groups-rings-fields

[golden pasture]

stop this fake structures

ahh that makes sense

[chilly ocean]

you should give an example instead of doing this complicated symbol pushing lol

you know

two actual matrices from those groups

rings without unity 🤢

[old lava]

lol, ya, you can often just construct matrices with 0s and 1s (which are in every field)

to show that it doesn't work

[chilly ocean]

i think that the GL(2, Z_2) is really trying to push that point lmao

[golden pasture]

we need to stop using Z_2

and start using F_2

[chilly ocean]

no, use pi_1(RP^2)

[golden pasture]

i agree

but that doesnt have a natural ring structure

sad

[chilly ocean]

oh just slap one on it

simple

[golden pasture]

:chad:

[old lava]

Prove that if $H$ is an abelian subgroup of a group $G$ then $\langle H, Z(G)\rangle$ is abelian.

[cloud walrus]

PorosInMyAshe

[old lava]

why am I so hardstuck on this

[delicate mauve]

what do < > mean?

[chilly ocean]

subgroup generated by the stuff inside

i.e. smallest subgroup containing H and Z(G)

[delicate mauve]

Let x be an element of the group, the homomorphism f_x (y)=xyx^-1 is identity on the generators hence is identity on the whole group

@chilly ocean what's wrong?

[chilly ocean]

(the react means nothing)

🤨

i saw that middle finger react

[old lava]

wait, why is it identity on the generators, if x is in the group, but is neither in Z(G) or H, then is xyx^{-1} still guaranteed to be y

[delicate mauve]

it's easier if you give an explicit description of the group

[old lava]

I mean ya

[delicate mauve]

as finite products of elements in any of the groups

[old lava]

it already intuitively feels right, because I checked with some test groups, but I just can't prove it

[delicate mauve]

elements in H commute with elements in H and elements in Z(G)

elements in Z(G) commute with everything

[old lava]

ya, but if x is an element that's in the group, but not in H or Z(G), then it's not guaranteed to commute with elements of H

that's where I'm stuck

[scarlet estuary]

why's that matter

[delicate mauve]

it can be written as a product of elements in H and Z(G)

permute one element in the product at a time

[old lava]

ah I see, thanks

[astral galleon]

Is $(R_1 \times R_2)/I \cong I$ ? note $I= {(0_R, r): r \in R_{2} }$ and $R_1 \times R_2$ is the product of two rings

[delicate mauve]

For a more rigorous proof, you can just look at the center of <H, C(G)>, prove that it contains H and C(G), and since it's a group it must be the whole <H,C(G)>

[cloud walrus]

Zophike1

[delicate mauve]

Shouldn't it be R_1?

Take the homomorphism pi(x,y)=x from R1 x R2 to R1

Your I is the kernel

[astral galleon]

ahhh makes sense thanks

[delicate mauve]

apply the first isomorphism theorem

[sturdy marsh]

[old lava]

"almost obvious"

of course, how could I not see it

[leaden finch]

can someone help me with this one

[uncut girder]

Why did grothendiek all Serre a bastard for including computations? 😢

[wind steeple]

bc grothendieck sucks at computations and it was an friendly "bastard"

[vital quail]

@old lava i mean literally all your elements in <H, Z(G)> are made up of things that commute with each other

the parts in H commute with each other, and the parts in Z commute with everything

[old lava]

that's not how it works, you can have elements in <H, Z(G)> that are not part of H or Z(G)

you have to prove those rest commute too

which was easy enough, I was just brainfarting for a while

[oblique river]

I mean that is kind of how it works lol

anything in <H, Z(G)> is built up of pieces which all commute with each other

so take two such things, and they together are also built up of pieces which all commute with each other

[old lava]

ya, that's what I used at the end

I was referring to

the parts in H commute with each other, and the parts in Z commute with everything

which doesn't account for every element in <H, Z(G)>

[oblique river]

I think by "parts" he meant "parts of every element"

[old lava]

oh I guess ya then it makes sense

I took it to mean the part of the subgroup that is in H, and the part of the subgroup that is in Z(G)

[vital quail]

no

your elements of <H, Z> are products of things in H and Z, and all those things commute with each other

that's what i meant

[old lava]

ya, that's what I ended up using

[vital quail]

👍

[devout crow]

i read that if every element of a group has order 2, then it's a direct sum of copies Z/2Z

which is the same as the direct product if the group is finite, but not the same when it's infinite

my question is why is an infinite group of that form necessarily isomorphic to the direct sum rather than the direct product?

[oblique river]

the direct product of infinitely many copies of Z/2Z does satisfy that property

but it also happens to be isomorphic to a direct sum of (even more) copies of Z/2Z

every direct product of some number of (possibly infinitely) many copies of Z/2Z is isomorphic to some direct sum of (possibly more) copies of Z/2Z

but the reverse isn't true

so "direct sum" is more general

[devout crow]

:o

thanks

[next obsidian]

Wait what? How is that true?

[oblique river]

axiom of choice

all vector spaces over F_2 have a basis

[next obsidian]

I’m not sure what you’d map to the elements in a direct product which have infinitely many non-zero stuff

Thinking

Thinking

[oblique river]

why do your basis elements have to have only finitely many nonzero entries?

[next obsidian]

I meant what do you map to those from a direct sum

[oblique river]

assuming AoC all vector spaces over all fields have a basis

[next obsidian]

🥴

[oblique river]

I'm not sure what your point is, I'm not saying that a countable direct product of Z/2Z is isomorphic to a countable direct sum of Z/2Z

[next obsidian]

I mean using that sure

I know that lol

[oblique river]

I don't get your question then

[next obsidian]

I just thought maybe there was some explicit construction of an isomorphism

And I couldn’t see how tf you’d make that

[oblique river]

no

[next obsidian]

Whack

[devout crow]

yh im finding it hard to intuitively picture e.g. how an uncountable direct sum is isomorphic to a countable direct product

[marsh fractal]

so basic question

a^n = e for any group

i get the concept but not so sure of the proof

there must be a k <= n such that a^n = e otherwise too many elements in G

[scarlet estuary]

n is?

[marsh fractal]

then im a bit lost from there

[scarlet estuary]

the order of the group?

[marsh fractal]

yes

yes

yes

Im ok with the proof for that

ive done it, but dunno why this is weird

ah

a subgroup must divide n

yes

so we have a group {a,a^2,....a^n}

but its not a group unless it has an e

and if you add the e then the order is n+1

so some a^k = e

i dunno does this make sense

yea im not sure

[scarlet estuary]

hint: the order of a subgroup divides n, and as you mentioned, for some k, a^k = e

[marsh fractal]

that is it pretty much

[scarlet estuary]

if you can justify from the first fact that k divides n

thats your proof

since that means kb = n for some integer b, so we can just write a^k * a^k * a^k * ... * a^k, b times

which is e * e * e * ... * e

and is also a^n

so you already know that there exists a k <= n with a^k = e, so now prove that k divides n

hint: note that {a, a^2, a^3, ... a^k} is a subgroup

(do you see why?)

then observe that this subgroup has k elements and apply lagrange's.

[marsh fractal]

ok i see

it doesnt matter which element in the subgroup a^k = e holds for

[scarlet estuary]

i mean

i'm taking k to be the least element such that a^k = e

you justified that there must be some such k

since otherwise {e, a^1, a^2, ... a^n} are all distinct and our group has "too many elements"

so let k be the smallest number (1 <= k <= n) with a^k = e

then we can consider the subset {a, a^2, a^3, ... a^k}

can you see why this is a subgroup?

[marsh fractal]

yes

a^k is e

[scarlet estuary]

what would the identity be? what would inverses be? how do we know associativity holds?

[marsh fractal]

and k = n

so it divides n

[scarlet estuary]

no, we don't necessarily know k = n

we picked k to be some element that might be smaller than n

i dont know what examples of groups you've seen before but

[marsh fractal]

yes

[scarlet estuary]

if you're familiar with the dihedral group of the triangle

it has 6 elements, but the element corresponding to one rotation (call it r) has order 3

that is to say, r^3 = e

[marsh fractal]

im not sure what the inverses would be , but the identity must be a^k and by normal integer multiplication associativity holds

[scarlet estuary]

wdym "normal integer multiplication"?

the elements of the group might not be integers

the answer is that it inherits associativity from the parent group

[marsh fractal]

yeah i guess i mean the exponentiation

[scarlet estuary]

we know that the parent group follows associativity

since, well, it's a group

[marsh fractal]

ah ok right

[scarlet estuary]

and every a^x, a^y is from the parent group

anyway, as for inverses

note that our identity is a^k

[marsh fractal]

but what group is it a subgroup of?

thats the part i dont understand

[scarlet estuary]

so for example, the inverse of a^2 would be

a^(k-2)

since a^2 * a^(k-2) = a^k = e

[marsh fractal]

yeah

[scarlet estuary]

well, you're proving this for an arbitrary group, so it could be any group

[well, any finite group]

the point is that, no matter what group you take, if a^k = e

then {a^1, a^2, a^3, ... a^k} is a subgroup

by our above justifications

[marsh fractal]

but a^(k-2) is not in the group right?

because its of size n

[scarlet estuary]

uh

[marsh fractal]

thats what i dont understand really

[scarlet estuary]

okay lets say k = 7

[marsh fractal]

true i should just spend some time on this and come back

[scarlet estuary]

then a^k = a^7, right?

and a^(k-2) = a^5

now we know this must be in the parent group

since if a^5 isnt in the parent group

that means we can multiply a by itself

over and over

[marsh fractal]

lol ok

[scarlet estuary]

and eventually get something that ISNT in the group

which violates closure

a should be in the group, so a*a = a^2 is also in teh group

hence a^2 * a = a^3 is also in the group

and so on

all the way up till a^k (and beyond, in fact; this applies for ANY integer)

[marsh fractal]

ah ok theres just inverses because a^k = e

[scarlet estuary]

so a^(k-2) is certainly in the group

[marsh fractal]

so when you reach a^k then it is the identity

a

?

ok

so it just loops

alright

understanding the inverses helps a lot

ok

yes i havent encountered any other finite fields other than integers mod n

ok but then yes you show {e,a,a^2....a^k} is a subgroup

and so it divides G

yes right

yea i was thinking that

i guess they cancel each other out since a^k = e

by set theory laws

[scarlet estuary]

note that we're implicitly noting that {e, a, a^2, ..., a^(k-1)} has k elements

which is "obvious" but worth mentioning

since thats what tells us that k divides n (combined with lagrange's theorem)

[marsh fractal]

ah

yes that is also a missing link

mind readers!

im not sure

so yeah a^k = a^ka^m

so a^m must be a^k s inverse

and so a^k = e?

is it not equal to (a^k)(a^m)

ok yes

[scarlet estuary]

anyway, we know a^n = a^km = (a^k)^m = a^k * a^k * a^k * ... * a^k

[marsh fractal]

yeah i see that

[scarlet estuary]

where there are m "a^k"s in the far right of that equation

now what's a^k?

[marsh fractal]

so it can only be e

for that to be true

[scarlet estuary]

no

i mean

we already know what a^k is

is my point

[marsh fractal]

lol yes sorry

[scarlet estuary]

remember that we started this argument by letting a^k = e

so a^n = a^k * a^k * ... * a^k = e * e * ... * e

and of course if we multiply the identity by itself a bunch of times

its still the identity

so a^n = e

[marsh fractal]

so then a^n = eee m times which is = e

[scarlet estuary]

and that concludes the proof

[marsh fractal]

soz weird * thing there

ok thank you sorry for the long procedure

but yeah it makes complete sense, it isnt just a method, so tis great thank you

[leaden finch]

can someone help me with isomorphism

im lost 😦

[smoky cypress]

Is it just me but this question doesn’t make sense

[delicate bloom]

seems to make sense to me since it's just modding integers

it's a homomorphism, since addition and multiplication are mapped onto just fine, but it's not bijective so it's not an isomorphism

I'm probably being too lenient in my interpretation since multiplication of elements will mix Z with Z_6 so I'm figuring they're both meant to be in Z_6

that, or it's just not a homomorphism then lol

[oblique river]

I think it's clear that it should just say "for all a + bi in Z[i]"

[leaden finch]

yeah i got confuse

can someone help me get started on how to start injective

[carmine fossil]

Do the usual thing(f(a)=f(b) should imply a=b)

[leaden finch]

would it be like be f (a + b i) = f(c +di) ?

[carmine fossil]

Yes

[dense pecan]

Can anyone help me? All I know is that an order is the lcm of lenghts of the cycles

Like, do u need to find all the elements of s17...? or what 😓

[thorn delta]

so, cycles can have length 2 through 17

[dense pecan]

oh wait would it be 72 then?

wait

uh would i take the prime factorization of the answer choices?

cause if so 65/72 have factorizations using 2-17 👀

[sturdy marsh]

but you have only 17 letters

[thorn delta]

well, let me ask u this, knowing that the order is the lcm of the length of cycles, how would you construct an element of order 72 for instance

[dense pecan]

3 2-cycles and 2 3-cycles?

and then some 1 cycles?

[thorn delta]

wouldn't the lcm of those lengths be 6 then?

[dense pecan]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

i was multiplying them 😅

wait would it be 65? like a 5 cycle and 13 cycle

[thorn delta]

the lcm of 5 and 13 is 65, yes

[carmine fossil]

Do ||1-8cycle and 1-9 cycle|| for 72

[thorn delta]

you have to make sure the cycles you find can be taken to be disjoint tho too

the order of a product of disjoint cycles is the lcm of their lengths, not just any cycles

[dense pecan]

wait this might be a dumb question but s17 has 17 elements right? so then could it be 5 and 13 since they add up to 18?

im going to think about the disjoin part a bit more 😪

[thorn delta]

nono S17 has 17! elements

[dense pecan]

okay okay thats what it thot but just double checking :D hehe

[thorn delta]

but cycles are made by choosing elements from {1,2,...,17}

so, for example (1 2 3)(3 4 5) are not disjoint cycles while (1 2 3)(4 8 7) are

[dense pecan]

hm okay so if u have a 5 cycle and 13 cycle since they would have 18 nums total and theres only 17 nums to choose from (1-17), they wouldnt be disjoint?

[thorn delta]

right, they can't be disjoint

[dense pecan]

ooo interesting so then that means it cant be 65 right since the lcm isnt the order? so the answer would be none of these options?

[carmine fossil]

No

[dense pecan]

wiat what

[carmine fossil]

Consider (1 2 3...8)(9 10 11... 17)

[dense pecan]

the lengths of the cycles are 8 and 9 OH

and the lcm is 72

and they are disjoin

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

[carmine fossil]

In general if you can write n=lcm(m1,m2...mk) such that m1+m2+...mk<=a then there is an element of order n is S_a

[dense pecan]

i shall write that in my notes tysm @carmine fossil @thorn delta 🥺 😭

[thorn delta]

npnp

[carmine fossil]

Do you understand why that works?

[dense pecan]

yes i think so cause the lcm is the order and if the sum <= # elements then they are disjoint?

[carmine fossil]

Yes

[dense pecan]

😁

[dense pecan]

oops have another question 🙈

What I've got so far is:
(a) if u have 1-cycle, 2-cycle, 3-cycle, 5-cycle then you can get order 30
(b) order of u70 = phi 70 = 24
(c) 26
So so far a is the biggest, but idk how to check (d)?

[thorn delta]

cycles with even length are odd and cycles with odd length are even

[dense pecan]

wdym by that? all i know is that a is the alternating group by idk hwo to check order

[thorn delta]

ehh so, there might be an easier way, but you know that a permutation is in the alternating group if it is even, i.e. a product of an even number of transpositions.

take a permutation like (1 2 3). This is an even permutation since (1 2 3) = (1 3)(1 2). Now, starting from (1 2 3), you can make a permutation like (1 2 3 4) by applying the transposition (1 4) on the left: (1 2 3 4) = (1 4)(1 2 3). So (1 2 3 4) must be an odd permutation. It follows from induction that cycles of even length are odd while cycles of odd length are even

So the point of this suggestion is to have a way to determine which products of disjoint cycles belong to A_12, so u can compute orders easily

[dense pecan]

ah okay that makes sense, so basically the alternating group can only use cycles with an odd number of values? so having a 1-cycle, 2-cycle, 3-cycle, 5-cycle like in (a) won't work, but having a 1-cycle,3-cycle,7-cycle would work? which in that case the answer would be (a)?

[thorn delta]

the alternating group can only use cycles with an odd number of values?
cycles of odd length are even, yes, but you also have that the product of two odd permutations is even. For example, (1 2 3 4)(5 6 7 8) is even, even though its cycle components are odd permutations

idk if its clear but that's true because (1 2 3 4) is a product of a number of odd number of transpositions and (5 6 7 8) is a product of an odd number of transpositions, and odd + odd = even.

[dense pecan]

hm so having 2 6-cycles would fall under a_12? and yeah i think it makes sense ty for the explanation :D

[thorn delta]

hm so having 2 6-cycles would fall under a_12?
ye, but to be clear, two disjoint 6 cycles wouldn't have order 36

[dense pecan]

ye since they're not relativley prime correct?

[thorn delta]

yep

their product would have order 6

[dense pecan]

okay great tysm you've been so helpful 🥺 😭

[thorn delta]

npnp

[old lava]

this should be a typo right? There's only 16 elements in D16, and so there possibly can't be 16 such elements

[carmine fossil]

Ye

It should be 8

[old lava]

ya, that's what I got

just wanted to make sure I wasn't going crazy

[viscid pewter]

@old lava

watch out

there's two different conventions for Dn

some ppl use Dn to mean the dihedral group with n elements

and some ppl use Dn to mean the dihedral group of an n-gon

[old lava]

ya, they use D_{2n} format here

so D_16 is 16 elements

[viscid pewter]

oh

weiiird

maybe they copypasted the q from a different context lol

[old lava]

lmaoo, maybe

[vital quail]

D_2n weirdchamp

do they also say S_n!

[dense pecan]

The answer is (b) but I'm not exactly sure how/why? I see that the factorization of 385=5711 so i assume it has something to do with sharing common factors but not exaclty sure

also idk how to check (a) (c) and (e)

[viscid pewter]

order arguments cut through half of these

do you know lagrange's?

being explicit, consider the subgroup of the numbers of the form 5i in Z385

[dense pecan]

like if u have a finite group , the size of a subgroup divides the size of the group?

[viscid pewter]

yeah

[dense pecan]

wdym by this? sorry 😅

[viscid pewter]

so like s6 is very big

too big to fit

neither 15 nor 121 does not go into 385

[dense pecan]

wait just to back up for two groups to be isomorphic the size of the groups have to be the same correct?

[viscid pewter]

yes

it's a bijective thingy, one-to-one covering the whole thing

[dense pecan]

isnt the size of s6 , 6! or is that the formula for another group

[viscid pewter]

it is

so that's not gonna work, lol

[dense pecan]

im so sorry can u pls explain why bc like it says u can take any subgroup of z385 no?

[viscid pewter]

yeah

but all the subgroups of z385 are gonna be smaller than 720

so it's not gonna work

s6 does not fit in z385

[dense pecan]

oh my goodness pls excuse my stupidity

i thot 6! was 120

[viscid pewter]

been there done that

[dense pecan]

a;skldfj;laksjdf;lkajsdf

im smarter than that pls 😭

[viscid pewter]

i careth not

onwards

[dense pecan]

okay i def see how that is gone

[viscid pewter]

anyway c is the one that's a little interesting bc 35 does divide 385

but now considering the orders of the elements not the orders of the groups it does not work

if you see why cool, if not ask

[dense pecan]

im sorry but i don't see why, im not v clear on that last half?

[viscid pewter]

ok so for things to be isomorphic all the orders of the elements have to be equal

like taking any element in one there's an element in the other with the same order

because intuitively like isomorphism means all the elements behave the same in both groups

[dense pecan]

okay so for isomoprhism it has to 1. have same number elements and 2. those elemnts have to have the same order?

[viscid pewter]

an isomorphism is a surjective homomorphism

do you know what those words mean

[dense pecan]

yes i think so, like surjective and bijection?

[viscid pewter]

wait yeah it should be a bijective homomorphism

pain

[dense pecan]

haha 😅

[viscid pewter]

okay but like in a homomorphism it's like

so call the homomorphism f: (G, *) -> (H, #)

then for any a, b in G, f(a*b) = f(a)#f(b)

[astral galleon]

Dumb question is the units of z mod n its multiplicative inverse?

[viscid pewter]

this is the definition of a homomorphism, it basically says that elements act the same way whether mapped or not mapped

so it's like they act the same way in both groups

and then the bijectivity says the two groups are equal sizes

so they're basically actually the same group

[scarlet estuary]

@astral galleon what is "its" here?

[dense pecan]

ah okay that makes more sense i just wrote that down in my notes haha 😅 ty

[viscid pewter]

so anyway yeah

it's pretty easy to show from all of that vagueness that if you have an element with order n in G, then when you map it it has to have order n in H

and in particular if you have an element with order m in H and no elements with order m in G then G and H cannot be isomorphic

[astral galleon]

@scarlet estuary to be clear I'm asking that if the units of z mod n are the multiplicative inverse of the group

[viscid pewter]

and then you just have like

so what do you notice about orders in z35

and in u35

[scarlet estuary]

how does a group have a multiplicative inverse?

[astral galleon]

Its elements cancel out and equal 1

[dense pecan]

@viscid pewter just give me a few min to comprehend pls haha

[scarlet estuary]

one of us is confused by definitions here

groups themselves dont have inverses

elements of a group do

[astral galleon]

Ahh okay

[scarlet estuary]

moreover, multiplicative inverses are unique; each element will be the inverse of precisely one element

(sometimes elements are inverses of themselves, such as the identity being its own inverse)

[astral galleon]

Ahh okay

[scarlet estuary]

anyway, in the multiplicative group of integers modulo n, the inverse of an element a is an element b such that ab = 1 (mod n)

[astral galleon]

Oh so you just solve it like a linear equation

[scarlet estuary]

well its a modular congruence

[astral galleon]

Ok makes sense so would the solutions be the same as the units of z mod n ?

[scarlet estuary]

as an example, $(\bZ/8\bZ)^\times = {1, 3, 5, 7}$ under $\cdot$

[cloud walrus]

Chillin' with Quillen

[scarlet estuary]

so the inverse of $3$ is the number (from this set) such that $3 \cdot x = 1\pmod{8}$

[cloud walrus]

Chillin' with Quillen

[scarlet estuary]

of course this is 3 itself

since 3 * 3 = 9 = 1 mod 8

[astral galleon]

Ahhh ok so my earlier question was correct

[scarlet estuary]

and in fact every element of this group is its own inverse

[dense pecan]

wait do you mean z15?

[viscid pewter]

ah yes i skipped a step there

so z15 can't be in z385 bc it doesn't divide

we're considering c rn

u35

[scarlet estuary]

[but this doesnt necessarily hold for every (Z/nZ)*]

[viscid pewter]

if it's isomorphic to subgroup of z385

[scarlet estuary]

[consider, say, (Z/9Z)*]

[it's good practice to determine its elements and the inverses thereof]

[viscid pewter]

then it's isomorphic to z35 in z385, bc all the subgroups of a cyclic group are cyclic intuitively and rigorously, and isomorphism means the orders have to be the same

[dense pecan]

wait are you saying u35 is cyclic? bc i thot it wasn't? or am i like way off base

[viscid pewter]

it's not

that's why it can't be a subgroup

that's one way of showing it

[chilly ocean]

wait hold up

how can groups of different orders be isomorphic

[viscid pewter]

they can't?

[astral galleon]

@scarlet estuary that makes sense

[chilly ocean]

nvm

sorry didnt read the question.

[viscid pewter]

i'm saying if it were isomorphic to a subgroup of z385 it would be isomorphic to a cyclic subgroup, as all subgroups of z385 are cyclic, and specifically z35 because the sizes have to be the same

[dense pecan]

ah okay that makse sense! and u121 isnt cyclic as well correct? which leaves only b

[astral galleon]

@scarlet estuary you didnt answer my earlier question

[viscid pewter]

sure, that's one way to do it

so what i did was i considered the orders of the elements explicitly

so in z35 all the elements have order 34, right? apart from 0

but in u35, very easily like 34^2 == (-1)^2 == 1

so order 2

contradiction, right?

[dense pecan]

yes since the orders have to match up for it to be homomorphic right?

[viscid pewter]

yes

[dense pecan]

that makes way more sense now, tysm!!!

[viscid pewter]

np

[dense pecan]

If you have an alternating group, such as a_23, and you have a 2 cycle, 4 cycle, and 17 cycle, what would the order be? would it be 4*17? idk since theyre not all relatively prime

[next obsidian]

Not all of those are even

They won’t even all exist

[dense pecan]

wdym? like if u have (1,2)(3,4,5,6)(7,...23)

[next obsidian]

Oh you meant the product of them sure

But if they aren’t disjoint then you’re boned

Anyway if you have disjoint cycles the order is the lcm of lengths of the cycles

Since they commute

So like in this case(abc)^n = a^nb^nc^n

So you need the smallest n for which all of them are e which is just the lcm of their orders = length of cycles

This is kind of unique to symmetric groups this kind of thing isn’t true in a general group

[dense pecan]

that makes sense ty for the info!!!

[dense pecan]

Idk what n it would be? The answer is true?

I checked all cyclic groups up to u19 and their orders

but i cant find two ajdacent numberes that have the same size

[next obsidian]

Do you know what the size of U_n is in general?

Don’t quote me, but I feel like it should be impossible

But im@not a number theory guy

I just feel like the sizes literally couldn’t line up

But idk

[chilly ocean]

I have tried to do it by an argument about size of Un, and not that my number theory is great, but it doesn't seem obvious

[next obsidian]

Idk it just seems really odd...

[dense pecan]

That's what I thought and like i even looked at the table of euler totients and i cant find one that works

But my teacher sent me the answer key (since these are practice exams) and he said it was true

but he also said that there might be an error 👀

[next obsidian]

Okay so apparently totient(15) = totient(16)

[dense pecan]

dont the groups have to be cyclic?

[next obsidian]

Totient(3) = totient(4) yeah?

[smoky cypress]

yeah but that doesn't mean they are isomorphic

[chilly ocean]

But also I think U15 is cyclic and U16 is not, or maybe the other way around

[dense pecan]

i dont think either of them are cyclic

[next obsidian]

Sure

[chilly ocean]

Oh

[next obsidian]

16 is

[smoky cypress]

I think U_p and U_(2p) are isomorphic

[dense pecan]

wait isnt it 2,4,p^a,2p^a are cyclic

[next obsidian]

There’s a classification for when you have a primitive root

[dense pecan]

so then how is 16 cyclic

[smoky cypress]

16 is not cyclic

[next obsidian]

What?

Uhhh

[chilly ocean]

Ah I got them confused, I thought it is 2^n that is cyclic

[dense pecan]

wait do the groups hvae to be cyclic in order for them to be isomorphic?

[next obsidian]

Isn’t 16 = 2^4

[smoky cypress]

Yeah but remember it's number theory and 2 is weird

[next obsidian]

Oh shit right

[chilly ocean]

But is it weird in which direction? This is hard for me to memorize

[next obsidian]

2 is weird

Ducking odd primes

Yeah I have no clue

[smoky cypress]

If I remember correctly $U(2^p)=C_2\times C_{2^{p-2}}$

[next obsidian]

There are infinitely many for which they agree

[cloud walrus]

Whoever

[dense pecan]

oh wait and theyre both cyclic right

[next obsidian]

I think

Oh

lol

[chilly ocean]

Oh nice

[dense pecan]

oopsie 🙈

[smoky cypress]

Also

[dense pecan]

thank you

[smoky cypress]

Like I said

[dense pecan]

also is this true or false: do groups have to be cyclic in order for them to be isomorphic?

[next obsidian]

Yeah whoever

[smoky cypress]

Ah I'm so sorry

[next obsidian]

What lol

Obviously not

Take any non cyclic group

Isomorphic to itself

[dense pecan]

oh 🙈

oopsie i was getting everything confused hehe

thank you

[next obsidian]

I feel like there's no good way to check the others

since sadly 3,4 are the only "1 away" primes

I mean

wait

kill me

4 is the first non-prime prime

Wait

am I Grothendieck

[nova plank]

Lol.

[sonic parcel]

i don't think so

(Z/104Z)*
≅ (Z/8Z)* × (Z/13Z)*
≅ (Z/3Z)* × (Z/3Z)* × (Z/13Z)*
≅ Z/2Z × Z/2Z × Z/12Z
≅ Z/2Z × Z/2Z × Z/3Z × Z/4Z

(Z/105Z)*
≅ (Z/3Z)* × (Z/5Z)* × (Z/7Z)*
≅ Z/2Z × Z/4Z × Z/6Z
≅ Z/2Z × Z/2Z × Z/3Z × Z/4Z

[next obsidian]

It looks like just applying Chinese Remainder Theorem

Or really, the fact that the group of units of G x H is the products of the group of units in both factors

I don’t see how you go from (Z/8Z)^* to (Z/3Z)^* x (Z/3Z)^* though

[sonic parcel]

phi(8) = phi(3)*phi(3), and both are isomorphic to the klein group

[golden pasture]

probably like narrow it down to phi(n)=phi(n+1)

so we can write some decently efficient code (ofc can improve computing phi(n) but wtv:

sage: for i in range(1,1000):
....:     if euler_phi(i)==euler_phi(i+1):
....:         #print(i)
....:         if Zmod(i).unit_group().is_isomorphic(Zmod(i+1).unit_group()):
....:             print(i)
....:
1
3
15
104
495
975

the first instance that the first condition is satisfied but not the second is 164

[golden merlin]

OKAY

okay

can someone here define me the "Centralizer" of a group

[thorn delta]

the centralizer of g in G in is the set of elements in G that commute with g

[golden merlin]

that's the thing.

@thorn delta what you just defined is the Center of a group

which is different than a Centralizer

[thorn delta]

nope. the center is the set of elements of G which commute with every other element of the group. not just g

[golden merlin]

i got this definition

[viscid pewter]

A is a subset

not an element

[golden merlin]

and it conflicts with what u told me

[thorn delta]

no, it doesn't

[viscid pewter]

the centraliser of A in G is the set of all elements g in G that commute with every element a in A

[thorn delta]

the center would be C_G(G)

[steep hull]

No it doesn’t, A is just {g} here

[thorn delta]

for example

[viscid pewter]

the center is the centraliser of G in G

[golden merlin]

right

[viscid pewter]

we good?

[golden merlin]

but centralizer doesn't talk about commutativity

[viscid pewter]

it's equivalent

[golden merlin]

it talks about conjugates

[viscid pewter]

if gag-1 = a, ga = ag

it's equivalent

[golden merlin]

WHAT

SINCE WHEN i never learned that

show me

[viscid pewter]

multiply both sides by g on the right

done

one step

it's very trivial

[steep hull]

Um this shouldn’t be a surprise if you’ve discussed normal subgroups

[viscid pewter]

if the conjugate of a with g is a again, g commutes with a

[golden merlin]

woah

if the conjugate of a with g is a again, g commutes with a

woah

no one told me that

[viscid pewter]

ok

[scarlet estuary]

why do you think we care about normal subgroups

[steep hull]

Are you learning on your own

[golden merlin]

:P

[golden merlin]

yea

[viscid pewter]

i love jumping in head-first as much as anyone

but it really is worth going through all the theorems related to normal subgroups

[old lava]

ya

[viscid pewter]

they are very important conceptually

many equivalent definitions

[golden merlin]

ok

[old lava]

I'm learning abstract algebra on my own as well right now, but even if it seems really boring, I did all the intro chapters

[golden merlin]

list me theorems about normal subgroups

[old lava]

where they throw a million definitions

[viscid pewter]

okay so like

you can quotient by a subgroup iff it's normal

[golden merlin]

yee

ik what else

[old lava]

probably a pretty complete theorem

[viscid pewter]

there we go

[old lava]

for normal subgroups

[viscid pewter]

that's some

[golden merlin]

ok any more theorems

[viscid pewter]

there's the homomorphism thing

[old lava]

there's a million theorems lol

[golden merlin]

ik that 1 too

[old lava]

you should read a textbook

to properly understand all of it

[golden merlin]

gag^-1 on my cock bich

lol

[scarlet estuary]

...

could you not

[golden merlin]

lol itsa joke!!!! a pun

[chilly ocean]

Thank you, I enjoyed the joke

Nami was just born without a funny bone

[old lava]

I do be agreeing with nami here

[cursive basin]

WHERES the joke

[golden merlin]

IT''S CONJUGATION!!! OF COURSE DUH

in abstract algebra when gag^-1 = a

but then i made it into a pun

DO I REALLY HAVE TO EXPLAIN THE JOKE 4 U

[golden merlin]

yea mathbath gets it he thinks it's funny ❤️

[scarlet estuary]

[carmine fossil]

I still don't get why it's funny

[chilly ocean]

[fading wadi]

How do you go about showing a group of order n is simple or not

when it's possible to do so

[golden merlin]

A_n is simple ik tht

but i don t know much elsee yet

[thorn delta]

for n!=4 it is. The proof is not simple though

[sonic parcel]

lol

[fading wadi]

what is A_n?

[golden merlin]

alternating

[viscid pewter]

avocado's number

[golden merlin]

lel

[fading wadi]

So for example, I have a group of order 1722 and a question says it's not simple. I think it has to do with the prime factors 1722 = 2 * 3 * 7 * 41, but idk what exactly it is

[thorn delta]

not simple is often easier

[next obsidian]

You use Sylow

QED

It has a normal Sylow 41

This is basically how you approach a lot of these

At least those which will be assigned

[fading wadi]

oh

any other way to approach this if you don't learn sylow?

[next obsidian]

Or I guess theoretically it could have 42 Sylow 41’s

Lol probably not

You have nothing you could do

[fading wadi]

alright, thanks I'll look it up

[next obsidian]

With a ton of effort you could probably prove the stuff that Sylow would imply

Why do you ask?

Is this assigned in a class?

I find it unbelievable you’d be assigned something that large without knowing Sylow

And if you’re just doing random problems you should just learn enough group theory to naturally get to Sylow

[fading wadi]

yea it's part of a class

it's a homework that we were assigned but maybe we haven't got to that point yet

[scarlet estuary]

yeah these types of problems are like

the most stereotypical and computational questions in a group theory course

in that they are very rote and basically always rely on the sylow theorems

[sturdy marsh]

LHC2012, do you go to UCLA by any chance

yea can we please not write anymore hints on the 1722 thing

[fading wadi]

I apologize for this. Our professor did not write that we cannot ask others online, but I should not have asked before consulting him.

[scarlet estuary]

understandable; i dont think the help you got here would be much different from what you'd get with a quick google search honestly

so its probably not a big deal

but for the time being, yeah, let's drop this questioning

[cloud walrus]

bdobba

[untold sapphire]

Struggling with this one

[cloud walrus]

bdobba

[fading wagon]

You can try to prove that G has a normal subgroup of index 2. Consider the group as permutation group over the elements (Cayley). Let there be an element a of order 2 (pair elements and their inverse, then identity pairs itself, but total number of elements is even OR cauchy), then a corresponds to an odd permutation because a has 3*7*41 transpositions. Hence, consider the set of elements that correspond to even permutations. Clearly, it's a subgroup of index 2.

[sage fjord]

Can anyone help explain what I've got to do to answer this question. "Describe the classification of all irreducible unitary finite dimensional representations of the simple lie algebra su(3) of the lie group SU(3)". I understand and know the irreducible representations and classification theorems (Dynkin diagrams etc) but not sure what exactly this question is asking me to do

[old lava]

Lmao it took like 2 pages of proofs to prove sylow stuff, and it wasnt just computationally hard, would be impressive to independently prove sylow stuff

[sage fjord]

<@&286206848099549185>

[next obsidian]

This is true in any abelian category

Not sure how categorical the stuff you know is, but you could show they preserve limits and colimits respectively

[leaden finch]

determine if the polynomial is irreducible $x^4 +2x^2+1$ in $Q$

[cloud walrus]

SUNSHINE

[leaden finch]

can i use eistein criterion?

i get confused when to use it

[next obsidian]

No

You could try making a substitution and then applying t but I don’t imagine it will work

[leaden finch]

should i do rational root test?

[delicate bloom]

that won't be enough either

it could have no rational roots but factor into two irreducible quadratics

notice it's a quadratic in y=x^2 and think about that

@leaden finch

[sturdy marsh]

aight

[delicate bloom]

ty ty

[untold sapphire]

Limits are next on my list

I was wondering if there was a simple way to show that left adjoints preserve cokernels but perhaps not

Guess you can't really avoid category theory in homological algebra

[next obsidian]

I think

You can probably do it manually

But I imagine the proof is actually just doing a special case of preserving limits and colimits

I think?

So i feel like it’s more fruitful to just prove the general result

[barren sierra]

I'm stuck on this

so I know by trichotomy that either 0 <= 1 or 1 <= 0

I proved that if 1 <= 0 then we have a trivial ring

so now I'm stuck on this other case

[sturdy marsh]

There's an 'infinity' element in your ring

what can you say about 1 + infinity

[barren sierra]

how do we know that there is such an element

that infinity element is the supremum right?

[sturdy marsh]

yes

[barren sierra]

how do we know it's in the ring?

Cause I had an original proof that assumed that

but we can't assume that

[sturdy marsh]

is that not what they're saying, every subset has a supremum?

[barren sierra]

R is a subset of R

[sturdy marsh]

yes

[barren sierra]

so R has a supremum

but that doesn't say said supremum is in R

[sturdy marsh]

im pretty sure they're saying the supremum is in the ring, otherwise the statement doesnt really make sense (unless R is a sub-thing of something else)

[barren sierra]

I don't think they're saying that

[sturdy marsh]

okay, then what does 'has a supremum' mean?

or really, what is your definition of supremum

with all quantifiers

[barren sierra]

A supremum of a set A is an object s such that s >= a for all a in A

it may or may not be in A

[sturdy marsh]

what is this object s?

[sturdy marsh]

right, over here b lives in P

[barren sierra]

sure?

yea

[sturdy marsh]

read

it doesnt make sense to say 'an object s'

[barren sierra]

hm ok

so can we conclude that the supremum is in R

in my problem?

[sturdy marsh]

im pretty sure that's what they mean, unless the ring sits inside something else (and if the something else is just a set, then it is wrong)

is there any more context to this?

if not, then yes the supremum is in R

[barren sierra]

there is no more outside context

[sturdy marsh]

then the supremum is an element of R

[next obsidian]

Also your definition of supremum is wrong

That’s an upper bound

The supremum as they defined is also <= any other upper bound

[barren sierra]

oh I see

[leaden finch]

hi

im completing a hw problem if i post it here can someone please check if i am on the right track?

i completed as much as i know

if anyone can kindly check my work

ill make a donation if someone can please check my work

[leaden finch]

can someone please help me

[sharp sonnet]

eh

isn't there missing some stuff

for the integral domain part

[leaden finch]

can you please help me

fix my work?

ill make a donation

[sharp sonnet]

other than that it is fine

the nonempty part you wrote down weird

[chilly ocean]

something something financial transactions are looked down upon

[leaden finch]

are you able to draw on my work so i can fix it?

i apologize terra.

[sharp sonnet]

i mean, idc about donations or wtv

for the nonempty part, just observe that $0 \in S$

[cloud walrus]

Lochverstärker

[leaden finch]

ahuh

[sharp sonnet]

you state "let ..." which is weird

you can't decide what is in S

[leaden finch]

ahuh

so hoq can i fix it

how*

[sharp sonnet]

fix what

i just told you

[leaden finch]

i mean the wording

do i remove let?

[sturdy marsh]

Yo SUNSHINE what's up with the donation thing, did you always offer donations?

[sharp sonnet]

yes

just state that 0 = 3*0 is in S

[leaden finch]

well i appreciate people's time i thought it would be kind to do that.

[sturdy marsh]

it's just kinda weird as it is finals season ya know

[sharp sonnet]

other than that is fine, just 'sloppy'

[leaden finch]

not in my country

i see

[sharp sonnet]

you don't state where your x's and y's and n's and w's are from

[leaden finch]

is that all you think i need to fix?

ok

[sharp sonnet]

and then you need to show that the product of nonzero elements is nonzero

[leaden finch]

hmmm

im not sure what that means

ill look again...

[sharp sonnet]

take two elements 3n, 3w from your set S

multiply them

then show that if the product is 0, one of the factors must be 0 as well

[leaden finch]

let me check

[barren sierra]

is the answer to both of these the empty set?

[thorn delta]

Yes for 39. The empty set doesn't belong to X*

[barren sierra]

?

wdym

it says "empty sequence {} in X*"

[thorn delta]

im pretty sure they are different set theoretically.

maybe not tho

[sturdy marsh]

yup, the empty sequence is the only invertible element in the free monoid on a set

[barren sierra]

cool

ty both of u

[main quiver]

what book is this? @barren sierra

[barren sierra]

my teacher's notes

[main quiver]

thanks

[barren sierra]

do you want the rest of the notes?

[main quiver]

sure

is this an algebra 1 course?

[barren sierra]

no

it's an intro to proofs course

@main quiver

I'm not gonna send notes 1 cause it's just "this is what the for all symbol is. This is what => and <=> means" etc etc

[main quiver]

Intro to proofs goes over monoids?

Dang

[barren sierra]

this one does ;-;

from intro set theory to the construction of the reals

nice progression but it was hard

I have my final tomorrow evening ;-;

[barren sierra]

Would it not be easier to prove b and then use b to prove a?

[celest nebula]

If anyone here has a sec and wouldnt mind lending me a hand I would appreciate it. I know the formula is xNx^-1 for a normal subgroup but im not sure how to apply it.

[viscid pewter]

all subgroups of ZxZ are normal as ZxZ is abelian

[celest nebula]

I think for this example my prof wants us to use the formula xNx^-1 cause we havent learned that all subgroups of abelian groups are normal

[old lava]

(a, b) is some arbitrary element in Z x Z, (a, b)^{-1} is (-a, -b)

(a, b)(2nx, 3my)(-a, -b) = (2nx, 3my)

since any arbitrary element acts on any element of the subgroup trivially

it's a normal subgroup

[viscid pewter]

so for any subgroup N of an abelian group, for all x, xN = Nx, as for all n in N xn = nx

it's very short

[celest nebula]

thank you very much @old lava @viscid pewter

[scarlet estuary]

yeah you can view being normal as kind of being "locally abelian"

since xax^-1 = a is the same thing as saying xa = ax

obviously the catch is the quantifiers (ie where x, a are coming from)

but still

[wheat musk]

How can I compute the ideal in C[x,y] that is generated by x^{3}-y^{2} and x^{5}-y^{2}. C are the complex numbers

[latent anvil]

what do you mean by "compute"

like what should your answer look like

[wheat musk]

I want to know which polynomials are in the ideal and which are not.

Need to figure out dim C[x,y]/(x^{3}-y^{2},x^{5}-y^{2})

[chilly ocean]

I guess every monomial is of the form x^a y^b, where a=0,1,2,3,4, and b=0,1, although more than this I cannot say

So, no more than dimension 10, and I guess it divides 10

[latent anvil]

I agree with that

[chilly ocean]

I am not that good at algebra to know how to show if it is smaller than 10 though

I thought there is something about grobner bases, although tbh I don't know if they are relevant here

[latent anvil]

I think it should be exactly 10, but I'm not sure what the best way to show it is. Maybe you can argue that any polynomial in I cap C[x] has degree >= 5, so all of the x^i y^0 are linearly independent

I mean those two statements are equivalent

And similarly if you have a polynomial involving as its own term in I then the y degree should be >= 2

that's not exactly precise lol

solution: think hard about the degrees of elements of your ideal

[old lava]

Wouldnt that be center/ centralizer rather, because normal is like a weaker condition than being locally abelian

[next obsidian]

Uh

[old lava]

Usually all the texts I've seen just are more informal and just refer to it as S, the operation is either implied or stated somewhere, and not written again as (S, *)

[next obsidian]

Well if you know literally nothing about * it’s called a “magma”

[old lava]

and from then on, they just refer to S as a set

until it's shown to be a group

[chilly ocean]

How would you describe to somebody what a ring/group/field is in English, rather than mathematically?

[viscid pewter]

i'd do it in terms of symmetries

vaguely

[old lava]

ya, using the dihedral and symmetric groups as examples lends itself nicely to explaining groups as symmetries

for rings and fields, I feel like you kinda have to go a bit into some math-y stuff

like polynomials at least

[sturdy marsh]

tbh it took me a long long time to buy the whole "groups are symmetries of stuff" thing

and dihedral groups did not help at all

[old lava]

I feel like dihedral groups are the second most basic symmetries possible (geometrically)

[sturdy marsh]

sure, but they just felt stupid when I started learning about them

reading about applications of groups helped me ig, and eventually I started to enjoy learning group theory for it's own sake

reading about stuff like this helped:
https://en.wikipedia.org/wiki/Particle_physics_and_representation_theory

https://en.wikipedia.org/wiki/Wigner's_theorem

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

[scarlet estuary]

i think the most natural perspective to approach groups/rings/fields from is as generalizations of Z

or Z/nZ

[viscid pewter]

i had the exact opposite experience brof

[tiny pagoda]

It can also help to abstract from groups as elements sometimes

[viscid pewter]

the symmetry thing really really helped

[scarlet estuary]

i mean maybe im weird, i hate working with like

specific groups

like i enjoyed reasoning about groups abstractly

but despised things like finding the sylow subgroups of blahblah

or determining the centre of bleurgh

[viscid pewter]

fair enough

[sturdy marsh]

this was fun for me too, but just the same amount of fun as solving a super-contrived puzzle

[scarlet estuary]

i think i've said this on this server before

but when i finished group theory i knew 2 things:

• i wanted to spend the rest of my life doing algebra
• i absolutely did NOT want to spend it doing group theory

[sturdy marsh]

I wouldn't mind doing group theory

[scarlet estuary]

admittedly the theory of lie groups was a bit more compelling than finite groups

but still

[sturdy marsh]

I also remember reading an article on the SO(4) symmetry of the hydrogen atom, which was also super cool

[next obsidian]

I love groups

literally never have I been motivated by "groups are symmetries"

At a certain point I started thinking of groups far more in terms of how they act on things, which I guess you can say is some sort of symmetry but...

idk it doesn't feel very motivating to me

[golden pasture]

ye i found it as well by accident when trying to get lrl vector to qm xd

cool stuff

[old lava]

oh my god that's so annoying

the only time I liked those questions was when I had to use lattices to basically easily just look it up

[leaden finch]

can someone please help me with this one

[delicate bloom]

this looks like a #linear-algebra question, depends on what beta is

[leaden finch]

oh yes linear algebra

[marsh fractal]

is there a nice book to read on the development of abstract algebra?

kind of the story of axiomatic systems, why they were made, by whom, and how it developed etc.

[thorn delta]

my understanding is that the development of algebra happened mostly separately from the development of the axiomatic systems used to define things like groups, rings, etc... The latter is more a result of work done by the Bourbaki group

[sturdy marsh]

I don't know if there's a good book on algebra, but Dieudonne wrote a pretty good article on the history of algebraic geometry

Also, the introduction section of Kleiman's article on Picard schemes has ~15 pages of history, going from Bernoulli to Grothendieck

[chilly ocean]

I guess some (a lot?) of group theory was developed for galois theory, although I think the formalization came after

Or more generally for the study of roots

Eg I think I recall that Lagrange came up with his theorem about order of subgroup divides order of groups while studying roots

[sturdy marsh]

the orginal form of Lagrange's theorem was something like if you had a polynomial in n-variables, say p(x_1, ..., x_n), then the number of different polynomials you can get by permuting the variables divides n!

the permutations which fix p is a subgroup of Sn

[rain tulip]

Might be a silly question but are the left-invariant vector fields of a Lie group precisely the generators of that group? I would think yes since they are both said to be the elements of the Lie algebra, but I can't figure out how to reconcile the two. Take for example U(1), it has a single left-invariant vector field ∂/∂θ but its generator is the imaginary unit i. How are these two the same (or how are they connected)?

I can't help but feel I'm getting a little bit ahead of myself when it comes to this stuff

[sour plume]

It depends on how you represent your Lie group and the generators, but yeah, the left-invariant vector fields are in 1:1-correspondence with the generators, or the Lie algebra, if that is a familiar term to you

One way to see that is by noticing that a left-invariant vector field on a Lie group is fully determined by its value at the identity, since if you have the value at the identity, you can just extend the vector field to every other point by the left action. That's exactly what left-invariance means. So left-invariant vector fields are in 1:1-correspondence to the elements of the tangent space of the Lie group at the identity. But that's how people often define the Lie algebra of a Lie group. That's the space where the generators live

The generator of U(1) is equal to i if you think of U(1) as the circle in the complex plane. But of course, you could also view it as the circle in R^2, and then the vector i from the complex plane is precisely mapped to the vector (0,1), which is exactly ∂/∂θ at the identity

@rain tulip bepis

[rain tulip]

Ahhh brilliant. So it's just the vector field evaluated at the identity, and that characterises the whole vector field. That makes a lot of sense, thanks!

[mellow forge]

It is exactly that: group G is symmetries of space M <=> M is a representation of G
So you care very much about symmetries 😛

[next obsidian]

Yeah, I just don’t really see representations as symmetries at the moment lol

[dim escarp]

My AA introduction Exam ended >.<

Idk why they were obsessed with functions not easily irreducible

[carmine fossil]

Let F be a field and F' be an algebraically closed field containing F. Take polynomials p(x),q(x) in F(x). If p(x) divides q(x) when both are seen as elements of F'[x] ,is it true p(x) divides q(x) in F[x]?

[sturdy marsh]

yes

there are several ways to prove it

one is is to work with q = pf and prove it term by term

another way is embed everything into F(t) and F'(t) and work over there

another way is that p divides q is the same thing as gcd(q,p) = p, and the gcd is computed by the euclidean algorithm, which never leaves F

[golden pasture]

isnt it direct cuz
p divides q -> exists k in F[x] q=pk
since k in F[x], k oso in F'[x]

[carmine fossil]

I meant to ask if k exists in F'[x] is k in F[x],so it's not that direct

[delicate bloom]

maybe I've made some kind of terrible mistake or not idk. We know q(x)=p(x)r(x) in F'(x) and we have q(x), p(x) in F(x) and let's suppose r(x) is not in F(x). But r(x) = q(x)/p(x) is in F(x). So we're pretty much done.

actually maybe I just said what a cute cat just said but more long winded lol

[sturdy marsh]

no, a cute cat assumed p divides q over F

[sturdy marsh]

you did this

[delicate bloom]

oh she went backwards

[sturdy marsh]

there's a lot of these neat field extension results

if two matrices over F are similar of F', then they are similar over F

the nicest proof of that I know uses RCF

ig another way to do it is using the fact the field extensions are faithfully flat

[golden pasture]

oh oops misread lol yh went backwards xd

too tired these days to england

[thorn delta]

random question, but when we have a group with a presentation G = <X | R>, we know that G is the image of the free group on X with kernel equal to the normal subgroup generated by R. About that normal subgroup: is the subgroup generated by R always normal, or do you generally have to take the normal closure to get the kernel of this map?

[sturdy marsh]

let G = < x, y | y>

the subgroup generated by 'y' of the free group on x and y is not normal

in particular, xyx^-1 is not in the subgroup generated by y

[thorn delta]

ah yea, okay. thanks

[next obsidian]

Please forgive for the crosspost (in the algebra and geometry channel), but I think it’s justified. A lot of people here have heard about the moduli and stacks course that Alper is running this winter. Here is a link to the course website, in particular note at the bottom he’s included a way to contact him if you would like to informally join. Many people have asked me about this, so here’s the info: https://sites.math.washington.edu/~jarod/math582C.html

In particular I believe @sturdy marsh (I hope that’s brofibration) and @fierce perch have asked me about this so here’s that

[fierce perch]

Thanks!

Wow the notes look extremely down to earth

Do you know if he recommends any prereqs?

[next obsidian]

I think they’re mainly just a general first course in scheme theory, and he says knowing resolution of singularities is important

There’s some stuff at the end about like recommended texts

One thing I forgot to mention: he said that he will record lectures and post them to the website as well. I don’t believe that’s stated on the course page.

[wind steeple]

for me, the ramification index is, when you consider number fields extensions K subset L, A and B their number rings and if p is a prime ideal in A and P a prime ideal in B, the P-ramification index of p is the exponent of P in the decomposition of pB

and de f_P is the inertia degree, it's the dimension of the vector space B/P over A/p

[bronze trench]

ok quick maybe dumb question I should really know the answer to

if for a group G and a prime p there's a single sylow p-subgroup P of G then P is normal in G right?

[carmine fossil]

Yes

[bronze trench]

oh yeah nvm it's direct, the only conjugates of a sylow p-subgroup is another sylow p-subgroup and since there's only one, every element conjugates P to itself hence normal

is that it?

[carmine fossil]

Yes

[bronze trench]

ok yeah dumb moment thanks. even just writing the question helped 😛

[cyan marten]

Generally if there is a unique subgroup of a given order, then it's characteristic and hence normal.

[bronze trench]

hm, interesting. That's because automorphisms send subgroups to isomorphic subgroups?

[carmine fossil]

Yes

[bronze trench]

ok nice, never even thought of that, but it's simple (pun intended)

thanks for giving me one more tool for my GT efforts 😄

[cyan marten]

An exercise given in D&F is to prove that Z[i] / (a) has order N(a), where N is the norm (x^2 + y^2). How does this generalize? (I know almost no algebraic number theory).

[carmine fossil]

Do you know if a and b are non associate irreducibles a^n and b^m are coprime?

[cyan marten]

Yes.

[carmine fossil]

So,you can use crt to reduce it to the case where alpha is p^n,p is irreducible

[cyan marten]

True, that's also the method given in D&F.

[carmine fossil]

Which follows if you prove it for when alpha is a irreducible (by the previous part of question)

And that's literally the preceding exercise

[cyan marten]

Thanks. So the fact that Z[i] is a PID is irrelevant.

[carmine fossil]

You use Z[i] is a Euclidean domain to show |Z[i]/(pi^n)|=|Z(i)/(pi)|^n(Ok,You can't,I made a mistake here)

[cyan marten]

I showed it by using the first lemma and applying
[R : J] = [R:I] [I:J]
repeatedly.

The first part*

My question was about the integer ring of Q[sqrt(-n)] and its norm N(...) = x^2 + n y^2. Do we always have the same result?
I know that it isn't usually a PID.

[wind parrot]

You can generalize this to arbitrary orders in number fields: for a field extension $K$ of $\mathbf Q$ of finite degree $n$ and $x\in K$, define the map $\mu_x\colon K\to K$ by $\mu_x(a) = xa$. This map is $\mathbf Q$-linear, and we call its determinant the norm of $x$ (you can check that for $K=\mathbf Q(\sqrt{-n})$ this agrees with your definition of the norm). An order in $K$ is a ring $R\subset K$ that is a free abelian group of rank $n$. Now, if $R$ is an order and $x\in R$, then the index of the ideal $xR$ in $R$ is exactly the norm of $x$.

[cloud walrus]

NielsK

[carmine fossil]

Yes,It is irrelevant

[cyan marten]

I think the norm is defined differently depending on n mod 4..

[cyan marten]

Neat, thanks!

It is easy to find all the orders explicitly, right?

[bronze trench]

ok more dumb "I should really know this" questions

what's Gl(2,p) like? is there an easy description?

more specific. Are there elements of order q for a prime q less than p? If q=2 there are, it's just picking a basis for our vector space and permuting the bases elements. Are these the only ones? And if q is an odd prime? My intuition tells me there aren't any but tbh I don't know how to develop this