groups-rings-fields

[coral shale]

I will try that

But what exactly am I meant to deduce

Normal subgroups from 1st iso?

And so it tells me how to quotient?

[sharp sonnet]

there are 4 diagonals, right?

[coral shale]

yh

So I can view them in S4?

[sharp sonnet]

so your action gives you a group hom phi: G -> Sym(4)

[coral shale]

yh agreed

So I then try to figure the kernel

[sharp sonnet]

order of sym(4) is uhh

[coral shale]

24

[sharp sonnet]

i think thats a pretty good hint

show its injective or surjective

then your G is Sym(4)

i think this can be done as a purely geometric argument

just get a rubiks cube and rotate it

[coral shale]

i literally have one kek

uh

so what I can do is go through the 3 types of rotation

And show none of them leave the diagonals invariant

hence the kernel must be trivial

but uhhh its a bit hard to convince myself in some cases so ill have a think

[sharp sonnet]

you can also try rotating around a diagonal

by uhh 120(?) degrees

and then do something orbit stabilizer

im also bad at imagining things

(i would let sage do the work )

[barren sierra]

sage 🥰

[coral shale]

well i certainly managed to convince myself with a prop

thanks

kinda surprised it is S4

[tribal moss]

What is our goal here?

[coral shale]

figure out group of rotations of cube

[tribal moss]

Ah, okay.

[coral shale]

initially i was expecting to figure this out in terms of semidirect product or something

to decompose via finding normal subgroups

cus idk - I thought it would be related to C4, S3 or something

[sharp sonnet]

they live in S4, so it is related i guess

[tribal moss]

Rotating 180° around an edge midpoint transposes two diagonals, and you're free to choose which two. So you get at least the entire S4.

[coral shale]

right thats a clever way to see

looking it up, I see S4 is V4 semi S3

V4 being the normal

maybe theres some way to see this

[tribal moss]

The group acts on the three face-center axes.

[coral shale]

👌 will have a go that way

Does this diagonals approach work for higher dimensional cubes?

and/or polyhedra

It seems to be a good technique

but I cant see it guaranteed to give an answer

[sharp sonnet]

you want to let the group act on something

that will always help

if you choose the correct something you are probably done

[pastel cliff]

may i interject

[coral shale]

This seems fun

ofc

[pastel cliff]

about G/H again

[coral shale]

im done

just admiring group actions

[pastel cliff]

so is it basically like

the group G if we think about everything in H as being in an equivalence class with identity

[coral shale]

alright

you glue everything in H together

it is now one object

[pastel cliff]

just like gluing multiples of k in Z/kZ

[coral shale]

yes

[pastel cliff]

yes?

yes

[coral shale]

And now you want this H

to act as the identity

in the quotient group

This is why we require the normal condition

gH = Hg

(gH)H = H(gH) = gH
So H is an identity

[pastel cliff]

coset multiplication between H and any coset gH yields gH yes

i get how it becomes a group

[coral shale]

H(gH) is only gH because (Hg)H = (gH)H
I skipped some steps above

[pastel cliff]

i think im just missing some intuition about G/H notation-wise and first iso-wise

[coral shale]

yh sure

well notation wise

youve glued things together

you are working with sets

the elements of the quotient group have to be sets

As for this, my intuition isn't quite there for why it works

I suppose

[tribal moss]

I think the intuition about the notation is simply that G/H has |G|/|H| elements.

[coral shale]

Well mathematically we write quotients this way - all quotients correspond to the same kindof idea

[pastel cliff]

my first instinct first seeing it was like

G sans H

which now seems kinda true?

but not really

[coral shale]

The slash goes the other way

G\H = G - H

[pastel cliff]

yeah ik lol

[coral shale]

so no not really

it really should be thought of as G divided by H

G mod H

[tribal moss]

It still smells of something useful.

G/H is G where we have killed everything in H as dead as we can, namely by mapping it to the identity.

Also if G happens to be a product group and H is one of the factors, then G/H is (isomorphic to) the other factor.

[coral shale]

The moons rotate about the planets which rotate about the sun.

We can consider the movement of the moons to be like the elements of the group.

We can then 'quotient' the movement of the moons around their respective planets. And now we are basically only considering their motion around the sun - ie only the planetary motion.

An analogy that sorta works. Its like you draw all the circles for the orbits (big circles per planet, small ones per moon), and you erase the small circles of the moons orbitting planets by 'quotienting'.

[delicate orchid]

oh my god what have I walked in on

it's just the group defined by g_1Hg_2H = g_1g_2H

[coral shale]

feed intuition why this is true

f : G -> H
G/ker f === im f

[delicate orchid]

trying to provide intuitive explanations for something nitezba barely has a mechanical understanding of is absolutely counter productive

[pastel cliff]

no i think i do have the mechanical understanding

[coral shale]

yh I think u clocked it

[pastel cliff]

ive done the lame operations

[coral shale]

Well alright - at least it should hopefully be clear ker will be normal?

using the homomorphism property

[delicate orchid]

sorry I shouldn't've reacted so hastily

I just saw orbital mechanics and shat my pants, proverbally

[pastel cliff]

i understood this first actually

[coral shale]

yh ok good

so it remains to see

what exactly G/ker f looks like

and how might it relate to the image of f

[pastel cliff]

YES

ive been having trouble even putting words on what im missing

[coral shale]

So we want to construct an isomorphism

[pastel cliff]

but i think that's it

ok hold up

back to the very beginning

for showing that G/{1} is isomorphic to G

[coral shale]

yh

[pastel cliff]

can i just show that G/{1} is actually just G and that's it

[coral shale]

no

its not just G

G is a box of chocolates
G/{1} is a box of chocolates, each of which is in a wrapper

They are not the same set - there is an isomorphism to find

[pastel cliff]

oh oh oh

[coral shale]

G = {g : g in G}
G/{e} = {{g} : g in G}

[delicate orchid]

me looking for the isomorphism

[pastel cliff]

wait

G/{1} is still a set of cosets

[coral shale]

yh

[delicate orchid]

yes, quotients always are

[coral shale]

{g} = g{e}

You need to be clear what the group operation defined on G/H is

(aH)(bH) := (ab)H

[pastel cliff]

isn't it always coset multiplication

[delicate orchid]

it is

[coral shale]

thats what i wrote

[delicate orchid]

but being explicit about it can help when you're trying to find isomorphisms

[coral shale]

for this particular example, it isnt too hard to guess this is sensible

f : G -> G/{e}
f(g) = {g}

It remains to show this is an isomorphism

[delicate orchid]

love how well definedness is trivial

[pastel cliff]

ok im just gonna shart out a proof

brb

my brain

pooped

this shouldnt even be that hard i just overcomplicate

i feel like im writing something awfully longwinded to show an isomorphism between two things that are clearly isomorphic

though i guess "clearly" is awfully arrogant of me

[delicate orchid]

nah it's pretty clear lol

[coral shale]

well its clearly bijective ok

u just need to do the homomorphism properly

[delicate orchid]

holy... hell

[coral shale]

and by properly i mean a 1 line algebraic proof

[delicate orchid]

chatters I just realised

[delicate orchid]

how have I not noticed

[pastel cliff]

not done since i was gonna do injective too, though ig that's enough reasoning for both inj and surj

[coral shale]

just state its bijective there is no need

that appears fine

[pastel cliff]

just to be sure, it is accurate though right

[delicate orchid]

G and {{g} : g \in G} are clearly the same size (and finite... wait fuck)

[coral shale]

kek

[pastel cliff]

shhhhh

tomorrow i will read first iso thread

[delicate orchid]

here imo you just show that g -> {g} is the inverse map

[coral shale]

i dont see anything no good

[delicate orchid]

much easier

[coral shale]

show

state

[delicate orchid]

I'd just state but for a first year might as well show

[pastel cliff]

good thing im a third year

[delicate orchid]

no fuckin way this is a third year course

are you an engineer or like mathematical physicist or something?

[pastel cliff]

i started taking relevant classes late

also i took a gap semester so this is really my fifth semester

but classes only being offered in the spring and shit like that messed me up

[delicate orchid]

blunderrrrrrrr

[pastel cliff]

im also a CS major, if it wasn't for that i would've spedrun my requirements by now

[delicate orchid]

double blunder

[pastel cliff]

[frail zealot]

are there any uses for vector spaces over finite fields or is there anything interesting about them? i can't find much material on them but maybe i just don't know where to look

[delicate orchid]

Representation theory over finite fields is the obvious example that comes to my mind

[frail zealot]

i figured that would be a thing

i was thinking about it it seems interesting

[delicate orchid]

Loch is doing something on elliptic curve cryptography which has finite field vector spaces in it

F_p[x] seems important to like... everything lol

[pastel cliff]

are antihomomorphisms a thing

or did my professor pull this extremely far out of his ass

[umbral shale]

Hey I am a bit confused with the intuition in problem 6 regarding the whole idea of a sphere of radius k and how t is defined to show that x is not an element of both s_t(a) and s_t(b)

this is in Pinter's Abstract Algebra book chapter 3 exercises.

[astral galleon]

Should have put The art of computer programming in there as well

[pastel cliff]

https://github.com/cat-milk/Anime-Girls-Holding-Programming-Books

much algebra

[kind temple]

they reverse the order of multiplication

[full panther]

I dont see why the representatives being the same implies that the cosets are equal

and i thought cosets are described as equivalent instead of equal for this exact purpose

[chilly ocean]

What would the inverse of a direct product be?

Like, say I'm studying the group 2/2Z x S_3

What would (1, (2,3))^-1 be?

[hidden haven]

Coordinatewise inverses, because multiplication is defined coordinatewise

[chilly ocean]

Makes sense

[trail stump]

so the answer says B.6

but i couldn't get my head into thinking how it is not A.5

[terse crystal]

Order of (123)(45)

Is 6

[chilly ocean]

its not asking subgroup

its asking a single element

[trail stump]

i know that

for example

for permutation on set (1 2 3 4 5)

an element would be to change that to (2 3 4 5 1)

which would be order 5 right?

[weary terrace]

Second try: Let N be the group of nilpotent nxn matrices. Let GLn be the general linear group over the complex field.

Let A be in N.
Is it true that for every C in N there exists an X in GLn such that C=XAX^-1?
If so, how do you prove it?

[terse crystal]

I just told you an element of order 6

[terse crystal]

No

[trail stump]

wait how is that element order 6?

[terse crystal]

You can have two different matrices with different Jordan forms

[terse crystal]

Lcm(3,2)=6

What do malg and mgeom mean?

[fickle brook]

why is the same question, word for word, posted both here by @olive monolith and in #linear-algebra by @chilly ocean ?

[terse crystal]

Oh I see. You answered this in linear algebra

[fickle brook]

yes i did

and i have a question of my own now

though it's more conceptual

[terse crystal]

Which is?

[fickle brook]

what does the algebraic closure of Z/pZ look like, informally?

[hidden haven]

Direct limit of F_{p^n}s

[terse crystal]

Splitting field of all irreducible polynomials in Z/pZ

Direct limit? I thought it was inverse limit… I mean colimit

[fickle brook]

for real?

[hidden haven]

Direct limits are colimits

[fickle brook]

i thought that was something like the p-adics

[hidden haven]

Inverse limits are limits

[terse crystal]

Oh I see, I got the name wrong

Thanks

[hidden haven]

p-adics have characteristic 0

[fickle brook]

wait so hold on

what are the maps we take the limit along

[hidden haven]

Inclusion of F_{p^n} into F_{p^m} when n | m

[terse crystal]

Inclusion F_p^m to F_p^n where m|n I think?

[hidden haven]

It is essentially the union of all the finite fields

[fickle brook]

also is a direct limit like stuff -> limit or limit -> stuff

[hidden haven]

stuff to limit

[fickle brook]

so what does an element of this field look like

[hidden haven]

It lies in some F_{p^n}

[broken stirrup]

our instructor was going to prove it but then left as an exercise and he wrote that A,B are not necessarily unitary

[next obsidian]

Wtf is a unitary module

[proud bear]

One where 1*x=x i think?

[broken stirrup]

so am I supposed to show that if P is unitary projective module, then A,B must be unitary too

right?

[next obsidian]

Isn’t that just a module…

[proud bear]

Yeah . But i think i saw somewhere in D&F that some people don't require that

[next obsidian]

This is fricked up

But if M is a unitary module and there’s any map M -> N

[broken stirrup]

we never assumed every ring has an identity

[next obsidian]

Hmm

This requires surjectivity

Idk I’ve never seen this

Also that’s hurbed

Rings deserve a1

[broken stirrup]

and so not every module contains identity

[next obsidian]

A module always contains an additive identity

A module also can’t contain a multiplicative identity

So I don’t get what that means

[broken stirrup]

why?

[next obsidian]

There’s no multiplication defined…

You can only add the elements

[broken stirrup]

if a ring has an identity and 1*a=a then you call it unitary

I mean the action by the multiplication

[next fulcrum]

We say a module is unitary or not, but we don't say a module "contains" the identity ... so that's probably why CHMOLUMBIA didn't get what you've said

[broken stirrup]

ooh yeah why did i say that

i probably wanted to say if R contains identity and then 1*a=a

[next fulcrum]

yeah

[coral shale]

ah its so beautiful

Now I understand why finite fields aren't boring

[vestal snow]

You can think of it as the union over n of F_p^n

[lethal dune]

F_p^infty

[stoic rose]

p-adics are the inverse limit of Z/p^nZ, while the algebraic closure of Fp is the direct limit of F_{p^n}, not the same thing at all

[coral shale]

We have group actions...

Ring actions?

[sharp sonnet]

a ring acting on an abelian group is a module

more generally you can look for maps R -> End(X) for some abelian group X i think?

when does End(X) have a canonical ring structure

what does X need to satisfy

[coral shale]

we cannot consider sets any more? 🤔

idk

[sharp sonnet]

the thing about group actions is that for every set X, Aut(X) has a group structure

so you can consider homs G -> Aut(X)

i dont think you can endow every X with a ring structure

at least not canonically

[coral shale]

ic

[sharp sonnet]

so you need more special X

[coral shale]

btw can u write Aut X?

For just a set X

I have seen it done before and 🤔

[sharp sonnet]

whats the problem?

[coral shale]

X is just a set

Is it an automorphism in the categorical sense i suppose?

[sharp sonnet]

automorphism in Set are bijections are they not

[coral shale]

yah ok

[sharp sonnet]

you can work in other categories

if you work in Vect, you get representation theory

for groups

traditionally all kinds of representations were studied

group actions are permutation representations, whats today called representation theory are called linear representations

but you can study other kinds of things

so in this more general sense i guess a group action is just a group hom X -> Aut X, but automorphisms can be in Set, Vect or wtv

if you want ring representation theory you need some X you can endow with ring structure

it might just turn into module theory

somethin Ab-enriched categories?

[coral shale]

some thinking to be done. But I suppose so for module

when i looked it up i could only see that

hmmm

[sharp sonnet]

ah nice, moldi is here to explain the cat memes to me

[hidden haven]

Ring is a preadditive category with 1 object, so its actions are functors into preadditive categories

Similar to how monoids/groups are categories with 1 object, so their actions are functors into categories

An R-module is a preadditive functor from R as a preadditive category to the preadditive category of Ab groups

A more down to Earth explanation for why we don't consider actions on sets is

Groups and monoids arise as Auts and Ends of sets, in the sense that the symmetric group is the prototype for a group (and Cayley says that all groups arise as that) and endomorphisms of sets are the prototype for monoids

Similarly, the prototype for a ring is End(abelian group)

So it makes sense to talk about actions of rings on abelian groups

[coral shale]

This is... profound

[hidden haven]

[sharp sonnet]

oh nice, so i was correct

[coral shale]

'Monoids arise as Ends of sets'

i have yet to clock this hmmm

ah ok i think i have a clue

Surjections cannot be undone when they are not injective

I can see the endomorphisms under composition will result in a monoid now yh

[coral shale]

So Aut(abelian group) would be prototype for field?

vector spaces

[hidden haven]

Not quite, you also have 0 in field lul

But yeah field is like aut(abelian group) along with 0 map

Rather division ring is like that

Field is commutative division ring

[tribal moss]

And conversely every monoid embeds into End(X) for some set X. Namely, you can take X to be the underlying set of the monoid itself, and let each element act by pre-composition.

[coral shale]

yh I'm trying to clock that right now

We certainly have finite monoids (that aren't groups)...

But we have to consider infinite X

To find which X the monoid embeds into its End(X)

right?

[tribal moss]

Yeah, but that can just be the set of elements of the monoid itself.

It's a very direct parallel to Cayley's theorem.

[coral shale]

hmm I feel like I'm missing something

End(X) are all the surjective morphisms from X to X right?

[hidden haven]

No they are all morphisms

[coral shale]

Ah.

Epimorphism
Endomorphism

I always mix these

[hidden haven]

[coral shale]

Endomorphism is just f : X -> X =.......=

[hidden haven]

=........=

[coral shale]

Yeah ok that makes a lot more sense

You can compose any 2 maps, but there not necessarily any way to undo it

No wonder I was thinking in circles

[tribal moss]

I don't think that quite works. For example, if our abelian group is Z×Z, then the maps id and f(x,y) = (-x,y) are both automorphisms, but their sum is neither 0 nor an automorphism.

(Even id+id over Z is neither, too).

[hidden haven]

True

[tribal moss]

So for a division ring we need the special case where Aut(G)+{0} is closed under addition.

[hidden haven]

This happens when G is a simple module

By Schuri's lemma

[coral shale]

So all rings embed into End(G) for some abelian G

Fields are a bit trickier 🤔

[tribal moss]

The direction that's tricky to generalize is "If G is abelian, then End(G) is a ring".

[coral shale]

Because we need the compatability of operations right

or wait hmm

The endomorphisms of G have to preserve the structure of G
f(x+y) = f(x) + f(y)
So we have this

[coral shale]

mmm I don't get why this might not be true

[tribal moss]

One way to look at it is that when S is some structure, then End(S) naturally inherits all the operations on S (in this case addition). However, Aut(S) as a subset of End(S) is always closed under composition, but not necessarily closed under those inherited operations, which makes it problematic to view Aut(S) as "a thing of the same kind, but with an additional operation, namely composition".
For automorphisms of sets this is not a problem, because there are no inherent operations on sets to be closed under.
Automorphisms of groups gives us the problem: Aut(G) is not closed under the addition it inherits from G.

[coral shale]

Right, as highlighted by this example, I see

[tribal moss]

Similarly, if M is an R-module, then End(M) gives us a prototype example of an unital associative R-algebra, but Aut(M) is difficult to give more than a group structure.

[tribal moss]

Hmm, on further thought this is not quite correct. If G is a non-abelian group, End(G) doesn't inherit an analogue of the group operation, so it's more subtle than that. The group operation does inherit to the set of all functions G->G (not necessarily group morphisms), though. This produces a "near-ring" which lacks distributivity from the left.

[coral shale]

A new world of abstract objects I have yet to delve in

[untold basin]

I would like some help (don't give me the answer) please, on :
gcd(n,m)Z include nZ + mZ

where Z is the set of integers

[sharp sonnet]

consider the minimal element of $(n\mathbb{Z} + m\mathbb{Z}) \cap \mathbb{N}$

[cloud walrus]

Lochverstärker

[untold basin]

Isn't just the ppcm(n,m) ?

lcm(n,m) I mean

[sharp sonnet]

no

maybe check some examples

[untold basin]

Oh you changed

[untold basin]

I did

[sharp sonnet]

ye, i had a typo before i edited

[untold basin]

But I struggle to generalize it

I took 8Z + 12Z = 4Z

[untold basin]

I'll think about that
If I don't find anything I'll ask again if you're ok

[sharp sonnet]

the minimal element of the set i defined above generates nZ + mZ

so you have to show that it is in gcd(n, m)Z

[stoic rose]

Do you have trouble understanding why gcd(n,m)Z contains nZ+mZ or the other way around (gcd(n,m)Z is contained in nZ+mZ)?

[untold basin]

Second one

[stoic rose]

Then it's basically Bezout's theorem

[untold basin]

gcd(n,m) = 1 <=> there exists u,v in Z, s.t : nu + mv = 1

This one ?

[stoic rose]

The version I had in mind was "there exists u,v such that nu+mv=gcd(m,n)", but that follows easily from the one you stated

[untold basin]

Okay okay
I'll think about what you told me guys and I will try

Thank you

[sharp sonnet]

Let $I$ be an ideal in an integral domain $R$ with field of fractions $K$. We can consider the set $X = {x\in K : xI \subseteq I}$. Clearly $R \subseteq X$. Is there a name for when $X = R$?

[cloud walrus]

Lochverstärker

[sharp sonnet]

(alternative question: how should i call it)

[chilly radish]

In what context did this come up?

Haven't heard of something like this, but all the names I'm coming up with are taken

Maybe something like indivisible ideal?

That doesn't fully capture it either tho

Hmm

[sharp sonnet]

context of the ideal class group, when is the inverse of an integral ideal also an integral ideal

might just be equivalent to that and i call it invertible (i.e. dont give it a name)

was just wondering if there is some other context where this appears and if it has a name there

[next obsidian]

It definitely isn’t equivalent to invertible

Or uhhh

Oh maybe it is

Idk

[sharp sonnet]

i mean inverse being an integral ideal as well

i work in dedekind domains anyway, so considered as fractional ideals everything is invertible

[chilly radish]

What does integral ideal mean here

[sharp sonnet]

a normal ideal, i.e. not fractional

at the end of the day i define an equivalence relation ~ on the set of ideals where I ~ J iff (a)I = (b)J for some a, b in R with ab != 0
now [R] is the equivalence class of all principal ideals and call I invertible if there is an ideal J such that IJ = [R]

question: when is I invertible

[fickle brook]

@prisma ibex @tribal moss @stark sigil update on the thing i came here with yday: turns out i should have looked at & attempted to generalize a result in a paper my advisor sent me, as opposed to trying to prove it entirely from scratch

[stark sigil]

😆

[prisma ibex]

yeah see haha I told you this is like, massively nontrivial

how did the meeting go?

[fickle brook]

it went ok

we talked about related works in the same field

[tribal moss]

Heh.

[chilly ocean]

Even if some factorization of c doesn't have any unit, but still p is an irreducible.

Then why is the sentence true

[south patrol]

i'm not really sure what your sentence means

[thorn delta]

the point is that if p = uc and c = ab, then if a is not a unit, then p = uba implies ub is a unit, and therefore b = u^-1 ub is a unit. so c is irreducible

[thorn delta]

i.e. a factorization of c must have a unit

[chilly ocean]

Why ub is a unit

[thorn delta]

p = (ub)a and since p is irreducible, a or ub is a unit. Since a is a not a unit, ub must be a unit

[chilly ocean]

Why b is a unit...@thorn delta

[thorn delta]

b = u^-1 (ub), a product of units. More explicitly,
b^-1 = (ub)^-1 u

[chilly ocean]

I see

[toxic zephyr]

when solving a polynomial equation in a modular ring (of a composite number) how do you find all of the solutions? (without just checking every possibility)

working with $x^3-x=0$ in $\bZ/8$

[cloud walrus]

nix

[toxic zephyr]

which factors to $x(x-[1])(x+[1])$ which gives solutions $x=[0],[1],[7]$

[cloud walrus]

nix

[toxic zephyr]

im assuming you need to use the zero divisors to get the others, but is there some sort of algorithm/method to do that reliably?

[coral shale]

factor theorem

as you demonstrated

Since you're in a finite field, you can even brute force through all possible numbers

and find all possible linear factors

[toxic zephyr]

"factor theorem"? i havent heard of that. unless you mean like the basic basic algebra factor theorem.

[toxic zephyr]

is there a non-brute force method that would work if it was something like Z/128?

[coral shale]

yeah for polynomials in C

factor theorem or even remainder theorem applies

Polynomial rings over fields are Euclidean Domains

Euclidean algorithm exists in them

(I can't remember/be asked to remember if you need the polynomial ring to be a Euclidean Domain for the factor theorem to apply, or can you take a slightly weaker premise)

But anyways - euclidean alg exists, and factor thm applies.

[coral shale]

Some clever tricks that are unlikely to generalise for all cases?

Factoring polynomials isn't a trivial thing over non-finite fields in the first place.

I don't know why you feel it should be easier just because the field is finite 🤔

If anything I feel it gets harder (except for the fact you can brute force)

idk

[toxic zephyr]

not "easier" per se. i just mean a more algorithmic approach that isn't just brute force computation. i assume a method of that sort would involve the zero divisors.

[coral shale]

I give you a polynomial in Z[x]

how do you propose to factorise this over C with an algorithmic approach?

[toxic zephyr]

its not the factorization part that im talking about. its finding the zeros. because when it isnt an integral domain, you cant just set the factors equal to zero and call it a day

[coral shale]

Oh in fact I was talking crap earlier --- you aren't even in a finite field

You're just working in Z/nZ[x] for n that might not be prime 🤔

🤔 yh I don't know anything special to help --- it really is brute force to my knowledge

And you almost certainly won't be asked for something in Z/128Z unless it's obvious

If you aren't even in an integral domain though...

the polynomial ring that is... factorisation may not even be unique???

[toxic zephyr]

yeah what i mean is that i got the solutions 0,1,7. but 3 and 5 are also solutions (because it isnt an integral domain), and im not sure how i could come to that conclusion without having brute forced it. if there is no method to do that then... well that kinda sucks...
im going to try setting each factor equal to a zero divisor and seeing if i can make any deductions about which ones could or could not work

[coral shale]

and like what are irreducible factors aren't obvious

[toxic zephyr]

right

[coral shale]

not all linear factors are necessarily irreducible

sounds very not fun

[toxic zephyr]

haha

well... personally i think its kinda cool. i like that all of the solutions arent obvious by inspection. i just wish there was a "smarter" way to find them.

[coral shale]

So anyways - I don't even see the point of factoring much?

[toxic zephyr]

in Z/8, all of the zero divisors are just the even numbers, right? 2,4,6?

[coral shale]

for solving your polynomial equation

[coral shale]

yh

Like if u see a factorisation, use it

But regardless, u kinda have to brute force

cus you have have non-zero factors solving it

[toxic zephyr]

it gives you the obvious factors, at least. which is a good start.

[sharp sonnet]

if you have a root in Z/p^nZ that maps to a root in Z/p^{n-1}Z, so you can look for roots in Z/pZ first and then go in the other direction

if your polynomial is nice, hensel lifting gives you a unique root

if you check polynomials mod n and n isnt a prime power, you use the chinese remainder theorem first

@toxic zephyr

[coral shale]

My number theory weak I see 👀

[toxic zephyr]

oh interesting. so in Z/4 (x-[3]) is a linear factor (from the (x+[1])). is it not a coincidence that x=[3] is also a solution in Z/8?

[sharp sonnet]

its not, 3 in Z/(4) must lift to a root in Z/(8), either 3 or 3+4 = 7

since 3 and 7 get mapped to 3 under Z/(8) -> Z/(4)

so you have a little less cases to check

[toxic zephyr]

wow thats so cool 👀

[sharp sonnet]

and if you look at the hensel lifting theorem, it can be better

[toxic zephyr]

man now i really wish i had taken number theory this semester. its only offered in spring at my college, but im already taking a bunch of other harder classes 😦

[toxic zephyr]

looking into it now

[sharp sonnet]

this is usually taught in like a second course in number theory though

its important for the theory of the p-adics

~~but also yes, both of you should take number theory ~~

[coral shale]

i did. i forgot it.

[delicate orchid]

oh I actually know the hensel lifting theorem

[coral shale]

Was not a good student before this year

[sharp sonnet]

the weird thing is you can do newton method to compute the lift

or rather, they are formally the same thing

so a lot of people know hensel lifting

[coral shale]

https://i.imgur.com/tIUESSm.png

yep.

I uh- do not remember much of Number theory from 2 yrs ago

Some relearning to do when I look into p-adics ig

[sinful mirage]

how to see that $(\Theta^{-1}(m)(u,v,w))(z)=z(m(u,v,w))$?

[cloud walrus]

ProphetX

[next obsidian]

this might be better in #diff-geo-diff-top

[sinful mirage]

ah sure,sorry

i thought to post it here since it's a purely algebraic problem

composition of maps

if $\theta(f)=g \circ f$, where g is some map

[cloud walrus]

ProphetX

[sinful mirage]

is $\theta^{-1}(f)=g^{-1}(f)$?

[cloud walrus]

ProphetX

[sinful mirage]

I'd say no

[delicate orchid]

what makes you say no

[sinful mirage]

intuition

[delicate orchid]

let $\theta^{-1}(f) = g^{-1}f$
then $\theta(\theta^{-1}(f)) = gg^{-1}f = f$
$\theta^{-1}(\theta(f)) = g^{-1}gf = f$

[cloud walrus]

Wew "Peenie Wallie" Tbh 🦟

[delicate orchid]

I believe your intuition might be wonky

[sinful mirage]

fair,so seems like so

now i'd somehow have to apply it xd

[chilly radish]

Follows trivially from functoriality of Hom(A,-)

[sinful mirage]

functors

i'm a poor physicist

[delicate orchid]

I don't speak.... hmm... I think that's ancient Mesopotamian

[chilly radish]

I'm mostly memeing

[delicate orchid]

tbf, yes you're right (I think I get what you mean by that anyway)

[sinful mirage]

ok a pure algebra problem

Let Y iso Y'. Then Mult(X,Y) is iso to Mult(X,Y')

how to prove this?

[delicate orchid]

what's Mult again

[sinful mirage]

Multilinear maps

[delicate orchid]

ah ok

[sinful mirage]

I should prove that theta is an isomorphism

I claimed it,but no proof

[delicate orchid]

the iso I think will work is just taking f in mult(x, y) and then composing it with the iso from y to y'

[sinful mirage]

that's what i did

[delicate orchid]

would that work ShiN, if you're still here

[sinful mirage]

Can circ f

[sinful mirage]

that's why i asked what is the inverse of that map,since it should be iso

[sinful mirage]

this is basically it i think,no?

[delicate orchid]

that's the proof of what you asked me to prove, composition of two isomorphisms is always an isomorphism so should work

[chilly radish]

You need to conjugate by the iso

[sinful mirage]

wdym

[chilly radish]

That should work

[sinful mirage]

so Theta(f)=iso circ f is not good?

[chilly radish]

Oh wait nvm

I was getting confused with smth else

That should worj

Lemme.think for a sec

[sinful mirage]

psi is invertible

[chilly radish]

Yes that should work, if the iso is linear then this preserves multilinearity and exhibits an obvious inverse

It's also clearly linear itself

[sinful mirage]

that's the question

what's the obivous inverse

is it Can^{-1}(f)?

[chilly radish]

what wew lads said

Just composition with the inverse of the iso

Oh lemme look at the sc

Yes

[delicate orchid]

yeah the one thing I was worried about was preserving linearity

[sinful mirage]

i'm going crazy

[chilly radish]

It should be composition with Can^-1

[sinful mirage]

i've been on this proof for 5days

[chilly radish]

Well I think we're assumig a linear isomorphism

[sinful mirage]

18 pages still not finished

[chilly radish]

So this shouldn't be a problem

[sinful mirage]

pre or post composition?

[chilly radish]

Jesus

[sinful mirage]

that's the question

i'd say pre

[chilly radish]

To the left

Like wew said

[sinful mirage]

yes

[chilly radish]

I don't like saying pre and post

Those mean diff things to diff people

[delicate orchid]

oh yeah DUH x and y are some modules/vec spaces

I need to get out of "group mode" sometimes

[sinful mirage]

finite dim vector spaces

physicists go easy mode

no general

[sinful mirage]

how to show that V and V^{**} have same dimensionality in finite dim?

if I can show that the dimensionalities match, then i'm mostly done with the profo

[chilly radish]

You showed you have a linear bijection didn't you

That implies equality of dimension

Always

[sinful mirage]

wdym I showed I have a linear bijection

I showed surjectivity by dimension arugment

I only showed injectivity properly

[chilly radish]

Oh my b

I missed that

You can prove that V and V* always have the same dimension. Choose a basis for V, construct the dual basis for V* and show that it's indeed a basis

[sinful mirage]

i'll try this. then the argument would be: V and V* have same dimension. But V* and V* * have the same dimension too. therefore, V and V * * have the same dimension

right?

[chilly radish]

Yes

Exactly

[barren sierra]

ok so I think first thought is it can't also be a right ideal right?

I came up with I = 2 x 2 real matricies with 0's in the right column

that's a left ideal

and also not a right ideal

but now i'm right multiplying by units

and am not getting ideals

[thorn delta]

Huh? How is that possible Ia should always be an ideal

[barren sierra]

right

uh it's always the same ideal

my bad

[thorn delta]

Take a to be the matrix which switches the columns under right multiplication

[barren sierra]

that's not a left ideal tho right?

wait

fuck me it is I would have been done so much earlier

ugh

that was literally my first thought 😭

[thorn delta]

I know the feelin

[lavish nexus]

in a noetherian module a map u: M-> M surjective implies bijective

we have the chain ker u \subset ker u^2 \subset .... \subset ker u^n so it must terminate

then ker u^n = ker u^(n+1)

but ker u^(n+1) = {x | u(x) \in ker u^n}
so the map u restricted to ker u^(u+1) maps onto ker u^n

if u restricted is injective then I'm done
but I don't know ker u^n is finite

[next obsidian]

That’s not what you need

You can show that it can’t terminate

u^n is still surjective

So for some nonzero x in ker u, write x = u^n(y)

But now y is in ker u^{n+1} \ ker u^n

[lavish nexus]

👀 cool

[chilly ocean]

Is every field UFD?

?

[upper pivot]

yes

ufd is unique factorization upto units right

everything is a unit in a field

so its kind of vacously true

[lavish nexus]

$A = k[x]/(x^2)$
$A \rightarrow_{f} A \rightarrow_{f} A \rightarrow_{g} k \rightarrow 0$ is exact. $f$ is multiplication by $x$ and $g$ is evaluation at 0.
Denote the sequence as $E$, WTS $E \otimes k$ is not exact

[cloud walrus]

Iteribus

[lavish nexus]

tensor is right exact so the problem should be at the second A but I can't find why it is not exact there

$\mbox{Im}f \otimes 1 = (x) \otimes k$

[cloud walrus]

Iteribus

[lavish nexus]

$\mbox{ker} f \otimes 1 = {(x+a)\otimes b | f\otimes 1 ((x+a)\otimes b) = 0} = {(x+a)\otimes b |(ax \otimes b) = 0} = {(x+a)\otimes b |a = 0 \mbox{or} b = 0} = (x) \otimes k$

[cloud walrus]

Iteribus

[next obsidian]

I mean this should turn into multiplication by x on k, and I’m assuming you’ve defined x to act as 0 on k

So the induced map is the 0-map between k and k and this breaks injectivity or some Shit

[lavish nexus]

but after tensoring doesn’t a map f changes by default to f tensor 1

where does x act on k

[next obsidian]

When you tensor with A over A it doesn’t do anything

A (x)_A k = k

[upbeat swift]

Hi um in confused about what to do with Euclidean Algorithm up to this point

I'm*

[lavish nexus]

x^3+1 is not irreducible

[upbeat swift]

It should be if there's an answer for it's gcd

[lavish nexus]

what is the common root of the two

[upbeat swift]

The factors are (x+1)(x^2-x+1)

Gcd of both are x+1

[lavish nexus]

so you’re done

[upbeat swift]

I can't

I have to show it using Euclidean Algorithm

[lavish nexus]

oh that… that is not a fun calculation

[upbeat swift]

Ikr

I'm struggling so much even when I know the answer

[lavish nexus]

we’re tensoring over A not k?

[next obsidian]

I mean

If you tensored with k over k

Then it does nothing

So it stays exact

So it only made sense to tensor over A

[lavish nexus]

👀I see
then x is like a scalar and I can just move it over to the right

and define that to be 0 map on k

[lethal dune]

is this what u guys call a long exact sequence?

[lavish nexus]

ty

It looks very long indeed

[lethal dune]

maybe longer

[muted lintel]

Im trying to understand rings of fractions, and there is this lemma, where I dont understand the equalities. I post a picture, since its a little easier, than typing it all out

Im sorry, its in another language

[next obsidian]

Uhhhh

I think this is just saying that the addition and multiplication is well defined

[muted lintel]

yea, I understand what the lemma says, but the proof and those equalities that lead to the conclusion, I'm having trouble with

[next obsidian]

Yeah idk how to help you with that, I can’t read the shit lol

[muted lintel]

aaah I got it

[next obsidian]

Pretty sure it just comes from distributing everything out and then replacing the stuff you know is equal

[muted lintel]

they are just multiplying out the parenthesis

[next obsidian]

Yeah

[muted lintel]

yea, thank you!

[tough raven]

If we have $\forall c \in R, ac = 0 => bc = 0$ for some $a, b$ in a commutative ring $R$, does it follow that $a$ divides $b$? If not, is there a counterexample?

[cloud walrus]

Raghuram

[delicate orchid]

sorry, just have to ask

ah nevermind

[hidden haven]

good handwriting

[lethal dune]

what font is it? xkcd?

[lethal dune]

,tex wait did they change the font?

[lethal dune]

[delicate orchid]

lol

[delicate orchid]

Nah that isn't true

pick your favourite number n with two non-equal prime factors p, p'

then in Z/nZ p and p' are both zero divisors but don't divide each other - I think

[lethal dune]

ac=bc => (a-b)c=0

[delicate orchid]

I'm assuming you're saying I'm wrong

cause I have absolutely no idea what you mean by that

[bitter glacier]

Does $ H(A,B) = TrAB^T $ give an inner product over $ V:=mat(n,\mathbb{R}) $ ?

[tribal moss]

Yes -- basically it's just the sum of products of matching entries from A and B.

So the same as the dot product of the vectorizations of A and B.

[bitter glacier]

im confused on notation here does TrAB mean trace of AB or trace of A times B

[tribal moss]

I understood it as Tr(AB).

[bitter glacier]

ahh damn yeah i think that right gonna have to redo it all lol

[tribal moss]

(Tr A)B can't be an inner product in the first place -- it isn't even a scalar.

[bitter glacier]

yeah thats what i thought but i was just confused over no brackets

chhers

[tough raven]

Right. The hypothesis is that b is “more” of a zero divisor than a in the sense that it gives 0 when multiplied with more elements.

[median pawn]

could someone explain the last two lines?

"this cannot be true for a non-zero free R-module"

why is that?

[median pawn]

suppose I is a free R-module, so it has a basis {f_1, ... , f_n}. We took f in I, so f = r_1 f_1 + ... + r_n f_n. If there exists g in R such that fg = 0, we would have r_1g f_1 + ... + r_n gf_n = 0. by linear independence of the basis, r_1g = r_2g = ... = r_ng = 0.

where is the contradiction?

[tough raven]

I think this should be true in any quotient of a PID, so counterexamples would have to more complicated than Z/nZ.

(Consider R = P/nP where P is a PID and n in P. For any a, a = g (a/g) where g = gcd(a,n). gcd(a/g, n/g) = 1 so we can find x, y st
x a/g + y n/g = 1 => xa = g (mod n)
Thus a | g and g | a (mod n), so upto units (which in particular don't affect the set of c st ac = 0), all elements of P/nP are factors of n.
For g dividing n, gh = 0 (mod n) iff n/g | h. So if we have a,b factors of n and ac = 0 => ab = 0, then n/b | n/a => an | bn (multiplying by ab) => a | b.)

[next obsidian]

Take 2x2 matrices over Z

And consider the elements which are
2 0
0 0
and
3 0
0 0

I think this probably works

I hope

[tough raven]

Yep
Nice

Thanks

[south patrol]

what stops you from just taking non zero divisors a,b which don't divide one another? or am i just confused by the wording

I read it as 'Suppose there are a,b such that whenever ac = 0 from some c, we have bc = 0. Show that a|b'

[next obsidian]

Yeah so

I was gonna also suggest that

[south patrol]

Or is it saying that for all a and c such that ac = 0 there is b such that bc= 0 (but then just take b = a)

Quantification seems a lil unclear

[next obsidian]

This makes it trivial

But it’s not as interesting

¯_(ツ)_/¯

[fickle zealot]

what is it called when you extend a ring with a singular element

like you'd call $R[G]$ a group ring (where $R$ is a ring and $G$ is a group) but idk what you'd call like $\Z[\sqrt{3}]$

[cloud walrus]

56

[next obsidian]

I don’t know that there’s a term for this

Usually when things are generated by one thing they’re called cyclic

[coral shale]

adjoin. You adjoin elements to rings to make new ones

[next obsidian]

But I’ve never heard anyone use that for algebras

[fickle zealot]

ah ok

[delicate orchid]

I've seen R[G] for a group ring but not for a general algebra

[next obsidian]

There’s two meanings

It could be a polynomial ring on the elements of G

So it’s just a super formal thing

[tough raven]

fix a, b; then take the hypothesis (which is specific to that a and b) to be forall c.

[next obsidian]

And if G is a subset of an R-algebra S it could the R-subalgebra generated by G

[south patrol]

Oh well then yeah just take for example a =2,b = 3 in Z

Nice

[ember field]

shld I go with gallian or dummit foote as a beginner.

[chilly canyon]

Hi! I'm not much of a group theorist, so maybe this is very trivial, but: do you know of an example of a group that is both finitely generated and finitely related, but which is not finitely presented?

(that is, we can have a finite number of generators or relations, but not both at the same time)

[coral shale]

finitely generated, related but not presented?????

i dont understand

<G | R>

is the form right

and you are saying G and R are finite

[tribal moss]

I don't think there can be one. Such a group would need to be the free product of a finitely presented group and a free group of infinite rank.However, a finite number of generators can never allow us to make all of the generators of the free factor.

[chilly canyon]

Yes, you can find two presentations G=<S | R> and G=<T | P> with S finite and P finite, but never a presentation G=<A | B> with both A and B finite

[tribal moss]

He's saying it has one presentation where G is finite and a different presentation where R is finite.

[coral shale]

ohhhh i understand

sry

[chilly canyon]

That makes sense, but is it really true?

[tribal moss]

On further thought, no, that doesn't work.

[chilly canyon]

Maybe some wild presentation with a sh*t ton of relations can yield that group with infinite relations

[coral shale]

this doesnt look easy

[glass grail]

i dont think its possible

i think you can pull something with schanuel

[tribal moss]

It needs to be finitely generated before we start adding our shitton of relations.

[chilly canyon]

Like, Schanuel's conjecture?

[glass grail]

schanuel's lemma

[chilly canyon]

Ah I was scared at first because I couldn't see the link x')

[coral shale]

can i have an explicit example of a finitely generated infinitely related

[chilly canyon]

Yeah, Z \wr Z for instance

It has presentation < a, b | [t^n a t^-n, t^m a t^-m] for all m,n in Z >

And by Baumslag, it is not finitely presentable

[coral shale]

I see, ty

[chilly canyon]

(actually, here's a link with that example: https://math.stackexchange.com/a/547149/259363, and it doesn't use Baumslag)

(but it uses HNN thingies and Britton lemma, both I never heard about before xD)

I don't quite understand the connection between this and the group-theoretic version... 😢
I know if we take R=Z in your linked MSE, then we're dealing with abelian groups, but is that sufficient in general?

[glass grail]

oh wait im having a stroke

sorry

[chilly canyon]

(as I disclaimed earlier, I'm not much an algebraist sorry :D)

[glass grail]

i thought that was on groups

[chilly canyon]

I am with groups indeed

[glass grail]

ok i think my brain is just being idiot rn schanuel's lemma isnt even on groups

[chilly canyon]

Seems like it's for modules 🙂

[glass grail]

yeah

[chilly canyon]

But it looks like it makes use of a free product with amalgamation, maybe there's a group-theoretic version?

(looking at the Proof section of Wikipedia)

[chilly ocean]

Y does this hold

[chilly canyon]

Anyways, thanks for the insight folks! I'll post a MSE question that won't get drowned under the flow of messages, and maybe someone will come up with a counter-example later (I suspect there is one, group theory is wild xD)

[tribal moss]

Actually I think my argument from before does work after all.

Suppose our group is <x1,x2,x3,... | finitely many relations>.

Since there are only finitely many relators, there is a maximal variable they even mention -- suppose that is xk.

Now can this group be finitely generated? Each generator would need to be given as a finite word over {x1,x2,x3,....} (and their inverses), so all in all there are finitely many xi's mentioned by those generators. Suppose the highest variable mentioned by any of the new generators is xm.

Now let p > max(k,m). How can xp be generated by the new generators?

Simply sticking the generator words together in any combination will not give something that mentions xp.

And reducing using the original relators cannot introduce an xp that wasn't there already.

Contradiction!

[coral shale]

Should I care to learn about the Wreath Product at this point in time

Doing Galois theory

[tribal moss]

Just as long as you have it ready in time for Christmas decoration.

[coral shale]

where is it used in

[chilly canyon]

Well, maybe this is because of the infinitely many relations you have in the presentation with finitely many generators you considered?

Unless I'm missing something in the argument :\

Oh, but yes

Instead of asking "can it be finitely generated", just ask "can it be finitely presented", and you have a finite number of relations x')

Mmmh, I still need to write it down properly because I'm not 100% sure there's not shady stuff under the hood... But thank you very much for the insight, I'll see if I can make the argument work!

[tribal moss]

Not really -- I'm saying we don't even get to the point where we begin picking relations in the new presentation.

Oooh, I see what you mean.

Right. Changing my mind once again, then. :-p

Actually, no, I'm not changing my mind.

[chilly canyon]

Hehe I think it's not an easy question afterall x')
Yeah what happens is you express the new finitely-many generators in terms of the old, but the old ones express as the new ones also

[tribal moss]

Even if we can add infinitely many relators in the new presentation, they can't introduce relations between elements that were different according to the original presentation.

[chilly canyon]

Ooooh no no no okay I have it, your argument does work!

Thank you very much!

(I was trying to translate Schanuel's lemma in terms of groups only, it also looks like some fun actually, and seems like it's feasible if you replace direct sums with free products and if you make use of the free product with amalgamation to almost copy-paste the proof)

[chilly ocean]

Why <c>⊆N_r

[tribal moss]

Because c is in N_r.

And N_r is an ideal and therefore closed under the operations that make <c> from c.

[chilly canyon]

Wait @tribal moss, what this shows is the following:

If G=<S | R> with R finite and S infinite, then G is not finitely generated.
But how does this show that [finitely generated and finitely related] => finitely presented? (I feel like a stupid first year student now, because I have the impression I'm missing something very obvious here...)

[tribal moss]

My claim is that you can't have all three of these at the same time:
a) G is finitely generated
b) G is finitely related
c) G is not finitely presented
To prove this I show (c) and (b) => not (a).

Namely by (b) we have G=<S | R> for some finite R. If S were finite, then G would be finitely presented, which would contradict (c), therefore S is infinite. Now insert the argument from above, which concludes not-(a).

[chilly canyon]

Ah, yes, thank you, my universe just re-stabilized itself!

We need to do something similar for (a) and (c) => not (b) too x')

[tribal moss]

No; by propositional tautology "C and B implies not-A" is logically equivalent to "A and B implies not-C".

Since either of these is equivalent to "not-A or not-B or not-C".

[chilly canyon]

There are clever people on this Earth! Thank you sooo much, I'll sleep well tonight!

[broken stirrup]

Hi everyone, I'm trying to show if that short exact sequence is split exact, the P is projective. Is it correct?

[next obsidian]

You need B to be a free A algebra in order to get that to work

[barren sierra]

(3) means that the characteristic polynomial has the same roots as the minimal polynomial?

just different multiplicities?

[chilly ocean]

why b_1 would be a unit

[thorn delta]

yes, and that's exactly what the last sentence says

[languid walrus]

if a and a1 are associates, a1 = ua for some unit u. then, a = a1 b = a u b

since a is nonzero, ub = 1

[chilly ocean]

Why a is a unit?

[languid walrus]

if the ideal generated by a equals the ideal generated by a1

then a1 belongs to the ideal generated by a

so it can be written as ua for some u

[chilly ocean]

@languid walrus
Why a=aub implies ub=1..........

[languid walrus]

a is nonzero, and D is an integral domain, right?

so we can cancel nonzero elts

[chilly ocean]

But there can be a nonzero element which is neither unit nor 0 divisor....@languid walrus

[next obsidian]

a - aub = 0

a(1 - ub) = 0

Either a = 0 or 1 = ub

[chilly ocean]

Tq

[barren sierra]

How does this property connect to like invarient factor form or something

so I get that (5) is a subset of Ann(M) but does that give anything?

[broken stirrup]

Can you elaborate please?

[next obsidian]

Just because a map splits

B -> A

Won’t imply A is projective

If B is a free module

Then it’s true

Also I misspoke oops, you want a free module

But a free algebra is a free module so I guess it doesn’t matter

[broken stirrup]

Did you mean P?

[next obsidian]

Uh, sure

Yeah

Anyway the point is, a direct summand of a free module is projective

[broken stirrup]

But we can take an identity function right?

[next obsidian]

So you just want to surject by some big enough free module

And then splitting tells you P is projective

[broken stirrup]

Oohhh i see

Damn

Conversely

If P is projective then I can take identity function so that i get gh=1 which means B is splitting

I messed up the order

Yeah thanks

[eternal furnace]

if you have a group G and a subgroup H, if you apply an element of G which is not in H to an element of H will it always take you out of the subgroup H?

[chilly ocean]

yeah because like if x in G\H and y in H, if xy=a was in H then x=ay^-1 would be in H

@eternal furnace

[broken stirrup]

is every vector space over a division ring D both projective and injective?

I know that it's also a free D-module. If D had identity, then it would also be a projective module, and every module is projective if and only if every module is injective

but we don't have identity necessarily, so there must be another way of showing it

[tribal moss]

Doesn't "free" automatically make it projective?

[tribal moss]

Ah sorry, hadn't understood your ring was non-unitary.

[broken stirrup]

yes

[tribal moss]

How can a division ring not have a unit?

I thought that was integral to even defining what "division ring" means.

[broken stirrup]

im sorry

im dumb

lol

yes of course its already a ring with identity in which every element are invetible

my bad

[tribal moss]

Even if we define "division ring" by saying that for every a and nonzero b, the equations a=bx and a=yb always have solutions, that automatically makes R\{0} a group under multiplication.

[broken stirrup]

yeah you are right i should probably get some sleep

lots of assignments

i have to meet the deadlines and i start to make stupid misteakes

[barren sierra]

Anyone have any ideas about this?

[lethal dune]

5M = 0 implies that M is a module over Z[i]/(5)

since 5 splits in Z[i] by 5 = (2+i)(2-i)

[barren sierra]

oh

and that gives me invarient factor form

[lethal dune]

and use CRT

[barren sierra]

right?

[lethal dune]

maybe

kinda

[barren sierra]

or elementary divisors actually

[lethal dune]

Z[i]/(5) is not PID tho

[barren sierra]

yea

[lethal dune]

not even an integral domain

you can't use primary decomp

[barren sierra]

oh hm you're right

ok I'll think about it from there

[lethal dune]

M is finitely generated meaning it's some quotient of a free module over Z[i]/(5)

[barren sierra]

yea

[lethal dune]

actually

ok I'll leave it here

otherwise I'll give away the solution

[barren sierra]

ye I think I'll try to work on this from here

thanks tho

[lethal dune]

use FTOFGMOPID

[barren sierra]

ok nvm I'm quite stuck still

I'm trying to find the elementary divisors

Structure Theorem isn't really helping me find what the divisors actually are

So I know M is finitely generated, say with n generators

so there is a homomorphism from R^n to the module

and then so is the kernel (5)????

not necessarily right?

[lethal dune]

ye not necessarily

[barren sierra]

Hm

ok yea then I'm struggling to use the structure theorem

[untold basin]

Let G be a group.
Show : Aut(G) is cyclic => G abelian

Can someone help me ? (indication pls)

My prof already told us that G abelian <=> G/Z(G) cyclic

[lethal dune]

hint Inn(G) normal in Aut(G)

(also a subgroup)

@untold basin

[untold basin]

ok thank you

[lethal dune]

ok if Aut(G) were of prime order my proof could hv worked

anyway

[untold basin]

I don't really know how does it help me

I wrote what does it mean, but when I want to show that G/Z(G) is cyclic I have to choose a x in G such that G/Z(G) = <x.Z(G)>

I don't see where is the link

I know that there is tau : G -> G an isomorphism such that Aut(G) = <tau>

[lethal cipher]

Question: is the only difference between a normal and separable extension that the min. poly of separable extensions have distinct roots?

[woven delta]

Take the map from G to Inn(G) by sending g to the map congugation by g

[next obsidian]

No

A normal extension means that if a polynomial has any root in that extension, all of its roots exist there

A separable extension has that the minimal polynomial of every element has no repeated roots

There’s no implication in either direction

[lethal cipher]

But don't separable extensions also need min poly's to split completely

[next obsidian]

No

Any extension in char 0 is separable

So just find a non-Galois extension of Q

That gives a counterexample

[lethal dune]

any subgroup of cyclic is cyclic

[next obsidian]

Yes

Or an easier example

Take k = Q(t)

And L = Q(t^1/3)

This is not a normal extension basically because Q contains no primitive third root of unity

But it’s separable because it’s char 0

[untold basin]

I realise that I use too much the definitions

I should use every properties I proved

[hidden haven]

I will come with time 😌