groups-rings-fields

[median pawn]

i don't understand this notation

[woeful sage]

no problem, ur asking an actual question go ahead

[median pawn]

for instance, what is A for Q(x1,x2) = x1^2 + x2^2 + 2x1x2?

the matrix is n x n, and (x_1,...,x_n) is 1 x n

so the output is 1 x n, not 1 x 1 (which we need)

so that's weird

[chilly ocean]

Wrong

[median pawn]

show me the light pls

[chilly ocean]

We multiply it from both sides

1 x n times n x n times n x 1

[median pawn]

OH it's x^TAx

so it's a typo in the book?

or is it common to use this notation?

[chilly ocean]

There's a typo? I don't see it

[median pawn]

they haven't multiplied on both sides here

[chilly ocean]

They multiplied by transpose of (x_1, ..., x_n)

From the right

[median pawn]

oh crap, sorry

i thought the matrix A is a function of (x_1,..,x_n)

feeling dumb

[chilly ocean]

Oh, no

It consists of some elements of your field

[woeful sage]

(x_1, .... ,x_n) is a row vector then

[chilly ocean]

This would be a matrix consisting of 1's btw

[chilly ocean]

How do you properly study group theory/algebra?
I'm struggling because it feels like studying random definitions/abstractions, each one taking a lot of time to understand. It also seems to be without any clear direction or objective, which makes it really hard to get myself to do it.
Is there anything I could learn about to get more sense of why I am learning all these things? Or any tips for studying in general?

This is my first abstract algebra course

So far we've only been doing group theory, up to group actions, Sylow groups, solvable groups

[main needle]

How to find the normalizer of subgroup of upper triangular matrices with 1's in the diagonals inside GL_3(Fp)

[atomic python]

personally i've learned to treat abstract math like this a bit like how i would treat things as a kid: "why is the sky blue?" that is, you ask questions just because you can
while it can feel really out of pocket in a textbook, i think the perspective you should have is that many of these things are pursued after staring at these objects collectively over centuries

for example take a rubik's cube. you should be able to convince yourself that a corner piece can be in any corner or that an edge piece can be in any corner, but after a few years you might ask yourself if you can place oriented stickers and if it's possible for the sticker to be pointing in any of the four direction on the same location.
(while that's the end of the example you might also see that probably the most intuitive way to try and prove if this is true or not is orbits and group actions)

however i also wanna point out that

1. hopefully you can see lots of intuition/motivation to why the idea of group actions can be important (you can imagine group actions as things that change states of objects/etc)
2. Sylow groups are important at decomposing/analyzing a group (and therefore understanding things whenever groups arise) hopefully you can understand the motivation for this as more of a generalized tool you can use to understand structure
3. Solvable groups will have some motivation much later. though right now they're probably extremely out of pocket

solvable groups will come up later in Abel-Ruffini Theorem, aka the infamous "There is no degree 5 quadratic equation" theorem
and they arise somewhat naturally

also for (1) you might not find many examples at first but remember that a group need not be finite, nor does it need to be the only property about a group.
for example the Lorentz Group is a group that is also a differentiable manifold. and it has a lot to do with the structure of spacetime in Special Relativity.
it acts as isometries on spacetime through, you guessed it, a group action. which leads to all sorts of things

[molten viper]

So, I need to show that if every left coset is a right coset for a subgroup H, then H is normal. Any pointers?

[cunning dust]

What is your definition of a normal subgroup?

[molten viper]

ghg^-1 \in H for all g \in G, h \in H

or you can swap the order

g^-1 h g

so rn I have aH = Hb, so should I write like ah = hb?

oh hmm....

[uneven folio]

You don't know that it's the same h

[cunning dust]

What I think you can show is that Hb=Ha

[molten viper]

I don't, but there is an h with that property no?

[uneven folio]

Oh, maybe

[molten viper]

I feel like I'd need to already know H is normal for this

[next obsidian]

Cosets partition

[uneven folio]

My thought was to try and show a = b

[cunning dust]

This is not necessarily true

[next obsidian]

You ultimately will end up using that if they intersect they’re equal

To show that aH = Ha or something

[uneven folio]

Well up to an element of H.

Sure

What you said

[next obsidian]

Like if every left coset is a right coset you have a bijevtion between the two giving something like

aH = Hf(a)

[molten viper]

So cosets partition, but what does that mean for showing H is normal?

[next obsidian]

For f just some set function

[molten viper]

could I do something with aha^-1?

[next obsidian]

It should be something like this

[cunning dust]

You also know that a is in aH, so a is in Ha and Hb (since Hb=aH)

[molten viper]

cause if that's in H then we're done

[uneven folio]

That's it

[molten viper]

same with b^-1 h b

[cunning dust]

Phil, since cosets partition and Ha and Hb have a non empty intersection, they must be equal

[molten viper]

I see...

[white grotto]

u an endomorphism
if deg(P0(u) )=d
then (ui) for i in [0,d-1] is a basis for K[u]
P0 minimal polynomial

[cunning dust]

Do you see how to continue from there?

[white grotto]

can someone help me make sense of this proposition

[molten viper]

I'll give it a shot and come back if I get stuck

[cunning dust]

[next obsidian]

Didn’t you finish the exercise?

Like what you said is the end of the proof Slurp

[cunning dust]

Yeah basically, they’d just need to justify why that means aha^{-1} is in H, since that’s the definition they’re using

[molten viper]

(I'm actually gonna use b \in aH but I appreciate the help)

[cunning dust]

Yeah it’s the same thing dw

[molten viper]

so if aH = bH, we have a few facts known
ab^-1 \in H seems most useful?

[cunning dust]

No the most powerful fact is that you’re given bH=Ha

And since bH=aH (this is what we proved above), you know that aH=Ha

[molten viper]

Can we conclude there's an h such that ah = ha?

cause if so I think we're done

[cunning dust]

You don’t need to

[rotund aurora]

The identity works

[atomic python]

no, in fact that's not true in general

[cunning dust]

You just need to show that aha^{-1} is in H

[atomic python]

well

other than 1

[molten viper]

oh wait

ok I think I see what's going on...

so h \in H gives ah \in aH

[rotund aurora]

(Just wanted to point out that the observation above wasnt useful even when true)

[molten viper]

don't we also have that ha^-1 \in Ha?

[cunning dust]

ah is in aH, so it is in Ha, so ah=h’a, and therefore aha^{-1}=h’ for some h’ in H. So aha^{-1} is in H.
Alternatively, since aH=Ha, you have aHa^{-1}=H, which is the same thing

[rotund aurora]

Slurp already said all there is to it

[molten viper]

the answer is taking shape, however slowly

[cunning dust]

It’s all good take your time

[atomic python]

sure, maybe, but if you think ha = ah, it makes me suspsect you're not familiar enough with normal groups to understand that they're different from being in the center.

[chilly ocean]

Let M be the set of all nxn matrices. Let A, B in M be related iff det(A) = det(B). How would you describe the equivalence class using quotients?

[rotund aurora]

What are you talkig about? "if you think ha=ah", thats completelly meaningless

[molten viper]

I see why this is true, but I have no clue how I'd come to that answer myself

[rotund aurora]

Unless you mean that I think a and h commute

Actually I dont know if you said "you" generically or if you were reffering to me. I dont see the point in the comment anyways

[chilly ocean]

I meant all equivalence classes if it wasn't clear

[cunning dust]

I think the more intuitive reason is the second one I gave there: aH=Ha so aHa^{-1}=H. And that means that aha^{-1} is in H. (The first reason is very similar, but a little overcomplicated)

[cunning dust]

It’s basically the set of equivalence classes of matrices with a single \alpha on the diagonal and the rest of the diagonal being 1

[atomic python]

rhetorical you. and imo abstract algebra (and math in general) is best learned with what holds and doesn't hold in general, trivially or nontrivially.
i think being aware that the center and a normal group can be trivial is important

[cunning dust]

Essentially the idea is to find a simple family of matrices which can get you any determinant

And since determinants are multilinear, it makes sense to think of “multiples” of the identity (where by multiple I mean you just multiply a single row/column)

[chilly ocean]

I'm not sure if I follow because shouldn't it intuitively be all nxn matrices with the same determinant in each equivalence class?

[cunning dust]

Oh well yes

That is what an equivalence class is

Then what is your question? Wdym by describe?

[chilly ocean]

Describe, as in write a set

[cunning dust]

Well like if you need to find a representative for each equivalence class, these matrices give you that

[chilly ocean]

Why the diagonal though?

[cunning dust]

Simplest type of matrix to deal with

The determinant is just the product of the values on the diagonal

[atomic python]

$$E_d = {\forall m \in M | \det(m) = d}$$?

[cloud walrus]

Quarky

[atomic python]

well Ig the intended (?) answer could just be R/whatever ring/field now that I think about it

[chilly ocean]

What about as a quotient?

[cunning dust]

Okay I’m not sure I understand. What do you want exactly? To describe the quotient set? This usually means to give a representative of each equivalence class, which I just did

[atomic python]

are you trying to solve a problem given to you? or ask one of your own questions

if it's a problem, post the exact wording

if it's one of your own questions, then yea, what Slurp said

[chilly ocean]

I basically want to write a set of the representatives of each equivalence class

[atomic python]

then what Slurp had before was fine

[cunning dust]

Sure, that’s what I gave you then

[chilly ocean]

I didn't really follow it

[cunning dust]

(ams) Matrices of the form
[ \begin{pmatrix} \alpha \ & 1 \ & & \ddots \ & & & 1 \end{pmatrix} ]

[cunning dust]

You see how the determinant of that is \alpha?

[chilly ocean]

What about the other entries?

[cunning dust]

So using matrices of this form, you can get any determinant you’d like

[cunning dust]

0

[chilly ocean]

So in the 2x2 case, it would be like \begin{pmatrix} \alpha & 0 \ 0 & 1 \end{pmatrix}

[cunning dust]

Yep

[restive birch]

okay, so i read this answer, and i understand everything up to the last step- that sigma(S) ≠ S shows that the action is faithful. i dont understand this

[cloud walrus]

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

[atomic python]

the reason it's erroring i think is because there's hidden \ before the [

markdown character escaping, if you know what that is

[chilly ocean]

So then $E(\alpha) = { \begin{pmatrix} \alpha & 0 \ 0 & 1 \end{pmatrix} }$ and the actual thing that we are looking for would be $\bigcup_{\alpha \in \mathbb{R}} E(\alpha)$?

[cloud walrus]

n/c

[cunning dust]

(ams) By $E(\alpha) = { \begin{pmatrix} \alpha & 0 \ 0 & 1 \end{pmatrix} }$, do you mean the equivalence class given by this matrix? If so it should not be a union, it’s simply the set of all such $E(\alpha)$s.

[atomic python]

that union would be the set of all representatives*

[cunning dust]

The union would be the set of all matrices

[atomic python]

i took E(a) to be a singleton---not an equivalence class

[chilly ocean]

No, I want the union of all representatives for each equivalence class

[cunning dust]

Oh then sure

[chilly ocean]

But yes E(alpha) is just the one matrix

[cunning dust]

Then yes the union gives you a set of representatives

[chilly ocean]

Thanks. Is there a way to do this using quotients though?

Like, taking M_2(R) / something to get what we want?

[atomic python]

"what we want"

[chilly ocean]

(set of representatives for each equivalence class)

[atomic python]

then that already is a quotient

[chilly ocean]

I meant, instead of explicitly constructing a matrix and taking the union of those matrices

[cunning dust]

I think they want a subgroup of M_2(R) where the quotient group is that partition

[atomic python]

would you rather a homomorphism?

[chilly ocean]

I know we can do something like M_2(R)/~ to get the (set of) equivalence classes, but how do we take exactly one representative from each class from then on without explicit construction?

Or do we have to do some kind of explicit construction?

[restive birch]

?

[cunning dust]

It may be better if you’d send a link rather than an image

[atomic python]

it sounds like you want a homomorphism
if so then just map all matrices to their representative (or even better map it to the determinant in the base field)

[restive birch]

sorry, herehttps://math.stackexchange.com/questions/2692303/for-which-values-of-k-the-action-of-s-n-on-k-element-subsets-is-faithful

[chilly ocean]

So you mean mapping all matrices with det d to (alpha 0 \ 0 1)?

[atomic python]

with det(m) to ...

[chilly ocean]

How would I write that though?

[atomic python]

$$m \in M \mapsto \big(\det(m), 0, 0, 1\big)$$
or
$$m \in M \mapsto \det(m)$$

[cloud walrus]

Quarky

[atomic python]

the latter is not a representative (nor technically an explicit quotient)

but it is an isomorphism of the quotient

[chilly ocean]

I would still have multiple representatives of the same determinant though?

[cunning dust]

So I’m not familiar at all with group actions, but I quickly skimmed the wiki article. And from what I understood a group action is faithful if for every g\neq e, gx\neq x for some x. And since they found such an S for each \sigma where \sigma S\neq S, then the action is faithful

[atomic python]

name one?

[chilly ocean]

The representative of each equivalence class is a matrix, so I'd have multiple matrices that have the same determinant

[atomic python]

with
$$L~ : ~m \in M \mapsto \big(\det(m), 0, 0, 1\big)$$
?

no, if $A, B \in M$ with $\det A \neq \det B$ then $L(A) \neq L(B)$

If $\det A = det B$ then $L(A) = L(B)$.

[cloud walrus]

Quarky

[chilly ocean]

first I'd have to do L[M_2(R)] and then this is the set of representatives?

[restive birch]

this is not the definition im used to- the one my book gave is that an action is faithful if every element in G acting on B represents a different permutation of B

maybe the definitions are equivalent, but i dont see how

[south patrol]

Same thing because these are equivalently saying the map G -> Sym(S) has trivial kernel

Slurp basically said trivial kernel whilst you said the map is injective essentially

[sinful mirage]

If L is al Lie algebra,I an ideal, L/I is nilpotent and the adjoint representation restricted to I is nilpotent, then can I conclude that L/ker ad_I is nilpotent?

[chilly ocean]

Is this right?

[restive birch]

remind me- how do you show that these things imply each other?

[south patrol]

let f: G-> H be a map of groups. If f is injective then if f(x)=1, x =1; conversely if the kernel is trivial and f(g)=f(h), then f(gh^-1) =1, so...

[chilly ocean]

When I see $(12)(123) \in A_n$, are these two different permutations?

[cloud walrus]

Iced Sugar

[chilly ocean]

or one permutation that is represented as a product of two permutations

[cunning dust]

This

It’s very common to see permutations written as the composition/product of cycles

Also this permutation isn’t even in A_n, it’s not even

[south patrol]

Isn't even

Lol

Nice

[cunning dust]

Oh lol I didn’t even realize that

[chilly ocean]

yeah I am just reading the book

[cunning dust]

I see

[chilly ocean]

for $(1,2) \in S_n and (123) \in A_n$

[cloud walrus]

Iced Sugar

[chilly ocean]

wait how do you tell if a permutation is even?

i know that it is even if it is an even number of transpositions but is there a way you can just look at a permutation and know that

without having to actually figure out what transpositions it equals

[unique valve]

i mean you could write it in terms of disjoint cycles which is pretty easy

and cycles of odd length are even

cycles of even length are odd

[chilly ocean]

so this is just always true

if it has 3 elements, the permutation is even

(Abc) is even always

[unique valve]

3-cycles are even

[chilly ocean]

(Abcd) is always odd

[unique valve]

yes

[chilly ocean]

ok

thanks

[unique valve]

because (a b c d e) = (a e)(a d)(a c)(a b) i think

and you can write cycles like this for any length

[rotund aurora]

What are examples of extensions F<K<E with F<E purely transcendental but F<K not purely transcendental?

Is the example Q<Q(x^2+y^2,x^3,y^3)<Q(x,y) an example?

[chilly ocean]

Thanks so much for the write up! I will check out the things you mentioned 🙂

[glossy crag]

Isaacs disagrees

I don't know transcendence bases yet, but thanks.

Shit, what I meant was "finite order => 2 or less", my bad.

[prime sundial]

the parity of a permutation is the parity of the number of even cycles

i.e. a permutation is even (odd) <=> it has an even (odd) number of even cycles

[molten viper]

Alrighty, part 2 of my confusion

I know that |H| = |G|/2. I need to show every left coset is a right coset.

With H a subgroup, to clarify

[next obsidian]

There’s not enough space

Use that both sets of cosets partition

[molten viper]

Hm

[next obsidian]

You know how many cosets there are right?

[molten viper]

There's only 2

[next obsidian]

Right

So let gH be a non-trivial coset

What elements form that?

[molten viper]

Well it's all the elements of G not in H yeah?

[next obsidian]

Right

What about Hg’ a non-trivial coset

In fact actually let’s just do Hg

This is non-trivial if gH is non-trivial

[rustic crown]

[molten viper]

Give me a few minutes to figure out how to prove this

so if I take some arbitrary gh \in gH

hm.

[next obsidian]

My dude

You told me gH = H^c

Because it’s everything in G not in H

Why doesn’t the same logic apply for Hg?

[molten viper]

It totally does 😅

[next obsidian]

Anyway this is all you need

There’s literally not enough space in G for it not to be true

[rotund aurora]

Any hints?

The case were the polynomial may have a linear factor is easy

[south patrol]

don't they give you the non-linear (i.e. a = -1) one though

[rotund aurora]

what Im asking is to show its irreducible

[south patrol]

Oh okay so if a isn't any of those then it's irreducible, sure

[quiet pelican]

There’s a long simultaneous eqn way to do it (suppose you have a factorisation (x^2 + bx + c)(x^3 + dx^2 + ex + f) and then expand our and equate, showing there’s no integer solution to the simultaneous eqns you get)

[rotund aurora]

nvm I think I got it

yeah

the restrictions are actually pretty strong, but for some reason I was just looking at one of them (since actually, its probably enough, but without the other equations is harder xD)

[rotund aurora]

Anyone please? Thanks.

[rustic crown]

is there even a nice way to say when something isn't purely transcendental?

i remember that Luroth's theorem says that any non-trivial intermediate extension of k --> k(x) is automatically purely transcendental

[rotund aurora]

yeah

and simple too

[next obsidian]

I think the example Croqueta works

Q(the three things) could only have transcendence degree <=2

[rotund aurora]

because additivity

[next obsidian]

And then you should be able to say that if s field extension is generated by a_i, then you can take a transcendence basis out of that

[rustic crown]

and it equals 2 as the other extension is algebraic

[next obsidian]

Yeah, something like this

The point is then like something is algebraic over Q(some subset of the 3 generators)

Or something

But I’m not 100% sure

[rotund aurora]

I was trying to come up with an example with two generators, though this must be impossible no?

in the case of an intermediate subfield of Q<Q(x,y) I mean

[rustic crown]

yea, i think that would be a transcendence basis, but i'm not sure this will be enough to show that it's not pure. by definition a transcendence basis only has to generate the field up to an algebraic extension

[rotund aurora]

My intuition was that Q(x^2+y^2, x^3,y^3) cant be generated by less than 3 elements. I think that's the case no?

[rustic crown]

[next obsidian]

Yeah, I’m not sure

I wanted to say something like something with a transfer or something like

Because there’s some algebraic aspect over the purely transcendental extension

It can’t be purely transcendental

[rustic crown]

i was looking at some stackexchange answer about certain extension not being purely transcendental, it wasn't like completely easy and people are using some clever stuff there

[next obsidian]

Which like isn’t true because Q(x^2) < Q(x)

Yeah

[rustic crown]

https://math.stackexchange.com/a/5285
it's this one if you're wondering

[coral shale]

nice pfp

[rustic crown]

if we can fit this inside a purely transcendental extension, you'll at least get an example you're looking for

[rotund aurora]

people using elliptic curves lmao

[rustic crown]

uwu

[warm wyvern]

What are presentations useful for?

[rotund aurora]

Why do you think thats possible?

[rustic crown]

not sure lol :p

[rustic crown]

for groups?

they let you easily construct maps out of that group

[warm wyvern]

Yeah

[rotund aurora]

I was thinking about the following problem: Given extension F<K such that all elements of K not in F are transcendental over F, can you find a field E such that F<K<E and F<E is purely transcendental?

[rotund aurora]

For presenting groups

[rustic crown]

Like giving a map from D_n --> G is equivalent to giving two element r and f in G such that r^n = f^2 = (rf)^2 = 1 in G

[rotund aurora]

As stupid as it sounds, that's why they are useful xD

[coral shale]

concise way to express the group relations

[rustic crown]

yea i was thinking something similar but didn't get any far

[rotund aurora]

I was gonna say that the presentation tells you everything you need to know about the group, effectively speaking. But I think that's not true, or at least the question is way more subtle

See the word problem

You can read Keith Conrad's exposition, its pretty nice

@warm wyvern

[rotund aurora]

Transcendence is difficult really

[rustic crown]

apparently luroth's problem is solved for n=2

[warm wyvern]

Imma read that tomorrow

I have questions about the word problem too but I think I need to read a bit more first

[rotund aurora]

Think about this problem: The Burnside problem asks whether a finitely generated group in which every element has finite order must necessarily be a finite group.

So beautiful

(maybe you know it already, idk)

[warm wyvern]

I do not, actually

Hmm

[rotund aurora]

but dont look the solution!

[rustic crown]

me wanna spoil 🙈

[warm wyvern]

I really need to sleep tho

I'll think about it tomorrow

Thx every1

[tribal furnace]

gn stack

[warm wyvern]

GN zane lel

[true cairn]

Can someone help me with part a?

[rustic crown]

first list down all the subgroups

and check what all among them are normal

(you can save some trouble by using standard propositions like, subgroup of index 2 is normal, the center is normal, trivial and full group are normal etc)

[lavish nexus]

How do you show φ1-ψ1 maps everything to Imδ^0

Or is it not true at all

[indigo ravine]

Probably dumb question : is every module a quotient of 2 free modules ? It's obvious that every module is a quotient of a free module by a module but I don't see a reason for both to be free.

[next obsidian]

Hmmm

No this is not true

If this was the case then every module would have projective dimension bounded by 1

I can’t really give a better justification without invoking some non-trivial homological algebra

But it is true over any PID for example, but this is because submodules of free modules are free in that case

You can come up with explicit examples over a polynomial ring in 2 variables though, for reasons I also again can’t quite prove rn, k considered as a k[x,y]-module cannot be the quotient of a free module by another free module

[indigo ravine]

Thanks! What knowledge of homological algebra would this require ? Might not know it but I can reflect on that when I get there

[next obsidian]

Hmmm

Technically the definition of projective dimension

Do you know what an exact sequence is?

If so, I can define stuff, I just can’t provide an example of why there’s something with projective dimension > 1

[indigo ravine]

Yes

[next obsidian]

Okay so the projective dimension of a module M is the smallest integer n such that there’s an exact sequence

0 -> P_n -> … -> P_0 -> M -> 0 with the P_i projective modules

Or infinity if none exists

It’s the minimal length of a projective resolution

If you don’t know what projective means, just pretend I said free

So if M was the quotient of two free modules you have an exact sequence
0 -> P_1 -> P_0 -> M -> 0

So the projective dimension of M is bounded by 1

So it would suffice to find an M with a projective dimension > 1, and the easiest way (for me to think of) to prove one exists is to find modules M,N such that either

Ext^2(M,N) ≠ 0 or Tor_2(M,N) ≠ 0

Because one can show that if the projective dimension of M < n then Ext^k(M,N) = 0 and Tor_k(M,N) = 0 whenever k>= n

[next obsidian]

So in particular, if the projective dimension is <= 1, then these dudes have to be 0

The reason this is (IMO) easier is that one has tricks to compute Ext and Tor

So it’s really easy to find M,N such that this is true

[indigo ravine]

Alright I see. I think I should have the tools to understand this, just need to familiarize again with stuff I had forgotten from my homological algebra class, thanks!

[next obsidian]

Np!

Also:

I ran an argument through my head and I’m pretty sure if A is Noetherian and finite dimensional, if dim A > 1 then it’s impossible to have this property be satisfied

I imagine it would also be impossible for infinite dimensional A as well, but I can’t justify it, it just would be weird if it existed IMO

(Then again an infinite dimensional Noetherian ring is very weird)

(Looked up a theorem, I can justify it when A is local Noetherian, without needing to assume it’s finite dimensional)

Ooo, don’t need local either

Sweet

So for a Noetherian ring of dim > 1, this property cannot be satisfied

[slender hamlet]

Ok yeah if it has finite order then it is true

By some galois cohomology stuff

[tender wharf]

Do + and * share the same identity or different?

[tender bough]

"if xy = 0 and x is nonzero, then y is zero"
In what algebraic structures is this property true?

[rustic crown]

domains

[tender bough]

https://en.wikipedia.org/wiki/Domain_(ring_theory)
I see, thanks!

[chilly ocean]

those are called rings without zero divisors

[tender bough]

"domain" is just cooler

[lethal dune]

Domains without zero divisors

[tender bough]

more confusing too...

[lethal dune]

[chilly ocean]

Mathematical literature contains multiple variants of the definition of "domain"

I think it's domain that's the confusing one?

[tender bough]

I know, that's my point. I'm an elitest and I like to confuse people 😈

[rustic crown]

[tribal furnace]

I didn't know integral domains have to be commutative

[chilly ocean]

[lethal dune]

Some authors don’t require commutativity for IDs

[tender bough]

wikipedia >> some authors

[rustic crown]

wikipedia is written by the community :p

[tender bough]

but also, in a vector space, when x is a scaler and y is a vector

[rustic crown]

ah but here x and y live in different things >.<

i assumed both x and y have "same footing"

[chilly ocean]

they both live in some vector space

[rustic crown]

[tender bough]

Do vector spaces exist by themselves or do they only start to exist when people construct them a.k.a. think about them in their minds?

[chilly ocean]

neither? They always exist as long as someone thinks about them

[tender bough]

hmmmmmmm

did they not exist before 1888?

[chilly ocean]

they didn't

[rustic crown]

[unique valve]

Actually I knew about vector spaces before 1888

I just didn’t publish

[chilly ocean]

vector spaces are so passe

[tender bough]

I didn't even exist back then 💀

[rustic crown]

you did, you just changed your form

[chilly ocean]

no because it wasn't them

it was particles that make them up

what makes us humans is not only what are we made of but also how are we made of

[rustic crown]

humans aren't sets

[unique valve]

I’m made of better than you

[tender bough]

wtf aren't everything sets

[unique valve]

Yeah

That

I am the set of myself

[chilly ocean]

humans are sets, but they inherit a structure of a human being

just like how groups are sets but they have the structure of a group

[unique valve]

Humans are vectors

[tender bough]

where is the #math-philosophy channel?

[unique valve]

Also things exist before you think about them

[chilly ocean]

nope

[unique valve]

Something can be happy without thinking about happiness

Same thing

[rustic crown]

.<

no confuse me

[chilly ocean]

it's an illusion because if you think about something then you already visualize so before you even think about something you think that it exists

but it only comes to motion after it appears in your brain

[unique valve]

What if you don’t have a brain

Like a plant

A yellow flower

Why can’t the flower be happy

[rustic crown]

🌻 😦

[tender bough]

🌻 😃

[chilly ocean]

what does it mean for a flower to be happy?

why are we talking about happiness in the first place

[chilly ocean]

but happiness as a concept only exists because you think about it

it's you that gives it a description

[rustic crown]

my brain hurt

[unique valve]

You give it a description but it exists before it’s described

Just like with physical objects

Ok let’s say someone comes up with the concept of vector spaces and then dies

[chilly ocean]

not as how you describe it

[unique valve]

Does the concept still exist

[chilly ocean]

that's a complicated question and the answer depends on multiple factors

which your situation doesn't exhaust

[tender bough]

you guys should stop hurting det

[unique valve]

It is not the suns fault if you stare into it and blind yourself

What might these factors be

[chilly ocean]

well first of all let me mention how your question is unclear in the way you phrase it

[unique valve]

They don’t tell anyone else

[chilly ocean]

it might exist but in what way does it exist

if you think about it, you can mean multiple things and you can say what I explain here is wrong based on what you are currently thinking about

[unique valve]

As a concept?

[chilly ocean]

but if we dive deep into it there's really nothing there apart from semantics

[chilly ocean]

that's still not clear enough

[unique valve]

What’s an example

[chilly ocean]

it can exist to someone, it can exist as something written down

it's all different

but it could still be described as existing as a concept

[unique valve]

Well we’ve been discussing whether there is an existence of things that is beyond thought

If no one has thought of something, you’re saying it doesn’t exist

So I’m asking if you’d say it exists if someone thinks of it and then dies and no one thinks of it anymore

[chilly ocean]

if you want to be technical with it then of course there can exist such things

but did anyone saw 2 walking around?

in some, maybe obvious to some meaning, in the platonic meaning of the word

no I don't think it does, it's complete crap

[chilly ocean]

no, that was in specific context

[unique valve]

Ok well what’s the distinction for that

In what contexts does that apply

[chilly ocean]

no

if it's not written down anywhere etc. etc.

it doesn't exist in other peoples conscious for sure

[chilly ocean]

in the contexts that I decide I don't need to specify meaning of the word

[unique valve]

So if someone writes down the rules for a vector space on paper and then dies and no one thinks about it or sees it?

[chilly ocean]

but the meaning is still important of course

[chilly ocean]

it doesn't exist for anyone

but it exists as a piece of paper

[unique valve]

The concept of a vector space

Does not exist as a piece of paper

[chilly ocean]

what's the point here

[unique valve]

I would like to understand your perspective

[chilly ocean]

alright but now I'm going to post an exercise to change the course of discussion

Exercise: Show that any subset of A^omega can be recognized by a (possible infinite) Buchi automaton in which all the states are final

[unique valve]

I have done it

What do you like about topology

[chilly ocean]

https://tenor.com/view/forget-algebra-mari-takahashi-atomicmari-still-watching-netflix-ignore-algebra-gif-26017210

[unique valve]

Why

Would one ever wish to forget algebra

[chilly ocean]

we've already stretched it by discussing philosophy
let's talk about algebra or not

[unique valve]

What is your favorite algebra

[chilly ocean]

real numbers

[unique valve]

R u a grad student?

Finite Fields vs Infinite Fields

[rotund aurora]

this kind of conversation is completelly useless. To be able to think you need a body, that was formed by X processes, etc. Oh, but if you don't think about it (or no one else does) then it doesn't exist? Well, then that's Descartes paranoia, and you cannot say anything interesting from that position

also I think Descartes, in his method of perfect reasoning and after the cogito ergo sum, he asserted the existence of god. LMAO

[chilly ocean]

you've missed the point and you think you're smart

[restive birch]

so, if im understanding correctly, the orbit of an element of a set under a group action is just all of the other elements that elements can "reach" using the given group action?

[chilly ocean]

also why revive a discussion that I said doesn't fit in this channel already

[chilly ocean]

yep

good intuition for it

[chilly ocean]

I'm sorry for being rude with my first response. I didn't think about it clearly and I just said what was on my mind.

let's just not discuss philosophy in this channel

[restive birch]

is it true that the centralizer of r in D_8 is {1, r, r^2, r^3, sr^2}?

[tender bough]

D=====8

[restive birch]

thank you for your helpful insight, matt

[spice whale]

sr² doesn't work idt

[restive birch]

ah, so it doesnt

thanks

what about the centralizer of s in D_8? is it {1, r^2, s, sr^2}?

[restive birch]

of course, my answer made no sense anyways because of Lagrange's theorem

5 doesnt divide 8

[spice whale]

this looks right

[chilly ocean]

double checked it's correct (not counting that sr^2 in the first one)

[restive birch]

is there a term for when two separate terms have the exact same centralizer?

[next obsidian]

I’m pretty sure this is true iff g and h commute

[restive birch]

hmm, but r^2 and r commute but r^2 commutes with s and r doesnt

in D_8

[chilly ocean]

that's true
for a centralizer of an element there can be multiple representants but not everything serves as a representant

[chilly ocean]

I don't think there's a term but I found this characterization https://math.stackexchange.com/a/799971/476484

[sweet echo]

What am I missing here: I have an algebra A, (finite, associative, unital) over a alg closed field and an A-module M. Since A is an algebra, M is a vector space. What is preventing M from being a free module? I have a linearly independent subset of M, its basis. And it appears to be as if this generates the module as I have a copy of the field sitting inside A that I can act by. But I know that not all modules are free so what is the obstruction here?

[spice whale]

all vector spaces are free

so by definition any module over an algebraically closed field is free

[sweet echo]

right but I have an algebra, and not all algebra modules are free, since they would then all be projective, so something im writing doesnt make sense

[spice whale]

oh wait

hm

[wooden ember]

M isn't necessarily a vector space here unless im mistaken

[spice whale]

no it's not nvm

[wooden ember]

ah no it is

[elder wave]

it is

[wooden ember]

but it's not a module over that field

or at least that's not what youre considering

free over the base field doesnt mean free over the algebra

[spice whale]

M isn't a vector space over A because A doesn't necessarily have inverses or commutativity

[wooden ember]

yeah but they were saying vector space over the base field of A im pretty sure

[sweet echo]

okay i think this is what im missing

we don't get linear independence over A right? I might act on a basis element and end up on another basis element

[wooden ember]

the issue is with linear independence

yeah

[sweet echo]

aha okay

[wooden ember]

the extra elements in A can add relations

[sweet echo]

thought i was going crazy

thankyou^

[pastel cliff]

does Z[sqrt(d)] being noeetherian follow directly from Z being noetherian

by hilbert basis theorem

im asking bc the theorem is about Z[x] but im not sure if that x has to be some "variable" or it can be anything, such a square root

[chilly ocean]

It's Z[x]/(x^2 - d), and Z[x] is Noetherian and that's preserved by quotienting.

It's a quotient of the polynomial ring Z[x].

[pastel cliff]

oh so im kinda right but not enough

[normal oasis]

What is the Pythagorean theorem really? How does it create a field extension?

[restive birch]

how would you show the normalizer part of this? i still dont fully understand normalizers

[formal ermine]

the normalizer contains all elements whose right cosets are equal to their respective left cosets

$N_G(A) = \Set{g \in G | gA = Ag}$

[cloud walrus]

yes yes yes no

[formal ermine]

one thing that you might already instantly see that if $G$ is abelian then $N_G(A) = G$

[cloud walrus]

yes yes yes no

[formal ermine]

because you can just do $$gA = Ag \iff gAg\inv = A \iff gg\inv A = A \iff A = A$$

[cloud walrus]

yes yes yes no

[formal ermine]

one thing that you might already immediately see is that*

[warm wyvern]

What's an algebra?

Is it a well defined term? Or is it just a blanket term for an algebraic structure that doesn't have a proper name?

[rotund aurora]

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

I think they are reffering to this

[warm wyvern]

Ah

[warm wyvern]

Can you link me which paper you're referring to?

I found a list of their expository papers but non of them had "presentations" in the name

[rotund aurora]

https://kconrad.math.uconn.edu/blurbs/

Section group theory

"Why word problems are hard"

@warm wyvern

[warm wyvern]

Thank youu

[warm wyvern]

Nooooo

I got spoiled

[rustic crown]

nope

that's not what Corqueta said

here the element

[1 1]
[0 1]

has infinite order

Croqueta was asking if you know every element has finite order and that the group is finitely generated, then does it imply that the group is finite

[south patrol]

Oof

[rustic crown]

so you didn't get spoiled

[warm wyvern]

Read further

[-1 -1]
[0 1]
And
[-1 0]
[0 1]
Also generate that group

And both have order 2

[rustic crown]

yea, but Croqueta says you want all elements to have finite order, not just the generators

[warm wyvern]

Wait, wat

Oh, yeah

Why did I misread the problem like that? Lol

[rustic crown]

happens :p

[cinder onyx]

quick gutcheck, S_0 is isomorphic to S_1 right

as they're both just the trivial group with one element

just checking im not being really stupid

[formal ermine]

how are you defining S_0?

[rustic crown]

yea that should be right

[cinder onyx]

I mean, I'm assuming the case here is asking about permutations of an empty set

of which there is one

{ø}->{ø}

[rustic crown]

nuuu

ø -> ø

not {}

[cinder onyx]

ah yes

[formal ermine]

how can we show that Tor(M) where M is an $R$-module over a domain is closed under M's group operation (let's call it +)?
like for $a,b \in \operatorname{Tor}(M)$ and $r,s$ being their respective elements that send them to $0$, there must exist $x,m \in R \setminus \Set{0}$ such that $0 = mr.a + ls.b = x.a + x.b = x.(a + b)$. but this gives us no actual way of finding the $x$, does it? like we have $x = mr$ and $x = ls$ but that doesn't really help us, no?

[cloud walrus]

yes yes yes no

[rustic crown]

oh what's your definition of Tor?

i only know the left derived functor definition

[dusky birch]

If p ∈ F[x] is irreducible, then the field G = F[x]/(p)
contains F as a subfield, and p has a root in G
Can someone please explain what p has a root in G means.
Like if I say 5 has a root in Q
wtf does that mean.
suppose, for the sake of a concrete example, G = Z2[x]/x^2+1 = {0,1,x,x+1}
what does (x^2+1) has a root in {0,1,x,x+1} mean?

oh frick

x^2+1 isn't irreducible

give me a sec.

suppose G = Z2[x]/x^2+x+1

the elements are still the same right?

still, my question is, what does it mean?

[rustic crown]

5 is not an irreducible polynomial, and what would even be the field here

[dusky birch]

oh.

what does it mean for p to be a root in G.

I don't understand that part.

[rustic crown]

p is a polynomial with coefficients in F. you can also pretend that it's a polynomial with coefficients in G as F is a subfield of G

so all it's asking you is if there is an element r in G such that p(r) = 0

[dusky birch]

So it's G[Y]

and you're saying Y^2+Y+1, that x is a root

[rustic crown]

yep

[dusky birch]

ok why is it not stated as that then?

[rustic crown]

because if you plug in x (more precisely the coset of x in F[x]/(x^2+x+1))

[dusky birch]

Why is it just "p has a root in G"

[rustic crown]

cause that's what a root is

a element of G such that p evaluates to 0

[dusky birch]

could you give an example with fields I'm more familiar with

like the exact statement but in reals

if that's possible.

[rustic crown]

sure

R

[dusky birch]

ok

[rustic crown]

and consider p = x^2+1

[dusky birch]

OK

[rustic crown]

p is irreducible, in particular has no real roots

[dusky birch]

OK

[rustic crown]

but the extension R[t]/(t^2+1) does have a root of the polynomial x^2+1

in fact R[t]/(t^2+1) is isomorphic to C, the field of complex numbers

if you plug in x = [t]

it evaluates to [t]^2 + 1 = [t^2 + 1] = [0]

(by brackets, i just mean the coset containing that element)

[dusky birch]

are you able to do an example without the / polynomial thing or is it not possible

like if it was a number of something

Z / 7

Z mod 7.

[rustic crown]

didn't get you. the goal of doing that is to get a bigger field where your given polynomial does have a root.

[dusky birch]

First, am I right to say the elements of this G is precisely {[0],[1],[x],[x+1]}

[rustic crown]

if you don't give me a polynomial, then what are we even doing :p

yep that's right

[dusky birch]

ok so if I say p is a root in {[0],[1],[x],[x+1]}

what does that mean

[rustic crown]

p "has" a root

[dusky birch]

like I don't know where the Y comes from.

Ok x^2+x+1

has a root in {[0],[1],[x],[x+1]}

I don't know what that means

to have a root in a field.

can I pretend {[0],[1],[x],[x+1]} that these are like numbers

[rustic crown]

it means to have an element such that your polynomial evaluates to 0 >.<

in Z/2Z, the only elements are 0 and 1, and none of them satisfy x^2+x+1 = 0 when you substitute them for x

but in the bigger field you have some extra elements now

and [x] there does satisfy our given polynomial

because [x]^2 + [x] + 1 = [x^2 + x + 1] = [0] = 0

[dusky birch]

Ok

Where can I read more about this.

[rustic crown]

any abstract algebra book should have this

[dusky birch]

is there anything online

my book only mentions it once

and there's nothing else.

[dusky birch]

I want to make a substitution of what you're saying.
So you said in Z/2Z, the only elements are 0 and 1, and none of them satisfy p(x)=0. Eg p(1) != 0 and p(0) != 0
but you say in G = {[0],[1],[x],[x+1]}, p([x]) = 0, so there's an element in G such that p(something in G) = 0

I think I kind of understand now.

I think my question now is what is the point of that.

Is that not like saying 7 = 0 mod 7

[rustic crown]

yes it is!

but the point is that F[x]/p is a field

as p is irreducible

what you've proves is that you can always add an element to your field satisfying a certain irreducible polynomial

for example, what was the first definition of C that you saw?

it was elements of the form a + bi, where a and b are reals

and then you formally defined how to add and multiply

from this theorem, you easily get that you can add an element to R such that the polynomial x^2+1 now can be factored

[dusky birch]

O

I see what you mean

Thanks for your help

[rustic crown]

so this is the simplest way to get extensions of a field

[dusky birch]

if you still know any resources online could you send it.

[rustic crown]

http://abstract.ups.edu/aata/fields.html

maybe this?

idk any online books really >.<

[dusky birch]

what would you say is a good book on abstract algebra in general that includes what we were talking about

[rustic crown]

any book that works for you is good lol

[dusky birch]

well idk any books

what did you read

[rustic crown]

#book-recommendations message

[dusky birch]

wait I have question if I do Z/x^2-2 do I get Z[sqrt2]

[rustic crown]

yep

but you mean Z[x]/(x^2-2)

[dusky birch]

O

Ok so Z[x]/(x^2-2)

gets me Z[sqrt(2)]

so it gets me the set S = {a+bsqrt2: a,b E Z}

[rustic crown]

yep

"gets me" should be replaced with "is isomorphic to" to be more precise

[dusky birch]

Oh

:(

[rustic crown]

oh 😦 what happened >.<

[dusky birch]

what's Z[x]/(x^2-2)

before I had a G = {0,1,x,x+1}

[rustic crown]

it's the polynomial ring quotiented by x^2-2

so the elements look like [a + bx]

where a, b are integers

[dusky birch]

wait Z isn't even a field

how can I do anything.

The theorem needs it to be a field

[rustic crown]

you don't need to look at higher powers as dividing by x^2-2 will leave a remainder which is a linear polynomial

[rustic crown]

right, but as long as you have a monic polynomial, quotienting the polynomial ring by that gives you a bigger ring where the polynomial has a root

[dusky birch]

🤯

[rustic crown]

i say this because, if you work with say Q[x]/(x^3-2) then it's isomorphic to Q(cbrt2), Q(cbrt2 * w) and Q(cbrt2 * w^2) where w is a complex cube root of 1

so it's not really clear what Q[x]/(x^3-2) "gets you"

[dusky birch]

Ok ignore the integer case because idk what's happening. What if I look at rationals.
if I have W = Q[x]/(x^2-2) = {Q U (ax +b)} and you say this is isomorphic to Q[sqrt2] = {a+bsqrt2: a b in Q}

[rustic crown]

idk what {Q U (ax +b)} means

[dusky birch]

like all the elements of Q

with ax +b , where a b are in Q

is that right?

[rustic crown]

you're already counting them when a = 0, right?

[dusky birch]

like for G before I had {0,1,x,x+1}

OH

you right

so just {ax+b} like you said before

[rustic crown]

cosets of them, but yea

[dusky birch]

but there's an x here

does the x

turn into sqrt2

[rustic crown]

yea

[dusky birch]

i c

[rustic crown]

more precisely, there is an isomorphism Q[x]/(x^2-2) --> Q(sqrt2)

[dusky birch]

kinda makes sense

[rustic crown]

which sends [x] to sqrt2

(there is also an isomorphism which sends [x] to -sqrt2)

[dusky birch]

I appreciate your time. thanks!

I understood a lot better than when I asked the question

I will still look into it further, but thanks for the headstart.

[rustic crown]

[rustic crown]

i mostly read Aluffi

[dusky birch]

didn't mean to send that

[rustic crown]

(oh dw, i didn't even read it :p)

[dusky birch]

it was among us gif

which book did he write with abstract algebra?

this one?

[rustic crown]

yea that one

[dusky birch]

Ok thank you

[rustic crown]

[dusky birch]

I will obtain it in a particular way

[rustic crown]

don't get it if you don't wanna learn category theory :p

hewwo walter

[agile burrow]

hi det

det I am learning group cohomology

[rustic crown]

ooooh

me wanna learn too

[agile burrow]

but I am taking a small break as I study for finals

Yes I find it very interesting, and eventually I want to understand interactions with algebraic geometry

[rustic crown]

i am learning about etale cohomology, i think it has relations to group cohomology for galois groups

but at the moment i only know the definitions :3

of etale morphishm, etale sites, etale sheaves etc

[agile burrow]

Yeah, my prof told me that Galois cohomology is the specific case of etale cohomology applied to a point (?) I kinda forgot :p

[rustic crown]

yee, that makes sense

[agile burrow]

i don't really know any algebraic geometry yet, but when I eventually learn it everything will make sense 😃

[rustic crown]

[tender bough]

why do ad_E and ad_F look transposed to me?

[rustic crown]

they aren't transposes tho

[tender bough]

right, thanks!

[rustic crown]

[tender bough]

but uhhh sanity check: the first column of ad_E should be all zero, because ad_E (E) = [E, E] = 0?

[rustic crown]

right, they seem to be using rows instead of columns

dunno why

[tender bough]

ikr...

[tender bough]

how is a "row representation" formally distinguished from a "column representation"? I thought "column representation" was the only one used...

I mean, do they have formal names?

[rustic crown]

it only depends on how you wanna write your functions

usually we write f(x) for, f applied to x

that's why matrix * column vector = column vector is used

if you write (x)f, as the image of x under f

then you would prefer row vector * matrix = row vector

(that's actually a good notation >.<)

(only that you're not used to seeing it)

one argument for its favor is that you don't have to always flip back and forth between drawing diagrams and writing equations

as we usually draw diagrams left to right

A --f--> B --g--> C

[tender bough]

wow, neat

[rustic crown]

in this notation the image of a, would be (a)fg

[tender bough]

this really sells

[rustic crown]

another thing which you can do to break that weird symmetry is that

notice that an element a in A, is really a function a : * --->A

from the one element set

[tender bough]

😂 true, and?

[rustic crown]

so if you treat an element a, as the function then, the image of a under f : A --> B

is just the composition af

[tender bough]

composition as fuck indeed!

[rustic crown]

that gives you a function * --> B

🙈

ofc you can do this last thing in the usual notation as well

f(x) is just f composed with the function x : * --> X

but the order is reversed as usual

anyway, these are soft issues. ignore them :p

[rustic crown]

for matrices, you could always take transpose of the whole equation to convert from one to other

[pastel cliff]

stupid question - given a ring R, is the fraction field Frac(R) what you get from localizing the whole ring...?

[chilly ocean]

fraction field of some arbitrary ring?

you mean integral domain for sure?

[pastel cliff]

yes integral domain

[rustic crown]

yep, and in that case the non-zero elements form a multiplicative set

(equivalently localizing at the prime ideal (0))

[pastel cliff]

my next silly questions is how a localization that isnt Frac(R) differs from Frac(R)

like localizing a ring R at a subset S isn't necessarily a field is it

[rustic crown]

yea, if R is an integral domain, it won't be a field unless S = R \ {0}

for integral domains, localizations are subrings of Frac(R)

for non-integral domains, you need to be careful with inverting zero-divisors as say if you invert 3 in Z/6Z you get Z/2Z

one way to see that is by writing Z/6Z as Z/2Z * Z/3Z

inverting 3 is now same as inverting (1,0)

so inverting 0 in a ring kills that entire component

and you're left with Z/2Z

[pastel cliff]

ok i see that

but im still having a bit of trouble understanding Frac(R)

localizing a R - {0} means that that has to be multiplicatively closed

[rustic crown]

yep

[pastel cliff]

oh well i guess it already is by defn of a ring

my qualm was gonna be with the fact that not every r in that necessarily has an inverse

[rustic crown]

nu, definition of an integral domain

[pastel cliff]

ok but still - not every element has an inverse so what happens to elements that dont

[rustic crown]

in general complements of prime ideals are multiplicatively closed

[pastel cliff]

they get lumped into an equivalence class i guess?

[rustic crown]

didn't get you

[pastel cliff]

when localizing at R - {0} i mean

[rustic crown]

you adding reciprocals of every element (except 0)

so that does make every non-zero element invertible

r/s is non-zero if and only if r is not 0. in this case s/r makes sense and is an inverse

[pastel cliff]

ohhhhhh i keep forgetting this is technically just R x R

or well

R x R - {0}

[rustic crown]

and quotient by an equiv relation

:p

[pastel cliff]

my immediate question is then why i care about this

in my profs lecture notes, UFD's immediately follow this

not sure if they're related at all

[rustic crown]

they are

first fraction fields are nice

they completely formalize the construction of going from Z to Q

one place where they are really useful is as you say... for UFDs

you have this theorem that if D is a UFD then so is D[x]

and proving this directly can be hard

what you do is look at F = Frac D and compare D[x] with F[x]

and since F is a field, F[x] is a euclidean domain

which is ofc a UFD

[pastel cliff]

:O

[rustic crown]

then there is this one important lemma by gauss which tells you how to go back and forth between factorizations in D[x] and those in F[x]

[pastel cliff]

love how you just know this off the top of your head

[rustic crown]

i've studied this a couple of times by now lol

like at least 4-5 :p

[pastel cliff]

oh goddamn lmao

yeah this is my second algebra coursee

[rustic crown]

later ones are when i taught someone else :p

[pastel cliff]

first one was fake tho

[rustic crown]

oh i can see another reason to care about fraction fields

so do you know how to prove that det is a multiplicative function?

so the setting is that you have a commutative ring R, and so you define the map det : Mat(n, R) --> R in the usual way

why does this satisfy det(AB) = det(A)det(B)

tbh, i don't like this proof much. but it can be a really useful way to think

[pastel cliff]

i mean i know it's a homomorphism

[rustic crown]

of underlying monoids?

[pastel cliff]

eugh

[rustic crown]

but to prove that, you need to prove that det is multiplicative >.<

the cool thing is that proving it for fields is enough to prove it for arbitrary rings!

which may seem surprising

here i'm not even asking that R is an integral domain

(but hopefully for those, you already see a proof as you can embed them into Frac R)

the idea is that, you look at det in the integral domain D = Z[a_ij, b_ij] where i,j vary over {1, .., n}, so in total there are 2n^2 variables

pick A = (a_ij) and B = (b_ij)

if you prove det is multiplicative for this specific D with these specific matrices, then you've actually proven it for all rings and all matrices!

because given any R and matrices (a'_ij) and (b'_ij)

you can define the map D --> R

which sends a_ij to a'_ij and b_ij to b'_ij

and since det in the end is defined by a huge polynomial formula, it will commute with ring homomorphisms

(another way to say that is det is a natural map between the functors Ring --> Monoid, given by R --> Mat(n, R) and R --> R)

so it reduces to show it for this D and these A, B

but now D is an integral domain!!

so you can embed it into its field of fractions F and prove it there

(in fact D is a UFD, which is why this method can be really powerful)

so all this above stuff reduces to prove that det is multiplicative for matrices over fields

and for this you can follow your favorite proof. the first one i saw is the one i don't like much :p. it reduces to prove it when A is an elementary matrix and checks it by properties of det and how actions of elementary matrices work

notice that this elementary matrix proof only works for fields, as over general rings you can't decompose any invertible matrix as a product of elementary ones

okie me will sleep soon

[chilly ocean]

good night

[rustic crown]

Good night uwu :3

[unique valve]

math is wild

[twilit steeple]

yea, it can be very awesome behind all of its complex patterns

[chilly ocean]

I don't know what that means

Complex patterns?

[twilit steeple]

I don't know I guess it means something about how apples grow on bookshelves

[pastel cliff]

trying to prove this - my idea is to take the kernel of phi and define a chain of ideals containing it, and then show that it's contained by the empty set (like the maximal ideal has to be the empty set) so it has to be empty too but i cant seem to get this to work, any ideas?

[pastel cliff]

also det sorry i went afk still super helpful as always tho

[next obsidian]

show that the kernel has to be trivial

idk, if I say what to do the entire problem will be trivalized

think harder I guess? Idk

I don't want to be rude

If you think about it for >=2 hours and still can't get it then ping me and I'll give you a hint (which will kinda spoil the problem but oh well)

[pastel cliff]

i moved on to other problems i'll come back to it in a bit

was my idea somewhat right tho? ngl i was just kinda forcing in some usage of the fact that it's noetherian

[next obsidian]

I don’t think understand what you mean by the maximal ideal being the empty set

Like, that doesn’t make sense

At some point you do have to use that R is Noetherian

[pastel cliff]

mm fair enough

i'll be back

[pastel cliff]

ok different question - being asked to prove that any field must contain $\Q$ or $\Z/p$ for some prime $p$.

[cloud walrus]

sebbb

[pastel cliff]

idek where to start there tbh

[thorn delta]

start with 1 and look at the subfield it generates

or if you prefer, start with the unique ring homomorphism Z --> F, and examine possible options

[delicate bloom]

another hint: ||what's the characteristic of your field to start with?||

[rustic minnow]

If we say that something is integral over a ring is this the same as saying that there exists an integral extension of a ring?

[chilly ocean]

the first one is specifying what is integral over the ring, the second one is just saying there is something integral over it

[rustic minnow]

yes, on further reading I believe if we have an integral extension of a ring then all the elements of the ring are integral over the subring and like you said something being integral over a ring just describes that something being integral over a subring.

[chilly ocean]

"if we have an integral extension of a ring then all the elements of the ring are integral over the subring"

this is just the definition of an integral extension

[wooden ember]

If you want to find a polynomial subring of a polynomial algebra in general, is the most reliable way to do it to just follow the proof of Noetherian normalization and construct a sequence of integral extensions until we get down to a polynomial ring, or are there contexts where there are quicker methods

[chilly ocean]

What?

[rotund aurora]

Is this how you characterize inversion of elements in a monoid? Ofc then you would have to show that such a G exists

[rustic crown]

Yee looks good

[rotund aurora]

The situation for rings makes it confusing, because you have zero

that extends to rings just fine no?

[rustic crown]

Yep it does.

Commutative rings more specifically

[rotund aurora]

for noncommutative setting that's still fine no?

[rustic crown]

Maybe. Idk I never encountered it in the wild till now.

We probably can't use the usual fraction construction ig

[rotund aurora]

mmh I think Im still confused

wait nvm Im not

uhm no Im still am

the idea is that if as=bs does not imply a=b, and you want to invert s, then a and b should have the same image

but what is that image? Like is it necessarily the identity?

[rustic crown]

This isn't really a problem because when you forcefully invert weird elements like zero divisors, the ring just collapses. For instance if you inverted 0, then the localisation you'll have 0=1, which means it's zero

[rotund aurora]

in a ring you would consider the ideal rs=0 for r in R, and then send such r to zero, so if as=bs, then the image of a and b are the same. But what is it?

[rustic crown]

Just a/1

Which is same as b/1

a/1 = as/s = bs/s = b/1

[rotund aurora]

yes

but you give no other information as to how a/1 and b/1 interact with other elements? Like, would a/1 x=x for some other x necessarily or something like that?

ok no

I guess the image of a is independent of everything else and is only subject to the relation a=b whenever as=bs in the orignal monoid. And that's the whole point

[rustic crown]

Yep

[rotund aurora]

and subject to the relations in the original monoid ofc

[rustic crown]

Wait we were doing rings >.<

[rotund aurora]

its the same

xD

[rustic crown]

hehe :p

[rotund aurora]

I was only worried about the multiplicative structure of the image of the ring R, and these are monoids

thats why I said monoid, but could have said "ring" ig

[chilly ocean]

not a big distinction though there are less monoids coming from a ring than there is monoids, so some theorems might not hold true

[rotund aurora]

uhh never thought about that

nice observation

Thanks

[rustic crown]

Ah yes

[rotund aurora]

For a moment I forgot psi (above) is a monoid homomorphism from M, so how the images of a and b will interact with the rest is given precisely by psi, and thats it

[wooden ember]

Im asking this cause I’m struggling to find a polynomial ring of which k[x1,x2,x3,y1,y2,y3]/(x1x2x3 + y1y2y3) is an integral extension

There’s probably a smart change of variable to make but I’m struggling

So far I’ve got that the residue class of y3 is integral over k[x1,x2,x3,u,v] with u and v given by y1-y3 and y2-y3 (and here I’m taking the residue classes obviously)

But not sure how to express that algebra as a quotient of a polynomial ring

[foggy merlin]

Guys I need a reality check
Is being a UFD equivalent to
(integral domain + (prime <=> irreducible))
I know being a UFD implies this, but does the reverse direction hold?

[slender hamlet]

For a non-commutative ring we can localise with respect to a multiplicative subset satisfying either the left Ore condition or the right Ore condition

Point is, there is a general construction called localising a category C with respect to a left multiplicative (or a right multiplicative) system S of morphisms which is basically a universal way to turn morphisms in S invertible. So basically the category S^-1C we obtain, say we denote it D, has a functor F from C->D which is essentially surjective and turns morphisms in S into isomorphisms, along with the universal property that any functor G: C->E turning morphisms in S into isomorphisms, factors uniquely through F.

Now treat a ring as an additive one-object category and localisation of categories reduces to localising a ring

For a non-commutative ring, the left multiplicative system condition is just left Ore condition and similarly for right Ore condition

Ok I guess I should state what the right Ore condition is:
A subset S of a ring R, closed under multiplication, satisfies Right Ore condition if:

1. For a in R, s in S, there exists t in S, r in R such that at=sr.
2. (Not required when R is a domain) If a in R, s in S such that sa=0, there exists t in S such that at=0.

Left ore condition defined similarly

[formal ermine]

,, \begin{align*}\operatorname{Tor}(M) &= \Set{m \in M | m \text{ is a torsion element}}\ &= \Set{m \in M | \exists r \in R \setminus \Set{ 0 } : rm = 0}\end{align*}

[cloud walrus]

yes yes yes no

[slender hamlet]

Is R commutative here? If so just take x=rs

[formal ermine]

omg I'm so fucking stupid lmao

yeah

R is commutative

I forgot that lol

thanks

[next obsidian]

No

You also need the ascending chain condition on principal ideals

Or alternatively (equivalently) that every element can be written as a product or primes

[foggy merlin]

gracias chair person

[formal ermine]

this seems more than simple but how can I show that the inverse of a torsion element is also a torsion element?

[elder wave]

Use the same ring element

[formal ermine]

yeah I thought about that already but then I'd have r.(-a) (for a in Tor(M) and r in R its w/e)

is r.(-a) = -(r.a)?

it'd seem too good to be true

[slender hamlet]

Yes

[formal ermine]

ok thanks

lemme try proving that

[elder wave]

it's directly from the properties you demand of r.m