#groups-rings-fields | Mathematics | Page 436

mild laurel Jul 19, 2019, 5:17 PM

#

you can definitely put some group structure on G/N

#

Like just make it into a cyclic group or something

#

But this group operation has nothing to do with the group operation of G

#

So G/N isn't related to G at all anymore

inland bough Jul 19, 2019, 5:18 PM

#

yeah

#

just to clarify

#

the fact that aH*bH=abH comes from the fact that G is homomorphic to G/H

mild laurel Jul 19, 2019, 5:19 PM

#

that's a way to think about it sure

#

If we have a group homomorphism from G to G/H

#

that sends a to aH

#

Then, since it's a group homomorphism, we must have that aH * bH = abH

inland bough Jul 19, 2019, 5:20 PM

#

yep

mild laurel Jul 19, 2019, 5:20 PM

#

because that's the requirement that all group homomorphisms must satisfy

inland bough Jul 19, 2019, 5:21 PM

#

say that c and a belong to the same coset, and d and b belong to the same coset

#

so then aH * bH = cH * dH

#

then it must be true that cd and ab belong to the same coset

#

my question is

#

assuming that aH * bH = abH

#

is there any way that cd and ab belong to different cosets?

mild laurel Jul 19, 2019, 5:25 PM

#

if c and a are in the same coset

#

then you have that aH = cH

#

and similarly, bH = dH

inland bough Jul 19, 2019, 5:27 PM

#

yeah

#

so no

mild laurel Jul 19, 2019, 5:28 PM

#

yep it's always true that cdH = abH

inland bough Jul 19, 2019, 5:32 PM

#

with the same variables as above

#

say there is a homomorphism from G to some group F with kernel H

#

is it true that K is always isomorphic to G/H

#

where G/H is the set of the cosets

#

i'm asking this bc when say say that H is a normal subgroup of G, this means there exists a homomorphism from G to some other group with kernel H. i'm just not sure how this implies that there exists one from G to G/H.

mild laurel Jul 19, 2019, 5:41 PM

#

This is only true if your homomorphism from G to F is surjective

inland bough Jul 19, 2019, 5:43 PM

#

hmm yeah

#

but then why is it said that G/H exists when H is a normal subgroup?

#

if being a normal subgroup is not sufficient

mild laurel Jul 19, 2019, 5:45 PM

#

What do you mean

#

G/H is always a group when H is normal

inland bough Jul 19, 2019, 5:45 PM

#

right

#

H is normal is the same as saying there exists a homomorphism from G to some group with kernel H

#

right?

#

but we specifically need a homomorphism from G to G/H in order for G/H to properly by a group

#

i don't see how this follows from H being normal

mild laurel Jul 19, 2019, 5:53 PM

#

because the function that sends a to aH is a homomorphism

inland bough Jul 19, 2019, 5:54 PM

#

how

#

this is before we assume that G to G/H is a homomorphism

mild laurel Jul 19, 2019, 5:56 PM

#

no

#

you can check that this is a homomorphism

inland bough Jul 19, 2019, 5:57 PM

#

how so

#

(ab)H=aH * bH

analog oracle Jul 19, 2019, 5:58 PM

#

Do you know how a group homomorphism is defined? You just need to use the definition

inland bough Jul 19, 2019, 5:58 PM

#

how do we prove this

#

i do

#

but how is this true

#

(ab)H=aH * bH

#

basically what i'm saying is don't assume that the mapping G to G/H is a homomorphism yet

analog oracle Jul 19, 2019, 6:02 PM

#

$ah_1bh_2 = ab(b^{-1}h_1bh_2)$ so it follows from normality

cloud walrusBOT Jul 19, 2019, 6:05 PM

#

Jamie:

📎 189069373158195201.png

patent turtle Jul 20, 2019, 6:02 AM

#

Hey, this is similar to a question I had asked yesterday (thanks @mild laurel). Given that G is a group what does G^0 represent?

tender mist Jul 20, 2019, 6:14 AM

#

If in general, talking about sets, G^n is the cartesian product Gx...xG, i.e. the set of ordered n-tuples of elements of G, G^0 can be defined to be the set consisting of only the 0-tuple or empty tuple: G^0={()}

#

This is also used to talk about nullary operations

#

If for example a binary operation is a function from G^2 to G and a unary operation is a function from G to G, a nullary operation is a function from G^0 to G

#

For a group (G,*), where * is a binary operation, one can also think about the inverses as a unary operation and the neuter element as a nullary operation

patent turtle Jul 20, 2019, 6:53 AM

#

@tender mist thanks a lot 😀

fickle brook Jul 20, 2019, 7:07 AM

#

the most reasonable interpretation for G^0 is as the trivial group

tender mist Jul 20, 2019, 7:57 AM

#

Any singleton can represent the trivial group, and G^0 is simply a set which happens to have one element

#

If one adopts the definition of X^n as the set of functions from the set with n elements to X

#

But it may be interesting to interpret through G^0 the neuter element of G as a nullary operation on G
n: G^0->G with n(emptyset)=e

#

I was taught this way, but I'd be curious to know if this interpretation is used somewhere

cloud walrusBOT Jul 20, 2019, 1:47 PM

#

idan:

📎 551139747054223388.png

stone forum Jul 20, 2019, 2:00 PM

#

what's nG?

#

@weary terrace

weary terrace Jul 20, 2019, 2:01 PM

#

You're right, it's weird.. i'll edit the question

stone forum Jul 20, 2019, 2:01 PM

#

it's not weird, I just don't know what you mean by that

weary terrace Jul 20, 2019, 2:05 PM

#

Ok, I deleted the previous version, edited question is:
Is the following correct? If so, why?

Let $H<\mathbb Z_p$ (the p-adic integers)
If $[\mathbb Z_p:H]=n$ then $n\mathbb Z_p\subset H$.

cloud walrusBOT Jul 20, 2019, 2:06 PM

#

idan:

📎 551139747054223388.png

hollow peak Jul 20, 2019, 2:21 PM

#

No

#

lets take G=Z6 and H=<2>, you have [Z6:H]=3 and 3*Z6={0;3}!=H

#

wut

#

oh no [Z6:H]=2

#

okay so you can show that H=<n>

#

because H is cyclic as a subgroup of G and you know that a generator of H has order p/n

#

so it shows what you want

weary terrace Jul 20, 2019, 5:02 PM

#

Thanks , but I wasn't referring $\mathbb Z/6\mathbb Z$. I was referring the p-adic integers $\mathbb Z_p$ where $p$ is prime.

hollow peak Jul 20, 2019, 5:03 PM

#

oh

#

mb I thought Zp was a notation for Z/pZ

cloud walrusBOT Jul 20, 2019, 5:08 PM

#

idan:

📎 551139747054223388.png

inland bough Jul 20, 2019, 9:17 PM

#

what is the motivation for defining the product of cosets
aN * bN = (ab) N
i don't see why we couldn't have defined the product of cosets as aN * bN = (a^-1 b^-1) N

#

i know that this forces the quotient group to be homomorphic to the original group

#

but that does not answer the question

magic owl Jul 20, 2019, 9:47 PM

#

I’m not sure I understand your thinking?

#

I suppose you could do that, but as far as I can tell you’re just taking the automorphism induced by -1

#

Or sorry not by -1 but by sending g to -g

#

Oh wait that’s only an automorphism if your group is abelian

#

Yeah I’m not sure what you’re getting at with this construction, and I def don see how you get an isomorphism from the original to the quotient

mild laurel Jul 20, 2019, 9:56 PM

#

@magic owl if ab gets sent to b^(-1) a^(-1) then that's an automorphism

#

Even for non abelian

magic owl Jul 20, 2019, 9:59 PM

#

Yeah that’s what made me realize what he said wasn’t

#

But either way the fact that it works for abelian disproves his isomorphism claim

tender mist Jul 21, 2019, 12:07 AM

#

So basically the question becomes: what happens if given a group (G,•) one defines another operation on G by a*b := b•a ?

woven delta Jul 21, 2019, 12:52 AM

#

It becomes the opposite group G^op, which is isomorphic to G

#

Every left action by G corresponds to a right action by G^op, and vice versa

wind steeple Jul 21, 2019, 9:06 AM

#

a -> a^-1 is an isomorphism

magic owl Jul 21, 2019, 12:55 PM

#

only in abelian groups because (ab)^-1 = b^-1 a^-1 not a^-1 b^-1

#

Not a homomorphism

tender mist Jul 21, 2019, 2:25 PM

#

Thanks Liquid

#

I think Zak is saying that a -> a^-1 is an isomorphism between G and G^op

brisk granite Jul 21, 2019, 4:41 PM

#

Are there only four isomorphism theorems?

oblique river Jul 21, 2019, 4:44 PM

#

there are more but you don't unlock them until you get to grad school

#

and the Final Isomorphism Theorem is only unlocked once you become a prof

wind steeple Jul 21, 2019, 4:52 PM

#

yes, I was answering to the previous question

#

a->a^-1 is an isomorphism between G and G^op

bleak abyss Jul 21, 2019, 6:53 PM

#

~~The final one is one that would've made your thesis much easier~~

topaz solar Jul 21, 2019, 10:53 PM

#

~~the final one is what your thesis is on~~

abstract sandal Jul 22, 2019, 4:33 PM

#

(riley hobson and bence) uses the fact that two groups have the same number of elements and the same number of elements of any particular order exclusively to prove that they are isomorphic. is this correct?

hollow peak Jul 22, 2019, 5:05 PM

#

if they are finite yes

#

hmm it is if they are abelian for sure

#

Im not sure for non abelian groups

cloud walrusBOT Jul 22, 2019, 5:48 PM

#

creamy shits:

📎 210924359240384513.png

hot lake Jul 22, 2019, 5:56 PM

#

isn't associativity an axiom of categories

#

I think you just have to check that the composition of isomorphisms is an isomorphism

#

and the inverse of one is one

#

both are very straightforward

#

yeah the identity is free

#

I don't think you did actually

#

I would expect to see things about the definition of what is an isomorphism

#

it's not just any morphism from X to X

#

yeah

#

yeah

#

and f^-1 ° f = idX and f ° f^-1 = idY

#

what ?

#

i'm just completing your incomplete definition of isomorphism

#

f : X -> Y is an isomorphism iff there is a f-1 : Y -> X such that f°f-1 = idY and f-1°f = idX

#

with that in hand you have to show that if f : X -> X and g : X -> X are isomorphisms then f°g is also one

#

by explaning who is the inverse of f°g

#

and checking the related equalities

cloud walrusBOT Jul 22, 2019, 6:05 PM

#

creamy shits:

📎 210924359240384513.png

#

creamy shits:

📎 210924359240384513.png

#

creamy shits:

📎 210924359240384513.png

#

creamy shits:

📎 210924359240384513.png

#

creamy shits:

📎 210924359240384513.png

#

creamy shits:

📎 210924359240384513.png

#

creamy shits:

📎 210924359240384513.png

gentle pendant Jul 22, 2019, 6:16 PM

#

@abstract sandal @hollow peak the answer is no for non-abelian groups.

Eg/
Consider
G = Z/4Z x Z/4Z and H = Z/2Z x Q where Q is the quaternion group of order 8.

Both of these have 1 element of order 1, 3 elements of order 2 and 12 elements of order 4.

But they are non-isomorphic, as can be seen by comparing the cardinalities of {x^2: x in G}

hot lake Jul 22, 2019, 6:41 PM

#

you should explain how (g°f)°(f-1°g-1) = idX follows from the axioms and properties that you have

magic owl Jul 22, 2019, 6:44 PM

#

Tbh

#

If you’re doing category theory I feel like composition of isomorphisms is an isomorphism

#

Is generally something you’ve had mentioned in class before yeah?

cloud walrusBOT Jul 22, 2019, 7:26 PM

#

creamy shits:

📎 210924359240384513.png

#

creamy shits:

📎 210924359240384513.png

#

creamy shits:

📎 210924359240384513.png

hot lake Jul 22, 2019, 7:31 PM

#

no

#

you haven't justified that f-1 ° g-1 is the inverse of g°f because you haven't showed that (g°f)°(f-1°g-1) = id

#

you only said "it is the inverse, trust me on that"

#

it is sad to do category theory and never use the axioms of category theory

#

on the other hand maybe it feels too obvious to warrant a proper proof

#

you have to show that (g°f)°(f-1°g-1) = id follows from f°f-1 = id, g°g-1 = id, and the axioms of category theory

#

(i'm not putting the subscripts on id cuz I would have to look up on which object they are)

#

(and they aren't exactly relevant for the proof to work)

magic owl Jul 22, 2019, 7:38 PM

#

(and no one ever remembers to use them)

#

It’s almost as bad as the subscripts on the differential maps of a complex

hot lake Jul 22, 2019, 7:40 PM

#

I never really managed to do anything nice with cohomology :(

cloud walrusBOT Jul 22, 2019, 8:16 PM

#

creamy shits:

📎 210924359240384513.png

magic owl Jul 22, 2019, 9:09 PM

#

Yeah that’s fine

#

Grammatically weird

#

But fine

worthy kindle Jul 22, 2019, 9:15 PM

#

OhNo_cat

brisk granite Jul 23, 2019, 12:36 PM

#

I'm trying to show that $N \triangleleft G \implies n_{p}(G/N) \leq n_{p}(G)$. So far, I know that I can construct a bijective function $f$ from the subgroups $G/N$ to the subgroups of $G$ containing $N$ such that, for any subgroup $K$ in $G/N$, $|f(K)| = |K||N|$. \

If we say $|N| = p^{\beta}m$ and $|G| = p^{\alpha}n$, then $|G/N| = p^{\alpha-\beta}\frac{n}{m}$. Hence, if we consider any sylow $p$-subgroup $K$ of $G/N$, then $|f(K)| = p^{\alpha}m$, and, from sylow one, we know that $f(K)$ must have at least one sylow $p$-subgroup which, in this case, turns out to be sylow $p$-subgroup of $G$. My only problem now is that I don't know how to show that, for two distinct sylow $p$-subgroups $H$ and $K$ of $G/N$, there are no common sylow $p$-subgroups of $f(H)$ and $f(K)$ (I don't actually know if this is true).

Should I continue with this reasoning? Does this look promising?

cloud walrusBOT Jul 23, 2019, 12:37 PM

#

∀ScoopityPoop, Scoop ≠ Poop:

📎 375866732226543616.png

hot lake Jul 23, 2019, 12:47 PM

#

what is np ?

brisk granite Jul 23, 2019, 1:10 PM

#

number of sylow p subgroups

#

n_p(G) is the number of sylow p subgroups in a group G

torn gate Jul 23, 2019, 5:22 PM

#

$G = \mathbb Z_{10}, H = {0, 5}$ (Explain why $G/H$ is isomorphic to $\mathbb Z_{5}$).

The operation table can be renamed. However, the additive group of integers is cyclic (the generator being 1), an isomorphism must preserve such the cyclical nature so what is the generator element of $G/H$?

cloud walrusBOT Jul 23, 2019, 5:22 PM

#

birb:

📎 248887037447503872.png

torn gate Jul 23, 2019, 5:23 PM

#

There must be a generator right? Did I make a shoddy assumption somewhere or am I missing the obvious? (Sorry if I don't reply, the internet here is spotty)

mild laurel Jul 23, 2019, 5:24 PM

#

There is a generator?

hollow peak Jul 23, 2019, 5:25 PM

#

is Z10=Z/10Z here?

torn gate Jul 23, 2019, 5:28 PM

#

Yeah, it's the additive group of integers modulo 10

#

If it's isomorphic surely it has a generator since it must also be cyclic?

mild laurel Jul 23, 2019, 5:31 PM

#

That's true yes

hollow peak Jul 23, 2019, 5:35 PM

#

yeah it must be cyclic

#

but why? (you have to argue here)

torn gate Jul 23, 2019, 5:36 PM

#

Nah, I get why, but what's the generator?

#

I can't figure out what it is

hollow peak Jul 23, 2019, 5:36 PM

#

any element

#

any element different from 0

torn gate Jul 23, 2019, 5:36 PM

#

Aww... I'm slow

#

That's obvious now that I think about it

hollow peak Jul 23, 2019, 5:37 PM

#

that's Bezout

torn gate Jul 23, 2019, 5:37 PM

#

Sorry to bother you lot 😄

mild laurel Jul 23, 2019, 5:41 PM

#

Do you understand why any of those generates the group

torn gate Jul 23, 2019, 5:45 PM

#

Yeah

#

This is impart due to my inexperience, I know they all do I just can't write the proof why if that makes sense?

mild laurel Jul 23, 2019, 6:00 PM

#

Maybe you don't understand the definitions well enough to work rigorously with them?

torn gate Jul 23, 2019, 6:19 PM

#

Most likely

#

I've been going through a book in my own time so I've been flying solo 😄

magic owl Jul 23, 2019, 8:48 PM

#

Wait @hollow peak that’s not correct

#

You new Zp for prime p to have all elements as generators

#

For example take 5 in Z_10

#

It won’t generate 2

hollow peak Jul 23, 2019, 8:49 PM

#

10 is not prime

magic owl Jul 23, 2019, 8:49 PM

#

Or anything other than 0

#

I know

#

That’s my point, they were talking about Z/10Z

hollow peak Jul 23, 2019, 8:50 PM

#

they quotiented

#

by a group of order 2

#

so it was isomorphic to Z/5Z

magic owl Jul 23, 2019, 8:50 PM

#

Oh I’m so sorry

#

You’re correct

hollow peak Jul 23, 2019, 8:50 PM

#

np

magic owl Jul 23, 2019, 8:50 PM

#

My b, that’s what I get for walking and typing lol

stone fulcrum Jul 23, 2019, 10:28 PM

#

@torn gate
Important you see the conversation after
An element generates Zn iff it is coprime to n

#

2 does not generate Z10
3 does generate Z10

analog oracle Jul 24, 2019, 1:07 AM

#

mood

#

though i'm also open to salvation now if someone wants to spare me the ongoing torture of trying to understand a 100 year old paper that even my supervisors are complaining about :p

magic owl Jul 24, 2019, 1:14 AM

#

What’s the paper

analog oracle Jul 24, 2019, 1:15 AM

#

@magic owl

📎 Miller.pdf

magic owl Jul 24, 2019, 1:15 AM

#

Why are you reading it out of curiosity

#

Are you interested in finite group theory?

analog oracle Jul 24, 2019, 1:19 AM

#

yeah, we're working on classifying all groups that have alpha = 3/4 where alpha is the number of cyclic subgroups divided by the order of the group, we think we have all of them and from a remark on a paper about it it seems like this paper might classify the nilpotent 2-groups we're dealing with. We've practically done it with GAP but we now have to deal with proving it. And tbh this would probably be over my head with modern language but this paper is really pushing me towards insanity

#

it also doesn't help that since I'm transferring from the UK to the US my summer was cut in half so what was going to be a 16 week project is now an 8 week project with 2 weeks left

#

though it's probably safe to assume this will turn into my master's thesis topic

magic owl Jul 24, 2019, 1:22 AM

#

It’s a cool topic for sure

placid pond Jul 24, 2019, 1:22 AM

#

is it

magic owl Jul 24, 2019, 1:22 AM

#

Are you in any of the higher math discord’s?

#

There are people who can help you better in there probably

analog oracle Jul 24, 2019, 1:22 AM

#

nope, this is the only math discord I'm in

woven delta Jul 24, 2019, 1:22 AM

#

I don't know if there are any group theorists on hopf

#

Who would want to be a group theorist lmao

magic owl Jul 24, 2019, 1:23 AM

#

Let me shower real quick and I’ll invite you to one with an algebraist or two

analog oracle Jul 24, 2019, 1:23 AM

#

me!

magic owl Jul 24, 2019, 1:23 AM

#

People who get lonely

placid pond Jul 24, 2019, 1:23 AM

#

finite group theorist

#

so it's even worse

analog oracle Jul 24, 2019, 1:23 AM

#

thanks :)

magic owl Jul 24, 2019, 1:23 AM

#

Because they prefer to be in groups

analog oracle Jul 24, 2019, 1:23 AM

#

eh that's still me :p

magic owl Jul 24, 2019, 1:23 AM

#

It was a pun, I think the field is cool

#

If a little out of my wheelhouse

analog oracle Jul 24, 2019, 1:25 AM

#

i'm not sure it's what i want to do long term but i'm really enjoying this even though i feel 100% out of my depth. I also loved real analysis when I did it like 2 years ago (complex analysis not so much) and I enjoyed IMO level number theory so I might do something along those lines too, idk. I have a while to figure things out since I'm graduating in 2021 and then hopefully going into a PhD

placid pond Jul 24, 2019, 1:27 AM

#

@analog oracle it's difficult for me to bring myself to care about these kinds of problems at all, but good for you if you enjoy it i suppose

#

it's always been hard for me to imagine people enjoying finite group theory

analog oracle Jul 24, 2019, 1:30 AM

#

yeah i don't really have an answer for when people ask me why do we care and I got rejected from a research grant because it's not a "central problem" or whatever which basically sounded to me as "who cares" but I thought it would be a fun summer project, now we're all realizing this is a way bigger project than just a summer thing and I guess while it's still fun I might as well stick it out and see what happens

#

it's not like I have a better idea besides just vague "these areas are cool" :p

placid pond Jul 24, 2019, 1:34 AM

#

@analog oracle sure, i have no intuition for which kinds of problems in finite group theory are easy and which ones are hard

#

but picking problems that are easy and then working on them should be fine even if nobody cares about the problem

#

at least you get an experience of what the field is like

analog oracle Jul 24, 2019, 1:36 AM

#

i don't think my supervisors do either. Though surprisingly we expected the non-nilpotent groups to be way harder than the nilpotent case because none of the papers about it went anywhere near that but nope! we are literally getting a single group crossed with powers of C2

placid pond Jul 24, 2019, 1:38 AM

#

@analog oracle i'm already confused by his description of what seems like a semidirect product in the second paragraph

#

no, i'm wrong

analog oracle Jul 24, 2019, 1:39 AM

#

it took us like 2 hours to get through the first and second paragraph and then we were annoyed about how simple it was compared to the explanation. But I don't have my notes on me and I'm too lazy to go fetch them

brisk granite Jul 24, 2019, 1:40 AM

#

um, so, I don't wanna bother anyone, but I kinda need help with something I posted earlier. It got buried and nobody responded; so, can I post it again?

analog oracle Jul 24, 2019, 1:41 AM

#

fine by me!

brisk granite Jul 24, 2019, 1:42 AM

#

ok, then

cloud walrusBOT Jul 24, 2019, 1:42 AM

#

∀ScoopityPoop, Scoop ≠ Poop:

📎 375866732226543616.png

placid pond Jul 24, 2019, 1:47 AM

#

@brisk granite it's promising enough, i suppose, though the last statement is not how you should be going about it

brisk granite Jul 24, 2019, 1:48 AM

#

what should I do?

placid pond Jul 24, 2019, 1:48 AM

#

think about the transitive action of G on the sylow-p subgroups of G versus those of G/N

#

see what you can make of that

brisk granite Jul 24, 2019, 1:49 AM

#

Do you mean I should consider the transitive action of G/N on its sylow p-subgroups as well?

placid pond Jul 24, 2019, 1:49 AM

#

well, the action of G/N lifts to an action of G

#

so yeah

#

not sure how to give any more hints without giving away the entire solution

brisk granite Jul 24, 2019, 1:50 AM

#

ok thankyou

bleak finch Jul 24, 2019, 2:25 AM

#

Let R be a commutative ring, N(R) be the nilradical. Then N(R/N(R)) = 0.

#

Could you give me a hint?

placid pond Jul 24, 2019, 2:26 AM

#

@bleak finch i don't know how to, this is too simple to give a hint for

#

i can tell you the solution

bleak finch Jul 24, 2019, 2:26 AM

#

Don't

#

I know, generally speaking, that if take the quotient of a ring by an ideal, then every element of that ideal will map to zero in the quotient ring.

#

But that isn't in proofese

placid pond Jul 24, 2019, 2:28 AM

#

proofese?

#

ah

#

that doesn't prove this statement, no

bleak finch Jul 24, 2019, 2:28 AM

#

Was my statement at least correct?

placid pond Jul 24, 2019, 2:28 AM

#

try to reformulate this statement as a statement about the arithmetic of R itself

#

yes, it was correct

#

try to get rid of the nilradical and the quotient entirely, just try to write it as an arithmetic statement about R

#

what does it mean to say that N(R/N(R)) = 0?

bleak finch Jul 24, 2019, 2:30 AM

#

It means that for any a in R/N(R), if there exists an n such that a^n = 0, then a = 0.

placid pond Jul 24, 2019, 2:30 AM

#

yes, so for any a in R, if there exists n such that a^n \in N(R), then a \in N(R)

#

does that make it easier to see why it has to be true

bleak finch Jul 24, 2019, 2:34 AM

#

a^{n+1} in N(R). a^{n + 1} = a^n a = 0 a = 0

#

Wait

#

That was obvious

#

Nevermind.

#

So N(R) = {0, a, a^2, ..., a^{n-1}}

#

The kernel of R with respect to the projection operator is {r in R : phi(r) = 0}

uncut girder Jul 24, 2019, 2:56 AM

#

The elements of R/N(R) are literally elements of R

#

You just have equivalence classes

#

So any nilpotent element in R/N(R) is nilpotent in R

#

Therefore N(R/N(R)) =N(R) =0 as an equivalence class

bleak finch Jul 24, 2019, 3:08 AM

#

painful death by integration, for any rin g R, it is not necessary that N(R) = {0}

#

lots of concepts floating around are mixed up I'm sorry

#

painful death by integration, the elements of R/N(R) are a subgroup of R, yes?

#

but Q[x]/(x^2 + 1) is a subring of Q[x]

#

Q[x]/(x^2 + 1) is a subset of Q[x]?

#

but 1 + x is an element of both Q[x] and Q[x]/(x^2 + 1)

#

because 1 + 1 + 1 = 0 in the latter but not in the former?

#

does that matter?

#

so if 1 behaves differently from 1 in Z then the latter does not have the same elements as the former?

uncut girder Jul 24, 2019, 10:24 AM

#

Yes, I was not careful in my wording. Elements in the quotient are not elements of the ring. Elements in the quotient are represented by elements in the ring.

#

@bleak finch

brisk granite Jul 24, 2019, 10:43 AM

#

@placid pond Could you give me the answer?

brisk granite Jul 24, 2019, 11:00 AM

#

Oh wait

#

I think I got it

#

Suppose you have a sylow $p$-subgroup $P$ of $G$ that is also a sylow p-subgroup of two subgroups $H$ and $K$. Then, $PH/H \cong P/(P \cap H) \implies PH/H \cong P$ and $PK/K \cong P/(P \cap K) \implies PK/ \cong P$. Thus, $PH/H \cong PK/K \implies H = K$.

cloud walrusBOT Jul 24, 2019, 11:04 AM

#

∀ScoopityPoop, Scoop ≠ Poop:

📎 375866732226543616.png

brisk granite Jul 24, 2019, 11:08 AM

#

oh, um, wait

#

Does this make sense?

chilly ocean Jul 24, 2019, 11:10 AM

#

Isn't $P\cap H = P?$

cloud walrusBOT Jul 24, 2019, 11:11 AM

#

Obama Llama:

📎 603498934433873920.png

brisk granite Jul 24, 2019, 11:25 AM

#

@chilly ocean yea, I see that now. There's more stuff wrong with it too...

placid pond Jul 24, 2019, 11:40 AM

#

ignore the hint i gave earlier, that doesn't work i think

#

@brisk granite i think you show that the map {sylow-p subgroups of G} -> {sylow-p subgroups of G/N} given by P -> PN/N is surjective

#

if you have any sylow-p subgroup of G/N that lifts to a subgroup H of G, it contains some sylow-p subgroup of G, say P; and it also contains N, so it contains the composite group PN which lifts to a sylow-p subgroup of G/N, since its index is coprime to p

#

and that implies PN = H

#

so i think this argument is good

uncut girder Jul 24, 2019, 3:29 PM

#

Is the ring of power series in z with coefficients in C with an infinite radius of convergence Nothereian?

#

Sidenote: can you give an example of a subring of a noetherian ring which is not noetherian?

placid pond Jul 24, 2019, 3:32 PM

#

@uncut girder any integral domain embeds into its field of fractions, which is noetherian

#

so just pick any non-noetherian domain and its field of fractions

uncut girder Jul 24, 2019, 3:33 PM

#

Any field has only the 0 ideal so its noetherian, sure

tender mist Jul 24, 2019, 3:42 PM

#

(0) is the only maximal ideal

placid pond Jul 24, 2019, 3:42 PM

#

@uncut girder for your first question, you can find entire functions on C with prescribed zeroes via the weierstrass factorization theorem, so take N as embedded into C on the real line and consider the ideals I_n of entire functions vanishing at {n, n+1, n+2, .....} for n = 0, 1, 2, 3.....

#

I_n is an ascending chain that doesn't terminate

#

so the ring is not noetherian

#

but the germs of analytic functions at any point are noetherian, this is a consequence of the preparation theorem

#

so in some sense the non-noetherian behavior comes from the global properties of the ring

uncut girder Jul 24, 2019, 3:48 PM

#

Do you need Weirerstrass factorization theorem for this? Because I never learned it

placid pond Jul 24, 2019, 3:49 PM

#

no

#

preparation theorem is for the second part

#

the first part you can do explicitly too, you don't need the factorization theorem

#

the factorization theorem shows all of the inclusions in the chain I_0 < I_1 < I_2 .... are proper

#

which you need to do ,but you can do it in other ways also

#

try thinking of how you might construct an entire function that only vanishes at natural numbers, for example

#

it's a good exercise

#

once you have that function you can get all of the other ones by just translating it to the right

uncut girder Jul 24, 2019, 3:50 PM

#

I didn't even know that was possible

#

Is sin entire?

placid pond Jul 24, 2019, 3:50 PM

#

sin is entire

#

but sin vanishes at integer multiples of pi

#

so you can make it Z, but not quite N

uncut girder Jul 24, 2019, 3:51 PM

#

So do sin^2

#

And then rescale

placid pond Jul 24, 2019, 3:51 PM

#

sin^2 still only vanishes at integer multiples of pi

#

do you mean sin(z^2)

uncut girder Jul 24, 2019, 3:52 PM

#

Oh that's right

#

Yes

placid pond Jul 24, 2019, 3:53 PM

#

you're close

#

this one doesn't quite give you N

#

it gives you some pure imaginary zeroes

#

and pure real ones

#

on both sides of the origin

#

little weird shape

#

well, the obvious idea is, let's ignore N for the moment, since for the problem we don't need to make it work for N

#

consider a product like (1 - z/exp(0))(1 - z/exp(1))(1 - z/exp(2)).....

#

infinite product

#

it's clearly convergent for all z, and vanishes only when z is equal to one of exp(0), exp(1), exp(2)....

#

i use exp of natural numbers instead of natural numbers directly because otherwise it's not so obvious it converges

#

in fact generally it doesn't converge in that case

#

to get N you need to be more clever about how you construct the product

uncut girder Jul 24, 2019, 3:56 PM

#

Why does it converge even

placid pond Jul 24, 2019, 3:56 PM

#

take the logarithm

#

to turn it into a sum

#

and use the asymptotic log(1-z) = -z + O(z^2)

#

you get a geometric sum

#

so it's convergent

#

and nonzero so long as z doesn't equal any of the exp(n) for n \in N

#

it's also clearly entire since on compact subsets it's the uniform limit of entire functions and blabla

uncut girder Jul 24, 2019, 3:59 PM

#

"and use the asymptotic log(1-z) = -z + O(z^2)"

#

I dont follow

placid pond Jul 24, 2019, 4:00 PM

#

it's the first order taylor expansion

#

of log(1-z) around z = 0

#

you take the logarithm of the partial products, and that gives you some partial sums where the terms are log(1 - z/exp(n))

#

you approximate this by -z/exp(n)) and pick up an error term of O((z/exp(n))^2)

#

and that second guy is summable, in fact it decays really fast

#

so that means your original product is convergent and nonzero

#

i didn't think i would have to explain so much for this construction, if you really don't know any of this i should think of a different solution

#

but this one seemed very obvious to me

#

@uncut girder yeah, you can do your argument using sin alone actually

#

define I_n = { entire functions which vanish at 2^n Z }

#

then I_n clearly form an ascending chain of ideals

#

and you can show the inclusions are all proper by using sin(pi z/2^n)

#

which is an entire function vanishing exactly on the set 2^n Z

#

so it's just the same idea but simpler execution

uncut girder Jul 24, 2019, 4:06 PM

#

Is there a reason you are using 2^n Z and not nZ

placid pond Jul 24, 2019, 4:06 PM

#

if i use nZ there is no chain of ideals

#

3Z is not a subset of 2Z

uncut girder Jul 24, 2019, 4:06 PM

#

Oh right

#

I like this solution, I will work it out for myself now

uncut girder Jul 24, 2019, 5:01 PM

#

Thank you @placid pond

placid pond Jul 24, 2019, 5:01 PM

#

no problem

uncut girder Jul 24, 2019, 5:02 PM

#

The next question is about whether or not the following ring is noetherian:
The ring of polynomials in z whose first k derivatives vanish at 0.
My question does this mean p(0)=0, ie do we count the 0'th derivative as well?

placid pond Jul 24, 2019, 5:03 PM

#

i doubt it'll make a difference

#

you can just use the idea of the division algorithm, no?

uncut girder Jul 24, 2019, 5:07 PM

#

Yeah over C

#

Ok I'll try on my own first

placid pond Jul 24, 2019, 5:08 PM

#

hm, actually it's not quite so simple

#

let me think about this

#

@uncut girder ah, this is a nice problem

#

the answer is yes, for p(0) = 0 at least

#

you need to think of a cute idea

#

in retrospect it's a bit obvious but whatever

uncut girder Jul 24, 2019, 5:27 PM

#

First of all, this ring is simply (x^(k+1)), correct?

placid pond Jul 24, 2019, 5:27 PM

#

yes

uncut girder Jul 24, 2019, 5:34 PM

#

It's a theorem in the book that if A is Noetherian and f is a homomorphism of A onto a ring B, then B is Noetherian.

So C[x] is noetherian (which is proved in the book as Hilberts Basis theorem, also its PID by division algorithm)

Now the map f:C[x]->(x^(k+1)) given by
f(p) =px^(k+1) is a surjective ring homomorphism so (x^(k+1)) is Noetherian.

#

Is this what you had in mind? @placid pond

placid pond Jul 24, 2019, 5:36 PM

#

that is unfortunately not a ring homomorphism

#

so no

uncut girder Jul 24, 2019, 5:36 PM

#

Wat

placid pond Jul 24, 2019, 5:36 PM

#

it's a module homomorphism

#

it's not multiplicative

#

f(pq) = pq x^(k+1), but f(p) f(q) = pq x^(2k+2)

#

not equal

uncut girder Jul 24, 2019, 5:37 PM

#

Right

placid pond Jul 24, 2019, 5:37 PM

#

also, i think the argument i have in mind requires you to have C as a subring

#

if you impose f(0) = 0, let's say you take, idk, k = 1

#

so you're dealing with (x^2)

#

what do you do with an ideal like (x^2, x^2/2, x^2/3, x^2/4, .....)

#

i don't think this one is finitely generated

#

obviously if you have the elements of C as a subring then you have no such problem

uncut girder Jul 24, 2019, 5:40 PM

#

f is a polynomial?

placid pond Jul 24, 2019, 5:40 PM

#

yeah

#

as in

#

in the derivatives you require to be 0

#

you also require the zeroth derivative to be 0

#

that is, the polynomial itself

#

if you do that, i.e. you're working with (x^(n+1)) C[x]

#

then my thing seems like an ideal that's not finitely generated

#

but i think it's different if you work with C + (x^(n+1)) C[x]

uncut girder Jul 24, 2019, 5:43 PM

#

I'm hungry, I'm gonna eat lunch

placid pond Jul 24, 2019, 5:43 PM

#

okay

bleak finch Jul 24, 2019, 5:45 PM

#

Suppose I have a set of left cosets of a group by a subgroup that is not normal (so leftcosets != rightcosets).

#

How do I show that no group structure is possible on this set?

placid pond Jul 24, 2019, 5:46 PM

#

@bleak finch you can't, because a group structure is possible on the set

#

it's just not compatible with the action of G on the cosets

#

any set can be given a group structure

bleak finch Jul 24, 2019, 5:47 PM

#

What is the action of G on the coset?

placid pond Jul 24, 2019, 5:48 PM

#

say your subgroup is H

#

you have a natural map G -> {left cosets of H in G} given by g -> gH

#

this is a function, and if H is normal and you give the set its quotient group structure, it's a group homomorphism

#

that is, it's true that (g_1 H)(g_2 H) = (g_1 g_2) H for any g_1, g_2

#

this property requires H to be a normal subgroup

#

which you can see by, say, setting g_2 = g_1^(-1)

#

but if you just want to put a group structure on the set {left cosets of H in G}, you can easily do that

#

doesn't require H to be normal

bleak finch Jul 24, 2019, 7:37 PM

#

Starfall,

#

How is (g_1 H)(g_2 H) defined?

placid pond Jul 24, 2019, 7:37 PM

#

when H is normal, it's defined as (g_1 g_2) H, and you show that this is well defined

#

that is, it doesn't depend on the choice of g_1 or g_2

#

but if you just want to put a group structure on the coset space, you can do that however you want

bleak finch Jul 24, 2019, 7:39 PM

#

What happens if I say (g_1H)(g_2H) = (g_1g_2)H when H is NOT normal?

#

Don't you know what emerges?

placid pond Jul 24, 2019, 7:41 PM

#

you can't say it

#

because it's not well defined

bleak finch Jul 24, 2019, 7:41 PM

#

Starfall, do you know what an emergency is?

placid pond Jul 24, 2019, 7:41 PM

#

it depends on the choice of representative you pick for each coset

#

not in this context

bleak finch Jul 24, 2019, 7:43 PM

#

An emergency is when something breaks and undefined behavior occurs. In medicine when a body part breaks and symptoms occur we say "this is an emergency!" Emergency = emergent phenomenon.

placid pond Jul 24, 2019, 7:44 PM

#

i'm lost

bleak finch Jul 24, 2019, 7:45 PM

#

In order to define an action on a left coset, we need an agreement on the choice of representatives beforehand.

placid pond Jul 24, 2019, 7:45 PM

#

not when H is normal

#

which is the entire point

#

if H isn't normal then you're correct

#

though it does depend on what you mean by "action", the action of any group of the left cosets of any subgroup is well defined

#

the action by left multiplication, that is

#

this one doesn't depend on any choice of representative

bleak finch Jul 24, 2019, 7:47 PM

#

So what you're saying is that if (g_1H)(g_2H) = (g_1g_2)H when H is not normal, an emergent phenomenon happens where there might be a different action (g_1H)(g_2H) = (g_1g_2)H for two different choices of representatives.

#

I am going to work out A_4 and A_3 (latter is not normal) to test this idea out.

placid pond Jul 24, 2019, 7:49 PM

#

uh, okay

#

i'm saying g_1H = g_2H may not imply (g_1 g_3)H = (g_2 g_3)H in general

bleak finch Jul 24, 2019, 8:06 PM

#

Why does that g_1H = g_2H may not imply (g_1 g_3)H = (g_2 g_3)H mean it's not well-defined?

bleak finch Jul 24, 2019, 9:12 PM

#

Starfell,

#

typically when a function is not well-defined it means that x = y does not imply that f(x) = f(y)

#

So when you say that the morphism g --> gH is not defined for unnormal subgroups, that typically would mean that g_1 = g_2 does not imply g1_H = g_2H. But that is not the case.

mild laurel Jul 24, 2019, 9:32 PM

#

@bleak finch it works fine as a set function

#

But not as a group homomorphism

chilly ocean Jul 24, 2019, 10:00 PM

#

hmmmm

placid pond Jul 24, 2019, 10:02 PM

#

@chilly ocean ?

chilly ocean Jul 24, 2019, 10:03 PM

#

yes

placid pond Jul 24, 2019, 10:04 PM

#

i was wondering what your "hmmmm" was about

chilly ocean Jul 24, 2019, 10:13 PM

#

feigned pensiveness

placid pond Jul 24, 2019, 10:13 PM

#

ah

unreal raptor Jul 24, 2019, 10:49 PM

#

could someone explain what a free quadratic module is (in relation to free modules)?

bleak finch Jul 26, 2019, 4:30 AM

#

I don't understand why abstract algebra textbooks keep saying "ideals in ring theory are like normal groups in group theory"

uncut girder Jul 26, 2019, 4:31 AM

#

Because

#

You can quotient by ideals

#

You cant quotient by subrings

mild laurel Jul 26, 2019, 4:33 AM

#

@bleak finch What don't you understand

bleak finch Jul 26, 2019, 4:34 AM

#

For a subring S, let rS, Sr, represent the left and right cosets.

#

If rS = Sr for all r then is S an ideal?

mild laurel Jul 26, 2019, 4:39 AM

#

For rings, if you take all the elements and take the multiplication element on them

#

It's not always a group

#

So properties of left/right cosets you have for groups don't hold if you try to do it with the multiplication operation

bleak finch Jul 26, 2019, 5:43 AM

#

So ideals are a tenuous analogue to normal groups

mild laurel Jul 26, 2019, 5:44 AM

#

It's really not tenuous at all

#

Instead of looking at multiplication like you did

#

You just let your cosets be r + S

bleak finch Jul 26, 2019, 5:45 AM

#

what do ideals have in common with normal groups besides the fact that both are used for quotient/coproduct constructions.

mild laurel Jul 26, 2019, 5:45 AM

#

And then, since your additive group is an abelian group, S is a normal subgroup, so this forms a quotient group just fine

bleak finch Jul 26, 2019, 5:45 AM

#

"since your additive group is an abelian subgroup, S is a normal subgroup"

#

An abelian subgroup is not necessarily normal

mild laurel Jul 26, 2019, 5:46 AM

#

sorry, typo

#

normal subgroups are kernel of group homomorphisms

#

just like ideals are kernels of ring homomorphisms

bleak finch Jul 26, 2019, 5:47 AM

#

first isomorphism theorem and that's just a rephrase of "normal/ideals are used for coproducts"

#

😃

mild laurel Jul 26, 2019, 5:48 AM

#

I mean sure

#

if you want to say it like that

bleak finch Jul 26, 2019, 5:49 AM

#

Now I have a question

#

Composition series when restricted to cyclic groups are a kind of unique factorization right?

mild laurel Jul 26, 2019, 6:13 AM

#

I'm not sure what you mean by restricted to cyclic groups

#

Composition series are unique, it's called the Jordan Holder theorem i think

bleak finch Jul 26, 2019, 7:31 AM

#

Composition series of a cyclic group of order n = what you would get if you took the prime factorization of n

sonic hatch Jul 26, 2019, 12:12 PM

#

Hi, I got a bit of an odd question for any abstract thinkers out there -
What would a universe be like with ' i ' spacial dimensions?

#

Its more geometry than algebra

brisk granite Jul 26, 2019, 12:23 PM

#

it'd be pretty neat

magic owl Jul 26, 2019, 1:47 PM

#

This question doesn’t really make sense as normally we only define dimension to make sense for whole numbers

spring gate Jul 26, 2019, 3:46 PM

#

Was reading this smbc http://smbc-comics.com/comic/a-new-set-of-numbers
and the property described seems like some generalization of irreducibility to structures without identity, wondering if this was interesting enough to explore

Saturday Morning Breakfast Cereal -

#

Undergrad student who only just took basic ring theory last quarter so my understanding isnt that deep

#

An element p in a structure <S,*> such that for all a, b in S, a*b ≠ p

#

Seems to me that such elements do exist, i.e. in 2Z 2 isnt a product of any other elements of the set but idk what properties there are to be gleaned from this

weary terrace Jul 26, 2019, 6:08 PM

#

Consider the additive group $G$. For each $g\in G$, let $g+g+...+g=ng$ be an element of $nG$.

How does one calculate $[G:nG]$?

cloud walrusBOT Jul 26, 2019, 6:08 PM

#

idan:

📎 551139747054223388.png

mild laurel Jul 26, 2019, 6:12 PM

#

Are you assuming that G is abelian?

weary terrace Jul 26, 2019, 6:14 PM

#

Yes, thank you @mild laurel

mild laurel Jul 26, 2019, 6:14 PM

#

If G is infinite, this index is always n I'm pretty sure

weary terrace Jul 26, 2019, 6:16 PM

#

How come?

mild laurel Jul 26, 2019, 6:17 PM

#

well the cosets are [0]nG, [1]nG, .... [n-1]nG

weary terrace Jul 26, 2019, 6:18 PM

#

G is additive

#

So g+nG will equal nG only if g is n times the sum of some g' in G

oblique river Jul 26, 2019, 6:29 PM

#

that's certainly not true (@ zopherus's last statement)

#

take G = (Z/pZ)^N and n = p

#

here N = natural numbers

#

then pG = 0

#

it doesn't make sense to talk about [k]nG as a coset

#

because [k] isnt an element of G

#

@weary terrace @mild laurel

mild laurel Jul 26, 2019, 6:33 PM

#

yeah you're right

steep hull Jul 26, 2019, 6:35 PM

#

F_(p^k) seems like the classic counterexample.

weary terrace Jul 26, 2019, 6:38 PM

#

I agree.. my approach was assuming each g in G is a sum of some amount of g', say k.

So in order to calculate the order of the quotient, we can take a general element and calculate its order.

So taking g in G, we say g=kg' where g' in G and k is natural.
If gcd(k,n)=1, then the order of g+nG is n.
Otherwise the order is less then n.

Now my problem is proving the existence of this element..

oblique river Jul 26, 2019, 7:40 PM

#

I don't think there is a general answer, it will depend on exactly the structure of the group

#

for example, if G is a finite group of odd order, then every element is a square (in multiplicative notation) or a "double" (in additive notation) so G/2G would be 0

weary terrace Jul 26, 2019, 7:54 PM

#

You're correct! Thanks for the example..

chilly ocean Jul 26, 2019, 11:58 PM

#

whats the channel to ask about prime numbers?

#

galois?

#

I will have the intro to Abstract Algebra book on Wednesday. Any ideas on which of the big topics I should read about before starting the class?

bleak abyss Jul 27, 2019, 12:04 AM

#

Probably good to have your basic NT down for sure

magic owl Jul 27, 2019, 2:25 AM

#

Huh? What basic NT is involved?

bleak abyss Jul 27, 2019, 2:38 AM

#

By basic I mean like, ultra basic, stuff that might get referenced in passing

scarlet estuary Jul 27, 2019, 2:39 AM

#

that stuff can be picked up along the way

bleak abyss Jul 27, 2019, 2:39 AM

#

Knowing gcd and Euclidean algorithm and shit like that

scarlet estuary Jul 27, 2019, 2:40 AM

#

i self-studied D&F with absolutely no ENT

#

i knew what mod was, and thats about it

#

and ftarith

bleak abyss Jul 27, 2019, 2:40 AM

#

I think D&F has a bit of that stuff in chapter 0 anyway

scarlet estuary Jul 27, 2019, 2:41 AM

#

it defines like, z/nz, yes

bleak abyss Jul 27, 2019, 2:41 AM

#

If there's anything that's useful to read up on before starting algebra it's to make sure you remember that stuff

#

I guess maybe reviewing some linear algebra stuff can't hurt either

scarlet estuary Jul 27, 2019, 2:41 AM

#

linear algebra + actually knowing proofs are the most useful things.

chilly ocean Jul 27, 2019, 3:00 AM

#

I can already tell this will be a painful class 😂 😭

bleak abyss Jul 27, 2019, 4:42 AM

#

Oh yeah you didn't like linear algebra as much, right?

#

In group theory at the beginning it won't matter as much honestly, just that matrices lead to some examples of groups

#

But it basically boils down to multiplicativity of stuff

#

For example, determinant is multiplicative, product or orthogonal/unitary matrices is orthogonal/unitary

magic owl Jul 27, 2019, 1:22 PM

#

I didn't like linear algebra until I learned more applications of it

chilly ocean Jul 27, 2019, 1:32 PM

#

how long does abstract algebra usually take to learn, well more specifically, learning from dummit and foote, it looks pretty long and I'm just curious.

magic owl Jul 27, 2019, 1:42 PM

#

Um most people learn it

#

in a 1 year sequence of undergrad

#

but that's learning about 1 years worth

#

becomming an expert would take at least a decade

edgy charm Jul 27, 2019, 2:28 PM

#

It takes a decade to master most things

#

Once you get over the steep slope of learning a certain topic, learning is much more enjoyable

magic owl Jul 27, 2019, 2:32 PM

#

I'll add that Dummit and Foote is not my recommended starting point if you are self learning

#

its very much a great reference text

#

but not super pedagogical imo

#

pinter or similar is better I think for starting out

gentle pendant Jul 27, 2019, 4:53 PM

#

my favourite "first course" in algebra is Herstein's topics in algebra.

analog oracle Jul 27, 2019, 4:59 PM

#

I don't like D&F either and I'm annoyed I have to deal with it this upcoming semester. My intro to abstract algebra course at UC Berkeley used this textbook and I thought it was way better than D&F (I had used it before for a different class elsewhere) http://homepage.divms.uiowa.edu/~goodman/algebrabook.dir/download.htm

uncut girder Jul 27, 2019, 5:10 PM

#

Were using D&F in the fall for grad alg 1

#

Kinda hype

analog oracle Jul 27, 2019, 5:11 PM

#

same here \o/ though i'd much prefer to use Lang tbh

chilly ocean Jul 27, 2019, 5:21 PM

#

Is using D&F for an intro like reading munkres or baby rudin for an intro to the topic?

magic owl Jul 27, 2019, 5:27 PM

#

No

#

The second two are fine intros

#

DF is just too terse and technical to be a good intro

#

Pinter is fairly conversational

uncut girder Jul 27, 2019, 5:40 PM

#

I see

#

I have the book lemme look into it

#

DF doesn't use end of proof symbol scoff

placid pond Jul 27, 2019, 5:51 PM

#

why do people not like DF

#

i liked it precisely because it's terse and technical

chilly ocean Jul 27, 2019, 5:51 PM

#

they didnt say they didnt like it

uncut girder Jul 27, 2019, 5:51 PM

#

I see

chilly ocean Jul 27, 2019, 5:51 PM

#

they said its bad for begineers

placid pond Jul 27, 2019, 5:51 PM

#

i disagree, but to each their own

uncut girder Jul 27, 2019, 5:52 PM

#

You could just skip the unnecessary parts of DF?

placid pond Jul 27, 2019, 5:52 PM

#

@uncut girder sure, DF is a fairly large book

uncut girder Jul 27, 2019, 5:55 PM

#

DF is very easy to read so far

#

So composition series is like unique factorization for groups

#

DF motivated composition series very well in 2 pages

placid pond Jul 27, 2019, 6:03 PM

#

yes, you can say that

uncut girder Jul 27, 2019, 6:06 PM

#

Is proof of Jordan Holder actually "fairly straightforward"

placid pond Jul 27, 2019, 6:07 PM

#

yes

bleak abyss Jul 27, 2019, 7:33 PM

#

Lol I dunno DF just puts me to sleep so fast

#

I was originally using it one time in my REU to try to learn about group actions and Sylow stuff

#

And I just started reading it and things just dragged on forever

#

Eventually I was like aight I'm gonna need something much faster than this

#

Said thing was Herstein, much as I wouldn't recommend it now since while I philosophically agree with x(f) notation, it's non-standard and thus inconvenient to learn. Also it doesn't cover enough

#

I am liking Jacobson's Basic Algebra I quite a lot (using it to review for quals)

woven delta Jul 27, 2019, 7:37 PM

#

What do you think of Serres rep theory book?

bleak abyss Jul 27, 2019, 7:38 PM

#

Used it briefly when I was taking rep theory and it seemed real nice

plain sequoia Jul 27, 2019, 9:39 PM

#

Read Lang and git gud doggo

#

Lang loves categories tho, it can actually be benefitial to know some very very basic defs and stuff in category theory before starting to read Lang.

#

Whatever, who reads Lang as an intro to abstract algebra?

placid pond Jul 27, 2019, 9:40 PM

#

lang would be an extreme choice, but i suppose it's doable

#

i always thought lang's books are meant to be used as reference books

plain sequoia Jul 27, 2019, 9:43 PM

#

Maybe

#

I bought it on sale some months ago

#

I think it is a must have for every mathematician

analog oracle Jul 27, 2019, 9:43 PM

#

UC Berkeley's first grad algebra course usually uses Lang but you're assumed to have done at minimum 2 semesters of abstract algebra before that. I definitely wouldn't recommend Lang as an introduction but it's very comprehensive and in my opinion better than D&F (but again, wouldn't use D&F as an introduction so :p )

plain sequoia Jul 27, 2019, 9:43 PM

#

I did Fraleigh for intro to algebra, and I do not recommend that book

analog oracle Jul 27, 2019, 9:44 PM

#

that's the one book I can think of that I consider worse than D&F :p

plain sequoia Jul 27, 2019, 9:44 PM

#

For the Rings and Modules course we’re doing Bhattacharya, Nagpaul and some other guy which seems good

#

To illustrate Lang’s addicition to categories; already on page 8 where he is giving examples on groups, he describes the automorphism group as: «For any object A in a category, the automorphisms form a group; Aut(A)»

#

Lang seems to have a bunch of good problems tho

#

Theorems I’ve never heard about

placid pond Jul 27, 2019, 9:49 PM

#

@analog oracle why is D&F bad

#

in this respect

analog oracle Jul 27, 2019, 9:54 PM

#

to be honest my opinion of it might change significantly now that I'll be using it again but using it as an introduction when I was self teaching and later on auditing a class at Stanford I found it very dense and dry and things weren't explained in a way that was intuitive to me while everything just clicked the second I started reading the Goodman book I sent earlier. The exercises are good, I'll give it that. I worked through them even after giving up on the book itself

#

but it's that way with most things, ask 20 different people what the best book for some specific field is and you'll hear at least 3-5 different ones being recommended and people having very strong views about them. So nowadays when I'm picking textbooks I'll gather up recommendations and then try to find them at the library and see which one is more suited to my intuition and learning style

bleak finch Jul 28, 2019, 12:17 AM

#

Let phi : R -> R/Nil(R). So for every x in Nil(R), phi(x) = 0. Let x, y be two nilradicals of R/Nil(R). So there exists a, b such that phi(a) = x and phi(b) = y. But because these are homomorphisms, and homomorphisms preserve multiplication, (e.g. if x^m = y^n = 0 for some m,n then a^m = b^n = 0 for same m, n), therefore a, b are nilpotents, so, phi(a) = phi(b) = 0. But by definition phi(a) = x and phi(b) = y. Therefore x = y = 0. Is the above proof sound?

oblique river Jul 28, 2019, 12:43 AM

#

It's not wrong but why do you need x and y?

#

I don't really understand what you are trying to prove

#

and why you need both x and y

bleak finch Jul 28, 2019, 12:50 AM

#

I trying to prove that Nil(R/Nil(R)) = {0}

#

The most direct way to that is to take x, y in Nil(R/Nil(R)) and show that x = y

oblique river Jul 28, 2019, 12:50 AM

#

no, the most direct way is to take x in Nil(R/Nil(R)) and show that x = 0

#

which you've done, although now I see a mistake in your proof

#

I was confused by all the names: you know phi(a) = x and x^n = 0, so you know phi(a^n) = 0

#

that doesn't mean a^n = 0

#

because there can be nonzero things that map to 0 under phi

bleak finch Jul 28, 2019, 12:52 AM

#

I don't see your logic. if x^n = 0 (i.e. after being annihilated by repeated multiplication n times) and phi preserves multiplication, then why shouldn't a^n = 0 after being multiplied n times?

oblique river Jul 28, 2019, 12:53 AM

#

because all you know is that phi(a^n) = 0

#

phi preserves multiplication which just means phi(a)^n = phi(a^n)

bleak finch Jul 28, 2019, 12:53 AM

#

You know that phi(a^n) = x^n

oblique river Jul 28, 2019, 12:53 AM

#

absolutely, and x^n = 0

#

so phi(a^n) = 0

#

you know that phi(0) = 0, but it is NOT TRUE that if phi(blah) = 0 then blah = 0

bleak finch Jul 28, 2019, 12:54 AM

#

Yes I can see that.

oblique river Jul 28, 2019, 12:54 AM

#

okay so how are you concluding that a^n = 0

bleak finch Jul 28, 2019, 12:54 AM

#

I have no idea.

oblique river Jul 28, 2019, 12:55 AM

#

I mean that was exactly your argument. you said phi(a^n) = 0 and thus a^n = 0

#

and i'm saying that that step is not correct

bleak finch Jul 28, 2019, 12:55 AM

#

Yes I know that was my argument.

oblique river Jul 28, 2019, 12:55 AM

#

so as a hint: in this problem, you won't be showing that a^n = 0 because that's not true in general

#

you will show that some other power of a is 0

#

second hint: a correct proof should use the word "kernel" (or at least the idea of kernels)

bleak finch Jul 28, 2019, 1:03 AM

#

Let phi : R -> R/Nil(R). So for every x in Nil(R), phi(x) = 0. Let x, be a nilradical of R/Nil(R). So there exists a such that phi(a) = x. Let n be the integer such that x^n = 0. But because phi is a homomorphism, and homomorphisms preserve multiplication, therefore phi(a)^n = 0. But because a is in the kernel of phi, therefore there must exist an m such that a^m = 0 (by the first isomorphism theorem, the kernel of phi is the nilradical of R). Because phi(a^n) = 0 and phi(a^m) = 0 and phi(a) = x...

#

I need a hint on how to proceed.

oblique river Jul 28, 2019, 1:04 AM

#

so you should specify how you found that m

#

also the whole point is you prove that a is in Nil(R)

#

so x = phi(a) = 0

bleak finch Jul 28, 2019, 1:16 AM

#

Let phi : R -> R/Nil(R). So for every x in Nil(R), phi(x) = 0. Let x, be a nilradical of R/Nil(R). So there exists a such that phi(a) = x. Let n be the integer such that x^n = 0. But because phi is a homomorphism, and homomorphisms preserve multiplication, therefore phi(a)^n = phi(a^n) = 0.We know that phi(a^n) = 0. So a^n is in the kernel of phi.But the kernel of phi is the set of elements such that there exist some m such that said element raised to the power of m is 0. So there exists m such that (a^n)^m = a^mn = 0. But that means that a itself is in the kernel of phi, which means phi(a) = 0. Because phi(a) = x and phi(a) = 0, therefore x = 0.

oblique river Jul 28, 2019, 1:43 AM

#

looks good

bleak finch Jul 28, 2019, 2:22 AM

#

That's wonderful.

#

😉

chilly ocean Jul 28, 2019, 6:52 PM

#

Hello, ya’ll.

placid pond Jul 28, 2019, 6:52 PM

#

hi

chilly ocean Jul 28, 2019, 6:52 PM

#

:0

#

@placid pond Hello, good sir. I’m trrying to learn AA with a book in the #resources and the book mentions the universal set U, but I thought this wasn’t allowed

#

Due to Russell’s Paradox?

placid pond Jul 28, 2019, 6:53 PM

#

what does it mean by universal set

#

there is no "set of all sets"

#

but there's nothing wrong with "set of all sets with cardinality < aleph_{whatever}"

#

which may be the sense in which you have a universal set

stone fulcrum Jul 28, 2019, 6:55 PM

#

The book likely means "universal set" as "all sets you're currently working with"

chilly ocean Jul 28, 2019, 6:55 PM

#

Universal set is defined as that which, A C U means: A’ = {x: x is an element of U but not an element of A}

placid pond Jul 28, 2019, 6:55 PM

#

that's not a definition of universal set

#

that's a definition of relative complement

chilly ocean Jul 28, 2019, 6:56 PM

#

Sometimes we will work within one fixed set U , called the universal set. For any set A ⊂ U, we define the complement of A , denoted by A ′, to be the set A ′ = { x : x ∈ U and x ∉ A }.

#

I’m going to go with Kaynex’s interpretation of it I guess

stone fulcrum Jul 28, 2019, 6:59 PM

#

A' is everything that is not in A. However, "everything" is something you define

chilly ocean Jul 28, 2019, 6:59 PM

#

Is that the definition of a compliment?
(The book has not mentioned it yet)

#

Oh — I get it now, :c

#

Now that I think of U as all sets currently defined other than A, the compliment of A makes sense to be that

stone fulcrum Jul 28, 2019, 7:00 PM

#

Indeed, Russell's paradox shows that some "things" can't be sets

#

So you can't work with these "things" with set theory alone

#

Just don't include them in U

magic owl Jul 28, 2019, 7:02 PM

#

For a concrete example

#

Let U be all real numbers

#

And A be (0,1)

#

The complement in this context would just be everting less than or equal to zero or greater than or equal to one

#

But I could also make U=(-1,2)

#

Then the complement changed

#

Changes*

chilly ocean Jul 28, 2019, 7:03 PM

#

And so A’ = {-1, 2}

#

?

magic owl Jul 28, 2019, 7:03 PM

#

Well no it would include 0 for example

#

It would be (-1,0] union [1,2)

chilly ocean Jul 28, 2019, 7:05 PM

#

But the complement is defined as all elements not in A

#

When all sets under consideration are considered to be subsets of a given set U, the absolute complement of A is the set of elements in U but not in A.

stone fulcrum Jul 28, 2019, 7:07 PM

#

You may not know the notation
(0, 1) is the set of all real numbers between 0 and 1, not including 0 and 1

chilly ocean Jul 28, 2019, 7:07 PM

#

Oh.. lol

#

I thought he just meant {}

stone fulcrum Jul 28, 2019, 7:08 PM

#

If U is all real numbers,
And if A = (0, 1)
Then A' = (-inf, 0] ∪ [1, inf)

#

Note that [ means to include the endpoint

chilly ocean Jul 28, 2019, 7:09 PM

#

Oh! I was going to ask ;c

stone fulcrum Jul 28, 2019, 7:09 PM

#

But yeah you have the idea of universal sets and complements down, the way you just described it was lit

chilly ocean Jul 28, 2019, 7:09 PM

#

I get it now, thank you!

stone fulcrum Jul 28, 2019, 7:09 PM

#

Np. Feel free to ask if you have anything else!

uncut girder Jul 28, 2019, 7:11 PM

#

Btw this basic set theory probably fits better in #proofs-and-logic

bleak finch Jul 29, 2019, 1:38 AM

#

So

#

An R-algebra is an R-module that is also a ring where the multiplication is compatible with the scalar multiplication.

#

But what would it mean if the multiplication were incompatible with scalar multiplication?

topaz solar Jul 29, 2019, 1:41 AM

#

So, tell me, what is the compatibility with scalar multiplication?

#

preferably in a concise statement like $(rs)\cdot x = r\cdot (s \cdot x)$ or some such

cloud walrusBOT Jul 29, 2019, 1:41 AM

#

Darkrifts:

📎 218876477146398720.png

bleak finch Jul 29, 2019, 1:42 AM

#

Well, I know scalar multiplication is associative. Let's denote scalar multiplication by •

topaz solar Jul 29, 2019, 1:42 AM

#

Yeah iirc that's what I just put down, so what other compatibility things are there?

bleak finch Jul 29, 2019, 1:45 AM

#

Let u, v, w be module elements and a, b, c ring elements. Because u, v, w are elements of an R-Algebra, we have uv being valid and closed (unlike in a vector space like Reals^3 or structures like (Reals^3, inner product)).

#

So I would guess that compatibility would mean r•(uv) = (r•u)v

topaz solar Jul 29, 2019, 1:46 AM

#

so, how would $(rx)\cdot (sy)$ look like?

cloud walrusBOT Jul 29, 2019, 1:46 AM

#

Darkrifts:

📎 218876477146398720.png

topaz solar Jul 29, 2019, 1:46 AM

#

x, y elements of your module

bleak finch Jul 29, 2019, 1:47 AM

#

(rx)•(sy) =r•(xsy)?

oblique river Jul 29, 2019, 1:49 AM

#

You should straighten out what the dot means

bleak finch Jul 29, 2019, 1:50 AM

#

Oh wait

oblique river Jul 29, 2019, 1:50 AM

#

is it multiplication in the R-algebra? Or is it scalar multiplication of R acting on the R-algebra?

topaz solar Jul 29, 2019, 1:50 AM

#

on one side it's algebra mult, on the other it's scalar x alg

bleak finch Jul 29, 2019, 1:50 AM

#

• is scalar multipication

topaz solar Jul 29, 2019, 1:50 AM

#

you might wanna pick

oblique river Jul 29, 2019, 1:50 AM

#

in that case, (rx)*(sy) doesn't make sense

bleak finch Jul 29, 2019, 1:50 AM

#

So what we have is (r•x)(s•y)

oblique river Jul 29, 2019, 1:50 AM

#

ok

bleak finch Jul 29, 2019, 1:51 AM

#

And (r•x)(s•y) = r•(x(s•y))

oblique river Jul 29, 2019, 1:51 AM

#

the right hand side doesn't make sense

#

wait nvm my b :)

topaz solar Jul 29, 2019, 1:51 AM

#

it makes sense

oblique river Jul 29, 2019, 1:52 AM

#

yes that is correct but you don't need the s there

topaz solar Jul 29, 2019, 1:52 AM

#

doesn't really use the R-bilinearity

oblique river Jul 29, 2019, 1:52 AM

#

just r*(xy) = (r*x)y

bleak finch Jul 29, 2019, 1:52 AM

#

Is that it?

topaz solar Jul 29, 2019, 1:52 AM

#

I was tryna get you to go for R-bilinearity as the property there but aight

oblique river Jul 29, 2019, 1:52 AM

#

that's what it means to be an R-algebra yes. It should be an R-module, and also a ring, and those structures should be compatible

bleak finch Jul 29, 2019, 1:53 AM

#

Can you derive bilinearity from r*(xy) = (r*x)y

topaz solar Jul 29, 2019, 1:53 AM

#

i don't think so

#

since that's just linear in the first part

oblique river Jul 29, 2019, 1:53 AM

#

what do you mean exactly by bilinearity

#

scalar multiplication is acting on the left here and it distributes over the addition in the R-algebra (which I'm just going to call S from now on)

#

and you need the condition that r*(xy) = (r*x)y = x(r*y)

bleak finch Jul 29, 2019, 1:55 AM

#

(r•x)(s•y) = r•(x(s•y)) = s•((r•x)y)

oblique river Jul 29, 2019, 1:55 AM

#

idk why you need the s, you can just state it how I did above

topaz solar Jul 29, 2019, 1:56 AM

#

iirc algebras are your modules with your multiplication f: S x S -> S such that f(rx, sy) = rs f(x, y)

bleak finch Jul 29, 2019, 1:56 AM

#

I don't understand how you're swapping in the last equality

oblique river Jul 29, 2019, 1:56 AM

#

sure, it suffices to also say f(rx, y) = rf(x,y) = f(x, ry)

topaz solar Jul 29, 2019, 1:56 AM

#

yeah

oblique river Jul 29, 2019, 1:56 AM

#

what do you mean? that's the definition of an R-algebra

topaz solar Jul 29, 2019, 1:57 AM

#

i just put both r, s since i'm way too used to seeing it as such

oblique river Jul 29, 2019, 1:58 AM

#

I mean the easy definition of an R-algebra is "a ring homomorphism R --> S"

#

(assuming everything is commutative)

bleak finch Jul 29, 2019, 1:58 AM

#

What was S again

topaz solar Jul 29, 2019, 1:59 AM

#

your algebra boy

bleak finch Jul 29, 2019, 1:59 AM

#

Oh okay

oblique river Jul 29, 2019, 1:59 AM

#

S is the R-algebra

topaz solar Jul 29, 2019, 1:59 AM

#

yeah that definition works, I prefer the "monoid in Ring" version though tbh

oblique river Jul 29, 2019, 1:59 AM

#

get your categories out of here, there are children present

topaz solar Jul 29, 2019, 2:00 AM

#

i mean, the way I learned it originally was as a module + a multiplication anyhow

bleak finch Jul 29, 2019, 2:00 AM

#

So the ring homomorphism φ(rs) = φ(r)φ(s)

topaz solar Jul 29, 2019, 2:01 AM

#

but also a compatibility with addition

#

don't forget that one

bleak finch Jul 29, 2019, 2:01 AM

#

So also φ(r + s) = φ(r) + φ(s)

oblique river Jul 29, 2019, 2:01 AM

#

yeah if phi: R --> S is a ring hom

#

and R and S are commutative

#

that turns S into an R-algebra

#

where you define the scalar multiplication by: r*x = phi(r)x

#

for r in R and x in S

bleak finch Jul 29, 2019, 2:02 AM

#

What do you mean by a commutative module? And in an R-algebra the ring structure of S need not be commutative

oblique river Jul 29, 2019, 2:03 AM

#

R had better be commutative

#

and the definition I gave was in the special case that S is commutative

#

if S is not commutative, then add the condition that the image of phi lies in the center of S

bleak finch Jul 29, 2019, 2:04 AM

#

Can I still say that r*x = phi(r)x?

oblique river Jul 29, 2019, 2:04 AM

#

what do you mean

#

yes that is how you define the scalar multiplication

bleak finch Jul 29, 2019, 2:04 AM

#

If S as a ring is not commutative, is it still the case that r•x = φ(r)x?

topaz solar Jul 29, 2019, 2:04 AM

#

yes

bleak finch Jul 29, 2019, 2:04 AM

#

That is good

#

You guys helped a lot.

oblique river Jul 29, 2019, 2:05 AM

#

you're kinda saying that like it's some absolute fact. the statement should be "If R is a commutative ring and phi: R --> S is a ring homomorphism whose image is contained in the center of S, then you can turn S into an R-algebra via the scalar multiplication r*x = phi(r)x"

bleak finch Jul 29, 2019, 2:06 AM

#

Hold on there's one more step I need to understand.

oblique river Jul 29, 2019, 2:06 AM

#

and conversely, if S is an R-algebra (under whatever other definition you want) then you can define a ring homomorphism phi: R --> S whose image lies in the center of S via the definition phi(r) = r*1

bleak finch Jul 29, 2019, 2:06 AM

#

Why is it the case that if the image of φ does not lie in the center of S that r•s = φ(r)s is not well-defined?

#

Oh wait

topaz solar Jul 29, 2019, 2:07 AM

#

if it's not in the center, then it doesn't commute with them, throwing off your bilinearity

bleak finch Jul 29, 2019, 2:07 AM

#

I get it

oblique river Jul 29, 2019, 2:07 AM

#

it's perfectly well-defined, it just doesn't give you bilinearity

#

so it doesn't satisfy the other definition of an R-algebra

bleak finch Jul 29, 2019, 2:07 AM

#

If the image is in the center, then φ(r)s = sφ(r)

#

Because the center commutes with every element

oblique river Jul 29, 2019, 2:07 AM

#

that is true but not the relevant part

#

like that by itself isn't really relevant for making S into an R-algebra

#

you need the commutativity in order to show bilinearity

bleak finch Jul 29, 2019, 2:09 AM

#

Buncho Bananas I am reading this handout

#

https://faculty.math.illinois.edu/~r-ash/Algebra/Chapter4.pdf

#

We don't know that the R-module S is a commutative ring. All we know is that it's a ring.

oblique river Jul 29, 2019, 2:10 AM

#

yeah i've been saying that

#

basically everything I've said about homomorphisms is section 4.1.8 in those notes

bleak finch Jul 29, 2019, 2:11 AM

#

But then there's this seemingly arbitrary identity that r•(xy) = (r•x)y = x(r•y)

#

I don't know where that identity came from

oblique river Jul 29, 2019, 2:12 AM

#

what? that is part of the definition!!!

#

also I'm not sure what you mean by "but then"

bleak finch Jul 29, 2019, 2:12 AM

#

If you could derive that identity from something else (e.g. algebraic structure) I would understand what it meant

oblique river Jul 29, 2019, 2:13 AM

#

it's literally the definition

#

it's not possible to derive from something else

topaz solar Jul 29, 2019, 2:13 AM

#

it's one of the things you derive from

oblique river Jul 29, 2019, 2:13 AM

#

unless you take some other definition as your definition

topaz solar Jul 29, 2019, 2:13 AM

#

assume r * xy = (r * x)y = x(r * y), then .....

oblique river Jul 29, 2019, 2:13 AM

#

but the notes you sent take that as the definition

topaz solar Jul 29, 2019, 2:13 AM

#

if something is the definition, then you'd be pretty hard pressed to derive it

bleak finch Jul 29, 2019, 2:14 AM

#

So I just have to memorize it and say "well, that's just some identity. So there."

topaz solar Jul 29, 2019, 2:14 AM

#

imagine deriving that a ring has an multiplication that distributes

oblique river Jul 29, 2019, 2:14 AM

#

I mean that's what you had to do with the associativity law for rings, didn't you?

#

it's literally the definition of an R-algebra

#

if you want an example, think of matrices

#

if A and B are real matrices

#

and r is a real number

#

then r(AB) = (rA)B = A(rB)

bleak finch Jul 29, 2019, 2:15 AM

#

Let me put it to you like this.

oblique river Jul 29, 2019, 2:15 AM

#

it literally just says that if you wanna multiply r by a product xy, then you can either multiply it to x, or multiply it to y

bleak finch Jul 29, 2019, 2:15 AM

#

I can never remember the component-wise definition of a cross product

oblique river Jul 29, 2019, 2:15 AM

#

that's bad :(

#

idk what to say other than you need to practice remembering definitions

#

wait a minute

bleak finch Jul 29, 2019, 2:16 AM

#

But I can remember that it's what you get if you multiply two vectors written in xi + yj+zk treating i, j, k as elements of the group H.

oblique river Jul 29, 2019, 2:16 AM

#

okay nvm thats not as bad as I thought

#

I thought you were trying to talk about products of rings

#

yeah cross products are harder to remember

bleak abyss Jul 29, 2019, 2:16 AM

#

Yeah same

oblique river Jul 29, 2019, 2:16 AM

#

but this is literally one line

bleak abyss Jul 29, 2019, 2:16 AM

#

I just remember the cross product as just like

#

f_{a,b}(x) = det(x,a,b) is a functional on R^3

oblique river Jul 29, 2019, 2:17 AM

#

and it doesn't derive from anything. all you have to do is remember that "an R-algebra is an R-module S along with a compatible ring structure"

#

and there is literally only one thing that that statement could mean

#

and it's r*(xy) = (r*x)y = x(r*y)

#

or just remember the definition in terms of ring homomorphisms

bleak abyss Jul 29, 2019, 2:18 AM

#

So Riesz rep theorem gives some vector v such that f_{a,b}(x) = <x,v>. Call v = a\times b

bleak finch Jul 29, 2019, 2:18 AM

#

So my question is, why is it that the only thing "compatible ring structure" can mean is r*(xy) = (rx)y = x(ry)?

oblique river Jul 29, 2019, 2:18 AM

#

"An R-algebra S is a ring hom phi: R --> S whose image lies in the center of S"

#

what else would it mean lol

#

the ring structure on S

#

has to be compatible with the scalar multiplication of R

#

so if you're computing r * (a product of things in S)

#

it shouldn't matter when you do the "r *" part

#

either first, or last, or in the middle

#

you have to get the same answer

bleak finch Jul 29, 2019, 2:19 AM

#

Well it seems that in the first and last equality you're swapping r and x

oblique river Jul 29, 2019, 2:19 AM

#

exactly!!!

#

like I said

#

it shouldn't matter when you do the "r *" part

#

you can do it to x

#

or to y

#

or to xy

bleak finch Jul 29, 2019, 2:20 AM

#

So what the identity is saying is...

oblique river Jul 29, 2019, 2:20 AM

#

if you had 3 elements in S you could write r*(xyz) = (r*x)yz = x(r*y)z = xy(r*z)

#

or for shits and gigs: (r*(xy))z

topaz solar Jul 29, 2019, 2:20 AM

#

stop

oblique river Jul 29, 2019, 2:20 AM

#

yes sir

bleak finch Jul 29, 2019, 2:21 AM

#

Any permutation that makes sense (fits the domain of the operations) of r, x, y are equal

bleak abyss Jul 29, 2019, 2:21 AM

#

Problem 1 in algebra: verify associativity of abcd

oblique river Jul 29, 2019, 2:21 AM

#

no, as you said earlier S isnt commutative

#

so you can't switch x and y

#

but you can do the scalar multiplication part on either x or y

topaz solar Jul 29, 2019, 2:21 AM

#

it's saying that you can move r, everything else needs to remain in the same order

bleak finch Jul 29, 2019, 2:21 AM

#

Wait

#

The three element definition makes perfect sense

oblique river Jul 29, 2019, 2:21 AM

#

it's the same as the 2-element one :(

topaz solar Jul 29, 2019, 2:22 AM

#

:(

bleak finch Jul 29, 2019, 2:22 AM

#

I don't know it just seems obvious "what's happening" in the 3 element definition while it seems somewhat occult in the 2 element definition.

oblique river Jul 29, 2019, 2:22 AM

#

well whatever works for you

bleak finch Jul 29, 2019, 2:22 AM

#

There's a clear pattern

oblique river Jul 29, 2019, 2:22 AM

#

like I said, just remember how matrices work

#

r(AB) = (rA)B = A(rB)

bleak finch Jul 29, 2019, 2:23 AM

#

OH

#

That makes perfect sense

oblique river Jul 29, 2019, 2:23 AM

#

I said that like 10 minutes ago :(

bleak finch Jul 29, 2019, 2:23 AM

#

I didn't know what you meant by "just remember how matrices work" until you spelled out that identity that I used one thousand times before.

oblique river Jul 29, 2019, 2:23 AM

#

I spelled it out 10 minutes ago :P

bleak finch Jul 29, 2019, 2:24 AM

#

Well I just couldn't process it until just now.

#

Perhaps I had trouble seeing the connection because I was staring at a different perspective. Ring and modules are rings and modules helping us generalize ideals. Matrices are linear transformations used for functional analysis. And never the twain shall meet.

oblique river Jul 29, 2019, 2:26 AM

#

sry im having trouble parsing that sentence

#

the second one

bleak abyss Jul 29, 2019, 2:26 AM

#

Wait matrices aren't really used much in functional analysis

bleak finch Jul 29, 2019, 2:27 AM

#

Rings and modules are things we use to study ideals, short exact sequences notherian rings, etc... and matrices are linear transformations that we use for something else generally speaking.

oblique river Jul 29, 2019, 2:27 AM

#

I wouldn't really say that we use rings to study ideals

#

also there's a difference between "a matrix as a linear transformation" and "a matrix as an element of a matrix ring"

#

in the sense that in my examples I was only thinking of the matrix ring as an algebra, not as some kind of set of linear operators

#

also, and this isn't meant to be shady, but how do you know what noetherian rings and short exact sequences are but you dont know the definition of an R-algebra?

#

I'm just confused by what your background is

bleak finch Jul 29, 2019, 2:30 AM

#

I took a rings and modules class over a year ago and I vaguely remember the professor saying something about noetherian rings and short exact sequences when I was slacking off in class and aping the homework.

#

Okay?

#

I could barely pass the class because it was just a bunch of random definitions that didn't make sense.

#

So consequently it all became defenestrated.

#

...so I am asking you so I can rebuild from first principles just what exactly he was saying.

#

Look sorry for being so thick headed but you really did help so your patience was not in vain

#

And that's the important thing right?

oblique river Jul 29, 2019, 2:41 AM

#

I wasn't trying to be mean, I was honestly curious what your background was

bleak finch Jul 29, 2019, 2:41 AM

#

oh okay

oblique river Jul 29, 2019, 2:41 AM

#

but let this be a lesson to you about slacking off in class :P

bleak finch Jul 29, 2019, 2:41 AM

#

yeah

#

but I had a sadist for my other professor which made it really hard to focus on this professor's class.

#

because he loved giving us waterboard-level assignments

#

Near the end of his class me and some other student were the only ones regularly attending because everyone else was so brutally tortured by the assignments.

#

And so we triaged

bleak abyss Jul 29, 2019, 3:42 AM

#

Is your professor's name "Souganidis" by any chance?

#

jk he wouldn't be caught dead in an algebra class

uncut girder Jul 29, 2019, 3:43 AM

#

mnoop

random forge Jul 29, 2019, 3:50 AM

#

Why not? @bleak abyss

bleak finch Jul 29, 2019, 3:57 AM

#

I am a Master's student at DePaul University so no

#

If I is an ideal of the ring R, show how to make the quotient ring R/I into a left R-module.

mild laurel Jul 29, 2019, 3:59 AM

#

do it exactly like how you'd think to do it

solemn hollow Jul 29, 2019, 4:00 AM

#

~~sorry but I thought souganidis was a sougandis nuts joke~~

bleak finch Jul 29, 2019, 4:02 AM

#

What I would think to do it is to just take R/I as a left R-module itself

#

(Because any ring can be interpreted as an R-module)

#

But this seems really, extremely dull.

oblique river Jul 29, 2019, 4:05 AM

#

wait

#

how do you interpret any ring as an R-module?

#

how do you interpret the integers Z as a module over the complex numbers?

bleak finch Jul 29, 2019, 4:07 AM

#

Any ring is an R module because for any n R^n is an R-module including n = 1.

oblique river Jul 29, 2019, 4:07 AM

#

what???

#

any ring R is always a module over itself

bleak finch Jul 29, 2019, 4:07 AM

#

Yes

oblique river Jul 29, 2019, 4:07 AM

#

R is always an R-module

bleak finch Jul 29, 2019, 4:07 AM

#

Yup mmhm

oblique river Jul 29, 2019, 4:07 AM

#

so what does that have to do with R/I as an R-module

bleak finch Jul 29, 2019, 4:07 AM

#

Just take R/I as an R-module over itself.

oblique river Jul 29, 2019, 4:08 AM

#

NOOOOOOOOOOOOO

#

NO NO NO

#

when you say "R-module"

#

R is a SPECIFIC RING

topaz solar Jul 29, 2019, 4:08 AM

#

consider Z/(2)
how is this a Z module over itself

oblique river Jul 29, 2019, 4:08 AM

#

like a Z-module

bleak finch Jul 29, 2019, 4:08 AM

#

OIC

oblique river Jul 29, 2019, 4:08 AM

#

or a C-module

topaz solar Jul 29, 2019, 4:08 AM

#

b r u h

oblique river Jul 29, 2019, 4:08 AM

#

did you even look at the notes???

topaz solar Jul 29, 2019, 4:08 AM

#

C - module

oblique river Jul 29, 2019, 4:08 AM

#

that is pretty clear from those notes you sent

topaz solar Jul 29, 2019, 4:08 AM

#

smh

bleak finch Jul 29, 2019, 4:08 AM

#

I just thought the R was a decoration.

topaz solar Jul 29, 2019, 4:08 AM

#

decoration

#

no

oblique river Jul 29, 2019, 4:08 AM

#

I mean

bleak finch Jul 29, 2019, 4:09 AM

#

I didn't think it acted like a bound variable.

oblique river Jul 29, 2019, 4:09 AM

#

in the definition of an R-module

#

it literally talks about the ring R

bleak finch Jul 29, 2019, 4:09 AM

#

Let me explain

oblique river Jul 29, 2019, 4:09 AM

#

I get what you mean

#

it's just wrong

#

in this specific problem

#

R is a FIXED ring

#

and I is an ideal of R

topaz solar Jul 29, 2019, 4:09 AM

#

usually it's implied, but like
if you have a ring R, and then you see the symbol R-module
i'd personally think it's the same R being referred to but that's just me

bleak finch Jul 29, 2019, 4:09 AM

#

Okay, so R isn't actually a free variable dependent on the context.

oblique river Jul 29, 2019, 4:09 AM

#

and you have to turn R/I into a module over the ring R

bleak finch Jul 29, 2019, 4:09 AM

#

It's a fixed variable

topaz solar Jul 29, 2019, 4:09 AM

#

$\forall R$

cloud walrusBOT Jul 29, 2019, 4:10 AM

#

Darkrifts:

📎 218876477146398720.png

topaz solar Jul 29, 2019, 4:10 AM

#

just throw that in front or something

#

boom it's bound now

oblique river Jul 29, 2019, 4:10 AM

#

for example, R = Z and I = 2Z

#

turn Z/2Z into a Z-module

bleak finch Jul 29, 2019, 4:10 AM

#

OIC

oblique river Jul 29, 2019, 4:10 AM

#

Turn C[x]/(x^2 - 3) into a C[x]-module

#

R is the ring in question

bleak finch Jul 29, 2019, 4:10 AM

#

Well, here's what I would do

#

Take the cosets of Z/2Z as an abelian group

oblique river Jul 29, 2019, 4:11 AM

#

how about do it for R and I

bleak finch Jul 29, 2019, 4:11 AM

#

Take the elements of Z as the ring

oblique river Jul 29, 2019, 4:11 AM

#

instead of Z and 2Z

bleak finch Jul 29, 2019, 4:11 AM

#

check if the axioms hold

oblique river Jul 29, 2019, 4:11 AM

#

so your strategy is "do the thing i'm supposed to do"

#

which is I guess good

topaz solar Jul 29, 2019, 4:11 AM

#

do it in general, and preferably with a touch of rigor

bleak finch Jul 29, 2019, 4:11 AM

#

Well I misunderstood the question 😛

topaz solar Jul 29, 2019, 4:12 AM

#

of course, simple mistake

#

just a grave one

oblique river Jul 29, 2019, 4:12 AM

#

given s + I in R/I and r in R, how do you propose to multiply r*(s+I)?

#

I'm going to bed so darkrifts you will have to take over

#

good luck

topaz solar Jul 29, 2019, 4:12 AM

#

aight

bleak finch Jul 29, 2019, 4:12 AM

#

r•(s + I) = rs + I

topaz solar Jul 29, 2019, 4:13 AM

#

So, how would you make sure that plays nice with different representative elements?

bleak finch Jul 29, 2019, 4:14 AM

#

I thought that choice of representative element didn't matter.

topaz solar Jul 29, 2019, 4:14 AM

#

It doesn't, so how would you show your multiplication respects that?

bleak finch Jul 29, 2019, 4:15 AM

#

if r•(s + I) := rs + I then it respects that because rs + rI = rs + I for any left ideal.

topaz solar Jul 29, 2019, 4:16 AM

#

is the ideal I two-sided?

#

or are we assuming comm. ring?

#

just asking before I throw shit at it

bleak finch Jul 29, 2019, 4:16 AM

#

I think it has to be two-sided if we're speaking of R/I as being well-defined.

topaz solar Jul 29, 2019, 4:16 AM

#

just making sure they aren't pulling some shit

#

personally, I wouldn't take that and instead would do

Given $a \sim b \in R$, then $a - b \in I$ so $r(a + I) - r(b+I) = ra + I - (rb + I) = r(a-b) + I = ri + I = I$ or some such