that gcd(k, n) = 1 iff

Z/nZ is generated by k

oh wait

[south patrol]

well should be Z/7Z without 0

[coral shale]

multiplication

sry thought it was addition

in which case no it isnt easy to find the generators, in general

[south patrol]

(Z/7Z)* or U(Z/7Z) or (Z/7Z)^x or whatever you prefer lol

[tribal drum]

ah fair enough

was just curious, thanks for clarifying

[coral shale]

this is the problem of finding

primitive roots

mod n

[south patrol]

and ys finding a generator is uh

well it's kinda interesting that primitive roots even exist

but once you've found one you've found them all (as is the case in any cyclic group)

[solar glacier]

can I check my solution to proving nonzero prime ideals in PIDs are maximal. Let P be a prime ideal generated by p and let I \subset R be any ideal such that P \subset I \subset R we must show either I = P or I = R.

since I is an ideal in a PID there is an m \in R such that I = (m)

if p \in I then p = rm for some r \in R

But P is prime so either r \in P or m \in P

if m \in P then I \subset P forcing P=I

if r \in P then r = sp for some s \in R but this becomes r=srm thus 1-sm = 0 and therefore m is a unit forcing I = R

[woven obsidian]

Looks correct

[solar glacier]

cool thanks ^

[rotund aurora]

Why did he change notation from bigoplus to bigotimes ?

[chilly ocean]

typo

[rotund aurora]

ah ok

yeah doesnt make any sense lol

thanks

[rustic crown]

.<

[formal ermine]

it's ok I got it c:

[rustic crown]

oh okie uwu

[chilly ocean]

[rustic crown]

[pastel cliff]

can someone eli5 semi-direct products to me pls

[chilly ocean]

yes

[rustic crown]

eli5?

[chilly ocean]

explain like i m 5

[pastel cliff]

explain like im 5

[rustic crown]

oooh

[chilly ocean]

do you know about group actions sebbb?

[rustic crown]

arigatou

[pastel cliff]

yeah i understand

[chilly ocean]

semidirect products are basically a way to encode a group action (of a group on another group) as a group

[lethal dune]

[chilly ocean]

So let's take a simple example

there's an action of C_2 on C_n given by sending n to -n, right?

[pastel cliff]

cyclic groups i assume?

[chilly ocean]

yes

[pastel cliff]

yeah sure

[ripe basalt]

D_n!

[chilly ocean]

so first we take the set C_n x C_2

and we want it to contain copies of both groups

those are C_n x 0 and 0 x C_2

now, in order to encode the action

we will have the first coordinate of (0,1) + (m,0) be -m

[pastel cliff]

this is C_2 acting on C_n right?

[chilly ocean]

yes

[pastel cliff]

ok just making sure i usually write em the other way around but doesnt make a differenc e

[chilly ocean]

C_n x C_2 is the cartesian product of both sets

[ripe basalt]

well hes going to write C_n x| C_2

idk how good that ascii notation is

[pastel cliff]

i dont wanna bog down with notation it's no biggie

[chilly ocean]

the action is given by $\phi_1(m) = -m$

[cloud walrus]

Uwu math girl

[ripe basalt]

isnt it important that that action is an automorphism on C_n

[chilly ocean]

that's what action means

an action of $A$ on $X$ is an homomorphism from $A$ to $\textrm{Aut} X$

[ripe basalt]

actions don't have to be injective do they?

[cloud walrus]

Uwu math girl

[chilly ocean]

no

[rustic crown]

functor from G as a category to Grp

[ripe basalt]

oh i see

[chilly ocean]

there are some subleties with thinking that way, because there are non-trivial natural transformations

anyway let's not get out of track with sebbb about this

[rustic crown]

right, but natural transformation would then be a map of G-objects right?

[formal ermine]

inner or outer?

[ripe basalt]

--continues to get off track--

[chilly ocean]

you still here sebbb?

[pastel cliff]

reading my own notes at the same time

[chilly ocean]

you can also think of them as a "twisted" direct product

in direct product you have $(x,y) + (x',y') = (x+x', y+y')$

[pastel cliff]

trying to see how what you're saying fits with the defn i have

[cloud walrus]

Uwu math girl

[chilly ocean]

in semidirect product the second coordinate of first summand will act on the first coordinate of the second

[coral shale]

noo i missed my chance

[pastel cliff]

it's in my notes as a way to get information about a group G given subgroups H, N with certain properties

[ripe basalt]

i love how my course covered semidirect products pretty in depth and then there was just not a single question about them on the exams..

[chilly ocean]

$(x,y)+(x',y') = (x+ \varphi_y x', y+y')$

[cloud walrus]

Uwu math girl

[chilly ocean]

as you can see, if the action is the identity for all y, then this is a direct product

[coral shale]

Did we talk about the bestest example with affine spaces 😵‍💫

[ripe basalt]

thats not a very rigorous definition now is it

[pastel cliff]

not a definition that's the motivation

[chilly ocean]

it can be a definition if you say the right properties

[ripe basalt]

ok well you said "its in my notes as..."

[pastel cliff]

fuckin uhhhhh

i remember this

[chilly ocean]

do you know about group extensions?

[ripe basalt]

uwu do you make the distinction of "different types of semidirect products"

[pastel cliff]

N is normal, HN = G, trivial intersection

[pastel cliff]

no

[ripe basalt]

i know d&f classify like 3 different types

sebbb just said the "internel view" i think

[chilly ocean]

you mean inner and outer?

[ripe basalt]

i mean like synthetic, internal, external i think?

[chilly ocean]

idk what synthetic is

[ripe basalt]

so in synethetic you take two groups and make a new group

[chilly ocean]

don't you do that in external too?

[ripe basalt]

internal i think is G = HN thing

i actually dont remember

[chilly ocean]

you will still have G = HN in external

[ripe basalt]

you will yes

[chilly ocean]

if you identify H with 0xH and N with Nx0

[pastel cliff]

fuck imma go read keith conrad and i'll be back in a bit

but thank you uwu

[ripe basalt]

but in internal you start off with identifying subgroups

[pastel cliff]

idk how i feel saying uwu

[ripe basalt]

and i dont think you need to talk about the homomorphism

[coral shale]

you're in the right channel

[pastel cliff]

[ripe basalt]

the internal view is like (a,b) (c,d) = (ac dbd^-1, bd) or something

since N is normal it works

err im sure i typoed something there

[coral shale]

chm remember when u explained semidirect to me

[next obsidian]

Yeah, this is why I kinda think internal/external distinction is kinda silly. It makes sense and i use them, but it’s more confusing than anything else for ppl learning it

[coral shale]

its stuck 💛

[next obsidian]

I did?

[coral shale]

yh

[next obsidian]

That’s good

[ripe basalt]

yeah i dont think i like the distinction

[next obsidian]

It’s a twisty direct product

uwu

[chilly ocean]

ikr

[ripe basalt]

its easiest to remember just the "external" and then the internal view is when your automorphism is conjugation

and you have subgroups etc

[next obsidian]

But that’s kinda not even the point

[coral shale]

internal and external direct products

[chilly ocean]

Also for ppl who know short exact sequences: semidirect products are equivalent to split short exact sequences of groups

[next obsidian]

Like that’s how you make it yes, but I think it’s more useful in that we know if you have the case where you have HN = G, H\cap N = 0, and N is normal but H isn’t that you can STILL recover G as built up from H and N

The way this is done is via some little twisty bit with conjugation

[ripe basalt]

thats the point of the distinction, sure

[next obsidian]

But it’s the fact that G can be built up out of its parts is what’s important when like classifying groups of small order for example

[ripe basalt]

hm yeah

[next obsidian]

Like the “can be realized by conjugation” is meaningless

Because if you have any abstract map into the automorphism group

Once you identify N and H with subgroups of N semi direct H

It is realized via conjugation inside N semidirect H

Like in a sense any construction of a semidirect product is via a conjugation

I dunno uwu

[ripe basalt]

chmonkey how do u pass galois theory final in 3 easy steps

[coral shale]

All this uwu. I can't 😆

[ripe basalt]

yeah tell him to change his name idk

[coral shale]

[ripe basalt]

we can just say math girl

[next obsidian]

Idk I suck at Galois theory

[ripe basalt]

oh good

[chilly ocean]

?

[next obsidian]

I’m doing a det

[coral shale]

you arrange a duel right after your final

[pastel cliff]

chmonkey how do u pass groups + rings final in 3 easy steps

[ripe basalt]

yeah i hate this course and everyone who i ask who is good at math just goes "idk i sucked at that"

[next obsidian]

Step 1: learn group theory
Step 2: learn ring theory
Step 3: Chmonkey

[coral shale]

and spend all night writing up your final comprehensions in it

[formal ermine]

chmonkey how do you pass groups + rings + modules + fields + galois final in 3 easy steps

[pastel cliff]

and it's as simple as that

[rustic crown]

what did i miss >.<

[coral shale]

[next obsidian]

The secret to getting good at Galois theory is to do algebraic number theory

[rustic crown]

hi chmonkey

[next obsidian]

This is my observation based on multiple people

Hi det

[ripe basalt]

to be fair it's basic

[formal ermine]

algebraic number theory more like x + 3 = 2

[ripe basalt]

intro galois

[next obsidian]

Yeah idk, just like compute examples

[ripe basalt]

ok actually let me rephrase

how do u pass a final on group theory

[pastel cliff]

[chilly ocean]

answer all the questions in a way that gives you enough marks

[chilly ocean]

Ah, a diophantine equation

[ripe basalt]

good strat

[next obsidian]

Honestly for these u should just do lots of problems

[pastel cliff]

classify all groups on your exam

[next obsidian]

Basic algebra is learned IMO by doing shit tons of exercises

Basic meaning like intro level stuff

[ripe basalt]

i just hate group theory man you do some new question and it's just like "apply random counting technique #5,738"

[coral shale]

i did, but that didn't help fuck

that stuff got me lost

its more like the other way round

[next obsidian]

That’s the point of doing lots of exercises

You discover the counting techniques

Gimme a second

[pastel cliff]

derive sylow thms for yourself

[ripe basalt]

it's never enough

id rather die

[unique valve]

just be smart

[ripe basalt]

idk the thing is the counting principle isnt obvious a priori

[pastel cliff]

homeless? buy a house!!

[ripe basalt]

so it feels like luck in a timed setting

[unique valve]

could not agree more

[ripe basalt]

like u just happen to think to count the right thing

[rustic crown]

[unique valve]

its about intuition

[next obsidian]

It becomes pretty intuitive

[unique valve]

which is developed through practice

[next obsidian]

https://cpb-us-e1.wpmucdn.com/sites.uw.edu/dist/9/16099/files/2021/12/WS7.pdf

Familiarize yourself with these counting methods at least

These will get you pretty far in most situations

[ripe basalt]

ok

[next obsidian]

You should also be trying to look for anything to act on which is small

You get maps into S_n for small n in that way and this affords you a lot

[pastel cliff]

this is most things on my final

[ripe basalt]

my final is tomorrow and i think it will be more galois than group

so im unsure

im actually ok at galois but bad at groups

[pastel cliff]

i think there will be groups on my final tm

maybe some rings

[rustic crown]

galois theory is uwu

[ripe basalt]

this course was groups and fields mostly

[chilly ocean]

rings btfo

[next obsidian]

FUCK YOU

[rustic crown]

whut is btfo?

[chilly ocean]

blown the fuck out

[rustic crown]

souka

[ripe basalt]

rings are just those things u mod by a maximal ideal to get a field

[pastel cliff]

fight fight fight

[chilly ocean]

weeb

[chilly ocean]

noncomm rings btfo

[pastel cliff]

who would win

terra or monke

[ripe basalt]

no no no

u are mistake

all rings are unital and commutative

thats the definition u see

and then u mod by a maximal ideal

[chilly ocean]

😬

[ripe basalt]

and u get a field

[coral shale]

meow meow

[ripe basalt]

hahaha

[rustic crown]

hewwo shuri

mrow

[pastel cliff]

uwu out

[next obsidian]

What does ur active developer badge mean

[coral shale]

means im an active developer

[next obsidian]

For what

[pastel cliff]

discord user

[coral shale]

or more accurately, a person who created a temporary community server, created a bot with 1 slash command and ran it there, then yeeted said server

[next obsidian]

Bruh

[ripe basalt]

nice

"10 years leadership experience" on this guys resume

it's okay my brother once told people he was doing machine learning because he wrote some javascript which sorted an array of integers

[rustic crown]

[pastel cliff]

yeah im working on the millenium problems

[coral shale]

theres no cat or algebra in those

oh wait, ive never actually come across hodge, maybe this counts

[glossy crag]

Hodge

[glossy crag]

isn't BSD also a MP?

https://en.wikipedia.org/wiki/Birch_and_Swinnerton-Dyer_conjecture

[coral shale]

yh these are the ones which were skipped in my super basic intro to them i read in hs for 'some reason'

[rotund aurora]

BSD is very close to what Andrew Wiles did (does ? idk what he is doing now)

or thats what I have heard at least

I suppose theres tons of algebra lol

[rustic crown]

how you people know so much about these problems >.<

[rotund aurora]

I actually read some parts of the oficial book on the millenium prize problems lmao

• Richard Borcherds has a video on BSD

[rustic crown]

[shell agate]

they crave the million dollars

[coral shale]

win the lottery or do some math. i know which one id go for

[rotund aurora]

The main reason why we care about tensor algebras is that a linear map V to A, where A is any associative algebra, factors uniqueliy through T(V) by an algebra homomorphism T(V) -> A

Is this correct?

[rustic crown]

won't say that's a reason to care

but that's definitely its universal property

[rotund aurora]

mmh

isnt it important?

[rustic crown]

yea, but does that imply you care about it

[rotund aurora]

mathematicians care about important things

[rustic crown]

ig then you're right :p

[rotund aurora]

and important things are car"ed" by mathematicians lmao. Thats the jargon I use

when talking about math

like I tell one of my classmates "you don't care about this blah blah" haha

[rustic crown]

idk i like tensor algebra because non-commutative polynomials can be sometimes useful

but other than that, idk >.<

[rotund aurora]

Is the tensor space isomorphic to some polynomial algebra?

[rustic crown]

maybe you like them because you also get the universal property for wedge algebra and symmetric algebra by quotienting by suitable relations :p

[rotund aurora]

yeh

[rustic crown]

nah, tensor will be non-comm

the symmetric algebra of a rank n free R-mdoule is the R-algebra R[x1, ..., x_n]

but for the tensor algebra, the element x1⊗x2 would be different from x2⊗x1

[rotund aurora]

but you said noncommutative polynomials

[rotund aurora]

thats the point no? I was asking about this, because Im not sure what you meant

[rustic crown]

yea it's like when you say gaussian integers

[burnt flower]

Does Q not finitely generated as an abelian group imply that it is not finitely generated as an algebra over Z?

[rustic crown]

you don't mean integers that are gaussian

[formal ermine]

let K be a field and K(X) the field of rational functions with coefficients in K. let f = X^3/(X^2+1) in K(X). show that X is algebraic over K(f).
I'm unable to find a polynomial for that, any hints?

I already tried using (a - b)(a + b) = a^2 - b^2

with x^4 - 1

and other things

[rotund aurora]

its easy

[formal ermine]

I don't see it

[rustic crown]

yea more precisely i would say we're interested in it because it's large enough and you can get a lot of interesting algebras by quotienting it

[rotund aurora]

I c I c

[burnt flower]

We can send 1 in Z to 1 in Q so Q is a Z-Algebra. However Q is not finitely generated as an abelian group because if it were, as every finite subgroup of Q is cyclic as an additive group, that would imply that there's a cyclic generator q but for example, q/2 cannot be expressed in terms of q but is in Q.

[rotund aurora]

By this I mean tthat believer in mathemagic is probably overthinking it

[rotund aurora]

First thing you would need to do is kill the denominator, try doing that with the allowed operations

[coral shale]

so using the defn of algebraic you need to find a polynomial in X with coeffs in K(f) equal to 0 correct?

[formal ermine]

ok sec

[rustic crown]

right... but these are sort of different questions. you can have stuff like the algebra k[x] over k which is finitely generated as an algebra, but not finitely generated as a module

[formal ermine]

is it X^3/(X^2+1) x^2 + X^3/(X^2+1) - (1 + 1)x^3

[burnt flower]

okay

[coral shale]

thats not a polynomial in X

[burnt flower]

thank you for your help!

[rustic crown]

but there is one nice theoerem

[coral shale]

the polynomial is meant to have coeffs in K(f)

[formal ermine]

but I have to find a polynomial in x over K(f), no?

p(x) such that p(X) = 0?

[rotund aurora]

That polynomial is fx^2+f-2x^6, its fine

[rustic crown]

if F and k are fields and F is a finitely generated k algebra, then it's actually finitely generated as a module

[rotund aurora]

unless Im missing something

[burnt flower]

OHHHH? that's very helpful

thank you

[rotund aurora]

okay, so fx^2+f=x^3, then subtract x^3

[coral shale]

yh but its not written explicitly as such

[rotund aurora]

okay you edited

thats it

[rustic crown]

yea it's called Zariski's lemma or something simlar

[coral shale]

i dont think they realise it

[rotund aurora]

it is

[formal ermine]

yeah I made a typo :v

thanks for the hint!

[burnt flower]

Thank you!

[rotund aurora]

I figured that was a typo lol.

[burnt flower]

I'm reviewing for my final

so I might talk a little more here too

[rotund aurora]

btw, it is not -2x^3

fx^2+f=f(x^2+1)=x^3

[formal ermine]

oh I forgor how to do basic algebra 🤦

[rotund aurora]

yes you did

Btw this is what I was asking for when I asked about non-commutative polynomials lol, so tensor product not commuting is the whole point

[formal ermine]

is K(X)/K(f) algebraic? (just need a yes/no so I can first try showing it myself)

[rotund aurora]

you dont have to think too hard for this one

it is a consequence of what you have already done

[rustic crown]

try showing either one, if you can't continue the proof you'll realize what's gonna happen

[formal ermine]

oke

[rotund aurora]

Given that I already know the importance of determinants, should I immediately infer that symmetric/antisymmetric multilinear maps are important in general ?

[hard flame]

im curuos if my work follows, im showing $\mathbb{Q}/\mathbb{Z}$ has no ring isomrophism with identity

[cloud walrus]

Optimism

[peak hornet]

i'm trying to prove that the set of continuous functions, together with pointwise addition and multiplication, is a ring. i'm having a little bit of trouble with associativity of multiplication and addition. consider $f(g + h)(t) = f(t) \cdot (g(t) + h(t))$. is $f(t)$ still an element of $C[a, b]$? or can i claim that $f(t)$ is the element $f$ of $C$ evaluated at $t$ and so $f(t) \in \mathbb{R}$?

[cloud walrus]

failingphysics

[chilly ocean]

latter

[peak hornet]

perfect and then i can use the associativity of multiplication in $\mathbb{R}$?

[chilly ocean]

yes

[cloud walrus]

failingphysics

[chilly ocean]

yes

[peak hornet]

ok wonderful thanks

i was just psyching myself out lol

[pastel cliff]

isn't this example wrong?

messing around with openai

[barren sierra]

it literally gives an example in paragraph 2 and then says that example is a non-example in paragraph 3

lmfao

[pastel cliff]

this is the future

[burnt flower]

I was thinking about something. So Gauss' lemma would imply that if a polynomial ring in n variables over a field is a UFD then a polynomial ring in n+1 variables over the same field is also a UFD. But then what about the polynomial ring in countably many variables?
In particular we have an ascending chain ordered by set inclusion so every polynomial belongs to a polynomial ring in a finite number of variables so by Gauss' Lemma and the above inductive argument, we have unique factorization of polynomials.
Is this argument valid?

[pastel cliff]

are R and C isomorphic as groups? as rings?

groups i think probably but idk about rings

[barren sierra]

iirc there's an isomorphism from R^2 to R as groups

but it's nonconstructive or something

something something Zorn?

[tall mountain]

hint: what property do certain elements of C have that no element in R has?

[chilly ocean]

as vector spaces over Q they have the same cardinality, so their additive groups are isomorphic. this is non constructive and uses choice

(what spamakin said)

[pastel cliff]

non constructive

[barren sierra]

Ah right right that's the proof. You view them as vector spaces

[pastel cliff]

i shall choose not to view them for now

t minus 10 hours until the final

[burnt flower]

If it were #point-set-topology you'd be wrong in a way

haha xD

[pastel cliff]

could this be a typo

[south patrol]

Oh, lol i autocorrected it in my brain and thought it was fine

yes, it should be b=c not c=a

otherwise lol a.e =e.a implies a= e so you have the trivial group

So I mean the statement is correct as stated but for trivial reasons XD

[pastel cliff]

ok thank god i thought i was losing my mind

[hard flame]

do yall have any concrete example of this? i cant find one that seems t owork

[coral shale]

let M = N

[next obsidian]

I guess try Z and nZ, then take m coprime to m

[pastel cliff]

being asked to construct a field of order 27 - can i just say that since (Z/3 x Z/3 x Z/3)[x] is an integral domain (bc i said so) that product must be a field

[chilly ocean]

what's its identity element

what if i told you (1, 0, 0)(0, 1, 0) = (0, 0, 0)

zerodivisors

[pastel cliff]

im admittedly trying to force a thm in lol

"Let R be a comm. ring with id. Then R[x] is an integral domain iff R is a field"

[chilly ocean]

is that a theorem, now?