groups-rings-fields

[torn warren]

right , I think they miss this logic chain as you said

[agile burrow]

Yeah, they seem to jump the gun a bit right below where they write ex = x

Like very explicitly you could say *(e, x) = x for all x in X

But the standard notation for a group action is just gx

sometimes there's a dot in between, you can usually figure it out from context

[torn warren]

so they really mean this, right?

[agile burrow]

Yes, that's correct

[torn warren]

great, it is clear now, I am confused that initially the map is defined from $G\times X \rightarrow X$, but here it becomes $X\rightarrow X$

[cloud walrus]

Witness

[agile burrow]

Yeah, so once you fix an element g in G, you now just have a map X -> X that sends x to gx

Glad I could help clear it up, group actions are very fun

[torn warren]

thank you so much!

[torn warren]

I am one step closer to understand the quintic eq is not solvable

[agile burrow]

That's a great goal to have in mind

I think that's what originally motivated me to start studying algebra lol

[torn warren]

yes, exactly, me too!

[rustic crown]

walter

[torn warren]

which one is right?

top or bottom?

[tender wharf]

depends on whether your textbook does clockwise or counterclockwise rotations

[torn warren]

if we look at this, it means the label 1 is replaced by label 2 ,right?

[hollow shore]

Wow pigeon u transformed!

[barren sierra]

I wouldn't really treat it as labels

this is a bijection from ${1, 2, 3, 4}$ to itself, so $p_1(1) = 2$ and $p_1(2) = 3$

[cloud walrus]

Spamakin🎷

[barren sierra]

etc etc

permutations are bijections, that is all

[torn warren]

my question is how do you label them

[barren sierra]

there's no label, these are elements of a set

[torn warren]

I know p(1)=2, but how do you label me on square?

yes, but if we want to label them on square, how to do

[barren sierra]

I mean if you want to number the corns of a square then you can do it either clockwise or counterclockwise

no way is more correct than the other

depends on your text / professor

[torn warren]

if P is a function, then input 1 gives 2

so should I label it in this way?

[barren sierra]

if you want

P is a function on a set

if you want to give extra meaning to the elements of the set

you can, but that has no bearing on how P works

[prisma bluff]

you could, but keep in mind a group by itself is just a closed system of maps that are all invertible

the order in which you label the square is a separate matter, the group only comes into play once you devise a way to have the group act on the labelled squares

the labels could be emoji if you wanted it to be

you just have to make sure your action on those emojied squares make sense

[tender wharf]

sure

[prisma bluff]

also, don't look at group elements as "elements" usually, think of them as maps that are not intrinsically tied to any concrete model - for example, the symmetric group S2 is a system of (two) maps: an identity map id, and an involution f (such that f^2 = id). now this group does not come attached with an intrinsic model; in fact we could describe it as a freely-generated group with one nonidentity map where we arbitrarily impose the relation f^2 = id because we feel like it

but there are a TON of contexts where we have a system of maps that behaves the same way – i.e., there is an identity map that does nothing, but there is also a map that is its own inverse. for example, in any vector space, we have an identity map id(v) = v that does nothing, and a linear map f(v) = -v that multiplies by the scalar -1, and we have f(f(v)) = f(-v) = -(-v) = v, or f^2 = id. or on the complex plane, we have an identity map id(z) = z of course, but we also have f(z) = z * exp(pi * i) such that f(f(z)) = z * exp(2pi * i) = z, that is, f^2 = id. with group theory, we can isolate these systems of maps from their context and just study those, which also allows for any general results about these systems of maps. you can tie these results to a concrete context via group actions

[torn warren]

if rho(1)=2, then only the top one is correct, the bottom one is rho(1)=4

[torn warren]

why you said it depends on the rotation clockwise or counter-clock?

[solar shore]

y'all

im struggling rn

what even is this proof

[solar shore]

<@&286206848099549185> do you guys have a clue

:0

[white oxide]

what's an example of a decomposable group?

[rotund aurora]

I havent read that proof, but iyou can write (ab)=(1a)(1b)(1a)

Try to finish the proof with this observation

[solar shore]

can 1 just be replaced with any other element?

or

[rotund aurora]

No

Ah

Yes sorrt

Its irrelevant the 1 there, but the idea is that you should choose just one element

Then all transpositions are of the form (xa) with x fixed, and that equals the identity

[solar shore]

i see

idk im just

completely lost on the proof and im like

my brains refusing to work now

i see what your saying though but i honestly am struggling trying to apply it here

[rotund dragon]

try considering the direct product of Z_n and Z_m where n and m are relatively prime

[kind jacinth]

Let φ : G1 → G2 be a homomorphism of groups
For any element g ∈ G1 and n ∈ Z, φ(g^n) = [φ(g)]^n

can i get some hints for this?

g^n makes me think of generators but i dont know if its of use

[north sand]

this is just exponentiation. repeated multiplication

do n positive and n negative separately

what does the formula mean in the case n=2

[kind jacinth]

yeah i managed to do it with induction i think!

φ(g^n * g) = φ(g^n)*φ(g) = [φ(g)]^n φ(g) = φ(g)^(n+1) right?

used IH to replace [φ(g)]^n

huh? what is pi? dont really get it

i thought it was supposed to be the the cycle so for (0,1) pi(1)=3 because 1 -> 3

[north sand]

yes

pi is some permutation

yes and because pi(1)=3 you have M_31=1 and the other entries in the column are 0

for pi=(2 1 3)

[kind jacinth]

ohh.. i started counting from i=0 j=0..

alright thanks!

[kind jacinth]

so no matter the permutation we always get the same matrix back right? so even with (1 2 3) ...

so for (a) the answer is the example they gave?

[north sand]

no for pi=(1 2 3) you would get another matrix. in that case eg M_21 = 1 cause pi(1)=2 and M_31=0

[kind jacinth]

ohh right, and for (a) they want us to give a general expression matrix right?

[north sand]

well S_3 has only 6 elements so maybe they want you to give all 6 matrices

considering they write "explicitly"

[kind jacinth]

oh alright

[kind jacinth]

Show that ψ is a group homomorphism.
any tips on how i can do the above?

φ(g1 · g2) = φ(g1) ◦ φ(g2) this is what i have to apply

and i did it on a small example where it gives me the identity (i took (123)(213))but idk how to generalize it

[north sand]

where do the matrices send the unit vectors

[kind jacinth]

to pi(j)

[north sand]

that would be a number

but you are thinking of the correct thing, yes

[kind jacinth]

M_pi(j)?

by that I mean a matrix with 3 rows 1 column

[north sand]

bad notation

in other words, a vector

[kind jacinth]

yes

[north sand]

e_pi(j)

[kind jacinth]

so id need to show that LHS and RHS same transformation happens right?

[north sand]

so you have to show that $\psi(\pi\circ \sigma) = \psi(\pi)\cdot \psi(\sigma)$. where does the LHS and the RHS send the vector $e_1$

[cloud walrus]

Denascite

[north sand]

yes

[kind jacinth]

the notation is a bit confusing for me, why is it pi o sigma? are they just elements taken from S3?

[north sand]

pi and sigma are elements in S_3, yes

the circle is composition, aka the multiplication in S_3

[kind jacinth]

so in LHS its gonna result in a new element in S3, say f and then ψ(f) would give us a matrix

[tender wharf]

pi circ sigma is in S_3 yes

psi(pi circ sigma) might not be in S_3

[rotund aurora]

Suppose the identity is a product of an odd number of transpositions. Write each transposition (ab) appearing in the product as (1a)(1b)(1a). Then the identity is a product of an odd number of transpositions of the form (1a). Written: $\text{id}=\prod_{i=1}^m (1a_i)$ where $m$ is odd and $a_i\in [n]$.

Show that this is impossible (this is way easier than what you had before)

[cloud walrus]

Croqueta

[rotund aurora]

(note that 3*odd=odd)

[tribal moss]

Hmm, it is not clear to me how that is any easier -- and the proof @solar shore posted initially certainly doesn't go that way.

[rotund aurora]

yeah I said I haven't read the proof of blanket

In my opinion its way simpler

say you map 1 --> a

then you must map a --> 1 again

and since all transpositions are of the form 1-->x, if (1a) appears once, it must appear twice

etc

[tribal moss]

But so what?

[rotund aurora]

am I missing something?

[tribal moss]

It could appear three times.

[rotund aurora]

well nuh, but you use the same argument

[tribal moss]

How?

[rotund aurora]

cuz you can delete the first two, as they are irrelevant

[tribal moss]

The first two are not necessarily neighbors in the sequence.

[rotund aurora]

if you send 1-->a

then it will be unaffected until it arrives at a transposition of the form (1a)

[tribal moss]

Otherwise you're saying that 1a)(1b)(1a) = (1b) which is definitely not true.

[hollow shore]

can someone help me with part c?

[rotund aurora]

i mean im not trying to make a rigorous arugment here

In my opinion this is way simpler, and certainly intuitively clear

but maybe its just me

[tribal moss]

You're opining that something that as far as I can see makes no sense is "way simpler" than a proof you also say you haven't even read?

Which basis do you have for making such a claim?

[rotund aurora]

because I have read similar proofs?

I mean

like

its certainly not more complicated, right

but you just have to keep track of the 1 here

(and btw, by "didnt read", I meant not to follow the details, but I glanced at it to know that the proof didnt use what I said)

[tribal moss]

To the extent that handwavy nonsense is "not more complicated" than a proof that actually works ...

[rotund aurora]

have a nice day

this is futile

[tribal moss]

Futile indeed.

[rotund aurora]

you just have to keep track of one element (namely, 1)

[tribal moss]

You keep repeating that. That doesn't make it make sense.

Where would your purported proof then say anything about the number of transposition that don't move the 1 element?

[rotund aurora]

you agree that a transposition (1a) with a fixed cannot appear just once right

[tribal moss]

Yes.

[rotund aurora]

then it appears at least twice

[tribal moss]

There are odd numbers that are >= 2.

[rotund aurora]

so you have $\prod_i (1a_i)=P(1a)P'(1a)P''$ where there are no transpositions $(1a)$ in the products $P$ and $P''$

[cloud walrus]

Croqueta

[tribal moss]

For example, (12)(13)(12)(13)(12)(13) is the identity and contains three copies of (12).

[rotund aurora]

what

wait you are right

Im speaking nonsense

I apologize lol

My bad, tho that's the argument my professor used, and he kinda waved hands to be fair, and didn't really check rigorously.

[hollow shore]

halp pls

part c

[rotund aurora]

(btw, it is written in the notes, so not that I misinterpreted him)

Thanks for the patience, and making me realize I made a mistake 👍

[tribal moss]

Apologies for the impatience.

[rotund aurora]

Hopefully blanket didn't try to pursue my hint, or reads this

[tender wharf]

try using your answers to a and b to explain why the subgroup is as such

[hollow shore]

I did it thanks pigeon

XD

[tender wharf]

oh its been 1h

I just realized

[rotund aurora]

Isn't K here just the composite LE (the field generated by L and E) ?

[rustic crown]

yep

[rotund aurora]

seems weird that he uses an additional letter K

[tribal moss]

It has to be ... but the theorem can't quite be stated that way because LE isn't really well defined unless we already have a common field for L and E to embed in.

[rotund aurora]

ah

well

uhm yeah I see

Thanks

[tribal moss]

For example, Q[x]/<x³+2> and Q[y]/<y³-2> are both perfectly good field extensions of Q. But it doesn't make sense to speak of the field "generated by" them, because it's not clear how arithmetic on combinations on x and y should work. Each of the two fields embeds into C in three different ways, and depending on which embeddings we choose, we might either end up with x+y = 0 or x+y != 0.

[solar shore]

i ended up understanding it

after a bit

[kind jacinth]

i tried giving this an attempt now but i am stuck on what u mean by sending the vector e1. Maybe i can see it better with an example?
so we can take pi as (2 1 3)

and in this case vector e1 is sent to e_pi(1)?

[north sand]

yes

[kind jacinth]

so even if the general case its always sent to e_pi(1)

but the result vector will be different depending on pi(1)

wait no

ok i have no idea how to generalize this. vector e1 is sent to the vector e_pi(j) if pi(j) = i for the element pi

and for the function with sigma as element (In RHS) would be the same thing e_sigma(j) if sigma(j) = i

[north sand]

what are i and j

e1 is sent to e_pi(1) like you said

or to e_sigma(1)

[kind jacinth]

but in the example case e1 is sent to e_pi(1) because for i=pi(j) this holds true forM13

thats what i mean by i j

[north sand]

M_31 if anything

[kind jacinth]

sorry yes m_31

but e1 isnt always sent to e_pi(1) this is only true in the example case right?

like i could also have e1 to e_pi(2) or e_pi(3)

[north sand]

e1 is always send to e_pi(1)

[kind jacinth]

or nvm i am getting a little confused. e1 is alwaysa sent to e_pi(1) yeah

its just that yeah the pi(1) will have a different value

[kind jacinth]

in that case yeah we have this and for e2 e 3 e_pi(2) e_pi(3)

then we would multiply them with the sigma ones

[north sand]

well the way I wrote it down earlier you would do sigma first. but yes

so e_pi(1) is again a unit vector, so where does psi(sigma) send it to

[kind jacinth]

what is psi?

should be sent to sigma_pi(1) in the end?

[north sand]

bad notation

psi is the function which takes in a permutation and gives you the matrix

[kind jacinth]

ohh ok

e_sigma(1)*?

[north sand]

no

the pi shouldnt vanish like it does nothing

[kind jacinth]

e_sigma(pi(1))?

[north sand]

yes

so that is $\psi(\sigma)\cdot \psi(\pi) \cdot e_1 = e_{\sigma(\pi(1))}$. what about $\psi(\sigma\circ\pi)\cdot e_1$

[cloud walrus]

Denascite

[kind jacinth]

so in the part that we solved $\psi(\sigma)\cdot \psi(\pi) \cdot e1 = e{\sigma(\pi(1))}$
computation starts from right to left correct?

[cloud walrus]

tom310

[kind jacinth]

so we first transform it to the e_pi(1) then sigma sends it to e sigma(pi(1))

[north sand]

well just like normal matrix vector multiplication

and I mean in general composition of function. most of the time you start at the right and then successively apply the next and next function

[kind jacinth]

ok for the next part we have for the function sigma(pi(1)) sigma(pi(2)) and sigma(pi(3))

thats how our new element in S3 will look like

oh and from this thats how we get e_sigma(pi(1))?

and to complete the proof we would do e2 and e3?

[north sand]

well how about we first actually finish looking at e1

you havent answered my question

[kind jacinth]

for this one the sigma o pi will be the new element in S3 that looks like:
pi(1) -> sigma(pi(1))
pi(2) -> sigma(pi(2))
pi(3) -> sigma(pi(3))

so if we would have (1 2 3) as sigma and (213) as pi
sigma(pi(1)) = sigma(3) = 1 (3->1) and so on

i hope this was the question i am supposed to answer

[north sand]

the question was, where does psi(sigma circ pi) send e1

[kind jacinth]

and to do this i first computed the sigma circ pi (the above) and e1 will be sent to e1_sigma(pi(1))

[north sand]

ah ok that's how you meant that

ok

so we see that both sides send e1 to the same thing

then yes, now we can do e2 and e3

[kind jacinth]

oh alright and the process would be really similar

[north sand]

yes

[cloud walrus]

tom310
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[north sand]

yes

[kind jacinth]

alright thank u very much for the help!

i do understand it

and thanks for being patient with me! took me a while

[north sand]

youre welcome

[pastel cliff]

linking myself to this convo dont mind me

[high fable]

I'm bad at group theory so I'm probably being dumb. But is (Z, +) isomorphic to all rational points on the complex unit circle?

[lethal dune]

no

[high fable]

Why not

[lethal dune]

for one, all rational point of S¹ has finite order unlike Z

[high fable]

What's S1

[lethal dune]

also by rational point, do you mean p/q + i r/s or points with angle a rational multiple of π

[lethal dune]

complex unit circle

[high fable]

Rational multiple of π

[lethal dune]

exp(2πi p/q) ^ q= 1 so all rational points have finite order

[high fable]

Err

I think I'm too dumb to understand stuff

I have a similar question then

Is Cn isomorphic to x^n=1?

[lethal dune]

yes

if your cⁿ means cyclic of order n

[high fable]

Yeah

So why does that stop working for infinite cyclic groups

Sorry if you explained it already. Like I said I know very little group theory.

[lethal dune]

the group you described is collection of all roots of unity and not a single xⁿ=1; so you can't expect them to behave the same. i.e. still remain cyclic if you take their union

[empty rose]

this is essentially (Q, +) quotiented by the integers

which contains for instance the element 1/2 != 0 with 1/2 + 1/2 = 0
there is no nonzero element of Z that is "half of zero"

in terms of the complex numbers the "1/2" counterexample corresponds to the complex number -1

which, if you do it twice, you get the identity point (1), whereas there's no integer that can be added to itself to get 0 (the identtiy)

[pastel cliff]

anyone know an an example of a finitely generated module over an integral domain R which is not a direct sum of cyclic R-modules.

i dont think that should exist? but i’m probably wrong lol

[rustic crown]

the structure theorem for pids says that you can't do this when R is a pid

so if you're looking for a counter example, try something like k[x, y]

can you find a f.g. k[x, y] module which is not a direct sum of cyclics?

[white oxide]

wait but then it would be isomorphic to Znm, and aren't Zn and Zm not subgroups of Znm or am i trippin

[tender wharf]

they arent subgroups immediately

but they are isomorphic to subgroups

for example in Z_20 we have a subgroup of order 2 which is isomorphic to Z_2

[torn warren]

why the permutation can give two results, I mean the bottom one, how ?

[south patrol]

I mean, must be a mistake then lol

[torn warren]

yesterday someone here said, it can be both, depends on rotate clockwise or counterclockwise

but I thought this permutation can only give clockwise result, since p(1)=2

the bottom one is corresponding to p(1)=4

[south patrol]

Well then it isn't the same permutation

The top is correct

[torn warren]

yes, so I think the textbook has some problem here

[south patrol]

The bottom of the first image is wrong

[torn warren]

the textbook takes the left square, and says $\rho$ will rotate this square counterclockwise by 90 degree

[cloud walrus]

Witness

[torn warren]

but it should be clockwisely

[tender wharf]

that was me

[torn warren]

[tender wharf]

both ways are fine so long you insist your rotations are ALWAYS in that direction

thats what I said which evidently you seem to have ignored

[lament dawn]

derpz stealing rycs pfp lmao

thats so funny

[tender wharf]

abstract chill moment

[lament dawn]

oh shit we're in abs-alg my b

[tender wharf]

no this is chill

[rotund dragon]

take a look at what derpz wrote (ty derp )

[tender wharf]

I did not say it can be both

I said either direction is fine so long you fix one

[torn warren]

I didn't ignore, I put this picture to show rho(1)=2 so should be only the top case, but you didn't reply

so I have to ask again

[white oxide]

ohhh ok got it thanks!

[white oxide]

oh ok cool thx

[tender wharf]

your textbook mandates a counterclockwise rotation

so you'll use thst one

yeah sorry I sometimes answer qns while im at work and I don't follow up in time

[south patrol]

Wait how

The permutation they gave maps 1 to 2 etc

There isn't any ambiguity as to whether it is clockwise or not

[tender wharf]

as in

I didn't know

because the user didnt present the textbook info at first

they just showed me the permutation in both directions

so I said if the textbook has a convention or similar use that

if not you choose one (typically counterclockwise)

[south patrol]

Hm how has this got anything to do w conventions tho

[tender wharf]

I did not see at first that

[south patrol]

Hm ok I dont understand but fair

[tender wharf]

oh sorry

I just ssw

saw

yeah the first one would be wrong then

[south patrol]

Huh

[tender wharf]

the first one seems to send 1 to 4 rather than 2 which is wrong

[torn warren]

the textbook only gives this square and they mark the corner in this 1234 order. So my confusion is if accordint to this , after applying rho to this square, what is the new labels at the corner

[south patrol]

First is correct

[tender wharf]

wait now I am confused

what

oh yeah it is mybad the textbook goes counterclockwise

[south patrol]

What im so confused lol

[tender wharf]

Now I am confused too

[south patrol]

The permutation maps 4 -> 1, 1 -> 2 etc

So it must be the first one in the first image sent

[torn warren]

..............

[tender wharf]

oh mb

yep thats it

i think i need more sleep

[torn warren]

this is what I interpret this permutation, it is clockwise

[tender wharf]

p = (1234) yeah

p(4) is 1 so it goes down

it correponds to a 90 degree counterclockwise rotation

[torn warren]

why this pic is incorrect?

[tender wharf]

you did a 90 deg clockwise rotation

p(1) = 2 not 4

[south patrol]

What does the rho_{(2)} etc mean

[torn warren]

it should be clockwise

[tender wharf]

why would it be clockwise tho

[torn warren]

as I said in pic

[tender wharf]

congrats on graduate+

[torn warren]

rho(1)=2

[south patrol]

Isn't this fine

Lol

[delicate orchid]

Yes that’s fine

[tender wharf]

Now I am confused

[delicate orchid]

Counterclockwiseheads btfo

[south patrol]

P(1) =2 so 1 goes to 2...

Idk why this is being overcomplicated

[torn warren]

ah, I konw why he is confused

[tender wharf]

Oh ur going clockwise

Right

[south patrol]

What

[torn warren]

let me draw

[delicate orchid]

What

[south patrol]

Wjsjsjdnfnwsnsb

Being clockwise or counterclockwise isn't a convention

[tender wharf]

Oh nevermind

I get it

[south patrol]

You just read where the permutation sends the vertices

[delicate orchid]

Wdym “oh you’re going clockwise” the counter clockwise rotation is (1432) it’s a completely different thing

[tender wharf]

I just realized

lmao

[tender wharf]

thats fine I think your question was already answered

I instead will go take a nap

[torn warren]

way1 gives clockwise 90degree rotation, way2 gives counterclockwise

[south patrol]

Okay that makes more sense lol

Though the first seems more natural lol

[torn warren]

yes, I like the way1

[pastel cliff]

isnt a finitely generated module also just a PID tho?

[south patrol]

Huh

[ebon gyro]

what

[delicate orchid]

Thank goodness you replied now I can start saying stupid shit

[pastel cliff]

[south patrol]

Idk what you mean

[pastel cliff]

i blaspheme

[delicate orchid]

Consider Z/8Z as Z/16384Z-module

[south patrol]

You can be a fg module over a ring without even being a ring

[pastel cliff]

id really rather not

[south patrol]

Or like

[tender wharf]

ok yes thank you

[south patrol]

any ring R is a f.g. R module

[pastel cliff]

yeah i didnt realize what i was saying lol

[tender wharf]

I use way 2

[south patrol]

I think the first is what like

[ebon gyro]

I also use way 2 lmao

[south patrol]

I've always seen

But idk

[tender wharf]

Bring on the sullies

[south patrol]

This a meme

[tender wharf]

Well its notation

And notation get sullies because its funny

[torn warren]

ah, I catch a fly, by hand!

[tender wharf]

[torn warren]

[pastel cliff]

mfw hw doesn't do itself

[south patrol]

Why doesn't it

That's not very nice

:*

[tender wharf]

man I hate when homework doesnt do itself

[pastel cliff]

ok starting simple

i can do this just by matrix operations right?

like smith normal form

[delicate orchid]

Indeed you can

Le free abelian group has arrived

[white oxide]

wouldn't this only be true if H was the kernel of some homomorphism? what if H was just a normal subgroup?

[tender wharf]

every normal subgroup is a kernel

[white oxide]

oh damn

okay

thx

[torn warren]

wait, very kernel is normal subgroup, but every normal subgroup may not be a kernal, depends on how the homomorphism is defined, right?

[tender wharf]

every normal subgroup is the kernel of the natural homomorphism

the proposition is that given any normal subgroup, there is a homomorphism such that it is a kernel

[torn warren]

but we can find some non-natural homomorphism to make some normal subgroup fail to be the kernal of this non-natrual homomorphism, right?

[south patrol]

Hm well

"Non-natural" is a bit odd and yeah I mean the normal subgroup isn't the kernel of every map out of the group

[tender wharf]

I mean sure you don't have to try very hard

[south patrol]

But N is the kernel of the quotient map G -> G/N

[tender wharf]

but why

why would you do that

[south patrol]

por ejemplo

[tender wharf]

Like for example I could always be like

Ok let f(x) = e

Then yeah your normal subgroup probably won't be the kernel

[torn warren]

but in this trivial case the normal subgroup H is inside the kernel, I mean is that possible the normal subgroup not inside the kernel at all, except the identitiy e

H is normal, but $ker(f)\cap H={e}$ and $H$ is not inside $ker(f)$

[cloud walrus]

Witness

[torn warren]

is this possible?

[south patrol]

For some groups sure

But like

Since the kernel is a normal subgroup

You can just ask like can you have two normal subgroups N and H with trivial intersection and neither contained in the other

And the answer is yes

E.g. take the subgroups 1 x Z and Z x 1 of Z x Z

[torn warren]

what I mean is this

[torn warren]

ah, yes

[glacial token]

Hi! I am currently trying to show that $\mathbb{Z}_m\times\mathbb{Z}_n$ being cyclic implies $m$ and $n$ are relatively prime. So far, I have worked out that if we have generator $(a,b)\in\mathbb{Z}_m\times\mathbb{Z}_n$, then the order of $(a,b)$ must be $mn$ since $\mathbb{Z}_m\times\mathbb{Z}_n$ has that many elements. Is it okay from here to make the following assertions? \

The order of $(a,b)$ is the smallest integer $k$ such that $(ka,kb) = (0,0).$ So $ka$ divides $m$ and $kb$ divides $n$. The smallest such $k$ for which $ka$ divides $m$ is $\mathrm{ord(a)}$ and the smallest such $k$ for which $kb$ divides $m$ is $\mathrm{ord(b)}$. So then
\begin{align*}
\mathrm{ord}(a,b) = \mathrm{lcm}(\mathrm{ord}(a), \mathrm{ord}(b)) = \mathrm{lcm}\left(\frac{m}{\gcd(a,m)}, \frac{n}{\gcd(b,n)}\right),
\end{align*}
where we consider the orders of $a$ and $b$ in $\mathbb{Z}_m$ and $\mathbb{Z}_n$, respectively. We know that this is less than or equal to $\mathrm{lcm}(m,n)$ so
\begin{align*}
\mathrm{ord}(a,b) \le \mathrm{lcm}(m,n) = \frac{mn}{\gcd(m,n)}
\end{align*}
But since $\mathrm{ord}(a,b) = mn$, it follows that $\gcd(m,n)$ must be $1$.

My main confusion is whether it is okay to make the jump that ord(a,b) = lcm(ord(a), ord(b)) or if i need to justify that better

[cloud walrus]

Haya

[tender wharf]

yes

[rotund dragon]

(to be more clear derpz means yes your proof is fine)

[tender wharf]

man I thought already implies it

[rotund dragon]

the fact that you take lcm guarantees you do in fact get the smallest such k for which (ka, kb) = (0, 0)

[glacial token]

tyy

[torn warren]

no, if ka=0 mod m, it means ka=mq, so m divides ka, but you said ka divides m

for exmaple, take m=5, a=4

[glacial token]

OH true . Sorry, I should've thought of that. If I rephrase to that we know its the smallest positive integer for which ka = 0 mod m (and likewise for kb in mod n), then the overall argument should hold?

[torn warren]

I stopped after I read here

[rotund dragon]

yes

[torn warren]

why? @glacial token

[glacial token]

The cardinality of Zm x Zn is mn so if (a,b) is a generator, then the subgroup it generates should have order mn, right?

[torn warren]

oh, I didn't see it, yes if (a,b) is generator, then no problem

[rain patio]

Is there anybody that knows about drazin inverse matrix?

[cloud walrus]

MutatedCluster

[wooden ember]

anyone have an example of a non-noetherian ring and an ideal in that ring not containing a product of prime ideals?

[rustic crown]

ring of algebraic integers works i think

yea it does

say you have prime ideals p_1, ..., p_n in Zbar, such that p_1 * p_2 ... * p_n is contained in (2). you can intersect this with the ring of integers A of Q(2^(1/m))
to get another containment,
p'_1 * p'_2 * ... * p'_n inside (2)
these p'_i are prime in A, but now you can use the dedekind domain stuff and factorize (2) in A, since (2) = (2^1/m)^m for very large m, say m > n, (2) has more than n prime factors but it divides a product of n primes!

[wooden ember]

im a bit confused here isnt Zbar noetherian (note: first learned about dedekind domains yesterday so i may be saying smth stupid)?

[rustic crown]

it isn't

you can take squareroots as much as you want

so you have a infinite nested sequence of divisors

or better, since you know ACCP is equivalent to existence of factorizations into irreducibles, notice that there aren't any irreducible elements in Zbar

[wooden ember]

Right yeah

Huh that’s a surprisingly nice example of a non noetherian ring im shocked I never had it in the back of my mind before

[kind jacinth]

Find all the subgroups of D4. Which subgroups are normal? What are all
the factor groups of D4 up to isomorphism?

how do I do this?

for now i wrote all generators of D4:

<e> = {e},
<r> = {e, r, .., r^3} = <r^3>
<r^2>= {e, r^2}
<s> = {e, s}
<sr> = {e, sr, r^2, sr^3} = <sr^3>
<sr^2> = {e, sr^2}

[tender wharf]

it looks like you know group presentations

D_4 is small enough to just compute this by hand tho lol

[kind jacinth]

so for the subsets I generated do I do gN = Ng?

[tender wharf]

there's one normal subgroup which is really trivial

[kind jacinth]

but did i generate all subgroups? are there more?

[kind jacinth]

<e>

[tender wharf]

oh I meant beside that one

[kind jacinth]

normally id say its the ne that generates the full set

that one?

[tender wharf]

D_4 should have 8 proper nontrivial subgroups

[kind jacinth]

hm so i currently have 6

well 7 would just be D4

[tender wharf]

well I meant besides the obvious normal subgroups like D_4 and {e} lol

[kind jacinth]

ohhh lol

r?

[tender wharf]

the rotation group?

[kind jacinth]

no

[tender wharf]

are you asking if <r> is normal in D_4

[kind jacinth]

yes

[tender wharf]

yeah

[kind jacinth]

but i dont think it is

oh

[tender wharf]

do you know why

are you aware of the proposition that if a subgroup has index 2 then it is normal?

[kind jacinth]

no, might have missed that when reading

[tender wharf]

you can try proving it

It shouldn't be too painful

[kind jacinth]

ohh

ok the first element will always be the identity

[tender wharf]

I guess you could painfully check the rest by hand

[kind jacinth]

so in this case <r^2> <s> is also a normal subgroup

and <sr^2>

[tender wharf]

it should be a 2 element normal subgroup

[kind jacinth]

ohh

<r, s> ?

[kind jacinth]

(i havent done cosets and group actions in case that theorem appears there)

but ive done group homomorphisms, isomorphisms and normal subgroups and kernels

[kind jacinth]

which subsets are missing here? should i try doing generators with more than 1 element to find them all?

but how do I know when i found them all?

[tender wharf]

^

happy computing...?

[kind jacinth]

yeah my question is how did u know there are 8 proper nontrivial subgroups? like what is ur stopping point for computing subgroups

[tender wharf]

I know there are because I looked at my book

where I computed this before

and now you know too

[kind jacinth]

oh so there isnt really a way unless they tell us

[tender wharf]

I mean....

[formal ermine]

check all possible subsets

[kind jacinth]

ahh

[tender wharf]

please don't

[formal ermine]

you can immediately rule out a bunch of subsets though

[tender wharf]

there's a lot of them that aren't closed

firstly we know any subgroup of D_n must either be all rotations or half are

secondly the rotation group is cyclic so we don't have to worry too much about that

[kind jacinth]

shouldnt cyclic generate the full set? because <r> doesnt generate the full set D4

[tender wharf]

D_4 is not cyclic

but the subgroup <r> is

[kind jacinth]

oh so <r> is cyclic in that it generates all rotations

[tender wharf]

yeah

[kind jacinth]

and so is r^3

but not r^2

[tender wharf]

you know about cyclic groups right

[kind jacinth]

yes so if i have a set there is an element in that set that can generate the full set

so for {e, r, r^2, r^3} there is <r>

[tender wharf]

r^3 generates <r> because it's coprime to order of r

but yeah I assume you know this

then all you have to do is

just start sticking in the other stuff

sr, sr^2 and whatnot

[kind jacinth]

ohh ok so like <r, sr>, <r, sr^2>

[tender wharf]

sure

[kind jacinth]

alright ill write all the subsets down

[tender wharf]

not all

,calc 2^8

[cloud walrus]

Result:

256

[tender wharf]

there are 256 of them

just use what we said and toss out everything that can't be a subgroup

[kind jacinth]

sorry subgroups

not subsets

but when u said start sticking other stuff do u mean with r? because this just generates the set D4 (<r, sr>, <r, sr^2> gives the set D4)

so would <r, sr^3>

or do we stick everything besides r/r^3? (<sr, sr^2> etc)

[opal osprey]

Let's fix $V$ a vector space over $\mathbb{K}$ and $g$ a symmetric, non-degenerate bilinear form on $V$. I am trying to find an explicit formula for the Hodge Star operator on $V$, but it doesn't agree with the one I find in the literature. By fixing an orthonormal basis ${e_{1},\cdots, e_{n}}$ of $V$ and writing for $\beta \in \Lambda^{k}(V)$.
$$
\beta = \dfrac{1}{k!} \beta^{i_{1},\cdots, i_{k}} e_{i_{1}} \land \cdots \land e_{i_{k}}
$$
Using the Einstein summation convention. We have that:
$$
\star \beta = \dfrac{1}{k!} \beta^{i_{1},\cdots, i_{k}} \star (e_{i_{1}} \land \cdots \land e_{i_{k}})
$$
So we need to compute:
$$
\star (e_{i_{1}} \land \cdots e_{i_{k}})
$$
In order to find an expression for $\beta$ by linearity. I have found that (no Einstein convention here)
$$
\star (e_{i_{1}} \land \cdots \land e_{i_{k}})
$$
$$

$$
$$
\varepsilon^{i_{1}, \cdots, i_{k}, i'{1}, \cdots, i'{n-k}} g(e_{i_{1}} \land \cdots \land e_{i_{k}}, e_{i_{1}} \land \cdots \land e_{i_{k}}) e_{i'{1}} \land \cdots \land e{i'{n - k}}
$$
Where ${i'{1} < i'{2} \cdots < i'{n-k}}$ is the complement of ${i_{1}, \cdots, i_{k}}$. Is this correct?

I think there's a slight problem here because I am not assuming the i_1, ..., i_k to be in increasing order.

[cloud walrus]

MisterSystem

[opal osprey]

Well

[opal osprey]

Is this computation correct?

For some reason

most other sources I find online actually have the last formula with a 1/k! at the end

for some reason I am not getting it

Here they are using a different index convention for which components and subscripts and which are superscripts.

So our formulas are pretty much the same except for that 1/k! factor.

So what's happening here?

[opal osprey]

Also, I don't even think I am using the fact that we are working with an orthonormal basis here.

[kind jacinth]

lets assume (a) and (b) are true and i wanna prove group homomorphism

cant i just say that because the RHS is commutative i get
φ(g1g2)(x) = φ(g1) (φ(g2)(x)) = (φ(g1)oφ(g2))(x) and thats the formula for group homomorphism ?

or thats not allowed?

[lapis trail]

Can someone help me with tensor product basics? I'm trying to show the usual construction of the tensor product satisfies the universal property. Here is my work so far:

[opal osprey]

You made a conceptual mistake

Which is understandable since the notation doesn't really help much

but in $R^{M \times N} / U$ we do not have generally that:
$$
[(x,y) + U] + [(x',y') + U] = [(x+x',y+y') + U]
$$

[cloud walrus]

MisterSystem

[opal osprey]

Using a more standard notation, by denoting:
$$
x \otimes y = [(x,y) + U]
$$
What this is really saying is that:
$$
x \otimes y + x' \otimes y' \neq (x + x') \otimes (y + y')
$$
In general

[cloud walrus]

MisterSystem

[lapis trail]

Ohhhh

Yeah I took for granted that operation

[opal osprey]

Yeah so

The intuition for why we do not want the tensor product to satisfy that

is the following

you can think of a tensor $x \otimes y$ really as an abstract notation for a certain bilinear map $f(x,y)$.

[cloud walrus]

MisterSystem

[opal osprey]

and we want $x \otimes y$ to satisfy the properties that a bilinear map satisfies

[cloud walrus]

MisterSystem

[opal osprey]

namely:
$$
(x + ax') \otimes y = x \otimes y + a x' \otimes y
$$
Which is really """just""" the condition:
$$
f(x+ ax', y) = f(x,y) + a f(x',y)
$$
I.e, $f$ is linear in the first entry.

[cloud walrus]

MisterSystem

[opal osprey]

The same linearity condition holds for the second entry

but ofc

if you have a bilinear map $f$

[cloud walrus]

MisterSystem

[opal osprey]

we know in general that:
$$
f(x+x',y+y') = f(x,y) + f(x,y') + f(x',y) + f(x',y')
$$
$$
\neq
$$
$$
f(x,y) + f(x',y')
$$

[cloud walrus]

MisterSystem

[opal osprey]

And since the tensor product really satisfies the same formal properties as a bilinear map

[opal osprey]

this should correspond to this fact

[opal osprey]

yeah so

the sum of equivalence classes of pairs (x,y) and (x',y') is not defined to be the sum component wise

to reflect the properties of bilinear maps correctly

instead

the sum of these equivalence classes is just a formal sum (x,y) + (x',y')

i.e, we are really working with the free module over pairs of the form (x,y) or x in R^N and (x',y')

[opal osprey]

What you really have to do here is to define φ([(x,y) + U] = f(x,y) and extend it linearly to all of R^(M+N)/U

and so φ is by definition linear if it is well defined

so the whole problem here is just checking this map is well defined

and we win

[velvet lake]

So I saw this recently

And it doesn't really make sense to me. The lie algebra su(2,C) already has entries in C. Why do this? Does its complexification yield something different?

I thought the idea was su(2, R) is not isomorphic to sl(2,C) but if we first complexify it by tensoring with C over R then we obtain an isomorphism? But the picture says its actually su(2,C) that we need to complexify?

It's probably something basic that I am misunderstanding here.

[lapis trail]

Thanks for your help. I think I'm slowly getting it

ok

When we mod out by U we are essentially declaring (x,y+y') = (x,y)+(x,y') and (x+x',y) = (x,y)+(x',y)

[prisma ibex]

this is the same as showing sl(2,C)=su(2,C) x su(2,C). Concretely, for matrices A and B in su(2,C) you have that A+iB is in sl(2,C), and if you have a matrix X in sl(2,C) then you can find matrices A and B in su(2,C) such that X=A+iB

[prisma ibex]

the Lie algebra already has entries in C

yes, but you're regarding it as a Lie algebra over R in a natural way. So it's natural to ask what this Lie algebra becomes over C.

the complexification of su(2,C) is sl(2,C). The complexification of sl(2,R) is also sl(2,C). So here are two Lie algebras over R which become isomorphic over C, but are not isomorphic over R.

When you go to classify (semisimple) Lie algebras over R this same issue arises. You first classify (semisimple) Lie algebras over C, where the classification is rather simple, and then you have to answer this question about classifying the different real forms of your Lie algebra

[velvet lake]

This all makes a ton of sense 🙂

[velvet lake]

but my fundamental misunderstanding here I guess is what we mean by a lie algebra over R or over C. So su(2,C) here is a lie algebra over R? despite having complex entries.

[prisma ibex]

right, despite having complex entries you're regarding this as a real vector space

like all you're demanding in this example is that you have a 3-dimensional R-vector space and some bracket operation on this

[velvet lake]

Okay, so to be clear, something else that is related that I was unsure of was the lie algebra of u(1). Right naively this is i\mathbb{R} with the commutator but is this a complex lie algebra or a real lie algebra? I was never sure.

I guess now I would say its a real lie algebra right? despite the elements being ir where r \in \mathbb{R}

[prisma ibex]

again this is a real Lie algebra, this extra "i" is sort of a red herring but it does help you keep track of things when you pass to complexification

the Lie group U(1) is the circle group, so its tangent space at the identity ought to be a 1-dimensional R-vector space!

[velvet lake]

Right! Yeah it seemed bizarre that we also had this i and well, so on...

But no, that all makes sense now.

Thanks nGroupoid 🙂

[prisma ibex]

no problem! I find this stuff confusing sometimes too

[opal osprey]

nG, could you verify some computation I did with the hodge star?

[prisma ibex]

#❓how-to-get-help

[opal osprey]

I think I am getting confused with some 1/k! convention

[elder wave]

Lmfaoo

[prisma ibex]

sure, although maybe Hodge star stuff belongs in #diff-geo-diff-top

[opal osprey]

Yeah, but I am only doing stuff over vector spaces, so not really anything related to differential forms.

[opal osprey]

here

this is the computation

[prisma ibex]

ahh okay let me stare at it

[prisma ibex]

bro idk I'm not reading all this indexing lmao

[opal osprey]

I understand your pain

But yeah so

The problem is that I am not getting that 1/k! at the end

and some materials even have some sqrt(det(g)) or something

[prisma ibex]

right so the sqrt(det(g)) will only show up when you have a non-orthonormal basis

[opal osprey]

And I just don't know why I am not getting these constant factors

[prisma ibex]

the factorial comes from like

so in here

you're indexing over i_1<...<i_k

if you instead indexed over all i_1,...,i_k

you would be overcounting, and you would need a factorial to deal with this

any differential can be written like this

the difference is in this indexing and overcounting

I think what you have is basically correct you just have to think about these differences in indexing, and then think about where the extra metric term is coming from when you don't have an orthonormal basis

[opal osprey]

Oh

I will take care with that now

[opal osprey]

Yeah, I thinkI might have counted things wrong here

[prisma ibex]

right you're counting with increasing indices, so you shouldn't end up with a factorial in the end

I mean there's nothing "wrong" with this, it's just a choice of how you're counting indices

[pastel cliff]

what is this asking

to show that V is the kernel of something...?

[elder wave]

that for every element you get an element of K[t] such that applying your multiplication gives 0

find a polynomial such that you get 0 when you plug in your endomorphism

wait am i misreading this

hol up

[pastel cliff]

timo someone messed up your roles

[elder wave]

ok what i said covers the torsion part

[opal osprey]

What is V_T in this notation?

[elder wave]

probably V as k[t] module wrt to T

since you're fixing one endo to define this

yes ok the torsion part is covered by lin alg

[pastel cliff]

got interrupted at a bad time, brb

[elder wave]

i'll be gone in a bit

i should be doing numerics

[pastel cliff]

gotta move from the library im at, silly spring break

[opal osprey]

The fact that it is finitely generated too

I am pretty sure it follows from the existence of a minimal polynomial for T or Cayley Hamilton

Isn't V_T just the quotient of K[t] by the ideal generated by the minimal polynomial?

[elder wave]

okay i gotta go sry

[pastel cliff]

it's okay ty nonetheless timo

might bother latere

[pastel cliff]

technically havent learned minimal polynomials at this point

[south patrol]

lol talking about torsion modules but not covered min polys

Bruh

[pastel cliff]

these are corrections for an old hw

we covered it the week after this was originally due

[lethal dune]

[kind jacinth]

what do they mean by g in the notation lambda_g?

[void cosmos]

a fixed element

[kind jacinth]

both a and g are elements of G so what is this supposed to mean?

[void cosmos]

in G

not alwayws

in this case yes

[pastel cliff]

going back to some concepts, this should be working up to structure theorem, but what is this thm saying

[kind jacinth]

not always u mean for a or g?

[void cosmos]

for a

for a*

[kind jacinth]

could u maybe explain with an example what this lambda_g is supposed to do though?

as im not understanding what exaclty it does

[pastel cliff]

more context is probably needed

this looks like group action stuff tho

[void cosmos]

this is the homomorphism induced by a group action

[kind jacinth]

currnetly reading about cosets

[void cosmos]

the group action here being on the group itself with the operation

do u know about group actions @kind jacinth

[kind jacinth]

no

[void cosmos]

okay forget about those for now

this function is identified with a specific element in G

namely here,g

and multiplying it by the input

so this function image is just all of the elements that differ by the fixed element g

[kind jacinth]

yes this makes sense

[celest furnace]

other notation would be $\lambda(g, a)$

[cloud walrus]

Geopchad

[celest furnace]

but the other way makes thinking about it easier (getting a 1 variable functoin when you fix g)

[void cosmos]

yes

first one implys g varies

which is not cool 😦

[kind jacinth]

so its a fixed g but a varies

[celest furnace]

yes

it really should say "for fixed g in G"

[void cosmos]

@kind jacinth if ur having trouble understanding something try to get your hands dirty with concrete examples

[kind jacinth]

and all it does is it maps all elements a to a ga

[void cosmos]

take G any group ur comfy with and can do operation by hand inside

[celest furnace]

yeah e.g. f(x) = 2x

[void cosmos]

yea see where this leads

[pastel cliff]

take Z

easiest example

[kind jacinth]

ok ill try this with Z_4

and i need to define a lambda_g for it right?

[pastel cliff]

dont overcomplicate it

cosets partition a group

[void cosmos]

try with whatever u feel compy

comfy

try it with amtrix groups

doesnt matter

[void cosmos]

wdym

do u want a proof

[pastel cliff]

not a proof

just

[void cosmos]

or intuition

[pastel cliff]

english lol

yeah intuition

[void cosmos]

hahah sorry cant help with that lmfao

[pastel cliff]

[kind jacinth]

since u mentioned cosets im not sure if this is what u mean but something like Z_4
subset {0, 2} and for this the cosets would be
1 + {0, 2} and 2+{0, 2}

[pastel cliff]

ok what is Z_4

that's some cringe notation for Z/4

[kind jacinth]

{0, 1, 2, 3}

[void cosmos]

idk its just oilke brute forcing

[kind jacinth]

oh

[void cosmos]

like any ideal is jsut a submodule

of the ring itself ig

so its free and has a basis but its an ideal so it must have dim 1

[kind jacinth]

is this what the proposition is saying but im not getting it?

[void cosmos]

not dim

but like only one basis element

so its principal

[pastel cliff]

why not 3+{0,2}

[kind jacinth]

3+{0, 2} would give the same thing as 1 + {0, 2} right?

[pastel cliff]

yeah it would

so you've basically partitioned Z/4 using the subgroup {0,2}

do you know quotients of groups?

[void cosmos]

btw @pastel cliff what i said isnt enough cuz you still have to prove R is an integral domain

first

[pastel cliff]

all rings are integral domains

[void cosmos]

oh then its done

[kind jacinth]

no

[void cosmos]

the intuition ig is that when the action is multiplicatoin the linear independant subsets must have only one element

thats it XD

[kind jacinth]

i have quotients and rings as the next chapter

[pastel cliff]

weird but ok

[kind jacinth]

what are the partitions? the cosets 1 + {0,2} and 2 + {0,2}?

[pastel cliff]

yeah

but remember that that 1 is actually [1], the equivalence class of 1 mod 4

you noticed earlier that with this subgroup, 1 + {0,2} and 3+{0,2} are the same

what element of Z/4 have you not considered

[kind jacinth]

0 which gives {0, 2}

[pastel cliff]

which is the same as 2 + {0,2} right

[kind jacinth]

yes

[pastel cliff]

so we have a new group here actually

can you tell what it is

[kind jacinth]

a group that gives a different set?

so u mean not Z/4 anymore?

[pastel cliff]

what you've done is really quotient Z/4 by {0,2} to get Z/2

[kind jacinth]

ohh because of the cosets 0 1?

[celest furnace]

Yes there’s only two elements

And there’s only one group of order 2 (in fact, only one group of prime order)

[void cosmos]

@kind jacinth you are in for a fun ride

[kind jacinth]

and this is true for every group when i find the cosets?

i get another group with order = nr of cosets

but why is this important?

[void cosmos]

sometimes you can say somethings about the original group

quotienting the right things.

[kind jacinth]

i dont really understand how the example is related to the proposition

[void cosmos]

i think he was responding to this

[kind jacinth]

ahh i see

[torn warren]

Is this a typo of the textbook?

anyone here?

[glossy crag]

Given the fraction field functor R->Q(R) for integral domains, is there an analogous construction producing a division ring out of a noncommutative R w/o null divisors?

[torn warren]

nobody today

[torn warren]

https://tenor.com/view/hhgf-gif-25031041

[sonic coral]

Did i do okay on these first two?

[white oxide]

why is the factor group of order 2? Shouldn't it be of order 4, since we have $(0, 0) + H, (1, 0) + H, (2, 0) + H, (3, 0) + 4$?

[cloud walrus]

okeyokay

[barren sierra]

yea I think so

I mean explicitly giving the mappings is overkill but yea I think what you wrote is correct

[prime sundial]

all field homomorphisms are monomorphisms
how?
isn't the trivial homomorphism not monic?

i assumed it was all injective homomorphisms, so i assume this is talking about nontrivial homomorphisms

[barren sierra]

Yea

Every nontrivial field homomorphism is injective

Not hard to show

[prime sundial]

right

im just asking if nontrivial was assumed, because clearly the trivial homomorphism is not monic

[barren sierra]

Ye

[coral shale]

i dont think 'monic' is a word

the way u use it

[prime sundial]

i'm using it like my author uses it

[coral shale]

so synonymous with injective???

monic polynomials are a thing

as well

[prime sundial]

monic in the sense that it is a monomorphism

[coral shale]

first time i seen this

[prime sundial]

https://math.stackexchange.com/questions/2558702/how-exactly-is-a-field-monomorphism-defined

a monomorphism f:X -> Y satisfies that for g,h in Hom(W,X)
fg = fh implies g=h. it's from a category theory perspective, but i figured the question was actually about algebra

i have seen monic polynomials, don't know if there's any relation there

[coral shale]

there isnt.

[coral shale]

confused what this comment means

monomorphisms are injective homomorphisms?

[prime sundial]

no

they are saying they arent

but in the category of fields they do happen to be just injective homomorphisms

that is, a mono in the category of fields is an injective homomorphism

riehl (the author) does say that monomorphisms and epimorphisms are categorical analogues to injective and surjective maps, but they aren't really the same

take what im saying with a grain of salt, it's my first time seeing this stuff as well

[torn warren]

can anyone help check my proof

[tender wharf]

Z6/Z3

[tender wharf]

Z6/H has order 2 because H has order 3

H = <(0, 2)>

[rough spindle]

There is no trivial field homomorphism. The function which sends every element to zero is a morphism of abelian groups, but not of rings

[torn warren]

thank you , but what do you mean overkill, to what...

[barren sierra]

idk I wouldn't give the actual isomorphism

I would just say "compositions of two isomorphisms is an isomorphism qed"

That's just me being me it honestly doesn't matter

[torn warren]

hmmm

what about the proof?

[white oxide]

Ohhh ok thanks!

[prime sundial]

the zero homomorphism is not a homomorphism?

[chilly ocean]

Why I=(b)

R Is domain

[lament dawn]

Would be easier to say if we saw the statement, but one inclusion is obvious and the other follows since a was generic

[rough spindle]

In the category of unital rings, 1 must go to 1. If you want commutative algebra to match geometry, this is pretty important

[prime sundial]

fair enough

thanks for the input

in like a non sarcastic way, i actually appreciate the new perspective

[void cosmos]

ig its R is a pid iff its a dedekind-hasse domain

[chilly ocean]

do people do analysis first or abstract algebra? typically

[coral shale]

seeing the latter first might be marginally more helpful, but it doesn't really matter

[rotund aurora]

At an introductory level, none is a prerequisite of the other of course. But doing one before the other will build maturity, etc. Personally, I think analysis is really good for building rigorous thinking and proof skills. But if you want to do algebra before analysis, just go for it. You can also do both at the same time btw, if you have enough time

[formal ermine]

don't get too analysis pilled

[coral shale]

In terms of rigorous thinking I would think more Algebra?

Pretty much nothing is blackboxed

In analysis you at least blackbox R's construction from the start

[void cosmos]

doing analysis first is better

imo

as for me who taught my self this i would have been better if i started with harder analysis

proofs in (basic) algebra are VERY misleading

most of the time they are just literally definition chasing

[chilly ocean]

ah

[void cosmos]

proofs in analysis involve a trick or a tricky but cute argument

[chilly ocean]

yeah i plan on self studying analysis bit um not sure

[void cosmos]

i would suggest starting with analysis tbh

[chilly ocean]

sure

[coral shale]

What is wrong with this

[coral shale]

Is this referring to picking delta for epsilon? or did u have another example in mind

[void cosmos]

not really good setting up for harder stuff

[void cosmos]

any significant theorem in analysis really

minkowsi's inequality for example

every measurable function has a monotone increasing sequence of simple functions converging to it

etc

[coral shale]

But then that sounds like you think algebra is easier

in which case it surely makes sense to take it first

[void cosmos]

i would advocate for doing hard stuff first

in anything

[coral shale]

🤨

[void cosmos]

yea it just makes you better prepared

[coral shale]

If anything, I would recommend taking number theory before either of the 2, but maybe OP already has

[void cosmos]

ig NT is cool

i just think problems in like intro algebra are just very misleading and/or easier

(the problems one would see in a basic textbook for first time algebra)

not easier

but just the way they get solved is just idk

[coral shale]

misleading for the difficulty of math at a higher level?

[void cosmos]

ig yea

[coral shale]

I just can't figure how taking analysis and then algebra would be any better with the case you're making

in the end, you're taking both

[void cosmos]

yes but when i learnt algebra b4 anallysis for the first time i was surprised i couldnt do most of the problems (in analysis) and i was just weirded out by the fact that u cant definitoin unfold to the solution

i was mislead

thats just my experience

[rotund aurora]

I guess its personal opinions/experiences at this point. But I learnt analysis form Terence Tao from the first time, and he doesn't black box anything. He constructs R by quotiening in the space of Cauchy sequences

[coral shale]

based

[void cosmos]

yes its just opinions after all

[frigid lark]

Doesn't algebra give you quite a bit of number theory for free? (I haven't done much number theory so I could be really wrong)

[coral shale]

been too long, not necessarily but certain things perhaps

Like the idea of a quotient explains modular arithmetic ig

But usually its understood the other way round

Modular arithmetic is the usual introductory example to get ppl to understand quotients

[frigid lark]

I was thinking along the lines of Z/pZ being a field and the NT equivalent.

[rotund aurora]

Lagrange is like Fermat little theorem

[tender wharf]

tbf I think fermat's little theorem is better understood with lagrange

yeah

[rotund aurora]

But I would do modular arithmetic before groups, that's how I did it and really liked it

[coral shale]

I would say number theory is more focused on the results related to primes

thats the 'difference'

[livid willow]

I am doing both of them simultaneously

Well its self study, so my rules lol

[coral shale]

most ppl do take them at the same time

i believe

[sharp sonnet]

yes, a lot if results in elementary number theory are examples in abstract algebra
abstract algebra was developed to attack questions in number theory really so this is no surprise

a european math curriculum would do analysis first always

the problem with algebra is you need some examples first and its hard to come by them

so probably at least some familiarity with polynomial rings

[hollow shore]

how would i find the identity element of the set Q* (set of non-zero rational numbers)?

[coral shale]

recall defn of identity

and you basically want to 'solve' this equation

or at least, think of it that way

If such an identity doesn't exist, then you'd have to prove there are no solutions

[hollow shore]

hm

i thought about 2

[tribal moss]

A set doesn't have an identity element unless you choose a binary operation to go with it. Then you can ask if that operation has an identity element.

[hollow shore]

say p, q is element of Q*

[coral shale]

good point, I thought it was implicitly multiplication

[hollow shore]

isn't p x 2 = p

[coral shale]

ok, so whats the definition of an identity, first of all

like state it

[hollow shore]

the element that is unchanged in the set when apply to all the other elements?

like p x e = p

[coral shale]

so for any g in your set under the operation *

An identity e satisfies
g * e = e * g = g

Not just one direction

[tribal moss]

That sounds vaguely like you're defining an absorbing element rather than an identity. When you say "p×e = p", do you mean that p or e is the identity?

[hollow shore]

e is the identity

[tribal moss]

Okay good.

[hollow shore]

for us thats how we been taught

[coral shale]

its important to quantify

what each symbol means when u use it

so going back u suggested 2

So if 2 is an identity, for any p in Q*

2 x p = p x 2 = p

[hollow shore]

would that be true?

[coral shale]

Well do you think it is?

[tribal moss]

Try plugging in, say, p=5 and see what you get.

[hollow shore]

2x any non-rational p is a non rational p

[coral shale]

Like did you try a few p to check

no, its the same p

[hollow shore]

why it the same p?

[coral shale]

the statement would be written differently if so

[coral shale]

check this

We would instead write
g * e and e * g is a member of G

We wouldn't reuse the same symbol to mean something different, no.

[tribal moss]

The definition of "e is an identity" is "for every p in the set, computing e×p gives p as a result, and computing p×e gives p as a result too".

[hollow shore]

wait im a bit confused about the e * p = p * e = p part

if i classify p as any non zero rational

say 10

[tribal moss]

So if 2 were an identity, we would need to have, among other things, "computing 2×5 gives 5 as a result, and computing 5×2 gives 5 as a result".