groups-rings-fields

[coral shale]

and the elements need not be numbers

So I gave a hint above

If f = ab, a and b disjoint

f^n = (ab)^n = a^n b^n

Or more generally

For disjoint cycles $a_i$, $$\left(\prod^k a_i\right)^n =\prod^k \left(a_i^n\right)$$

[summer path]

||(meow owo uwu)||

[coral shale]

So try to use this to relate orders of cycles to orders of disjoint products of them. If you're struggling to see, try a small example like (1 2)(3 4 5)

(An aside - its also a result that all permutations can be decomposed into a product of disjoint cycles)

[primal tusk]

ahh that makes so much sense

[coral shale]

You got it? So whats the order of
(1 2)(3 4 5)
and
(1 2)(3 4 5 6)

Just to check

[primal tusk]

6?

[coral shale]

first one is 6, yes

[primal tusk]

and for the other one 4?

[coral shale]

yes good

Can you tell me what it is in general, then?

[primal tusk]

least common multiple

[coral shale]

yep, now u can do your original Q

[primal tusk]

nicee

2

thank u so much

[wild solar]

Hi, I'm wondering could anyone help me with this? I think I'm supposed to prove it by showing that p is irreducible over Z[x] so it must be irreducible in Q[x], right? But I'm not sure how to show it is irreducible over Z[x]....

[white oxide]

can i get hint on how to show that HN is the smallest subgroup containing both N and H? i felt like it may have to do something with maximal groups, but that was wrong since that would require HN to be always normal. I'm trying to construct a subgroup X that contains H and N and am trying to showing that it contains HN, but i would appreciate a hint with regards to that, have been stuck on it for a bit

im assuming i have to use the fact that N is normal

I already showed that HN is a subgroup

[tender wharf]

you can try contradiction bashing

[white oxide]

what's that lol

[tender wharf]

assume we somehow have a subgroup that contains both H and N

and is somehow smaller

[white oxide]

hm ok

i'll try that

[lethal dune]

try replacing x by x+1 or something

[wild solar]

sorry how does that help

[lethal dune]

say K be any subgroup containing both H and N then what kind of elements K must contain?

[lethal dune]

what do you get after replacing?

[white oxide]

i said this, but it feels a bit hand-wavy; Let hn in HN. If X were to be a subgroup h in H and n in N, it must be closed under its operation; that is, hn in X

i feel like i need a more precise argument

or maybe that's sufficient idk

[lethal dune]

that's precise enough

[white oxide]

ok cool i was hella overthinking

cuz they told me N is normal and so i was like

i have to use that for something right

nope turns out just for showing that HN is a subgroup

[lethal dune]

you already showed HN ⊂ X for any subgroup X containing H and N

H or N being normal gives you that HN is a subgroup. It's not true otherwise

[white oxide]

right

[lethal dune]

ok that's not the precise statement, HN is a subgroup iff HN=NH. one of them being normal gives you that HN=NH and hence a subgroup

infact you are asked to show HN is a subgroup in the provlem

that's where you use the fact one of them is normal. Your argument for second part is perfectly fine

[white oxide]

Since $gN = Ng$ for all $g \in G$, $hN = Nh$. Let $n_1h \in Nh$, such that $n_1h = hn_3 \in hN$. For $h_1n_1, hn \in HN$, we have [h_1n_1hn = h_1(n_1h)n = (h_1h)(n_3n)] so $HN$ is closed under its operation. Clearly $e \in HN$, since $e \in H$ and $e \in N$. For each $h \in H$, $h^{-1} \in H$, so $nh^{-1} = h^{-1}n$. Choosing $h^{-1}n^{-1} \in HN$, we see that [hnh^{-1}n^{-1} = hh^{-1}nn^{-1} = e = h^{-1}hn^{-1}n = h^{-1}n^{-1}hn] which shows that $(hn)^{-1} = h^{-1}n^{-1} \in HN$, so $HN$ is a subgroup. Suppose $X$ is a subgroup containing both $N$ and $H$. Let $hn \in HN$. Now, if $X$ were to be a subgroup containing both $h \in H$ and $n \in N$, it must be closed under its operation; that is, $hn \in X$. Hence, we see that $HN \leq X$ for any subgroup $X$ that contains $H$ and $N$, so $HN$ is the smallest subgroup that contains both $H$ and $N$. \qedsymbol

This is fine right

[cloud walrus]

okeyokay

[white oxide]

oh ok got it

[pastel cliff]

a submodule of a free module isnt necessarily free right

wait that's silly

omg det

[rustic crown]

Over PIDs that true

[pastel cliff]

that's actually what im thinking abt

im speedrunning some notes

[torn warren]

🐧

[pastel cliff]

i was doing this a while ago but never finished

working up to the classification of modules over a PID

lemme post thm one sec

[rustic crown]

But in general not, because if it were then every ideal of an integral domain would be principal :p

[pastel cliff]

this is basically what u said

[rustic crown]

yea, but i didn't wanna think about non-zero divisors lol

also for PIDs that true in the infinite setting as well, just need some AoC magic

[pastel cliff]

do u have a counter example for the general case

[rustic crown]

just pick a ring which is not a pid lol

like k[x,y] and the ideal (x,y)

[pastel cliff]

oh i had this as hw at some point i think

(x,y) isn't free then right?

[rustic crown]

yep

if you pick any two elements, they're dependent

like for a, b you have b*a-a*b=0

so if it were free, then the basis has either size 0 or size 1

i.e. it's a principal ideal

[pastel cliff]

k is a field right

[rustic crown]

yeap

if you want easier counter example, pick any non-integral domain

say Z/4Z, the ideal 2Z/4Z is not free because a free module will have cardinality that is a power of 4

[pastel cliff]

,ti

[cloud walrus]

The current time for stμ₂dying is 04:48 AM (EDT) on Sun, 19/03/2023.

[rustic crown]

[pastel cliff]

kinda tired so not processing but i read in morning :)

[elder wave]

[pastel cliff]

hey it's not homework this time timo

that's tomorrow night

im tired but i also couldnt sleep

so here i am

[rustic crown]

[pastel cliff]

but you'll probably see me in #point-set-topology in the next 24 hours

prof was so kind he gave us both homework and an exam in the first two days back from break

love him

[rustic crown]

[lethal dune]

ye great prof

[pastel cliff]

all around based prof

those Free Abelian Groups went crazy that day

[lethal dune]

aren't all abelian groups free

[pastel cliff]

oh it was finite*

misremembered

[rustic crown]

yea cause you don't have to pay

[pastel cliff]

[lethal dune]

great answer

you pay with your sleep schedule

[rustic crown]

[pastel cliff]

i have a abelian grape deficiency

[pastel cliff]

in any case, this significance of this thm is that we get immediate information about all modules over a PID right

[elder wave]

[pastel cliff]

like im assuming the forward is more useful than the backwards

[next obsidian]

all modules

[pastel cliff]

CHMONKEY

[lethal dune]

btw @pastel cliff do you think $\prod_{n=1}^{\infty} \mbb Z$ is free?

[rustic crown]

all submodules (of R^n) >.<

[pastel cliff]

it's not right?

[lethal dune]

free of zero divisors, got'em

[elder wave]

[pastel cliff]

https://media.discordapp.net/attachments/496784958430380033/1080246261929750558/caption-2-1.gif

[rustic crown]

https://media.discordapp.net/attachments/796976320067272704/818150000482582528/612933479730249728.gif

[elder wave]

On the pic?

[pastel cliff]

ye

[elder wave]

Should be free if you’re talking about homology

The C_n are the chain groups I suppose

Simplicial homology I guess so C_0 is free on the vertices

[pastel cliff]

shurg, this man didnt say anything until someone had the balls to ask what his fun abbreviation meant

[pastel cliff]

to rephrase we get immediate information about submodules of a module over a PID

[rustic crown]

nu

every module is a submodule of itself :3

[pastel cliff]

we get immediate information about submodules of a module R^n over a PID R

[rustic crown]

you also get information about every finitely generated module over a pid

because they're cokernels of R^m --> R^n

but that requires a little more work, which you'll do soon

[pastel cliff]

the next step here is to use that proposition to describe how to choose generators of those submodules of R^n over a PID R

[rustic crown]

[pastel cliff]

is this phrased correctly

im trying to get much more manageable restatements of these thms before going back tm

[rustic crown]

yee

[pastel cliff]

and that's essentially a change of basis like we do with matrix operations

this proof is long

i shleep

[rustic crown]

https://tenor.com/view/pat-gif-26359797

[lethal dune]

better proof would be to use SNF

[lethal dune]

sure

it's just notation anyway

[south patrol]

Yes

Well same operation__s__

[lethal dune]

I wonder why you need to differentiate between the 2 operations. They are obviously literally the same

unless like that's what you are trying to show

[rotund aurora]

proof by observation

[pastel cliff]

[lethal dune]

our prof literally used to give arguments like "stare at it" and qed

[pastel cliff]

my prof frequently says "you can stare at this long enough and convince yourself that ___"

[south patrol]

Lol I had that a lot in physics

Doesn't that just mean they can't explain it well lol

[pastel cliff]

my prof just tends to leave nitty gritty details for us

to his credit he doesnt test us on those details much

[livid willow]

All of my high school and college math used to skip proofs

[pastel cliff]

damn how?

[livid willow]

they assume its true

[pastel cliff]

meanwhile this is on a practice exam for a test i have on tuesday

[livid willow]

I am self studying rings and real analysis right now so I am doing proofs these days

Proofs can be hard to understand at times

but when it clicks, it feels so good iyk

[elder wave]

i mean this is like 3 or 4 lines if you can assume those things

as soon as you know pi_1 of D^2 and S^1 you're good to go

(for the n=2 case)

[lethal dune]

those the exact words of my prof,, who's your prof

[pastel cliff]

where do you go ryu

it couldn't be

[lethal dune]

[elder wave]

my diff geo prof and even his assistant used those exact words as well

[pastel cliff]

oh ig it might be common then

[elder wave]

common terminology i suppose

[pastel cliff]

my prof is andrew harder tho

[lethal dune]

I see

mine is Yogish holla

[pastel cliff]

mfw math genealogy is down

[rotund aurora]

truu its down

[south patrol]

Ye that is common lol

[rotund aurora]

Let R be a ring and let m be the smallest integer such that every ideal in R is generated by <=m elements (this integer m might not exist, but assume it does). What can you say about this m? Also, how is it related to the dimension?

For example, let k be algebraically closed, what would m be in k[x1,...,xn] ?

[solar glacier]

does anyone have example problems of field extnesions, min polynomials, ect... ones i could work on to prepare for an exam in field theory

i.e., finding inverses of elements in field extensions

[rotund aurora]

if you are asking for more computational exercises, you could try to come up with your own examples

[glossy crag]

If you define the discriminant $\Delta(f)$ of a polynomial $f\in K[x]$ with roots $\alpha_i$ in some splitting field $L$ as $\Delta(f)=\prod_{i<j}(\alpha_i-\alpha_j)^2$, how do you then show $\Delta(f)$ is a polynomial in the coefficients of $f$? I know how $\Delta(f)\in K$ and how its independent of $L$, but I don't see how this statement holds.

[cloud walrus]

Ocean Man

[rotund aurora]

I think the standard way to see this is to see that this is a symmetric polynomial in the roots of f

and apply Newtons theorem on symmetric polynomials

[glossy crag]

Oh, lmao, wait a minute, I forgot its symmetric in \alpha_i

Thanks

[solar glacier]

can you verify if i did a problem correctly

[cloud walrus]

MyMathYourMath

MyMathYourMath

[rotund aurora]

you can ask this stuff to wolframalpha

[solar glacier]

really?!

[rotund aurora]

mmh yes

[solar glacier]

it does this type of algebra?!

[rotund aurora]

its nothing too fancy

in the end is extended gcd for polynomials or either finding roots of polynomials

here you can just do

,w x^3+8x+2=0

[cloud walrus]

Results provided by WolframAlpha

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

Requested by Croqueta#3385

[rotund aurora]

,w 1/7*(x^2-x+9)(x+1)=1

[cloud walrus]

Results provided by WolframAlpha

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

Requested by Croqueta#3385

[rotund aurora]

and you see the solutions for x coincide

so gg

actually you just see that the polynomials are the same lol

[solar glacier]

ah thanks!

[glossy crag]

So we know that for R commutative R[x_1,...,x_n]^{S_n}=R[s_1,...,s_n] (any symmetric polynomial is a polynomial in the elementary symmetric polynomials s_i). Now say we have that b in R is symmetric in some a_i in R, i.e. b=f(a_1,...,a_n) for some f in R[x_1,...,x_n], and there are no assumptions on R (not an integral domain, not a field, nothing). Do we then know that b is a polynomial in {s_i(a_1,...,a_n)}? What I mean is, is it not possible for b to be symmetric in the a_i by virtue of there being some relations between them, not because the original polynomial f itself happens to be symmetric?

[rustic crown]

what about Z[x,y]/(x-y) and the symmetric element being x = y

it doesn't look like a polynomial in x+y and xy

i.e. 2x and x^2

[glossy crag]

You mean taking b=x+(x-y) and f(s,t)=s, then b is symmetric in x+(x-y) and y+(x-y)?

Yeah, that works I think

[rustic crown]

yee

[glossy crag]

Actually you can make it much simpler by taking a1=...=an=b and f(x1,...,xn)=x1, then b=f(a1,..,an) even though f is clearly not symmetric.

OK, and what if we assume a_i not all equal to one another?

[glossy crag]

NVM, you can take a_i\neq a_j and rest equal to one of them, take something like f=x_i+x_j, then b will be symmetric in them w/o f being symmetric (in all the variables).

OK, and what if ALL the a_i are distinct?

[rustic crown]

ig you can cheat again

take F2[x,y]/(x^2-y^2) and the symmetric element x^2

i made the char 2 just to make computations a bit easier

cause x+y would be nilpotent

[barren sierra]

I haven't got the faintest idea how to do this

I presume this has something to do with the character table

considering the next question is "what property of the character table of a group shows the existence of three different conjugacy classes" with this property for A_5

but I've never worked with character tables in any of my prior courses and I think the prof is assuming we have

so I'm kind of lost

I mean ok I can write a sage script, generate all the computations, and just declare "yea here are all the ways, look it's all the same number" but there are 60 ways and 60 * 24 elements is alot to write and definitely not the intended approach

but idk how to actually do this

[white oxide]

Is it always true that if H is a normal subgroup, then there exists h in H, that for any g in G, gHg^{-1} = H implies ghg^{-1} = h?

Oh yeah I suppose that's true

you can just take h = e

oops

[white oxide]

can i get another hint in this question lol

i know it's trying to get me to show that hkh^{-1} = k

but that simply isn't true for all h in H and k in K

i suppose that i'm supposed to use the fact that their intersection is trivial

ok well i suppose that's the same thing as trying to prove hk = kh lol

nvm i got it

nvm

i was going to say that since their intersection is again a normal subgroup, gxg^{-1} is in their intersection for all g in g and x in their intersection

so that hkh^{-1} is in their intersection for k in their intersection

but that doesn't make sense since not all k in K are the identity

i'm stuck lol

[formal ermine]

the hint is pretty much the entire trick to it

[white oxide]

right

but i'm struggling to see how to use that hint because all the ideas i come up with are just not true

[formal ermine]

show that hkh'k' is in the intersection of H and K

using that reparenthesizing in the hint

[white oxide]

that was my idea but are we allowed to let h and k in the intersection of H intersect K? because doesn't the statement have to follow for all h in H and k in K and some h in H might not be in the intersection

[formal ermine]

huh?

[white oxide]

like

i know their intersection is normal

[formal ermine]

hkh' is in K so hkh'k' is also in K, you can do the same thing for H

[white oxide]

so gxg^{-1} is in their intersection

for any g in G

[white oxide]

oh ok got it

[formal ermine]

#computing-software ?

[simple bobcat]

Guys, shell ring need to satisfy the distributive and the additive commutative?

[formal ermine]

what is a shell ring

[simple bobcat]

I mean shall

like shall ring need to satisfy the distributive and the additive commutative?

I'm sorry I typed the wrong letter

[rotund dragon]

well yes thats by definition of a ring

[simple bobcat]

got it, thanks so much

[formal ermine]

oh lmao I was gonna ask what a shall ring is then I realized that you meant the english word 😵‍💫

[barren sierra]

Just cause this got buried, still no idea

[white oxide]

why can no two of the cyclic subgroups of order 10 have an element of order 10 in common?

[sonic coral]

I think each of them must generate a different cyclic subgroup

[white oxide]

ye but how do we know that

why though?

[frigid lark]

For part b do we require that $\lambda^n(1) = 0$ for $n \geq 2$?

[cloud walrus]

Parrot Tea

[white oxide]

ahhhh ok thank you so much!

[frigid lark]

Consider a map from $\bar{f}: F_{ab}(M) \to M, [x] \mapsto f(x)$, then the kernel of that map includes all of the generators for B, thus the natural hom from K(M) to A is well defined

[cloud walrus]

Parrot Tea

[white oxide]

Is it true that if A is finite and B and C are groups, then $A \oplus B \simeq A \oplus C$ if and only if $B \simeq C$?

if so why?

[cloud walrus]

okeyokay

[white oxide]

does it follow from isomorphisms giving an equivalence relation?

[chilly ocean]

Yes. Let f be an isomorphism from $A \oplus B$ to $A \oplus C$, then consider also the natural projection maps from $A \oplus B$ to $B$ and $A \oplus C$ to $C$. Then from this construct a map from $B$ to $C$ making the diagram commute, last check it is a well defined isomorphism from B to C

[cloud walrus]

LoganB

[white oxide]

got it, thanks!

[chilly ocean]

Is there a reason you need A to be finite?

[tender wharf]

probably not

[white oxide]

that was what the book said lmfao

[tender wharf]

i think its just something given in the hypothesis

[white oxide]

that A must be finite

[chilly ocean]

I can’t think of a counterexample rn

[tender wharf]

you could also bash it with finite abelian group classification

if its abelian

but no need

Q oplus Z ?

[white oxide]

is associativity of multiplication in the set of integers where multiplication is defined in the usual way taken as an axiom? (assuming you haven't shown commutativity and the distributive laws)

[wooden ember]

A does need to be finite

Take A=B an infinite countable direct sum of Z and C=Z (should be able to get away with A fg though)

[white oxide]

moreover how do we even know that the set of integers is closed under multiplication? how can we even prove that? is that again axiomatic?

i always find myself using in proofs, and clearly ab is an integer since a and b are integers, but i never bothered to ask myself why

because idk if i'm right but there's no obvious choice as to what to express an integer as

[wooden ember]

Over integers multiplication is just defined as repeated addition

[white oxide]

if that would even apply

[wooden ember]

And addition of integers being an integer is axiomatic by definition of addition over the integers

[white oxide]

is this related to the peano axioms or is that something entirely different

[wooden ember]

At least that’s how I’d view it, maybe there’s a better way to see it though

When it comes to axioms it’s probably better to ask in #foundations

[white oxide]

hmmm ok yeah it just came up when i started to really think about why the set of integers under the usual operations of addition and multiplication is a ring

but thank you!

[spice whale]

i think it's better to think of the integers as the free ring on 0 generators

in that case multiplication is repeated addition as well

i suppose

[torn warren]

Suppose $H<G, K<G, H^\vartriangleleft H, K^\vartriangleleft K$, it is easy to show $H^\cap K \vartriangleleft H\cap K$, but it is not true that $H^\cap K \vartriangleleft K^*(H\cap K)$, right?

[cloud walrus]

Witness

[flint crater]

Is a simple module the same as an irreducible module?

[elder wave]

yes

[flint crater]

Okay thanks. Was a bit confused since at class we always use simple but when reading up more on it I saw irreducible get used in the same way

[elder wave]

simple is more common afaik

[trim grove]

Some one please help with question 3 , How to find Character?

[trim grove]

,rotate

[rotund aurora]

To give topological structure to linear groups, say to GL_n(C), you can view GL_n(C) as a subset of C^(n^2) by mapping (say for n=2), the matrix
a b
c d
to the point (a,b,c,d) in C^4, correct?

and then take the inherited topology

I guess you can do it naturally without having to choose a particular embedding tho

[rotund aurora]

well I guess you also have some norm or something, which is probably easier. But still wondering about the approach I outlined above

[royal cosmos]

Does anyone know what the circular arrow in the upper left corner is?

[rotund aurora]

I guess is G acting on that space of polynomials

and then you can talk about the G-invariant polynomials

[flint crater]

Am I correct in saying that C_3 x C_5 is isomorphic to C_15?

[delicate orchid]

yes

[tribal moss]

The CRT says (1,1) generates the group.

[feral agate]

For (b), I cannot seem to figure out what G(i) may be. This is what I have so far:
Since, $G$ is a ring homomorphism. We have that $G(1) = 1$. Moreover, for $m \in Z_{>0}$, it holds that $$G(m) = G(\sum_{i=1}^m1) = \sum_{i=1}^mG(1) = m.$$ When $m \in Z_{\leq0}$, it still holds that $$G(m)=m$$ because $G$ is a homomorphism w.r.t. to the additive operator, implying that $m^{-1} = -m$. Moreover, we have that \begin{align*}
G(i^2) = G(-1) = -G(1) = -1 = i^2 = (-i)(-i) = G(i)^2
\end{align}
Note that $G(-1) = -G(1)$ due to G being a homomorphism w.r.t. additive operator. Let $a,b \in \Z$. Consider $G(a+ib)$. Then, we get
\begin{align*}
G(a+ib) &= G(a) + G(i)G(b)\
&= a + G(i)b\
&=
\end{align*}

[cloud walrus]

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

[empty rose]

"we have that $G(i^2) = G(-1) = -G(1) = -1 = i^2 = (-i)*(-i) = G(i)^2$" \
well looking at two of the things here that you said are equal, $G(i)^2 = -1$

[cloud walrus]

bee

[feral agate]

Yes, the obvious thing is to take the square root, but I don't think that is allowed here

[south patrol]

Z[i] is a subring of C

[feral agate]

Yeah that is true

[empty rose]

yeah G(i) is a gaussian integer, and therefore a complex number, which when you square it, you get -1

[feral agate]

Yeah, but from operation's perspective how should I write it down. Because I can multiply both sides of $G(i)^2 = -1$ by $G(i)^{-1}$ which is the operation of taking the square root but that gives $G(i) = G(i)$.

[cloud walrus]

FrankF

[empty rose]

...i don't think multiplying by $G(i)^{-1}$ is square root

[cloud walrus]

bee

[empty rose]

if you want you could expand out what $(a+bi)^2$ is and then consider when the result can be equal to 1

[cloud walrus]

bee

[feral agate]

Why would that be useful though?

[empty rose]

... it isn't, because i'm stupid
but if you consider when the result can be equal to -1 then that would be more useful here

because we know G(i)^2 = -1

[feral agate]

I am stuck on $a^2-b^2+2abG(i)=-1$

[cloud walrus]

FrankF

[empty rose]

why is $G(i)$ here?

[cloud walrus]

bee

[empty rose]

just work out generally what the square roots of -1 are in this ring, then we'll look at G again after we've finished doing that

[feral agate]

I get two solutions, when b = 0 and a^2=-1 and when b=2 and a^2=5

I still don't really get why this is helpful to solving what G(i) is here

[empty rose]

well the idea is

$G(i)^2 = -1$

[cloud walrus]

bee

[empty rose]

that's a lot of information about G(i), there aren't many numbers that give you -1 when you square them

[feral agate]

ohg

[solar shore]

is there any discernable pattern to $D_n$ so that you can deduce what appearance any product will have?

[cloud walrus]

blanket

[feral agate]

Would this be correct?
To find $G(i)$, we need to find all the $x \in \Z[i]$ such that $x^2 = -1$ because $G(i)^2 = -1$. Consider $(a + bi)^2 = -1$. This is solved as the following.
\begin{align*}
a^2-b^2+2abi = -1\
a^2-b^2=-1 \land 2ab = 0\
a = \sqrt{b^2 - 1}\land 2ab = 0\
\end{align*}
When $a = \sqrt{b^2 - 1}$, this gives $\sqrt{4b^4 - 4b^2} = 0$, which means that $4b^2(b^2-1)=0$, which gives $b = 0 \lor b^2=-1$. Assume $b=0$, this gives $a^2=-1$. Assume $b^2=-1$, this gives $a = 0$. This means that $G(i) = \sqrt{-1} \lor G(i) = i\sqrt{-1}$.

[cloud walrus]

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

[empty rose]

$b^2 = -1$ would not make $b^2 - 1$ zero

[cloud walrus]

bee

[empty rose]

also a and b are integers so they can't be sqrt(-1)

[feral agate]

It should be assume $b=0$, this gives $a^2=-1$. Assume $b^2=-1$, this gives $a = \sqrt{-2}$. So there are no possible mappings because the solutions are not integers.

[cloud walrus]

FrankF

[empty rose]

what if $b^2 = 1$?

[cloud walrus]

bee

[feral agate]

That works with a =0. How did you come up with that? Because I thought I exhausted all the possibilities by merging two conditions into 1.

[empty rose]

when is $b^2 - 1 = 0$?

[cloud walrus]

bee

[feral agate]

Okay, I see

[formal ermine]

Dn is generated by rotation and reflection. is that what you meab?

[solar shore]

no imean is there a pattern to any combination of reflections and rotations

[formal ermine]

wdym?

[solar shore]

like if i gave you the i-th reflection composed with a j-th rotation can you come up with what index reflection it would have to then be

[formal ermine]

"ith reflection"?

[solar shore]

like im pretty sure r_ir_j is just r_(i + j)

like rotations would add up

i think im not explaining it right hold on 🦫

[feral agate]

Thank you, I fully get it now, G(i) is either i or -i for G to be a ring homomorphism.

[empty rose]

yep

[solar shore]

i will be right back and organize my thoughts

🫡

[feral agate]

For (b), I don't really know what to do. I mean I know that all elements are of the form $a+b\sqrt{2}$ such that $a+3b$ is a multiple of 7, but what is the general tactic for finding generators?

[cloud walrus]

FrankF

[rustic crown]

find generators, and calculate the quotient, if you get Z/7Z you done, else you try more

like there are two obvious generators here :p

probably in general you can convert it to a problem about the polynomial ring Z[x]

[feral agate]

Is that because of the first isomorphism theorem?

[rustic crown]

yee

[feral agate]

I have no idea how to start with finding the generators though, what are you thinking of when you're trying to do so?

[rustic crown]

can you tell me some explicit elements in the kernel?

[feral agate]

b = 0, a = n7 with n as integer

[rustic crown]

yep, so 7 works

can you tell me one more, but which is not an integer

[feral agate]

b=1, a = 4

[rustic crown]

yee

so let's check

Z[sqrt(2)]/(7, 4+sqrt(2))

that is same as Z[x]/(x^2-2, 7, 4+x)

(by third iso thm)

now simplify

first going mod 7 would be nice

[feral agate]

I don't think we treat that in our course

Only the first one is treated

[rustic crown]

oh interesting

[feral agate]

It's an introductory course lol

[rustic crown]

the iso thms are pretty introductory

[feral agate]

Would it work if I just show that (7, 4+sqrt(2)) generates the kernel by showing inclusion from both sides?

[rustic crown]

yea but that's going to be a little messier,

usually computing a quotient is eaiser to carry out than actually thinking about elements in the kernel

[feral agate]

So that is usually done with the aid of the isomorphism theorems

[rustic crown]

yee, iso thms are bread and butter 🙈

[feral agate]

Even if you were to conclude that the size matches, you still have to show that the set generated would be equal to the kernel right?

[rustic crown]

yea, and that follows because if the kernel would be any bigger then the quotient would be strictly smaller

which it can't

[feral agate]

I am not sure if I get that 😦

[rustic crown]

okie lemme continue with that example

[rustic crown]

if we go mod 7, this is same as (Z/7Z)[x]/(x^2-2, 4+x)

but here (x^2-2) = (x^2-9) = (x+3)(x-3)

so that ideal is just (x+4)

and the quotient is (Z/7Z)[x]/(x+4) = Z/7Z

so you proved that the ideal (7, 4+sqrt2) lies in the kernel right
and so first iso gives you a morphism
Z[sqrt2]/(7, 4+sqrt2) --> Z/7Z

and this is an iso by that computation

this is injective, which means (7, 4+sqrt2) was actually the whole kernel

or equivalently, like i said above, if the kernel was any bigger than (7, 4+sqrt2) then the quotient Z[sqrt2]/ker would have less than 7 elements, which it can't by first iso thm

[feral agate]

Oh okay, thank you for the help. I will try to prove it via inclusion though because this involves the third isomorphism theorem

[rustic crown]

okie :3

[formal ermine]

hi det

[rustic crown]

hi illu

[feral agate]

I've managed to prove that the generated set is a subset of ker(f). However, I got stuck on proving the other direction. In particular, I cannot seem to prove that b is a multiple of 7, but rather that 3b is a multiple of 7.\\
\newcommand{\Z}{\ensuremath{\mathbb{Z}}}
Note that $7, 7\sqrt{2} \in ker(f)$. We claim that $(7, 7\sqrt{2}) = ker(f)$. Let $x,y \in \Z[\sqrt{2}]$. By definition, there exists $a,b,c,d$ such that $x = a + b\sqrt{2}$ and $y = c + d\sqrt{2}$. We see that
\begin{align*}
ker(x7+y7\sqrt{2}) &= ker(7a+14d+\sqrt{2}(7b+7c))\
&= 7a+14d + 3(7b+7c) + 7\Z\
\end{align*}
We note that $7a+14d + 3(7b+7c)$ is divisible by 7. Therefore, we get
\begin{equation}
ker(x7+y7\sqrt{2}) = 0 + 7\Z.
\end{equation}
This gives that $(7, 7\sqrt{2}) \subset ker(f)$.\\
Let $x \in ker(f)$. By definition, there exists $a,b \in \Z$ such that $x = a + b\sqrt{2}$. Moreover, $f(x) = 0 + 7\Z$. Hence, it must be the case that $7$ divides $a+3b$. Equivalently, 7 divides $gcd(a, 3b)$. So $a$ and $3b$ are both multiples of 7. Therefore, we may find $u,v \in \Z[\sqrt{2}]$ such that $7u = a$ and $7v\sqrt{2} = 3b\sqrt{2}$. We need to prove that $7v\sqrt{2} = b\sqrt{2}$ instead here.

[cloud walrus]

FrankF

[feral agate]

To recapitulate, 8b is what I'm trying to prove here

[rustic crown]

(7) = (7, 7sqrt2) tho

[feral agate]

I just changed it to $(7, 4+\sqrt{2})$. The second direction to prove is still problematic though.\\
\newcommand{\Z}{\ensuremath{\mathbb{Z}}}
Note that $7, 4+\sqrt{2} \in ker(f)$. We claim that $(7, 4+\sqrt{2}) = ker(f)$. Let $x,y \in \Z[\sqrt{2}]$. By definition, there exists $a,b,c,d$ such that $x = a + b\sqrt{2}$ and $y = c + d\sqrt{2}$. We see that
\begin{align}
f(x7+y(4+\sqrt{2})) &= f(7a+4c +2d +\sqrt{2}(7b+4d+c))
&= 7a+4c +2d + 3(7b+4d+c) + 7\Z
&= 7a + 7c + 14d + 21b + 7\Z
\end{align}
We note that $7a + 7c + 14d + 21b$ is divisible by 7. Therefore, we get
\begin{equation}
f(x7+y(4+\sqrt{2})) = 0 + 7\Z.
\end{equation}
This gives that $(7, 4+\sqrt{2}) \subset ker(f)$.\
Let $x \in ker(f)$. By definition, there exists $a,b \in \Z$ such that $x = a + b\sqrt{2}$. Moreover, $f(x) = 0 + 7\Z$. Hence, it must be the case that $7$ divides $a+3b$. Equivalently, 7 divides $gcd(a, 3b)$. We certainly can find a $v \in \Z[\sqrt{2}]$ such that $v\sqrt{2} = b\sqrt{2}$, because $b \in \Z$. However, is it possible to find $u,v \in \Z[\sqrt{2}]$ such that $7u+4v = a$ and $v\sqrt{2} = b\sqrt{2}$ both holds?

[rustic crown]

in equation (1) you mean "f" not "ker"

and 7 | a+3b doesn't imply 7 | a and 7 | 3b

[cloud walrus]

FrankF

[rustic crown]

btw, have you seen the definition of an ideal?

what i mean is, the first part of the proof follows from the basic properties of any kernel, so you don't need to write it out so specifically with all the computations

7 and 4+sqrt2 are in the kernel, then so are 7x and (4+sqrt2)y since ideals absorb anything from the whole ring, and now this would mean the sum 7x+(4+sqrt2)y is also in the kernel

[feral agate]

Oh yeah that is true

It's not updating, but this should be the corrected last part:\\
\newcommand{\Z}{\ensuremath{\mathbb{Z}}}
Let $x \in ker(f)$. By definition, there exists $a,b \in \Z$ such that $x = a + b\sqrt{2}$. Moreover, $f(x) = 0 + 7\Z$. Hence, it must be the case that $7$ divides $a+3b$. We certainly can find a $v \in \Z[\sqrt{2}]$ such that $v\sqrt{2} = b\sqrt{2}$, because $b \in \Z$. However, is it possible to find $u,v \in \Z[\sqrt{2}]$ such that $7u+4v = a$ and $v\sqrt{2} = b\sqrt{2}$ both holds?

[rustic crown]

that's not equiv >.<

7 divides 4+3 but it doesn't divide gcd (4, 3)

[feral agate]

[cloud walrus]

FrankF

[white oxide]

what exactly do they mean by "collapse" here?

[summer path]

They belong to the same equivalence class

[rustic crown]

say x = a+bsqrt2 is in the kernel, then so would be x - b(4+sqrt2)

[white oxide]

oh to the same coset right

[feral agate]

a-4b is in the kernel?

[rustic crown]

and that's then a multiple of 7 :3

say 7c

so x = b(...)+7c

[feral agate]

I don't understand

a-4b is in the kernel so it is a multiple of 7 so b*something + 7c is x?

wh is a-4b in the kernel though

[rustic crown]

oh i was being lazy >.< (...) = (4+sqrt2)

[rustic crown]

x - that thing = 7c

[rustic crown]

because x and 4+sqrt2 are

[feral agate]

Okay thanks!

[white oxide]

how can a group have no elements of finite order greater than 1 besides the identity? because if g in G such that g is not the identity, then the order of g is at least 2, since it has to have {g, e}

oh i suppose i could find elements of infinite order

which could be the set of integers

lol $\mathbb{Z}/\mathbb{Z}$

[cloud walrus]

okeyokay

[celest furnace]

Exactly what I was thinking

[pastel cliff]

ok continuing my 3 am ramblings the other day at a reasonable time

can this be done/proven with change of basis and matrix stuff

[rustic crown]

that is the standard way to do it

[pastel cliff]

proof omitted

[rustic crown]

[pastel cliff]

yeah imma skip it i wanna move on to other stuff

my class is "fields and modules"

we moved on to fields today so i need to finish tidying up module material

[rustic crown]

oh

[pastel cliff]

yeah it's unfortunate but i have two more weeks of material to tidy up

[rustic crown]

that particular proof is not bad, but could look a little computational

[pastel cliff]

multilinear maps, RCF and JCF, and tensor products left

[pastel cliff]

yeah i'll settle for the gist of it

the only application ive seen is for structure thm of fin generated men over a PID

and that im just taking the same way i took classification of abelian groups

*finitely generated ones

bc in a sense it should just be a generalization of that anyways no?

[rustic crown]

yee

also the proof isn't any different, so i would just call it the same theorem

[pastel cliff]

[sonic coral]

If two numbers are relatively prime, then their least common multiple is just their product, right? So could this Theorem be generalized to lcm(|G|,|H|)?

Theorem 2.

[celest furnace]

Which theorem?

Ah, the answer to your question is no

Z/2 x Z/2 is not cyclic ( and 2 is not relatively prime to 2 )

[pastel cliff]

remember that in general LCM depends on the unique factorization of an integer

[celest furnace]

And order of product is always product of the order

[sonic coral]

good counter example

[sonic coral]

what do you mean?

[celest furnace]

A helpful identity is lcm(x,y) * gcd(x,y) = xy

[sonic coral]

oh that’s neat. yeah that helps.

[pastel cliff]

i was addressing your first question, given two integers their lcm depends on how many prime factors they share

[rustic crown]

enslaved second isomorphism theorem

[pastel cliff]

[torn warren]

Suppose $H<G, K<G, H_1\vartriangleleft H, K_1\vartriangleleft K$, it is easy to show $H_1\cap K \vartriangleleft H\cap K$, but it is not true that $H_1\cap K \vartriangleleft K_1(H\cap K)$, right?

[cloud walrus]

Witness

[pastel cliff]

so many symbols

[sonic coral]

another question, is there any “trick” or intuitive or “smart” way to list all groups of to isomorphism of any given order. A group of order 20 was easy bc there wasn’t many prime factors. But doing 360 was a bit tougher. I know that i can count the number of partitions of the exponents of the prime factors and take the product of those to find how many there are, but writing them all down by exhaustion doesn’t seem like the best way to go about it

or many it’s just one of those things that comes w experience and a lot of examples?

[tender wharf]

mfw classification of finite simple groups

[pastel cliff]

mfw classification of finitely generated abelian groups

mfw sylow theorems

[south patrol]

Yes it is pain with Sylow

[rustic crown]

potato

[south patrol]

Or at least can't do it all at once lol

[tender wharf]

hello sebbbbbbbb

bbbb

[south patrol]

Helo det

[pastel cliff]

abstract chillgebra

[tender wharf]

lmao you posted this like twice and also in my server

[warm urchin]

abelian groups are cool

[rustic crown]

Z has a server? >.<

[pastel cliff]

det howd you get so cracked at algebra

and topology too

[tender wharf]

studying

[next obsidian]

He took the -pill

[tender wharf]

by stμ₂dying

[rustic crown]

[pastel cliff]

[rustic crown]

(me no know topology tho)

[pastel cliff]

o

you've saved me in topology sometimes tho

[next obsidian]

That’s just cuz ur mega noob so det is still able to help

[rustic crown]

.<

[pastel cliff]

true true i agree with your statement

i really wanna post a certain gif but i'll get muted

[next obsidian]

Didn’t stop you from posting sus in obsidian that one time

[rustic crown]

https://tenor.com/view/anime-girl-blush-cute-shy-gif-18113807

[torn warren]

how to find a counter-example?

[feral agate]

[rustic crown]

does * mean something or is it just notation

[torn warren]

just a notation

you can take it as K_1 is a normal subgroup in K

Suppose $H<G, K<G, H_1\vartriangleleft H, K_1\vartriangleleft K$, it is easy to show $H_1\cap K \vartriangleleft H\cap K$, but it is not true that $H_1\cap K \vartriangleleft K_1(H\cap K)$, right?

[cloud walrus]

Witness

[rustic crown]

so say we pick H1 = H and K1 = K

then want to find H such that H n K is not normal in K

simplest example of a non-abelian group you know is S3

and <(12)> there is not normal

but you want H and K to be proper

take G = S4

H = H1= <(12)> (= S2 lol)

K = K1 = S3

[rotund aurora]

lol never thought about that

now I'll never forget (again) the second iso theorem

[rustic crown]

hehe :p

[rotund aurora]

actually, the second isomorphism theorem is pretty useful

[rustic crown]

pretty useful anytime you have chains

[rotund aurora]

ye

[feral agate]

Are you a graduate math student?

[rustic crown]

me finished my first sem of masters

[feral agate]

pure math major?

[rustic crown]

[feral agate]

How ar eyou so good at algebra?

[rustic crown]

am i? >.<

me is literally trying to learn commie alg for so long now, but i always end up procrastinating :p

need to learn some higher alg too

[feral agate]

I plan on refreshing my memory on analysis and linear algebra, then read a book about measure theory over the summer, just before going into my master's study. I don't think I'll ever have time for brushing up on algebra except for when I really need it 😦

Which I don't really see the need for, but yet it feels as an important of a subject as analysis

[next obsidian]

[rustic crown]

D(monkey) sounded so cool >.<

[pastel cliff]

higher algebra

[tender wharf]

reported for glorifying violence

[feral agate]

I dunno but algebra feels so dry from time to time. But that video series by mathemaniac on group theory was really instructive for me towards making it not as dry as reading off the text book

[rustic crown]

yea measure theory cool

[feral agate]

Guys, I have been postponing this proof for a while now. It is for my thesis. What would be the approach in proving that $dist_d$ is indeed the minimal distance in the following setting?

Consider the metric space $(\R^d, dist_d)$ with the random geometric graph $G^d_n(r) = (V,E)$ as defined in \autoref{sec:defRgg} Let $x,y \in [-\frac{1}{2n},\frac{1}{2n}]^d$. We define the distance function
\begin{equation}
\label{eq:dist}
dist_d(x,y) := \min{[dist'_d(x,y), dist'_d(x',x), dist'_d(y,y')]},
\end{equation}
where, $dist'_d$ is the euclidean distance:
\begin{equation*}
dist'd(x,y) = \sqrt{\sum{k=1}^d \abs{x_k-y_k}^2},
\end{equation*}
and $x'$ and $y'$ are respectively points where the half-lines $y + \lambda x$ and $x + \lambda y$ with $\lambda \geq 0$ intersects with the boundary of $[-\frac{1}{2n},\frac{1}{2n}]^d$.\\
Geometrically, we have defined $dist_d$ such that it is equal to the shortest euclidean distance,
\begin{itemize}
\item in the case of $d=1$, after having connected the two ends of the line segment $[-\frac{1}{2n},\frac{1}{2n}]$.
\item in the case of $d=2$, after having connected both
\begin{itemize}
\item the line segments $[\begin{bmatrix} -\frac{1}{2n}\ \frac{1}{2n}\end{bmatrix}, \begin{bmatrix} -\frac{1}{2n}\ -\frac{1}{2n}\end{bmatrix}]$ and $[\begin{bmatrix} \frac{1}{2n}\ \frac{1}{2n}\end{bmatrix}, \begin{bmatrix} \frac{1}{2n}\ -\frac{1}{2n}\end{bmatrix}]$, as well as,
\item the line segments $[\begin{bmatrix} -\frac{1}{2n}\ \frac{1}{2n}\end{bmatrix}, \begin{bmatrix} \frac{1}{2n}\ \frac{1}{2n}\end{bmatrix}]$ and $[\begin{bmatrix} -\frac{1}{2n}\ -\frac{1}{2n}\end{bmatrix}, \begin{bmatrix} \frac{1}{2n}\ -\frac{1}{2n}\end{bmatrix}]$.
\end{itemize}
\item in the case of $d>2$, this generalizes in the same fashion as the cases $d=1, 2$, which is that 'opposite' hyperplanes that forms the boundary of $[-\frac{1}{2n},\frac{1}{2n}]^d$ gets connected to one another.
\end{itemize}

I remember someone saying that group theory would be the way to go

[rustic crown]

[cloud walrus]

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

[feral agate]

You can disregard the random geometric graph thing, it is not really important except for we consider a graph

[rustic crown]

gomen det too sleepy to read column vectors

[pastel cliff]

eepy

[rustic crown]

https://tenor.com/view/pokemon-sleep-eevee-gif-25414908

[rotund aurora]

[pastel cliff]

eepy eevee

[rotund aurora]

[feral agate]

These symbols are not helping, the idea is that you connect opposite sides of the boundary of the closed set $[-\frac{1}{2n}, \frac{1}{2n}]^d$. In 2d case you have a torus, in 1d case you have a closed curve

[rotund aurora]

[cloud walrus]

FrankF

[rotund aurora]

[feral agate]

@rustic crown What time is it for you now?

[rustic crown]

,ti

[cloud walrus]

The current time for det is 01:24 AM (CET) on Tue, 21/03/2023.

[feral agate]

wow same

I might just be your neighbor who knows

[rustic crown]

[rotund aurora]

5% chance, no way

[rustic crown]

where you live?

[feral agate]

The netherlands

[rustic crown]

oh me is in germany

[pastel cliff]

oh i was about to say

60% chance it was india

[feral agate]

Mannheim?

[rustic crown]

whut that?

[pastel cliff]

(isnt most of the server indian?)

[feral agate]

[rustic crown]

netherlands reminds me of violet evergarden

https://tenor.com/view/violet-evergarden-gif-24397172

[feral agate]

I dunno what that is

[rustic crown]

so pwetty

[feral agate]

waifu

[barren sierra]

No idea how to start (a) or (b)

[tribal moss]

Hmm. For (a) it is enough to prove it for one representative of each conjugacy class in A5.

(But that just seems to lead to pain anyway).

[barren sierra]

do you just mean computationally?

I mean I have a sage script where I've verified everything lol

that yes it indeed is the same

you can write every element g in 60 ways in the desired form

which curiously is the order of the group

didn't get anymore info from that

[tribal moss]

When I wrote that, I was hoping there would be a simple ad-hoc computation for each of the conjugacy classes, but I'm not getting anywhere.

[barren sierra]

yea I had no luck either

My prof gave a hint which I haven't found helpful

so let $C$ be the character table for a group $G$ as a matrix. Let $D_\omega$ be a diagonal matrix whose entries on the diagonal are the sizes of the conjugacy classes of $G$. Then $C D_\omega C^* = |G| I$

[cloud walrus]

Spamakin🎷

[barren sierra]

or at least idk if that was the hint, it's just something we did in class today

and he said something we did in class today would help

💀

y'all think I can turn in a sage script? /s

[barren sierra]

anyone else have any ideas lol

[barren sierra]

ok I have a different question, probably easier to answer

Let $\ell$ be a polynomial in $\mathbb{F}[x_1, \ldots, x_n]$ of degree $\leq 1$ and $f = \ell^d$ for some $d$. I want to compute the dimension of ${\partial_{\overline{x}}^{\overline{a}} (\ell^d) | \deg {\overline{x}}^{\overline{a}} = k}$ as a $\mathbb{F}$ vector space. Basically this set is the set of all $k$th partial derivatives.

[cloud walrus]

Spamakin🎷

[barren sierra]

This should be less than or equal to 1 right?

[next obsidian]

Bro your notation makes no sense

Wtf

[barren sierra]

since for example $\partial_{x_1} (\ell^d) = \text{coeff}_{x_1}(\ell) \cdot d \ell^{d - 1}$

[cloud walrus]

Spamakin🎷

[barren sierra]

gimme better notation plz

this is what my prof uses

How do I talk about arbitrary multi-partial derivatives

[next obsidian]

The dimension of the partial derivatives of x1x2 is 2 dimensional

[barren sierra]

degree of x1x2 = 2

so this is irrelevant here

[next obsidian]

Oh

Idk this seems chmonka

[barren sierra]

kinda is

[next obsidian]

[barren sierra]

algebraic complexity moment 🤡

[next obsidian]

Okay what about x1^2 + x1x2 + x2^2

2x1 + x2 and 2x2 + x1

I don’t think these are linearly dependent

Oh wait

Oh nvm

It should be 2x1x2

Okay maybe you’re right

[barren sierra]

[barren sierra]

this ain't even the worst of it

hm ok if I am right then I solved my problem for this HW which is neat

partial derivative spaces suck man

so much notation to carry around

no mortal was meant to comprehend this

[somber thorn]

when is $m(\mathbb{Z}/n\mathbb{Z})$ a ring?

[cloud walrus]

sean

[somber thorn]

i'm not really sure how to even interpret m(Z/nZ)

like for example if it was 2(Z/6Z), then would this be equal to ${0,2,4,6,8,10}$ where the numbers are treated as different elements, or just ${0,2,4}$ where we reduce those numbers mod 6

[cloud walrus]

sean

[oblique river]

I think the latter

[somber thorn]

ah i see

because if it was the former, we'd have multiple additive identities

[oblique river]

I think the notation just means “the multiples of m in the ring Z/nZ”

Equivalently, the ideal generated by m

[somber thorn]

ah right

[formal ermine]

to be a ring an ideal needs a unit

[south patrol]

Careful, some D&F readers will get annoyed

[somber thorn]

as in an element with multiplicative inverse?

[formal ermine]

ye

[somber thorn]

what if u define rings to not necessarily have identity

[south patrol]

Then you don't

But the typical convention in algebra is that they do, I guess

[somber thorn]

hmm i see

[formal ermine]

🤨 if it contains a unit then it contains 1

[rustic crown]

nope

if m is (non-zero) idempotent in Z/nZ then it's a ring

with the unit m

for example in Z/6Z the elements 3 and 4 are idempotents

and {0, 3} and {0, 2, 4} are rings with units 3 and 4

[south patrol]

Okay true

[rustic crown]

[south patrol]

Well

Like

Ehhhh

[rustic crown]

just to be precise, these are not subrings of Z/6Z, they're just rings on their own with the induced operations + and *. to be a subring, it would need to contain 1 like everyone said :p

[chilly ocean]

Typical convention in algebra is that conventions in ring theory are all over the place

Sometimes subrings have 1... sometimes homomorphisms preserve 1... sometimes rings are even commutative

Why does the homomorphisms in the book I'm reading sometimes prsserve 1 and sometimes they don't

[lethal dune]

[rustic crown]

det has never encountered rings without 1, and i'm happy about that :3

[chilly ocean]

2Z

[rustic crown]

that's a submodule, not a ring :3

[chilly ocean]

It has multiplication

[rustic crown]

ideal, happy? :p

🙈

[chilly ocean]

Here 2Z is a structure on its own

[rustic crown]

[empty rose]

(are you referring to yourself in third person here or is "det" also the name of some other person or mathematical object)

[rustic crown]

(third person >.<)

[empty rose]

(and then later in the same sentence you said "i")

[rustic crown]

(mistakes happen >.<)

(forgive det >.<)

[chilly ocean]

No forgiveness, I'm going to embedd you in a non-unital ring

[rustic crown]

[warm urchin]

a detrimental mistake

[rustic crown]

don't detain det after detecting degeneracy >.<

(me tried )

[broken stirrup]

How does this argument work?

[lethal dune]

deteriorated

[lethal dune]

wait till you see a subring with a different unity than the parent ring

gore maths

[rustic crown]

do you have some finiteness assumptions? else tensoring would be a better thing to do

notice that Hom(F_i, Z/2Z) = Hom_{Set}(X_i, Z/2Z) = 2^X_i (this comes directly from the universal property of the free object)

so if F1 and F2 were isomorphic, then the hom-sets are bijective

so you get 2^|X_1| = 2^|X_2|

[lethal dune]

How the Hom(F_i, Z/2) is even defined?

[rustic crown]

yea the cat wasn't specified but the argument works in both Grp and Ab

[rustic crown]

from this you can't conclude |X_1| = |X_2| directly. if one is finite then sure, else need some generalized continuum hypothesis stuff or something

[lethal dune]

I presume F_i free on X_i means free as a module so I just not getting what the Hom means? Hom as X_i mod can't be right

[broken stirrup]

No X is just any non-empty set

[broken stirrup]

F_i is free group generated by X_i

[hot lake]

you probably want to assume that Xi is finite yeah

[lethal dune]

oh

[rustic crown]

if you don't have finiteness assumptions, i would reccomend to do this instead
Grp --> Ab --> Vect(F_2)
where you first abelianize then tensor by F_2 over Z.

that way, you send F --> Z^(⊕X) --> (F_2)^(⊕X)

and then linear algebra takes over

another idea is to use AoC and prove that if X is infinite then |X|^2 = |X|, so |F| = |X|

but iirc this is a little harder to do directly, need some theory on well ordered sets

[broken stirrup]

I'll just assume that X_i is finite

good enough for me

[rustic crown]

[barren sierra]

What is this script G

[rustic crown]

S_n

[barren sierra]

Some group of size n?

Oh that’s a fucking S?

[rustic crown]

yes

[coral shale]

[barren sierra]

Jeez

[coral shale]

,,\mathfrak S

[barren sierra]

Why

[delicate orchid]

common notation for the symmetric group

[delicate orchid]

it's mathfrak

[barren sierra]

That’s a G tho

[delicate orchid]

I think

[coral shale]

HAHAHA

[rustic crown]

[cloud walrus]

det

$\mathfrak{S} \quad \mathfrak{G}$

[rustic crown]

[broken stirrup]

@rustic crown just sharing

[rustic crown]

[pastel cliff]

1. is just saying that injective maps preserve the generating set, and that homomorphisms preserve relations and 2) the image of any point is determined by the image of the generating set (couldnt come up w a nicer/simpler way to say this part)

[delicate orchid]

I'm confused as to what your question is, are you asking if your shorter definitions are equivalent?

[pastel cliff]

pretty much yeah

scooped

also how is the coproduct different from the weak product?

[white oxide]

is there any notion of exponents in a ring? i.e. a^k = a multiplied with itself k times. my book didn't define any of the sort so i'm wondering if it's appropriate to use in a proof

my book defined ak, which is a added to itself k times

but not a^k

[void cosmos]

yea why not

[white oxide]

just don't want to be sloppy with notation lol

[formal ermine]

yeah it's common notation

even with groups

[void cosmos]

thing is sometimes u can use a^k notation in groups also

[white oxide]

oh yeah with groups it makes sense

[void cosmos]

so a^k in a ring would be ambigious? cuz which operation are yo ureferring to

[white oxide]

cuz only notation is addition

[void cosmos]

yea

[white oxide]

oops

not notation

operation

anyways

[white oxide]

oh multiplication

cuz ka is addition

well

[void cosmos]

ik

i meant thats one thing to say about it

but since u were asking about the "notion"

then yes

[white oxide]

ah ok

thanks!

for this question, i was going to define an "isomorphism" from 2Z to 3Z in the obvious way, namely multiplication by 3. however, even if i show that this isn't an isomorphism in that it's not a homomorphism, how do we know that it's the only isomorphism from 2Z to 3Z? wouldn't i have to show that all "isomorphisms" from 2Z to 3Z are not really isomorphisms?

[next obsidian]

Your setup doesn’t work

For one multiplication by 3 only hits multiples of 6

[white oxide]

well yeah that's the point

[next obsidian]

Secondly, your attempt won’t pan out into a proof

how do we know it’s the only isomorphism from 2Z to 3Z

For one this isn’t an isomorphism, and there’s no reason this is the only potential one either

[white oxide]

i suppose i could construct an "isomorphism" that sends 2 to either 3 or -3

because if phi were to be an isomorphism

it would have to satisfy that property

and show that it doesn't work

[hot lake]

you can argue that there are 2 isomorphisms between their additive groups, and then check that they don't give ring morphisms

[white oxide]

ohh that makes sense

thanks!

i'll try that

[young fiber]

how is 2Z a ring? what's the multiplicative identity?

[formal ermine]

2Z isn't a ring for that exact reason

it does not have a multiplicative identity

[young fiber]

yea so I'm referring to this

[formal ermine]

maybe their book defines rings without a 1

[young fiber]

ah right

[formal ermine]

the only convention in comm alg is that there is no convention

[summer path]

Rngs

[rotund aurora]

Is SU(2) isomorphic to O(3,R) ?

[south patrol]

SU(2) is simply connected whilst O(3,R) isn't even connected

[rotund aurora]

lololo

[south patrol]

I don't actually know how to do it for discrete groups but I assume SU(2) doesn't have a normal subgroup of index 2

[white oxide]

wtf

[rotund aurora]

mmmh good observation

[white oxide]

i didn't know that rings are defined to have a multiplicative identity in some texts

[rotund aurora]

based af

[formal ermine]

in most texts they have a multiplicative identity and are commutative

[white oxide]

that's crazy wtf

i thought ring and commutative ring were separate

🤯

[formal ermine]

yeah but "most" things build on top of comm alg

the study of commutative rings with identity

[white oxide]

true

[formal ermine]

and modules

[rotund aurora]

non-commutative algebra and ring theory is a thing tho

[rotund aurora]

but is this sufficient to assert that they are non-isomorphic? Because you could have an isomorphism not respecting the topological structure at all (?)

[south patrol]

Yeah you could have discontinuous isomorphisms I guess

[formal ermine]

this is the correct channel, dw

[molten viper]

So, I'm working on this problem:
I need to find the degree of $\mathbb{Q}(\sqrt{3},\sqrt{21}$ over $\mathbb{Q}(\sqrt{7}$, how should I go about starting this?

TeXit please 😭

[formal ermine]

the most stupid approach would be to compute minimal polynomials

but maybe there's a smarter way

how much galois theory do you know?

[summer path]

is bot ded again? :c

[formal ermine]

oh wait

I thought 21 was prime for a second

21 = 3 * 7

[molten viper]

very little, we're doing intro stuff

[formal ermine]

so sqrt(21) = sqrt(3) * sqrt(7)

[molten viper]

Yeah that makes sense

[formal ermine]

thus [Q(sqrt(3), sqrt(21)) : Q(sqrt(7))] = [Q(sqrt(3), sqrt(7)) : Q(sqrt(7))]

[rotund aurora]

no

thus [Q(sqrt(3), sqrt(21)) : Q(sqrt(7))] = [Q(sqrt(3), sqrt(7)) : Q(sqrt(7))]

Q(sqrt 3) is not an extension of Q(sqrt 7) >.<

[formal ermine]

typo

[molten viper]

Thanks folks

[formal ermine]

now it really is just a matter of computing the minimal polynomial

[molten viper]

Next question: For a finite extension K subset F, and u in F, I want to have [K(u):K] odd, I need to show K(u^2) = K(u), and I think I can do this by using facts about the minimal polynomial but idk

[formal ermine]

do you "want" [K(u) : K] odd

or is that given

[summer path]

im pretty sure it's given

also i remember this as a problem from d&f

[formal ermine]

looking at the minimal polynomial would work, there's a simpler one via the tower law though

[summer path]

you can try to consider how a tower consisting for K(u), K(u^2), K would look like

and then use that [K(u):K] is odd

[formal ermine]

shhhh you're spoilering everything already

[molten viper]

given sorry

[rotund aurora]

How does O(1,1) have four connected components

If you set the equations, I get that its like the union of the circle and the hyperbola x^2-y^2=1, but these are only three components?

[molten viper]

And what's the tower law?

nvm

I'm not sure what that tells me about the degree of K(u^2) : K

So I think that K(u^2) is a subset of K(u)

[rotund aurora]

yeah made a mistake lol

[summer path]

tower law says that if you have a "tower" of fields K \subset L \subset H, then [H:K] = [H:L] [L:K]

[molten viper]

All I can really deduce from this is that [K(u):K(u^2)] and [K(u^2):K] are odd

I guess u is algebraic over K(u^2)

[summer path]

now you can look at the min polyn and you should be able to conclude

[molten viper]

so the degree of the minimal polynomial is odd

of u over K(u^2)

idk actually

of u^2 over K specifcially

so deg(u) is odd, and deg(u^2) is odd

[pastel cliff]

what is this saying

like what does it mean for sigma to be in the subscript

[rustic crown]

sigma is a bijection from {1, 2, ... n} to itself

[pastel cliff]

det

[pastel cliff]

so is it just a reordering of the elements?

[rustic crown]

yee

[pastel cliff]

and it's saying that reordering elements and putting them through a multilinear map affects the sign of the result right

[rustic crown]

multilinear map just means it's linear in each entry

you say it's alternating if swapping two of the entries changes the sign by 1 (assuming char is not 2)

(in general you define alternating by saying that if two entries are repeated then it evaluates to 0, in char not 2 this is equiv to the previous because swapping the repeated entries would both preserve and reverse the sign)

[pastel cliff]

wait how do you change a sign by 1

[rustic crown]

oh i meant -1 >.<

flip the sign >.<

[pastel cliff]

ok just making sure lol

char wasnt defined at when we got lectured on this so imma just ignore that for now

but i know char 2 is weeird

[rustic crown]

hehe

[pastel cliff]

also what's the difference between viewing the determinant as a map Mat(R) -> R

and a map R^n x ... x R^n -> R

where you product it n-times ofc

prof's notes treat them as different things but they seem the same to me

like does one give us something the other doesnt

[south patrol]

I mean, this is tautological, but they are "the same" under the identification of R^n x ... x R^n with Mat(R). The difference is when you think about the structure involved - you probably wanna think of it as a map out of n copies of R^n (or similar) if you're viewing the determinant as an alternating multilinear map

But there are other ways of stating even that lol

[molten viper]

so is the \ here meaning set difference or like a quotient ring of some kind?

I assume it's K adjoin a single letter

element

goodness I'm tired tonight

Oh wait is K(x) the field of rational functions

[oblique river]

set difference

[molten viper]

I see

[oblique river]

"if u is a nonconstant rational function, then u is transcendental over K"

[molten viper]

Got it

That seems... really obvious?

[oblique river]

not all exercises are meant to be difficult

[molten viper]

ok I'mma rewrite what I have

I'm already turning it in late so like, bad student hours

[oblique river]

(but also, some exercises are meant to look obvious but then be difficult!)

[molten viper]

yeah

so I guess what this means is that for a rational function f in K, there is no polynomial g in K such that g(f) = 0?

[oblique river]

that's right

(unless f is constant; if f(x) = 7 then f satisfies the polynomial g(t) = t - 7)

[molten viper]

and specifically, g(f) is not the zero polynomial

it's fine if g(f) has roots

[oblique river]

that's right, g(f) should not be the zero element of K(x)

[molten viper]

huh ok

hm.

I guess the question is if we had such a g, then what would be its degree?

because like, say g has degree m, and if u = p/q, and the degree of p is n, then I guess the degree of f(u) would be m*n

well maybe not necessarily

I mean, just to play with ideas, if f = x^n and u = p/q, then f(u) = p^n/u^n which is gonna have a bigger degree

Am I thinking on the right lines?

[oblique river]

how do you define the degree of a rational function?

[molten viper]

deg(p) - deg(q)

[oblique river]

so what if deg(p) - deg(q) = 0?

then deg(u) = 0 so deg(u^n) = 0 as well

[molten viper]

that's true

[oblique river]

(just pointing out, if you go along this line of thinking that's something you might have to contend with)

[molten viper]

well let's say...

deg(u) = m, an integer

then deg(f(u)) = m*deg(f) yeah?

except deg(0) is -inf

so the product of 2 integers will never produce that

[oblique river]

so u can't satisfy a polynomial of the form x^n

[molten viper]

I don't think it can satisfy any polynomial

because all you're left with is lower order terms

nothing to "cancel out" the leading terms

like if it's x^n + ax^n-1, then you have u^n (degree nm) + (lower degree terms)

and so on and so forth

[oblique river]

sure -- but you want to be careful if deg(u) = 0

[molten viper]

but if deg(u) = 0, we're fine are we not?

Oh I seeeee

[south patrol]

I would say it is easier not to worry about degree tbh lol

[molten viper]

that's really the only way I know how to worry about polynomials

well, k(X) is not a finite extension

[south patrol]

Hm, well my advice would be to imitate the proof that no (non-integer) rational satisfies a monic equation over Z

[molten viper]

Well, I have that x is transcendental over K

does that mean that polynomials in x are?

[south patrol]

1 is a polynomial in x

What I'm basically saynig is like

[molten viper]

sure but it's constant

[south patrol]

Say you have p/q a rational thing and f in K[x] is non-zero with f(p/q)=0

We'll be done if you can show that q is a unit (1 wlog lol)

(you see why?)

[south patrol]

And so you just need a similar idea to this

Have a think with this