unless im not understanding how the powers should work
#groups-rings-fields
obsidian sleet
obsidian sleet
okok nice
next obsidian
Now we want to induce any map from S (x)_R R[x]
Which we can verify is this inverse
We literally only need it to be well-defined as a set map
And the fact that inverses of ring maps are ring maps automatically tells us it is multiplicative
This is good because the tensor product univ property only gives us linearity
But we would have to manually verify it’s multiplicative, except we don’t
So what’s the bilinear map you were trying to use?
obsidian sleet
i was going to take [s (x) Sum r_ix^i] and send it to Sum sr_ix^i
next obsidian
Yah so why is that bilinear
You don’t have to like write out the details for me
But like just ponder it
obsidian sleet
oki
next obsidian
Yah
From the formula
The question tho is
How does that equation make sense in S[x]
Cuz it got r_i in it
If u can answer that then like I will believe u understand what the map really is
obsidian sleet
o
next obsidian
Oh lmfao
Okay
Well so this is true more generally if S is an R-algebra
Aka a ring map R -> S
But like literally the exact same formula holds
obsidian sleet
except its the one u get from the defn of R algebra
next obsidian
Cuz if you’re an R-algebra u know how to multiply by things in R yeah?
next obsidian
Yeh
obsidian sleet
yea
obsidian sleet
Yes
next obsidian
So we know the map from tensor product is additive yeah?
And the map from S[x] is additive?
obsidian sleet
Yes
next obsidian
So to see that they’re inverse, you only need to consider simple tensors
So like
When you compute Tensor product -> S[x] -> Tensor product
You only need to see this is identity for a simple tensor
Cuz now for a general one you just use additicty
Break it up into a sum of simple tensors
Omg its id on all of those
😱
That make sense?
obsidian sleet
wait let me try that
oh yeah cus i can move the r_is to the left and right of the (x)
obsidian sleet
Yes
obsidian sleet
right
next obsidian
If u only did over Z you can’t do that
So the tensor product is wayyyy bigger
Cuz if u remember the construction
You did free abelian group on like M x N
Then modded out by relations to make like, moving that scalar work
obsidian sleet
Yes
next obsidian
U did it
next obsidian
No
obsidian sleet
o
next obsidian
😈
obsidian sleet
No more thoughts
next obsidian
obsidian sleet
next obsidian
Evil Chmonkey
obsidian sleet
🙀
what we did here was essentially write down an isomorphism of S algebras S[x] and S (x)_R R[x]
obsidian sleet
but we didnt have to check every single part of the hom
next obsidian
Confirmed it as rings
obsidian sleet
yea it seems straightforward to just
next obsidian
If u understand how the RHS is an S-alg
obsidian sleet
yeah
next obsidian
but we didnt have to check every single part of the hom
Wdym
next obsidian
Oh yah
I think umm
It suffices to check only that umm
The S[x] -> Tensor product one is
And that’s like
Obvious
Kekw
obsidian sleet
i see
next obsidian
Cuz umm
We defined it
Using the univ property
As an S-algebra
Like to have it work by only defining what happens to x
Uses that the latter is an S-algebra
obsidian sleet
ohh this is because we are using the correspondence of bilinear map from S \times R[x] to S[x] to S-alg maps from S (x)_R R[x] to S[x]
to not have to check too many things
we just get it cus it all matches
next obsidian
This is like
True but not something u know a priori
Like
What really is going on is
Maps S (x)_R T for two R-algebras S,T
Into an S-algebra W
Are the same as R-algebra maps T -> W
(Note that any S-algebra is an R-algebra via composing the ring maps)
So what you’re doing is there’s an adjoint between extension of scalars (tensoring by S)
And restriction of scalars (considering an S-algebra as an R-algebra)
The annoying part is, you can easily define a map from the tensor product using its univ property
But it isn’t a priori multiplicative
And it isn’t hard to verify it’s multiplicative but it’s just annoying
You have to write down stuff 
obsidian sleet
i see 
next obsidian
The way I guided you through this avoided that
Because you just checked the map from the tensor product is inverse to a ring map
But like
next obsidian
Are the same as R-algebra maps T -> W
this is obvious how to do it
I think
Based on what you did here
If you have phi:T -> W
Now you define S x T -> W by sending (s,x) to sphi(x)
It’s clearly bilinear
So you get a map from S (x)_R T into W
You see by definition of the map from the tensor product this is like, an S-algebra map
Then you manually compute it’s multiplicative
next obsidian
Anyway
From this
You get the iso here functorially
Namely, what is an S-algebra map S[x] -> W?
It’s a choice of an element in W
What’s an S-algebra map S (x)_R R[x] -> W?
Well it’s an R-algebra map R[x] -> W
What’s that? A choice of an element in W
So yah
next obsidian
So like the moral heee is like
S (x)_R - is like the “S-algebrafication” of an R-algwbra
next obsidian
Are the same as R-algebra maps T -> W
Which is validated by this property
obsidian sleet
mhm
next obsidian
Think of like what a groupificarion of a set is (the free grpup)
Abelianization of a group
Etc
It’s always about having a property like this
So the point is “S-algebrafication” sends a free R-algebra to a free S-algebra
Cuz both are characterized by like “a map into T is a choice of element in T”
In fact
Do u understand what adjoints are?
Cuz u have used Yoneda right?
obsidian sleet
yus i seen adjoints
next obsidian
Okay so
S (x) - is adjoint to R_-
The worst notation I know
But R_- is restriction of scalars
It’s when we consider an S-algebra as an R-algebra
It kinda a forgetful functor
So check dis out
obsidian sleet
hmmm
next obsidian
Hom_S(S (x)_R R[x], T) =
Hom_R(R[x], R_T) =
T
And Hom_S(S[x],T) = T
So this is just a like, “equational” version of what I said earlier
But like it’s Yoneda!
and it follows formally from that adjoint
So like my point here is this generalizes
Like the adjoint pair here is an instance of x-ization
And a forgetful functor
Then the whole poly algebra is a free functor
Like “free on one element”
next obsidian
Hom_S(S (x)_R R[x], T) =
Hom_R(R[x], R_T) =
T
So if you’re ever in a situation like this u can probably do this same thing to show the x-ization sends a free object to a free object
That make sense?
obsidian sleet
i think soo
next obsidian
Swag
obsidian sleet
This pretty cool stuff
next obsidian
So a few things:
1: that’s a super weird notation for the identity of a group lmao
2: your proof for (b) is on the right idea, but I don’t agree
Namely, your cases don’t make sense because the ring is commutative
So those cases are literally the same, those are the same thing
jovial kelp
we use epsilon in our class instead of e for groups. For rings we just use 0 😛
oh yeah you're right!
next obsidian
So your cases should be something like
Yeah, I guess I’ll let you think more but
It will be impossible to prove both a and b are zero divisors
The most you’ll be able to extract out of this is that at least one of them is
So that might lead to you doing a different sort of cases
If u wanted to prove it like that
jovial kelp
ah okay so showcasing for just one of them will be fine?
jovial kelp
If u wanted to prove it like that
tbh probably not necessary
next obsidian
Well you don’t know which it will be
It could be either
So I think you’re probably going to have to do cases tbh
jovial kelp
ah okay
next obsidian
Otherwise you aren’t gonna really be able to get what you want
I think
But maybe you should think about it yourself a bit
jovial kelp
yeah for sure. I'll chew on this a bit more, thank you!
north parrot
I see, thank you
tender bough
I'm perplexed
with this basis
a highest weight vector is some v_lambda such that E v_lambda = 0
and v_lambda^{(n)} is defined by F^n v_lambda
But then my question is:
tender bough
I'm perplexed
...........I think it's a typo. the note actually means v_lambda^{(dim V - 1)} != 0
hot lake
Yes that's a typo
white grotto
find all group morphisms from Q,+ to Z,+
zenith trellis
find all group morphisms from Q,+ to Z,+
Let f be such a morphism. If p is an integer and if p is non zero, pf(1/p) = f(1) so f(1) is a multiple of every non zero integer p... I let you finish.
white grotto
Let f be such a morphism. If p is an integer and if p is non zero, pf(1/p) = f(1...
it doesnt exist?
because if it did then f(p/q)=pf(1)/q which doesnt belong to Z
restive birch
super stuck on this problem. I have no idea how to show that every element can be written as x^{-1}sigma(x)
quiet pelican
super stuck on this problem. I have no idea how to show that every element can b...
Suppose that two elements had x^{-1} sigma(x) = y^{-1} sigma(y)
||Then prove they’re the same||
||And since it’s an injective map from a finite set to itself…||
restive birch
does sigma^2(x) here mean (sigma(x))^2 or sigma(sigma(x))
quiet pelican
does sigma^2(x) here mean (sigma(x))^2 or sigma(sigma(x))
The second
chilly ocean
If a subset A of the ring is contained in a prime ideal P, why is the ideal generated by A still a subset of P?
thorn delta
The ideal generated by A is the intersection of all ideals containing A. P is an ideal containing A, so the ideal generated by A certainly can’t be any larger than P
@chilly ocean
chilly ocean
The ideal generated by A is the intersection of all ideals containing A. P is an...
Oh of course
When you generate an ideal from a set, you're taking sums and products of elements of the set, so you're going to stay in any ideal containing the generating set.
zenith trellis
does someone know?
It's not that it does not exist. It's just that there is only the null morphism.
chilly ocean
I read this in my book, do you know why $Z_m$ and $Z_n$ each contain a subgroup of order 5?
cloud walrusBOT
Iced Sugar
chilly ocean
I figure it has to do something with Lagrange or that 5 divides n and m
So does a group have an element of order n for every n that divides the order of the group?
rustic crown
nope. this is a very special property of cyclic groups
consider any non-cyclic group G, then ofc you can't find an element of order |G| as that forces G to be cyclic
coral shale
If a generates Z_m, then a^m = e and <a^(m/5)> is your subgroup of order 5. For uniqueness... inspection 👀
rustic crown
hi shuri :3
coral shale
mrow
formal ermine
what is an element of a module called?
thorn delta
element of a module
rotund aurora
lol
thorn delta
don't call it a vector pls
formal ermine
no like the elements of a vector space are called vectors
what if we call them packages
because a module in programming usually contains packages
rotund aurora
is this trolling?
rustic crown
yee
formal ermine
trolling paired with seriousness
rustic crown
i think this has already been asked before :p
formal ermine
o
formal ermine
ok actual question now
we defined a module as an abelian group that defines an operation on a ring with a couple of axioms
rustic crown
found it :3
rotund aurora
haha thanks, I was trying to look for it
formal ermine
then later we had the example that every Z module is an abelian group
rotund aurora
"grouplicules" by far my favourite
rustic crown
formal ermine
wouldn't every module be an abelian group with that definition though lol
since it's literally in the definition
rotund aurora
no
rustic crown
yea, just liek every group is a set
formal ermine
or am I misunderstanding what they mean
rustic crown
but with more structure
rotund aurora
yea, just liek every group is a set
technically no
this is like the question "is R a vector space over R"? or similar stuff
rustic crown
yea, just liek every group is a set
more precisely, there is a forgetful functor from R-modules --> abelian groups, so you could talk about "underlying abelian group of a module"
formal ermine
more precisely, there is a forgetful functor from R-modules --> abelian groups, ...
what's a functor
rotund aurora
couldn't they have used any other ring
Try giving Z a module structure over Z/nZ
thorn delta
why did they use Z though?
what they mean is that the information included with a Z-module is exactly an abelian group
there is nothing extra to "forget"
formal ermine
Try giving Z a module structure over Z/nZ
this means Z/nZ-module with Z as the underlying abelian group?
formal ermine
what they mean is that the information included with a Z-module is exactly an ab...
what info included?
formal ermine
you take some object and produce another one
ah
rotund aurora
The abelian group is Z, and the ring is Z/nZ
then try defining a scalar product Z/nZ x Z-> Z
thorn delta
what info included?
the action of multiplication by elements of Z on the abelian group
rotund aurora
that satisfies the module axioms
formal ermine
then try defining a scalar product Z/nZ x Z-> Z
I'll try that and see where it fails tomorrow
it's already late here
but thanks
thorn delta
like if you have an abelian group A, x in A, and n in Z, then you already know what nx means: add x to itself n times.
This is the action of Z on A
formal ermine
the action of multiplication by elements of Z on the abelian group
isn't that 'included' with any other ring too tho?
formal ermine
like if you have an abelian group A, x in A, and n in Z, then you already know w...
ah
is that the only senseful action we can define
or are there other ones
rotund aurora
isn't that 'included' with any other ring too tho?
You should try these kind of things by yourself, since its about interiorizing definitions/axioms. That's why I made that suggestion
formal ermine
yeah you're right
thorn delta
is that the only senseful action we can define
sure there could be others, but this is just what we mean when we make the correspondence between abelian group and Z-modules
rustic crown
there is another way to say the same thing. If you know some group theory, you would know that group actions are a powerful tool to study groups. Similary, you could ask if there are objects on which rings can act on. And abelian groups nicely fit what you want.
If A is an abelian group, then the endomorphism on A, End(A) is naturally a ring under composition and the identity homomorphism is the 1 in this ring.
so the action of a ring is precisely a ring homomorphism R --> End(A)
thorn delta
is that the only senseful action we can define
actually depending on how you define things, the answer is yes
rustic crown
and what kxrider said about having a natural action of Z, is then same as saying that there is a unique map from Z to any other ring, in particular End(A)
thorn delta
if your ring action has to be a ring hom Z --> End(A), then 1 must map to 1
formal ermine
there is another way to say the same thing. If you know some group theory, you w...
do you have any good reading material on group actions specifically? my course introduced them once and then never did anything with them
restive birch
i want to clarify my understanding, because the book is rather brief on this- a set and an operation "generate" a group if using the operation in some order on the elements of the set and their inverses will produce all of te elements of the group?
formal ermine
i want to clarify my understanding, because the book is rather brief on this- a ...
"in some order" as in all possible combinations, yes
rustic crown
do you have any good reading material on group actions specifically? my course i...
i studied it from aluffi. and it really handled that properly i'd say
formal ermine
i studied it from aluffi. and it really handled that properly i'd say
mhm ok thanks
I'll take a look at that tomorrow. thanks for the help!
restive birch
"in some order" as in all possible combinations, yes
how do relations factor into this? thats what really confuses me
formal ermine
how do relations factor into this? thats what really confuses me
relations?
restive birch
as in, some kind of 'conditions' on the generators
formal ermine
those are conditions on the input set
$\gen{ab\inv | a,b \in G}$ is like saying \gen{\Set{ab\inv | a,b \in G}}$
cloud walrusBOT
yes yes yes no
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knotty frigate
$b\inv$
cloud walrusBOT
!Pymamba™
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knotty frigate
is that a preamble command?
next obsidian
No
Or uh idk what you mean
But if you’re trying to get ^{-1} that isn’t a default command
maiden heath
Hi I'm trying to prove the multiplicity freeness theorem from Martin Lorenz A tour of representation theory. I'm a bit uncertain about the last line I wrote. Because doesn't Schur's Lemma imply that dim End_G(V) = 1, since Hom_G(V,V) = End_G(V) and V is isomorphic to itself. In which case m cannot be 0. Thanks
kind temple
Hi I'm trying to prove the multiplicity freeness theorem from Martin Lorenz A to...
what book is this from
chilly ocean
Let IJ be ideals, if IJ is a subset of a prime ideal P, why is either I or J a subset of P?
charred bison
do normal subgroups go to normal subgroups under a homomorphism?
next obsidian
No
Here’s the stupidest example
Let G be a group and H a non-normal subgroup
H < H is normal
But the image of H in G is not normal
mighty spade
do normal subgroups go to normal subgroups under a homomorphism?
Normal as what subgroup?
charred bison
f : G -> H is a homomorphism, if K is normal in G, is f(K) normal in H is what i wanted to ask
next obsidian
I gave you counterexamples
charred bison
yeah understood
thanks
next obsidian
Cool
restive birch
im still kinda confused about how group operations being 'similar' connects to the official definition of an isomorphism
like how do you go from 'these groups look kinda similar' to 'these groups are isomorphic'
charred bison
by showing they are isomorphic? not sure what u mean
pastel cliff
yeah try and rephrase that question
an isomorphism isnt saying that something looks kinda similar, it's basically equivalence
vernal pine
pastel cliff
vernal pine
A labeling of D3 by 3 symbols such that any translation of all symbols by a fixed group element leaves the elements e and b with the same symbol
Yet, G acts faithfully on the set consisting of this labeling and its translations
pastel cliff
remind me what a faithful action is
vernal pine
For every non-identity g there is an x such that g(x) != x
I made this counterexample for a part of my current paper
Let X be a system of symbolic labelings of a discrete group G. Then G acts on X by translations. If X "separates elements" of G (for every g,h, there is an x such that x(g) != x(h)), then G acts faithfully on X, but the converse is not true in general
There is a partial converse though: if G acts faithfully on X, then there is a "higher block presentation" of X (replace each symbol by the pattern of symbols on its F-neighborhood, for some finite subset F) that separates elements of G
*provided that X satisfies a few hypotheses which I do have in the context of my paper
verbal harness
Say, Cayley's theorem states that any finite group is a subgroup of a symmetric group (or something along those lines) right?
Because if that is correct (and working up from first principles it does appear to be) then I'm curious which S group the Monster is a subgroup of.
next obsidian
S_|monster group|
verbal harness
S_|monster group|
Unfortunately neither Google nor Wolfram Alpha are familiar with that terminology. 😮
So S_(the order of monster group)? OK ty.
(less insight in that revelation than I was hoping might be had heh heh. 😉 )
next obsidian
I mean
Cauley’s theorem says that G embeds into S_|G|
That’s it
You act on yourself via left multiplication and see this is a faithful action (in fact it’s free), that’s it
verbal harness
I was running an ELI5 version of a definition for a group through my head in preference to waking up the other day, and came across a simplest definition that began from determining that every (finite, but maybe also infinite I just didn't check that far) group must be a subgroup of the symmetric group of it's "domain set" (I still don't know the correct terminology there.. any set of elements that the groups actions could conceivably transform)
And then I go searching my above question and basically find that Cayley made a theorem that says largely the same thing. 😁
vernal pine
There is a nontrivial question, what is the minimum n such that G embeds into S_n. By Cayley's theorem, n <= |G|
S_n embeds into S_n just as well as it embeds into S_n!
Seems like for the monster group M, the answer is n = 97239461142009186000
So the monster group can be realized as a group of permutations on 10^20 things
maiden heath
what book is this from
The book is A tour of representation theory by martin lorenz. And the image is an attempt at writing a proof
maiden heath
Hi I'm trying to prove the multiplicity freeness theorem from Martin Lorenz A to...
Hi I was wondering if someone could check to see what I did wrong here. Any help is much appreciated, thanks
simple mulch
Given a group such that $|G| = 12 = 12\times 1 = 4\times 3 = 6 \times 2.$Can we find an element for each number? i.e. $g_i\in G$ such that $|g_1| = 12$, $|g_2| = 1$, $|g_3| = 4$, $|g_4| = 3$, $|g_5| = 6$ and $|g_6|=2$ ?
We know $g_2 = e$
simple mulch
I think, $g_4$ and $g_6$ are given by Cauchy's theorem
cloud walrusBOT
mns
ionic spade
I think, $g_4$ and $g_6$ are given by Cauchy's theorem
This is correct. However, some of the others are not guaranteed because Lagrange’s theorem is an “if statement”. In other words, if an element of a group has a specific order, then its order divides the order of the group. However, the converse is not always true.
obtuse bear
Given a group such that $|G| = 12 = 12\times 1 = 4\times 3 = 6 \times 2.$Can we ...
this is true for cyclic groups but not in general. for a counterexample you could take Z2 x Z2 x Z3, which has no element of order 4
formal ermine
another counter example is A5 having no subgroup of order 15
ionic spade
another counter example is A5 having no subgroup of order 15
Or A4 not having a subgroup of order 6, if you want a smaller example.
restive birch
so on this question, we're given 2 specific values of phi for r and s, but if we arent given that phi is a homomorphism, how can we evaluate phi(rs) for example to prove that it is?
like, without assuming phi(rs) = phi(r)phi(s) (which we cant assume because it isnt a homomorphism, at least if my understanding is correct)
ruby sundial
I think thats the point of this question
You want to show it extends to a homomorphism
ruby sundial
Can you find another homomorphism phi’ such that phi’(rs)=phi(r)phi(s)=phi’(r)phi’(s)
restive birch
so the keyword here is 'extends'
ruby sundial
yeah
restive birch
so phi' would just be phi'(r^n*s) -> phi(r)^{n}phi(s)
formal ermine
for any commutative ring $R$, an $R^\times$-Module is automatically an $R^\times$-Vector space, yes?
cloud walrusBOT
yes yes yes no
rustic crown
what's R^{\times} here? the units?
(cause that is never a field)
and idk what a vector space over a group is
though a module over a group is by definition a module over the group ring Z[whatever group]
formal ermine
oh yeah nvm lol
this sounds more like #linear-algebra than #groups-rings-fields to me
cloud walrusBOT
loganb
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formal ermine
I don't quite understand why the chain isn't stationary
rustic crown
because at each point <a_1, ..., a_n> is strictly inside I, so you pick a_{n+1} in I which isn't already in <a_1, ..., a_n> and continue with life
this gives you a strictly increasing chain
formal ermine
because at each point <a_1, ..., a_n> is strictly inside I
why
rustic crown
that's our assumption right. I isn't finitely generated. if I = <a_1, ..., a_n>, then you just gave me finitely many elements that end up generating I
formal ermine
that's our assumption right. I isn't finitely generated. if I = <a_1, ..., a_n>,...
wouldn't that mean that I = R though?
cuz we have no restrictions on our elements a_x
so they can be arbitrary leading to x in I for any x in R meaning that I = R?
rustic crown
not sure what you mean, but at each point we're picking a_{n+1} from I itself, so now way our <a_1, ..., a_n> get bigger than I
maybe an example helps
take R = Q[x1, x2, .....]
I = <x1, x2, ....>
here we get the sequence <x1>, <x1, x2>, ....
rustic crown
they could be anything from I
but to show that R is not noetherian, all you have to do is give one strictly increasing chain which doesn't stabilize
formal ermine
ok wait so
I being non finite => R has a non finite amount of variables (?) => we can easily find a non stabilizing chain
rustic crown
okie lemme say the whole argument again
say the ideal I isn't finitely generated
pick an element a_1 in I
so you get an ideal <a_1> completely inside of I
this can't be the all of I as I isn't finitely generated
so pick a_2 which is in I but not in <a_1>
this gives you a strictly bigger ideal <a_1, a_2> which is still contained in I
delicate orchid
You’re welcome
formal ermine
since I is non finite we can always find an element that isn't in <a_1, a_2, ..., a_n> but still in I
rustic crown
hi wew
delicate orchid
Hello “det”
ionic spade
I love noetherian modules 
delicate orchid
All rings are noetherian
formal ermine
ok nvm didn't get it lol
I not being finite implies R not being finite. so how can we know that <x> is finite (for an x in I)?
rustic crown
(not for number theorists >.<)
delicate orchid
ok nvm didn't get it lol
I not being finite implies R not being finite. so how ...
<x> is clearly finitely generated
That’s what det is saying (I hope) when they say finite
rustic crown
(me never used the word finite >.<)
delicate orchid
Oh true
rustic crown
ig i only use finite to describe field extensions which are finitely generated as vector spaces
delicate orchid
Maybe an example will make things easier
formal ermine
does <1, 2, 3, 4, 5, 6, ...> count as non finitely generated?
rustic crown
in Z?
formal ermine
yes
rustic crown
nope, becuase that is same as <1>
delicate orchid
No cause that’s just <1>
delicate orchid
I’m doing an example now
formal ermine
yes
delicate orchid
This one is a classic: Take the ring of polynomials in infinite variables Z[x1, x2, x3, …], then the chain of ideals (x1) < (x1, x2) < (x1, x2, x3) < … is non-stabilising
rustic crown
take R = Q[x1, x2, .....]
🙈
delicate orchid
🙈
My fellow in the lords light. I will not be scrolling up on this side of eternity.
To retreat into the past is to sacrifice progress
south patrol
I followed wew's advice and now I have 500 billion in bitcoin
formal ermine
we can always find an element that is in I but not in <B = {a1, a2, ..., an}> because otherwise I would be finitely generated by B, yes?
delicate orchid
Yur
delicate orchid
Just means there exists a finite set that generates the dude
formal ermine
yeah I knew what it means but I didn't know that an ideal that is generated by a non finite set but also by a finite set is considered finitely generated
kinda sounds obvious when I type it out now lol
anyhow
thanks!
tribal drum
i cant visualise how these are equal, i read it as 1 -> 5 -> 1 -> 9, but the cycle is 1 -> 5 -> 9 -> 1 ...
south patrol
The second one is a composition
Under the "standard" convention, this means you apply (1,5) then (1,9)
So 1 is sent to 5 (which is then sent to 5 by (1,9)
5 is sent to 1 by (1,5) and then this is sent to 9 by (1,9)
And so forth
Also, note that "1 -> 5 -> 1 -> 9" wouldn't make sense, because it would make 1 be sent to 5 and 9
ionic spade
Does "ascending chain" and "ascending sequence" mean the same thing in the context of modules?
Or does "ascending chain" always mean that the ascending sequence satisfies the ascending chain condition (ACC)?
rustic crown
they should mean the same thing
ionic spade
OK, because the book I'm reading has two separate exercises that use both terms in separate instances
So I guess they are the exact same exercise, or there is a typo somewhere ;/
The exercises are : "Prove that the union of an ascending sequence of submodules of an R-module is also an R module" and "Prove that the union of an ascending chain of submodules of an R-module is also an R module"
If the second problem means that the ascending chain satisfies the ACC then the problem is trivial, no? Just take the first submodule S_k when the chain becomes constant
If not then they are literally the same problem, unless I'm missing something
ionic spade
LOL
rustic crown
next obsidian
-ass problem tbh
ionic spade
This is Advanced Linear Algebra by Steven Roman, it's actually been a decent book so far
next obsidian
Maybe he wanted to do the remix to make sure u got it the first time
ionic spade
😂
ionic spade
<:chmonkey:773795952125411338>-ass problem tbh
the problems have low key sucked ngl
The actual content is good though, I like his section on modules better than dummit and foote
next obsidian
🗿
next obsidian
Lowkey what u learn from kinda don’t matter
Unless the content sucks major cok n balls
Like, you get a similar enoigj experience in most cases
IMO
And if the book doesn’t suck deek n bollZ then u should know enough to be able to learn whatever little things it missed when u need them
ionic spade
yeah, that's why I like reading from a couple of sources when learning something I can soak it all in
just in case one author is a literal dum dum
next obsidian
Like ppl ask what the best calculus book is all the time and my answer is literally just whatever u can get ur hands on
Lol
rustic crown
i've learned way more from profs than books >.<
aluffi maybe one of the only books i've sort of read
ionic spade
aluffi maybe one of the only books i've sort of read
Chapter 0?
rustic crown
yea
ionic spade
why would you ever learn abstract algebra using category theory though smh
I feel like that is massive overkill
rustic crown
but its fun and cute
elder wave
ionic spade
I guess it's more of an appreciation book than a first time read
rustic crown
and also gave good motivation to learn cat
ionic spade
yeah for sure
next obsidian
I used it for my first run
It’s good
Category theory is the way you should be thinking about algebra
ionic spade
I'm confused about this problem: let S be a finitely generated submodule of an R-module M, and suppose that M/S is finitely generated. Prove that M is also finitely generated as well.
Why does S have to be finitely generated though, isn't that irrelevant?
If M/S is finitely generated, I don't see why M being finitely generated doesn't follow
rotund aurora
let M have basis (x_1,x_2,x_3,x_4,...) and let S=Span(x_2,x_3,...)
ionic spade
let M have basis (x_1,x_2,x_3,x_4,...) and let S=Span(x_2,x_3,...)
S is finitely generated
rotund aurora
but you asked if that was irrelevant
You need that hypothesis
I thought that was what you were asking
ionic spade
I thought that was what you were asking
oh yeah sorry
rotund aurora
In that case, M/S is finitely generated, but M is not
ionic spade
But why did you give M a basis
ionic spade
but M wasn't given to be free
rotund aurora
M is supposed to be an arbitrary module satisfying some hypothesis
rotund aurora
there are free modules that will satisfy those hypothesis
I think you should re read the problem or this conversation (maybe it was me who was not clear)
ionic spade
OK I'll latex it so that my question is succinct
next obsidian
Volkenborn
Consider this stupid af example
Let M be non-fg
S = M
Then M/S is fg
But M isn’t
So clearly S fg is necessary
elder wave
next obsidian
And instead of basis I think Croqueta meant “spanning set”
Or meant to give an example via a free module
rotund aurora
uhm no, because then that example is not necessarily true
oh okay I see where the confusion is coming
I said "let" M be free, as if M was given and it turned out to be free
I just made that module up
i just chose a module that is free
ionic spade
S = M
Do you mean that S is not finitely generated here, or that S is literally equivalent to M
next obsidian
I was giving an example of why S finitely generated is necessary
But it’s basically croqueta’s example
ionic spade
ok that makes sense
But how would I use the fact that S is finitely generated to show that M is finitely generated, when I was given that M/S was
because you can write M/S={m+S | m in M}
but that it is a weird form because you can't really use the generating set of S
next obsidian
You can
You need to think about it
Start with arbitrary m in M
Then all you can do is consider m + S
ionic spade
Oh wait S is a submodule of M
rotund aurora
what is an example of a separably generated inseparable extension?
next obsidian
That doesn’t sound possible?
rotund aurora
mmh I think it is? So far I only know it is impossible for algebraic and finite extensions, but idk about other tpyes
next obsidian
Separably generated is basically separable but makes sense for non-algebraic extensions
Like by definition it’s a purely transcendental extension followed by a separable one right?
There’s no place there that inseparability could come in
If I’m thinking about it correctly
ionic spade
Wait does S being finitely generated mean that I can use M being noetherian somehow?
Like establishing some ACC?
rotund aurora
wait I think that just doesn't make sense, because separably generated extensions must be algebraic no? Since separable elements are just algebraic
next obsidian
Wait does S being finitely generated mean that I can use M being noetherian some...
No
next obsidian
wait I think that just doesn't make sense, because separably generated extension...
No,
Like
If it’s algebraic it’s just separable lol
Separably generated literally means you have a separating transcendence basis
cloud walrusBOT
Volkenborn
ionic spade
I can't write m in terms of the generators of S though
because there are some m in M that are not generated by the elements of S
next obsidian
Yeah but that’s why you use that M/S is fg
Blitz if u say the solution I’ll kill you
ionic spade
Wait lemme try
chilly ocean
I guess it's more of an appreciation book than a first time read
It's more intuition based than most books tbh
next obsidian
Oh
chilly ocean
Blitz if u say the solution I’ll kill you
cloud walrusBOT
Volkenborn
ionic spade
just an idea
next obsidian
Yeah
But maybe let’s use like, m_i
So that s_i can generate S
It just makes more sense IMO
ionic spade
ok ok
next obsidian
So now take m in M yeah?
Arbitrary
All we can really do is look at m + S yeah?
So what can you tell me about m + S
ionic spade
m+S is in M/S
ionic spade
I don't know how to apply S after that
next obsidian
Well no
Even before that
Why did u bother writing down a generating set of M/S
If you don’t use it
ionic spade
just because m+S is in M/S doesn't make m a generator for M, no?
next obsidian
It doesn’t matter
rotund aurora
Separably generated literally means you have a separating transcendence basis
Let F be a field, then F<K is separably generated if there exists a set S consisting entirely of separable elements such that F(S)=K. Separable elements are algebraic, so F(S) must be algebraic. So you cannot have transcendental extensions that are separably generated, so my initial question didn't make sense.
ionic spade
Also how does that use the S being fg condition
rotund aurora
(this is a question about definitions)
next obsidian
Just use your generating set for M/S first
next obsidian
Let F be a field, then F<K is separably generated if there exists a set S consis...
No that’s not the definition
Separably generated means you have a transcendence basis S such that K/F(S) is a separable algebraic extension
next obsidian
Hurb
This is terrible because
Separably generated ALREADY IS DEFINED
IT’S A TERM THAT IS DEFINED
rotund aurora
you mean that is a standard term in algebra?
next obsidian
Yes
rotund aurora
if so, that's fucked up yeh
ionic spade
Just use your generating set for M/S first
Ok so I choose some m+S in M/S from my generating set
next obsidian
Uhhh
Well that’s not quite what a generating set is yeah?
You just can write m + S as a linear combination of the generators
ionic spade
Oh wait yeah so write m=r_1m_1+…+r_nm_n
ionic spade
Ok
ionic spade
Hmmmm
next obsidian
Not really
Like what does it mean that x + S = y + S
Like
Idk what else to say
ionic spade
That they are in the same coset
chilly ocean
maybe if those were groups and we had xH = yH
next obsidian
Right which means?
ionic spade
Or congruence class
next obsidian
So x - y is?
ionic spade
0
ionic spade
3-1= 0 mod 2 though
next obsidian
Yeah but
rotund aurora
3-1= 0 mod 2 though
and what does that mean
chilly ocean
what does it mean for (x-y)+S = S, say
ionic spade
I’m kind of confused, in the context of what?
next obsidian
For M/S
Like you have a notion of equality there
And it tells you things about the representatives
ionic spade
Right, m is equal to the linear combination I don’t know what else there is than that
next obsidian
Like when is x + S = y + S is what I’m asking
ionic spade
x=y
next obsidian
No, this isn’t true
ionic spade
Modulo S?
next obsidian
Yeah what does modulo S mean?
ionic spade
Means you take the remainder
chilly ocean
If 3 = 1 (mod 2), then it means that 3-1 = 0 (mod 2), which means that 3-1 is in 2Z
next obsidian
What does that mean?
I think you need to go back to the construction of the quotient modules
And review what equality means there
ionic spade
If 3 = 1 (mod 2), then it means that 3-1 = 0 (mod 2), which means that 3-1 is in...
Yeah so x-y is in S?
ionic spade
Thank goodness
next obsidian
m + S = r_1m_1 + … + r_nm_n + S
So wayyyyyyy back here
chilly ocean
next obsidian
What can we conclude?
ionic spade
m-(stuff) is in S
ionic spade
And S is finitely generated!!!!
next obsidian
:D
ionic spade
I can finally use that bloody condition
chilly ocean
3-space property for finitely generated modules?
oh wait. It's not a space. How do you call them?
next obsidian
3-space property for finitely generated modules?
It’s not true
ionic spade
m-(stuff) is in S
Since S is finitely generated, then this forms another linear combination of some element in S?
next obsidian
You’re asking about like Serre subcategories
Well sort of
Fix generators s_1 through s_m
Then it’s a linear combination of those
So now you move “stuff” back to the right
And now m = linear combination of m_i and s_j
Yeah?
chilly ocean
oh yeah. Serre subcategories is what I was thinking of (well, not by name since I didn't know)
next obsidian
Yeah but not true for fg
If the outer two are the middle is (hats this proof)
But the last two being fg doesn’t imply the first is
This is why you either need to work with Noetherian rings or more generally with coherent modules
ionic spade
Then it’s a linear combination of those
Wait are you saying m-(stuff) is a linear combination of the s_j?
next obsidian
Yeah!
next obsidian
Exactly
But now u have ur finite generating set
Once u move “stuff” back over to the right
ionic spade
So we have m=stuff + (generating set of s)
chilly ocean
But the last two being fg doesn’t imply the first is
is this part of Serre subcategory definition? I don't see it here https://stacks.math.columbia.edu/tag/02MN#footnote-1
next obsidian
is this part of Serre subcategory definition? I don't see it here https://stacks...
Yes this is actually a point of confusion
ionic spade
Since m is written as a linear combination of the m_i and s_j and m was arbitrary
next obsidian
This definition is non-obviously equivalent to asking the following
Given 0 -> A’’ -> A -> A’ -> 0 a SES that A is in it iff A’’ and A’ are
The latter is the more standard definition
chilly ocean
ah. I get it
ionic spade
Since m is written as a linear combination of the m_i and s_j and m was arbitrar...
Then every m is generated by that new generating set so we’re done?
YES!!!
Thanks @next obsidian and everyone else, that was really helpful.
next obsidian
ionic spade
I think I was stuck on getting passed M/S to use S
But that subtraction trick modulo S was really neat, I’ll definitely keep that in mind
(Probably should have thought of that honestly it’s kind of trivial in hindsight)
rotund aurora
Are non-perfect fields actually important?
rotund aurora
is it comparable to noetherian vs non-noetherian rings? Or nah
next obsidian
Sort of
Like if you do geometry
And stick to “geometric parts” where maybe you’re working only with alg closed
Or in char 0 (which reduces to C usually)
The classicism situations then neither Noetherian nor perfect are issues
But if you do NT
Both can show up
ionic spade
What are examples of non perfect fields in NT? Like Q_p?
rotund aurora
I just remembered, constructing non-perfect fields is not difficult at all. The field F_p(x) is not perfect
fading holly
is the U(10) group {1, 3, 7, 9} under multiplication mod 10?
im going through my notes and it looks like it must be modulus or something but im not sure
delicate orchid
yeah those are all the numbers < 10 coprime to 10
fading holly
so would it be correct to just show that i can get all of the elements by using 3^k mod 10
void cosmos
if u do that then u would be showing U(10) is cyclic
ionic spade
so would it be correct to just show that i can get all of the elements by using ...
Yes, that is right.
fading holly
ok ty
also for (Z, +) how do I generate 0 from <1>? is 1^-1 = 0 or 1^0 = 0
cus is it like 1^-1 = 1 + (-1) = 0?
rustic crown
its 1^0 which in the additive notation would be 0*1 = 0
so <g> = {..., g^-2, g^-1, g^0, g^1, g^2, ...} in the additive notation would look like <g> = {..., -2g, -g, 0, g, 2g, ...}
rustic crown
np
rotund aurora
If x is separable over F with char F=p it need not be true that x^p=x right? Even though we have F(x)=F(x^p). Can someone provide an example?
rustic crown
sorry, i'm a little sleepy rn
but x^p = x has at most p roots
and 0, 1, 2..., p-1 are definitely roots
so if x is anything outside F_p then x^p won't be x
ig the simplest example is F_4 = F_2[x]/(x^2 + x + 1), here x^2 = x + 1 and not x
true cairn
can someone help me with part a? how can I show that the elements are distinct?
solar glacier
quick question
chilly ocean
The answer is yes.
solar glacier
for shwoing if R is a commutative ring with ideal I
showing rad I is an ideal
I only need to show
if a,b \in rad I such that a^n,b^m \in I
then (a+b)^n+m \in I
chilly ocean
Is it?
solar glacier
and that for all r \in R that r rad I is contained in rad I
solar glacier
showing rad I is an ideal
contained in I
solar glacier
and that for all r \in R that r rad I is contained in rad I
this is the ideal property correct
ok jjust checking thanks!
ionic spade
So I want to show that if R is Noetherian and I is an ideal of R, then R/I is Noetherian
But I’m stuck on how to get started. I can either (1) show that every ideal of R/I is finitely generated or (2) show that an ascending chain of ideals of R/I eventually stabilizes
The problem I’m having is finding a general set of ideals for R/I
next obsidian
This should have been covered
Do u know what the submodules of M/N are?
Surely this was covered in the text
ionic spade
It’s not covered afaik
next obsidian
It should be
Like, I would bet 5 dollars it is despite never having read the book
Cuz it doesn’t make sense to include it otherwise kekw
ionic spade
It’s not in ALA by Roman
Checking Dummit and Foote, maybe it’s in there and I forgot
next obsidian
What page is this exercise on
ionic spade
In ALA by Roman?
next obsidian
Yeah
And it is, it’s Theorem 4.7 on page 102
Ideals are exactly submodules of R
So the theorem applies
ionic spade
so it would be r+I where r is in R?
next obsidian
No
You should go back to the theorem for vector spaces it is saying has an analogue and read the proof there for vector spaces
It works the same way for modules
But you’ll see why what you said isn’t the case
For one, what you wrote isn’t a set, just a single element of the quotient
So it doesn’t make sense to say that’s a submodule of M/S
Or I guess R/I
ionic spade
For one, what you wrote isn’t a set, just a single element of the quotient
This is very confusing. R/I={r+I | r is in R}
ionic spade
This is very confusing. R/I={r+I | r is in R}
But if this is the quotient what would the ideal be
next obsidian
There’s lots of them
And if you go back and read the proof for vector spaces you should see exactly what they are
I think it’s more instructive to let you go and find the result and read the proof
Than just me tell you
Since finding results is a useful skill to have
ionic spade
And if you go back and read the proof for vector spaces you should see exactly w...
Yeah I’m trying to pull it up in my book
next obsidian
I don’t mean to be mean, but you’ll have to do this time and time again
ionic spade
I don’t mean to be mean, but you’ll have to do this time and time again
Yeah it’s all good
I’m just trying to find the proof, that’s the problem lol
ionic spade
Yep just found it in the book
next obsidian
Swag
ionic spade
Thanks, I’ll take a look and reread that section again
next obsidian
ionic spade
OK, so the correspondence theorem (for modules) says that there is a bijective correspondence from the submodules N of M to the submodules of M/N
So if M is Noetherian, does this mean that M/N is Noetherian since the (finitely generated) submodules are in a bijective correspondence? Or is there more to justify?
I think this is true, just like how if M/N and N are finitely generated then M must be finitely generated as well, and if M is finitely generated then M/S is finitely generated.
south patrol
So if M is Noetherian, does this mean that M/N is Noetherian since the (finitely...
yes that is one way to do it
any ascending chain of ideals of M/N lifts to some ascending chain of ideals in M which then stabilises (and hence the original thing stabilises too)
ionic spade
Ok that makes so much more sense now
south patrol
or the posets behave in the same way and being noetherian can be characterised purely in terms of posets of ideals
elder wave
you can pull this back by the canonical surjection
south patrol
ye that was what i mean
elder wave
oh oops
a similar method gives the neat result with exact sequences and noetherian modules
i like that one
south patrol
yee it's cute
sullen sage
What book can I read to understand how to calculate projections onto irreducible representations of compact groups? Specifically I want to understand how to calculate the projection from C^n \otimes C^n \to Sym^2 C^n as representations of U(n)
Fulton-Harris do this for finite groups, but literally translating their formula to U(n) doesn't quite make sense to me as it involves taking an average of vectors, while in U(n) we have an integral that works on the level of functions and not vectors
restive birch
how can you show that the group binary operation of G is closed on the kernel of the action? given that the operation of the action is not necessarily related to the group action, how would you show this?
rustic crown
what do you mean by the second statement?
thorn delta
How do I even begin to show the reverse inclusion?
I don't know exactly off the top of my head, but there are relationships between prime ideals in S^-1R and prime ideals in R. And of course there is also a relationship between the radical and prime ideals. So that might be a place to start
rustic crown
a group action is a homomorphism G --> Perm(A)
rustic crown
dw you'll understand it soon enough >.<
restive birch
i hope so
past path
I'm puzzled because showing that S^{-1}(Rad I) is contained in Rad(S^{-1}I) isn't too hard, but the opposite is borderline impossible by the same approach.
thorn delta
the proof that rad(I) is the intersection of prime ideals containing I requires a somewhat nonobvious lemma iirc. That's my intuition for why showing your inclusion may be difficult
past path
I mean, I understand why that's the case.
I'm primarily a number theorist so that makes plenty of sense to me.
cloud walrusBOT
det
rustic crown
this is the easier direction ig?
restive birch
dw you'll understand it soon enough >.<
so its basically that every g ∈ G represents a bijection from set A to itself?
thorn delta
this is the easier direction ig?
yeah amizar is stuck on the other
cloud walrusBOT
det
rustic crown
isn't that equally easy?
past path
So, I looked at it slightly different.
I got hung up on trying to take an nth root of i/s, as you have defined them.
rather than multiplying x^n by the denominator and calling that an element of i.
I was trying to land on that element rather than use it as a way to show some other containment.
past path
Are we able to multiply elements of the quotient ring by elements of the actual ring like that?
Yeah, that's part of the issue.
rustic crown
lemme think again
past path
I mean, I know we can, but can we do it in a way that preserves equality other than that notion of multiplication by elements of S?
thorn delta
(gl yall i gtg 
)
past path
Does the xs thing work when you write x as a/b?
Because it looks obvious that x^ns is in I, but is a^ns/b^n in I?
What would that even mean? You're modding out by some element of S
rustic crown
i think what i wrote more or less works, but i should have been careful
the thing is i can just replace i/s with it/st and then the argument works
for some t, we will have (x * st)^n = (st)^{n-1} (it)
cloud walrusBOT
det
rustic crown
that's the second half of the argument then
past path
Hmm, I see.
I'll chew on it for a bit and make sure it all gels but it looks reasonable.
rustic crown
so its basically that every g ∈ G represents a bijection from set A to itself?
yep
given a g, we can define the bijection g * (-) : A --> A which sends a to g * a
rustic crown
Hmm, I see.
basically i'm picking t "large" enough such that xst is an element of R and the equality is in R and not in S^{-1}R
maybe it was a bit confusing as i mixed two steps into one
tough raven
Given a matrix m by n matrix A with entries in a ring R, is there a condition on the matrix equivalent to it being injective as an R-module homomorphism from R^n to R^m?
Same for surjective.
rustic crown
i don't know anything better than the n columns being R-linearly independent (or Spanning R^m)
which is pretty much spelling out what injective and surjective means 
maybe we can say something specific in case of pids as we have smith normal form
injective if all diagonals are non-zero with n <= m and surjective if and only if all are units with m <= n?
gritty sparrow
Given a matrix m by n matrix A with entries in a ring R, is there a condition on...
A linear map from R^n -> R^m is surjective if the mth determinantal ideal is (1). It is injective if the nth determinantal ideal has annihilator 0
This generalises the usual determinant criterion
tough raven
A linear map from R^n -> R^m is surjective if the mth determinantal ideal is (1)...
Oooh
rustic crown
how would one go about proving this 
or in general what can we say about the r-th det-ideal
formal ermine
can I get a hint on this?
Show that in an integral domain $a \in R$ is prime $\iff a \in R[x]$ is prime.
cloud walrusBOT
yes yes yes no
rustic crown
quotient and reduce to showing that R/a is a domain iff (R/a)[x] is
chilly ocean
Is there a name for the kind of algebraic structure that multiplication in a field is, i.e. like an abelian group except there is an element 0 such that 0 • x = 0 and 0 has no inverse?
gritty sparrow
Well it’s a monoid i guess
chilly ocean
Well it would be a special kind of monoid.
formal ermine
quotient and reduce to showing that R/a is a domain iff (R/a)[x] is
ah so I am using the fact that $(a)$ prime $\iff$ $R/(a)$ is an integral domain?
cloud walrusBOT
yes yes yes no
rustic crown
yep
gritty sparrow
Well it would be a special kind of monoid.
Monoid with zero i’ve heard it been called
formal ermine
yep <:eeveeKawaii:556858499930259471>
oke thanks
chilly ocean
Monoid with zero i’ve heard it been called
Ah, thanks!
tough raven
A linear map from R^n -> R^m is surjective if the mth determinantal ideal is (1)...
OK, what is a determinantal ideal?
tough raven
A linear map from R^n -> R^m is surjective if the mth determinantal ideal is (1)...
Also are these iff's or just if's?
gritty sparrow
Iff
rustic crown
OK, what is a determinantal ideal?
(my guess would be the ideal generated by det of all r x r minors)
gritty sparrow
(my guess would be the ideal generated by det of all r x r minors)
Yeah
(With the convention that it is zero if r>min{m,n}
The injectivity part is mccoy’s theorem
I don’t know if the surjectivity has a special name
formal ermine
quotient and reduce to showing that R/a is a domain iff (R/a)[x] is
is this correct?
$$
\begin{equation}
\begin{align}
a &\text{ prime in } R \
&\iff (a) \text{ prime ideal of } R \
&\iff R/(a) \text{ integral domain} \
&\iff (R/(a))[x] \text{ integral domain} \
&\iff (a) \text{ prime ideal of } R[x] \
&\iff a \text{ prime in } R[x]
\end{align}
\end{equation}
$$
I already shwoed that $R$ integral domain $\iff R[x]$ integral domain
cloud walrusBOT
yes yes yes no
rustic crown
yep
cloud walrusBOT
yes yes yes no
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tough raven
is this correct?
$$
\begin{equation}
\begin{align}
a &\text{ prime in } R \\
&\...
You might want to elaborate on the second-last step. It's because (R/(a))[X] = R[X]/(a) (why?), where the (a) on the left is the principal ideal of R generated by a, whereas the (a) on the right is the principal ideal of R[X] generated by a.
tough raven
maybe add one more step which says (R/a)[x] = R[x]/(a)
Right
So
Surjective <=> columns span R^m <=> A is right-invertible
rustic crown
yep