groups-rings-fields

[next obsidian]

Not always the same as isomorphism 😔

Topological space momener

If the morphisms are built on set functions, yea, because they have an inverse

[spice whale]

ohh endomorphism is for homomorphisms from G to G

[tawny pine]

[untold basin]

I just came in this server and finally I meet people who speaks algebra x)

[chilly ocean]

hello friends

[dull root]

I have a simple question: If I take the minimal polynomial m of a over Q, and extend Q by some other element b, is m still irreducible over Q(b)?

nvm its not necessary true i think

[weak matrix]

No. Consider a = b for any field extension Q(b) with [Q(b):Q] > 1

[pastel cliff]

what is necessary to prove that some group K is a subgroup of G

intuition tells me just to show that K is contained in G and that the usual properties hold right

[dull root]

yes, K is a subset of G, and K is itself a group

[pastel cliff]

then arent these just true by definition

[spice whale]

commutative square

[tender mist]

These two observations will be useful I think, together with the hint:

1. H,K both normal in G => HK normal in G
2. [G : HK] = [G/K : HK/K] = [G/H : HK/K]

[weak matrix]

K must be a subset that is also a group under the same operation of G

[thorn delta]

SL(2, R) and GL(2, Q) are defined as sets. You have to show they are actually groups under the operations induced by GL(2, R)

[chilly ocean]

when someone says X is subgroup of Y they really mean to say (X,*|X) is a group where X subset of Y and *|X restriction of * to X

[waxen hedge]

Oh sorry, it's just the quotient application E -> E/F, which maps x to his class modulo F, that is x+F

[chilly ocean]

whats the theorem again between intenral product groups and and isomorhpisms

[pastel cliff]

is there any notation to say that H "is a subgroup of" G

[thorn delta]

\leq sometimes

[pastel cliff]

oh bet ok

[weak matrix]

An illustrative example would be $a = b = i$ the imaginary unit in $\mathbb{C}$, then over $\mathbb{R}$ we have $p_a = x^2 + 1$, and over $\mathbb{R}(i) = \mathbb{C}$ the minimal polynomial is $x-i$

[cloud walrus]

Tₑ𝘖(n)

[weak matrix]

An example with a not equal to b could be done using Q as the rational numbers, b as some positive non rational square root, and a as the positive square root of b

[spice whale]

$H \trianglelefteq G$

[cloud walrus]

coalison (shyshu for honorable)

[chilly ocean]

that means normal subgroup

[chilly radish]

With the same operations

[spice whale]

oh

i didn't know that

[chilly ocean]

\leq for subgroup, \trianglelefteq for normal subgroup

[pastel cliff]

what does the normal mean

[spice whale]

hm

[chilly radish]

Trianglelelefteq?? Wtf??

[spice whale]

left cosets equal right cosets

[chilly radish]

It's \unlhd

I've never seen that tex command

[chilly ocean]

normal = quotient group really is a group

[pastel cliff]

$\unlhd \trianglelefteq$

[cloud walrus]

μ₂

[pastel cliff]

lol

[chilly ocean]

https://en.wikipedia.org/wiki/Normal_subgroup

[south patrol]

All groups are abelian

[chilly radish]

Normal means conjugation Inariant which is exactly what you need for an induced group operation on the quotient to be well defined

i.e. if n is an element of N normal, then so is gng^-1 for any g in G

[south patrol]

All groups arise from forgetting the multiplication of rings anyway /s

[chilly ocean]

can someone help me on 4b

[chilly radish]

Are you sure about your interpretation of a

of R^+

Since the nonnrgative reals don't form a group under multiplication

[chilly ocean]

part a wouldnt be isomorphic

am i wrong?

isnt r+

[chilly radish]

I'm saying that I think you misinterpreted R^+

[chilly ocean]

no no

isnt it addition of reals

[chilly radish]

Well they're saying it's a subgroup of R^x

I know

[chilly ocean]

oh

[chilly radish]

R^x doesn't contain 0 so R^+ can't either as a subgroup, they must mean the positive reals

i.e. without 0

Otherwise it's not even a group

[chilly ocean]

so my part a is wrong

fuck

[chilly radish]

Yea but the reasoning that brought you there is almost correct

You just need to reexamine whether HK=G

[chilly ocean]

okay so

which parts of it is wrong

[chilly radish]

Your interpretation of R^+ as containing 0

[chilly ocean]

then my whole logic is wrong

i base my whole answer on that lol

[chilly radish]

R^+={x\in R:x>0} in this context

Well try to reexamine if HK=R^x now

Can you reach any element of R^x from multiplying an element of R^+ by 1 or -1

[chilly ocean]

hmm

imma just go from the top

i know H int K = 1

right?

[white jackal]

@chilly radish then what would the identity be of R+ is R+ if a subgroup of Rx

[chilly radish]

1

[chilly radish]

Yes

1 is the multiplicative identity in any subgroup of R^times

[white jackal]

is R+ additive or just all positive reals?\

[chilly radish]

Positive reals under multiplication

Since it's presented as a subgroup of R^x

[chilly ocean]

yeah

[chilly radish]

Also, for what you wrote to be true you have to have that H,K are normal in G. In this case everything is abelian so it's trivial

[chilly ocean]

how would i go about proving this then

[chilly radish]

Actually, that might not be necessary in this case

[chilly ocean]

idek what my answer could be

[chilly radish]

You could try just constructing a direct isomorphism between G and HxK

I'm not sure how the theorem you're basing your answer on was formulated in your class

[chilly ocean]

d

is what theorem i was trying to use @chilly radish

[chilly radish]

Alright

[chilly ocean]

so idk

[chilly radish]

Note that it requires normality

But G is abelian in a

So all subgroups are nor al

Normal

[chilly ocean]

yes so that would work

part d

[chilly radish]

Yes

[chilly ocean]

so all i gotta show now is HK ?

[chilly radish]

Note that there still might be an isomorphism if this condition isn't satisfied, it just won't be the multiplication map

Yes

For a

[chilly ocean]

how would i show HK = G

or check for it

in this case

[chilly radish]

But actually the exercise wants you to give aj explicit isomorphism

It says 'give an isomorphism' not just show it's isomorphic. Finding the inverse to the multiplication map would be the same as finding the isomorphism in this case

[chilly ocean]

ohh okay

so i have to create one?

okay

[chilly radish]

Yes

For b, consider that both K and H are cyclic and hence abelian, and a product of abelian groups is abelian

[dull root]

If I have a,b are complex numbers with degree 2,3 over Q respectively, and I extend to Q(b), then the a must have either degree 1 or degree 2 over Q(b). Why can it not be degree 1 over Q(b)

[tribal moss]

Because the degree of Q(a,b) over Q needs to be an integer multiple of both 2 and 3.

[dull root]

It would just mean that a is some Q linear combination of 1,b,b^2, but I don't see why that would cause issues

oh I see because 2,3 both need to divide Q(a,b)

[tribal moss]

Yes.

[dull root]

That gives an lower bound, but where does the upper bound come from

[tribal moss]

The upper bound of 2 for [Q(a,b) : Q(b)]?

[dull root]

no, the upperbound of 6 for Q(a,b): Q. We know 2,3 must divide it, so it has lower bound 6

[tribal moss]

Do you know that [Q(a,b) : Q] = [Q(a,b) : Q(b)]·[Q(b) : Q]?

[dull root]

Yes, and [Q(b):Q] = 3 here. so we see [(a,b): Q(b)] must be some multiple of 2. In fact it must be 2 since a has degree 2 over Q \subset Q(b)

Oh i see

[broken stirrup]

Hi guys

its been so long since when i took topology class, should I take algebraic topology?

im afraid i forgot some stuff necessary for the course

what level of point-set topology does it demand?

[chilly ocean]

familiarity with the first few chapters of munkres should be enough

up to and including the chapter on separation axioms

2-4?

[broken stirrup]

thank you

[barren sierra]

Can someone check a proof for this for problem 6? It seems too easy.

If I is a subset of P, then we are done. Otherwise suppose that IJ is a subset of P and I is not a subset of P. Choose i not in P. For all j in J we have ij is in IJ which means ij is in P. But P is prime so either i is in P or j is in P. By hypothesis we have i is not in P so thus j is in P. Overall J is a subset of P.

[weak matrix]

I not being a subset of P doesn't mean I and P are disjoint, so not necessarily i is not in P

[chilly radish]

He chose a specific i not in P. This exists by the negation of being a subset

[barren sierra]

^

[chilly radish]

The proof is good

[barren sierra]

damn

[weak matrix]

Oh, sorry

[barren sierra]

seemed kinda easy lol

neat

[chilly radish]

No worries

Well it's not a very deep statement

[barren sierra]

ya idk the rest of this class has kinda murdered me

so I was expecting more

Tho so far this is just a review of undergrad level material

so makes sense

[frank fiber]

if $x$ is in all maximal subgroups of a p-group G, then how can i prove that $x$ is in $G^p[G,G]$?

[cloud walrus]

Or x1

[patent crescent]

if N is a normal subgroup and Na = Nb, you can not necessarily say a = b right?

But it means a and b are in the same coset?

[delicate bloom]

yup

[barren sierra]

why is N_K a sum of [g] in K

doesn't K consists of elements?

not equivalence classes of elements?

[next obsidian]

[g] means every element in the equivalence class, as a set

[tribal moss]

I think [g] means the element of the group ring that corresponds to g.

In the second sum, g runs over all elements of the conjugacy class K.

[next obsidian]

Oh

Maybe that true

[tribal moss]

So what the entire construction works out to is "any element of the center can be written as an integer combination of all the [g]s where conjugate elements have the same coefficients".

[barren sierra]

hm

idk that doesn't quite type check for me in my head

oh oh oh wait

nvm I get it now

it's written better in the textbook lol

this makes more sense to me

and is IMO clearer than this double sum lol

[tribal moss]

It is still a double sum, it just writes $K_i$ instead of $N_{\mathcal K_i}$.

[cloud walrus]

Troposphere

[barren sierra]

Eh I guess

yea

[rapid slate]

Normality can be used but like since we don't know what element commutes with x or its power , it's tricky to proceed

This basically

That method just didn't lead to anything

This one works:

But yeah, thanks for taking your time to look at it!

[trim grove]

Please Help Me with This.

[chilly ocean]

eisenstein looks relevant

[trim grove]

But shouldn't they tell over what they wants to show irreducibility?

[chilly ocean]

if it has integer coefficients then it's probably asking about irreducibility over Q, since that's what the classic form of eisenstein's criterion is concerned with

[trim grove]

Okk, so by Eisenstein's Criteria, option A should be right?

[chilly ocean]

[cyan raft]

how can i even comprehend something like R[x]/(x^2+x)? All I got is some basic property about x and some multiplication property for 2 elements in the ring, but other than that, i'm stuck.

(how does the entire ring look? is it isomorphic to something more comprehensible?)

[thorn delta]

what is R? the real numbers, or an arbitrary ring?

[cyan raft]

oh, it's the reals

[thorn delta]

do you know chinese remainder theorem?

[cyan raft]

umm... yeah?

[thorn delta]

alright good, that is relevant here

[cyan raft]

i dont know how we can use crt here though

[thorn delta]

you can write (x^2 + x) in a different way such that it is more clear how it applies

[cyan raft]

x(x+1)

oh 🤦🏻‍♂️

[tender mist]

Note that the second one I sent doesn't require commutativity, as I wrote I think it should work as a solution for the exercise. Anyway your method seems good as well to me!

[rapid slate]

Hey could you give me a hint for proving it's injectivity

I couldn't do it

[wooden ember]

in the proofs ive seen that the field of fractions is unique up to a canonical isomorphism, they use the universal property of a field of fractions. Like if $A$ has two fields of fractions $K$ and $K'$, we produce unique morphisms $\alpha$ and $\beta$ from $K\to K'$ and $K' \to K$ respectively and then use that to get the morphisms $\alpha\circ \beta : K\to K$ and $\beta \circ \alpha : K'\to K'$ that make the inclusion diagrams from $A$ commute, and then use uniqueness to argue they must be the identity

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

my issue with this is that i dont see how uniqueness of $\alpha$ and $\beta$ in their respective diagrams implies uniqueness of $\alpha\circ\beta$ and $\beta\circ\alpha$ in theirs

[cloud walrus]

𝓛ittle ℕarwhal ✓

[next obsidian]

The identity will also be a map that works

Namely, you look at the maps A -> Frac(A) which are both the natural inclusion and this induces a unique map Frac(A) -> Frac(A) making the triangle commute

The identity works

[wooden ember]

but the identity doesnt necessarily factor through K' (or K)

[next obsidian]

No but if you compose alpha and beta

[wooden ember]

yeah it makes the same diagram as the identity commute

[next obsidian]

You get a map from Frac(A) -> Frac(A) also making that same triangle commute

And so now you apply uniqueness

[wooden ember]

but we dont know that there's any uniqueness there

[next obsidian]

Yes you do

[wooden ember]

why should uniqueness of a and b imply uniqueness of a o b

[next obsidian]

No that’s not what I’m applying uniqueness to

I’m applying uniqueness to “map from Frac(A) -> Frac(A) making the triangle commute@

Where the other two legs are the inclusion A -> Frac(A)

[wooden ember]

ohhhhh

im an idiot i see

you just apply universal property to that triangle right

[next obsidian]

And the other composition is also the identity for the same reason

Yeah

[wooden ember]

okay yeah

thanks

[next obsidian]

[tender mist]

Hey, sure, I'd start from this
$$(x k_1)^n = (x k_2)^n \iff (x k_1 (x k_2)^{-1})^n = (x (k_1 k_2^{-1}) x^{-1})^n = 1$$

[cloud walrus]

Mat

[rapid slate]

But how xk_1 and (xk_2)^-1 commute

[tender mist]

@rapid slate Oh right, I fell for it again! Thanks for pointing it out.
I may have even stumbled upon a counterexample for the injectivity, so.. my input was totally useless, I'm sorry.
I found it a very nice problem though, that's why my attempts, and thank you then for sharing your solution

[rapid slate]

No problem!
Glad that u found it useful

[rapid bramble]

"Show that 0 can never be a unit in a ring"

My proof thus far is BWOC sps 0 is a unit and a is its multiplicative inverse
then 0*a is both 0 and 1, so 0=1
But 1 is unique, so contradiction

Is this good / at all true? lol

[delicate orchid]

you can have a ring where 0=1 it's just trivial

if your ring is non-trivial then you can just state that 0=1 contradicts with the fact that the ring is trivial

[rapid bramble]

What would a trivial ring be?

[delicate orchid]

just the set {0}

it's the ring on one element

[rapid bramble]

right, so if R isn't trivial, my proof is fine?

[delicate orchid]

yup

but say it contradicts with the fact that the ring is non-trivial rather than 1 is unique

because yes, 1 and 0 are unique but that doesn't mean they can't be equal

[rapid bramble]

right

[potent briar]

"Suppose M is a subspace of a vector space V. Two vectors x and y of V are congruent modulo M, in symbols x = y (M), if x - y is in M. Prove that congruence modulo M is an equivalence relation, i.e., that it is reflexive, symmetric, and transitive."

I've never proven something like that, I'm lost.

[chilly ocean]

do you know what reflexive, symmetric, and transitive mean?

you need to prove those three things

[potent briar]

i "know" what they mean sure

but im not sure how to prove it

[chilly ocean]

you prove them by checking that they hold

[lethal dune]

it's like a template where u just put the appropriate symbols and equality

[chilly ocean]

for example, reflexivity means that every element is congruent to itself. can you prove that?

[pastel cliff]

is the $\lang \rang$ notation for cyclic subgroups universal or textbook-specific

[lethal dune]

universal, but LA ppl use it for inner products also

[pastel cliff]

yeah that's why im asking cuz that's the only other place ive seen it

mildly related but when is $\lang \rang$ used vs C_n

[lethal dune]

what's C_n?

[pastel cliff]

prof used it in class as notation for a cyclic group

oh is the angled bracks specifically for subgroups

[lethal dune]

that's non standard

[pastel cliff]

[delicate orchid]

oh good lord

[lethal dune]

ruined your art

[delicate orchid]

I've found all the indecomposable 1 dimensional ones at least (hint it's all of them )

[pastel cliff]

what's C_p

[lethal dune]

cyclic group

lmao

[pastel cliff]

[pastel cliff]

[lethal dune]

(was a joke)

[pastel cliff]

https://tenor.com/view/superman-joke-gif-7319521

[delicate orchid]

mfs who use \bZ_n for cyclic groups make me ANGRY

[pastel cliff]

\Z > \bZ

[delicate orchid]

it doesn't matter because the latex bot is down

[pastel cliff]

sadge

so is C_n interchangeable with angled brackets or not

[delicate orchid]

if you write C_n = <a : a^n = 1> you're just specifying that a is the generator

I hope that's what you meant

[pastel cliff]

pretty much yeah merci

[chilly radish]

Angle brackets mean 'subgroup generated by'

Angle brackets with line in the middle usually mean presentation of group, but that's a different thing

[chilly radish]

You can say <S> when S is some subset. In the case that S is just 1 element, this is the cyclic subgroup generated by that element

[chilly ocean]

Correspondence theorem is very stronk
Just proved that any ideal of Z[x] containing x-a must be of the form (x-a, n) for an integer n
In fact, if we already have a list of generators for this ideal then n is exactly the gcd of all the non-kernel generators evaluated at a

Are there any interesting corollaries to being able to reduce lists of generators of an ideal?

[next obsidian]

There's a rich theory of trying to find minimal sets of generators of ideals, and the problem is incredibly hard. This forms a lot of like, computation commtuative algebra

Even in the case of ideals of k[x1,...,xn] the problem becomes complicated

[chilly ocean]

Woah cool

[chilly ocean]

Similar idea with correspondence theorem gives us that all ideals in Z[x] containing x^2 + 1 must be of the form (x^2 + 1, ax+b)

With the non-kernel generators of any such ideal evaluated at i in Z[i] having gcd ai+b

[chilly ocean]

can someone check if my answers are right?

pls

[chilly ocean]

Wait... why wouldn't the proposition you cited in a) guarantee the isomorphism for b)

[thorn delta]

one of the subgroups must be normal to have injectivity

[chilly ocean]

<r> is normal, no?

[thorn delta]

wait hold on

no, im wrong, i think both have to be normal. trying to remember why

[next obsidian]

You get a semi direct product otherwise

[chilly ocean]

I'm not making any claim about actually using this to prove a result for b), just sayin that even if all that is true then there are conditions missing from the proposition being cited in a)

Because if there weren't you would get b) for free

[chilly ocean]

By "is not commutative" you mean G right
Cause the product on the right is commutative and so is <s>

yah G

Hence them not being isomorphic

Ah cool just looks like you're saying <s> is commutative which might confuse a grader lol idk

fuck this class

fuck the grader too

im self studying everything

no teaching effort

absolute mess of a class

Fucc that I'm sorry :(

Answers are def right tho

[thorn delta]

H is normal, right?

in part b

[chilly ocean]

Yeah I think so since if you introduce two s anywhere in an expression with just r otherwise the two s goes away

[thorn delta]

also yea because it has index 2

[chilly ocean]

Which accounts for conjugating by all elements in G

Or that lol

[thorn delta]

anyway, this is enough to show phi is an isomorphism

i was right before, you only need one subgroup to be normal

wait no im wrong

oops

[chilly ocean]

Wait wat but (r^a, s^j)(r^b, s^i) = (r^b, s^i)(r^a, s^j)

Yeah

So it is both then?

That need to be normal

I don't remember this theorem tbh but I can see that one direction follows from the isomorphism theorem about products and intersections

So it kinda makes sense

[thorn delta]

yea both being normal is sufficient.

[chilly ocean]

for part a do i show it normal

Yeah but that follows right away from the entire group being abelian so showing they're normal is a one liner

dope dope

[cyan raft]

another dumb question: do you always have a homomorphism from Z[x] (Z = the integers) to a ring R?
i know that there are infinitely many homomorphisms from Z[x] to Z, so, are there any ways to determine the number of homomorphisms from Z[x] to a given ring?

[chilly ocean]

I dont know an exact answer and not sure how exact of an answer you need, but for easy cases obviously if R has copy of Z then its gonna be infinite, not sure about the cardinality. To a finite ring there for sure is finite amount homomorphisms

[cyan raft]

i kinda already figure that part out, but thanks

[next obsidian]

It’s exactly the cardinality of the ring

[chilly ocean]

wait yea

evaluation

[chilly radish]

The question asked you to give an isomorphism par. You didn't do that in a

[next obsidian]

The R-algebra maps from R[x] to any R algebra S is exactly in bijection with S

It’s given by sending x to some s, then extending

Because of how Z works, any ring map at all is a Z-algebra map

[next obsidian]

To use less fancy stuff

The point is, just because of additivity

You’re forced to send n in Z to n inside of R

If you send x to r, then x^n has to go to r^n

Now mx^n has to go to m•r^n

[chilly radish]

There's no free object in Ring, right

[next obsidian]

And then for any polynomial in Z, by additivity you only need to know what happens to monomials

[next obsidian]

What do you want to be the property?

[chilly radish]

That's a good question

[next obsidian]

And does Ring mean all rings, or commutative ones?

[chilly radish]

Either

[next obsidian]

Because at least for commutative ones, I think Z[S] acts as a free ring on S

[cyan raft]

oh okay i finally got it, thanks chmonkey

[next obsidian]

As in given any set map S -> R, unique map from Z[S] -> R making it commute

[chilly radish]

Ohh yeah that makes sense

For general rings this wouldn't work

[next obsidian]

So the thing about R-algebras is just you replace n with an element of R

So like if you have an R-algebra S, you can multiply things in S by things in R like they’re scalars

And now inside of R[x], you’re forced to send rx^n to r•s^n if x maps to s

It’s the exact same thing with Z, it’s just that with Z we’re forced to send nx^n to n•r^n

By additivity

[cyan raft]

so, like, vectors are a \bR-algebra? just wanna make sure i got it correctly

[next obsidian]

Wdym?

If by vector you mean like

Tuples

then yeah they form an algebra

Like a(b1,…,bn) = (ab1,…,abn)

But there’s nothing special about the real numbers here

You can replace the reals with any ring

[hidden haven]

Ye can use polynomials in non commutating variables

And you get free ring

And more generally free R-algebra similarly

[chilly radish]

I think that's only for commutative rings tho right

[hidden haven]

Non commuting variables

Like in R<x,y>, x and y don't go past each other

So xyxy ≠ x²y²

This is can also be expressed in terms of the tensor algebra

If you want free R-algebra on S, you can take the tensor algebra of R^|S|

@chilly radish

[chilly radish]

I meant the base ring moldi

That this is free in the category of commutative rings

[hidden haven]

I only know of R-algebras over commutative R

[hidden haven]

Oh not exactly

I guess you can take the tensor algebra of (R/[R,R])^|S|

@chilly radish does this work

Because R maps to the center of any R-algebra

Which means that the image of R under the structure map is a quotient of R/[R, R]

So what if we just kill it from the beginning

Wait doesn't the tensor algebra take care of that already

Since we take tensor algebra over R

Yeah I think this should work for algebras over non commutative rings

[hidden haven]

I was once again thinking here that R maps into the center therefore is commutative don't know how many times I've made that same mistake but yeah I think construction works for any R

[cloud walrus]

Kraft Macaroni

[lethal dune]

why are you sending sigm(\sqrt(a)) to +- sqrt(b)? that's not in general a endomorphism

[woeful flint]

why not?

doesnt the galois group permute the roots, why is sending sqrt(a) to -sqrt(b) not count as a valid member of the galois group

[lethal dune]

roots of same irreducible factor

[woeful flint]

wait so the galois group must send permute the roots of irreducble factors

so that must mean that sqrt(a) can only get sent to \sqrt{a} or -\sqrt{a}

[lethal dune]

example Q(sqrt(3)) is not isomorphic (field) to Q(sqrt(2))

[lethal dune]

yes

[woeful flint]

But in this case isnt F = Q(\sqrt{a},\sqrt{b})

[lethal dune]

take Q(sqrt(2), sqrt(3)) then sqrt(2)^2-2 is zero in this field

the map sqrt(2) -> sqrt(3) we get sqrt(3)^2-2 \neq 0 but it must be zero to be an isomorphism which is a contradiction

[woeful flint]

ohhh wait so the galois group has to contain isomorphisms

[lethal dune]

automorphisms to be exact

[woeful flint]

our professor didn't introduce it to us as that

he defined it as the set of embeddings from F to F fixing the base field, in this case Q

[lethal dune]

ya that's not wrong

[woeful flint]

I guess from linear algebra that has to be an isomorphism

[tribal moss]

He probably took "embedding" to implicitly require it to be a field homomorphism.

[woeful flint]

because its an injective linear map between two vector spaces of the same finite dimension

AHHHHHH

that makes a lot more sense

our professor is quite good but he has a mean way of not explaining to us the full picture and leaves it on the question sheet

[tribal moss]

(or, as Ryu said, field field automorphism because otherwise you don't get a group).

[woeful flint]

forces you to figure things out by urself

okok gotchu

thanks guys

I have a feeling as I get closer to exams Ill be on this chat a lot

[tribal moss]

So speaking of "embeddings" is slightly misleading because that word sounds like they'll usually not be surjective (which is dead wrong in this case).

[woeful flint]

homological algebra got me fucked up

[woeful flint]

thats what I'm referring to when I say he leaves stuff out purposefully

a quick follow up

I'm pretty sure this tells us that G is the klein 4 group then right

bc it has 3 elements of order 2

which the cyclic group of order 4 does not have

[tribal moss]

V4 sounds right, yes.

[woeful flint]

nice

cheers guys

[potent briar]

"If S and J are arbitrary subsets of a vector space (not necessarily cosets of a subspace), there is nothing to stop us from defining S + J just as addition was defined for cosets, and, similarly, we may define aS (where a is a scalar). If the class of all subsets of a vector space is endowed with these «linear operations», which of the axioms of a vector space are satisfied?"

how can I add to subsets?

adding cosets made sense intuitively to me because they were "parallel lines" in my brain

so there was a "natural correspondance" between elements

but arbitrary subsets?

[lethal dune]

$A+B = \left{ a+b | a\in A,; b\in B\right}$

[potent briar]

do you add the one element of S with all the elements of J, the "next" element of S with all elements of J, etc. ?

[potent briar]

yes so that's all the combinations?

ok

@lethal dune in R^2, if you sum two different lines, do you get all R^2?

[lethal dune]

yes

[potent briar]

thanks

so i guess that doesn't satisfy vector sum axiom

wait or does it

it takes two elements of the class and gives you another

it's commutative

[coral shale]

elaborate what is meant by this

[potent briar]

it's right there

sum of two sets

[spice whale]

do you mean product

[coral shale]

Let A = {(0, 0, k)}
Let B = {(1, 0, k)}

Do we get R^2 from A+B?

[potent briar]

but those are 2 lines in R^3

[paper flint]

They said in R^2

[coral shale]

aaa

ok mb

Let A = {(0, k)}
Let B = {(1, k)}

I still don't think this is R^2

[spice whale]

is a sum of sets the same as their product

[paper flint]

I don't like the term "two different lines" though, what about parallel lines for instance

[potent briar]

oh yeah, non colinear non parallel

[paper flint]

Guess it's just equivalent to saying that the span of any two linearly independent vectors in R^2 is all of R^2

[lethal dune]

[potent briar]

nono

these are sets of vectors

[coral shale]

it is though

almost

[paper flint]

Your "lines" are 1D subspaces

[coral shale]

it can be shown to be equivalent with a little bit of algebra

[potent briar]

not if they don't pass through the origin

!!

[paper flint]

That's correct

[potent briar]

what they are is an equivalence class

1 line that is

modulo some parallel subspace

[coral shale]

A = {mx + c : m in R}
B = {ny + d : n in R}
A + B = {???}

Check.

[potent briar]

yeah R^2

[coral shale]

You will find you use this fact to show its R^2

[paper flint]

I see what you're getting at, but the underlying idea will boil down to what I said (modulo translation/rotation/reflection of the plane)

[coral shale]

You consider this with possibility of affine shift

[potent briar]

okay but i haven't gotten to all that yet

[coral shale]

? try this

[potent briar]

it's a simpler problem

i can paste it again

[coral shale]

What is the set A+B?

Then you see it boils down to the above

[potent briar]

i'm being asked if the class of all subsets of a vector space, is a vector space

[coral shale]

Then try checking the axioms one by one

[potent briar]

that's what i was doing

so if the sum of two subsets is a subset, and it's commutative

i don't think there's additive inverse though

at least not unique additive inverse?

[coral shale]

You can't say that without even figuring out what the identity is

[potent briar]

the set {0}

[coral shale]

Prove it.

[lethal dune]

[coral shale]

wait now im confused

I said prove it because i thought u cant. But {0} works for almost everything

[coral shale]

since theres {}

maybe thats not so obvious lel

@potent briar ^

ohh wait

god now im getting confused

{} + A = {}

{0} + A = A

🤦 mb

[lethal dune]

[coral shale]

So yes, {0} is the identity

Then you can check inverses

Note the axiom for inverses does not require them to be unique

That is only a consequence if you have a vector space

[potent briar]

no, they have to be unique

i'm trying to check if it's a vector space

book says «unique»

[coral shale]

which of the axioms of a vector space are satisfied?

The inverses axiom can be satisfied without requiring the inverses to be unique

[coral shale]

Thats odd, you mean in the axioms themselves?

The usual convention is to take uniqueness as a consequence

[coral shale]

D:

[potent briar]

hey hey hey

[coral shale]

Well if your book says it is so, then it is so.

[potent briar]

take it up with halmos

so, in the R^2 case (just because it's the one where i have the most intuition)

[waxen hedge]

you could try to look up «abelian group» or «commutative group»
Basically a vector space is an abelian group with an «external law» acting as a «multiplication» between elements of a field (often the reals or the complex numbers) and element of the group

[potent briar]

i know

but i'm doing this book, so this book is law

@coral shale could I write all of R^2 as {ax + by}, with a, b any scalar, and x, y the canonical base?

[coral shale]

uhhh you could???

but whats the point

I thought you were checking axioms

[potent briar]

and then an arbitrary line in R^2 as S = {nv + c} with n scalar, v and c vectors?

[coral shale]

well yes....

[potent briar]

I'm just trying to write what you get when you do R^2 - a line

[coral shale]

🤔

Im confused

What exactly are you trying to do

I thought you were checking the inverses axiom

[potent briar]

well in general just understanding

but sure, if two non parallel lines gives you R^2, and then you don't get one of the lines when you do R^2 - the other line

there's no inverse

[coral shale]

uhhhhhhhhhhhhhhhhhhhhh

That is not how I would argue for inverses existing or not

1. Find your identity ✅

2. Use the defn of inverse to find it (or show it cannot exist for some elements)

[potent briar]

of course it's not general

but i'm trying to understand the way arbitrary sets behave

[coral shale]

Do it in R^2 if you have to, but that is not how to do it

[potent briar]

compared to spaces

[coral shale]

Look, use the defn of inverse

A + (-A) = {0}

Given a general A in R^2, then...

Also you realise that the idea of subspaces/lines is kinda irrelevant to this question

You are allowed arbritrary subsets

This includes the empty set as well as funny stuff like circles, anything

===
Basically - if you want to get an intuition for this/whatever, I would advise doing that after finishing up the question

Then checking which axioms didn't work (and why)

It seems to be confusing you for what you need to be doing to go towards the answer

[chilly ocean]

Trying to identify Z[x]/(x^2 - 3, 2x+4). Since correspondence theorem tells us order doesn't matter in taking quotients by principal/finitely generated ideals, we can view this quotient as one where we kill all higher degree than linear terms, then killing off any non-monic linear terms. This leaves us with constants and monic linear terms as elements and for the latter with addition rule (x+a)+(x+b) = a+b-4 and multiplication rule (x+a)(x+b) = (1-(-1)^(a+b))x+3+ab-4floor((a+b)/2). Wtf is this thing??? I must be going wrong somewhere right? If anything, if this is actually true then correspondence theorem provides a neat way to prove really strange things are rings without worrying about distribution proofs

[tribal moss]

Hmm, the multiplication rule doesn't look obviously wrong.

[chilly ocean]

Unfortunately the more time that passes the more I am convincing myself that this is fine... Which makes me feel weird LOL this ring is wacky af

[potent briar]

ok so @coral shale it's not a space. scaling works but addition doesn't because S -S is not {0}. That is ignoring the empty set, with the empty set nothing works.

[tribal moss]

However, note that you also have 2x²+4x = 0, which reduces to 6 - 8 = 0.

[chilly ocean]

Huh ok so I guess a good rule of thumb with these is to play around a bit to see if you can get other relations out? That closes the case on this particular problem then anyway lol thank you

[tribal moss]

I can't offhand suggest any better rule than "play around", at least.

Perhaps something could be formulated in the special case of quotients of Z[x].

[chilly ocean]

In this case you get GF(4) so that's cool

[tribal moss]

No you don't -- you get a still slightly weird thing where (x+1)(x+1)=0.

[chilly ocean]

Wh

Damn

[tribal moss]

We can reformulate it as F2[x]/(x²-1), so we're trying to adjoin a square root of 1 to F2. But when 1 already has a square root, that kind of thing makes zero divisors.

[chilly ocean]

Pain

How would one go about showing Z[x]/(x^2 - 3, 2x+4) and this wacky ring with four elements are isomorphic? The obvious map is surjective and both sets are of the same cardinality so that does it actually

And structure keeps ofc

[tribal moss]

The wacky ring is just the dual numbers over F_2, by the way -- x maps to 1+epsilon.

[chilly ocean]

Yeah true I can see that with Chinese remainder theorem

Another one: identifying Z[x]/(6, 2x-1) I used 3=3(2x) =6x= 0 to whittle this down to something with at most the structure of (Z/3Z)[x]

I might have found a non-obvious zero divisor though

Yess 0= 2x^2 -x -6 factors nicely into (2x+3)(x-2) which reducing the left factor leaves us with x=2 kekw GAMING

That felt good

So it's just Z/3Z since now we have two finite rings, obvious isomorphism blah blah

[chilly ocean]

Ahh I just had an idea that I think we've subconsciously been deferring to, computing the gcd of sufficiently small multiples of the elements (since Z[x] isn't a PID) and then knowing that it must be zero as well, and will, depending on how small said multiples are, give you a really small element to kill

I guess in general you won't be able to get something similar to the process of getting a gcd but it works a lot of the time

[tribal moss]

I'm not sure how you eliminate the possibility that 2x=0 here. But reducing 2(2x-1)=0 directly gives you x=2

[chilly ocean]

Oh I got overly eager and thought the relation I started with was 2x+2=0 rip

Thanks

Wait how @tribal moss

4x-2 = 0 and then

[tribal moss]

Hmm, I think I took your earlier conclusion that 3=0 at face value, but now I'm not sure about that anymore.

[chilly ocean]

That I'm certain about since 3=3(1)=3(2x)=6x= 0

[tribal moss]

Oh yes, that checks out.

[chilly ocean]

Oh

Wait lol

Then reducing 4x=x ok cool cool

And x-2=0 that way as you said

Ahah

Thanks again

[potent briar]

ok im confused again hehe

"Suppose M is a subspace of a vector space V. Corresponding to every linear functional y on V/M (i.e. to every element of (V/M)'), there is a linear functional z on V (i.e. an element of V'); the linear functional z is defined by z(x) = y(x + M). Prove that the correspondance y -> z is an isomorphism between (V/M)' and ann(M)"

I think I need to use complementary subspaces somehow.

[chilly ocean]

why not just show directly that it's injective and surjective

[potent briar]

im having a hard time even imagining them

[chilly ocean]

why

you have the map given to you

[potent briar]

what form do functional on a quotient space have?

[chilly ocean]

x + M -> y(x + M)

that's a general question whose best answer is the exercise you're supposed to prove

[potent briar]

ok can you help me show thats its injective and surjective?

well hmm, i already know that to each x in a complement of M corresponds a single x +M

[chilly ocean]

write down what it would mean for this to be injective and surjective

with care. you have the map

[potent briar]

to a single z there needs to correspond a single y

and this has to be true for every single z in (V/M)'

[chilly ocean]

for every single z in ann(M), you mean?

[potent briar]

yes sorry

[chilly ocean]

so you have a functional z which vanishes on M

can you define a functional y on V/M with y(x + M) = z(x)?

that's what your map does, after all

[potent briar]

is x any vector in V?

[chilly ocean]

sure

[potent briar]

i can conceive the existence of a function that takes a set in V/M and returns 0

I'm not sure how to define it though

[chilly ocean]

you need a linear function y: V/M -> F with y(x + M) = z(x) for all x in V

that looks like a definition, no? maybe you can show that you can do this and get a well defined function

[potent briar]

hmm

right

indeed that feels bijective

[potent briar]

but @chilly ocean don't i need to define a function that goes from (V/M)' to Ann(M)?

and see if that is a bijection?

[chilly ocean]

Before I text dump checking if the bot works again lol

$aaaa$

[cloud walrus]

zd

[chilly ocean]

Ayyy nice

Checking my logic here showing that $\mathbb{Z}[x]/(x^2 +7)$ and $\mathbb{Z}[t]/(2t^2 + 7)$ can't be isomorphic. If such an isomorphism existed, then it would be surjective so we have $\varphi(r) = t$, and $0=-7+7=\varphi(x^2) + \varphi(-2r^2) = \varphi(x^2 -2r^2) =0$ and we have $x^2 = 2r^2$ by injectivity.

But all the elements of the first ring are of the form $ax+b$ and we need an element such that $-7=2(ax+b)^2= 4abx-14a^2 +2b^2$

[cloud walrus]

zd

[chilly ocean]

And that's not going to be possible
I guess the big stinker in this is showing that all elements in the first ring really are of that form but I think it's fine? Aaaaa

[cloud walrus]

zd

[chilly ocean]

WAIT way easier way I just realized, since these are correspondence theorem related exercises: if they're isomorphic we could just kill x-1 in Z[x] before going down to the other two quotients respectively. In the first case you get Z/8Z and in the second Z/9Z by correspondence theorem

Jeez

That's pretty cool

[chilly ocean]

you did. it takes a functional y on V/M to a functional z on V vanishing on M. it's that which you want to show to be bijective

and to do so you show that to each such z there's one and only one such y

[potent briar]

hmmm but the questions says that z is a member of V', not Ann(M)

so how do i go from that to proving it's an isomorphism between (V/M) and Ann(M)

Is it because Ann(M) is isomorphic to the dual of a complement of M?

[chilly ocean]

sure, z is a member of V'. it's also a member of Ann(M), as you should show.

[chilly ocean]

(V/M)' and Ann(M). you prove that it's an isomorphism by proving that it's injective and surjective

(because hopefully linearity is clear)

[chilly ocean]

Consulting math.se this way of doing it is completely wrong since it kinda assumes the ideals (x-1) are the same in both cases which they are NOT since x doesn't even get sent to x in that (ends up being nonexistent) isomorphism

Ugh wtf is wrong with me

It's hard getting humbled like that

Also hard to avoid if you ever want to have confidence in a result and not just be begging around all day for validating the result

Cause you will inevitably make a silly mistake and simultaneously be somewhat confident in it

[trail stump]

how is the field of quotient /f thing different from field or quotient /L thing?

[tribal moss]

/_F is the division operation in the field F (which by definition is the field of fractions of D), whereas /_L is the division operation in the field L (an arbitrary field that happens to contain D as a subring). The original D is a subring of either of the fields, but two elements of D that are not divisible in D itself have one quotient in F and an different quotient in L, so two different symbols are needed.

[tender mist]

Your reasoning did make sense to me though

Mm I had sent a LaTex but actually now I see what they mean

Mm again, but we could use 2 instead of x-1, because 2 is fixed by an eventual isomorphism? Then we wouldn't have that problem

[trail stump]

Ok I guess I am not understanding how with one integral domain can have two different enlarged field of quotient for two different field. Can you give an example of it?

[tribal moss]

L is not necessarily a field of fractions, just a random field that happens to contain D as a subring. There's no telling how its elements look, and in particular there's no guaranteee that they will be the same mathematical objects as the elements of the field of fractions.

That being said, the very point of the theorem you're reading is that they can't be all that different, because F is automatically in bijective correspondence with a subfield of L. But of course you can't depend on that knowledge while you're still proving the theorem itself, so at last until the proof is concluded you need to distinguish which of the two fields' division operation you're speaking about.

For example D could be $\mathbb Z$, and then F would just be $\mathbb Q$. We could take L to be, ... hmmm ..., the field of meromorphic functions $\mathbb C\to\mathbb C$, where we have replaced every constant function with an integer value with that integer itself, such that it's genuninely a superset of $\mathbb Z$. However, in this L, the function you get by dividing 1 by 2 is \emph{not} literally the rational number $\frac{1}{2}$, but a \emph{function} that returns $\frac{1}{2}$ for every complex number you put into it.

[cloud walrus]

Troposphere

[tribal moss]

So $1 /_F 2$ is the rational number $\tfrac{1}{2}$, and $1 /_L 2$ is the function defined by $f(z)=\tfrac{1}{2}+0i$.

[cloud walrus]

Troposphere

[chilly ocean]

Oh clever yeah!!! Killing a constant removes that concern entirely

Leaves us with more to do though I think

But it probanly makes things slightly easier at least

Well wait, killing 2x^2+7 after means killing 1 so that one is the zero ring

And then likewise killing x^2 +7 after means killing x^2 +1 so you get something cool and nonzero, still contradiction NICE

AHAH I will write it up more clearly, edit, and undelete my math.se post muahahah

This angle on the problem isn't in the answers at all so that was my reason to even try

Thank you Mat!

[prisma shuttle]

that's the way i learned it, idk if there are other common ways of proving it

also in evan.sty does anyone know how i can change the font size

[lavish nexus]

what is a weight of a Lie group action on a vector space?

[hidden haven]

@brave trail

[chilly ocean]

Hey. I WTS given R-module M, M nontrivial iff Supp(M) nonempty.

M nontrivial implies there is element x in M, So localization of M at p a prime ideal has an element x/1 given by the homomorphism from localization.
We know x/1 not equal to 0/1 because otherwise this implies x*k =0 for some k in R-p. This is a contradiction because ?? Im not exactly sure but

[hidden haven]

Ye

[chilly ocean]

Oh wait

Maybe it would imply that k is in Annihilator of some subset of M which is an ideal

Nah not sure

There has to be easier way

Can k be an annihilator?

if not the other ideal is taking set of zero divisors as the ideal S so R-S cant have zero divisors

but i think my method should work with any prime ideal

I think thats safe

nvm

set of zero divisors isnt a prime ideal

Wait the statement I want to show is equivalent to M= 0 iff support is empty

which is easier to prove

M=0 implies support is empty immediately

if support is empty then the localization of M at any prime ideal is empty. And we know if localization of M at all prime ideals empty implies M is trivial. By a small proof. Pretty much We assume M is nonempty so there is x in M. We look at Ann(x) which is contained in a maximal ideal. And if we localize at this maximal ideal we know that it is empty. So x/1=0/1 this is a contradiction because this means xk=0 for some k in the complement of the Annihiltor of x.

[coral shale]

Which one is better notation for each case?\
Let $f\in K[x]$ or $f(x)\in K[x]$\
Ideals $(f)$ or $(f(x))$\
Quotient $K[x]/(f)$ or $f(x)\in K[x]/(f(x))$

[cloud walrus]

Shuri2060

[coral shale]

My galois lecturer chooses f in all cases I think

f(something) indicates the evaluation map.

On the other hand, my other lecturer (algebraic NT) uses the other notation f(x)

Personally I think the left makes more sense but 🤔

I guess it shouldn't matter as long as it's clear/consistent but...

[chilly ocean]

the function itself is better

looks cleaner

and if you use letters f,g,h its usually clear you are talking about functions

and you cant really mod out by functions

[coral shale]

I was thinking to myself if it was an abuse of notation, but no not really

[chilly ocean]

unless its a ring of functions

its exact opposite

[coral shale]

yh

[chilly ocean]

if you think of polynomials as elements of an infinite direct sum then it all makes sense

[coral shale]

could you elaborate on this?

as in sequences on K?

K[x] is

[chilly ocean]

not sequences because those are infinite

[coral shale]

K[x] is the set of finite ordered tuples of K

[chilly ocean]

yes

[coral shale]

K + Kx + Kx^2 + ...

is it this?

with circle plus

[chilly ocean]

mhm

[coral shale]

for direct sum

[chilly ocean]

so finite number of nonzero K

yeah. like this

[coral shale]

right ok

[chilly ocean]

$\bigoplus^\infty K$

i tried

[cloud walrus]

Toysem Teans

[chilly ocean]

what are u learning in algebraic nt class?

im very interested!

[coral shale]

im behind, so honestly not much clue yet lel

[chilly ocean]

same here!

do you know the upshot of the class?

[coral shale]

no

[chilly ocean]

what are the cool motivating theorems?

or are you learning bit by bit

[coral shale]

No, im probably near the starting line

[chilly ocean]

Can you tell me some things you learned that you found interesting?

[coral shale]

Ive yet to digest what the norm, tr, discriminant are

i have some indication

but havent properly sat down and figured it out

[chilly ocean]

trace and discriminant for what?

polynomials?

[coral shale]

uhh what was it

we have a number field

[chilly ocean]

oh cool which one

[coral shale]

uhhhh

[chilly ocean]

im taking class on iwasawa theory

and its first time im working with number fields

[coral shale]

Thats norm and tr

[chilly ocean]

ring of integers of K?

i forgot the definition 😣

oh

[coral shale]

L, K are number fields

[chilly ocean]

ring of all algebraic numbers of K

K embeddings of L in C im guessing are functions L->C that fix K?

[coral shale]

yes

[chilly ocean]

woooo

[coral shale]

O bar is the set of all algebraic integers

O_K is O bar intersect K

in my notes at least

[chilly ocean]

what the heeel

I dont see O bar in screenshot

[coral shale]

its elsewhere

[chilly ocean]

what textbook?

[coral shale]

my lecturers notes

[chilly ocean]

they look so nice

I wish I could learn algebraic nt like you

im taking a seminar class and the pacing is frightening

also im not sure if number theory is my thing

[coral shale]

im not sure either atm, feeling galois a lot more

[chilly ocean]

i enjoy the arguments but the results dont motivate me

[coral shale]

ant being applied results makes me go 🤔

[chilly ocean]

do you know of any?

[coral shale]

no i dont lol

but i get the feel from how the course is structured for me

A lot of statements could have been made for generic rings/fields, but instead are written specifically for number fields

[oblique leaf]

Can anyone tell me what map from F0 to M would make this sequence exact (and why)? I am assuming the map from F1 to F0 is just the inclusion

[hidden haven]

M is that direct sum

And F has the same number of summands

on the ith summand of F, you do the quotient map

[oblique leaf]

@hidden haven that's what i thought, but the kernel of that map would be $$(a_1)\oplus ...\oplus(a_m)$$ no?

[hidden haven]

By F I mean F_0

[cloud walrus]

alyosha

[hidden haven]

Ah yes

But those are isomorphic to R I think

Ye they are isomorphic to R as modules

Multiplication by a_i is the isomorphism R → (a_i)

[oblique leaf]

@hidden haven why are they isomorphic to R? because aren't those just the principal ideals

[hidden haven]

Yes but this is true for any principal ideal in a domain

[oblique leaf]

any principal ideal in a domain is isomorphic to the domain?

[hidden haven]

yes

[hidden haven]

by this

and

[oblique leaf]

oh ok thank you! related question: can we also construct the short exact sequence $$0\to M\to Q0\to Q1\to 0$$ if we apply the inclusion map to M, let Q0 be the injective module that the image of M is embedded in, and let Q1 be coker(i)? This is still for finitely generated M btw

[cloud walrus]

alyosha

[next obsidian]

You could certainly do this, yes

But showing the existence of Q0 is hard, and the only reason you can conclude Q1 is injective is to know that over a PID that injectivity is equivalent to divisibility

[oblique leaf]

but there is a theorem that any module is contained in an injective module

[next obsidian]

Yeah, but do you know the proof?

It’s quite involved

[oblique leaf]

we proved it in class

[next obsidian]

Oh okay then sure

[oblique leaf]

also for Q1, we know that over a PID, quotient of an injective module is injective

[next obsidian]

Ah okay well if you know that

Like I said, this follows from the fact that injectivity is equivalent to divisibility

As it’s clear the quotient of a divisible module is divisible

I just didn’t know if you knew these facts, if you have those two facts then this works

:)

[oblique leaf]

ok, thanks!

[next obsidian]

I assume you asked this because you know about global dimension?

Every module having a length 1 projective resolution means everything also has a length 1 injective resolution abstractly, but it doesn’t give you a way to turn a projective resolution into an injective one

Here you’ve given a direct proof that every module has a length 1 injective resolution

[oblique leaf]

oh yeah, i was mostly wondering because those constructions seemed trivial

[hidden haven]

oh yeah,

[lethal dune]

<g> is cyclic of order p (prime) then every non identity element has order p

[chilly ocean]

?

Oh in a domain

[lethal dune]

idk what u r trying to do

[chilly ocean]

why though?

cant domains have multiple generators

[hidden haven]

Multiplication by the generator is an isomorphism from the domain to the ideal

[hidden haven]

ye choose any one

[chilly ocean]

yes those generate the ideal as R mod

but you cant go other way around?

going from domain to R

rip

i meant going from principal ideal to R

as R mods

[hidden haven]

Multiplication by the generator is a bijective homomorphism onto the ideal

[bright pasture]

not the order of k

[hidden haven]

It's surjective by the definition of the principal ideal generated by an element and injective because domain

[bright pasture]

and i know that k^b = k

[hidden haven]

We assume that the ideal is non zero ofc

[chilly ocean]

yeah

its just so weird

where does domain definition come in?

oh

anihillators of the ideal?

nah im not sure

[hidden haven]

No, multiplication by a non zero divisor in any ring is injective

Because it is a module homomorphism with trivial kernel

[chilly ocean]

i didnt know this

but that makes a lot of sense

i need to level up my comm alg a lot

one more year and ill be confident 🙂

A little off topic

what do you recommend for someone who took a first course in AT and wants to continue but has weak understanding. i studied from hatcher

should I just spam hatcher exercises

or look to other resources

[hidden haven]

Yeah exercises are always good

Shin keeps recommending Rotman which you can try if you wanna switch books

Hatcher exercises are good

[chilly ocean]

i have time I think

im still in school

[hidden haven]

oh wow ok

[chilly ocean]

but im not a phd student or anything yet

so i think i can go at safe pace

[hidden haven]

Oh wait school in India means like high school or lower so I got confused

[chilly ocean]

i have no rush right?

oh yeah my bad

im 3rd year lol

[hidden haven]

I mean you can always aim higher

[chilly ocean]

im at weird spot where im not sure what area of math i love yet

[hidden haven]

You can try reading harder books since you have a basic idea about the subject

[chilly ocean]

i am really interested in ideas in AT

and the applications are super cool

yeah I think I should try harder books

but idk any clear pathways

like focusing on homotopy theory more

[hidden haven]

AT has plenty of hard books,
Moth is doing Fumenko Fuchs and seems to like it
I am reading May
Dieck is another good one

[next obsidian]

Haha Dick

[hidden haven]

I think all of them assume good familiarity with cat theory

[chilly ocean]

dieck no grothen?

[hidden haven]

Tammo tom Dieck

[chilly ocean]

yeah cat theory is something ive accepted ill pick up as i get older

same with number theory weirdly

but that isnt true

[hidden haven]

At some point you need to put in active effort lol

[chilly ocean]

yeah i think im at this point

[hidden haven]

With cat theory it is very worth it

[chilly ocean]

because i hear of all big important things and kinda want to commit to learning them

in my comm alg class something about commuting limits came up

and some theorem about adjoint functors

[hidden haven]

Ye these are good things to know

Knowing about adjunctions can make your life a lot easier

[chilly ocean]

yeah i dont think cat theory is too hard to digest either

but i didn’t see that much

[hidden haven]

Ye depends on the person

[chilly ocean]

i have been trying to think a lot in terms of universal properties though

[hidden haven]

Good first step 😌

[chilly ocean]

oh wait

May AT?

the one with a bunch of cat theory stuff

[hidden haven]

Ye

Concise course on AT

[chilly ocean]

that one always looked very hard

[hidden haven]

Ye it is

[chilly ocean]

how are you liking?

[hidden haven]

Definitely not recommended for first read

I'm liking some parts and hating others

I'm having to use other references while going through it

[chilly ocean]

hating like torus knots?

or less than that

[hidden haven]

And it is slow but rewarding

[hidden haven]

Similar I'd say lol but at least May does everything formally

[chilly ocean]

oh thats 😊

[hidden haven]

😌

[chilly ocean]

i kinda despise hatcher

but im not above using their book and floating around others

munkres has a book ive heard

but its really chonky

[hidden haven]

Munkres is for point set

[chilly ocean]

one for algebraic topology

[hidden haven]

Ye then you should try Rotman I think

[chilly ocean]

we referenced it in my class

[hidden haven]

Oh I haven't heard of this

[hidden haven]

ohh

[chilly ocean]

we referenced it a lot near end of course

[hidden haven]

Title sounds like it's supposed to be a reference book rather than textbook

[chilly ocean]

what is most recent thing you learned in mays or in AT in general?

or coolest to you?

[hidden haven]

Chain homotopy lmao I never knew that a chain homotopy between maps X → Y could be viewed as a map X ⊗ I → Y where I is the simplicial chain complex for the interval

If you're asking about specific theorems or cool things

Fibrations and cofibrations were really cool

Recent thing is axiomatic homology

[chilly ocean]

never got to learn about co fibrations

ppl in a lie groups course telling me a lot about them

or ig one specific instance of them being used to make a long exact sequence

[hidden haven]

That is also very nice, you can define a homology theory on just CW complexes then extend that to all spaces using the cellular approximation functor, and then derive all the main properties from simple axioms 😌 then you prove existence of such a homology theory by constructing one

I guess cellular approximation functor is the main cool thing here lol

[chilly ocean]

cellular approximation functor?

[hidden haven]

Any space X is weakly equivalent to a CW complex

And there is a functorial approximation

[chilly ocean]

weak equivalence i forgot the definition rip

it was a long one iirc

[hidden haven]

Meaning that for every X you have FX a CW complex

Weak eq means that there is a map that induces isomorphisms on all homotopy groups

[chilly ocean]

im curious for the proof that there exists a weakly equivalent cw complex for any space x

is it convoluted? or somewhat straightforward

because idk how i would start

[hidden haven]

I didn't find it that interesting, it was just induction

[chilly ocean]

oh so u start with X and you have what its homotopy groups are

and you artificially build up associated CW complex

that have same homotopy groups

[hidden haven]

Inductively define the n-skeleton of the FX such that at each stage, the nth homotopy groups agree

[chilly ocean]

oh ok

[hidden haven]

Or maybe (n-1)th homotopy groups agree

[chilly ocean]

i love waving my hands

[hidden haven]

👋

[chilly ocean]

but thats a good enuf argument

[hidden haven]

Lol

[chilly ocean]

this does sound like fun tho

when i grow up i wanna make a homology theory

[hidden haven]

[chilly ocean]

im also interested in motivating history behind different homology theories

because de rham seems like it came out of nowhere

but its super cool example

[hidden haven]

Ye the main idea ig is that integrals of differential forms turn out to be homotopy invariant

[chilly ocean]

i like general theme of finding failure of types of sequences to be exact also

[hidden haven]

ye it's very nice

[chilly ocean]

seems like i learn something fundemental

but pretty hard to conceptualize

[hidden haven]

I recently learned about simplicial objects and their homologies, and how any adjunction induces a simplicial object and hence a homology theory

Clerk keeps talking about this

And it is very cool

[chilly ocean]

idk what simplicial objects are. im guessing objects of simpicial category

[hidden haven]

lmao

[chilly ocean]

but thats something that idk what it is

[hidden haven]

Objects of the category of simplicial objects 😌

[chilly ocean]

lol

[hidden haven]

oh wait you weren't joking

There is this category called the simplex category

[chilly ocean]

yeah

i went on nlab deep dives

[hidden haven]

Simplicial objects are contravariant functors out of that

nice