groups-rings-fields

[silent bronze]

I’m not sure what’s lacking in the argument

It’s because there’s 2 maps isn’t it

I gotta use like f being bijective or smth weird

[round hull]

the argument is as it should be

but you weren't clear with the conclusion

[silent bronze]

Hmm

[round hull]

you need to keep track of the forall/there exists throughout the proof

and we started with "let g in G be such that θ(g) is the identity action. then... commutes for all x in G."

it doesn't make sense to then say, "hence x is in Z(G)"

[silent bronze]

Yea

The first sentence generalises x in G

The second only holds IF x lies in Z(G)

So we say that

[delicate orchid]

No. Your conclusion is that g is in Z(G)

[silent bronze]

For each g in G if the map (phi) sends x to itself then x lies in the ker (theta)

[delicate orchid]

No? g is in the kernel

[round hull]

raargh

Wew do the thing

[silent bronze]

Oh wait

Wait wait

[delicate orchid]

Because theta(g) is the identity map, which is the identity in Aut(G)

[silent bronze]

Kernel is all elements of G that map to the identity

[delicate orchid]

Hence g is in the kernel

[silent bronze]

What

[coral shale]

To do these kinda questions, I find it much easier if I write down every single definition I see symbolically

[silent bronze]

My head is gone

[delicate orchid]

Theta(g) itself is bijective and thus has trivial kernel

[coral shale]

It avoids this confusion

[coral shale]

At least to begin with for practice

[silent bronze]

I’m lost

[round hull]

the codomain of theta is the set of bijections

they're not homos and have no notion of kernel

[delicate orchid]

We’re not finding the kernel of theta(g). We don’t care about that

We’re finding the kernel of theta

[silent bronze]

Kernel of theta would be elements of G that map to the identity permutation in Sg

[delicate orchid]

“Group of bijections”

[coral shale]

Yes Ash

[delicate orchid]

The entire point of this exercise is to show that Inn(G) = G/Z(G) why would we map into Bij(G)

Anyway

[coral shale]

ker T = {g in G | forall x : g * x = x}
Agreed? * being conjugation

[silent bronze]

Yes

So Kernel of theta has elements of g, all good

And the centre of G

Also has elements of g

[delicate orchid]

That’s absoLUTELY not true

[silent bronze]

Oh man

BUT the kernel is elements of G

Is it nottttt

So g in G

[delicate orchid]

Omg

[silent bronze]

I’ve lost my minddddddd

[delicate orchid]

Ker T = {g in G : theta(g)(x) = x for all x in G} = {g in G : gxg^-1 = x for all x in G} = {g in G : gx = xg for all x in G} = Z(G)

[coral shale]

Since u already did the problem, see this quicker solution...

ker T = {g in G | T(g) = id}
= {g in G | forall x in G : T(g)(x) = id(x)}
= {g in G | forall x in G : T(g)(x) = x}
= {g in G | forall x in G : g * x = x}
= {g in G | forall x in G : gxg' = x}
= {g in G | forall x in G : gx = xg}
= Z(G)

oh lol wew

[silent bronze]

Wow

[delicate orchid]

Yours is better laid out

[round hull]

the actual source of confusion imo is between free and bound variables

but i could be wrong

[delicate orchid]

Nah it’s thinking about a map which maps things TO maps

[coral shale]

yes, so quantify everything (whether or not this is the confusion)

[silent bronze]

Possibly, I struggle with group actions on the group itself

[round hull]

agree to disagree

[delicate orchid]

I remember struggling with this in undergrad as well

[round hull]

they got that part immediately

[delicate orchid]

That seemed to be the only source of confusion for the past 5 minutes lol

[round hull]

.

[delicate orchid]

People can state things and not understand them

I do that with 95% of my messages

[tardy hedge]

truuuuueee

me

[silent bronze]

Oh man

[coral shale]

In undergrad, apart from quantification issues, maps to maps and sht confused me yh

just needs time to click

[delicate orchid]

Were you the summing over orbits guy I was talking with yesterday kiand

[silent bronze]

Maps are weird man

[coral shale]

The most straight forward way for me is to write down the set theoretical defns and plow thru

[tardy hedge]

no you were talking to me about fields yesterday

K(u) = K(u+a)

[silent bronze]

You wanna see a scarier question

[delicate orchid]

Correct and based

[coral shale]

like this. 1 step 1 step, every step is a small step i am sure about. Apply one defn or theorem at a time

[delicate orchid]

Sure

[silent bronze]

The final part

I legit did the entire question fine and just left (c) and (g)

[delicate orchid]

This follows immediately from what we’ve just done lol

[silent bronze]

Ye but it’ll take a second

[delicate orchid]

It’ll take a first ;)

[silent bronze]

For me to go into the headspace of like

Normal subgroup s

[tardy hedge]

I still find it cool that normal subgroups are kernels and vice versa

[delicate orchid]

No need for anywhere near that amount of thinking. This is my hint

[delicate orchid]

Kernels of CHARACTERS even (WHAAAATTT??)

[tardy hedge]

waaat

[delicate orchid]

That’s CRAZY mr lads tbh. I know chat I know settle down calm down

[tardy hedge]

me in 3 years will be so mind blown

[round hull]

chat 💀

i still can't get over it

[coral shale]

another confusion might have been letters

the order of the letters dont matter in that final definition

of Z(G)

Z(G) = {g in G | forall x in G : gx = xg}
= {x in G | forall g in G : gx = xg}

[silent bronze]

Isn’t it the first isomorphism theorem that’s like

[coral shale]

just because of the thing inside being symmetrical

[silent bronze]

The image of G is isomorphic to G/kernal

[delicate orchid]

I’m so happy with this hint it’s unreal

Yes. That is the first iso theorem

[silent bronze]

Hehe

[delicate orchid]

Oh look. The universal property of the quotient.

[coral shale]

discussions greatest moment...

[round hull]

oh fuck

that was good

[delicate orchid]

Ok here’s something you might actually find interesting: this is true of any subobject you can quotient out by

[silent bronze]

Oohh so for all normal subgroups

[delicate orchid]

Now we use part c

[low wyvern]

Is there a quicker way to prove a finite set is a subgroup than proving associativity for all elements?

[silent bronze]

Subgroup test

1. non empty
2. closed under inverses
3. closed under binary operation.

[low wyvern]

Ye, but I would still have to use the elements and it would still take ages

[silent bronze]

Uhhh, generalise the elements ?

[coral shale]

A subset B.
forall x in B : P(x)
=>
forall x in A : P(x)

its inherited.

[low wyvern]

oh ye ffs

lol

[coral shale]

thats what the subgroup test tells u

or how u prove it.

[low wyvern]

ye ik that but I have to find the inverses of each element then show they are in the set

which isn't much quicker tbh. Especially when theyre matrices

[coral shale]

show your specific problem

[delicate orchid]

If a, b is in H then ab^-1 is in H if you want to do it in one step

But yeah show the problem

[coral shale]

ur not makin sense

[delicate orchid]

They are. U ever tried to compute the inverse of a matrix?!? It makes me CRY

[coral shale]

wolfram does that

[low wyvern]

not in an exam

[delicate orchid]

Unless your group is infinite, your inverse will be some power of the matrix you started with

[low wyvern]

I can do this specific problem as it is the set of 2x2 rotation matrices. However my question stands. Ok thanks wew

[coral shale]

unrotate

[delicate orchid]

Well that’s an infinite group

[coral shale]

is inv

[low wyvern]

so it's easy but, I just wondered if it was weirder set, could it be done

[delicate orchid]

Yeah, and I bet those specific angles contain the rotation backwards as well

[low wyvern]

0,90,180,270

[delicate orchid]

Yup

[coral shale]

u said it. exam. they shouldn't give u stupid problems

[low wyvern]

Hopefully not but you never know

[delicate orchid]

180 backwards is 180, 90 backwards is 270

But at this point just show the whole thing is generated by the 90 degree rotation one

[coral shale]

if u have issue so everyone else

u do know, dont assume they could chuck u a 100x100

[low wyvern]

My aim is to get a high score though, and to do that, I need to rule out all negative possibilities

[coral shale]

well better start practicing on those 100x100's just in case

[low wyvern]

lol

[delicate orchid]

You don’t need to compute inverses for this problem if you just think about what these matrices are actually doing

[delicate orchid]

Or do this

[low wyvern]

That is quicker tbf

Thanks guys.

[low wyvern]

This was one of the ways they had it done, but they also said to draw a multiplication table and invoke the latin square property. Is this really a proof?

[coral shale]

why not

u need to show closed and has inverses

(and no empty)

[low wyvern]

it does do that, I just never really thought of it as a proof

[coral shale]

https://tenor.com/view/hypotenuse-a2b2-square-geometry-trigonometry-gif-18715277

[delicate orchid]

Dunno what a Latin is because I’m not a lawyer

[coral shale]

[low wyvern]

I never have seen this lol

[coral shale]

neither have i

[low wyvern]

Latin square property means that every element appears once in each column and row

so you get a group table basically

[coral shale]

is every latin square a group?

nah

every latin square

is a cyclic group

right

some groups are not latin squares then?

[low wyvern]

I think it is hold on tho theres a theorem

[coral shale]

yh, so i reckon all latin squares are groups

(cyclic ones)

but there are groups that arent

so this test is... well kinda useless in my view

cyclic groups are generally easy to spot

[coral shale]

eh am i being dumb

oh shit i am

. e a b c
e e a b c
a a e c b
b b c e a
c c b a e

k this is klein 4 which aint cyclic

nice talkin out my ass

[coral shale]

but then every latin square isnt necessarily a group

[low wyvern]

It's ok. It is nice to see my stupidity is spreading

[coral shale]

for starters you need the identity row/column

to be something precise

But beyond that... i dont feel sure

See this

https://math.stackexchange.com/questions/1350756/latin-square-property-sufficient

being latin square is necessary for a group

[low wyvern]

ye thats true

[coral shale]

but isnt sufficient

So this check is only good for seeing something isnt a group

[coral shale]

And so, this should be complete bullshit.

[low wyvern]

ye but remember we are talking about a subgroup so associativiy holds thus it works/

[coral shale]

uhh ok, fair enough

[low wyvern]

idk in general

[coral shale]

to check something is a subgroup, a latin square will tell you its closed

and everything has an inverse

so thats fine

e will be in every row

[low wyvern]

ye

[coral shale]

so that tells you x * ? = e has a solution

ite

but really it's like

more work than its worth

to complete the latin square ur computing the whole multiplication table

[low wyvern]

ye lots of multiplication still lol

[coral shale]

thats just brute force

(now i see why u asked your first question)

generally, we dont use this to check for subgroups

subgroup criterion way to go

[low wyvern]

I do use that, but I have only ever done it when a group is defined generally. I guess I have learnt somet from this though. Thanks

[tulip glacier]

Is there any intuition behind the tensor product of two vector spaces ?

[south patrol]

Yes - usually they are introduced in terms of bilinear maps

[prisma ibex]

beat me to it

[south patrol]

Given three vector spaces U,V,W, we can contemplate bilinear maps U x V -> W

[prisma ibex]

it's the universal way to turn bilinear maps into linear maps

[south patrol]

Now unfortunately bilinear maps aren't linear, so it'd be nice to rephrase it in terms of linear maps

And this space U (x) V has the property that bilinear maps U x V -> W correspond to linear maps U (x) V -> W

in a nice fashion

[south patrol]

Really that should be the important bit; you'll then see different constructions of the tensor product, but they basically just force this to be true

[tulip glacier]

Okay, this might sound like a dumb question, but how are bilinear maps not linear ?

[south patrol]

If you have any questions / want more details, i'm more than happy to expand on that

[prisma ibex]

they are not

[tulip glacier]

They are, by definition, linear in each of their two arguments.

[south patrol]

They're linear in each argument, but not overall.

So, for example, have you seen determinants?

[tulip glacier]

Hmm, what would it mean for them to be linear overall ?

(I've never really talked about linearity outside single-valued functions).

[south patrol]

Actually even determinants aren't really needed

[south patrol]

(be careful - these are all single-valued functions, but i guess yeah the point is multiple inputs)

[south patrol]

So uh here's a bit of intuition for example

Let's consider a function F which takes in the width, height and length of a cube and spits out your volume.

It's linear in each of its arguments

But it certainly isn't linear overall

Linear means F ( λ. (x,y,z)) = λ F(x,y,z)

whilst bilinear means you get that for each argument separately

[south patrol]

So for our F here, in fact like

F( λ.(x,y,z)) = F ( (λx, λy, λz)) = λ^3 F(x,y,z)

If you scale a cube up by some factor λ in terms of lengths, then the volume scales like λ^3 instead

(That make sense?)

[tulip glacier]

Okay, I see.

[south patrol]

So they are different concepts.

[tulip glacier]

So linear overall means that it reduces all arguments at the same time, so to speak.

Meanwhile bi/multilinear means that the "coefficient" is counted separately for each argument.

[south patrol]

Yeah like if you scale the input, the output scales accordingly

Rather than scaling each component separately

I think the example I gave of volume is a good one to keep in mind - in fact the determinant of a matrix / linear map is a multilinear map in the columns of the matrix for example for this reason

Anyway, so linear algebra is normally about linear maps, so it'd be nice to be able to talk about multilinear maps in terms of linear maps - and that's where the tensor product comes in

[slim kayak]

Maybe it also helps to think of V x W as just being the cartesian product of vector spaces for the moment

[delicate orchid]

Just realised the funny fact that bilinear maps out of modules over a Boolean ring are linear

[south patrol]

lol

[delicate orchid]

And in fact that’s a characterisation of Boolean rings

[slim kayak]

Category theory at it again, making ever more obtuse definitions for things. So sad

[tulip glacier]

Since you mentioned cartesian products, I actually initially thought that tensor products really were products in the category of vector spaces, but apparently not.

Which is what got me confused.

[tulip glacier]

Anyways, so is this the main motivation of a tensor product ?

Then, what do tensors have to do with elements of this weird vector space ?

[slim kayak]

Yeah no. But tensor products (of algebras) actually appears as an operation in a certain category

[slim kayak]

weird vector space means the tensor product?

[tulip glacier]

yup

I cannot see yet the connection between multidimensional arrays (my current intuition for tensors) and the elements of a tensor product of two arbitrary vector spaces.

[delicate orchid]

Tensors are elements of tensor products

[slim kayak]

They are elements/vectors of that space. Its more of a physicist thing to talk about tensors themselves

[delicate orchid]

By definition

[tulip glacier]

Yup, but I need a connection. Just like most vector spaces, while being abstract in definition, can intuitively be represented as sets of "arrows in space" (vectors), at least to a reasonable extent.

[slim kayak]

A tensor product, most naturally, is defined by its universal property. So you'd have to convince yourself that multidimensional arrays have the same property to see how they are "the same"

[noble lynx]

for a ring can ab = 1, but ba not be 1?

[south patrol]

Hm well one thing is that (if you also allow for taking duals) you can form lots of useful spaces like End(V) using duals and tensor products

[south patrol]

Yes

[noble lynx]

any examples?

[south patrol]

Here's a funny example - consider R, the group of sequences (a_0,a_1,...) of integers (under addition). Now let S = End(R) be the set of group homomorphisms R -> R. This forms a ring - you can add functions R -> R pointwise, but you can also multiply them just by composing functions. Here, the identity 1 is just the identity map R -> R

Now let b be the element of S which "shifts sequences to the right"

i.e. given some element (a_0,a_1,...) of R, b sends it to (0,a_0,a_1,...)

Let a be the element which shifts stuff to the left, so it sends (a_0,a_1,...) to (a_1,a_2,...)

Now ab is the function which shifts to the right then the left, so it is the identity function.

But ba sends (a_0,a_1,...) -> (a_1,a_2,...) -> (0,a_1,a_2,...)

So ba is not 1.

[tardy hedge]

Question 10. Does E have to be equal to F?

[delicate orchid]

I’m gonna be sick

[south patrol]

No.

e.g. K < K(x^2) < K(x)

[tardy hedge]

Ok

[lethal depot]

To see the relevance of multidimensional arrays in relation to tensor products, the following reference is useful to enable mindless number crunching. It's not the best way to conceptualise tensor product spaces but is useful to check a result on a computer or get a numerical result from a computer. https://en.wikipedia.org/wiki/Kronecker_product

[tulip glacier]

Thanks!

[tardy hedge]

Ok so basically do u need to show that E must be K(u^n) for some n, and then u can show u is algebraic over that

[south patrol]

Hm E needn't be of that form

[tardy hedge]

Dang it

[south patrol]

I think they are always of the form K(p(u)) though

for a rational function p

[lethal depot]

The dot product is a bilinear map so consider the dot product of (2, 2) = 2(1, 1).

[tardy hedge]

Elements of K(u) are like rational functions of u right

[tribal moss]

If the vector spaces are finite-dimensional, then recall that we can make a basis for the tensor product by choosing bases for the two vector spaces separately, and then taking the Cartesian product of those bases. So a basis vector of V otimes W is a pair of basis vectors v_i from V and w_j from W, which means that the coordinates of an arbitrary tensor in V otimes W can naturally be arranged as an multidimensional array indexed by i and j.

[tardy hedge]

Oh yea ok

[tulip glacier]

😮

That actually is really intuitive.

Thank you!

[lethal depot]

Take $\mathbb{R}$. 0 is not a basis element but 1 is. $1 \otimes 1$ is not a basis of $\mathbb{R}\otimes\mathbb{R}$, unless I've made a mistake?

[cloud walrus]

mikeliuk

[rose prism]

why not?

R tensor R is isomorphic to R so indeed

R tensor V = V in general

[lethal depot]

The Cartesian product is two dimensional but the tensor product space is only one dimensional?

[rose prism]

apparently

dimensions multiply

for tensor

like n-dim tensor m-dim is nm-dim

[tribal moss]

The cartesian product of {1} and {1} is {(1,1)} which is a one-element set.

The dimension of a vector space is how many vectors there are in a basis, not what each of the basis vectors looks like.

[delicate orchid]

this can easily be seen by what tropo said earlier about the basis of the tensor product

[lethal depot]

I guess $1\otimes 0$ is not an element of $\mathbb{R}\otimes\mathbb{R}$ ? I can't immediately see my blunder but guess I've blundered somewhere.

[cloud walrus]

mikeliuk

[delicate orchid]

it is but 1 (x) 0 is equal to 0 (x) 0

(1 (x) 0) = (1 (x) 0*0) = 0*(1 (x) 0) = (0*1) (x) 0 = 0 (x) 0

[lethal depot]

Ahh, the Kronecker product confirms as a nice sanity check too. 👍

[tulip glacier]

This actually made me ask about another intuition - that of an inner product.

I get that inner products somehow encode both the distance and orientation ("angle") between two elements of the vector space, but I've never seen how its universal property reflects that.

[delicate orchid]

oh I stopped thinking about them like that half a decade ago - they tell you how much two things "overlap". For example taking an inner product of something with respect to a basis element will tell you the multiplicity of that basis element in the decomposition of your thing

[tulip glacier]

Oooooh

This actually makes sense.

The most classic example being the dot product, it makes sense now why it has to do with projections etc.

I see, I see.

Thank you!

[coral spindle]

Might be worth mentioning that you really want your basis to be orthonormal in this interpretation

[delicate orchid]

yur

[coral spindle]

Which isn't a huge worry. Gram–Schmidt guarantees these things exist, and in nice circumstances you get it for free.

Like le epic character theory yee haw

[delicate orchid]

but even if it isn't the overlap interpretation holds

[coral spindle]

Yee (haw)

[cloud solar]

The number of zero divisors must be a power of a prime

[tulip glacier]

I have just one question - why aren't inner products commutative ?

A should overlap B just as much as B overlaps A.

(analogue to set intersection)

[delicate orchid]

it's symmetric up to conjugation

over all of the fields that matter

[tulip glacier]

Can inner products be defined on vector spaces whose field is not the complex numbers (or a subset of it, like the real numbers) ?

If yes, then in that case, how would the symmetry up to conjugation be phrased ?

[delicate orchid]

idk pick some automorphism from the Galois group

not my problem!

[tulip glacier]

Okay, I have no idea what that means. I'll probably leave this for later, I guess.

(The Galois group part)

[lethal depot]

It's interesting that we say symmetric in the arguments for inner products written with brackets but commutative for infix operators. Inner products are conjugate symmetric by definition and thus symmetric in the reals. https://en.wikipedia.org/wiki/Inner_product_space

[tulip glacier]

Okay, this made me think about another question (sorry if I am asking so many things!!!).

In the case of inner product spaces whose field is the complex numbers: what does it mean, intuitively, that the inner product of two vectors is complex ? How can vectors a and b overlap by exactly 3 + 4i, for example.

Similarly, why does, in this case, the inner product have to be conjugate symmetric, and not fully symmetric ?

[delicate orchid]

I think if you care about arbitary fields you just define some form on the space

[tulip glacier]

Like, theoretically, if A overlaps B by 3 + 4i, then B overlaps A by 3 - 4i. What does this mean ?

[delicate orchid]

like if you're going general on the field you might as well go general on the form as well

[tulip glacier]

Oh, I see. Thanks!

[tulip glacier]

(This must have a geometric / physical interpretation, especially given how much complex inner product spaces are used in physics)

But the funny fact is that in order to understand the physical interpreation of the mathematics (in this case, symmetry of inner products), I first need to understand the mathematics behind the physical interpretation. 😛

Any help would be appreciated.

Oh, it's obvious actually. 💀

If the vectors are complex, flipping them only actually means flipping them about their real part (that's the whole reason you would want a separation between C and R^2, I think) - so it makes sense that the complex part stays different, thus the inner product is not fully symmetric / commutative.

Is my intuition correct ?

[delicate orchid]

if it makes sense to you it makes sense

we're talking in heuristics here

tbh the only reason why we demand conjugate symmetry is because we're too CRAVEN to work with <x,x> being non-real

[tulip glacier]

Hmm, I see.

[delicate orchid]

"norms have to be real valued noooooo" silence...

[tulip glacier]

Though my brain still hurts about the overlap of two vectors being complex. And what hurts even more is that this is not just a mathematical thing done for fun, it actually has applications in physics (which I cannot understand yet 😐).

But I guess it is what it is, after working enough with inner product spaces I guess it will make more sense.

[delicate orchid]

it actually has applications in physics
oh right. I could NOT care less about that

the decomposition into a basis thing is more important

pick an orthonomal basis $v_1, ..., v_n$ using gram schmit then $w = \sum_{i=1}^n \langle v_i, w \rangle v_i$

[cloud walrus]

W;3w Lads Tbh

[tulip glacier]

What is w ?

[coral spindle]

any vector

[tulip glacier]

Ok, thanks

[crystal turtle]

Shouldn't that be <v_i, w> • v_i

[delicate orchid]

yeah

typo

[crystal turtle]

:swag:

[delicate orchid]

YOU are excused

[woeful sage]

this is the mathematician's turf kernel, we can't fight them there, let's get em next time

[thick carbon]

Let $G$ be a transitive permutation group of degree $n$ with permutation character $\pi$. Prove, for any irreducible character $\chi$ of $G$, $\langle \pi,\chi \rangle \le \chi(1)$, with equality iff the kernel of $\chi$ contains all the stabilisers of $G$.s

Can anyone help me get started on this

Oops one sec

[cloud walrus]

Faq

[delicate orchid]

oh this sounds fun

$\langle \pi, \chi \rangle := \frac{1}{|G|}\sum_{g \in G} \text{Fix}(g)\overline{\chi}(g) \leq |X/G|\overline{\chi(1)} = \chi(1)$

hmmm

that's |G| times <1, chi>

what did I do wrong here one sec

[rocky cloak]

First equality seems to assume chi(g) is constant

[delicate orchid]

yeah there we go

well wait no this can still work

|\chi(g)| \leq \chi(1)

[rocky cloak]

That's pretty slick

[delicate orchid]

transitive so the number of orbits is 1

still don't like this

[rocky cloak]

I think you mean to remove the sum there

Number of orbits equals average number of fixed points

[long obsidian]

I can't remember how to find a field that a ring sits inside. If I have a ring R, is the set R\setminus {0} a multiplicative subset and then I form the set of equivalence classes (x,y) and x in R and y in R with nonzero y!=0 via localization or whatever and this set of equivalence classes is a field?

Does this always work or do I need to work about stuff like whether R is an integral domain? I know I just have to repeat the construction of Q out of Z but Z is an integral domain to begin with.

[delicate orchid]

right, yes

[rocky cloak]

Any ring that sits inside a field is an integral domain

[delicate orchid]

it's so nice when someone asks a question in the same microfield my research is in it makes me look so clever

[long obsidian]

Oooh okay.

How about this say I had Z[x,x^1] is the smallest field this sits inside something predicable like Z(x)?

[delicate orchid]

hints! Burnside's lemma, and then try and bound \chi(g)

[rocky cloak]

It sits inside Q(x), which should be quite predictable yes.

For any integral domain it's the construction you described earlier. I.e. all fractions a/b with a and b in your ring and b nonzero.

[rotund aurora]

If H is normal in K, K is normal in G, and both K/H and G/K are abelian, does it follow that H is normal in G and G/H is abelian?

I don't think so?

[long obsidian]

Um okay well Z[x,x^-1] contains Z so I can see why it needs Q. Why isn't it all of Q(x)? Uhh so the elements of the field of fractions of Z[x,x^1] should be something like a fraction p(x,x^1)/q(x,x^1). Is the issue elements like (1/2)x+1 because it has an indeterminate and a non integer coefficient?

[rocky cloak]

The smallest field that Z[x^±] sits inside of is Q(x) yes.

(1/2)x + 1 is (x + 2)/2 for example

[long obsidian]

Ahh okay I see you can make the non integer coefficients work by adjusting stuff like that

[thick carbon]

Gotcha, I’ll try that. Thanks!

[tribal moss]

Let G be a (not too small) dihedral group, K the rotations, and H = {1}. That's normal in G alright, but G/H is not abelian.

[rotund aurora]

lol right

I was wondering about the definition of a solvable group. Does it follow that each G_i is normal in G?

[rocky cloak]

It does not follow

[long obsidian]

What is the algebraic closure of something like Q(x)? I've seen the term puiseux series before but I'm confused. They look like power series with fractional exponents. But I'm confused about stuff like whether there is abound on the exponents?

[rocky cloak]

One example could be
1 < C2 < V4 < A4
Where V4 is the Klein 4 group. Here C2 is not normal in A4. (Though here V4 is abelian, so maybe not the perfect example for what you're asking.)

[rotund aurora]

thanks

[rocky cloak]

The puiseux series look like

f(x^(1/n)) / x^m

Where f is a power series with complex coefficients, and n and m are arbitrary positive integers.

[long obsidian]

And this is what a general element of the algebraic closure of Q(x) is? Just to confirm the algebraic closure of Q(x) are infact these puiseux series?

[rocky cloak]

I believe so yes, just replace 'complex coefficients' with coefficients in the algebraic closure of Q.

[white oxide]

is the converse of this true?

namely if [L1L2: K] = [L1: K][L2: K] do we have [L1: K] and [L2: K] relatively prime

[south patrol]

Q(i)/Q and Q(sqrt(-2)/Q have degree 2, but Q(sqrt(-2),i)/Q is degree 4

I think the point is that n_1 and n_2 being relatively prime forces the intersection to be K again

But you could have intersection K for some other reason (such as in the example I gave)

[white oxide]

dammit

because if the converse held this wouldn't be so bad to prove i guess

[delicate orchid]

lol

[spice whale]

does this account for stuff like p⁵ - p - 1 which have solutions inexpressible in radicals

[rocky cloak]

Yes, the algebraic closure includes roots to so polynomials.

[white oxide]

can i get a hint for this one? i've been like playing with degrees and trying to show that [L1 n L2: K] = 1 but i've been stuck

[spice whale]

no i mean the puiseux series

would there be a puiseux series that satisfies the equation x⁵-x-1 = 0

[rocky cloak]

That would just be the constant sequence, equal to the root of x^5 - x - 1

[spice whale]

ok that's a bad example

[rocky cloak]

But yeah, the algebraic closure contains root of any polynomial equation

[spice whale]

no I'm asking can you really represent them all as puiseux series

that seems odd to me

[white oxide]

pleaseee

[delicate orchid]

I literally have no idea soz boss. I can see it if you pass to galois groups lol

[white oxide]

nah this book hasn't even gone over galois groups yet

[delicate orchid]

|HK| = |H||K|/|H \cap K|

for H and K subgroups of some G

I imagine the proof is similar here

[dim widget]

Try to prove the contra positive

[south patrol]

why lol

[delicate orchid]

because u cited the exact result he was trying to prove without knowing it

[white oxide]

okay i'll try that thanks

[south patrol]

oh lmao

I thought it was like a theorem rather than a problem

fair

[next obsidian]

You’re a contra positive

[summer path]

You're not a 🐔 🐒

[delicate orchid]

el pollo mono

[rocky cloak]

The field of puiseux series is algebraically closed

https://math.stackexchange.com/a/59837/306319

[white oxide]

oh wait yeah wouldn't this exercise follow immediately from the correspondence theorem

but i'm not allowed to use that

[slim kayak]

I didnt know field (extensions) had their own correspondence theorem

[delicate orchid]

find me the page in the united states constitution that disallows this

also I don't know if Gal(L_1L_2) is Gal(L_1)Gal(L_2). I do not know this

[white oxide]

oh i guess i was talking about the fundamental theorem of galois theory

[delicate orchid]

and now I know this

[white oxide]

i can't but it's unethical 100%

this is a stupid counting exercise

i hate counting exercises

because i can't count past 4

[dim widget]

What does the contra positive of this statement mean?

[white oxide]

huh

[dim widget]

You get to assume what and you’re trying to prove what?

[white oxide]

how is (*) true because [L2L1: L1][L1: K] is greater than or equal to [L1: K] no

[white oxide]

i mean just L1 n L2 = K and that the degrees aren't' equal to each other right after the prove that

[slim kayak]

check your earlier image

[dim widget]

No

In the contra positive you assume L1 \cap L_2 = E where E is not K

[white oxide]

Oops

I meant L1 n L2 not equal to K spacing out

[dim widget]

So if they are both E vector spaces what does that tell you about the compositum?

[celest furnace]

W

I’m so excited to learn this stuff in a couple weeks

Soon I will know why fifth degree polynomials aren’t solvable

[long obsidian]

Suppose I had a ring Z[x,x^-1] and another Z[y,y^1] both sit inside Q(x,y). Is there some natural way to consider the smallest ring inside Q(x,y) that contains both Z[x,x^-1] and Z[y,y^-1]? I wanna say the smallest ring should be Z[x,y,x^-1,y^-1] but I'm not sure

[slim kayak]

Well, can you show that such a ring must contain all monomials?

like the elements x^n y^m with n and m integers

Any subring containing Z[x,x^-1] and Z[y,y^1] contains Z[x,y,x^-1,y^-1], and Z[x,y,x^-1,y^-1] happens to contain the former two, hence it is minimum. And the intersection over a sets with a minimum is just said minimum. Assuming how you want to know how to formulate it

[south patrol]

Yes by definition essentially

Z[a] is the smallest subring of (bigger ring) containing a

And the same works for any collection of elements

[long obsidian]

Right okay I think I see what you mean it probably works out because any such subring would contain the generators of Z[x,x^-1,y,y^-1] as an algebra.

But ohh potato actually I think that is an exercise I've seen before in hungerford. That is the way there are defined. But is this notation Z[a] specific to subrings of a polynomial ring?

[slim kayak]

should just be the algebra generated by a

not literal polynomials, but if you have a you have a you have -a, then you can build up n times a. You also have a^2 and -a^2 ...

you get the gist

[cloud solar]

Silly question: the polynomial f=X^2011+2010X with f in Z/2011Z[X] is x(x-1)(x-2)...(x-2010)?

[long obsidian]

Okay that makes sense but the algebra generated by a has any power a^n by being an algebra and then is closed under sums so seems like it gives you any polynomial in a but yeah I think I get the jist thanks.

[slim kayak]

probably not?

nvm i was trippin

[cloud solar]

From little fermat theorem every element from Z/2011Z is a root of f right?

[tawny dune]

yes that is correct

[south patrol]

Yes this is correct

Both have the same roots 0,1,...,2010 and both are monic of the same degree, so they coincide @cloud solar

More generally, over $\mathbb Z/p$ we have an equality $x^{p} - 1 = \prod_{k=1}^{p-1} (x-k)$. Corollary: $-1 = (p-1)! \mod p$, by plugging in $x = 0$

:)

[cloud walrus]

potato

[slim kayak]

Why does a maximal multplicatively closed set (without 0) contain prime ideals in its complement?

[rose prism]

such a set S has the property xy in S => x,y in S

[slim kayak]

saturated right? But why?

[rose prism]

if not then it wouldnt be maximal

as you could just add x and y

and all their products with the other elts in S

like

suppose $xy\in S$ and $y\in S$ but $x\not\in S$. then $xS\cup S$ is a multiplicative set without $0$ that properly contains $S$

[cloud walrus]

memorylessfunctor

[rose prism]

i think this works

ok it does not

[slim kayak]

right, x cant be the zero divisor of anything in S

[rose prism]

but i think you get the point

something like that yeah

[slim kayak]

Okay thanks 👍

[rocky cloak]

Prime ideals in the localization exactly corresponds to prime ideals that don't intersect your multiplicatively closed set. So every multiplicatively closed set (without 0) contains a prime ideal in its complement.

[slim kayak]

Right, if they did intersect they would have units - hence not prime. And by zorn there are prime ideals in the localization, so just pull it back with the pre-image of the map A -> S^-1A?

[rocky cloak]

A short proof of existence:
(0) is an ideal that doesn't intersect S, so by Zorn's lemma there's a maximal such ideal. Call it p.

Consider x and y not in p. By maximality p + (x) and p+(y) must both intersect S. Let's say a + rx and b + sy are in S. Then
ab + sya + rxb + rsxy
is in S. Since ab + sya + rxb is in p, xy cannot be in p. => p is prime.

[slim kayak]

Yeah

I think I am maybe missing some intuition somewhere here, like thinking of checking inside S^-1A

[rocky cloak]

Yeah, if you have a multiplicatively closed set, your thoughts should wander toward the localization.

[slim kayak]

yeah...

[spark veldt]

in proof of (c) last line, how does cancellation law show they're distinct because n is the smallest integer?

[mighty kiln]

If x^a = x^b, then x^(a-b) = 1

[spark veldt]

Yeah i get it. Like, they're distinct because $x\neq x^{n-1}\implies x^{2-n}\neq 1$. How does $x^{2-n}\neq 1$ mean that they're not distinct?

[cloud walrus]

bluepianist

[spark veldt]

Can't u have a negative exponent?

[icy bear]

oh wait

[icy bear]

because if they were, x^a = x^b, then x^(a-b) = 1

if they were, it would be equal to one

[dire siren]

you don't have to deal with negative exponents
if x^a=x^b and a>b, then x^(a-b)=1 [in case a<b, just write x^(b-a)=1]
the key idea is that you found a positive integer k **smaller **than n with x^k=1, but n was the smallest positive integer with that property

[spark veldt]

ohh i see

contradicting the fact that a cyclic group with order n, n is the least integer for which x^n=1

got it, thankss!

[icy bear]

if two groups have the same amount of elements of each order, does that mean they are isomorphic?

or well, at least of each prime order

[winter shore]

very interesting question, it seems not

see https://mathoverflow.net/questions/39848/finite-groups-with-elements-of-the-same-order

[icy bear]

in the prime case I think yes cus

well no actually

I need to reformulate

well

if all of the elements have prime order (excluding the identity), like in a group of order 6 with no element of order 6, it seems to me that since the set of each element with a different order, like one of order 3 and other of order 2 generates the group, an easy homomorphism that turns each "basis" element to each "basis" element of the same order would be an isomorphism yk

I think they're actually called generators

in that context

[winter shore]

yes, that argument proves a much weaker statement though: that if a group is generated by a set of elements of order pi, with all pi primes and pairwise different, then G is uniquely determined
this is completely wrong

(in fact, it is the multiplication of the cyclic groups of order pi)

[icy bear]

what's pairwise different?

[winter shore]

pi =/= pj, whenever i=/=j

[icy bear]

oh

[icy bear]

wait so

I'm trying to classify the groups of order 6

[winter shore]

wait

I'm wrong

let me correct

[icy bear]

wouldn't that imply that if a group of order 6 has an element of order 3, than it is cyclic?

(which isn't true)

[winter shore]

So let G be generated by $x_i$ with each $x_i$ having order $p_i$. Let $G'$ be generated by $x'_i$ with each $x'_i$ having order $p'_i$.

[cloud walrus]

Shiranai

[winter shore]

We want define a map by $\prod x_j \mapsto \prod x'_j$

[cloud walrus]

Shiranai

[white oxide]

do we get r = 0 because we're assuming g is minimal with regards to f | gh?

also where do we use the fact that g's degree is less than f?

[rocky cloak]

If you just want to classify groups of order 6, then I'd show it has one element x of order 3 and one element y of order 2, then think about what yxy could be

[icy bear]

I am doing something similar right now

[winter shore]

I think that's not a homomorphism in general

which is why the argument doesn't work

the map is an injective homomorphism precisely on products of cyclic groups, which lead me to the previous wrong conclusion

anyways, as jagr mentioned, there's just three possible options on what y-1xy could be, one is contradictory so you don't even have to check the other two

[white oxide]

can somebody explain to me how 20 implies that either f_2 or f_1 is in K?

[winter shore]

if you know C6 is not iso to S3

[white oxide]

because we know that f|f_1 or f|f_2. but i'm not entirely sure how this implies that f_1 is a unit or f_2 is a unit

because there are nonconstant polynomials in the ideal (f)

[celest furnace]

I think that last line—degree argument should show that deg g is 0

[white oxide]

yea there's a lot cleaner proof involving prime ideals

[celest furnace]

Nice yeah I thought this was weird

[white oxide]

this book lowkey gives not like the other girls vibes

but mostly it's pretty good

[white oxide]

oh i think i see - we have f_1 divides f, if f_1(a) = 0, f divides f_1, but deg f_1 <= deg f which is the minimal polynomial for a

uhhhhh hence deg f_1 is a unit

nvm that doesn't work

lol

ah

f_1 | f, f_1(a) = 0 => f | f_1, hence f = f_1f_2 = ff_2 => f_2 is in K

[tardy hedge]

Wait why r u learning that now

When u were doing so much more advanced stuff b4

[icy bear]

are C6 and S3 the only groups of order 6?

according to my calculations they are

just wanna be sure

[rustic crown]

yee, the phrase is "up to isomorphism"

[icy bear]

AAAAAAA I know

[rustic crown]

:3

[icy bear]

specially in the case of groups isomorphisms are so much "relabeling" that I don't even really think of isomorphisms sometimes

[rustic crown]

true, but sometimes it can be nice because isomorphisms are not always canonical

[icy bear]

wdym canonical

[rustic crown]

so like, in \bC consider the subgruop mu_n of nth roots of unity

[slim kayak]

plays nicely with certain maps, its a categorical thing

[icy bear]

category theory?

[rustic crown]

it's a cyclic group with order n, but to get an iso to Z/nZ, you need to choose a generator

which amounts to fixing one primitive nth root of unity

and there isn't a natural way to pick one out of others

simiarly, there are times, when you would want to think of a group that's naturally a multiplicative group as an additive group, isos make it uwu

[icy bear]

makes sense ig

but I generally forget the nature of the operation lmao

which sometimes bugs me when doing modular arithmetic with the usual additive Z/nZ cus I forget exponentiation isn't adding

[rustic crown]

yee, confusing between additive and multiplicative order is

[icy bear]

sad

[icy bear]

thank god for the existence of Z/pZ multiplicative

[slim kayak]

That's a ring tho, no?

Or do you mean $(Z/pZ)^{\times}$

[cloud walrus]

Kerr

[rustic crown]

also by CRT, you can write down explicitly what each (Z/nZ)* looks like

[slim kayak]

CRT?

[rustic crown]

chinese remainder theorem

[slim kayak]

Critical ring theory 🤔

[rustic crown]

(Z/2^kZ)* is generated by 3 and -1 and looks like Z/2^(k-2)Z ⊕ Z/2Z

[slim kayak]

Close enough

[rustic crown]

(Z/p^kZ)* are cyclic for odd p

and if you know the generator of (Z/pZ)*, then you also know a generator there

[rotund aurora]

ok yeah

but what is a generator

[rustic crown]

that is hard :3

[rotund aurora]

indeed

[rustic crown]

if g is a generator mod p, then g or g+p works

[rotund aurora]

[rustic crown]

.<

[icy bear]

idk

I know it's a group tho

[icy bear]

prob

[merry dawn]

guys, can u describe in a few words how commutative algebra differs from a first course in groups/rings/fields (usually called abstract algebra), or if the number of differences outnumbers the similarities substantially, then what is it about?

There is a wiki article 'commutative algebra topics' and here is a sample of it

I mean, we've studied all of it in abstract algebra 🤔

[slim kayak]

you covered general rings, commutative algebra goes into deeper theory

you are more likely to see homological algebra shoved in somewhere. Deal with general module theory, localization, special classes of rings and some other stuff

[tribal moss]

My impression is there's no crisp dividing line, there's just too much of it for one course, so the first course has to stop somewhere.

[coral shale]

so its just more algebra

[slim kayak]

now it commutes

[coral shale]

no, all rings commute

and have 1

[slim kayak]

lol

[white oxide]

oh i'm just relearning it

it's part of a book titled fields and galois theory

it says it's also an algebra book but it's definitely not a first time algebra book if somebody's looking for one

i also want to refresh my knowledge on these things, i'm not that strong in it i think/definitely haven't mastered it

[icy bear]

what does classifying mean in that context

[white oxide]

what book is this

[icy bear]

algebra by artin

[white oxide]

oh nice

[icy bear]

maybe this is just a more abstract question maybe I should just investigate

🔎

[rotund aurora]

as semigroups

[icy bear]

what?

the identity doesn't need to be unique in a semigroup right?

wait no it does

need

e' = e'.e = e

[rotund aurora]

I mean, ig in part the exercise is to figure out what it means to classify semigroups

but it's a concrete question

[icy bear]

maybe yeah

maybe it's just the usual isomorphism thing

like, with the function preserving the law of composition

or not

[rotund aurora]

yes but you should also impose that the identity is brought ot the identity

[icy bear]

I'll just investigate 🔍

[icy bear]

wdym

oh right

no I don't need to impose it

[rotund aurora]

in the case of groups you can derive it as a corollary of preserving multiplication, I don't know if it is the case for semigroups

[icy bear]

it is

[slim kayak]

write down all up to isomorphism

[icy bear]

f(a) = f(a.1) = f(a).f(1)
and as proven above, the identity is unique, so f(1) needs to be the identity

[rotund aurora]

wrong

you assumed surjectivity

[icy bear]

ohh

yeah

[slim kayak]

90% of the time isomorphisms are all you care about. The other 10% of the time natural isos do the job

[icy bear]

what

wdym natural isomorphism

[icy bear]

well for an isomorphism I can assume that anyways which is what I care about

[slim kayak]

isomorphism with a little bit more spice, doesnt matter for now

[icy bear]

oh sure

[rotund aurora]

oh well yeah, in that case the identity is preserved

[icy bear]

I'm curious now, I don't need that for the exercise but I'm curious about the properties of a semigroup "homomorphism"

like, is the kernel a semigroup and stuff

[rotund aurora]

like, this is how in rings you say S is a subring of R if blah blah AND the multiplicative identity of S is the same as the multiplicative identity of R

[slim kayak]

you are meant to write down all associative laws of composition... actually have you introduced semigroup homomorphisms?

[icy bear]

I invented it rn

[slim kayak]

hold on thats just a monoid

[icy bear]

idk what a monoid is

[slim kayak]

its a set with a law of composition

It has associativity and an unique identity element

[icy bear]

oh

I heard this word before

monoid

[slim kayak]

they pop up here and there

[rotund aurora]

ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

[icy bear]

that's a lot of integers

are you ok

ok so by now I have proved that at least up to 1, s, s², finite cyclic semigroups are groups

[slim kayak]

good morning sunshin

[icy bear]

because of the uniqueness of the identity the compositions are quite limited

[rotund aurora]

I'm saying that monoids are ubiquitous in number theory, just that you don't call it like that, cuz there is no point

[slim kayak]

what about sending everything to 1

[icy bear]

oh

leme see

wait that'd just be {1} wouldn't it

1.s = 1
s would be an identity, so s is just 1

I think I have proven something look

[white oxide]

how would we show that L can expressed as in (22) if R is not of the form K[a]?

because doesn't the argument necessarily involve some simple extension

[icy bear]

if it's a semigroup {1, ..., s^n}, than if s^n*s = s^k for any k less than or equal to n and not 0, I can separate s^n as
s^k*s^(n-k)*s = s^k
which would mean s^(n-k+1) would be an identity, so if all of those s^something <=n aren't the identity, that's a contradiction, and by the definition of the set (as I'm defining it cus I can>:)) k can't be greater than n, so it's 0, which means s^(n+1) is the identity, which means these finite generated sets are actually cyclic

is my proof correct?

[icy bear]

by cyclic I mean they really cycle back to the identity

after some n

[slim kayak]

What L? The fraction field?

[white oxide]

yea like

in the general definition

[slim kayak]

As long as your R is an integral domain you can pull the same stunt

[white oxide]

i know the usual construction of the field of fractions

[slim kayak]

In this case one focuses on K[a] to later establish some results for algebraic and transcendental a

what is your question then

[static glen]

It's just the same proof, let $F={a/b : a,b\in R, b\neq 0}\subset L \subset E$ then $F$ is a subfield of $E$ with $R\subset F$. As $L$ is the smallest subfield of $E$ containing $R$, get $L\subset F$. Hence $L=F$

[white oxide]

i'm just curious as to how their proof works in that case

[cloud walrus]

Max

[white oxide]

oh

lol

nice thanks

[static glen]

no problem

[tardy hedge]

i see makes sense

[icy bear]

which means that s^k*s^(n-k+1) = s^(n+1) = 1, so all finite cyclic semigroups should be groups

[slim kayak]

hm

keep in mind that semigroups with identity must only have a unique two-sided identity

[icy bear]

ye

cus 1 = 1.1' = 1' still works

the argument

well the infinite 1-element generated can't in this context be groups cus otherwise they'd be finite

since you only have the positive exponents

ok sooo

the elements also can't repeat as usual

since s^k = s^n -> s^k-n*s^n = s^n which implies s^k-n = e which means k-n = 0 or would mean the semigroup is finite

so they're isomorphic to just the naturals

and if I'm correct I classified them

yey

[slim kayak]

Looks right

[icy bear]

ihuu

that was a fun exercise

just discovering stuff

that's why I like exercises which aren't just "prove x"

I think I made a mistake

I assumed that if for any a, a.b = a, b is the identity

need to check how that works for the cyclic semigroups

[rotund aurora]

This is wrong. s^(k-n)*s^n=s^n does not imply s^(k-n)=e

[icy bear]

ye I did realize that

hmmhmmhm

[rotund aurora]

For example, consider a nilpotent matrix and the semigroup it generates

[icy bear]

I have no idea what a nilpotent matrix is

[rotund aurora]

nilpotent means X^n=0 for some n

[icy bear]

oh

[rotund aurora]

I mean, it neednt be a matrix, you can just define a semigroup like that formally, but since nilpotent matrices are well known examples, I said that

@icy bear hint: ||consider the set of pairs of positive integers (x,y) such that s^x=s^y ||

[icy bear]

I suppose I could make some equivalence class with this

like with cosets

a + [x] -> s^a

then ig each semigroup would be related to some equivalence class

defined by the equivalence relation (x, y)

idk if all of them

I will find this out tomorrow

gn

[stable thistle]

is there any deep reason that the empty set should be a group but not a field?

[dull marsh]

The empty set is not a group though

[stable thistle]

i meant the singleton set mb

[dull marsh]

Fields are required to have additive and multiplicative identities which are also distinct

So in a field you should have at least two elements

[stable thistle]

yes but this doesn't rule out anything except for a singleton

i know what the axioms are i'm asking why should they be that way

[dull marsh]

So that the zero ring doesn't count as a field

[stable thistle]

sure but why

[prisma ibex]

Fields are required to have additive and multiplicative identities which are also distinct
this is why

[dull marsh]

They are asking why do we require that in the first place

[ivory trail]

because then you will have a lot of theorems that say "let F be a field that isn't the zero ring, then this happens"

[prisma ibex]

like even if you allow the zero ring to be a field, the usual theorems about fields don't apply to this at all

[dull marsh]

Looks like you would need to e.g. change how the theorems about maximal ideals are formulated

[prisma ibex]

all of Galois theory breaks, etc

[ivory trail]

for example, vector spaces no longer have a unique dimenison

[dull marsh]

And the zero vector space starts having ambiguous dimension

[prisma ibex]

there is supposed to be something like the field with one element where vector spaces over this are sets and so on but this isn't the zero ring

so yeah you don't allow the zero ring

[stable thistle]

i see

[cobalt heath]

What even is that

[prisma ibex]

what even is that
NOBODY KNOWS!

[cobalt heath]

Where can I read about this?

[prisma ibex]

nowhere because there isn't a good theory of this

you can read Manin and others for some explaination of what such a theory should satisfy

if you can find the right formulation of F_1 and prove enough things about it then you can prove RH

[rotund aurora]

Would it generalize to GRH?

[cobalt heath]

When something with one element is so nontrivial

[prisma ibex]

yes

the issue is that nobody knows what "Spec(Z) x_{Spec(F_1)} Spec(Z)" should be

by now we kind of know what this looks like "completed at (p,p)" but basically nothing else

[cobalt heath]

So basically, F_1 is discussed in terms of Spec(F_1)?

[prisma ibex]

not necessarily

[cobalt heath]

Wait wh

[prisma ibex]

I mean ideally if you know how to define one you should be able to define the other

[cobalt heath]

I did not expect Quantum integer here

Hmmm

[prisma ibex]

but for example there are lots of notions that work for F_q that make sense in the limit q->1

and there should be something which makes sense of this

the dictionary between function fields over a finite field and number fields suggests that number fields should be function fields over F_1

[cobalt heath]

This sounds like a good candidate for Gödel's incometeness

[prisma ibex]

no

that's stupid

[cobalt heath]

Wouldn't it be fun

[prisma ibex]

nah

[cobalt heath]

Why notttt

[prisma ibex]

because it has nothing to do with the actual goals of F_1 geometry

[cobalt heath]

I mean, it might quelch ppl's hopeful thinking

[prisma ibex]

you have your head on backwards

[ivory trail]

but is clearly 3 dimensional and not a curve

[prisma ibex]

apparently Scholze is "hopeful" but Clausen is "skeptical" that their new formalism for analytic geometry can say the right things about F_1

[cobalt heath]

There is a new formalism on this?

[prisma ibex]

yes

all the previous formalisms of analytic geometry cannot treat Archimedean and non-Archimedean geometry on equal footing

Scholze and Clausen can

[cobalt heath]

Damn, mathematics can go pretty deep

[prisma ibex]

it's a fairly deep formalism yeah

in particular it requires doing away with topological spaces

and replacing them with something better

[cobalt heath]

I regret my decision to become a mathematician

[cobalt heath]

Hopefully simpler, I guess?

[prisma ibex]

not simpler

but better

[ivory trail]

they're condensing F_1? i haven't heard about this actually

[prisma ibex]

they're trying to

idk if it will actually pan out

Scholze right now is trying to define a sort of "global Fargues Fontaine curve"

which might be some insane analytic space in this formalism

we already know what this should look like at the p-adic places and at the Archimedean places

[rotund aurora]

bruhh stop lying this is not geometry

[prisma ibex]

it ought to be!

one should be able to geometrize the global Langlands correspondence over number fields like this

but that's very pie in the sky at the moment

[cobalt heath]

Is mathematics ever worthy to do for me at this point, like there is too much to study before any trivial publication

[prisma ibex]

no just get good

[cobalt heath]

Someone might actually "finish" math, it feels

[rotund aurora]

I think this kind of mindset will just draw you back

[prisma ibex]

nah

[cobalt heath]

True..

But it would be so hard to get a paper published at this point, I mean.

[prisma ibex]

there are lots of things in math that people consider "finished" but are absolutely not finished

a bulk of the stuff I work on was stuff that was "finished" in the 90s but nobody actually made explicit

[stable thistle]

i like abstract mathematics for its interestingness i don't really care about applications whether internal or external

[ivory trail]

"finishing math" or even coming "close" would involve a theory of everything for physics, but it is impossible to "finish math" imo

[cobalt heath]

Yea, it's just my feelings

[rotund aurora]

if it were possible to finish math, then you could ask to finish metamathematics

and then metametamathematics

etc

[stable thistle]

i'm a radical mathematical anti-realist, i think math is just the study of interesting consequences of sets of axioms

[rotund aurora]

but meta....metamathematics is still mathematics

[ivory trail]

there are so many unsolved puzzles left in physics that have yet to be studied deeply/mathematically

[cobalt heath]

Hmm

Even as a grad 1st yr student, I struggled with figuring out how a circle (from complex plane) can be described in R^2 inner product space. The one so trivial yet I had hard time with

Idk why I am suddenly ranting, sorry

Simply that there is massive skill gap between my level and the frontline of research

[celest furnace]

Ok

[rustic crown]

does anyone know how to cutely think of Emil Artin's argument about [F:F^G] <= |G| when G is finite subgroup of Aut(F) for some field F?

say |G| = n, and a0, ..., an in F are elements. we want to show these are linearly dependent over k = F^G. so consider the vector space V = F^(n+1) and then those elements define a vector a in V. we want to find x in V such that <x,a> = 0 and x is G invariant. Since g*x = x, these also satisfy <g*x, a> = 0 which means <x, g*a> = 0 for each g in G.

so the place to look for this x is the subspace W of V which is orthogonal to every g*a. since these are n vectors in an (n+1) dimensional vector space, W is non-trivial. so we want to find a non-zero vector x in W which is fixed by the action of G, hence lies in k^(n+1).
how do i say such an x exists without a lot of by hand tricks?

E. Artin's argument was this:
look at x in W with minimal number of non-zero coordinates. WLOG x0 is non-zero and so dividing by x0, we have an x in W with x0 = 1. claim is this x works. if it didn't, there will be some coordinate which isn't fixed by G, WLOG it's x1 say g*x1 != x1, then x - g*x is another vector in W with smaller number of non-zero entries which is a contradiction >.<

me get whenever these people use clever induction/well-ordering arguments

det was wondering if there is some k[G] or F[G] semi-simplicity happening behind the scenes which i don't see because of the clever-ness >.<

[round hull]

not really an answer but you can alternatively define a field as being a cring where an element is invertible iff it is nonzero

then one direction is the 0 ≠ 1 axiom and the other one is nonzero implies invertible

[summer path]

Do you consider classification of finite simple groups finished, out of curiosity

[prisma ibex]

Yes and no

It’s certainly finished in the sense that we know the correct statement and essentially have a complete proof of it (spread out between many papers but whatever)

There is still a lot of work to be done streamlining this proof

[delicate orchid]

Good evening chat

[summer path]

Hewwo det

[wraith cargo]

https://math.stackexchange.com/questions/2386626/intuition-for-artins-lemma

Does this possibly help?

🥺

[rustic crown]

ooooh

det should look at MSE more often ><

will read it uwu

ty

[wraith cargo]

It turns out people generally have no idea what the proof does so this wasnt hard to find >-<

[rustic crown]

Hewwo tubu

[rustic crown]

okie it didn't really say anything new than what i had already thought :c

[wraith cargo]

Aw :(
Sowwy

https://everysingleproblem.wordpress.com/2019/06/15/descent-and-artins-lemma/

What about thissssss

[slim kayak]

Can you explain what [F:F^G] means? Isn't the latter just a vector space of homomorphisms into the multiplicative group of F? How do you give it field structure?

[rocky cloak]

F^G is the fixed field of G, and is a subfield of F. [F:F^G] is the dimension of F as a F^G vector space.

[coral shale]

its noon u snoozler

[teal vessel]

so I'm trying to show that all proper subgroups of a group Z are both finite and cyclic. I've shown that all finitely generated subgroups (of Z) are cyclic, which is sufficient for the second condition, since all finite groups are finitely generated, but now I need to show that all proper subgroups are finite, which is kinda annoying to do. The group in question is the group Z of all "p^n"th roots of unity for some fixed prime p (with n being any positive integer). That is, all complex numbers c such that c^(p^n)=1 for some positive integer n.

my first instinct is to try and show that any infinite subgroup must be equal to Z, but I'm not sure how to word that correctly

wait hold on.... this group's subgroups are ordered by inclusion...

[slim kayak]

Z the integers under addition?

[teal vessel]

$Z={z \in \mathbb{C} | z^{p^n}=1 \ \text{for some} \ n \in \mathbb{Z}^+ }$

[cloud walrus]

GoldenPhoenix

[teal vessel]

this is pulling directly from the question, which has access to the double-stroke Z vs just script Z, sorry

grouped under multiplication

[delicate orchid]

this is a limit of cyclic groups

[teal vessel]

because n is a positive integer, it seems it must be the case that no matter what infinite subset of n I get I should be able to "absorb" the conditions of the subset into the complex base itself

idk how to articulate that meaningfully tho

[icy bear]

how is the concept of a set with a law of composition formally defined as a set?
is it like, ordered 3-tuples with (a, b, c) representing axb = c?

a function?

[teal vessel]

probably the set of pairings of ordered pairs (a,b) with singletons c

or, a subset of that set, anyway

[delicate orchid]

It's clear that if the nth graded elements of Z are not in a subgroup H, then all higher gradings must be gone too to satisfy closure (or else we could have z | z^{p^{n+1}} = 1 in our subgroup but not z^p, a contradiction)

the grading I'm referring to is from the isomorphism $Z = \underset{\leftarrow}{\text{lim }} \bZ/p^k\bZ$

[cloud walrus]

W;3w Lads Tbh

[icy bear]

and I suppose if it had 2 laws of composition it would be like {((a,b), (a,b)), c}

and so on

[teal vessel]

what do you mean by law of composition here?

[delicate orchid]

this would be a valid way of encoding the data of the composition map

a strange one, but it would work

[icy bear]

a function from A x A -> A in a set A

that is, all pairs of elements can be composed to go to another element

[teal vessel]

sure, within ZFC (the system I'm familiar with) you can always make pairs of elements, and the collection of objects "nominated" by members of a set according to some function or principle is always itself a set according to the axiom schema of replacement.

then you just take a subset of the cartesian product of (A x A) x A where the first member of each ordered pair is itself the pair which is associated with the second member

i.e. x(a,b)=c means that the object ((a,b),c) is a member of your function. (don't ask me to write the ordered pairs as pure sets, I don't feel like writing that many curly braces)

actually, because I'm curious: || {{a,{a,b}},{{a,{a,b}},c}}|| don't look if you value your sanity (this only works given the axiom of regularity is true).

[icy bear]

ig I could also represent it as (A, f)

just one ordered pair

set

law of composition

[teal vessel]

where A is acting as a stand-in for your original ordered pair?

[icy bear]

no A is the set

f is the function

[teal vessel]

you could, sure

[icy bear]

that is a way of encoding the information

[teal vessel]

yeh

[icy bear]

I have seen stuff like (G, *, +) before so maybe that's the common way

the standard

tho idk if I like it

[teal vessel]

I've seen that, I've seen just "grouped under +"

[delicate orchid]

The actual answer is we don’t care

[teal vessel]

"G is a group" is sufficient

[delicate orchid]

And this would be more akin to a ring

[teal vessel]

what's the grouping operation? unless it has some special properties like commutativity, we don't really care.

[icy bear]

Ik probably that's where I've seen it

[delicate orchid]

Well sometimes we care, in which case we just say “under [operation]”

[teal vessel]

yeh

but often it's enough to know it's a group in some sense.

[dull marsh]

I know something is wrong here because in the dihedral group the same relation holds for a rotation and a reflection, but the last equation is not true in that case

[rustic crown]

you can't show that by induction >.<

[dull marsh]

Ah yes I just realized

Thank you

[rustic crown]

say f(x) = b^-1 x b, then the assumption is f(a) = a^n. applying f again, you get f(f(a)) = f(a)^n = a^(n^2)

thus the correct statement you'll get is a * b^k = b^k * a^(n^k)

[dull marsh]

Oh

That's interesting

[winter shore]

Is there a characterization of finite rings? (similar to the one for finite groups)

[dull marsh]

I don't think there is a characterization of finite groups, are you perhaps talking about the classification of finite simple groups?

[dull marsh]

So if we let b have order m, then a = a^{n^m}, meaning the order of a divides n^m - 1

[rustic crown]

yep

(so the order of n modulo order of a divides the order of b )

[winter shore]

yeah mb

[next obsidian]

Yes, at least in the case it’s commutative

In that case the ring is Artinian and a structure theorem for Artinian rings tells you it decomposes as a product of Artinian local rings

This is already a ton of structure, but I think if you want an even more refined characterization there isn’t any known one

[winter shore]

cool, thanks!

[tardy hedge]

So why do we care about polynomials anyway?

[next obsidian]

Polynomial

[tardy hedge]

Is it just cause they happen to show up in different contexts?

Polynomial

[coral spindle]

Because they are elements of the free R-algebra

[tardy hedge]

Wooooooow

[slim kayak]

They are nice

[rotund aurora]

is this stuff in Matsumura?

[next obsidian]

I don’t remember, maybe not, but it’s standard stuff. Easy enough to prove

[rotund aurora]

okie

[glossy crag]

Can this be derived from the semisimple decomposition somehow? Do finite rings have 0 Jacobson radical?

[next obsidian]

No

[rocky cloak]

You can derive it from the semisimple decomposition in the sense, that if R is commutative artinian with Jacobson radical J, then R/J is semisimple and commutative, so R/J is the product of fields. Since idempotents lift modulo J, and in a commutative ring all idempotents are central, R is the product of rings such that modding out the radical gives a field. I.e. R is the product of local rings.

[void cosmos]

king

similar to that problem u did with me

in my past exams

king of kings

[topaz solar]

Well, apparently if it’s finitely generated and either Noetherian + PI or commutative, then you can determine it by first order formulas

But that is

[rotund aurora]

why do people bully noncommutative rings? Aren't they literally more fun?

[glossy crag]

That's precisely what I was wondering about, how the R/J decomposition might lift to R. Thanks.

On a related subject, does anyone have a link by any chance to that one short paper that proves Artin-Wedderburn on 2 pages using the Brauer lemma (eRe)? Can't seem to find it anymore.

[rocky cloak]

Depends what you find fun. When you have more constraints, you also have more tools to state and prove theorems. But at the same time if you're too constrained you'll run out of interesting things to say.

In the field of representation theory of algebras, finite dimensional (noncommutative) algebras is in some sense the "next" level of difficulty after commutative Noetherian rings. So even with this very strong constraint of being finite dimensional, I'd argue noncommutative algebras is more complicated or similarly complicated to commutative Noetherian rings.

Still it is a very interesting field. And I would agree they are more fun 😊

Commutative algebra have this strong connection with geometry and number theory though. So some people (masochists) might argue that makes it more fun.

[rotund aurora]

I was kind of baiting, but I don't get why there is so much memeing of denying the existence of noncommutative rings, when they are not at all rare

[rocky cloak]

I guess it all comes down to "ring" being a cool word, and those pesky commutative algebraists wanting it for themselves.

Not liking non-commutative rings is just cring(e)

[rustic crown]

what about liking rngs?

[prisma ibex]

I mean yes and no

you can't really set up algerbaic geometry with noncommutative rings the same way you can with commutative rings

there's lots of things that break for general rings that make commutative rings extremely nice