groups-rings-fields

[delicate orchid]

nvm I like this now

[white oxide]

omg i got wew's approval

[delicate orchid]

except, I don't think that's exact

[white oxide]

oh

[delicate orchid]

trying to think

A{i+1} -> A{i+2} -> 0 exact implies that f_{i+1} is surjective, which would mean that f_{i+2} has to be the zero map

but there's no reason f_{i+2} has to be the zero map

[white oxide]

oh right yea i guess it's for any exact sequence huh

hm

[wraith cargo]

Is the right homology supposed to be H_0?

The proof seems to imply it

[white oxide]

for the second part of the proof, they're implicitly referencing the canonical epimorphism from R to R/M right

[delicate orchid]

yes, that's a homomorphic image of R into a field

"into"? not sure what word to use there

[white oxide]

why not onto lol

[delicate orchid]

sure, that works

[solemn dew]

how to find a homomorphism from dihedral group = {e, x, y, ...}, where x^2 = e, y^{n-1} = e, xy=y^-1x to W = {-1,1}?

how should one think in this context ?

[chilly ocean]

just any homomorphism?

[solemn dew]

i'm supposed to show that G/N ~= W, where N = <y>

[delicate orchid]

y^(n-1) = e
odd

[solemn dew]

i guess any homomorphism

[delicate orchid]

you'll want a surjective one to use first iso

[formal ermine]

f(x) = 0

easy

[delicate orchid]

otherwise you'll get people who think they're funny

but honestly? it's easier just to construct the two cosets in this case

[solemn dew]

alright a surjective one lol

i've shown N is normal in G

[delicate orchid]

yeah that's immediate from it being index 2

[solemn dew]

really?

maybe my proof is wrong then

[delicate orchid]

so you know that G/N is a group, and that the size of G/N is 2

there's only one group of order 2

[delicate orchid]

I'm not going to read this and just assume it's correct

[solemn dew]

sure

[delicate orchid]

so any groups of order 2 have to be isomorphic

[solemn dew]

how did you draw the conclusion that the size is 2?

[delicate orchid]

D_n is of order 2n and <y> is clearly order n

do you not have the order of D_n yet?

[solemn dew]

haven't come that far in the book yet

[delicate orchid]

ok, that's fine

map y to 1 and x to -1

that should be a group homomorphism, as it's basically just the map G -> G/N but relabelled

but again, just writing down the cosets is easier here imo

[delicate orchid]

what about things that are in neither of those subgroups, like xy

[solemn dew]

im listening to the expert so i wont argue

[formal ermine]

LMAO

calling wew an expert

so true so true

[solemn dew]

true, this one does not hold that argument

[delicate orchid]

N = {1, y, y^2, ..., y^(n-1)}
xN = {x, xy, xy^2, .., xy^(n-1)}
is it clear that these are all of the elements of D_n?

[solemn dew]

yes sir

[delicate orchid]

sorry where's your first class masters degree?

[rocky cloak]

Replace g in <x> with "otherwise"

[solemn dew]

sorry i have to sleep now

i'll try to dream about this

and maybe it will solve itself in my dreams

hahah

goodnight ppl

[elder wave]

where is his high school degree

[delicate orchid]

lol

[stuck fiber]

Okay so this prop is kicking my ass, I'm seeing why everything works now except why the last element in G's tower is X

[night onyx]

Probably cause the preimage of the identity in G/X under the canonical projection G -> G/X is X

[stuck fiber]

See I thought that but I didn't see why G' must end in {e}

[charred crescent]

is it just a coincidence that, for the natural projection pi : G -> G/K, ker(pi) = K and for the map, phi : G/H -> G/K, ker(phi) = K/H ?

i was able to prove ker(phi) = K/H, but I am lacking intuition as to why this is true or how/if "modding" by H has this apparent congruency

is it as simple as ker(pi) is a subset of G, thus G/H results in ker/H ?

[topaz solar]

That sounds a lot like the uhh

Was it third? Isomorphism theorem

[charred crescent]

yes, i noticed this in proving the 3rd iso

i was able to prove it but again i lack intuition

[topaz solar]

Consider (G/K)/(H/K)

I think that’s the right order?

[charred crescent]

K/H

[topaz solar]

Oops im dumb

[night onyx]

I guess the intuition (very unrigorous lol) is that (G/H)/Ker(phi) is isomorphic to G/K via the first isomorphism theorem, and you can do a sort of "cancellation" by writing G/K = (G/H) / (K/H), which is the kernel of phi as you said

[charred crescent]

"cancellation" like fractions

[night onyx]

lol I think of it just as a cancellation result

yeah

[topaz solar]

I mean yeah I think of it like that too pretty much

Maybe a lil sillier structurally tho

[night onyx]

lol yeah of course

[charred crescent]

see but it sure looks like cancellation of fractions but they're groups so idk i guess i was wondering about the deeper level stuff "under the hood" if you will

[topaz solar]

Swapped the two oop, but (G/H)/(K/H) ~ G/K which is kinda exactly your desired map?

[topaz solar]

Well, you proved it so presumably you saw under the hood

[charred crescent]

yeah, to be able to use the 1st iso, i constructed a map phi : G/H -> G/K and showed the ker(phi) = K/H

[charred crescent]

well once i knew the trick from the second isomorphism theorem it was pretty obvious that the crucial step was to show K/H was the kernel of phi

to then use 1st iso of course

does my question even make sense? i think i need a sanity check

[topaz solar]

Well it’s certainly iso to the image since haha G/ker

So if it’s surjective then that’s a whole iso

[night onyx]

Try thinking of it in terms of dividing G up into pieces (which are cosets, or elements of G/K). If H \subgroup K, and you divide G into n copies of K, and then divide K into m copies of H, the result is the same as dividing G into mn copies of H. In terms of cosets you can say |G/H| = |G/K| * |K/H|, or rearranging you get |G/K| = |G/H| / |K/H|.

[coral shale]

i remember trying to prove the 3rd iso from 1st principles 100% algebraically. Fucking god.

[night onyx]

so there's an obvious analogue between cancellation of fractions, and cancellation when chopping G into copies of K and then K into copies of H, but it's super handwavy

but I think it's the correct intuition to keep in the back of your head so it's not just manipulating symbols lol

[charred crescent]

these are all very interesting points

thank you all for your input

[night onyx]

Another analogous idea is looking at the connection between the unique factorization of numbers into powers of primes, and the unique factorization of finite abelian groups into subgroups of prime powers. There are a lot of relationships between properties of numbers and properties of groups like that

[livid grove]

Does anyone know how to prove this actually using D&F's hint?

I know how to show it from the fact that G acts transitively on the set S={O_1,..,O_r} so |S|=|G:stab(O_1)| and stab(O_1)=HG_a

Since gO_1=O_1 iff there exists h s.t ga=ha iff there exists h s.t h^{-1}g\in G_a i.e iff g\in HG_a

[wraith cargo]

Are you asking for b)?

Ah yes you are

[livid grove]

Indeed

The eternal pain of studying for the algebra qual

[wraith cargo]

Did you show the first equality?

[livid grove]

Yes, O_1 is the orbit of a in H and so |O_1|=|H:H_a|. We have that H_a=H\cap G_a as H\subseteq G.

[wraith cargo]

Oh lol okay

They wanted you to write |O_1|=|HG_a:G_a| by the second iso theorem
Then r = |A|/|O_1|
But |A|=|G:G_a| by orbit stabilizer
So r = |G:G_a|/|HG_a:G_a| = |G:HG_a|

@livid grove

[livid grove]

Ahh yes that makes sense

Thanks

[stuck fiber]

So why does the cyclic tower in G' have to end in {e} here?

[night onyx]

induction on the size of the group, G' has a smaller cardinality than G because X isn't trivial

[stuck fiber]

I'm not really seeing how that implies G' ends in {e} though

[topaz solar]

Well, you decrease in size strictly

What else could it end in

It’s not going on forever since finite thing decreasing

[night onyx]

The proof is just skipping a few steps, but it's proving the result by induction on |G|. If |G| = 1 then the result is trivial, and then to prove the result for G when |G| > 1, they reduce to G/X which has a smaller size than G. So by induction there's a tower ending in {e} in G/X, then if you look at the premiages of those subgroups you get a tower in G ending in X (because the preimage of {e} in G/X under the projection G -> G/X is X), and then then you just add {e} onto the end of that tower

[stuck fiber]

So it's just some tower of G/X that ends in {e}? Because from the definition I have a tower is just a sequence of subgroups so I wasn't seeing why it couldn't end in the subgroup before {e}

[night onyx]

This is from Lang right? I think it's actually missing that "it suffices to prove that if G is finite and abelian there is a cyclic tower ending in {e}", otherwise the proof doesn't really work

I could be wrong lol, but I'm pretty sure that should be part of it

[stuck fiber]

Yeah this is Lang I don't think he mentioned that

I'll double check when I get home it might be like a little line somewhere that I've missed

[night onyx]

cause the tower in G' needs to end in {e}, otherwise the preimage of the tower under the canonical projection G -> G/X doesn't necessarily end in X

lol this is definitely a lang proof, very terse!

[stuck fiber]

Yeah I had worked out most of it I actually like that aspect of the book a lot because I'm usually not a very active reader

[void cosmos]

yo

i literally just realised

that the first iso theorem is just universal property of quoitients

lol

[topaz solar]

Makes sense, considering they’re quotients

[lyric furnace]

wait until you realize abaout the other 3

[untold turret]

if we consider nonzero unital rings and require ring homomorphisms to map to multiplicative identity, all kernels of homomorphisms are strict (i.e. proper) ideals, right?

[topaz solar]

Well, multiplicative identity isn’t in the kernel

So proper subset

Can you show it an ideal?

[untold turret]

yeah that part isn't a problem, i'm just trying to rule out the kernel being the entire ring

[topaz solar]

Well, f(1) = 1

So, is 1 in the kernel

[untold turret]

no, and we also have 1= 0 only in the zero ring

ty!

[small bramble]

aight stupid question time: if I is an ideal in F[x1,..., xn], why is tI an ideal in F[t, x1,..., xn]? My first instinct to show p(t, x1,..., xn)tq(x1,...,xn) is in tI by combining p and q but we only know I is an ideal of F[x1,..., xn] so that won't work. I don't have any other idea rip

Oh

Bruh

It's an error in the book

Just checked errata of dummy and feetpics and they removed the entire part claiming that it was an ideal as a fix

[topaz solar]

I will assume this isn't commutative?

[small bramble]

No F is actually a field

[topaz solar]

well, whst if I just like

t p q

by commutativity

[rocky cloak]

It's clearly not closed under multiplication by t

[topaz solar]

ah right there's no t^2

pea brain moment for me

[small bramble]

book even had the gall to say "Clearly" to make me question my existence

[frigid lark]

Lang?

Oh, a Dommy's feetpic, didn't read

[rotund aurora]

so (12) and (12...n) generate S_n

(ab) and (12...n) generate S_n if and only if gcd(b-a,n)=1 (see kconrad "generating sets")

so in general, the subgroup generated by (ab) and (12...n) will be a proper subgroup

what is it's size?

[rotund aurora]

looking at the proof of this, it's probably all permutations that respect the structure modulo d=gcd(b-a, b), since it is "trivial" (tho I didn't see it immediately) that the permutations you get will respect the structure modulo d (by this I mean that i=j mod d implies sigma(i)=sigma(j) mod d)

ok to count this is actually interesting

[delicate bloom]

I might have it

since we have (12...n) that cycles around, we can imagine a regular polygon with points on it and imagine that this rotates (ab) anywhere we want around the edges, which exchanges markers located at distance b-a around it

so we have d=gcd(b-a,b) partitioning them into d separate disjoint sets, each now looks like S_{n/d}

so it's a direct product of d copies of S_{n/d}, so it should have size (n/d)!^d

I think we can go through the orbit stabilizer theorem by thinking of these d disoint sets as the orbits and these S_{n//d} as the stabilizing group. Maybe it even helps to think of these as like drawing stars in your regular polygon

oh I think I might have swapped n/d with d above

like n=6 with (1,3) you have d=2, so you can imagine drawing two triangles where each one now looks like S_3

hopefully that's clear, by using (123456) we can "rotate" (13) anywhere we want around the hexagon to make (24), (35), (46), (15) and now we have the disoint triangles we could color red and blue which are fixed by {(13),(35),(15)} and {(24),(46),(26)} respectively

|S_3 x S_3| = 3!^2 = 36
as a special case of
|S_{n/d} x ... x S_{n/d}| = (n/d)!^d

as a sanity check is n!/(n/d)!^d an integer? I think that's a multinomial coefficient so it is

I suppose visualizing them this way also gives us more to count about them, we have phi(n)/2 subgroups of this form

or somethin like that lol w/e

well number of divisors, I'm thinking inside out here, well some stuff to say on both sides of that I guess

[solemn dew]

Any hints on how to prove this?

All i've come up with so far

[lusty marlin]

If G is a group and the quotient group G/Z(G) is cyclic, then G is abelian.

[solemn dew]

i'm only at chapter 2

i don't know these things yet

[prime sundial]

do you have lagrange's theorem?

[solemn dew]

yes

[prime sundial]

what are the possible orders for the subgroups

[solemn dew]

9,3,1

[prime sundial]

so what are the possible orders for the elements

if one of your elements has order 9 you're done, so just consider the case where you have only elements of order 3

[solemn dew]

{e, x^3,x^6} ?

since x^9 = e

from <x>

[prime sundial]

so let's suppose you have an element of order 9

what does that mean in this case

you have a group of order 9, and an element of order 9. how can you describe that element

[solemn dew]

don't know 😅

[prime sundial]

it's a generator, your group would be cyclic

[rotund aurora]

yes, but can't you premute the triangles too?

sorry for late response (very noisy out of my home so I had to stop doing math 😤 (actually angry lol))

[prime sundial]

are cyclic groups abelian? @solemn dew

[solemn dew]

i think so

[prime sundial]

they are. you should convince yourself of that, it's fairly straightforward

so then if we have an element of order 9, our group is abelian

so the only other case is for elements solely of order 3 (and 1 for the identity)

[delicate bloom]

they're independent of each other, you're not really permuting them as part of the group

[solemn dew]

but my subgroup that is generated is order 3, so how does this have anything to do with a group of order 9 is abelian

in my head you only showed that the subgroup of a cyclic group must be abelian

[prime sundial]

i think you're mixing up the concepts a little bit

[solemn dew]

i think so too

haha

[prime sundial]

so let's backtrack for one sec

the subgroups of a group of order 9 can be of:
order 1
order 3
order 9

if we have a subgroup of order 9

those are also the possible orders for the elements

do you agree

[solemn dew]

yes i agree

[prime sundial]

then let's knock out one of the cases. suppose we have an element g of order 9

like we said above, this element must generate G

[solemn dew]

yes

[prime sundial]

so G is cyclic -> abelian

so if we have an element of order 9

G is abelian

so that case is done

now we need to check what happens when we only have elements of order 3 (and the identity of order 1)

do you understand the argument so far

[solemn dew]

is it that if you have elements of order 3 in our group G, we need to find a argument that they are also abelian?

[prime sundial]

the argument now is that if all nonidentity elements are of order 3, then the group is abelian

yes

[solemn dew]

alright i got you so far then

[prime sundial]

ok

so let's do this in full generality

let's take two arbitrary elements in G

g,h

now our goal is to show gh = hg

[rotund aurora]

@delicate bloom btw d=gcd(b-a, n) I made a mistake while writing xD, you wrote the same. Im sure you know it should be that anyway

[solemn dew]

alright

so now the only possibility for G is G = {e, g, h, g^2, h^2, gh, gh^2, g^2h, g^2h^2} since ord(G) = 9

and g^3 = h^3 = e

[prime sundial]

yup

show then that gh has to be hg

[solemn dew]

alright i think i got it

thank you 😄

[delicate bloom]

yeah haha

[rotund aurora]

mmh

pretty sure you are missing a d!

well not pretty sure

but sage agrees with me

so

if we look at permutations sigma with the property that i=j mod d then sigma(i)=sigma(j) mod d, then the number of this permutatinos is d!·((n/d)!)^d

[rocky cloak]

Order would be d * ((n/d)!)^d right?

[rotund aurora]

and I think this is just generated by (1,d+1) and (1,2,3,..., n), which is what we want

[rocky cloak]

Or why d! ?

[rotund aurora]

sec

[rocky cloak]

I would think it's the semidirect product of C_d and (S_n/d)^d

[rotund aurora]

C_d?

as cyclic?

dude I dont find my calculator lmao

so

take d=4

and n=8

sage yields

[rocky cloak]

(1, 4) is a difference of 3

[rotund aurora]

ah yeah mb

yeah this didnt make sense

I should of taken 1, 5

64

so yeah it seems you are right

[rocky cloak]

I guess it's not quite a semidirect product, but (S_n/d)^d is a normal subgroup with quotient C_d

[rotund aurora]

so I think the point is, you can obtain the transposition (1,d+1) with d dividing n. Then, if we let X_1,...,X_d be defined by X_k={t in [n] : t=k mod d}, then the n-cycle (1,2,...,n) cyclically permutes the X_1,....,X_d, and each X_k is stabilized (as a set) under (1,d+1) (so that's the C_d of jagr2808). Then, inside each X_k, we can permute the elements arbitrarily. For this, you want to show that you can obtain the transposition (1,1+dk) for every k.

To show this last step, look at the set {1+dk} where k=0,..,n/d-1. Then composing the n-cyclie d times, takes 1+dk to 1+d(k+1), and (1,d+1) takes 1+d*0 to 1+d

but this situation is the same as (1,2), (1,2,..., r) where r=n/d (look at the ks)

which we know is the whole S_r

[delicate bloom]

ah yeah I missed a d good catch

[rotund aurora]

yeah my X_k's are your triangles

[delicate bloom]

cause the rotation itself, I left it out cause I was so focused on fixing that to focus on the internal stuff

cool

basically these pentagram type shapes made out of sticks, where you can rotate the entire picture or flip the sticks that make up the edges of the stars

something you could draw out and ask a kid to play with I guess

[rotund aurora]

btw

what would be a presentation of this group?

[vague granite]

Let M/K be a field extension, is it the case that End_K(M)=Aut_K(M)?

[vague granite]

So do you reckon this is a mistake?

penultimate paragraph

[uneven jackal]

oh wait a sec

Aut_K(M) is field automorphisms

[vague granite]

yes

[uneven jackal]

not M as a K vector space

[vague granite]

Nonono

End(C/R) = {id, conjugation} = Aut(C/R)

[formal ermine]

Yes

in general injective endomorphisms of artinian modules are surjective

[vague granite]

artinian module?

[formal ermine]

oh wait my bad

it's not a yes

[uneven jackal]

wouldn't C((x)) -> C((x)) mapping x to x² be a counter example ?

[formal ermine]

your vector space needs to be finite dimensional for this to apply

[uneven jackal]

in the infinite dimensional case

[vague granite]

no, because it doesn't preserve R

[formal ermine]

dcc

[uneven jackal]

what ?

it does preserve C

C((x))/C

[vague granite]

It's not an R-endomorphism

[uneven jackal]

?

[formal ermine]

pearl is taking the base field to be C

[uneven jackal]

I think you're misunderstanding my mapping

[vague granite]

oh right

[uneven jackal]

it maps the variable x to x², but it sends a constant to itself

[formal ermine]

but yeah in the fin dim case it's (trivially) true

[vague granite]

So

the lectures are wrong

Do people study endomorphisms of field extensions?

As an extension of solely studying automorphisms

[formal ermine]

ig it could be easier to view them as tensors but I have no idea lmao

[rocky cloak]

The "hint" also says "use rank nullity", which is a theorem about finite dimensional spaces. So it seems they assumed it to be finite dimensional at some point, but then maybe rewrote the notes...

[south patrol]

Is this Damien Rössler by any chance

But actually this is true for algebraic M/K

Just for different reasons

[rocky cloak]

I guess people are most concerned with algebraic extensions, in which case every endomorphism is an automorphism.

[south patrol]

But yeah I'm p sure I read these same notes and thought the same thing when I read it lol

These notes are a bit rough iirc and not really the best to learn from lol especially since the course has changed since

[formal ermine]

ong wait I mixed up finite dimensional and finitely generated lmaoo

I hate these conditions

a module over an artinian ring is artinian iff it's finitely generated

[south patrol]

Well depends what you mean by finitely generated lol

[formal ermine]

I think that's the correct one

[delicate orchid]

generates finitely nothing personell kidd

[formal ermine]

french

[south patrol]

maybe I am missing context lol

[formal ermine]

#groups-rings-fields message

[south patrol]

oh is this true nice

Well it must be just dual in some sense to surjective endo of a noetherian module being injective

Nice

Oh yeah sure

[delicate orchid]

it's a nice result

cause otherwise you have an infinite chain of dudes

[south patrol]

Sure yeah I guess it's like

if φ is ur endo then you consider either the ascending chain (ker φ^i) or the descending (im φ^i)

[delicate orchid]

it's to do with the kernel of successive applications of the map isn't it

[south patrol]

and conclude those are zero in the Noetherian resp. Artinian case

[delicate orchid]

I am very glad I came up with this on the spot

[south patrol]

This came up in my exam lol

Though surjective endo of a Noetherian module has an easier proof, since it's true in the f.g. case I'm p sure

[delicate orchid]

well that makes sense, noetherian is artinian but for babies

[south patrol]

lol isn't it the other way round

[delicate orchid]

I have artinian as the stronger one in my head even though they don't imply each other at all (for modules)

[untold turret]

for any ring homomorphism f:A -> B, do we have if f(a) = 0, a is not a unit?

[delicate orchid]

take B to be the trivial ring and f to be the zero map

[formal ermine]

no

[delicate orchid]

then consider A a field

[formal ermine]

fucking beat my wife to it

[vague granite]

It is

[south patrol]

nice

[untold turret]

hm.. add the following condition: unital rings, unity preserving homomorphism

[delicate orchid]

you need to stop

[formal ermine]

YOU need to STOP

[vague granite]

Oxon?

[delicate orchid]

my example still works lol

[south patrol]

oh lol i mean considering for rings artinian => noetherian

oh stronger

sure nvm

[delicate orchid]

yus, that'll be why

[south patrol]

Yes

[vague granite]

Sure?

[wraith cargo]

They do???
Artinian implies noetherian

[south patrol]

for rings

not for modules in general

[delicate orchid]

there are artinian modules that are not noetherian

[wraith cargo]

I think it works for modules of finite length or something right

[untold turret]

ring homomorphisms redux electric boogaloo 2: unity preserving ring homomorphism f:A->B, B nonzero. f(a) = 0 implies 'a' nonunit?

[south patrol]

finite length implies artinian and noetherian

so uh

[delicate orchid]

I buy it

[south patrol]

lol

[wraith cargo]

I know Eisenbud has this sort of equivalence

[delicate orchid]

a \in ker(f) which is an ideal of A, so if f is not the zero map ker(f) cannot contain any units otherwise it would be all of A

think this works?

assuming f is surjective

because why wouldn't it be

[untold turret]

that's what i'm thinking, but f need not be surjective

[elder wave]

module has finite length iff it is noetherian and artinian innit

[delicate orchid]

corestrict to img(f) then do the same thing

[untold turret]

i am blind to any words starting with co

[elder wave]

so the module already is both

[wraith cargo]

Ye
I was thinking of rings lol

[delicate orchid]

basically all this means is that, yes your statement is correct unless f is the zero map @untold turret

[untold turret]

thanks!

for an arbitrary nonzero ring, there's always a nontrivial homomorphism to an algebraically closed field. any other class of ring hom. codomains have this property?

[delicate orchid]

regular fields too I think

wait are these unital

[untold turret]

yes, and commutative

[delicate orchid]

ok, this could be true for non-unital too but

[formal ermine]

take embedding into fraction field

[delicate orchid]

by zorns every ring R has a maximal ideal M, so R/M is a field

hence there is a map into there

[delicate orchid]

this is good alsobeit

[untold turret]

hm right, R/M field is the fact we use to embed into algebraically closed field

ty for answers

[untold turret]

could this depend on being integral domain?

[delicate orchid]

even for non-integral domains you can just consider localising like (R-{0})^-1R

[formal ermine]

R^-1R

so true

so true

[delicate orchid]

quipcel

[untold turret]

hm i see, and then we also get embedding into algebraic closure from this. ty!

[south patrol]

i mean R - 0 isn't multiplicatively closed if R has zero divisors

so i'm not sure what you mean

[delicate orchid]

you know what I mean, search within yourself

you will find the truth

[south patrol]

well i don't

[mortal mortar]

https://math.stackexchange.com/questions/2550381/localization-with-respect-to-a-multiplicative-set-containing-zero-divisors

[south patrol]

I'm not really sure what posting this answers

[delicate orchid]

yeah me too lol

[mortal mortar]

Thats why i posted it

In agreement

[south patrol]

Oh okay sure

[delicate orchid]

just take the field of fractions you stinkers

[south patrol]

Of a non-integral domain?

[delicate orchid]

yeppers

[south patrol]

You can't do that is what I'm saying

At least not conventionally

[delicate orchid]

yes, you can ?

[south patrol]

How are you defining it

[delicate orchid]

you have to mod out by an extra relator where instead of just like x/y ~ a/b iff xb-ya = 0 you have to multiply it by a unit I believe

I'll check my notes

[south patrol]

That's the normal definition of localisation wrt a multiplicatively closed subset, sure

(a,b) ~ (c,d) iff there's s in S such that s(ad - bc) = 0

[mortal mortar]

yes

[south patrol]

But then what subset are you taking as your S

Cause you will end up producing the zero ring according to what you're doing (if you're doing what I think you're doing)

[formal ermine]

why?

I give you a (a, b)

how do you show that's ~ (0, 0)

[south patrol]

Well what is your S

[formal ermine]

oh wait

[south patrol]

The idea hasn't been given

[delicate orchid]

no potato is right, if you just do it a priori you get 0

[formal ermine]

I see

[south patrol]

If you try to localise a zero divisor you wind up with 0

[mortal mortar]

https://math.stackexchange.com/questions/4044579/localization-with-respect-to-set-that-is-not-mutliplicatively-closed

[delicate orchid]

I can't find the actual construction I was thinking of so I'll take the L on this one

[south patrol]

because like 1 = (a/a)(b/b) = (ab)/ab = 0/ab = 0

if ab = 0 for a,b which are inverted

[delicate orchid]

yur

[mortal mortar]

as you say, it can not be made for non-multiplicatevly closed but one can extend to the smallest multiplicatve closure of the set

[delicate orchid]

so Illuminator's embedding is bad but mine is good :iwon:

[formal ermine]

who is Illuminator

[untold turret]

do we have: two rings map to the same algebraically closed field if one of them is a subring of the other?

[rocky cloak]

Yes, if f:R -> K is a map from a ring to a field and S is a subring of S, then you can simply restrict f to S.

The converse is not true though: both Z/2 and Z/4 map to F2bar, but neither are subrings of each other.

[untold turret]

i see, thank you!

[formal ermine]

S is a subring of S
impossible

[rocky cloak]

*improper

[pure mulch]

ok so this is my question

[delicate orchid]

yus

[pure mulch]

what would firstly the commutator subgroup be like, i have a hard time imaginging a group generated by multiple things

[delicate orchid]

well lets start by picturing a group that's generated by multiple things then

The group of symmetries of a triangle, S_3, is generated by two elements

rotating by 2pi/3 radians, and a reflection (or (123), (12) if you want to think of them as permutations)

[pure mulch]

[delicate orchid]

lets call that rotation r and that reflection s

so all elements in S_3 look like some "word" with "letters" r and s (and r^-1, s^-1 if you want to be technical)

like rrssrr or rsrsrr, or r^-1s^-1sr

[pure mulch]

hmm so they look like r^2s or srs or smth like that?

okie dokie im following

[delicate orchid]

yes

that's it

[pure mulch]

so a group generated like this just has all combinations of all the elements r and s and their inverses?

[delicate orchid]

obviously, you can simplify most of these words down

[delicate orchid]

yes

[pure mulch]

[delicate orchid]

it's like a basis but not linearly independant (because linear doesn't really make sense here)

[pure mulch]

icic

yeah gotcha

[delicate orchid]

ok so now we take our "r" and "s" to be things that look like xyx^-1y^-1

so we consider the group of "words" that look like (x_1y_1x_1^-1y_1^-1)^2(x_2y_2x_2^-1y_2^-1)^-1 or whatever

[pure mulch]

my first thought was that that would turn the comm subgroup into the trivial group

[delicate orchid]

it is trivial you're correct

[pure mulch]

Oh

[delicate orchid]

but why

[white oxide]

i get the part about f(x) dividing g(x), but how do we know that g(x) has all zeros of multiplicity one? because K is a finite extension of E, but that doesn't mean that E has to be a finite extension of F right (which would imply that K doesn't have to be finite). because to use the perfectness of F we're assuming that K is a finite extension

or maybe it's obvious or i'm trippin idk

[pure mulch]

becuase since each element is of the form xyx^-1y^-1, and all combinations of each of these
then since G is abelian, xyx^-1y^-1=xx^-1yy^-1

=1_G

[delicate orchid]

yup

[pure mulch]

so all elements would be of the form the identity iterated multiple times

[delicate orchid]

and <1_G> is the trivial group

[pure mulch]

yeah

Noice, for a second i was confused, ty wew!

[delicate orchid]

np np

[hidden haven]

You can reduce to the case where E is a finite extension of F

Standard trick: you look at one element of K at a time to prove that that element is separable. You then consider the extension generated by this element over F(all coefficients of the minimal polynomial of this element) instead of E, essentially throwing away the irrelevant parts of E.

[pure mulch]

another question, does the commutator group always happen to be a subgroup of the original group G?

[delicate orchid]

yes, it's always a subgroup

[hidden haven]

Actually can you send your definition of perfect @white oxide?

[white oxide]

wait why can we do this? algebraic doesn't imply finite right

[delicate orchid]

G is closed under multiplication, and all of the elements in [G,G] are products of stuff in G, so it must be contained in G

[white oxide]

every finite extension is a separable extension

[hidden haven]

ok I see

[white oxide]

yea that prolly changes things

[pure mulch]

[hidden haven]

No, the stuff I said under that lol. You can instead look at only one element of K and whatever elements of E are needed to make this element algebraic.

[white oxide]

oh so i could just consider E($\alpha$) is what ur saying

[hidden haven]

I am assuming a lemma: An extension E/F is separable iff every element of E is separable over F

[white oxide]

yea

[cloud walrus]

okeyokay

[hidden haven]

Yes but you will still have the issue that E/F may not be finite

[white oxide]

ah right

weird proposal: i could consider irr(alpha, E) and then the coefficients of irr(alpha, E), and try to deduce something from the fact that they're algebraic over F

[hidden haven]

Yep exactly

[white oxide]

oh wait wtf

[hidden haven]

You replace E by the subfield generated by the coefficients of that polynomial

That is a finite extension of F so it allows you to use the perfectness of F

And the minimal polynomial of alpha over this subfield is the same as the minimal polynomial over E, so alpha is separable over one if and only if it is separable over the other

[white oxide]

okay i won't read the last part u just sent only as a last resort lol (to try to solve it with minimal hints)

but thanks! will try

[hidden haven]

[white oxide]

haha okay got it

that was fun

[white oxide]

help i need showing that if K is a possibly infinite separable extension of E and if E is a possibly infinite separable extension of F then K is a separable extension of F

i let alpha in K and considered E(a) which is a separable extension of E, my idea was that if i could show that E(a) is a separable extension of F then F(a) would be separable over F by definition

this was how fraleigh defined a separable extension in the broader case

ok i split it up into cases; if E is a finite separable extension of F then it's pretty easy, now i'm assuming it's an infinite extension of F

oh wait nvm i got it

[untold turret]

given a nonzero unital commutative ring A with two nontrivial homomorphisms f,g to nonzero commutative unital rings B,C, is it possible to have f(a) = 0 and g(a) \neq 0?

[formal ermine]

non unital rings aren't real, they can't hurt you

[next obsidian]

Wait you said f(a)

Yeah you can totally have this happen

[topaz solar]

I mean, as long as they’re not just like

The Z->A or Z->B

[formal ermine]

mfw non injective morphisms exist

[next obsidian]

Find a ring with any endomorphism with a kernel

Compare id and that endomorphism

[topaz solar]

Then yeah you’ll usually have that

[next obsidian]

You have your example

[untold turret]

Z -> Z/(2) Z -> Z/(3), take f(2), g(2)

[next obsidian]

Take Z/4Z and multiply by 2

[topaz solar]

Mfw Z x Z

[next obsidian]

Oh wow you’re even letting the codomain vary

Then you can do this for any nonunital a

Consider A -> A id

And then A -> A/(a)

The projection

[untold turret]

yeah this is a sad day for this hypothesis

ty for help

[next obsidian]

I don’t know why you’d expect this to be true to be honest 😭

This is wayyyyy too strong

[topaz solar]

“Yeah bro there’s only one function”

[untold turret]

its too powerful

[next obsidian]

Too much power for one man

[formal ermine]

if this thing is true then erh is true ong

I showed this implication

[ivory trail]

https://xkcd.com/1310/

xkcd: Goldbach Conjectures

The weak twin primes conjecture states that there are infinitely many pairs of primes. The strong twin primes conjecture states that every prime p has a twin prime (p+2), although (p+2) may not look prime at first. The tautological prime conjecture states that the tautological prime conjecture is true.

[untold turret]

i'm trying to explore the extent to which f(a) = 0 implies a nonunit

[formal ermine]

what 💀

[ivory trail]

if f is a map of unital rings, then it preserves units

[next obsidian]

It implies nonunit always, unless the codomain is 0

[untold turret]

important, ty

[next obsidian]

A unit is sent to a unit

[ivory trail]

it's a short calculation: 1 = f(1) = f(a a^-1) = f(a) f(a^-1)

[hidden haven]

Based

[south patrol]

Assuming consistency of ZFC

[hidden haven]

You can prove it with Falso

[ivory trail]

i can also prove falso with it

[topaz solar]

axioms: 9

🤓 schemas

[hidden haven]

Second order ZFC

Tbf it is the number to memorize not the number itself 🤓

Actually I'm not sure if even second order ZFC would be finitely axiomatized

Probably not what relations could you even add to the language

[median pawn]

alright basic question about presentations

[hidden haven]

Haha that's funny because presentations involve bases

[median pawn]

want to show <S|R> with S = {x} and R = {x^n} is isomorphic to Z/nZ using the universal property of generators and relations. I define f: S to Z/nZ by f(x) = 1, so that f(x^n) = 1 + ... + 1 (n times) = 0. by the universal property, there is a unique map g from <S|R> to Z/nZ such that g composed with the canonical map (S to <S|R>) is f.

to show the isomorphism, i should define h from Z/nZ to <S|R> (= F(S) quotiented by the normal subgroup generated by R in F(S)) also. h(1) = x is the obvious choice

[delicate orchid]

swubsituwion twest

[median pawn]

h(1) = x really means the equivalence class of x in the quotient

everything seems right so far? just g o h = id and h o g = id should do it

[delicate orchid]

is it part of the question that you have to use the universal property

[median pawn]

yes

[delicate orchid]

cvringe!!!!!!!

[median pawn]

that's the whole point lol

[delicate orchid]

that map should work though

[hidden haven]

How do you solve a problem without using universal properties when the terms in the problem are defined using universal properties

[median pawn]

just trying out the first approach lol

[delicate orchid]

or via the explicit construction
ahhhhh there's the based...

[median pawn]

oops didn't mean to reply

[hidden haven]

Why are you

[delicate orchid]

so cool? dunno bro

[hidden haven]

[untold turret]

K field, R some subring, t not in R, R[t] has t in a maximal ideal M. why is there a valuation ring V with maximal ideal M' that contains R[t] and has M \subset M'

[formal ermine]

did you not use K???

[untold turret]

i did but secretly

valuation rings are subrings of K, t is in K but not in R

[white oxide]

can i get a hint for b pls

i suck at image subset problems lol

i'm trying to use the fact that it's a direct sum but struggling f

Okay; say $b \in f(B)$ so that $b = f(x + y)$ for $x \in B, y \in C$ since $A = B \oplus C$

[cloud walrus]

okeyokay

[white oxide]

Then $b = f(x) + f(y)$

[cloud walrus]

okeyokay

[white oxide]

Kinda blanked out from there

f(C) \subset C1 is clear enough

[rocky cloak]

I don't think this is true. Let K be k(x, y) and let R be k[y + 1/x, 1/y + 1/x, x], then I'm fairly certain x is not a unit in R. But if you adjoin either y or 1/y it makes x a unit.

[median pawn]

ofc it worked but imo not the best way to show isomorphisms between a presentation and a given group? because F(S) may not always be "simple" and dealing with cosets can be painful

this particular example was just too easy

[rocky cloak]

Guess a hint is, what happens if you multiply A by 1

[median pawn]

also uniqueness of g didn't play a role

[white oxide]

oh oops right you get 0

thanks

at least i think that's right

nvm

that was C

i'm trippin

[rocky cloak]

Say you have a = b+c

And you multiply by 1, what happens?

[white oxide]

well it just becomes a = b right

ohh

well i know that all elements of C are mapped to 0

does it have to do with the fact that f(b + c) = f(b) + f(c) and since that's in A1 = B1 \oplus C1 we can conclude that f(b) is in B1?

[rocky cloak]

So imagine f(b) = b' + c'

And remember that f is a homomorphism

What you want to show is that c' is 0.

[white oxide]

ye i know that multipliying by 1 in C makes c go to zero but i'm not sure what properties C1 has

[rocky cloak]

C1 has that same property by assumption right

[white oxide]

i thought it didn't

because then it wouldn't be so bad

oh wait

i guess rc1 = 0

oops

and 1R obviously in R

ah there we go

got it thanks!!

[rocky cloak]

Guess this exercise shows that studying nonunital modules over unital rings is just the same as unital modules except you just keep an extra abelian group around.

[white oxide]

oh yea that makes sense now

[untold turret]

cool counterexample! thank you

[charred crescent]

can we say, for a group G, and N < A < G, with N normal in G, that (A/N)(A'/N) = (AA')/N

[rocky cloak]

Did you read somewhere that this should be true, or did you come up with it yourself?

[untold turret]

I'm trying to distinguish the key components of the proof that the intersection of the valuation rings containing a ring A is the integral closure of A, and I'm testing out to what extent we can extend valuation rings containing A

[rocky cloak]

Maybe you're able to write that what the elements are for each, and see if they match?

[charred crescent]

well the thing is, i have an idea for what an element of (A/N)(A'/N) looks like but im not sure

i think it should look like aNa'N for some a,a' in A, which is (aa')N iff N is normal and N < A

if that's the case, then im inclined to say that the equality does hold

but if im misunderstanding the definition of (A/N)(A'/N), then im not so sure @rocky cloak

[rocky cloak]

No I think you got it.

[white oxide]

how does this exercise show that f(P) is a subset of B1? i can't think of any direct sum that we can include P in to align with the statements of (b) in exercise 17

oops wait

referencing this

idk could we just form $P \oplus A$?

[cloud walrus]

okeyokay

[white oxide]

in exercise 17, i'm letting g: A --> B be the given module homomorphism in (b). but all exercise 17 says is that f(A1) \subset B1 and f(A2) \subset B2, i don't understand at all where P comes in

does it have to do with the factorization?

[warm ember]

im so dumb bruh

i got that $f(h):=g'^{-1}hg$ is a bijection

[cloud walrus]

kevinyang2.71

[warm ember]

where i assumed Hg=g'H for g \neq g'

can somoen help pls

[rocky cloak]

What is an element you can be sure to find in Hg?

[warm ember]

g

o

ty

[untold turret]

btw how did you construct this example? just finished verifying it and it wasn't obvious heheh

[rocky cloak]

I just came up with it in the spot. But I'm having some doubts, as it seems to contradict Atiya Macdonald

Or maybe I'm misunderstanding something

Or maybe x actually is a unit

[untold turret]

it does have that adjoining either y or y^{-1} makes x a unit

will keep working with your example to see what exactly makes it work

[untold turret]

the theorem in Atiyah-Macdonald requires that whatever t we adjoin to R to make R[t] must be the inverse of a nonintegral element

so i don't think it contradicts it

second part of corollary 5.22

oh btw if you're reading 5.22 be sure to check the errata, forgot to say

[white oxide]

anyone?

[rocky cloak]

P = P + 0

[white oxide]

ah right oops thanks

[charred crescent]

thanks 🙂

[untold turret]

hmmm but i'm actually not seeing the part integral dependence plays in 5.22

yeah, in fact, it seems 5.22 only uses the fact that t is in a maximal ideal of R[t]. @rocky cloak i'm trying to figure out if x is a unit in your example

[rocky cloak]

It seems

(y + 1/x) + (1/y + 1/x) + x - x (y + 1/x)(1/y + 1/x) = 1/x

So x is a unit and all is right with the world

[untold turret]

oof, close call. almost discovered inconsistency of ZFC

[rocky cloak]

I will go to bed now

[untold turret]

cheers, thanks for the help

i wonder if there's some sort of commutative algebra software we could use for this

[formal ermine]

idk what exactly you're trying to do right now but this looks like the software whose name I forgot could do this

sage

that was the name

[south patrol]

Sadge

[formal ermine]

[warm ember]

can help pls

so my progress is there is a primitive root gZ(G)

and let k=|G/Z(G)|

so g^k h=hg^k for all h in G

but idk how to use the primitive root

btw Z(G) is the set of elements commutative with every element in G

[wraith cargo]

Why does the quotient with the center have to be finite?

[chilly ocean]

G = (G/Z(G))Z(G) and each is abelian so G is abelian qed

[warm ember]

o

[warm ember]

what

what is that

multiplication

[chilly ocean]

it's bullshit

[warm ember]

what

[empty rose]

wouldn't this imply that all finitely generated groups are abelian

[warm ember]

i dont understand

[chilly ocean]

this got out of hand

[warm ember]

(G/Z(G)) x Z(G)?

[wraith cargo]

It's called a "joke"

[empty rose]

this was intended as a joke

[warm ember]

o

[chilly ocean]

i'll put the /s next time

[warm ember]

o

[formal ermine]

do you intuitively understand why it's true?

[warm ember]

how to do

[warm ember]

uh

quotient groups arent intuitive

unlike rings

[delicate orchid]

Tbh, I don’t understand it intuitively either

Anyway, so you have a generator for G/Z(G), which you’ve named gZ(G)

[frigid lark]

so let G/Z(G) = <xZ(g)>, then any element in G, g, can be written as g = x^ih for some h in Z(g)

[delicate orchid]

The cosets of Z(G) partition G, so we can write any element of G as x^i*h as parrot says

[warm ember]

oh

wait what

why x^ih

[delicate orchid]

(And what terra was joking about by writing G = (G/Z(G))Z(G))

[frigid lark]

well {x^i} is the set of enough? coset representatives of G/Z(G)

[delicate orchid]

Any coset of Z(G) in G is of the form x^iZ(G) by our assumption that the quotient group is cyclic

[formal ermine]

this might be really wrong but I like to think of it as like, if we look at the non commuting elements they're all x^n so they intrinsically commute with each other again, and so we get abelianess

[warm ember]

ye

[delicate orchid]

These cosets partition G, so every element is in one of these x^iZ(G)

[warm ember]

yes

o the h is outside the exponent

[delicate orchid]

Which is equivalent to saying it’s equal to x^i*h for some h in Z(G)

[warm ember]

ok

ok ty for hints

[delicate orchid]

So now use this form to show G must be abelian

[delicate orchid]

Right yeah, the “non-central” components commute

Reminds me of uhh p-parts and p’-parts of elements in a weird way

[formal ermine]

wait when is G/Z(G) not just empty when G is abelian lol

[delicate orchid]

Yes

Not empty but trivial

[formal ermine]

same thing

[delicate orchid]

Almost like Z(G) cyclic => G = Z(G) implies G is cyclic or something

[formal ermine]

I need sleep

[frigid lark]

is it 2 am again?

[formal ermine]

2:38

how did you guess lol

[frigid lark]

you're on discord

[formal ermine]

are you b'i'*sh? 🤢

[formal ermine]

so true

have you found a gf yet?

[delicate orchid]

Ur GMT+1 don’t drag me into this

[frigid lark]

no, a former colony

[formal ermine]

it's been like what, 7-8 months?

[frigid lark]

That isn't known for being nice or shooting schools

[formal ermine]

I just stepped on the fucking head of my fan

[frigid lark]

Or being big

[delicate orchid]

Ah… the glorious nation of Fiji…

[formal ermine]

it's almost as bad as stepping on a lego

[frigid lark]

fuck

[formal ermine]

south africa?

[frigid lark]

England had too many colonies

[formal ermine]

I'll be in the us starting friday

[frigid lark]

populationwise, not landmass wise

[formal ermine]

australia?

[frigid lark]

ye

[formal ermine]

oi m8 bloody cunt

[delicate orchid]

The glorious territory of Tokelau

[formal ermine]

where's @wraith cargo

asleep????

[frigid lark]

can we not wake up more people in the middle of the night?

[formal ermine]

it's 10 am for you

[frigid lark]

yes

but around vampireland, it's not 10 am

[delicate orchid]

Southern Hemisphere… so cringe

[formal ermine]

I email'd my prof a couple of mins ago

hope he responds soon 🤞

[frigid lark]

and lemme guess, they responded

[formal ermine]

I had my oral exam today

I forgot the MOST EASY THINGS

and couldn't think at all

I wanted to say that the coordinate ring describes Mor(X, C)

but just couldn't think

[frigid lark]

today?!

[formal ermine]

I was too nervous

[frigid lark]

at 1am?

[formal ermine]

so I emailed him this as a fwi

[formal ermine]

16 hours ago

[frigid lark]

idk, in Australia, we call that yesterday

[formal ermine]

today is when I haven't gone to bed yet

[delicate orchid]

And you also say December is in summer so excuse me if I ignore you

[formal ermine]

yesterday is what was before I gone to bed

[formal ermine]

SO TRUE

guys do I flip a coin whether I go to sleep

[delicate orchid]

Go to sleep

[frigid lark]

so what happens when you pull an allnighter

[formal ermine]

it's one big day

[delicate orchid]

Skips a day

[frigid lark]

yeah, heads sleep, tails sleep

[formal ermine]

it's addicting

I feel young again

[frigid lark]

that thing that happened in ngnl, also sleep

[glossy grotto]

Suppose you have a group isomorphism between two groups, are isomorphisms basically permutations on the generators of these groups?

[lethal dune]

There’s no canonical generator, not even cardinality is unique for a generating set so no in general

Quick example: Z with generating sets {1} and {2,3}

[glossy grotto]

I see, does this mean that even isomorphisms do not map generators to generators on presentations?

[topaz solar]

Well, if S generates A, then f(S) is gonna generate B

[delicate orchid]

The problem is the presentation of a group is not unique

[topaz solar]

But that’s about it

[topaz solar]

Using this can get you two different presentations for Z

My favorite presentation is the unholy one where we use <G> = G

[hidden haven]

Nothing unholy about it

It's the only natural one

[topaz solar]

Certainly not minimal in any sense, that’s for sure

[glossy grotto]

Ah there's that.

[topaz solar]

But it’s probably natural in a categorical sense

[hidden haven]

Maximal nondegenerate that's pretty special

[topaz solar]

It’s degenerate in another sense though

[hidden haven]

It's a comonad

[topaz solar]

But, of course, generating sets -> generating sets

And isomorphisms in particular have inverses

So you get that at least

[glossy grotto]

We have this same problem of non uniqueness of the generating sets for different algebraic structures no? In particular for algebras, we should be able to show that any algebra is some quotient (encapsulating relations) of the free algebra on some generators. The problem is that these generator-relations is not unique, as you have said.

[median pawn]

What's this notation? What's the square bracket doing on the right?

The relations seem to be a bit weirdly written

[topaz solar]

after all, feely adjoin an extra generator and quash it to 0

[delicate orchid]

the commutator

and yeah the double | is weird

I'd write this as $\langle s, t | [t^nst^{-n}, t^mst^{-m}] \text{ for all } n, m \in \bZ \rangle$

[cloud walrus]

wew ladz

[median pawn]

ohhh okay! alright

apparently this group is interesting because it has no finite presentation

[delicate orchid]

I believe it

[teal vessel]

⟨s,t | ∀m,n ∈ ℤ ([tⁿst⁻ⁿ,tᵐst⁻ᵐ])⟩

Just to prove I can with this keyboard

[untold turret]

if V is a valuation ring of field K and t is in K but not in V, is V[t] = K?

[next obsidian]

No

Wait maybe?

[formal ermine]

chmonkey is indeed correct, the answer is either yes or no

[solemn dew]

Is the normal subgroup the image?

Or what is that symbol+++

[wraith cargo]

Inner automorphisms

[solemn dew]

A(G) is just S_{n] then?

[chilly ocean]

it's a \mathscr{I} i think

[wraith cargo]

A(G) is the group of automorphisms of the group G

[solemn dew]

oh oke

thanks¨'

[untold turret]

t is not integral over V since V is integrally closed

we do know valuation rings are maximal wrt. extensions of homomorphisms mapping to algebraically closed fields

[next obsidian]

No this can’t be true

Yeah this is not true

It’s true if V is a DVR

But look at eg

Thm 10.1 of Matsumura

So this doesn’t quite exactly have the statement written down immediately

But there’s a bijection between primes of a valuation ring and valuation superrings

You can look at this in terms of convex subgroups of the value group too, this is all written down in Bourbaki

But the point is, you can have valuation rings which are not maximal subrings of K

So in that case, just pick an element t living in a larger subring of R of K, and look at V[t]

You have V < V[t] <= R < K

In fact you can even use this exercise to see that most valuation rings are not maximal

10.5

[untold turret]

excellent argument 10/10

yeah makes sense, thanks for the answer

[untold turret]

is the fact that a homomorphism f:A->B from a ring A to an algebraically closed field B can be extended to a ring f:C->B a strong statement about the structures of A,C?

[solemn dew]

does anyone know what this lemma is called? I don't understand it

don't know what " i(H)! " means

[coral spindle]

Maybe index of H, factorial?

[solemn dew]

you are right

didn't think of it

[coral spindle]

Really weird notation overall.

[solemn dew]

🤷‍♂️

[coral spindle]

I don't know if this has a name.

[solemn dew]

lemma 2.9.1 in topics in algebra

[rocky cloak]

So a homomorphism from A to a field is basically the same as a prime ideal in A, by looking at the kernel.

So this extending to C would mean to find a prime in C lying over the prime in A. In other words we need the extension A -> C to satisfy the "lying over property".

It's a theorem that any integral extension satisfies the lying over property (the going up theorem).

[untold turret]

oooooooooh! that makes a LOT of sense!

I was missing the explicit connection between homomorphisms and prime ideals

so extensions of homomorphism have a very intimate relationship to the structure of prime ideals lying over each other

I guess we're interested in homomorphisms to specifically algebraically closed fields because these are so easy to find?

there's also my question that you've already answered before: if a ring A contains a ring B, we can easily find an algebraically closed field to which both map to

[rocky cloak]

I think it's more a nice "maximal" situation to be in.

Like if p is a prime in A and q in C lies over it, then the fraction field of C/q is a field extension of the faction field of A/p. And if A -> C is integral then the extension is algebraic. So if A/p already sits inside something algebraically closed you never need to switch up your codomain.

[untold turret]

that makes sense, thanks a lot!

[solemn dew]

i've shown that the subgroup must be cyclic

how to continue from here?

how will every element in this group commute?

[delicate orchid]

you're right in that G has to be abelian

do you know about the fact that if the quotient G/Z(G) is cyclic then G is abelian?

because I think the easiest route here is to just show that G has to be abelian and then conclude the result

[rocky cloak]

This is probably intended to be a step in proving that every group of order p^2 is abelian.

[delicate orchid]

very odd proof of that fact then

[solemn dew]

So if G/Z(G) is cyclic => G is abelian?

Does this work the other way around too? G is abelian => G/Z(G) is cyclic

[delicate orchid]

if you consider the trivial group cyclic then yes

[solemn dew]

cool fact

hence every subgroup must be abelian

so subgroup must be in center

[delicate orchid]

ok but you still need to show that G/Z(G) is cyclic

[solemn dew]

yes i'll try to do that

[solemn dew]

i don't understand what i am doing at this point

lol

any hints on how to show G/Z(G) is cyclic?

ahh sorry didn't copy everything

[delicate orchid]

perhaps there is another way

but that will require me to think

[delicate orchid]

||let H < G be a normal subgroup of order p, then consider C_H(G) - you claim to have shown that H must be cyclic, which is true, so we know that H <= C_G(H) so C_G(H) cannot be trivial. If C_G(H) is a proper subgroup of G then take N_G(H)/C_G(H) which is isomorphic to G/C_G(H) because H is normal in G. This quotient must be cyclic of order p as 1 < C_G(H) < G <=> C_G(H) \cong C_p. By the N/C theorem G/C_G(H) is isomorphic to a subgroup of Aut(H), which is a contradiction, as the automorphism group of C_p (and therefore the automorphism group of H) is C_{p-1}, which is smaller than G/C_G(H) so cannot possibly contain it, so C_G(H) must equal G, and so H <= Z(G)||

this proof is a uhh mess but I didn't want to think too much

[solemn dew]

i appreciate it

[karmic moat]

u’re very welcome :)

[solemn dew]

i am not at this level yet so i will have to try something else

[summer path]

Did you just use C for both centralizer and cyclic group

[delicate orchid]

yeah and I'll do it again

[summer path]

[celest cairn]

Is there a polynomial who’s splitting field is non-Galois? If so, would this imply the polynomial does not have a Galois group associated with its roots?

[delicate orchid]

do you know about the orbit stabilser theorem/group actions?

[karmic moat]

arent all splitting fields galois

or am i mistaking it

[solemn dew]

i think i do

f be a homomorphism, so f(x) = x is the stabilizer

group action is just a action in the group?

what i've done so far:

[delicate orchid]

ok so you don't know about group actions

[solemn dew]

😅

"action" is only mentioned 11 times in my book

i'll look it up on the internet

[summer path]

The proof of orbit stabilizer is really trivial for how useful the theorem actually is

[delicate orchid]

indeed it is

[formal ermine]

wtf???

why is that 50 pages???

[summer path]

Maybe it's on the big paper

[solemn dew]

very good book

difficult exercises tho

[formal ermine]

[untold turret]

that font has set my computer on fire and i had to throw coffee over it, rip these are my last seconds on the internet

[solemn dew]

isn't an orbit all the permutations of a group action?

for a element a in G, suppose o(a) = 3, then the orbit would be {e, a, a^2}?

[delicate orchid]

kind of, but you still haven't convinced me you know what a group action is

[solemn dew]

i don't blame u

[celest cairn]

Are all splitting fields Galois over Q or are there polynomials without Galois groups?

[barren sierra]

I can tell from the way these fonts look that it's already going to be a hard text

[south patrol]

I assume you mean a splitting field of a polynomial f(x) in Q[x}?

In which case yes, in fact for finite extensions K/F, K/F is Galois iff it is the splitting field of a separable polynomial

And every polynomial over a field of characteristic zero - in particular over Q - is separable

[summer path]

Galois theory is nice

And fun

[untold turret]

correct statement, I think: either V[t] = K, or t is not in the maximal ideal of V[t]

[next obsidian]

What do you mean “the” maximal ideal

[untold turret]

If V[t] is not K, then it is a valuation ring too (let me think through why)

[next obsidian]

t will be a unit in V[t] yes, because t^-1 is in V

Also you’re right

I was being dumb lol

It’s because of the condition that you’re a valuation ring iff either x or x^-1 is in V for any x in K

They can both be in V

Not an exclusive or

[formal ermine]

oh lmao

I thought it was exclusive or too

[next obsidian]

That’s not how or is used in math

[untold turret]

i always consider them exclusive because they're special tho

[formal ermine]

yeah I realized too

[warm ember]

whats Inn(G)

[delicate orchid]

the inner automorphisms of G

[solemn dew]

@rocky cloak

U were right:

[delicate orchid]

basically, all automorphisms of G that look like conjugating by some fixed element

[warm ember]

o

[solemn dew]

@delicate orchid what about doing two subgroups of G, in which "a" is in one of the subgroups and "b" is in the other. We know the first subgroup is generated by (a), and due to subgroups being normal in G, the other group that has that contains "b" has to be normal too. Thus a,b must be in Z(G), and so o(Z(g)) > p since the first subgroup has order p? then due to Lagranges theorem o(Z(G)) = p^2 thus Z(G) = G?

Does this work?

[delicate orchid]

how do we know that all of the subgroups must be normal

[solemn dew]

true i just contradicted myself

sorry

[delicate orchid]

and there are normal subgroups which are not central, so the following implication doesn't hold either

although, hmm

if we know the centre is non-trivial that implication holds

but the centre being non-trivial is equivalent to G being abelian in this case, so kinda missing the forest for the trees

[solemn dew]

yea...

abstract algebra is so hard

i love it

[formal ermine]

I find analysis to be harder