groups-rings-fields

[chilly ocean]

And this, this has always been helpful to me

[void cosmos]

any reducible algebraic sets must have finite components right?

over a field or any noetherian ring correct?

cuz u go like V = V1 U V2 and if thats not it then u can decompose V2 or V1 into further pieces ( with each being proper subsets of V )

but each V1 is algebraic so its the zero set of some ideal

so each V_i gives an ideal?

and this gives us a chain?

which must terminate?

[coral spindle]

It must have finitely many components, but not typically finite components.

[void cosmos]

why

how does the proof go

[coral spindle]

as you said it does.

[void cosmos]

yea cool

tysm

[formal ermine]

(cuz hbt)

[void cosmos]

what

whats that

[formal ermine]

Hilbert's basis theorem

[void cosmos]

yeah yeah

op and has relatively easy proof ( to follow not to come up with ofc haha )

[coral shale]

🤨

n = qd + r ???

n = (q-1)d + (r+d) ?

[viral haven]

Can someone give an example of a short exact sequence of $\mathbb{Q}[x]$-modules that does not split? Would $0\rightarrow \mathbb{Z}\rightarrow\mathbb{Z}\rightarrow \mathbb{Z}/p\mathbb{Z}\rightarrow 0$ be one such sequence?

[cloud walrus]

ceramic.sky

[south patrol]

Not really sure how you're turning Z into a Q[x]-module

That is an example for Z-modules, sure

[viral haven]

I don’t know either, tbh.

[next obsidian]

Virtually all quotients of Q[x]

If it split then Q[x] would contain some torsion

You can say the following:

For any ring A, the module A/I is projective iff I is a principal ideal generated by an idempotent element

Note that A/I projective would be equivalent to asking that the natural map A -> A/I splits

[somber sleet]

If I take any Subgroup $H < D_4$, why do I know that $D_4/H$ is a simple subgroup?

[cloud walrus]

damn_guuurl

[somber sleet]

I mean, I know that it is of order 2(prime order), which means that it is normal and I also know that it is cyclic, but to say that it is simple I need the subgroup to be abelian right? Why is it abelian?

[south patrol]

Simple subgroup of what? Lol

I assume you mean simple group but then it is false

[somber sleet]

I show you gimme a sec

[elder wave]

Potato understands German btw

[somber sleet]

then I'm gonna send a screenshot

[rocky cloak]

There is only one group of order 2

[somber sleet]

I do not see why H is simple

[rocky cloak]

It isn't

[somber sleet]

then why is there written "einfach", is it meant for D_4/H? If yes, then why is it wnough that it has primeorder? shouldn't it also be cyclic?

[elder wave]

It is cyclic

[somber sleet]

okay any group of prime order is cyclic

[elder wave]

It has order 2

[rocky cloak]

Yes, there is only one group of order p

[somber sleet]

I don't remember algebra I

then it makes sense, thanks guys

[rocky cloak]

My German skills are not great, but I think the point is that there are no simple groups of order 4 or 8, so the first step in the composition series is finding a subgroup of index 2.

Then they find all those subgroups, and after that all composition series

[stone bane]

hey, does this multiplication table mean that i have an abelian group?

*| a|b|c|d|e

a| a|b|c|d|e
b| b|e|d|a|c
c| c|d|b|e|a
d| d|a|e|c|b
e| e|c|a|b|d

[south patrol]

Well you just gotta check if its symmetric about the (anti)diagonal to see if it is abelian

So yes

(Also every group of order 5 is abelian as it happens)

Tho I've not checked if you actually get a group lol

[stone bane]

how do i do that?

[summer path]

You do that in the natural way, e.g. you check if ac = ca, by reading off the cayley table

[stone bane]

How do i know if it is assosiative

[summer path]

Yes you need to also check you indeed have a group, but you can also do this by looking at the cayley table

[livid willow]

Yes, to go further, you can try proving something more interesting that every group of order 5 or less is abelian

[celest cairn]

Anyone wanna read my answer to a precalc problem using Galois Theory?

[solemn dew]

no

[coral shale]

If you wanna share something just share

[rough spruce]

Let $M$ be an $R$-module and let $K$ and $L$ be submodules of $M$ then $L/(K\cap L) \cong (K+L)/L$. This is the statement of the second isomorphism theorem for modules, but since $(K+L)/K = L/K$ we can simplify it to $L/(K\cap L) \cong L/K$. I didn't see anyone write it in this simpler form, why?

[cloud walrus]

ZornHadNoChoice

[crystal turtle]

(K+L)/K ≠ L/K in general

[rough spruce]

How?
$(K+L)/K = {(l + k) + K: l \in L, k \in K} \ = {l + K: l \in L} \ = L/K$

[cloud walrus]

ZornHadNoChoice

[crystal turtle]

If K is not contained in L, it does not make sense to talk about L/K anyways

[rough spruce]

Oh that's right!

[crystal turtle]

There's a 1-1 correspondence of submodules in M/K and submodules of M that contain K. The " + K" part essentially forces K to be contained in it

[rough spruce]

Thanks, this makes sense now.

But you know what, we can define $L/K := {l+K: l \in L}$ and this is an $R$-module (it's the image of L under the natural projection $M \to M/K$), further $L/(K\cap L) \cong L/K$.

[cloud walrus]

ZornHadNoChoice

[oblique river]

the automorphisms of that field are not rotations

And the size of the automorphism group of Q(zeta_n)/Q is not n

It’s phi(n), where phi is the euler totient function

[celest cairn]

Ok, thanks I think I got it now.

[solemn dew]

If H is a normal subgroup of G

is G/H cyclic?

[coral spindle]

No!

Very rarely!

[coral shale]

take G/1 for a noncyclic G

[solemn dew]

i was thinking about

some sylow subgroup of G that is normal

[coral shale]

You can choose G and H so that G/H is any group you like. example A = A x B / B

[solemn dew]

hence we can make a homomorphism

G/H_1 x ... x G/H_n for each p-sylow subgroup

[coral spindle]

If all of them are normal, yes.

[solemn dew]

this is a abelian group

so then G/H_i must be cyclic, i = 1,2,3...,n

[coral spindle]

Hmmm I don't see why this must be Abelian.

I also don't see why you say that follows; it doesn't.

[solemn dew]

alrighty

[rocky cloak]

So abelian groups need not be cyclic, modding out a sylow subgroup doesn't need to give you something abelian, and sylow subgroups arent usually normal

[coral spindle]

I think there is another mistake happening here

[coral shale]

intuition gone wrong

[coral spindle]

Jonathan seems to be confusing the fact that f.g. Abelian groups are products of cyclic groups with the incorrect idea that if an Abelian group is expressed as a product, all parts must be cyclic, which is simply false.

[solemn dew]

bruh

i meant that if G/H_1, and H_1 is a normal p-subgroup of, then G/H_1 must be cyclic

[coral spindle]

That is false, so I hope that clears that up.

[solemn dew]

yea it clears things up

[void cosmos]

what tf

yo

doesnt V being in this set literally say its irreducible?

ik if it were to be reducible it would have finitely many components ( over a noetherian ring )

[rocky cloak]

If V is irreducible, then it is the union of irreducibles

Just itself

[void cosmos]

yea but why is V not irreducible

im confused

[rocky cloak]

They are proving that every algebraic set is a finite union of irreducibles

[void cosmos]

oh

lmfao yeah so if V is irreducible then as you said its true

so we only look the redcuible case

im so stupid or idk the way its written

ugh

tysm

[rocky cloak]

Yeah, it's written a little akwardly maybe. But I'm not sure what would be a better way to put it, so...

[void cosmos]

imo it should have been at the very top

if V is irreducible : then we are done

if not

then ...

but its literally written "since its in the set,then "

even though i dont see how does it follow from being in the set

😦

[rocky cloak]

I think that would just be more confusing. V can't be irreducible, since then it wouldn't be in the set.

[void cosmos]

yea

ur right

but the thing is i didnt catch that

but yeah

[void cosmos]

for uniqueness

do i get to assume that both decompositions have the same length? ( number of components )?

[chilly ocean]

What are two fields that are isomorphic, but no isomorphic as extensions of some subfield?

[formal ermine]

you mean isomorphic as vector spaces over the subfield?

[void cosmos]

any two finite fields having the same number of elements are isomorphic

so think about those

yea as a vector space if you are extending over a polynomial of degree n then you would get a field with p^n where p is the number of elements in your finite field

choose another polynomial with degree m with m!=n and there u have it ig

[rocky cloak]

C and the algebraic closure of C(x) are isomorphic as fields, but not as extensions over C

[void cosmos]

oh lol i understood the question wrong

mb

[formal ermine]

another related fun fact

[rocky cloak]

The isomorphism is kinda cursed though

[formal ermine]

C and C_p are isomorphic but not homeomorphic

hey jagr, color change!

[rocky cloak]

Do the colors mean something?

[formal ermine]

dark blue = only slightly active

light blue = very active

[rocky cloak]

Ah, cool

[formal ermine]

advanced channels count more than e.g. #chill or #discussion

[rocky cloak]

Or, Im not sure that's a good thing hehe

[formal ermine]

if you go inactive again you get the emeritus role

and immediately get very active upon being active again

skipping the active role

you also get access to #feet

[rocky cloak]

Les cursed example is maybe

Q(x2, x4, x6, x8, ...) is isomorphic to Q(x1, x2, x3, x4, ...)

Or just Q(x2, x3, x4, ...) Isomorphic to Q(x1, x2, x3, ...)

[void cosmos]

how do you show that for any polynomial f in R[x,y] we have f = u(x,y)*(y^2-x^2)+r(x)

like euclidean algorithms arent really there in multivariate poly rings right?

[rocky cloak]

How would you write say f=y on this form?

[void cosmos]

yea exactly

i cant

right?

[rocky cloak]

Maybe you meant for it to say (y - x^2)?

[void cosmos]

omg yeah

im trying to show that any element of I({(t,t^2)| t in R}) is a multiple of the polynomial

as in show that V(y-x^2) = I(y-x^2)

which is prime so showing that this algebraic set is irreducible

i can show its prime easily by just quotining it out and it giving us R[x] which is a domain

i just dont know how to prove the other inclusion

[rocky cloak]

So the map R[y, x] -> R[x] sending x to x and y to x^2 has kernel (y-x^2) and a section given by the inclusion.

So f(y, x) - f(x^2, x) is in the kernel and thus is of the form u(y,x)(y - x^2)

So just pick r(x) = f(x^2, x)

[void cosmos]

damn

never would have thought about that like this haha

[void cosmos]

yea got this

so f(x,y) would be u(x,y)(y-x^2) + r(x) in x

so for this to vanish on the set r(x) must be 0

for all t

or all x

which is the zero polynomial ig

in R[x] ofc

right?

[rocky cloak]

Yes, assuming R is an infinite field

[void cosmos]

tysm

[rocky cloak]

np

[chilly ocean]

Is there an example with finite fields?

[rocky cloak]

No, you can find isomorphisms that are not isomorphisms of extensions, but the extensions will also be isomorphic.

It basically comes down to there being very few finite fields

In general if the field is algebraic over it's prime field I think you won't be able to do this.

Simplest example is perhaps Q(x) isomorphic to Q(x^2), but not as extensions over Q(x^2)

[coral spindle]

In fact as long as the algebraically closed field field of char 0 is uncountable, it is unique up to isomorphism (i.e., acf(0) is uncountably categorical). I don't know if the same fact holds for positive characteristic, but I wouldn't be too surprised.

I just found an article that claims acf(p) is uncountably categorical for all primes p, so there

[rocky cloak]

Yeah, should be something like an algebraically closed field is determined by characteristic and transcendence degree over prime field.

[coral spindle]

Exactly!

I should have underlined the point here in just saying that this happens due to the fields being mahoosively big and infinite, and with finite fields you get finite obstructions.

If F is finite then F(x) can't be isomorphic to F because one is finite and the other is not™️

[livid willow]

Do people actually solve like all the problems from D&F? Every time I flip 2 pages and there are like 40 problems 💀

[chilly ocean]

most people don't do every single one

[livid willow]

I tried to do so, gave up after the first section of groups , although not very hard, its really time taking to do everything

[crystal turtle]

there are like hundreds lmao. Diminishing return after a while

[livid willow]

I now do like 10 per subsection, at random, since many questions were more or less similar

[formal ermine]

do the ones where you don't immediately know the solution sketch

[livid willow]

I will, thanks

[teal vessel]

I'm working on it rn.

working on 2.2.12.e

[livid willow]

Very based

Although I don't have the time to go through it all

[teal vessel]

why not? what's your deadline?

I sometimes have a good week between solving single problems

spaced repetition will give a better understanding anyway

[livid willow]

I have no deadline, I just only get very few hours after uni

[livid willow]

Yeah I plan to do so, I marked the ones I have solved, would just return to unsolved ones once I have time to do them

[teal vessel]

I get a couple hours between projects at work. I usually read the section, then read the first problem and chew on it for a while while working, then use my whiteboard when I have a free moment.

rn I'm showing that a particular stabilizer in S4 is isomorphic to D8, which is interesting. I just need to figure out which of these elements is the flipper and I'll have it. I already got the spinner.

[teal vessel]

wait a second, shouldn't it be possible to take some dihedral group with generators s,r and form a distinct generating set x,r where x=sr^n, since the choice of what axis of symmetry to reflect over is technically an arbitrary choice? (i.e. in a square we can either generate all elements via a L-R reflection and rotations, or we can likewise generate them with a U-D reflection)

in other words, the only very important assignment in mapping between a dihedral group and an isomorphic group is finding which "orbit" the identity element is on and labeling that orbit the rotation subgroup.

[somber sleet]

guys what is a nice way on how to interprete semidirect product

[rocky cloak]

It's a very literal interpretation, but just the fact that when you have a normal subgroup N < G, then G acts on N by conjugation, and then

ng * n'g' = n (gn'g^-1) g g'

So the action of G on N apears when simplifying multiplications of elements.

A good example to have in mind is the dihedral group. Every element is either a rotation or a reflection, and the reflection acts on the rotations by inverting them. I.e. when you turn a shape over rotations now go on the opposite direction.

[karmic moat]

is this because Z is initial in Ring?

[rocky cloak]

All semidirect products are like this you have some operations that are like "reflections" (not necessarily of order 2 or anything like that), and some that are like rotations. Doing a reflection then a rotation is the same as doing a rotation then a reflection except the rotation is twisted somehow

[rocky cloak]

Yes

[karmic moat]

sick

[teal vessel]

so I just showed that for the polynomial x_1x_2+x_3x_4, there is a way to have S4 act on it in such a way that the stabilizer is isomorphic to D8. It's now asking me to show that the same is true for (x_1+x_2)(x_3+x_4) which I can already see will behave the same way. Is there a way to show directly that this structure of a pair of two pairs (of commuting operations) can be thought of as the same, or am I simply recognizing that internal D8 structure it's already telling me to find?

(the multiplication and addition in both cases are commutative)

[mossy lintel]

how do one argue the number of left cosets are the same as the number of right cosets? if the group is finite, i can use lagrange's

[somber sleet]

oh okay, kinda see it now, thank

[somber sleet]

you have to work with group actions

[rocky cloak]

Hint: what is an operation on groups that swaps the order of multiplication?

[somber sleet]

Why is $(\mathbb{Z}/8\mathbb{Z})^{\times}$ isomorphic to $C_2 {\times} C_2$

[wraith cargo]

\times

[rocky cloak]

Have you tried writing up the cayley table?

[mossy lintel]

Even when the group is not abelian?

[somber sleet]

yeah saw it, was looking on the internet haha

[cloud walrus]

damn_guuurl

[somber sleet]

we don't do it in my algebra course, my prof is fully against it

[rocky cloak]

So I assume you're just permuting the variables. In which case I guess you can create an isomorphism of S4 sets between the sums of degree 2 monomials and product degree 1 binomials.

[crystal turtle]

What is a necessary and sufficient condition for an element of Z/8 to be (multiplicatively) invertible?

[teal vessel]

yeah, but even more basic, I just associated both of them to the diagonals of a square labeled 1,3,2,4, which satisfies that "pair of pairs" notion

[somber sleet]

there has to exist an element n not equal to 0 mod 8, s.t. n*m = 1 mod 8

which is the case for 1 3 5 7

[rocky cloak]

That'll probably work yeah

[crystal turtle]

Once you have this, you can count the number of element (there are 4 ofc) that are invertible, and for this you can just check whether there's an element that has order 4

[somber sleet]

I also noticed that they all are of order 2

[crystal turtle]

Since if so, then it's iso to Z/4, if not, then it's iso to Z/2 \times Z/2 (these are the only groups of order 4)

[somber sleet]

OOHHH

see it now

[teal vessel]

the top left vertex is the first member of the first pair, the top right being the first of the second pair. All rotations and reflections then become very clear.

[somber sleet]

I get it, thank you!!

[rocky cloak]

If the group is abelian there's no distinction between left and right cosets, so then there is nothing to show

[somber sleet]

is somebody in the mood to help me working through an exercise? It's about finding the isomorphism classes of each group of order 16 s.t. it is a semidirect product of a cyclic group of order 8 with a group of order 2

[mossy lintel]

So something like Ha -> a^{-1}H?

[rocky cloak]

Exactly. Do you see why this is well defined?

[mossy lintel]

Yep

early i thought Ha -> aH, but quickly i realized it wasn't well defined

[somber sleet]

obviously we want to analyze $\mathbb{Z}/8\mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$

[cloud walrus]

damn_guuurl

[indigo ridge]

Im trying to use the rational root test to write this polynomial $2x^4 - 5x^3 + 3x^2 +4x - 6 \in \mathbb{Q}$ as a product of irreducible polynomials. so since the numerator $r$ is meant to divide the constant term we have that $r|6 \implies r = +/- 1,2,3,6$ and since the denominator s is meant to divide the leading terms coefficient we have that $s|2 \implies s = +/- 1,2$ so all the possible roots $ r/s = +/-1,2,3,6, 3/2, 1/2$ the only problem is I plugged all of these values in the induced function but none of these were roots and I did this twice over just to be sure. anyone know what to do?

[cloud walrus]

jayz

[chilly ocean]

check your work

[indigo ridge]

what do you mean?

I did

[chilly ocean]

,w 2x^4 - 5x^3 + 3x^2 + 4x - 6

[cloud walrus]

Results provided by WolframAlpha

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

Requested by vydrakko

[chilly ocean]

-1 and 3/2 are roots

[indigo ridge]

hmm weird ok 🙂 thanks

[indigo ridge]

see..

nvm

[chilly ocean]

[wraith cargo]

How do we know such a,b exist?
For reference R is the ring of p-integral elements of Q (i.e. no factor of p in the denominator)

[crystal turtle]

since p doesn't divide v, gcd(v, p^n) = 1, so take u * (bezout identity av + bp^n = 1)

[wraith cargo]

wow okay lol
I love it when the solution is something trivial

[crystal turtle]

(i think this works at least)

[proud spindle]

i've gotten into group theory over the past few months and i think i've discovered a somewhat interesting tidbit that i'd love to share but i feel like that requires TeX and TeX is completely alien to me

i'm not asking for TeX help or anything

i'm just curious if TeX is required for explanations

[topaz solar]

No?

Certainly helpful for symbols though

[frozen sequoia]

How do I even begin learning group rings and field?

[crystal turtle]

really useful for writing stuff up not necessary by any means for understanding

[proud spindle]

i'm kind of in the same boat as you and i feel like this every time i try to read up on stuff

[frozen sequoia]

Yh one of my modules are o n cryptography and group fields and rings is just in the way

[crystal turtle]

reading and practice problems. there's a pinned post in #book-recommendations (all the way at the bottom of pins) talking abuot some of the textbooks

[proud spindle]

the only real symbol i can think of needing is the double-stricken/blackboard Z and φ if greek letters are considered symbols

[topaz solar]

Anything you write is a symbol

[proud spindle]

if i did the math right

there are no isomorphisms with $(\mathbb{Z}_{2})^{8}$ as the domain or codomain

[cloud walrus]

nonpost

[crystal turtle]

identity

always an iso, identity on that group has it for both

[proud spindle]

i think i should've clarified; i meant groups of order 256 that are not that group

[crystal turtle]

What do you mean?

[crystal turtle]

(also yay latex )

[proud spindle]

i don't know how to put it in the aptest of words

but i found the group i'm talking about bc it's isomorphic to the set of values you can store in a byte, under the operation of XOR

[crystal turtle]

yeah, bitwise xor

[proud spindle]

and i wondered if there were any isomorphisms (bijective homomorphisms iirc) from different groups of the same order

[crystal turtle]

Well

[proud spindle]

to that XOR group

or vice versa

[crystal turtle]

We usually consider two groups to be the "same" if they're isomorphic (even if not equal)

[proud spindle]

but other than the group itself (so automorphisms) i don't think isomorphisms are possible given how the group elements relate with each other (i think people call that "group structure")

[crystal turtle]

Sure there are if you don't consider up to isomorphism. Just relabel the elements! I can call (10100100) = j, for all i want!

But if we consider groups to be unique up to isomorphism, then there certainly aren't any isomorphisms between unique groups

[proud spindle]

i always thought equality up to isomorphism was a sort of abuse of notation

turns out it is exactly what it says it is

[crystal turtle]

Nah. Once we start working with algebraic (and other) structures, working up to isomorphism really is the best way. For example, up to isomorphism, there are only 2 groups of order 4

[proud spindle]

yeah Z3 and K4

*Z4

[crystal turtle]

Because it doesn't really matter what I call each element. What might be x to you is y to me. That doesn't affect the group structure though

[proud spindle]

i find group theory to be extremely interesting but i find difficulty actually understanding a decent bit of it solely due to how jargon heavy and explanation-light things can be at times

[crystal turtle]

Yeah that's fair, it can take some time to get used to compared to HS and earlier maths. Once you do though, it's very fun indeed!

[frozen sequoia]

Anyone have exam paper questions and solutions on groups rings fields soo I can have a better understanding of it ,i would really appreciate it🫶

[toxic zephyr]

can someone help explain how this is a proof? i don't understand how this shows that cofactor expansion is the only function that satisfies these properties. this is from Artin

for context: 3.4-3.7

[coral steeple]

I understood it the following way (which I hope is right): any function d satisfying 3.5-3.7 is the map sending $A$ to whatever you calculate using 3.13. Since det satisfies 3.5-3.7, $\text{d}=\text{det}$

[cloud walrus]

person2709505

[lethal dune]

that's a very "unusual" uniqueness statement

[teal vessel]

so after doing some multiplication of matrices together, I have found that there appears to be a restriction on the center of the heisenberg group of an arbitrary field F such that a matrix with elements a,b,c is in the center only if it has the following relationship with arbitrary matrix (in the group with the elements of) x,y,z: az=cx. Because this is a restriction on the matrix containing a, b, c, we know that z and x can be any members of the arbitrary field F. This therefore implies that it must be the case that a=c=0.

Is this the full form of the center? all triangular matrices with a diagonal of 1 and the only free element being element 13 (the top right corner)? I see no restriction on that particular spot, and multiplying it out seems to work.

yeah, seems so. cool.

that would make sense, given the next part of the question was to prove that Z(H(F)) is isomorphic to the additive group F

[south patrol]

Oh lol

I googled your question and then it just gave me dummit & foote

so i mean yeah nice

[small bramble]

I think I found an error in D&F not listed in the errata

It's asking me to prove that sqrt(-n) is not prime in Z[sqrt(-n)] for squarefree n greater than 3...but if n is a prime then Z[sqrt(-n)]/(sqrt(-n)) is isomorphic to Z/nZ which is a field right? So then it would be prime

I used homo phi(a + bsqrt(-n)) = a mod n

And I think the ideal (sqrt(-n)) is equal to the kernel, i. e. all values a + bsqrt(-n) such that a = nk for some integer k

[south patrol]

I assume it should be always prime

[small bramble]

I don't think Z/nZ is an integral domain for nonprimes either though cus iirc that's iso to Z/p1 x... xZ/pn but for instance a \neq 0,(a, 0,...,0)(0,a,....,0) = 0 so it's not an integral domain & then sqrt(-n) isn't prime.

[formal ermine]

yeah

the iso obviously holds in general

[small bramble]

oh wait I was looking at errata from 2011 . But actually the errata from dummit's website which was last revised on 2021 doesn't have it either

Oh nvm it is in it now

there goes my million dollar bounty

[rocky cloak]

What was the exercise supposed to say?

[small bramble]

prove 2 is not prime

[rocky cloak]

Wut?

[small bramble]

in Z[sqrt(-n)]

[rocky cloak]

Ahh 2 is not prime in Z[sqet(-n)]

[small bramble]

For n >3

[rocky cloak]

Makes sense

Why n>3 though? Statement seems true for n=3, 2 and 1 aswell.

Don't think you need square free either.

[small bramble]

Oh well it's ultimately part of a part of a larger problem trying to get you to prove among other things that the quadratic ring of integers when -n is congruent to 2, 3 mod 4 is not a UFD.So for n = 1,2, 2 is not irreducible ((1+i)(1-i) and (-sqrt(-2)sqrt(-2)). for n = 3 it is but I guess it also doesn't include that since then -n is congruent to 1?

And when it's congruent to 1 the corresponding quadratic ring of integers is Z[(1 +sqrt(-n)) /2]

So proving there's a nonprime irreducible in Z[sqrt(-n)] doesn't apply

[void cosmos]

yo is this gaus's lemma

"implicitly"?

[chilly ocean]

what was in "Section 1?"

[glossy crag]

depends on what you call Gauss' lemma, but yeah

[void cosmos]

some algebra stuff

[void cosmos]

cool tysm

[glossy crag]

if R is an integral domain and K its FOF, then corpime in R[x] => coprime in K[x]: if p is a common prime divisor of f and g in K[x], then by factoring out its content you get a primitive irreducible R[x] polynomial that divides f and g in K[x] => divides in R[x] => contradiction

[void cosmos]

yea got it

[elder wave]

this exact thing is stated in section one and even explicitly mentions gauss

did you not even bother checking the referenced section before asking

[chilly ocean]

this is why i and

[void cosmos]

no

cuz it said you can skip it

[chilly ocean]

you missed timo's point

the text tells you to look at section 1

this has nothing to do with whether or not you actually read section 1 when you started the book

this has to do with you immediately coming here to ask instead of using the book's reference sections for their intended purpose

[void cosmos]

hmmm

its as if i asked if this is was gauss lemma just for confirmation

a simple yes would suffice :d

[chilly ocean]

the book gives you that "yes"

[glossy crag]

tbh, i'm reading Fulton atm as well and the way it's stated in Section 1 doesn't immediately imply the desired result (at least so far as i can see it)

[elder wave]

R is k[X]

[glossy crag]

Here's how it's stated. The way I understand the result you need the statement "f,g in R[x], g primitive, g divides f in K[x] => divides in R[x]" and I don't see it exactly as a trivial consequence of "irreducible over K => irreducible over R".

[glossy crag]

Hm?

[glossy crag]

What I mean is I had the same question he did at one point.

[topaz solar]

R is k[X], K is k(X)

[glossy crag]

I understand that.

What I mean is I was in the same spot he was at one point, wondering how exactly "irreducible over K => irreducible over R" implies "coprime over R => coprime over K". I had to pause and think for a moment and ended up using a result that's not in Section 1 (which although is deducible from it).

[topaz solar]

[glossy crag]

What I mean is, yes, this guy has a tendency of clogging up the channel with inane questions.

But in this one case I see his problem (assuming he is having the same problem I did).

[topaz solar]

This is altogether missing the point of him not just flipping back a page or 10

also are you sure your implication arrows are right there?

[glossy crag]

Bro, I just explained the situation. Yes, he could've bothered flipping to Section 1 and seeing the quoted result, what I'm saying is, personally, I had to pause and think how the quoted result follows from Gauss' lemma. If that's his problem, then it's understandable, although given his track record I could see how this is all just about him not bothering to check.

[elder wave]

the highlighted part is exactly what's written in section 1 and the question "is this gauss" is an immediate yes when you look at the section saying (by Gauss)

I'm not reading the same question into it as you are

[glossy crag]

w/e, this is all arguing over nothing, fundamentally we're all on the same page

[elder wave]

👍

[proud spindle]

quotient field???

[glossy crag]

some people call it that, yes

e.g. in German it's Quotientenkoerper

literally "quotient field"

i expect the corresponding English term is a calque of this

[proud spindle]

quotient groups i can understand but quotient fields? how do you do that

[crystal turtle]

I assume it's field of fractions here?

[glossy crag]

yes

[south patrol]

it's not a quotient in the same sense as with groups etc, but rather the elements of the quotient fields are formal quotients

[glossy crag]

it's just a taxonomic coincidence

[crystal turtle]

Then not quite like quotient groups

[glossy crag]

in German they don't have this problem, quotient objects are called "factor" objects

[crystal turtle]

In short: It's like considering fractions of elements in an intergral domain (special type of ring). Kinda like creating Q from Z. You'll learn the details in an abstract algebra course/book 🙃

[proud spindle]

ohhhhhhhhh

that actually makes a lot of sense

[crystal turtle]

There are also quotient rings (by ideals as opposed to normal subgroups), which under certain cirsumstances (maximal ideals) gives you a field as well

[proud spindle]

i assume ideals are somewhat like a ring equivalent to a normal subgroup

[topaz solar]

Yes, though a bit different in how you define em

They play the same role though

[crystal turtle]

fun stuff until you make the distinction between left and right ideals, which don't exist because noncommutative rings don't exist

[topaz solar]

Think uhhh multiples of 2 in the integers, or polynomials which have a factor of x^3

Not all ideals are so nice, but neither are all normal subgroups

[proud spindle]

"noncommutative rings don't exist" finally some good news

[crystal turtle]

A commutative algebraist at heart I see

[glossy crag]

i honestly don't jive with this meme, noncommutative rings are interesting, we wouldn't have CSAs otherwise

[proud spindle]

idk noncommutative rings not existing sounds nice to me

[coral spindle]

Me neither. Noncommutative shit is great.

Imagine not having rep theory of groups.

Imagine not having the quaternions.

[crystal turtle]

good

[topaz solar]

No matrices

Bye bye linear algebra

[proud spindle]

oh yeah matrices aren't commutative

[glossy crag]

this post makes William Rowan Hamilton very VERY mad

[coral spindle]

Anyway since I don't like this meme we should get rid of it uwu xP lol

[glossy crag]

to those who don't know, dude was so psyched about discovering the quaternions that he built his entire latter career around them, forcing them into absolutely everything

[glossy crag]

looks like Grothendieck

perfect target for the 2 minutes hate

[elder wave]

only problem with this is a lack of a word for the object itself when referencing it

like you can refer to X/A as "the quotient" in a sentence where said space was mentioned earlier

but the factor terminology kinda lacks that

if you get what i mean

[glossy crag]

i do

[elder wave]

i'm glad

[glossy crag]

that seems like a perfectly fine workaround

[proud spindle]

i know what a short exact sequence is but i don't know why or how they are used

[crystal turtle]

tbh I think you've got a while before you need to be worrying abot exact sequences

[proud spindle]

i suppose i should continue worrying about the semidirect product

[topaz solar]

Non commutative set theory

[elder wave]

you're 15 you should worry about homework instead

[proud spindle]

honestly i should be worrying about trig (i am horrible at shape-related math)

[glossy crag]

another one??

[elder wave]

i fear

[elder wave]

now THIS is a good timo sticker usage

exactly my reaction

[proud spindle]

are 15 year olds interested in group theory a common occurence

[topaz solar]

@formal ermine

[elder wave]

more common than one might think, which is also not hard though considering the expectation would be none

[topaz solar]

I mean it’s not uncommon

[glossy crag]

i've got the perfect reading recc for you, brb

[topaz solar]

Rubik’s cube -> group theory pipeline

Because every

Single

Book

Uses that example it feels like

[proud spindle]

that one video about group theory that got a million views -> group theory pipeline

[crystal turtle]

I have never actually seen that used as an example

[topaz solar]

what

Maybe I just had cringe resources

[elder wave]

you became a non commutative set theorist

[proud spindle]

no the rubiks cube is everywhere

[elder wave]

so yeah

[topaz solar]

There’s probably some unironic operator algebra thing that does it tbh

[crystal turtle]

noncommutative set theory for noncommutative geometry

[elder wave]

you can put any two mathematical terms together

somehow it will exist

[glossy crag]

This is about the unsolvability of degree >=5 equations.

[south patrol]

Put together algebra, cacilus and geometry and we get uh
Derived algebraic geometry

[glossy crag]

Based on the lectures Arnold delivered to Soviet highschoolers.

[proud spindle]

so like Galois type stuff?

[topaz solar]

Some commutative op algs subclass <-> stone spaces thing that implodes when noncommutative (maybe stonean too)

[glossy crag]

Yes, but not entirely, there's some topology there too.

The entire book is a collection of problems and their solutions that teach you all the necessary math from the ground up.

Starting with basics of group theory, continuing with some complex analysis and topology and so on.

And the intended audience of the original lectures was high-schoolers.

[topaz solar]

Then again, that feels like condensed math so

[proud spindle]

just got jumpscared by a sentence saying that good group theory resources exist

(i hate wikipedia articles so much)

[crystal turtle]

just read nlab then

[elder wave]

wikipedia articles have been your ressource

?

[proud spindle]

yes

[elder wave]

you're a demon

[crystal turtle]

what, it's not that bad once you get used to it

clearly we should be viewing addition as a monoidal product, I mean come on

[glossy crag]

@proud spindle instead of chasing disparate bits of math like the semidirect product or w/e (wtf are you going to do with the semidirect product at this stage, anyway), you should read some "programatic" math, i.e. something with a programme that builds towards something (e.g. like this book). Also try Borcherds' videos https://www.youtube.com/playlist?list=PL8yHsr3EFj51pjBvvCPipgAT3SYpIiIsJ.

[proud spindle]

to answer that rhetorical question: i start learning an area of math for shits and giggles and then later i do sensible things

hyperfixate first, reason later

[glossy crag]

Do what works for you, but I believe the organic Russian approach is better suited for learning and exploration.

They had a great series of booklets about different kind of topics (basically at a high-school level), where they introduce a concept and show what it's useful for. Shame they're probably not translatede into English.

[proud spindle]

it doesn't work at all but i've mastered doing the absolute wrong approach "correctly"

[glossy crag]

Then you should try this for a change.

I know the impulse is to flit from topic to topic like a summerfly, but you don't really learn anything that way.

[proud spindle]

if i can find a copy of that book, i certainly will

[glossy crag]

Meanwhile if you knuckle down and work at something a little bit, you'll see it pays off enormously.

[glossy crag]

IIRC I got a digital one, I can DM it to you

[proud spindle]

sure

[glossy crag]

At your level you could try Spivak's Calculus for example.

Great book and does interesting things with the math, e.g. actually deriving Kepler's laws of planety motion.

[proud spindle]

it's always better when they derive a formula rather than just giving it to you

[open sluice]

if only I had the patience to do this

[long obsidian]

Is there any analog of the fundamental theorem of calculus for a polynomial ring in several variables?

[coral shale]

calculus, what?

[toxic zephyr]

do you mean fundamental theorem of algebra?

[solar vessel]

what you mean by fundamental theorem of calculus

[coral shale]

what is calculus...

[solar vessel]

I have something in mind maybe but it depends what they mean

[toxic zephyr]

i think fto algebra makes more sense. since that's for like a polynomial in one variable ig?

[coral shale]

fta is a alg topological result sotrue

[toxic zephyr]

maybe they meant fundamental theorem of algebra but like... automatically said calculus bc i think that's probably mentioned more often at lower levels??

[solar vessel]

for fta there's Hilbert's Nullstellensatz

[long obsidian]

Yes sorry!

How about.. suppose I had a homogeneous polynomial in several variables. Can I factor it into a product of degree one homogeneous polynomials?

[coral shale]

x+y

now what...

ok nvm

xx+yz
whats this one

[solar vessel]

yep

that's prime

[long obsidian]

Thank you!

[indigo ridge]

Im learning about quotient groups, Im trying to undertand this statement **"a set of elements that have been, deemed equivalent to one another based of their relationship to the normal subgroup" ** I've vaguely heard about how quotient groups (say G/H) is a way to form a group from G using equivalent relations, my question is what exactly is the relation my guess is, is that if an elemet a of G is some distance from any element of G we group them together, is this the correct assessment?

[crystal turtle]

Have you heard of cosets? These are the equivalence relation you are looking for

(properly maybe the coset is the equivalence class but w/e)

[indigo ridge]

yeah I know what a coset is, so what is the equivalence relation?

[crystal turtle]

a ~ b if they're in the same coset

[indigo ridge]

Ok I see

[crystal turtle]

Yeah and you need normality of H for the left/right cosets to agree and form a quotient group

[indigo ridge]

on a more intuitive level is it like saying that they're in the same coset if they share the same 'distance' from any given element of H?

[indigo ridge]

gotcha

[next obsidian]

It’s saying their difference is in H

[indigo ridge]

right right

and this property gives it the effect of 'modding' H right?

[next obsidian]

I mean

That’s the definition yes

You’ve set things in H to 0

So if things differed by something in H they now must be the same

[indigo ridge]

Ok that makes sense

[rocky cloak]

The analog of fta might be the Nullstellensatz, which in particular implies that every non constant polynomial has a complex root.

Even more it says that if f1, f2, ..., fn are complex polynomials, then they either have a common root or there exists polynomials g1, ..., gn such that

f1*g1 + f2*g2 + ... + fn*gn = 1

[cloud walrus]

damn_guuurl

[wraith cargo]

@somber sleet try writing out what the product of two elements looks like in that case

[somber sleet]

tell me if I'm wrong, do I have to find homomorphisms, s.t. "the konjugation is the identity"?

[wraith cargo]

this is in relation to what?

[somber sleet]

wait, I'm gonna write it out

I'm trying to find for which prime numbers p,q, the semidirect product $\mathbb{Z}/p\mathbb{Z} \rtimes \mathbb{Z}/q\mathbb{Z}$ is a direct product

[cloud walrus]

damn_guuurl

[somber sleet]

for this to work I need the multiplication in the semidirect product to be componentwise multiplication right?

[wraith cargo]

yes

what homomorphism are you using here?

[somber sleet]

I'm just taking an arbitrary homomorphism from $\mathbb{Z} / p\mathbb{Z}$ to $Aut(\mathbb{Z}/ q\mathbb{Z})$

and what I also know is that $Aut(\mathbb{Z}/ q \mathbb{Z})$ is isomorphic to $(\mathbb{Z}/ q \mathbb{Z})^{\times}$

[cloud walrus]

damn_guuurl

damn_guuurl

[coral spindle]

Right so in the direct product, everything commutes. This is the same as this homomorphism being trivial (by which I mean everything goes to the identity), see?

[somber sleet]

I don't completely see it, I kind of not see the "konjugation" in Z

[wraith cargo]

conjugation isn't the only automorphism that exists

[wicked harbor]

I need help breaking this theorem down and understanding it.

[wraith cargo]

conjugations form a subgroup of the automorphism group known as the inner automorphisms

[somber sleet]

I see that it has to be the identity for it to become a multiplication

[coral spindle]

No, that's not true

In a semidirect product H \rtimes G, the map phi : G \to Aut(H) determines the way that elements of G act by conjugation on H

This map can be anything. It does not have to be the identity. You surely know this already!

Now what I'm saying is that in the direct product, everything commutes, so the conjugation automorphism is the trivial automorphism.

So in fact it H \rtimes G is the direct product, then phi must be trivial.

Is that clear?

[wraith cargo]

maybe break it down one statement at a time
The idea essentially is that there's a bijection between intermediate fields in some galois extension and subgroups of the galois group of that extension

[somber sleet]

yes I understand that

[wraith cargo]

Dummit and Foote have a nice section explaining everything with nice visuals of how the fields and subgroups work in tandem

[wicked harbor]

The first statement isn't too crazy because they went through the Galois correspondence before... But I'm just seeing how the other statements are related and how it applies to polynomials

But the fact that it is a bijective correspondence is pretty cool

[lusty marlin]

Wrong channel

[dim widget]

9 online and 21 members

[elder wave]

No unsolicited ads

[ashen heron]

why did i even join

[rocky cloak]

So the theorem is already broken down into parts. The first part establishes a bijection and the other parts are properties of that bijection.

Is there any specific point where you're struggling? Are you trying to understand the proof or just the statement?

[gusty thistle]

I remember being really overwhelmed by ftgt too. It helped me a lot to just start using it in exercises and like try per exercise what properties I could use. It really is such an amazing theorem that works exactly as you would expect when you are using it for problems.

[somber sleet]

if I have a group of order pqr with p<q<r and I want to analyze if G is solvable, I came up with the fact that it can have either one r-Sylowsubgroup or p*q. Now if I have exactly one Sylowsubgroup H, why is this normal?

is it because every r-sylowgruop is conjugated and thus, since we only have one, it holds trivially for H?

[wraith cargo]

all the r-subgroups are conjugate to each other so that means when you conjugate this one it stays fixed so it has to be normal

[somber sleet]

got it, thank you!!

[rotund aurora]

what's the difference between Matsumura "commutative algebra" and "commutative rings" ?

[rocky cloak]

There is something of an answer here
https://mathoverflow.net/q/25411/157483

[somber sleet]

I'm analyzing why a group G of order 8 * 5*7 is solvable, i concentrated by analyzing the 7-Sylowgroup. Either we have 1 /-Sylowgroup, and the proof follows easily, or we have 8 7-Sylowgroups. Now I'm looking at the solution and they take Norm_G(P) and say that the index of Norm_G(P) is 8 ([G : Norm_G(P)] = 8), now where does this come from, I have no clues

[coral spindle]

Are you aware of the orbit-stabliser theorem?

That should be enough of a hint.

[somber sleet]

uses the conjugation as an operation right?

[somber sleet]

omg this is so nice, thank you!!

[coral spindle]

Yes

It's so nice isn't it

Just immediate and so simple

really a beautiful result, that orbit-stabiliser theorem

[somber sleet]

I mean, simple is a big big word haha, but I don't use the orbit-stabiliser theorem really a lot, cause I never knoe which is the operation I need

but now I got it

[next obsidian]

The content is largely the same.

The arguments in CRT are often clearer, and CRT has many many more exercises. CA’s content basically subsumes everything in CRT, and the second edition of CA has an appendix which covers Kunz’s theorem, Japaneseness, and Excellence which aren’t covered in CRT. There’s also some stuff about eg pro-constructibility, but eh whatever.

My recommendation if you’re going to read either is to read CRT, and then after you read that to read the appendix of CA if you need that material.

[karmic moat]

is this just true because an element of $R^{\oplus n} / R^{\oplus (n-1)}$ is of the form $(0, \dots, 0, r_n)$, then you just have the projection $(0, \dots, 0, r_n) \mapsto r_n \in R$?

[cloud walrus]

ana(functor)mono(morphism)

[lusty marlin]

How does one prove that all elements of a root system of a finite reflection group can be written as an integral linear combination of simple roots?

[coral spindle]

Yup. First isomorphism theorem!

[karmic moat]

ohhhh yeah i was thinking abt iso thms

tyty

[lusty marlin]

The fact that they can be written as a linear combination with non-negative or non-positive real coefficients is obvious, but I don't know how to prove that the linear combination must be integral.

I am primarily interested in this because of the use of the 'height' of roots, in particular, the fact that certain proofs use induction on the height of positive roots, while not explaining why the height must always be an integer.

[dim widget]

The definition of a simple root is exactly that it cannot be written as a sum of two other positive roots. If you have another positive root, you can write it as a sum, then induct. Can you see why this process terminates?

[lusty marlin]

Yes, this makes it much clearer. Thank you

[white oxide]

yo i'm feeling absurdly stupid right now. only using the fact that F is a field, i need a hint on how i can show that (-1)(-1) = 1

tryna distribute something out or right 1 as 1 + 0 or some shit

and i also tried showing the identity property on (-1)(-1) and concluding from uniqueness that it's equal to 1 but obviously my brain doesn't have any enough power to proceed

[coral shale]

oh god no.

Also pls let F be a field

[white oxide]

ah yea my bad

[solar vessel]

consider

[coral shale]

no dont spoil it

[white oxide]

im doing some shit in my analysis class and we're just like verifying the basic properties of R but like yea it can obv be generalized to any field

[coral shale]

spoiler if anyones gonna write it

[solar vessel]

||(-1)(1-1)||

[white oxide]

and i was like oh yea this shit's easy, went back to verify and holy fuck i can't

[open sluice]

happens

[coral shale]

yay

ok thats simpler than what i did

[white oxide]

okay (-1)(1 - 1) = (-1)(1) + (-1)(-1) = 0, but then I have to assume (-1)(1) = -1 to get (-1)(-1) = 1 right

[crystal turtle]

definition of 1

[coral shale]

thats not much of an assumption

[white oxide]

wait rlly cuz my book has me prove that (-1)(1) = -1 in another exercise

[coral shale]

what.

[white oxide]

like they only want me to use the axioms of R

[crystal turtle]

[white oxide]

yea ikr

it's so annoying

[coral shale]

?????

[white oxide]

and my prof's like

[coral shale]

it makes no sense.

[white oxide]

sorry

[crystal turtle]

that's definitional tho

[white oxide]

field axioms

[crystal turtle]

1x=x1=x

[white oxide]

this requires that additive inverses are unique right

-1 + (-1)(-1) = 0 => (-1)(-1) = 1 by the uniqueness of additive inverses

or am i trippin

[crystal turtle]

That's always true

[night onyx]

you can just show that (-1) * r = -r for any r. That would imply (-1)(-1) = -(-1) = 1

[crystal turtle]

But you are just adding one to both sides

[white oxide]

ah yea i'm stupid

[white oxide]

ye that's true

[white oxide]

yo how does this argument show that U "satisfies all the axioms of a multiplicative group"? i know it's pretty easy to show that it is a group using standard methods but i'm not seeing how this one does

for instance, how does it show closure

[crystal turtle]

closure holds as it's the multiplication of a ring

Same with other requirements

oh wait, I see. Well it's easily verified that the product of two invertible elements is again invertible

[white oxide]

wait wdym

well yea

oh does he just assume

that you can show that

i'm just confused as to how the "therefore" follows from the sentences which precede it

[crystal turtle]

i forgot we were restricting to a subset lol

[white oxide]

lol nw

[crystal turtle]

But I think he just implicitly uses that, or has mentioned that earlier maybe. I see what you mean tho

[white oxide]

hmm yea i'm confused as to what it shows, luckily it's very easy to see that it's a multiplicative group lol

well now i have an excuse to go to OH

[crystal turtle]

ohio?

[white oxide]

office hours

okay fuck it i'm verifying everything in lang as i go

normally i just do a quick read through and then go back and verify but it seems like it's practically dependent on verifications

[crystal turtle]

oh this is lang? yeah he definitely just assumed you could show it lmao

[white oxide]

lolll makes sense

[topaz solar]

If you want to show something invertible the most direct approach is to give an inverse

[white oxide]

well yeah i'm just confused as to how it shows all the other axioms of it being a multiplicative group

but never mind like Ryx said it's prolly him assuming that I know how to do it

[topaz solar]

Well, just directly compute it

Show you have an identity (1), that you have inverses, and that you are closed under multiplication

[white oxide]

no yea i know how to do it

[topaz solar]

I mean it shows it pretty clearly imo, since 1 is a unit, inverses are nice, associativity inherited

[proud spindle]

they call them "fields" yet i cannot frolic through them

[white oxide]

ok now i'm confused as to why the elements of a ring which are left invertible do not necessarily from a group under multiplication, call it U

say a, b in U

then xa = 1 and yb = 1 for some x, y

then (yx)ab = 1 tada

so ab in U

and obviously each one is invertible and there's an identity

[crystal turtle]

2 out of 6 for monoids

[white oxide]

or for invertibility do you have to show that there's a two sided inverse for each u in U

lol

[crystal turtle]

group inverses have to be two sided

[white oxide]

oh yea

dude i'm trying to come up with a ring with elements that only have left sided inverses

[topaz solar]

Consider injective/surjective functions

[white oxide]

under multiplication

ohhhh

[crystal turtle]

a few weeks ago I had the revelation that this works on invertible elements 💀

[white oxide]

what's the nuance involved (if that's the word lol) for defining injectivity as functions which have left inverses vs. your more traditional definition of f(x1) = f(x2) => x1 = x2?

[topaz solar]

Weak equivalences, 2 out of 6

[white oxide]

Oops

I meant cancellation

Left cancellation I believe

[topaz solar]

Well, if you don’t have LEM or such not all injective have left inverses

Also, ya know, choice and surjectives having right inverses

[white oxide]

True

LEM?

[topaz solar]

Excluded middle

I think there’s some weirdness of the precise change you have to make

[white oxide]

strange

here is addition just the usual (f + g)(x) = f(x) + g(x) and obv multiplication is composition

[topaz solar]

Yeah

[chilly ocean]

yeah

[topaz solar]

What about it

Make an explicit inverse

[white oxide]

nothing

sir

[topaz solar]

Show the other direction fails because the requirements to be right cancellable

[white oxide]

what does excluded middle mean tho

[topaz solar]

huh, P v -P

Is true for all P

[crystal turtle]

excluded middle implies you can do stuff like proof by contradiction

some people reject it, so their proofs are strictly constructive

not sure why one would do so, but...

[topaz solar]

LEM is kinda about realizability of sets?

Doesn’t exactly give the explicit set though

[lusty marlin]

How does one show that no proper subset of a simple system of a finite reflection group can generate the group?

The book I'm following(Humphreys) has given a hint that I am not able to use properly. It tells us to assume that there exists an α in the simple system for which s_α isn't needed as a generator, and then to consider some reflection w for which w(-α) is a simple root

[long geyser]

just a sanity check, is this because x^p - 1 = (x - 1)^p by the binomial identity, and so 1 is the only pth root of unity?

[delicate bloom]

yep

[somber sleet]

when does a polynomial have real roots?

[delicate bloom]

we already know if it's odd degree it has a real root, so even degree is the one that matters. Every polynomial over R factors into ireducible quadratic or linear polynomials, so we'd need it to factor into exactly all quadratics for it to have no roots.

I don't know, for a single quadratic it's simple enough to check if the max/min lies above/below the x-axis, and for irreducible quadratics their roots will be complex conjugates, so I'm not sure if there's any magic trick we can use to leverage any of that

[somber sleet]

I have simple polynomials actually

like x^4 - 2 = (x^2 - sqrt(2))(x^2 + sqrt(2))

[delicate bloom]

oh, you can solve that with just calculus then

if it only has the form x^{2n} - a

well, that's also factorable ugh lmao

[somber sleet]

I'm asking because it is not the main focus of my exercise, cause I'm lookin for homomorphisms between Q(a) to C

[rocky cloak]

I'm not sure distinguishing between real and complex roots will help you in the regard

[somber sleet]

and I do not need to calculate the roots, I only need a motive to say that I have at least one real root, s.t. my field Q(a) goes into R, which does not depend on the root since the fields all are isomorphic to each other

[somber sleet]

this trick is enough, I guess

[chilly ocean]

the lie group would be the "most general algebraic structure"?

[delicate bloom]

@chilly ocean what leads you to ask this question

a boring but more general algebraic structure is a magma, but really nobody cares about this so I wouldn't advise going down some wikipedia rabbit hole on it. Your time is probably better spent reading an introductory abstract algebra book

[coral spindle]

Lie groups are quite specific, actually.

[rocky cloak]

The algebraic structure of a lie group isn't any different than the structure of the group. The interesting thing about lie groups is that they also have analytic / topological structure that is compatible with the algebraic.

[teal vessel]

so when describing containments, would it be sufficient to say that the group generated by x^15 is the intersection of the subgroups generated by x^5 and x^3? or would it be better to say that it is distinctly subgroups of both other groups?

[crystal turtle]

it is their intersection

quite litereally, as subgroups of Z/45

[teal vessel]

yes, but is that a fair way to describe its containment as a subgroup?

[crystal turtle]

yeah

[teal vessel]

k, so there should be a total of 4 distinct non-trivial subgroups, generated by the powers of 3, 5, 9, and 15. The trivial subgroups of {1} and Z_45 itself round it out to a total of six subgroups. The containments as follows:
{1} is trivially the universal subgroup
<x^9> < <x^3>
<x^5>
<x^15> = intersect(<x^3>,<x^5>)
all of the above < <x>

These all are using the natural generator for each subgroup, i.e. the smallest generating member of the group. All other elements in Z_45 generate one of the above subgroups, separated into equivalence classes according to their GCD with 45

[crystal turtle]

Yup, that works

A good sanity check for cyclic groups Z/n is that you can add one to each of the prime powers that divide n, and multiply them all together to get the total number of subgroups. I.e., here 45 = 3^2 * 5^1, so there are (2+1)*(1+1) = 6 subgorups

[teal vessel]

I just took the prime factorization, raised 2 to the sum of the powers to get an upper bound, then took away all repeats due to higher powers of prime factors, but yeh.

teaching math has really changed me from a "find the best solution" person to a "find a reliable solution" person, and idk how I feel about that.

speaking of dumb solutions:
suppose x is in cyclic group G, and |x|=|G|=n < infinity
prove that G=<x>

1. because |x|=n for some finite number, |<x>|=n=|G| which implies that <x> is isomorphic to G since all cyclic groups of the same order are isomorphic (theorem proven elsewhere)
2. Since x is in G, <x> must be a subgroup of G.
3. The only subgroup of G that is isomorphic to G (i.e. that has a bijection with G) is G itself, since G is finite. Therefore it must be the case that <x>=G

yes, there are easier ways to do this, but this one seemed funny to me.

also note that this only holds for finite groups, as (by counterexample) |2| in Z is infinite, but <2> =/= <1> = Z

[somber sleet]

I'm looking for the splitting field of the polynomials $X^3 + X^2 + 1$ over $\mathbb{F}_2$, I know that it does not have roots in it, how do I find it? I have no clues

[cloud walrus]

damn_guuurl

[somber sleet]

I was analyzing the field $L := \mathbb{F}_2 [X] /( X^3 + X^2 + 1)$ which is isomorphic to $\mathbb{F}_2(a)$ for an $a \in L$

[cloud walrus]

damn_guuurl

[coral spindle]

What do you mean 'find' it? Like you have a perfect description of the field L as a quotient. Can you be more specific about what you're looking for?

If you're looking for some value a, it's X. The value is X.

[somber sleet]

Okay fair, sorry for not being precise, yes but why? I don't understand how to find it?

[coral spindle]

I guess more specifically, the image of X in L.

What is 'it'

[somber sleet]

How did you find the X

[coral spindle]

It is X

Found it

I still do not understand what you're asking

[delicate bloom]

maybe try computing the multiplicative inverse of X idk

[coral spindle]

Are you asking why X in L is a solution to the polynomial?

[somber sleet]

YES

[coral spindle]

Finally ok

[somber sleet]

c'mon don't be mean, I'm already heartbroken thanks to algebra 💔

[coral spindle]

It's because X, in L, is really the coset X + (X^3 + X^2 + 1). I will write this ideal as P = (X^3 + X^2 + 1). Let's write this polynomial out in a different variable to be clear: f(t) = t^3 + t^2 + 1.

Then f(X + P) = X^3 + X^2 + 1 + P, but X^3 + X^2 + 1 is in P (by definition) so f(X+P) = P.

Oh and ofc P = 0 + P. It's the 0 in the quotient ring.

[somber sleet]

okay, I forgot that I was taking a class and not x by itself

[coral spindle]

Convince yourself of this.

Let f(t) be any polynomial in some ring R.
The image of X in the ring R[X]/(f(X)) is a solution of f(t).

Prove this.

[crystal turtle]

[teal vessel]

the study of abstraction is so beautiful.

[delicate orchid]

wrong

[teal vessel]

it makes people talk funny tho

[coral spindle]

hey wew

hope you're doing good

[delicate orchid]

yur

[teal vessel]

who be wrong?

[delicate orchid]

took a break from this hizzouse

[delicate orchid]

abstraction is for nerds

[coral spindle]

valid

[teal vessel]

guilty as charged. Still working through every example from D&F

[crystal turtle]

too much work

[next obsidian]

Bro said W abstraction and then said they’re working through examples. Make it make sense

[teal vessel]

ain't got no teacher to tell me to do it all in 3 months, so it's not like I'm doing 100 problems a night or something

[teal vessel]

never said I was good at it, just that it was beautiful

I love turning polynomials into literal squares that spin

[coral spindle]

Can I sacrifice some chickens to ba'al to get it all to end

[somber sleet]

guys I did 5 exams in 2 months, I have algebra next week and am literally desperate

[elder wave]

Most sane mathematician

[somber sleet]

still love it though

[formal ermine]

@crystal turtle when are you turning into god themselves

[coral spindle]

ur right I was being crazy. It has to be a whole ass COW

[teal vessel]

dihedral subgroups go brrrrr (because they're just helicopter blades)

[crystal turtle]

Oh hello

[formal ermine]

damn bro bojjiitjtye has me blocked

[crystal turtle]

I have ascended

[next obsidian]

Chmonkey

[formal ermine]

what did I do

I just got here

[elder wave]

Rightfully so

[formal ermine]

I hope he blocks you too

[teal vessel]

dunno, is 0 a natural number?

[formal ermine]

yes

[teal vessel]

correct answer, seems clearly unreasonable he blocked you then

oh, wait, hold on: define a trapezoid.

[delicate orchid]

trapezoid 😹 bro got trivial symmetry group 😹 basically just a point 😹

[formal ermine]

wtf is a trapezoid

[delicate orchid]

define... a functor...

[formal ermine]

functor? I hardly know her!

[teal vessel]

oof, you right. nvm, my bad, can't bring potentially irregular shapes in here.

(had somebody tell me the other day a trapezoid -- or trapezium -- had exactly one set of parallel sides)

[formal ermine]

I have honestly never heard of this before

was I supposed to know this

[teal vessel]

you never saw it in elementary school?

[elder wave]

Bruh

[formal ermine]

bro thinks we did geometry in elementary school

all we did was 1 + 1

and maybe some 4 * 2

[teal vessel]

it's just a square, but in projective geometry

[delicate orchid]

what the fuck

what being a hsct does to a mf

[formal ermine]

hsct?

[crystal turtle]

What the fuck

[crystal turtle]

He's not a hsct tho

[formal ermine]

what is a hsct

I'm dying to know

[delicate orchid]

true... bro don't even know what a Z_p-good space is 😹

[delicate orchid]

high school category theorist

[crystal turtle]

I don't even know that

[delicate orchid]

I can give you the definition but it's abstract nonsense that tbh I don't really understand either

[formal ermine]

gimme the definition

[formal ermine]

I know one of these

@hazy lion

[delicate orchid]

do you know what a simplicial space is

[teal vessel]

https://tenor.com/view/smg4-mission-failed-successfully-mission-failed-mario-luigi-gif-26027979

when you try to cause math drama but unintentionally cause entirely distinct math drama

[formal ermine]

I just looked it up

wew u ok?

[crystal turtle]

Simplicial space being simplicial object in a (convenient) category of spaces?

[formal ermine]

in top

[delicate orchid]

yus

[rocky cloak]

One thing that wasn't mentioned earlier is that, in general, to get the splitting field it is not enough to adjoin a single root, you have to adjoin all the roots.

With some Galois theory, you can prove that over a finite field adjoining a single root (of an irreducible polynomial) is the same as adjoining all the roots. Or in this example you can just manually check that there are two other roots in L.

[crystal turtle]

A convenient category of spaces is what I meant

Might be top

Might be ktop or cgwh

[delicate orchid]

I prefer to think of them as functors tbh

[formal ermine]

or zgygz

[crystal turtle]

Δ^op --> Top

[delicate orchid]

anyway a simplical space is R-good if it's equivalent in some sense to it's R-completion

[formal ermine]

where Delta is the full subcategory of Delta_a consisting of the free categories on finite and inhabited linear directed graphcs where Delta_a is the full subcategory of Cat on the free categories of finite linear directed graphs

[delicate orchid]

wtf

anyone know what this geezer is talking about

[topaz solar]

why would you think of it that way

That’s a cringe interpretation

[crystal turtle]

Lil bro learned about copy and paste from Wikipedia

[delicate orchid]

it's just finite sets with maps between them lol

[topaz solar]

It’s mf triangle

[formal ermine]

because that's the nlab definition

[crystal turtle]

Bro

[topaz solar]

nLab is never right

[crystal turtle]

Nlab is always right

[formal ermine]

I thought ryx pasted from nlab too

I was tryna pull a funny

[delicate orchid]

nlab is correct technically but not morally

[crystal turtle]

But nobody uses that definition

Like it's technically correct

[topaz solar]

No Ryx actually knows things

[crystal turtle]

But we use a skeletal version of that

[formal ermine]

really?

[topaz solar]

Yeah

[formal ermine]

whaaaaaaaaaat

[delicate orchid]

working in a non-skeletal category is asking for trouble tbh

[crystal turtle]

Believe it or not, in some things I do

[formal ermine]

ok anyway

[topaz solar]

The actual normal definition is the finite linear orders [n] = {0 < … < n}

[formal ermine]

who can help me understand this gibberish

I think there's a typo in it too

[crystal turtle]

But yes, we restrict to a category that's just finite linear orders with monotone functions as maps

[topaz solar]

As in, you have n < symbols

As in, you have n arrows

In a line

[formal ermine]

in the second equation

shouldn't the as be hs

[topaz solar]

Compare 0 -> 1 -> 2 to a triangle

[crystal turtle]

|[n]| = n+1

[formal ermine]

what

"compare 92 to a triangle"

yeah sure bro

let me just

calculate the triangleness

[delicate orchid]

the whole point of a simplitical set is to transform arbitary categories into simplices

that's the name

[topaz solar]

If you can’t see it as a triangle you need to forget about categories for now

0 -> 1 -> 2, but also 0 -> 2

[formal ermine]

herlp

[topaz solar]

Because composition

[formal ermine]

I know

[delicate orchid]

oh my gosh it's the wholesome face map

[formal ermine]

I was tryna be funny

[crystal turtle]

Fail

[formal ermine]

I always fail

[delicate orchid]

I don't know what a closed subgroup means here. C_p has no non-trivial subgroups

are we taking the discrete topology?

[crystal turtle]

What

[formal ermine]

Z_p not C_p

[delicate orchid]

oh!! if only we didn't overload the fuck out of \mathbb{Z}

[formal ermine]

????????

[delicate orchid]

maybe then I would understand what the hell is going on

[formal ermine]

bro

it's Z_p

wtf you on about

[delicate orchid]

which is a commonly used notation for cyclic groups

[formal ermine]

[crystal turtle]

Bro 💀

[delicate orchid]

the completion should be Z_{(p)} anyway

[topaz solar]

it’s ilum he doesn’t know anything except p adics

[delicate orchid]

you cannot get mad at me for making that mistake

[crystal turtle]

Anyways they're obviously ideals

[formal ermine]

that's the part I don't understand

I know how to classify the ideals

[crystal turtle]

Idk what this fancy argument is about tbh

[formal ermine]

what's your argument

[crystal turtle]

But they are obviously ideals

[crystal turtle]

Just calculate using the norm. Easy argument

[wraith cargo]

If you ever say obviously again I will have you thrown in the dungeon

[crystal turtle]

Yes please

[delicate orchid]

that's kinda what they're doing in the proof given

[wraith cargo]

the words trivial, obviously and just are officially banned from this server!!!!!

[crystal turtle]

(I didn't read it)

[delicate orchid]

me upon referring to the group with one element

[formal ermine]

can someone explain the proof pls

[crystal turtle]

What characterizes the norm of elements in Z_p? Their norm is less than....?

[hazy lion]

loool

i have no idea who you are but hi I guess

guilty as charged

[delicate orchid]

oh I know proffessor! pick me proffesor! I know Iknow!!

[formal ermine]

1

[formal ermine]

bruh

not again

[crystal turtle]

I want the child to answer

[rocky cloak]

I really dislike Z_n for cyclic groups, like just use Z/n. Then there can be no confusion

[delicate bloom]

p ackshully

[delicate orchid]

you and me both

[formal ermine]

What

[hazy lion]

😭 ok I do know who you are why tf do you keep changing your name

[delicate bloom]

they said less than, not less than or equal to

[formal ermine]

it's just

on this server

[delicate orchid]

the |1|_p in question:

[formal ermine]

@hazy lion #chill message

here's the context

[formal ermine]

broo

[crystal turtle]

Right, so also in p^n Z_p, it's less than 1/p^n
So just make an argument that its an ideal by considering what happens when you multiply by an element in Z_p

[formal ermine]

wgat?

[delicate orchid]

nvm that is a much cleaner argument than the one they gave

[formal ermine]

how do we use closedness

[crystal turtle]

If |x| ≤ 1/p^n and |y| ≤ 1, then |xy| ≤ 1/p^n

Done

[delicate orchid]

wait are we going the wrong way

[coral spindle]

https://tenor.com/view/its-as-shrimple-as-that-shrimple-clam-grass-jump-gif-25956447

[formal ermine]

why |x| <= 1/p^n

[delicate orchid]

because it's in p^nZ_p

[formal ermine]

but we don't know that yet

[coral spindle]

Wew, never change

[delicate orchid]

https://tenor.com/view/its-simply-resharkable-your-octopinion-is-invalid-its-as-shrimple-as-that-gif-26284133

[coral spindle]

I love u 🥺

[formal ermine]

we want to show that p^n Z_p are precisely the closed subgroups

[delicate orchid]

right so we are going the wrong way

[crystal turtle]

Bro fucking whay

Or am I misreading it

[delicate orchid]

we're going ideals => closed subgroups rather than closed subgroups => ideals

[formal ermine]

yes

we wanna do the second one

[crystal turtle]

Me thinks this is closed subgroup ==> ideal

[delicate orchid]

it is

[formal ermine]

it is tho?

but you are assuming that all closed subgroups are p^n Z_p

but we don't know that!

[crystal turtle]

Oh I'm being dumb again

[crystal turtle]

I misread this

[delicate orchid]

let a doodad be a closed subgroup, then every sequence in this mofo converges to something in the subgroup
say it converges to x', then |x_n-x'| goes to 0 as n goes to infinity, only element with p-adic norm 0 is 0 so x_n-x' -> 0 as n -> \infty
additionally we have for all a in Z_p that a(x_n-x') -> 0 as n -> \infty as |a(x_n-x')| \leq |x_n-x'|, so the sequence ax_n converges to something inside of our closed subgroup as well
consider the constant sequence?

does that work