groups-rings-fields

[toxic zephyr]

since that's always even (when it isn't trivially 1) i would conjecture that if n1...nt are all even, then it is isomorphic to some U(m) where phi(m)=n1...nt

[coral shale]

hmmmmm

[toxic zephyr]

so like i think C8, C2xC4, C2xC2xC2 are each isomorphic to some U(m) where phi(m)=8

[coral shale]

so the answer is probably no, but u now what nsc condition for when m exists?

[toxic zephyr]

yeah

[coral shale]

So you are given m (product of n_i)

and we need to know when m is a product of f(p, n + 1) = p^n(p - 1)

oh yh, so to be explicit most primes dont work

all but 2

Oh and i missed the 2^n exception

[coral shale]

doesnt seem any obvious way to me except clever brute force

prime fac m, then prime fac p-1 for each prime fac p of m. Then check which is possible

[toxic zephyr]

this is wrong. if phi (m)=8, the. U(m) is always isomorphic to C2xC4. interesting, i wonder why

[toxic zephyr]

what is f here?

[coral shale]

sorry, totient

[coral shale]

this is not right now u bring up your point

[coral shale]

You arent given m.

think im back to square one, misthought that

nvm ok, i think i had right idea just got to write it down correctly

[somber sleet]

why are these two ideals the same?

I understand that I1 is in I5

but why should be XZ in I1

[coral shale]

(X, XZ) = (X)

using this fact

[somber sleet]

why does this hold?

this is what I don't get

[coral shale]

use defn ideal

check what each are

if it helps, (2, 4) = (2) ?

[somber sleet]

(X,XZ)= aX+bXZ for a b in R right?

[coral shale]

yh go on

[somber sleet]

this is clear cause 2 is in R, so 2*2 also is

[coral shale]

(2, 6) = (2)

etc

[coral shale]

use same reasoning

[somber sleet]

oh but wait, Z is also in R right?

or in R[X]

[coral shale]

uh

X, Y, Z i was just assuming they were elements of your ring

that ur taking the ideal of

they are, are they not?

[somber sleet]

they are ideals over Q[X,Y,Z]

[coral shale]

(X, XZ) = aX+bXZ for a b in Q[X, Y, Z]

[somber sleet]

then it makes sense

I get it

may I ask you

this means that (Z,XZ)=(Z) for the same reasoning right?

[coral shale]

well yes

ideal of ring

[coral shale]

same

[somber sleet]

I get it, thanks a lot!

[somber sleet]

kinda dumb ngl

[coral shale]

https://media.discordapp.net/attachments/496784958430380033/1071361299570761748/321445119171756033.png

[coral shale]

order of G needs to be product of |Z/p^kZ|
which is product of
f(p, k + 1) = p^k(p - 1)

===
Now if such a product exists, check decomposition of cyclic orders, noting special case with 2

hmm..

doesnt look easy

[formal ermine]

Zm^times is cyclic

for uhh

prime m

wait nvm my argument doesn't work

[coral shale]

The decomposition of abelian groups is recursive right

making me see nothing clever

[formal ermine]

wdym

[coral shale]

https://media.discordapp.net/attachments/496784958430380033/1071353212029894687/Screenshot_20230204-085450.jpg

'which may further factor ...'

[formal ermine]

I thought about crt too

[coral shale]

bah nvm being confuzzled

being constantly confused switching between Z_n and C_(phi(n)) smh

[formal ermine]

I think this is not true

phi isn't surjective

[rustic crown]

uwu?

[formal ermine]

take C2 x C7

it's the smallest nontotient

[rustic crown]

lol just take C3

phi(m) is either 1 or even

[formal ermine]

oh yeah lol

[coral shale]

yh its not true, now we want nsc for it

originally asked by nix i was just bumping and thinking

So let me get this straight. If you want to decompose C_n completely, you decompose n into prime powers.

And C_n = prod C_{p^k} right?

And you can go no further than this

[rustic crown]

yeap

[coral shale]

So to decompose multiplicative group Z_m, you decompose m into prime powers.
Then Z_m = prod Z_q^j.

In turn, each Z_q^j = C_{q^{j-1}(q-1)}

[rustic crown]

put ^\times everywhere

[coral shale]

lazy

but ok

[rustic crown]

[coral shale]

So decomp of Z_m^x into cyclic groups is 2 steps

[rustic crown]

just into cyclic is one step

[coral shale]

whats the word

like undecomposable cyclic

,,\bZ_m^\times\cong\prod Z_{q^j}^\times\cong\prod C_{q^{j-1}(q-1)}

[rustic crown]

q = 2 behaves a little differently

[coral shale]

Then the last step to take is to factorise q-1

ah yes ofc

assuming m odd for now

[rustic crown]

(Z/2^(n+2)Z) = <-1, 3> = Z/2Z⊕Z/2^nZ

[rustic crown]

oh okie

[coral shale]

Let $m$ odd
$$\bZ_m^\times\cong\prod Z_{q^j}^\times\cong\prod C_{q^{j-1}(q-1)}$$
$$\cong\prod C_{q^{j-1}}\times C_{q-1}\cong\prod \left(C_{q^{j-1}}\times\prod C_{p^k}\right)$$

[coral shale]

q-1 chopped into product p^k

[coral shale]

Then... handle case q = 2 separately

[coral shale]

Then finally u can tackle when such an m exists.

So it boils down to inverting the totient function

[rustic crown]

not just that right

consider C2 x C3 x C3

[formal ermine]

so for Fp we have a nice field to look at instead: Zp, are there any analogs to this but for e.g. Fp^2?

[rustic crown]

phi(3^3) = 2*3^2

but U(27) = C2 x C9

[rustic crown]

splitting field of x^p^2-x over F_p

[formal ermine]

🙄

[coral shale]

yh i meant it in 2 steps.

1. list all possible m
2. check the decomps are isomorphic

[rustic crown]

there is one bad answer for that

so you know like F_p is teh residue field of Z_p (the p-adic)

you can also define the ring of witt vectors

then F_q would be the residue field of W(F_q)

but the right guy is ofc more complicated than F_q

[wooden ember]

For the specific case of F_(p^2) you could take a quadratic extension of F_p when p=/= 2 if that feels any better

[formal ermine]

no

I want like

an analog to know how the group operation beaves

[wooden ember]

I mean for a quadratic extension you can write out your field elements pretty explicitly in the same way you would for say complex numbers

[formal ermine]

too much work

[wooden ember]

So maybe that’s easier to get a feel for idk

Not sure I get what you want

[wooden ember]

If p = 3 mod 4 the analogy is even stronger since you can base your extension on a root of -1

[glossy crag]

@rustic crown so we were talking yesterday about how injectivity can fail when tensoring. How come in this case we know it's a subring (last line)?

[wooden ember]

Im pretty sure algebraic extensions over F are flat over F no? Or am I completely out of line

Feel like I’ve heard that somewhere before

[glossy crag]

Maybe it's that all vector spaces are flat? I think I remember seeing a statement in Jacobson that a direct sum is flat iff every summand is (and tensoring with K doesn't change anything).

[wooden ember]

Uhh actually might need to be finite dim cause infinite dimensional vector spaces aren’t in general flat (I think)

[wooden ember]

Yeah no I think you’re right

Since tensor preserves direct sums

[rustic crown]

It doesn't fail if the base ring is a field. All modules over a field are free, so in particular flat.

[wooden ember]

Oh lmao true of course they’re free

[rustic crown]

Oh that's already been said

[rustic crown]

You say Z/pZ is nice because you understand Z. What's so bad about F_p[x] for you? It's an equally nice pid. And you can write F_p^2 as quotient by some irres quadratic

You can always convert F_p[x] to Z[x]/p

But then Z[x] is not as nice as F_p[x]

[novel parrot]

how does (a,b) differ from (a) + (b) ?

[coral shale]

they are equal

?

[novel parrot]

i thought so

but then why different notation

[coral shale]

different concepts which turn out to be equal

[novel parrot]

ok

[coral shale]

you define addition and multiplication of ideals

it turns out they are ...

basically

[frank cosmos]

Lets say if a subgroup H of G has the same order of G

then (G:1)=(G:H)(H:1) -> (G:H)=1

so, the number of distinct cosets is 1?

[tender wharf]

if H subgroups G and H has the same order as G

then G = H

[frank cosmos]

yes thats what im trying to show

via lagrange

[wooden ember]

For finite G only

[frank cosmos]

oh yeah both r finite mb

[tender wharf]

yeah guessed it would be finite

because it wouldnt work if its infinite anyway

yes theres only 1 coset

[wooden ember]

Ofc, just wanted to be clear

[tender wharf]

I fully agree, it is important to be clear

[frank cosmos]

so then for all g, gH=eH=H so g is in H. does that proof make sense?

writing it out makes it make sense but intuitively i cant grasp it

[tender wharf]

I mean you already have G=H

so lol

[frank cosmos]

im trying to prove that, im not assuming it

all im asuming is H has same order

[tender wharf]

I mean

if G and H are finite like you said

and order of H = order of G

then H = G

like literally

this immediately follows from the definition

[frank cosmos]

yes i get that

but i want to show it for fun with this

[tender wharf]

what

[frank cosmos]

its fine

[tender wharf]

yes there is only one coset

because everything in G is in H

[coral shale]

it just seems incredibly pointless

[coral shale]

You will also have to specify finite G

else it isn't always true H = G

oh ppl got there first

[frank cosmos]

yes i just want to get a feel for the theorem

[coral shale]

ic ic

[frank cosmos]

like i just showed before that prime order groups are cyclic

like i get the proof pretty well, i just want it to make intuitive sense now

[tender wharf]

you might find it more interesting to

do some exercises

with it

[frank cosmos]

🧠

[tender wharf]

im assuming your textbook will have those

[frank cosmos]

yes

[tender wharf]

yeah do those

[frank cosmos]

of course

[tender wharf]

.

[rotund aurora]

If $S\subseteq R$ are rings and $I$ is an ideal of $R$ with $I\cap S=J$, is the ring $R/I$ a ring extension of $S/J$ ?

[cloud walrus]

Croqueta

[rotund aurora]

S is a subring of R

I guess if s and s' are in S then s=s' mod I if and only if s=s' mod J

is that it?

and J is an ideal of S

[rustic crown]

S/J is same as (S+I)/I

(second iso thm)

[rotund aurora]

thanks

[rustic crown]

btw what's a ring ext?

just an injective ring map?

[rotund aurora]

I guess yeah

the context I was looking at, the two quotients turn out to be fields

[rustic crown]

hm okie

i see

[rotund aurora]

for notation lol

what he calls little o I call \mathcal Z

[rustic crown]

souka

[rotund aurora]

little o is like the integers, K the rationals, etc

[novel plover]

ya, gokurou datta na. toTEMO yaku-ni-tatta yo ahahah, wahahaha

[rotund aurora]

[rustic crown]

"you had a hard time" is all i undersand

[novel plover]

Sorry, I should really reserve my Japanese Captain Hook impressions for the discussion thread

[rustic crown]

what's yaku-ni-tatta

[novel plover]

役にたつ to be unambiguous

[rustic crown]

something ni ta tsu

[novel plover]

It's how you say someone or something is "helpful"

[rustic crown]

aah

[novel plover]

Anyway, sorry

[rotund aurora]

today my brain is working so slow

Ill just do reading xd

[rustic crown]

me is really useful?!

[novel plover]

No I mean that's maybe how you should've responded to the people who helped you.
By saying basically "Thank you my dear, you've been MOST helpful..heheheh" before locking Tinkerbell in the cage

Anyway, please proceed

[coral shale]

waw meow

[rustic crown]

is your nose a fish or a submarine?

[tender wharf]

looks like a gopher

gopher from golang im guessing

[somber sleet]

guys do you know some tricks on how to show which factor ring is isomorphic to one another?

[south patrol]

Ig u mean isomorphic given the German but you wanna think about what is being identified in some of them for example

So for example for R5, you are identifying Y with Z and killing X^2 - which one does that look like?

[somber sleet]

yep oops, sry haha

[chilly ocean]

this looks tough

[somber sleet]

oof, I have no clue

[oblique river]

For ideals with multiple generatros, just do one quotient at a time

E.g. in R_6 you have z = 0

[somber sleet]

maybe something like R1

how are you so fast in seeing it guys

like no joke

[oblique river]

So first consider Q[x,y,z]/(z)

[south patrol]

Experience

[oblique river]

Etc

[south patrol]

with working w quotients

[chilly ocean]

intuition I'd say

[south patrol]

intuition gained from experience

[oblique river]

Which comes from experience…..

[delicate orchid]

I simply wrote down all of the cosets in my mind

[south patrol]

wew that made me exhale

[somber sleet]

do you have any trick for me? haha

like If I had to make an argumentation in an exam

[oblique river]

I gave you one

[somber sleet]

this one?

[oblique river]

Yeah

[chilly ocean]

the things in the quotient is what you identify to 0

[somber sleet]

yeah is like the kernel of the homomorphism of the universal property right?

[chilly ocean]

universal property?

[delicate orchid]

of the quotient, yes - but I wouldn't think about it like that

I'd recommend thinking of it as literally writing down "whatever" = 0 and seeing what happens to the ring

[chilly ocean]

I'm sorry I have a hard time understanding what you said

[somber sleet]

do you mean for example using the chinese remainder theorem?

[oblique river]

Not quite

Im talkjng about when your ideal has multiple generators

[chilly ocean]

this is what I do too, and I think everyone just thinks about it like that
that's why I said it's intuition, which comes from knowing how to understand it right

[somber sleet]

I have this universal property and the homomorphism theorem, and I thought about using them. I know is in german, as always

[somber sleet]

yes I'm trying to get there too

cause otherwise it's too complicated

[chilly ocean]

this looks like one of the isomorphism theorems

and formally proving that those rings are isomorphic to each other should probably use it

but to just identify by hand what rings should be isomorphic, you can just think about it like wew does

[delicate orchid]

and I think buncho's idea is just the third iso theorem - quotient by one generator first, then quotient whatever that is by the other one, and so on for each generator

[tribal moss]

It can sometimes pay to try to find a more concrete representation of the quotients. For example one could think of R2 as isomorphic to the subring of Q^Q generated by the constant functions as well as min(0,x) and max(0,x). That's also the ring of different polynomials to the left and right of 0 which fit together continuously at 0. And one can then think of different ways this ring could also have been generated. At least for me, that made it a bit easier to find a different set of generators that would behave like the X and Y in R4. Afterwards that can be boiled down to a slick symbolic argument that just pulls the isomorphism out of a hat.

[novel parrot]

how to solve this

since a(X) isnt in R[x], that means that it has coefficients as fractions,

[coral shale]

this yes

[coral shale]

what if you scale a and b so they live in R[x]

is my thought.

and proceed

[novel parrot]

i try that

but idk what to do after

we can scale a(x), da(x) = a'(x) which lies in R[x]

[coral shale]

call the scale factors c and d

then hmm

p'(x) = cd p(x)

[novel parrot]

we may not be able to scale b(x)

so d may be 1

[oblique river]

You now have two factorizations of cdp(x): (cd)(p) and (ca)(db). Can you show that these cant be refined?

[chilly ocean]

this is interesting. What does a point (x, y) correspond to in such subring?

[coral shale]

Also, the opposite of this is gauss lemma right

For Z, Q

[novel parrot]

yeah

[novel parrot]

whats (ca)(db)

[tribal moss]

I meant the ring of functions Q -> Q, not Q×Q.

[chilly ocean]

(x, y) as in R^2, right

[oblique river]

a and b are the polys we were given

C and d are the scaling factors needed to clear denominators

[chilly ocean]

oh wait. You meant R_2 as in the picture

[coral shale]

just checking, buncho - we want to show R is not a UFD right

not the poly ring

[oblique river]

Oh

My b

[tribal moss]

Ah yes.

[coral shale]

i only suggested this because I was thinking it might help with the 1st coeff or something

[novel parrot]

maybe we can show the converse instead

[coral shale]

Oh I think I have an idea.

[novel parrot]

R is a UFD and show that a(X) lies in R[x]

[coral shale]

I'm considering concretely in Z, Q

[coral shale]

Then this really is gauss' lemma isnt it

[novel parrot]

its very close it it

gauss' lemme just say that if reducible in F[x] then also in R[x]. if and only if primitve polynomial

i think i have an idea

[coral shale]

lets say p(x) = (x^2 - 1/2x + 1)(x + 1/2) in Z[x] (not actually true, but anyways)
Why can't this be in Z[x]? Because you have at least one fractional coefficient that needs to 'cancel' when multiplying out

This is my idea

you will multiply the 2 fractional terms together

And you want to show this has to be 0

Maybe doesn't work but just an idea

[coral shale]

Now I feel like I've shown its not even an integral domain
Have I made an error

[chilly ocean]

does lattices fit here

or is it more like foundations

[novel parrot]

which lattice

[chilly ocean]

order

[novel parrot]

idk about that

[wooden ember]

Combinatorial structures maybe?

[novel parrot]

this is algebra

[chilly ocean]

lattices can be thought of as algebraic structures

[wooden ember]

Eh ig just ask and if no one answers ask in #combinatorial-structures

Depends on the context ig

[novel parrot]

@coral shale do you know?

[chilly ocean]

So let $L$ be a complete distributive lattice. I'm wondering if the more general distributive law $x\vee \bigwedge Y = \bigwedge (x \vee Y)$ holds

[cloud walrus]

Blitz

[chilly ocean]

where Y is an arbitrary subset of L, and x is in L

[wooden ember]

So like maybe take the first fractional terms in a(x) and b(x)

With degrees n and m

Then the coeff of degree nm of p(x) will be a sum of stuff in R with just one term which is a product of two things in F

So that product is in R

Then maybe you can show something with that

Nah that doesn’t work

[novel parrot]

what about polynomials of degree 4? if it has a root, then we can also factor it has deg 3 poly x deg 1 poly

[rotund aurora]

it can factor as two quadratics that are irreducible

example, (x^2+1)^2 has no roots in R, yet it is clearly not irreducible

is there an example of a number field K and a rational prime p such that p is irreducible in O_K but (p) factors as a nideal in O_K ?

by rational prime I mean a prime of Z

[rustic crown]

Q(sqrt(-5)) works

since dedekind, there is always a unique factorization into prime ideals

2 in Z[sqrt(-5)] is irred but not prime.

you can check (2) = (2, 1+sqrt(-5))^2

[rotund aurora]

Nice, thanks

[rustic crown]

[formal ermine]

is every noetherian ring finitely generated?

[coral spindle]

Every unital ring is finitely generated

[formal ermine]

oh

I meant

module not ring

lmao

typo

[coral spindle]

Necessarily, yes. In fact one of the equivalent definitions of a Noetherian module is one whose every submodule is finitely generated.

[formal ermine]

ah I see

concluding that from the acc is the same as with a noetherian ring I suppose?

[coral spindle]

Yup

[formal ermine]

ok epic

[coral spindle]

R being a noetherian ring is just the definition of R being a noetherian module (as an R-module)

[formal ermine]

noetherian module <=> underlying ring is noetherian?

[coral spindle]

Not all modules are rings, so no

And you can have Noetherian modules over non-Noetherian rings

[formal ermine]

I see

[tribal moss]

As well as the other way around.

[formal ermine]

ok

another question

not sure if this belongs here or in #category-theory, but is showing that something is a universal property just like showing that the definition of a uni. property holds (constructed object will be unique up to iso) or is there an easier way?

[coral spindle]

Do you have a specific example? Because yes, in general, showing that something satisfies a definition is the easiest way to show it satisfies that definition lol

[formal ermine]

tensor product

this is the first time I've encountered a universal property

[coral spindle]

I daresay the universal property is the best way to understand the tensor product, yes

[void cosmos]

its the only way really

hahaha

like the only sensible way

the equivalence class one is too hard

[formal ermine]

hm ok

[void cosmos]

is this atiyah mcdonald

[formal ermine]

no

[rustic crown]

[void cosmos]

oh ok nvm

[rustic crown]

you would have already encountered universal properties of free objects or quotients, just not saying the word "universal" probably

[formal ermine]

what's a free object

[coral spindle]

It's when you don't have to pay for it

[rustic crown]

like free group, free vector space, free module

[formal ermine]

what's a free group

what's a free vector space

[coral spindle]

You also know about the free algebra k[x]

[formal ermine]

what's a free algebra

[coral spindle]

k[x]

[void cosmos]

so say G is a group

and X is a set

[rustic crown]

so idea is say you start with a set X, and a field k... what is the most efficient way to get a vector space out of X

[void cosmos]

for any function f :X--> G there exists a homeomorpihsm from F(X) --> G such that the diagram commutes

[rustic crown]

you don't wanna put any conditions whatsoever

i.e. the elements of X should be linearly independent

to get the vector space V, you then just consider abstract finite linear combinations of elements of x with coefficients in k

and that's called the free k-vector space of X

and good thing about free objects is that, you can easily construct maps out of them

to define a k-linear map V --> W, you only need to give a set function X --> W

[formal ermine]

what's the universal property here?

[rustic crown]

given X and k

you have a vector space V, together with a set function i : X --> V which is universal in the sense that given any other set function to a vector space W, f : X --> W, there is a unique k-linear map phi : V --> W such that the triangle X --> V --> W commutes

in linear algebra you just say "to define a linear map, it is sufficient to define on a basis"

does that make some sense uwu?

[formal ermine]

no

what is this universal property about

like

I don't understand how this is about a free k-vector space of X

[formal ermine]

"a vector V"?

"set function"?

[rustic crown]

kill me typos >.<

[rustic crown]

yee, just to emphasize that it's any function and doesn't not need to preserve any properties

[formal ermine]

ah

X is a subset of V and W?

[rustic crown]

you can make it a subset of V, but W is an arbitrary space and so the function f need not be injective

[formal ermine]

doesn't --> mean embedding

[rustic crown]

universal properties are nice because they characterize the object satisfying the properties up to (a unique) isomorphism

[rustic crown]

nope, just a map.

[formal ermine]

oh

[rustic crown]

people decorate the tail to indicate when it's an injection

and decorate the head to indicate when it's a surjection

[formal ermine]

I usually see people using --> in text to mean embedding

[rustic crown]

maybe you remember this from field theory where all maps are injective

[rustic crown]

what this means is that you never have to worry about the actual construction of that object, or ever have to construct maps element wise making sure that that the induced map, phi : V --> W is actually linear.

[formal ermine]

ahh I'm confused

when does the hooked right arrow mean injection and when does it mean inclusion

and when embedding

[rustic crown]

all 3 arrows are more or less the same thing categorically speaking

[formal ermine]

[rustic crown]

when you start working with arrows, an actual inclusion is in no way different from some arbitrary injective map

like in field theory you know that Q --> Q[x]/(x^2+1) is not an inclusion

but you still pretend that Q[x]/(x^2+1) is a field extension which "contains" Q

[formal ermine]

why

[rustic crown]

because say 1 on the left is sent to the coset containing 1, which is 1 + (x^2+1)

that's not "equal" to 1

[formal ermine]

I see

[rustic crown]

ofc the injective map identifies the two things together

embedding word is used in more geometric settings

in short, you zoom out

first you study actual groups

you see eleements

but after that you get bored

so you start talking about maps between groups

in the start there are not more than 2-3 groups at a time

but after a while there are too many of them

so keeping track of all the elements of all the groups is a pain

[formal ermine]

is there a word for an inclusion to an isomorphic structure

[rustic crown]

you just zoom out and pretend the actual groups are just point and the maps are arrows

[rustic crown]

embedding

[formal ermine]

so an embedding is an inclusion?

[rustic crown]

sort of?

if you're very much interested in looking at elements of an object, then you just use the word inclusion

[coral spindle]

I'd say the words are interchangeable

Most of the time we just specify that something is an inclusion for the sake of ease

E.g. a field extension is really just a nonzero homomorphism F -> F'. F doesn't need to be a subfield of F' – it only needs to be isomorphic to one

[rustic crown]

(wait do you define 0 ring to be a field )

[somber sleet]

When you have a field extension L/K and a finite set A in L, what is exactly K(A)

[formal ermine]

smallest supfield of K that contains A

[somber sleet]

Were is the difference between R[A] and K(A) ? First the fact that R is a Ring and K is a field right?

But I think there's more to it(?)

[coral spindle]

Some people do 🤷

[formal ermine]

yes

[formal ermine]

there really isn't more to it

[coral spindle]

1/A is not typically in R[A].

That's not true (message deleted)

[somber sleet]

so the difference is that we have rational polynomials in K(A)

[coral spindle]

This is true only when R is a field (message deleted)

[formal ermine]

o

[coral spindle]

That's right

[formal ermine]

mb

[rustic crown]

Z(sqrt(2)) = Q(sqrt(2))

(people will hate me for writing Z(sqrt(2)))

[somber sleet]

no worries

[coral spindle]

While we're on this subject

We use R[A] to mean one of two things:

(1) the ring of formal polynomials over a symbol A
(2) the image of this aforementioned ring

Let's say R is a subring of S, and we've chosen an element A of S

Then let's write R[x] for the ring of polynomials over R

[rustic crown]

(A was a set above btw)

[coral spindle]

I'll keep it as a single element for now

[rustic crown]

(oh okie :3)

[coral spindle]

Then for any choice of x in S, we then have an induced map R[x] -> S

So for instance if we choose x = A

we get an induced map R[x] -> S

What is the image of this map? It is R[A] — this is in some ways how to define R[A]

Illum – this is the universal property of R[x]

[rustic crown]

this sadly breaks for K(A) 😦

[coral spindle]

Aha true

[rustic crown]

lol thinking stretch

[somber sleet]

dumb question: Is this the inclusion map?

[coral spindle]

No

Let me put this another way

Let's say you have a polynomial f(x) in R[x]

and you have an element A of S

The induced map I'm talking about sends f(x) to f(A).

Will this induced map be an inclusion?

I think you should be able to come up with an easy counterexample

[somber sleet]

oopsiie

[coral spindle]

GIVE ME COUNTEREXAMPLE

[rustic crown]

(det will watch one piece tomorrow )

[somber sleet]

my mind went blank my dear

[coral spindle]

Bad news

I have 15 children

innocent babies

and my wife left me

so they're going to starve

except... you can help

just one counterexample will save them

Just one counterexample to the claim that the induced map R[x] -> S is injective

that's all they need

will you help my 15 babies?

[tribal moss]

No wonder wifey left if she had to pop out 15 kids in short enough succession that they're all still babies.

[coral spindle]

She had quindecuplets... it wasn't my fault!!!!

[tribal moss]

Boytjie is Rat confirmed.

[coral spindle]

🐀

Actually I am a chicken

[rustic crown]

are they still alive?

[coral spindle]

Let me ask them

ah

they're saying they've still got a bit left

they just need that counterexample!!!!!!!!!!!!!

[rustic crown]

@somber sleet you can do it!!

[coral spindle]

Maybe, if someone could just choose an appropriate value of A in S... it might all work out!

[rustic crown]

reminds me of dora the explorer

or mickey mouse clubhouse

[coral spindle]

I should make the algebra version of that

"And that's why we always work with Noetherian rings, kids!"

[tribal moss]

I hope for the kids' sake that we can assume 1 != 0 in S.

[south patrol]

All maps of rings are injective

[rustic crown]

poor babies

[south patrol]

For example, Z -> Z, x -> x^2

[coral spindle]

My children are giving up hope... my beautiful babies are despairing...

[somber sleet]

I feel so much anxiety I'm scared to try hahahaha

[coral spindle]

Try itttttttttt

If you try almost any example it'll be a counterexample

To produce a counterexample, we need to choose:

• A ring R
• A larger ring S, containing R, but it could be equal.
• a value A in S.

[rustic crown]

but it could be equal

[coral spindle]

Hey, it's all for the children

[somber sleet]

Maybe Z4 and Z2(?)

[coral spindle]

Z_2 is not a subring of Z_4, and vice versa.

[somber sleet]

pff

oops

[coral spindle]

What's the easiest possible example of a ring

Just any ol' ring

throw one at me

[rustic crown]

0

[coral spindle]

your favourite one

maybe not that one

[somber sleet]

Z

[coral spindle]

Nice

I like Z too

we'll say R = Z

Any nice rings you know that contain Z?

[somber sleet]

(I'm laughing without a right motive)

[somber sleet]

I'd say Q, but I don't think you like it

[coral spindle]

I like Q too

so we'll let S = Q

ok

so once we've chosen a number A in Q, we'll have a map Z[x] -> Q defined by sending the polynomial f(x) to f(A)

So we've got a concrete description of what this is

What would it mean for this map Z[x] -> Q to be injective? Can you describe it nicely?

[somber sleet]

for f and g in Z[x], f=g-> f(A)=g(A)

I feel so much pressure

[coral spindle]

Alright ok just a sec here

[somber sleet]

not even when I'm in an exam room

[coral spindle]

I want to tell you a very useful trick

If we're looking at groups (remember rings are, in particular groups under addition)

if we have a map f : G -> H, it's injective iff ker f = 0

So we can get a really nice condition in terms of this for this map Z[x] -> Q being injective

[somber sleet]

is it a ringhomomorphism?

[coral spindle]

Wanna give that a shot?

[coral spindle]

Well I was just talking about group homomorphisms, but remember that ring homomorphisms, if we just ignore the multiplication, is a group homomorphism too

[somber sleet]

wait, I pinged the wrong answer, is the map you gave me an homomorphism?

[coral spindle]

Every map we've been talking about is a homomorphism

Well – except the polynomials of course

[somber sleet]

ker phi = 0

0 polynomial in Z[x] of course

[coral spindle]

That's right

So can you translate that into saying something involving A?

Just a reminder: the map Z[x] -> Q is defined by sending the polynomial f(x) to f(A)

At this point, we've not defined A, but this might help us understand what values of A are helpful

[somber sleet]

I mean if your f(x) is the 0 polynomial, then it doesn't matter what value your A has, you always get 0 (?)

[coral spindle]

OK

But Z[x] doesn't just involve the zero polynomial, does it?

[somber sleet]

or in general by having a costant pol

[coral spindle]

The map sends any polynomial f(x) to f(A).

So what does it mean for the kernel of the map to be {0}?

Do you remember the definition of the kernel of a map?

[somber sleet]

yes

all the polynomials in Z[x] s.t. f(A)=0 in this case

[coral spindle]

Right, so if the kernel is {0}, what does this mean?

[somber sleet]

and because it's injective the only one is 0 right(?)

[coral spindle]

That's right

[rustic crown]

your babies will live

[coral spindle]

I would say this as: the only polynomial such that f(A) = 0 is f=0

[rustic crown]

we're almost there

[coral spindle]

Looking good so far!

[somber sleet]

It feels like I'm giving birth

[coral spindle]

OK so do you think you can choose some value A in Q such that this isn't true?

[somber sleet]

I'd say look at f(X)= X-A

[rustic crown]

[coral spindle]

Ah, slight problem there

Remember we're working in Z[x]

Is X-A always in Z[x]?

[somber sleet]

It was too easy

choos A=1

[coral spindle]

YES

🎉

[rustic crown]

[coral spindle]

MY BABIES WILL EAT TONIGHT!

[somber sleet]

I don't know how to feel

[coral spindle]

I will now talk about this a bit more

[somber sleet]

AHHHHHHHHHHHHHHHH

[coral spindle]

Algebraic number theory studies the maps Q[x] -> C, where C is the complex numbers

We talk about the numbers A such that the induced map is injective

They're given a special name

They're called "transcendental"

pi is an example

e is another example

[somber sleet]

oh wait a sec

it's this one right

[coral spindle]

:)

Yes, that's the restriction of this map onto its image

[somber sleet]

so when evala is not injective then a is algebraic

[coral spindle]

K[a] is the image of K[X] under the map which sends X to a

[coral spindle]

YES!!!!!!!! 🎉

[somber sleet]

you didn't work for nothing, see

[coral spindle]

Ok I'm going to go pick up my drying now

I mean – feed my babies

||to my man-eating plant||

[rustic crown]

that wasn't the part of the story!!!!

[somber sleet]

all this work to save the 15 babies

I don't know which death is more painful

[coral spindle]

It’s why my wife left me

[rustic crown]

i feel bad for what @somber sleet did now

maybe they could have escaped and fed themselves

[somber sleet]

yeah exactly

[rustic crown]

[glossy crag]

One direction is clear to me (F⊗L embeds into K'⊗L), but how does the other direction work? Is this some kind of (co)limit where K' is approximated by the finite extensions?

[rustic crown]

if K'⊗L is not reduced, then there is a nilpotent element, alpha = sum of k_i⊗l_i

what can you say about K(k_i | i = 1, 2, ...)

[glossy crag]

You mean that K(k_i)⊗L will have a nilpotent element, namely alpha?

But I feel like this hearkens back to what we were discussing yesterday

Does alpha have the same meaning in K'⊗L and K(k_i)⊗L?

I suppose you could say that K(k_i)⊗L embeds into K'⊗L, so alpha's contained in K(k_i)⊗L, but I'm still a little iffy about it (e.g. what if it embeds in multiple ways, etc.).

[chilly ocean]

can you guys ping me when you are finished

I have a question

[formal ermine]

@chilly ocean

[chilly ocean]

using lagrange's theorem for the first question, the possible orders are 1, 2, 4, 6, 8, 12

and 24 of course

correct?

[formal ermine]

it can't be 24 because then the group would be cyclic

[chilly ocean]

oh yea true

what about finding how many elements there are of order 2

[formal ermine]

what do the elements of S4 look like?

[rotund aurora]

just count

[chilly ocean]

permutations of the set {1, 2, 3, 4}

[formal ermine]

yes

when is a permutation applied to itself the identity?

so like you permute it twice and you get back to the original state

[chilly ocean]

yeah that's the question'

idk

[formal ermine]

try looking at some permutations

to see if you notice a pattern

like take (123), is (123)(123) = id?

[chilly ocean]

no

(123)(123) = (132)

[formal ermine]

is (1234)(1234) = id?

spoiler: no

what about (12)(12)?

[chilly ocean]

is there supposed to be a pattern

[coral spindle]

Why don't you try and see if there is one

[chilly ocean]

oh

[chilly ocean]

yes

(12)(12) is id

so as long as it's a transposition, its product will be id

that's the only pattern I see?

[rustic crown]

oopsie, i was waiting for a ping >.<

it doesn't matter what embedding you choose though. but here i used the one induced from the inclusion F = K(k_i) --> K'

again since we're working over fields, injective maps stay injective even after tensoring

F --> K' gets sent to F⊗L --> K'⊗L

now the alpha on the left is sent to the alpha on right

[formal ermine]

yes

that's correct

[rustic crown]

and since alpha was nilpotent on the right, it gives you a non-trivial element in F⊗L which is nilpotent, hence this is not reduced.

[rotund aurora]

I was trying to compute the quotient $\bZ[\alpha]/(5,\alpha^2+3\alpha-1)$, where $\alpha=\sqrt[3]2$. Any hints?

[cloud walrus]

Croqueta

[rotund aurora]

So the first thing I thought of is using the correspondence theorem, so maybe its easier to compute $\mathbb F_5[x]/(x^3-2,x^2+3x-1)$

[rustic crown]

yep that's the idea

[cloud walrus]

Croqueta

[rotund aurora]

both polynomials are definitely irreducible modulo 5

[rustic crown]

F5[x] is a pid, so that ideal is gen by a single poly

[rotund aurora]

ahh right its a pid

[rustic crown]

find gcd uwu

[rotund aurora]

ah lmao

[chilly radish]

This sounds off to me. What if you're tensoring with a field that's torsion

Like over Z flat iff torsionfree

[rustic crown]

modules over fields are vector spaces which are free

[chilly radish]

Oh wait you're tensoring over the base field

My b

[rustic crown]

and free => torsion free, projective, flat everything

[chilly radish]

I was thinking of tensor prod of arbitrary fields

[rustic crown]

ah oopsie, yea meant base field

[green shell]

ok I gave this another shot this afternoon and I think I got it

we write the id as a product of transpositions t_1 t_2 ... t_r

let t_r = (ab)

We have 4 possibilities for t_(r-1)

[rotund aurora]

2x^2+x+3, no? I had computed it earlier but wanted to check everything again

[rustic crown]

didn't you say they were irreducible?

(i didn't check though >.<)

[rotund aurora]

yes they are, since they have no roots

[rustic crown]

gcd of two distinct irreds

[rotund aurora]

I guess you are saying its 1 ?

[glossy crag]

What I mean is, do we know that alpha (element of K'⊗L) is contained in the image of K(k_i)⊗L in K'⊗L, as dumb as that sounds?

[rustic crown]

yeea lmao

[rotund aurora]

actually, 2x^2+x+3 has roots I think

[rustic crown]

does the image contain sum of k_i⊗l_i

[glossy crag]

I mean

The answer is yes, obv.

But for some reason I am unconvinced by it.

[rustic crown]

image containing alpha isn't a problem

what could have been an issue was that you know alpha^n in F⊗L is sent to 0 in K'⊗L for some large enough n

[rotund aurora]

ok I made a mistake, x^3-2 was actually reducible

[rustic crown]

but from this you can only conclude alpha^n = 0 on the left if you have injectivity

[rotund aurora]

so you get Z_5[x]/(x^2+3x+4), since x^3-2=(x+1)(x^2+3x+4). And x^2+3x+4 is definitely irreducible as 3 is not a quadratic residue modulo 5

the discriminant is 3^2-4*4=3 mod 5

[rustic crown]

yea mb, i was writing it as x^2-2x+1+3

which is (x-1)^2-2

so the final thing is F_5[1+sqrt(2)] = F_25

[rotund aurora]

yes, thanks

it is interesting tho that with the same polynomial we get two discriminants, 3 and -3, this shows -1 is a quadratic residue modulo 5

[rustic crown]

-3 isn't a discriminant

[rotund aurora]

wait I confused

yeah what I said didnt make sense

[rustic crown]

you calculated b^2-4ac

i calculated (b/2)^2-ac

4 just happens to be -1 here

[glossy crag]

we are not the same

[rustic crown]

[chilly ocean]

4 = -1

posted this problem before... can't figure out how to approach it

[barren sierra]

any ideas?

what have you tried

[chilly ocean]

Hint: S_3 is not commutative

[barren sierra]

also orders of elements

[chilly ocean]

yes anything above S_2 is not abelian

[chilly radish]

The main idea is to identify a feature of one group, one that is easily seen to be preserved by isomorphism, that is not true in the other group

[chilly ocean]

pj

oh

[rustic crown]

pj

[chilly ocean]

pj

[coral shale]

[chilly ocean]

can I ask what features are preserved by an isomorphism?

apart from cardinality

[coral shale]

uh

[rustic crown]

order of some element

[coral shale]

it is

[rustic crown]

for example maximum of all orders

[coral shale]

everything is preserved bar the labelling of elements

[chilly radish]

Order of elements for one

Certain commutativity relations

[coral shale]

everything !?

[chilly radish]

Not exactly

[coral shale]

if your isomorphism is f
Your group looks exactly the same

{a, b, c, ...} becomes {f(a), f(b), f(c), ...}

[chilly radish]

There's some subtle things that aren't invariant

But it's irrelevant for exercises at this level

[coral shale]

if your binary operation mapped (a, b) to c, then the new one maps (f(a), f(b)) to f(c)

[rustic crown]

(i think its a joke shuwi)

[coral shale]

im very curious - what?

[chilly radish]

Not exactly

[coral shale]

[rustic crown]

don't confuse det >.<

[coral shale]

im calling mods on this kid

[rustic crown]

det is not falling for that joke

[coral shale]

mod, ban this child

[rustic crown]

lmao

[chilly ocean]

guys can we stop the tomfoolery

we have serious business to attend to

[rotund aurora]

I guess you need to specify what type of isomorphism you are talking about

[rustic crown]

okie :3

[coral shale]

i gave a serious answer, dont look at me

[chilly radish]

There are subtle meta-properties that can fail to be preserved, but whether those are properties of groups is debatable

That was kinda my point

[coral shale]

An explicit example?

[rustic crown]

try listing number of elements of each order

[chilly radish]

Moreso the fact that treating them aa "the same thing" can be inaccurate. For example, in the context of the first isomorphism theorem, if you want to state it in the language of short exact sequence, then you say that if 0->A->B->C->0 is exact, then B/A is isomorphic to C, but this is inaccurate, because even though the image of A is isomorphic to a subgroup of B, they might not be the same thing. I guess instead of saying some properties fail to be isomorphism invariant I should have said that more emphasis should be put on the "up to relabeling" part

A might have been a subgroup.of B to begin with, but maybe A->B embeds it as a different subgroup

I guess what i'm saying is, an isomorphism of substructures might fail to respect a larger structure

Which on some level is a meta concern

[rustic crown]

but shuwi said "bar the labelling"

[chilly radish]

Sure, I just think people maybe don't always put enough emphasis on that

[rustic crown]

det is hungry

[coral shale]

Well it kinda has to be say in the context of an automorphism?

Maybe your point is more simply highlighted using non identity automorphisms as example?

or maybe i miss it

[rustic crown]

https://tenor.com/view/eevee-pokemon-go-cute-kawaii-gif-5988210

[chilly radish]

There doesn't even have to be an automorphism in play here

Maybe there's no way to switch two subgroups while preserving the bigger groups containing them

[coral shale]

i understand that point yh

@chilly ocean tldr, a good intuition for isomorphisms is a relabelling of elements

Including in the binary operation

So as has been mentioned, you can expect f(a) to have the same order as a

[chilly ocean]

hm

[coral shale]

if S is a generating set, so is f(S)

[chilly radish]

(My point is just, be mindful that relabeling sometimes amounts to changing your entire perspective of a thing)

[coral shale]

listed examples of some of the few things that are preserved (that are usually used), but theres much more

[chilly radish]

A good example is e.g. all separable hilbert spaces are isometrically isomorphic, but there's hardly only one separable hilbert space

Abstractly, isomorphism preserve everything, but in practice they should be chosen carefully

[chilly ocean]

wait

[chilly radish]

Ok that's all for my ramblings rn

[chilly ocean]

S_3 is not cyclic

C_6 is cyclic

thus

[chilly radish]

Correct!

[chilly ocean]

can we conclude

that there is no isomorphism

because there is no generator f(a)

[chilly radish]

Exactly

[chilly ocean]

can I ask why two groups must BOTH be cyclic

for an isomorphism

to be constructed between them

[coral shale]

.

you can prove this statement

[coral shale]

or first perhaps prove this simpler statement

[chilly ocean]

what about S_4 and D_12

neither S_4 nor D_12 are cyclic

[coral shale]

you can say something about generating set perhaps

Dihedrals groups can be generated by a reflection and rotation

so by 2 elements of orders 2 and 12 in this case

The same cannot be said of S_4

[chilly ocean]

OH

[coral shale]

Or instead you can count how many elements of a certain order there are

[chilly ocean]

D_12 contains an element of order 12

[coral shale]

How many elements are of order 2?

[chilly ocean]

S_4 does not

[coral shale]

that works

[chilly ocean]

thank you

but

how am I supposed to derive

that S_4 does not have an element of order 12

it's obvious w the dihedral group cuz

its just a rotation

[coral shale]

decomposition of permutations into cycles

[chilly ocean]

wdym?

[coral shale]

the lcm of cycle lengths will give you the order of the permutation

[chilly ocean]

LCM((a, b)(c, d)) = 4?

[coral shale]

2

[chilly ocean]

por que

[coral shale]

2 transpositions?

[chilly ocean]

into cycles or transpositions

[coral shale]

lowest common multiple

into cycles

transpositions are 2-cycles

[chilly ocean]

so 2 * 2 = 4

[coral shale]

decomposition into disjoint cycles i should say

[coral shale]

no thats not lcm

(012345)(6789) has order 12, say in S10

[chilly ocean]

wadafak

[coral shale]

u seen this notation for permutations right

[chilly ocean]

yes

(1 2 3) = 1 maps to 2, 2 maps to 3, 3 maps to 1

[coral shale]

yes

[chilly ocean]

but uh

[coral shale]

so if you decompose into disjoint cycles

[chilly ocean]

can u explain the LCM stuff

[coral shale]

the lcm of the cycle lengths

[chilly ocean]

pjh

[coral shale]

is the order of the permutation

ok

[chilly ocean]

hm

ok

so

[coral shale]

(012345)(6789) has order 12

i claimed this

[chilly ocean]

lcm((a, b, c, d)) = ?

[coral shale]

no, lcm(6, 4)

[chilly ocean]

ohj

[coral shale]

first cycle has length 6

[chilly ocean]

lcm(6, 4) is

uh

2

[coral shale]

thats gcd

[warm wyvern]

That's gcd

[chilly ocean]

oh

12

[coral shale]

lowest common multiple is lcm

[chilly ocean]

im dumb

it's 12?

[coral shale]

do u see why its 12

[warm wyvern]

Yea

[chilly ocean]

yes

cuz

12 = 6 * 2

12 = 4 * 3

ez

[coral shale]

no i mean like do u see why its lcm

[chilly ocean]

nope

but

[warm wyvern]

[chilly ocean]

ill use it without understanding

ig

[coral shale]

no its fairly simple

just have a think

(012345)(6789)

imagine 2 clocks

[chilly ocean]

the order of this permutation is 12?

[coral shale]

one has 6 ticks, the other has 4

they go tick tick tick

[chilly ocean]

yes

oh

[coral shale]

and you want to know how many ticks before they both go back to 0

[chilly ocean]

that's genius

okay so

the order of ANY permutation

is given by its lcm

[coral shale]

of disjoint cycle composition cycle lengths

[chilly ocean]

so

if I want the order of

say,

(a, b, c)

I can decompose this into a transposition

(a, c)(a, b)

[coral shale]

no because it wont be disjoint

[chilly ocean]

oh

[coral shale]

(123) is already a disjoint cycle

it has order 3

[chilly ocean]

how do I know it has order 3

[coral shale]

because its already a 'product' of disjoint cycles

lcm(3) = 3

[chilly ocean]

oh

so

[coral shale]

u only have 1 clock

[chilly ocean]

say we have a permutation $(a_1 a_2 \hdots a_n)$

[cloud walrus]

zzzzz

[chilly ocean]

the order of this permutation is just $n$?

[cloud walrus]

zzzzz

[coral shale]

yes

this is just a n cycle

[chilly ocean]

and the order of a permutation which is a product of disjoint cycles is the lcm

[coral shale]

assuming all a_i are distinct

yes

[chilly ocean]

you have taught me something new

i didn't know this

[coral shale]

so using this idea u can quickly find the order of everything in S4

[chilly ocean]

so we were talking about S_4 and D_12

D_12 evidently has something of order 12 (pi/6 rotation)

as for S_4

we are talking about permutations on the set {1, 2, 3, 4}

so to determine there's nothing of order 12

uh

[coral shale]

u can state it with brief explanation

[chilly ocean]

how do I determine there's nothing of order 12 agian?

i think im a bit confused on that part

[coral shale]

tbh its not that hard to list all permutation types in S4

List the 'shapes' of all permutations in S4

can you do this systematically?

The first is (), identity

then (ab)

then ...

There arent many

When I say shape - I mean the shape of its decomposition into disjoin cycles. Only considering disjoint form

For example (abc)(de) in S_5

[chilly ocean]

(a b c d)

(a b c)

(a b)(c d)

uh

[coral shale]

yep. Convinced thats all of them?

[chilly ocean]

well

[coral shale]

with the ones i began with

[chilly ocean]

() is id

(a)(b)(c)(d) = ()

do i have to consider all rearrangements of whats inside the permutation