groups-rings-fields

[velvet hull]

but then by the correspondence theorem, a composition series for G/K corresponds to a series in G that starts at K and ends at G

so you glue the two series together

[barren sierra]

I'm trying to use this to show that if G has a comp series then K, G/K have a comp series

[velvet hull]

hmmm, the other way around I see

[barren sierra]

yea

[distant summit]

Ok. What does he mean by "not claimed to be logically independent"?

Is he saying that some of the axioms imply other axioms?

[rocky cloak]

Take your composition series for G and intersect it with K.

And take the images in G/K.

Then this will give composition series except the same group may be repeated several times (so a trivial composition factor if you like)

[barren sierra]

oh just intersect with K

I see I see

and yea G/K that one I already got, thanks

[rocky cloak]

Actually, this doesn't work for non-abelian groups. I was thinking modules, but I'm guessing you're on groups

[barren sierra]

yea 😔

[charred iris]

Yes. Technically he's saying that they might ("not claimed to be"), but in fact they do

[rocky cloak]

Wait, no it works fine. You just need to use that K is normal

KnGi / KnGi-1 is a normal subgroup of Gi/Gi-1

[barren sierra]

yea just worked it out

thanks!

[distant summit]

So which axioms imply other axioms (and how)?

Isn't the point of axioms that they are essentially the minimum number of things you need to assume for a theory to be useful

[radiant knot]

What is meant by KnGi? Notation is a little vague.

[charred iris]

They're a small number, but no one claims they have to be minimal

[barren sierra]

$K \cap G_i$

[cloud walrus]

Spamakin🎷

[charred iris]

Even the usual presentation of the axioms of set theory isn't minimal

[distant summit]

Ok

[barren sierra]

where the G_i are the composition series for G

[distant summit]

So in this context which axioms are "surplus to requirements"?

[radiant knot]

Thank you. So it is an intersection of group?

[barren sierra]

yes

[charred iris]

I know 1 follows from some of the others

if you do (1+1)(x+y) then depending on which distributivity you apply first when expanding you get different expressions

and then axioms 2, 3, 4 on their own are enough to show (-x) + x = 0 and 0 + x =x for all x in V without needing 1 (with 1 these are obvious)

[distant summit]

1 under which letter?

[charred iris]

oh was it resetting

A

commutativity of addition

[charred iris]

and then this is enough to cancel anything extra from the two expressions you get here

[distant summit]

I'm not sure I follow your reasoning

(1+1)(x+y) can be expanded as

1. (1+1)(x+y)=(1+1)x+(1+1)y by distributing over vector addition
2. (1+1)(x+y) = 1(x+y)+1(x+y) by distributing over scalar addition

How would these imply commutativity of vector addition?

[glad osprey]

keep expanding

[rocky cloak]

x + x + y + y = x + y + x + y
subtracting gives
x + y = y + x

[distant summit]

Ah

Ok

[vivid kestrel]

https://en.wikipedia.org/wiki/Invariant_basis_number why would this proof that commutative ring have a well defined rank of their modules not hold in a non commutative setting? one would need to take I a maximal two sided ideal but since the rank over a division ring is well defined as far as i am aware, i dont see where commutativity is required (i know that commutativity is required, i just fail to see how its used in this proof)

[coral spindle]

It does not hold in a noncommutative setting, yes. Here is the standard counterexample.

[coral spindle]

I'm imagining the proof passes through a field quotient. A noncommutative ring may not have a field as a quotient, because noncommutative simple rings need not be fields (or even domains).

See e.g. M_n(k) the ring of nxn matrices over a field k

[vivid kestrel]

if we take a maximal two sided ideal, wouldnt the quotient still be a division ring

[coral spindle]

Exercise: M_n(k) has no nontrivial two-sided ideals

[vivid kestrel]

oh fair

hmm

is krulls theorem only for one sided ideals?

[coral spindle]

No, maximal left, right, or two-sided ideals still exist in noncommutative rings

that's not the problem

The problem is that simple rings aren't necessarily division rings

[vivid kestrel]

oh i see my mistake

[charred iris]

note that you need to subtract 'from the left' to do this, which is why I mentioned needing 2, 3, 4 to prove that you can actually do that

[vivid kestrel]

i had assumed that the quotient by a maximal ideal ought to be a division ring since no non zero element generates a real ideal but of course in a non commutative ring having one in the ideal generated by an element does not imply that element to be a unit

thanks!

[coral spindle]

Well yes I also gave you an example of M_n(k) so

[vivid kestrel]

yep, thanks

[spark veldt]

is it normal for undergraduate students to study this much algebra?

my university doesn't even offer finite field -> lie algebras

i've only taken abstract group theory, sylow theorems, some ring theory & field theory from Abstract Algebra and that's it

[thorn jay]

funny that there's a whole module just dedicated to the Nullstellensatz

[barren sierra]

I did not lie lie algebras in my undergrad or even my grad algebra courses

lie theory stuff I've only seen in separate courses

same with rep theory

and nullstellensatz only appeared in algebraic geometry which not every UG takes

[coral spindle]

This is pretty much everything I did in my undergrad in algebra, besides a few specialist things

[spark veldt]

T.T

[languid trellis]

are you classifying undergrad as 3 or 4 yrs?

[coral spindle]

oh

integrated masters

4 years

good catch

[charred iris]

I think at my university (not a good representation of average undergrad) Lie algebras are fourth year (integrated masters), but the rest should be undergrad (I did not take much algebra in this direction)

[thorn jay]

abstract field theory, finite field extensions, and galois theory are basically just "field theory", lol

[vivid kestrel]

finished my second year right now and did what i imagine is covered under abstract group theory, sylow theorems, abstract ring and field theory, galois theory and in an elective representation theory if finite groups so make of that what you will

[thorn jay]

what can you even say abouy fields thats not field extensions

[languid trellis]

you can classify finite fields

[thorn jay]

that uses finite field extensions though

[languid trellis]

oh i see what ur saying

[velvet hull]

how would you classify finite fields w/o galois theory lol

it wouldnt be a clean proof

[thorn jay]

absolute pain

[dull ginkgo]

I like how classical Galois theory is just like 3 lemmas lmfao

[languid trellis]

you can show all fields are char p or char 0 lol

[vivid kestrel]

showing that finite fields have order p^n is doable without galois theory and cant you appeal to the uniqueness of splitting field and the fact that the field with pn elements is the splitting field of the polynomial p^n? what would remain to be shown

[dull ginkgo]

The humble initial object Z

[velvet hull]

yeah, and I consider that to be galois theory already

[thorn jay]

what a course

[vivid kestrel]

oh okay, fair

[languid trellis]

you can look at the field of fractions of an integral domain if u want

[thorn jay]

and that would fall under finite field extensions anyhow

[languid trellis]

I'm being obtuse though. I love k[x]/ something

[vivid kestrel]

oh yeah i know i was just surprised by the statement that classifying finite fields would require galois theory. depends how you define the bounds of galois theory i guess lol

[thorn jay]

i agree with the sentiment that field extensions are Galois theory but i get it if you want Galois theory to just be about the study of automorphism groups of fields

[dull ginkgo]

Galois theory is basic field theory imo

[coral spindle]

Huh I always thought of that as a Galois theory thing

Interesting to hear that perspective

[vivid kestrel]

might just be because of the first exposition i got but for me galois theory starts when you look at the galois group which my prof didnt do until somewhat later

[dull ginkgo]

I mean it is pretty intricately related to Galois theory naturally

Finite field -> every function is a polynomial by interpolation -> Frobenius Endo/Automorphism

[rocky cloak]

R/I is a division ring iff I is a two sided ideal that is maximal as a left ideal (so not just maximal as a twosided ideal)

[languid trellis]

bro the frobenius map kinda goated like holy shit

its actually used so much

its insane

mfw x -> x^p

[dull ginkgo]

Division rings are left-simple :3

And right simple but :p

That’s how I interpreted them originally

Division rings are simple modules over themself :3

[rocky cloak]

Since R is called semisimple if it's semisimple as a module, we should call R simple if it is simple as a module.

Then simple ring = division ring.

Not sure what to rename simple ring, we'll think of something

[dull ginkgo]

bisimple

(I get the sarcasm though)

[rocky cloak]

Queersimple

[dull ginkgo]

Sesquisexual

[dull ginkgo]

I need to revisit algebra at some point

[thorn jay]

oh god more discourse between ring theory and UA
first two different notions of "algebra", now two different notions of "simple"

[rocky cloak]

Does UA have a different meaning of simple?

I guess you just meant if we change the definition of simple ring, cuz right now they're the same

[thorn jay]

yeah that

congruence lattice consists of two elements, stuff

[white oxide]

Does this work for one direction? Suppose that $S^{-1}M = 0$, and let ${m_i}_{i = 1}^n$ be a set of generators for $M$. Since $S^{-1}M = 0$, there are $s_i \in S$ with $s_i m_i = 0$ for $1 \leq i \leq n$. Set $s \coloneqq s_1 \cdots s_n$. If $m = a_1m_1 + \dots + a_n m_n \in M$, then
[sm = (s_1 \cdots s_n)(a_1m_1 + \dots + a_n m_n) = 0] whence $sM = 0$.

[cloud walrus]

okeyokay

[dull ginkgo]

I wonder if all this stuff works for ore localization

[white oxide]

I still don't get it

[dull ginkgo]

Ore localization is a generalization of localization for noncomm rings

But the multiplicative set S needs to be well behaved and have some “normal” esque properties

[elfin wraith]

No one does, just don’t worry about it, noncom localisation is silly and scary

[thorn jay]

I mean technically you can localise any noncommutative ring, but you just won't be able to use it at all basically

for every element s in S, add a noncommuting variable u_s, and quotient by the set of su_s - 1 and u_s s - 1

[languid trellis]

I always need to revisit algebra at all points. There's just far too much of it. I've decided to do other more interesting things and do algebra only when it helps with some number theory or geometry question, otherwise i feel its too dry and i forget it

[thorn jay]

"more interesting things" 😢

[languid trellis]

Maybe what I meant is 'less dry'

[white oxide]

Why is (p^n) primary in Z for n > 0 and p prime

I want to use Euclid's lemma

[thorn jay]

because by the fundamental theorem of arithmetic, a number divides p^n iff it itself is a power of a prime, and hence is nilpotent in Z/(p^n)

[white oxide]

Did you mean p^n divides a number?

[rocky cloak]

I mean Euclid's lemma just says "(p) is a prime ideal in Z"

[thorn jay]

no

[thorn jay]

hmm this argument doesnt work, nvm

[rocky cloak]

If xy is a multiple of p^n, then one of x and y is a multiple of p, by FTA

[vivid kestrel]

Ah that makes sense, thanks

[white oxide]

Is this because p | p^n | xy, so either p | x or p | y?

[thorn jay]

rather, if xy is a multiple of p^n and neither x or y are a multiple of p^n, then both x and y are a multiple of p

[dull ginkgo]

What are you working on rn

[languid trellis]

ireland and rosen. just finished chap 5 today

[south patrol]

Good book

I recognised the (p) notation from that aha

[dull ginkgo]

Prime ideal remote detonation

[languid trellis]

Yeah I've been enjoying it a lot. I love the way its paced and presented. super logical and each thing builds from the last. very satisfying. Hoping to write my second year essay on eisenstein reciprocity (chap 14 or something i think).

[white oxide]

I just started on ch 2 with this book, how many exercises did u complete from each chapter roughly

[potent condor]

stupid question: if a ring element f is in every maximal ideal containing a prime P, then must f be in P?

[proper parcel]

Not true in general

ℤ localised at the prime ideal 2ℤ is a good counterexample, take p=0, f any nonzero nonunit. (Any local ring of dimension ≥1 works the same)

[proper parcel]

Commutative rings which do have this property for all primes are known as Jacobson rings, sometimes Hilbert rings. Many properties of f.g. algebras over a field are also true of Jacobson rings, including a generalisation of the Nullstellensatz

[potent condor]

Thanks!

[kind temple]

just say you hate algebra 😂

[tardy hedge]

yea admit it bub

just admit it

you algebra hater

you al-jabr hater

[languid trellis]

lowkey i do

[languid trellis]

probably 10-12? Should probably do more though, and the early chapters have been on material i already know

[somber sleet]

Is there some rings, where the ideal (1) is prime?

[rapid cave]

this is a matter of definition

[elfin wraith]

I’m pretty sure every definition I’ve seen requires a prime ideal to be proper, and thus cannot ever contain 1 as an element

[rapid cave]

yes, but maybe some contexts want to use (1) as if it were a prime

[south patrol]

I don't think anyone would ever use (1) as a prime ideal as it breaks like everything

[elfin wraith]

Do such contexts exist? I’ve never come across it and I don’t see how it wouldn’t ruin every theorem about prime ideals

Or at the very least require youd have to add proper everywhere (in which case you just make it the definition anyway)

It’s the same reason we don’t allow 1 to be prime, it is supposed to be a generalisation of that after all

[thorn jay]

yes i want my zero ring to be embeddable into a field

what field you might ask?

well, i shant say...

[coral spindle]

Can you give an example of such a context?

[rapid cave]

no 🙂

[coral spindle]

Kudos for being direct

[karmic moat]

reminds me of this reply a prof sent me last year

[elfin wraith]

Smh, why can’t you people see the joy in pissing about with rings

[rocky cloak]

Especially when they see the joy in pissing about with shapes and curves

[karmic moat]

i don't drink enough water to be pissing about that much :(

[elfin wraith]

Why do you have to imagine a 5 dimensional hypersurface to feel the joy inherent in an ideal

[tulip otter]

is m supposed to be the least such natural number or something like that? because otherwise there would be multiple exponents, or is that fine?

[thorn jay]

you take the lcm of all orders of the elements

or periods, i guess, as stated here

[tulip otter]

yes, but if m is not required to be the minimum natural number which satisfies the property in the picture then any mutliple of the lcm would satisfy the def

thats why i am asking if the def of exponent must satisfy an additional condition (minimality)

[thorn jay]

then I guess that's fine, it says that the group is "of exponent m" rather than that "the exponent of the group is m"

it's essentially saying that the group satisfies the equational identity x^m = 1

[thorn jay]

it doesn't have to satisfy minimality

a fun thing to prove is that every group G is of exponent |G|

[tulip otter]

sure, but my question essentially was: is the exponent not unique

[thorn jay]

it's not the exponent of a group, a group is of some exponent

[south patrol]

I thought usually it is the minimal such m but maybe I am cooked

[thorn jay]

there's a subtle difference of wording there

[south patrol]

Ah are there two notions

[tulip otter]

i am not sure what the difference is

[south patrol]

Lol

[thorn jay]

the first means that it is a value assigned to the group, the second means that it is a property that the group has

i.e. a class that it belongs to

(the class of groups satisfying x^m = 1)

if they do mean the former, imo that is poor wording, as they have written it

[south patrol]

This feels inelegant given how english works aha

[rocky cloak]

I would use group of exponent m, to mean the group has exponent m, in the same way as "order".

In which case there is a "smallest" missing in the definition.

But you could do it differently...

[south patrol]

Saying "period" for order an element is crazy

[tulip otter]

thats the full context

[south patrol]

But probably better terminology lol

[thorn jay]

periodddd

[tulip otter]

i mean you have to talk to lang in that regard

[thorn jay]

file a formal complaint

[south patrol]

I mean Lang and some Bourbaki have various things odd from current POV

[rocky cloak]

I also disagree with a p-group necessarily being finite

[south patrol]

I have seen "homologism" for quasi-isomorphism

Cursed

[tulip otter]

thats what i thought of when i read it tbh, this is why i was confused as to why there is no minimum in the definition

[south patrol]

K[[X]]

[thorn jay]

what is a quasi isomorphism?

[south patrol]

Do you know about chain complexes

[thorn jay]

heard of it before but not sure you're referring to the thing im thinking about

[tulip otter]

but then how would you define a p-group if it is infinite?

[thorn jay]

yes

[south patrol]

Basically it is a weaker notion of equivalence than isomorphism for tjem

[rocky cloak]

Every element has order a power of p

[thorn jay]

oh okay so it was that

[south patrol]

Ah well then i can say like

[tulip otter]

isnt the "order is a power of p" the essence of the definition of p-groups?

[south patrol]

It is a map inducing an iso on all (co)homology groups

[tulip otter]

ohhh i see

[thorn jay]

pair of morphisms homotopic to the identity right

[south patrol]

Weaker than chain htpy equivalence

[south patrol]

Which is what this should mean

[thorn jay]

ah

[south patrol]

Analogous to situation in algebraic topology

[thorn jay]

that seems very weak

[south patrol]

Fairly weak but good for many things

[fading acorn]

.enpeace_music's homological algebra arc when

[south patrol]

I only work up to quasiisos personally lol

[thorn jay]

I get recommended papers on all forms of quandle/rack (co)homology lmao

[rocky cloak]

I can imagine cases where it's useful to talk about all groups with x^m = 1, so it's not that unnatural to not have minimum in the definition.

But I don't think it's standard.

[thorn jay]

like?

[south patrol]

Well derived functors stuff for example

[rocky cloak]

A module is quasi-isomorphic to its projective resolution. So convenient for dealing with derived functors

[tulip otter]

so to make sure i understand the difference between the 2 uses, one is for the minimum such natural number m and the other is the equivalence class of all natural multiples of m? so the second one is {km: k in N}?

[thorn jay]

how does that work? a single module being isomorphic to a complex

because it's an exact sequence?

[rocky cloak]

A module is a complex concentrated in a single degree (degree 0)

[rocky cloak]

The difference is just that for example Z/2 would be a "group of exponent 4" with one definition and not the other

So you might make statements like "this holds for all groups of exponent 4"

[tulip otter]

ohhh i see, tysm. I think what lang means will be clearer when exponents are used in the statements that he will make in this section.

[elfin wraith]

Every time I see anything about projective resolutions I get a slight cold sweat remembering that I forgot to fill in the proof that Mod-R has enough projective in my UG thesis

Just submitted it with an empty proof environment, still did well, but oof

[rocky cloak]

Tbf fair I feel like "obvious" would have been a sufficient proof.

[elfin wraith]

I mean that’s why I never bothered filling it in when I first wrote the section, I had plans to do a day of just filling in all of the easy and standard proofs and that one just slipped through the cracks

I’m sure if I had just stated it or said it’s known or clear I’d have gotten away with that too

Alas

[rocky cloak]

Could have just deleted the proof environment yeah

[south patrol]

Real

[elfin wraith]

There was so many errors like that that slipped through, I left myself no where near enough time for editing

Apparently where I defined projective in it I drew one of the arrows backwards

[delicate orchid]

Oh my chungus just like my heckin Kan lifts 🛗

[tulip otter]

why is n=p^k t

[lusty marlin]

k could be 0.

[rocky cloak]

Are you asking why a number can be factored into a product of primes?

[tulip otter]

no, i am asking why this particular p has to be in the factorization of n

[lusty marlin]

I think they assumed k must be positive and are asking why n is necessarily a multiple of p

[lusty marlin]

It doesn't. k could be equal to 0

[tulip otter]

ohhh i see. mb i am stupid for not considering that

[lusty marlin]

Happens to the best of us

[tulip otter]

ta_1 is a generator of A_1=<a_1> since (t,p)=1 and p^{r_1}ta_1=0 so that |ta_1|=p^{r_1}=|a_1|, but the author seemed to deduce this fact from something else (since he deduced that ta_1 has period r_1 after deducing that ta_1 generates A_1), so how does he deduce that?

[white oxide]

Does this prove the first part of the problem? 2. Let $\frac{a}{s} \in S^{-1} A$, $\frac{a'}{s'} \in S^{-1} \mathfrak{a}$. Then $s = 1 + b$, $s' = 1 + b'$ for some $b$, $b' \in \mathfrak{a}$. Hence
\begin{align*}
1 - \frac{a' a}{s' s} &= 1 - \frac{a' a}{(1 + b')(1 + b)} \
&= \frac{(1 + b')(1 + b) - a' a}{(1 + b')(1 + b)} \
&= \frac{1 + 2b' + b' b' - a' a}{(1 + b')(1 + b)}
\end{align*}
Since $\mathfrak{a}$ is an ideal, $1 + 2b' + b'b' - a' a \in 1 + \mathfrak{a}$, so that $1 - \frac{a'a}{s' s}$ is a unit in $S^{-1}A$. By Proposition 1.9, $\frac{a'}{s'}$ is contained in the Jacobson radical of $S^{-1} A$.

[cloud walrus]

okeyokay

[white oxide]

When we say an ideal q is p-primary, is p prime?

[karmic moat]

yes; the radical of a primary ideal is prime

[vast verge]

If $R$ is a ring and $I$ is an ideal of $R$, how do we know that every ideal of $R/I$ is of the form $A/I$ where $A$ is an ideal of $R$?

[cloud walrus]

Tropical Greens

[thorn jay]

if we denote [I, R] to be the set of ideals of R containing I, then we have maps A -> A/I and J -> \pi^{-1}(J) which can be shown to be inverses (where \pi : R -> R/I is the usual projection)

this is not very hard; \pi^{-1}(J) / I is J, essentially by definition:
\pi^{-1}(J) = { x in R where x + I in J }
so
\pi^{-1}(J) / I = { x + I in R where x + I in J } = J

[south patrol]

"Correspondence theorem" heh

[thorn jay]

my beloved

in any spectra a surjection induces a closed embedding 🔥

[south patrol]

Provided by spectra you mean spectra of rings gotem

[thorn jay]

actually talking about a secret fourth option that will probably be put on arXiv sometime in the next 50 years

[south patrol]

Lol fourth bruh

[thorn jay]

idk how many there are

i can think of three

four

fuck

secret n+1th option

[rocky cloak]

Not to be confused with ring spectra

[south patrol]

Yeah lol

[tall igloo]

Is rhere a spectrum of a ring spectrum

I know there are things that morally behave like this in some cases

[south patrol]

Spectral schemes

[delicate orchid]

Grok is this real

[radiant knot]

Hmmm yes indschemes that don't do anything

[dull ginkgo]

Before I try proving this because I’m lazy, is it true that if we have a ring R, that the multiplicatively-generated set of an additive subgroup is a subring (without unity)?

[white oxide]

Why does it suffice to show that M is the direct sum of the kernel and the submodule generated by the u_i (I've shown this)

[broken pollen]

the set $\left{\begin{pmatrix} 0 & k \ k & 0 \end{pmatrix} \ : \ k \in \mathbb{Z}\right}$ is an additive subgroup of the ring of $2 \times 2$ matrices with coefficients in $\mathbb R$. if you close this under matrix multiplication, you get $\left{\begin{pmatrix} 0 & k \ k & 0 \end{pmatrix} \ : \ k \in \mathbb{Z}\right} \cup \left{\begin{pmatrix} k & 0 \ 0 & k \end{pmatrix} \ : \ k \in \mathbb{Z}\right}$, which is no longer closed under addition

[cloud walrus]

GoslingGang

[velvet hull]

because if M = A+B such that M and B is finitely generated then A must also be finitely generated?

[white oxide]

Yeah, I guess I'm trying to see why that's true

[velvet hull]

actually you can even drop the B fg assumption

direct summands of finitely generated modules are finitely generated

https://math.stackexchange.com/questions/4244697/is-there-finitely-generated-module-which-has-direct-summand-is-not-finitely-gene

(just consider the projection map)

[tribal moss]

For each generator of M, subtract the appropriate combination of u_i's to make it map to 0 under phi.

Now if you have something that is already in the kernel and express it as a combination of the original generators, the total of the subtracted corrections must have a coefficient of 0 for each ui, since the value under phi doesn't change.

[white oxide]

I see, thanks

What's the point of the second sentence of the proof?

Doesn't the definition just require us to show that r(q) = p?

[rapid cave]

you need to show it is primary in addition to that it has radical p

the first line only shows that it has radical p

[white oxide]

I see, thanks

[velvet hull]

just because an ideal has radical p does not guarantee that it is p-primary.
in fact you will see in exercise 13 that even p^n is not guaranteed to be p-primary

[south patrol]

Wdym lol

[dull ginkgo]

Does the following map defined for simple tensors, $\bigotimes_{n = 1}^{N} F_n \left( \bigotimes_{m = 1}^{M} x_m \right) = \bigotimes_{n = 1, m = 1}^{N,M} F_n (x_m)$ give a map from $T( \mathrm{Hom}(A,B)) \rightarrow \mathrm{Hom}(T(A), T(B))$

[cloud walrus]

miz

[south patrol]

What is T lol

Tensor algebra I assume

[dull ginkgo]

Tensor algebra free functor

[south patrol]

What homs are those on the right

[dull ginkgo]

The latter Hom is R-alg

[south patrol]

Nice

[dull ginkgo]

inner Hom is R-mod

probably should've stated that but I'm lazy

[south patrol]

Is this just an explicit description of the map induced by the "obvious" map Hom(A,B) -> Hom(TA, TB)

[dull ginkgo]

o h yeah

free functor moment

[south patrol]

I think so and then yes

Hehe

[dull ginkgo]

I am trying to rectify a lot of this tensor bs for this engineering class

The great question

why even call it a tensor if you treat it like a matrix with shit that basically only works for the isomorphism between V (x) V* and Hom(V,V) for V finite dim

[velvet hull]

well even in infinite dimensions, V tesnor W is still "basically" the dual of the bilinear space

I've come around to realise that that definition is technically incorrect, but correct morally

[dull ginkgo]

Genie, for my first wish, I wish $\mathrm{Hom}_R(R,R) \not{\cong} R \otimes_R R$

[cloud walrus]

miz

[dull ginkgo]

It's an injection for inf dim iirc

[velvet hull]

yeah it is

but tensors are just "rules" for changing bilinear maps into linear maps

[dull ginkgo]

The humble biproduct

[velvet hull]

it just so happens that in inf dim these rules are no longer necessarily expressable as a finite linear combination

but they're still essentially just ""linear maps""

[dull ginkgo]

all of that isomorphism stuff seems to boil down to the fact that direct sum and direct product are a binary biproduct for R-modules

[dull ginkgo]

and this

So both sides of Hom and - (x) - distribute over finite direct sums on both sides

which gives that "matrix" isomorphism

[south patrol]

My condolences

[south patrol]

This is funny to me, like if they had said a 2nd order tensor is a matrix or smth it would be better aha

[dull ginkgo]

I think the tensor transformation laws for isomorphisms is just for iso F: A -> B we have

ISO H : T(A) (x) T(Hom(A,C)) -> T(B) (x) T(Hom(B,C)) by
T(F) (x) T(Hom(F^-1,C))

[south patrol]

I think it is related to how in applications to physics etc one often has an inner product (or at least a substitute like a nondegenerate bilinear form) which provides an identification between V and V*

So it is much more innocuous to conflate them

[dull ginkgo]

Characterizes the adjoint :p

Which you can then just sacrifice to the functor monster in the background

[south patrol]

Idk what you mean lol

[velvet hull]

functor for the functor god

[dull ginkgo]

Well like for matricies we have change of basis is "PMP^-1" but change of basis is kinda like the inverse of an automorphism so
PMP^-1 which you can identify with (Pv) (x) (P^-t u*)

[dull ginkgo]

which is basically what's here

that whole change of basis thing being the "inverse of an automorphism" really fucked with my head for a while

Since the basis you're changing into is the image of the bases of the automorphism describing it :p

I am doing engineering shit but how I learned about matricies and all that LA shit was via Jacobson before properly doing any LA in HS (BIG MISTAKE)

so now I am cursed to perpetually have to translate this shit into AA terms

Curse of $15 jacobson I textbook

[twilit wraith]

No matter how many times i take intro algebra at various levels i cant escape this problem that is asked on the first day

ive done it three times (self-study, undergrad, grad)

[velvet hull]

it's free points

something something gifthorse in the mouth

[twilit wraith]

hell yeah

i mean im not complaining it took 30 seconds to type it up

[south patrol]

just give funny solutions

[twilit wraith]

its just funny how popular this D&F problem is lol

[velvet hull]

"It's a vector space over Z/2Z, hence a module over Z, hence an abelian group"

[south patrol]

and then prove it is a vector space over Z/2Z by doing the usual proof ig lol

Prove it by introducing the opposite multiplication on a set and say taking inverses gives an isomorphism between G and G with opposite multiplication

then taking inverses is the identity here, so gg

[tribal moss]

Or: aba^-1b^-1 ... (bla bla) ... so the commutator subgroup is trivial, so G is isomorphic to its own abelianization, and therefore abelian itself.

[vivid tiger]

oh neat.

[vivid tiger]

if x is an involution and has a (left or right) inverse, then x = that inverse, which also means that x is its own unique inverse.

if x, y are involutions then xy and yx are inverses.

and these hold in rings too

[tribal moss]

I'm not sure "idempotent" is the right word here. Usually that means x²=x, but that would be absurd in a group (except for the identity).

[vivid tiger]

you are correct, i flubbed the meaning

[cursive spindle]

Grab involution by its... nvm

[vast verge]

Can someone explain why this is incorrect?

The correct answer is 5

[rapid cave]

(2-i)(2+i)=5 not 3

[vast verge]

Ohh thank you

Still, is this method good and rigorous enough?

[velvet hull]

working with divisors is kind of annoying

in this specific question, it's sufficent to show that 5=0

5 is prime, so if the actual characteristic were to be any lower then by CRT you get that 1=0, but the ring is not trivial

[south patrol]

Can also just show it is isomorphic to like Z/5 directly

[tribal moss]

I'm not sure that's actually more "direct" when the problem just asks for the characteristic.

[south patrol]

I guess given this sure

I just mean you can just prove in a line or two that it is Z/5

[rocky cloak]

I'm not sure I'm able to prove that it's Z/5 without first showing that 5 is in the ideal.

Pretty straight forward either way though I guess

[south patrol]

Z[i]/(2+i) = Z[x]/(x^2 + 1, 2 + x) = Z/((-2)^2 +1)

Ig

[rocky cloak]

Ig that works yeah

[tribal moss]

I suppose it depends on which prerequisites you have available. Once you have enough facility with ideals and helper theorems to carry out Potato's proof (and I'll grant that is not a particularly advanced set of tools), then sure.
But the original exercise also looks like something that could be given right after the definition of quotient rings, and considering Z[i] just to be a certain subset of C, so HChan's super low-tech solution seems relevant still.

[south patrol]

Yee fair

[south patrol]

These are ones where x^2 = x right

I think i did this a 2 or 3 years ago and have forgotten lol will try again

Isn't it like "consider this one element" and it'll be ok

[rocky cloak]

The solution is fairly close to how you do it for groups

[rocky cloak]

Yeah ||just consider (x+y)^2||

[south patrol]

Oh wait yeah

[rocky cloak]

||spoilers||

Yields ||spoilers||

[south patrol]

Ig you also need char 2 but that is easy cause 2^2 = 2 lol

Curious fact!

Like this shows ||xy + yx = 0|| right

But then also ||xy = -xy anyway||

[rocky cloak]

||(x+x)^2 = 4x = x+x = 2x
-> 2x = 0||

[south patrol]

Silly question: what happens if you have x^p = x

I suspect it breaks lol

[rocky cloak]

Iirc there's some horribly complicated proof that any ring such that for all x, x^n = x for some n (depending on x), then the ring is commutative

[south patrol]

Oh wait yes

Jacobson's theorem or smth

[rocky cloak]

https://math.stackexchange.com/q/2773974/306319

[south patrol]

Oh I mean the jacobson theorem was something I saw here

From mizalign

[rocky cloak]

@thorn jay probably digs this

[south patrol]

But also like often you may study smth years ago but then use it fairly often lol

Yeahh

Attempting anyway

Should be isomorphic processes

No I mean to the best of my knowledge it does not make a difference but idk

And I can only speak for one thing lol

One comment though is applying for each for phd was basically the same

And all the other unis took much longer to reply lol

Idk what you mean here like do you mean for postdocs or smth

I only applied within UK

To like these two and Warwick and Southampton

Ah ok

Thank

Yeah it is funny

[elfin wraith]

MMath is pretty common across the UK, people typically don’t do MScs because they’re far more expensive

Not me though, I enjoy spending 13k

[south patrol]

Mmath is usually like just part of a 4 year undergrad degree

And you pay as much as for ugrad

But if you do a masters separately it usually costs more

[elfin wraith]

What potato said, but you also typically get less time for your thesis during an MMath, as it’s a 9 month rather than 12 month course, so you need to write your entire thesis along side your classes and exams

But yeah as part of the undergrad it’s charged at the same rate and you can still get funding from the usual people for UG, but there’s very little funding for masters in the UK

I could only get a loan for half of my masters, the other half I had to pay out of pocket

[south patrol]

Further discourages doing a masters / phd in humanities vs sciences/math too oof

[elfin wraith]

Yeah even for the UG stuff it was rough, all my humanities friends had like 3 the entire year because of their thesis

I had 10

[south patrol]

Wdym had 3?

O £3k

[elfin wraith]

Oh, I thought you were talking about doing your thesis alongside your courses

[south patrol]

I meant funding

Like the fact you can't just do an integrated masters or smth

[elfin wraith]

I’m not sure I follow, are there not typically integrated masters in the humanities?

[south patrol]

Not rly in my experience

[elfin wraith]

Ah fair, I can’t say I’d ever really thought about it

[south patrol]

Fair ye

Was p noticeable here as all my humanities friends kinda disappeared whilst the science peeps stayed

[twilit wraith]

I've done that too

[elfin wraith]

I think I had that as a problem on my intro ring theory/LA exam

[twilit wraith]

Lol I just did a lot of algebra this summer

[thorn jay]

its even known to be true for all x^n

though i believe its an open problem to find a general formal deduction as a sequence of equational identities

[south patrol]

Just realised lol

"Abelian ring" 🥀

[elfin wraith]

It feels French but I don’t think even they say that

[south patrol]

anneau commutatif

[elfin wraith]

It is odd that everything other than groups are commutative

Like fair play to Abel but he really fucks up the whole system

[spark veldt]

Is Isaacs' Algebra a good book to deepen group theory?

[regal mango]

What does it trying to say in Q-18

Like what does it mean by G^(n)

??

[elfin wraith]

It’s the group that it describes right next to it

[charred iris]

It's defining that notation immediately

[regal mango]

Like what does it mean if we operate a group over itself

?

Like is it a set of all elements

Operated n times ?

[regal mango]

Group*

[charred iris]

It's the set {x^n : x in G}

[regal mango]

Ok let me try

[ebon prairie]

is {0,-1,1} a group??

[kind temple]

this question is not well-founded unless you tell us the group multiplication

[ebon prairie]

addition

[kind temple]

addition how?

[ebon prairie]

group under addition?

[kind temple]

1 + 1 = 2 is not in the set, so what type of addition do you mean?

[ebon prairie]

it's not a group under multiplication since 0 doesn't have inverse

[knotty badger]

yeah so the issue is that you don’t have closure

[ebon prairie]

yeah that was my doubt if in closure we consider a+a too

[knotty badger]

mhm, you do

[ebon prairie]

thanks

[kind temple]

it is a group under a certain type of addition

which is what i was getting at

addition is just too vague

imo

[knotty badger]

-# yes but I would recommend not confusing them at this stage

[knotty badger]

At this level I disagree

[ebon prairie]

oh no my doubt was basic

i didn't mean modulu addn or something

[knotty badger]

Mhm

[ebon prairie]

i wasn't understanding how nq belongs to H

[elfin wraith]

To make a more general point here, when you’re specifying an object like a group or a ring or a topology it isn’t enough to just give a set, a group is a set equipped with a binary operation, you need both

[ebon prairie]

set with operation?

[knotty badger]

Yes so the key takeaway is that closure has to apply for a + a as well

[ebon prairie]

thankss

[knotty badger]

The closure axiom is stated as $\forall a, b \in G, a + b \in G$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

Of course, in the axiom, you use different letters for a and b

But that doesn’t mean a and b have to be different!

Just like how I can say “let x = 2 and let y = 2”

[ebon prairie]

that makes sense

[elfin wraith]

Ok so you’ve seen that it’s not a group under regular addition because 1+1=2 and that’s not in your set. But it is a group under addition modulo 3, it’s not enough to just look at the set, you need a set and a binary operation on that set

[knotty badger]

(It’s a group under addition modulo 3, you mean)

[elfin wraith]

Yep lol was editing as you typed it thank you

[ebon prairie]

thanks, i forgot to add it, I'll take care from now

[south patrol]

Ngl lol closure is not really an axiom of a group

[knotty badger]

It’s still useful to keep in mind imo

Because of how groups often arise in practice

[south patrol]

Sure I just would like amplify it to saying this is part of what it means to give an operation on a set

[elfin wraith]

It’s less about adding it, because generally as mathematicians we’re lazy and don’t mention it, but that’s when there’s an assumption that the operation is clear. When you’re first learning it though you need to be more careful, and you should keep in mind that a group isn’t just a set, but it’s a set and an operation

[knotty badger]

Sure but that’s a more sophisticated viewpoint on it

It’s very common for groups to arise in practice in the following way:

[elfin wraith]

Yeah I mean I don’t count it as a group axiom, but it is in a sense, you’re just hiding it in the definition of a binary operation

[knotty badger]

You start with a monoid X

[elfin wraith]

I’m not sure starting with monoids is a less sophisticated approach though lol

[knotty badger]

You take some subset S of X, and restrict the operation to that, in order to turn X into a group

[knotty badger]

These are not the words I’d use when explaining to students obviously

I am just doing this for your benefit

[knotty badger]

But in order to make this work, you need to check that the restricted operation S x S -> X really does land in S

That’s what closure is for

In practice lots of group operations arise from restricting binary operations to chosen subsets

This is why I think including closure explicitly is useful

It’s less common in practice to just start with a set G and build a binary operation G x G -> G from scratch

[elfin wraith]

I don’t really take issue with having it as axiom in fairness, it does help make it more clear. We insist that a ring is a an abelian group under addition when there’s no need to specify abelian so

[glad osprey]

I'm not sure if you got a satisfying answer for this, but nq is in H because n is in H, and nq = n + n + ... + n (q times). So it belongs to H by closure of addition. The confusing thing here is that we think of nq as repeated addition, not as ring multiplication, though they happen to be the same for Z

[elfin wraith]

I don’t tend to mention it as an axiom, but I also don’t really think too much about the group axioms that often anymore

[ebon prairie]

why it's not a group under adn mod 2 btw??

[knotty badger]

Essentially cause your set has size 3

If you identify 1 and -1 then it’d work

[ebon prairie]

thanks

[knotty badger]

Abelian groups are Z-modules~

[ebon prairie]

under addn mod 2 it will be reduced to size 2?

[knotty badger]

Mhm

Because 1 = -1 mod 2

[ebon prairie]

(0,1,2) where 2 will work as -1 in + mod 3

[ebon prairie]

oh

[knotty badger]

Yep

[glad osprey]

Yep Z is the only ring where multiplication and repeated addition is the same AFAIK, and I think this is related to how Z is initial in the category of rings (?)

[knotty badger]

Yeah

[elfin wraith]

Interesting, I didn’t know that

Well I never put that together before at least

[kind temple]

rings with unit. otherwise, the zero ring is initial

[elfin wraith]

So rings

[glad osprey]

non-unital rings aren't rings 😡

[elfin wraith]

We have rng for a reason!!1!1!!

[kind temple]

just clarifying since the definition of a ring is not universal

[tribal moss]

I think where confusion creeps is that closure under the group operation is a defining property of subgroups.

[elfin wraith]

Rings are unital but not necessarily commutative, the other definitions are just wrong.

[glad osprey]

yeah, D&F uses non-unital rings I think, so many people can be confused by that

[elfin wraith]

(Half joking)

[knotty badger]

Yeah this is what I was talking about

[glad osprey]

actually, I'm not sure if this makes much sense what does it mean for multiplication and repeated addition to be the same in a general ring?

it's kinda true for Z by definition

[south patrol]

I guess it is just a bit funny saying we have a function f: X -> Y such that for all x, f(x) is in Y lol. Because that is what a closure axiom would be

[chilly ocean]

random number generator

[kind temple]

am = m + m + m + ... + m a times

[ebon prairie]

abelian and commutative are basically same thing right

[knotty badger]

As I laid out, closure is more about how groups arise in practice

[south patrol]

I agree lol

[kind temple]

they are synonyms

[glad osprey]

[1 0; 0 1] + ... + [1 0; 0 1] repeated [0 1; 1 0] times

[kind temple]

complex numbers be like

[elfin wraith]

Yeah, it means the same thing but we call groups abelian rather than commutative for one reason or another

[knotty badger]

Abelian categories are different to commutative categories

[elfin wraith]

That’s true but I’m going to guess diving into homological algebra is a bad call right now lol

[ebon prairie]

so group are abelian wrt * if they're
and ring are commutative wrt +,. if they're

[knotty badger]

Rings are actually always commutative under +

[south patrol]

Rings are called commutative if the multiplication is commutative

[elfin wraith]

Fun exercise, define a Band to be a ring without the axiom of commutativity under +. Show that Bands and Rings coincide.

[south patrol]

Is the name Band your making

[thorn jay]

whats a commutative category?

[elfin wraith]

My advisors, he taught the LA/intro ring theory course and that was the first question lol, I may have stolen it

[south patrol]

Ah nice

Yeah we had this as a first question too

Without the name

[elfin wraith]

Yeah I mean you absolutely don’t need to define a new object or anything, I just think it’s a fun way to frame the problem

[south patrol]

Maybe it is a consequence of eckmann hilton

[elfin wraith]

[tribal moss]

Are Bands related to Gangs?

[elfin wraith]

I’ve proven eckmann Hilton and know nothing about it at all, something something higher homotopy groups

[thorn jay]

god i knew a proof of this

ah

i remember

[tribal moss]

IIRC it's something like distribute (a+b)(c+d) from either the left or the right first, and cancel the identical end terms.

[elfin wraith]

It was a workshop problem in my quantum computing course to prove eckman Hilton but with no context beyond “this is a helpful argument something about monoids and higher homotopy groups”

[south patrol]

Nah not rly

[thorn jay]

(x + x) + (y + y) = 2x + 2y = 2(x + y) = (x + y) + (x + y)
as the + operation has a unit we have by Proposition 2.2 (https://ncatlab.org/nlab/show/commutative+algebraic+theory) that x + y = y + x, as desired

wait that doesn'tfolloe

[south patrol]

One form of it is basically the statement that like if you take the category Mon of monoids then a monoid internal to Mon is just a commutative monoid lol like the two operations you get coincide and are commutative

[thorn jay]

oh well idc

[south patrol]

But it is really cool how there are strengthenings

[elfin wraith]

Yeah this is exactly what we did as the exercise, but just by throwing around some string diagrams

[south patrol]

Well like in higher category theory there are different notions E1, E2,..., Eoo of commutativity, where E1 is just associative and E_oo is the most commutative

And the souped up version is essentially that if you have En and Em then you have E_(n+m) lol

For Sets, E_2 = E_oo

[elfin wraith]

I’m still so upset I started uni a year to late to take the higher category theory course at my UG

[south patrol]

If you put these together you see Eckmann-Hilton is a special case

Hm I have forgotten where Nope is

Was this Martin Gallauer or smth

[elfin wraith]

Barwick

[south patrol]

V cool

Yes now I recall

Big up Barwick

[elfin wraith]

It would’ve been, I audited the first few classes but then didn’t know enough category theory or have enough time to keep going lol

[south patrol]

Actually wanna catch up on what he has been doing recently lol

[elfin wraith]

But yeah I’m not at Edinburgh anymore in any case

[elfin wraith]

Quite a lot as far as I know, I think he’s currently got a pretty nice research grant, he isn’t teaching anymore

UG at least, I think he’s doing some AGQ stuff

[south patrol]

Yeah sure nice

What does AGQ mean sorry

😭

[elfin wraith]

It’s the new CDT at Edinburgh sorry lol, algebra geometry and quantum

[south patrol]

One of these is not like the other

I mean it makes sense w what Barwick has been doing and uh is Safronov at Edinburgh?

[elfin wraith]

Meh, Edinburgh has such a non com and mathphys slant that it’s not too insane, but yeah it is a little broad lol

Safronov is yeah

[south patrol]

O interesting

[elfin wraith]

I had him for Galois theory last year and promptly dropped Galois theory

[south patrol]

Safronov is cool like

Annoyed cause he visited Oxford on the week I was away aha

[elfin wraith]

He’s very intelligent and a lovely guy, but a very dry lecturer

[south patrol]

Fair ye

[elfin wraith]

And he’s also quite terse, his AG notes are uh, something

[south patrol]

Just I have a uh vague interest in deformation quantisation

[elfin wraith]

I hope I have a more clear idea of what my interests are soon, kinda need to find an advisor in the next 3 months

[south patrol]

Glgl

I forget lol are you applying this autumn

Maybe I am sillies.

[elfin wraith]

I’m starting my masters next month, then after that I’m unsure

[south patrol]

Yee sure

[elfin wraith]

I was very set on PhD for a while but I’ve been applying to jobs recently, I’m sorta tired of being poor and I’m not sure I’m cracked enough to make it in academia

[south patrol]

You and me both lmao

But no fair enough hmm I do hope it works out nicely

[proud vigil]

thats so valid 😭

[elfin wraith]

Thank you haha, I’m sure it will, I’ll end up doing something I enjoy in any case. Be that making money or doing cool maths

[proud vigil]

yesss and tbh life is so nonlinear things can change so much later on

[elfin wraith]

It’s why I’m still certain on the masters, it maximises my options

[south patrol]

I should learn some stats or smth

[elfin wraith]

And why I’m still doing just algebra in the masters, companies don’t give a shit what you know provided you have a good degree from a good uni

But if I end up going into academia they’ll care

[elfin wraith]

Yeah but have you done stats? It’s so boring

[south patrol]

I was in a funny situation where I swapped from a degree that does stats in 2nd year to a degree that does stats in 1st year

And bypassed all requirements lol

[proud vigil]

:0 im surprised they didnt make you just do it anyway

[south patrol]

I asked if I could avoid catching up on it cause I wasn't planning on doing anyway

Though tbh I did interact w a tiny bit of stats in physics labs lol

(45 +- 1.3). (1 +- 3%)

[elfin wraith]

Yeah I mean I technically did a bit of days analysis for my physics labs but we both know that doesn’t count

[proud vigil]

stats is kind of cool 3:

im pretty biased but i think probability theory is fun

[kind temple]

why

[elfin wraith]

The more advanced stuff does seem a bit nicer, my best friend is doing a PhD in stats right now and the things he talks about seems nice

[south patrol]

Some people like stats probability and some don't. It is binomially distributed

[proud vigil]

do u mean bimodal ;-;

[south patrol]

Like a sum of coin flips

[proud vigil]

ohh

[south patrol]

I guess for each person it is Bernoulli

However people are not independent

So it is a bad model

[proud vigil]

i like probability bc its very puzzly i guess

and i think stats is fun because linear algebra is so intrinsic to it

data science is kind of boring to me but i think its most tangential to things i do like

and some models are cool tbh

i really liked my monte carlo class

[karmic moat]

I think the hardest computations I’ve done thusfar were in statistical theory class

Had like sequences of integrals and n-fold integrals

Omg it was so dizzying

[tall igloo]

this has also been a conspiracy theory of mine but i havent given it any thought

[knotty badger]

One where every diagram commutes

[vast verge]

The fact that normal subgroups / ideals and homomorpisms have a one-to-one correspondance is probably my favourite fact from abstract algebra so far, although I'm not that good at those types of questions

I'm curious to see how this will get expanded on afterwards

[kind temple]

what is your favorite fact?

[vast verge]

Just fixed it now

It seems I'm not that good at grammar either

[knotty badger]

yeah this is the content of the first isomorphism theorem

it's a very very cool result imo

it tells you that studying homomorphisms is equivalent to studying subgroups and quotient groups

so that you can use one to study the other

[vast verge]

Well the first isomorphism theorem says that every homomorphism can be mapped to a normal subgroub (i.e. take its kernel)

[knotty badger]

the way i prefer to think of it is as follows:

[vast verge]

Then there is a different theorem a little later saying that every normal subgroub can be mapped to an a homomorphism (i.e. N -> gN)

[knotty badger]

every homomorphism can be decomposed into a quotient map (q), an isomorphism (alpha), and a subgroup inclusion (iota)

so $f = \iota \circ \alpha \circ q$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

thus, studying homomorphisms gives you information about subgroups and quotients, and vice-versa

[vast verge]

Yeah, so anytime you have a homomorphism question you can transform it into a normal subgroup question and vice versa

[knotty badger]

mhm

[kind temple]

you can do this in Set too

[knotty badger]

Mhm

[kind temple]

so studying set functions gives you information about quotients and subsets and vice versa

[knotty badger]

Mhm

[kind temple]

how far does this carry into other concrete categories

not sure what a concrete category is

but like

[knotty badger]

Regular categories is what you want I think

Abelian categories also

[swift tundra]

Interesting. I remember seeing that topoi accept epi-mono factorizations, but it seems that regular categories are more general?

[kind temple]

isn’t an epi-mono factorization more general than an epi-iso-mono factorization?

[knotty badger]

They’re equivalent categorically

[knotty badger]

Yes so every topos is a regular category

As is every abelian category

[tall igloo]

isn't this basically just model category nonsense

well, weak factorization system nonsense, but related

i guess this paper came out on friday so i can mention it lol, there are 6 WFS on Set:

prop 3.4 in https://arxiv.org/pdf/2508.15731

[vast verge]

Pullbacks in ring homomorphisms preserve prime and maximal ideals

Huh, that's pretty neat

I feel like that will be important later

[velvet hull]

oh god you've summoned them

[vast verge]

Pardon?

Who's 'them'?

[south patrol]

Not maximals

[delicate orchid]

kinda cool - are there any similar results for Grp or Ring?

[south patrol]

Provided by pullbacks you mean like preimage of maximal

[south patrol]

Nice

[vast verge]

Homomorphism needs to be onto, I forgot to mention that

[velvet hull]

if the homomorphism is onto then this is just the correspondence theorem

the interesting questions arise when you ask this about general homomorphisms

turns out there is a very large class of non onto homomorphisms where this is still true, and they are very interesting

[vast verge]

Do you know how I could do this question?

[velvet hull]

the kernel of any homomorphism is an ideal

[vast verge]

Is there anything I should know about an ideal in a field?

[velvet hull]

yes!

think about what kind of ideals are possible in a field

[rocky cloak]

What would happen if the ideal contained an element that wasn't 0?

[vast verge]

Is it that if an ideal contains the unity, then it must contain the entire ring?

Or any unit for that matter

But any non-zero element in a field is a unit

So that means:

• If ker(phi) is trivial then F/ker(phi) is isomorphic to F
• If ker(phi) is non-trivial then F/ker(phi)=F/F which is trivial

I'm starting to get the hang of it now, thank you

[south patrol]

You don't really need to go to first isomorphism theorem or anything, like it seems kinda backwards as written

You have a surjective map with trivial kernel, and those are isomorphisms

Also like to be pedantic rather than saying "is isomorphic" it is important to keep track of what maps you mean

Like the point is you already have maps F -> F/ker(phi) etc but they are isomorphisms

[vast verge]

Just something unrelated I just realised:
The fact that the only ideals of a field are the trivial sub-ring and itself is a lot like how the only normal subgroups of a simple group is the trivial group and itself

Does that mean that fields are essentially the simple groups of the ring world?

[velvet hull]

the converse is true, by the way, a (commutative unital) ring is a field iff it only has 2 ideals

[rocky cloak]

Fields are exactly the commutative simple rings

Noncommutative simple rings can be a little more complicated (just like non-abelian simple groups I suppose)

[vast verge]

Oh that's cool

[vast verge]

I tried redoing the proof so that it looks better

Just pretend that none of the text is cut off, ok?

I'm also stumped on this one

[rocky cloak]

A hint could be that a real number is positive if and only if it has a square root, which is a ring theoretic property

[vast verge]

I think I get what you mean

If x is zero then Φ(x) is zero
If x is positive then Φ(x) is positive
If x is negative then Φ(x) is negative

Using the properties of ring isomorphisms

[crystal vale]

It is an iff property

[vast verge]

How would I continue?

[warm dove]

This is not true in general - you need order preserving

If you have order preserving, try to show rational numbers must be preserved

[vast verge]

I remember doing a similar question for group automorphisms, but I forgot how to approach it

[warm dove]

Then use a property about arbitrary reals and relating them to rationals to complete the proof

[viscid parcel]

Hi

[velvet hull]

do you need order preserving? I don't think so

[warm dove]

I think you can prove order preserving using square roots

My bad

[velvet hull]

if that's what you meant then sure lol

[warm dove]

No I forgot u could do that

Oopsie

Thanks for the catch

[south patrol]

Ye nice

[tall igloo]

I am also wondering this but am too lazy to try to compute this lmao

[crystal vale]

Is there any R-module such that all proper submodules is finitely generated but that module is not finitely generated, give hint

[tribal moss]

How about something like the additive group of dyadic fractions? (Vibe suggestion, may or may not work).

(Hmm, no, that doesn't quite seem to work. Perhaps if we quotient out Z, though?)

[rocky cloak]

Z[1/p]/Z is an abelian group that works at leats

[crystal vale]

So submodules are subgroups of Z[1/p]/Z

[rocky cloak]

k((x))/k[[x]] should work as a k[[x]]-module

[tribal moss]

Yeah.

[crystal vale]

Why do you take quotient Z?

[rocky cloak]

I guess morally you want to think about it as the union of Z/p^n Z

but you think about it as <1/p^n> / Z, so the union makes sense

[tribal moss]

Otherwise, say, the set of all elements whose numerator is a multiple of 7 would be a proper subgroup that's not finitely generated.

(Any odd prime instead of 7 works too :-)

[rocky cloak]

I think the injective envelope of a simple module might work for any integral domain....

[crystal vale]

So here p = 7, so in lowest form the numerator is multiple of 7, yes then it is proper submodules and infinitely generated

[rocky cloak]

If p =/= 7, then 7Z[1/p] is a proper subgroup of Z[1/p]

[tribal moss]

No, my initial example was M = Z[1/p] with p=2. Then 7M is a proper submodule that is not finitely generated.

[crystal vale]

So here I have to show every proper submodule is finitely generated, what is the exact form here of proper subspace?

[tribal moss]

Hint: if some k/p^n with k not a multiple of p is in your submodule of Z[1/p]/Z, then 1/p^n is there too.

[crystal vale]

Yes p = 7, then 7Z[1/p] = Z[1/p]

[rocky cloak]

Hmmm, no probably doesn't work for all integral domains, but works for PIDs

[crystal vale]

Yes because k and p^n relative prime so 1/p^n will be in my submodule

[tribal moss]

This severely constrains which shapes submodules can have.

[crystal vale]

Okay say there exists polynomial f of degree n in Z[x] such that f(1/p) is not in submodule N

So there exists k ≤n such that 1/p^k is not in N.

It implies that 1/p^m is not in N for all m≥k.

But I have to show more, can you help me?

[rocky cloak]

So what you figured out is that if
a/p^n is in the submodule then 1/p^n is.

So the only question for which n it is that 1/p^n is in the submodule.

[tribal moss]

(where a is coprime to p, of course)

[rocky cloak]

If k is the largest number with 1/p^k in the submodule, what are all the elements in the submodule?

[vast verge]

When showing that a map is a homomorphism, when is it necessary to shoe that the map is well defined?

[crystal vale]

All 1/p^m, m≤k will lie there

[crystal vale]

If the map is not well defined how does it makes the definition of map?

[rocky cloak]

(and multiples of them), so how many elements is that in total?

[crystal vale]

Yes

[tribal moss]

The map is not even there until you're sure that its definition actually defines a map ...

[knotty badger]

this often happens if you're defining the map on a quotient set

[crystal vale]

(p-1)k

[knotty badger]

or if you're defining the map and need it to lie in some subset of the "natural" codomain

[rocky cloak]

Typically the cases where it's relevant to check well definedness is when you want to define a homomorphism on
R/I, but you define it on R instead.

[knotty badger]

mhm

[crystal vale]

How do I show there are no other elements?

[rocky cloak]

Well, that's not quite the right number. But the key take away is that it's finite.

[tribal moss]

Suppose there are. What could their denominators possibly be?

[rocky cloak]

What are all the elements of Z[1/p]/Z?

[vast verge]

How should I try approaching it?

[velvet hull]

hint: ||it is sufficient to show that conjugation respects addition and multiplication||

[crystal vale]

We have p-1 choices for numerator right? And there are 1/p,..,1/p^k possibility

[crystal vale]

f(1/p) + Z

Prufer group is a good example

[crystal vale]

p^n, where n>k

[vast verge]

So basically showing that C is its own homomorphic image through conjugation?

[crystal vale]

I see

So then the numerator is the relative prime to p

Then it implies 1/p^n will be in submodule

Which is contradiction

Right?

[tribal moss]

Yes. Which means the subgroup we're looking at has a very simple description.

(Or, following Jagr, we can just note that it is finite and therefore surely finitely generated too).

[crystal vale]

Yes

Thank you @rocky cloak @tribal moss ❤️

[vast verge]

So by showing that C is its own homomorphic image, I should maybe do something involving it's kernel I guess??

[crystal vale]

Now I am thinking about the converse, the module M is finitely generated but there is one submodule which is not finitely generated

[karmic moat]

Polynomial ring in infinitely many variables

[crystal vale]

Okay, if R = R[X1,....], and M is R, so it is finitely generated but its submodule N = R[X2,x4,....] is not, right?

[karmic moat]

Isn’t N still a fg R-module?

[south patrol]

Do you mean R = Z[X1,...] or smth

[south patrol]

Indeed it is a quotient of R

[karmic moat]

Yeah

Just take N to be the submodule of polynomials with no constant term, ie, the ideal generated by the variables

[crystal vale]

Yes

Actually I have ring R, and i am taking R[X1,...]

[south patrol]

And then S = R[X1..,] ig

[tribal moss]

Generally you'd just say: If f(a+bi) = 0, then expand the definition of f and conjugate all the numbers in that equation. Push the identity up to the top, and you get f(a-bi) = conjugate of 0. (Nothing happens to the coefficients of the polynomial because they are assumed real!)

[crystal vale]

Oh generator X2🙂

[crystal vale]

Yes, S is S-module

[crystal vale]

No

What is the generator here?

Oh it is not submodule because x•x^2 = x^3 is not there

[south patrol]

Oh wait yeah that too lol

[tribal moss]

Hmm, for pedantry the way that works is the other direction: $$f(a+bi) = 0 \implies \overline{f(a+bi)} = \overline{0} = 0 \implies f(\overline{a+bi})=0$$

[cloud walrus]

Troposphere

[vast verge]

I'd imagine there would be a more elegant solution using stuff from ring theory

[tribal moss]

This (applying a ring isomorphism) is stuff from ring theory.

[vast verge]

Nevermind the manual solution is just expanding it all out 😪

[velvet hull]

well, you are essentially proving that conjugation is a ring isomorphism

[south patrol]

yes, if f is a homomorphism S -> S of R-algebras then you get an induced map f: S[x] -> S[x] and p(f(x)) = f(p(x)) for any p in R[x]

[vast verge]

But you're right

Maybe I'm just over complicating things in my heat

[tribal moss]

As written you need to assume that f fixes the coefficients of p...

[crystal vale]

Why not generator just x, because I am taking the module over that ring itself

[south patrol]

Uhh yeah sorry yes

[crystal vale]

Any hint?

[tribal moss]

The polynomial y is not in the submodule generated by x.

[vast verge]

Just did it like this

[tribal moss]

Indeed. On the way to Galois theory you're going to repeat that argument with other self-isomorphisms than complex conjugation, but it will be the same basic structure (and it's generally considered so simple to do that I don't think it has a fancy name).

[crystal vale]

Sorry I was confused with the powers of x, my bad

[crystal vale]

@south patrol this doesn't make sense, my terrible mistake

But again, if submodule is N = R[X2,..] then x2•x3 has to be there

So I have to take the submodule generated by {x_2, x_4,...}

[vast verge]

So basically like finding other zeroes of polynomials over rings and fields with non trivial isomorphisms?

[crystal vale]

Oh ideal generated by variables

[crystal vale]

So it makes sense ?

[karmic moat]

Sorry was driving. Yes the ideal generated by variables

So really any ideal generated by infinitely many variables is not a fg submodule of your poly ring over infinitely many variables

Can you see why it’s not fg?

[crystal vale]

Yes

[karmic moat]

Good

Cheers

[crystal vale]

Ideals generated by finitely many elements whose constants are zero generates only finitely many variables x_i, right?

[tribal moss]

In a sense. Typically we have a field K (for example the rational numbers) and a field extension F. Then if there's a polynomial with coefficients in K that has roots in F\K, every automorphism of F that fixes K will take roots to roots, so it restricts to a permutation of those roots. If F happens to be generated by the roots, the permutation completely determines the original automorphism, and we can investigate the structure of F by studying the group of permutations of the roots.

(Galois theory ultra-condensed to 2½ sentences 😆)

[karmic moat]

Not really sure what you mean here. The reasoning is just that the elements of the basis are independent from each other

[vast verge]

Why is it that n^2 - n =0 in Z_2n only if n is odd?

[karmic moat]

Like if I'm looking at the ideal (x_1, x_2, ...), you can't get any x_i from any other x_j

[crystal vale]

what are your basis here?

[karmic moat]

the basis is exactly the generators of the ideal

x_1, x_2, ...

[crystal vale]

then x1 is not indepedent to x2, i am taking it is over R[x1,...] module

[karmic moat]

if this submodule were finitely generated, then i should have some relations among the x_i and x_j that would let me reduce (x_1, x_2, ...) to an ideal generated by finitely many variables

[karmic moat]

how do you get x_2 from x_1?

[crystal vale]

i mean i am taking S = R[x1,...] as S -module

[crystal vale]

x1x2 -x2x1 = 0

[karmic moat]

sure

[crystal vale]

so how they are independent?

[karmic moat]

this doesn't tell you how to get x_2 from x_1

this is essentially like a linear algebra observation now

[crystal vale]

yes this doesn't

[karmic moat]

suppose in R^2 i have my two standard basis vectors (1, 0) and (0, 1)

i'm essentially asking how you can get (0, 1) from (1,0)

[crystal vale]

we can't

[karmic moat]

right

[crystal vale]

i see your point

[karmic moat]

mhm

[tribal moss]

The pesky thing about modules relative to linear algebra is that "there's a linear relation between these generators" doesn't imply "we can do without one of them".

[crystal vale]

yes

[crystal vale]

here i mean say generators contains finitely number of variables so they will not generate infinitely many single variable x_i

[tribal moss]

Right.

[karmic moat]

yeah i'm just creating a loose analogy

[crystal vale]

I edited my message so please read it again, I assumed there are finitely many generators

[tribal moss]

Still agree with the edited post.

[crystal vale]

Okay thank you

[karmic moat]

hmm yeah maybe that's a better argument than mine tbh

[vast verge]

Could I get more hints for this question please?

[glad osprey]

Hint 1: ||An automorphism of a field maps squares to squares||
Hint 2: ||An automorphism of R maps positive numbers to positive numbers||
Hint 3: ||If f : R -> R is an automorphism of R, then x < y implies f(x) < f(y)||

[velvet hull]

1,2 were done already in the initial replies lol

[glad osprey]

I just shamelessly stole these hints from Fraleigh 🙈 I think hint 3 is enough to get him all the way tho

[vast verge]

I messed around but still couldn't do it, sorry

[elfin wraith]

The hints are essentially the entire problem, just combine those, and each hint implies the next

[vast verge]

Okay so the first hint comes from the fact that if $a \in \mathbb{R}$ then $\phi(a^2) = \left( \phi(a) \right)^2$, just a basic property of ring homomorphisms

[cloud walrus]

Tropical Greens

[vast verge]

The second hint comes from the fact that if $a > 0$ then $\phi(a) = \phi \left( \left( \sqrt{a} \right)^2 \right) = \left( \phi \left( \sqrt{a} \right) \right)^2 > 0$ since $\sqrt{a} \ne 0$ means $\phi \left( \sqrt{a} \right) \right) \ne 0$

[velvet hull]

a is positive iff there exists a b such that b^2 = a
but then apply phi on both sides to get that phi(a) = phi(b^2) = [phi(b)]^2

[cloud walrus]

Tropical Greens
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[vast verge]

Isn't it not an if and only if relationship since zero is not positive?

[velvet hull]

0 is an edge case that can be dealt with easily

[glad osprey]

just replace "positive" by "non-negative"

[vast verge]

Cool, cool

Now let me have a think about where the third hint comes from..

In order to show Hint #3, it would be sufficient to show that $x-y>0 \implies \phi(x-y)>0$ for an automorphism $\phi$

[cloud walrus]

Tropical Greens
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[vast verge]

But that's just the positive numbers map to positive numbers property from Hint 2

I think I've got it now

Suppose that $\phi$ is a ring automorphism on $\mathbb{R}$ that isn't the identity map. Then there exists $x \in \mathbb{R}$ such that $\phi(x) = y$ for some $y \in \mathbb{R}$ such that $x \ne y$.

[cloud walrus]

Tropical Greens

[vast verge]

We can suppose without loss of generality that $x > y$, so $x - y > 0$, meaning that $\phi(x) - \phi(y) > 0$, meaning that $y - \phi(y) > 0$ or $\phi(y) < y$ for this certain $y \in \mathbb{R}$.

[cloud walrus]

Tropical Greens

[vast verge]

Oh wait

Nevermind haha

I feel like I'm making it more difficult than it actually is 😅

So I guess that means $x > \phi(x) > \phi( \phi( x))$ and that gives an infinite sequence of decreasing terms

[cloud walrus]

Tropical Greens

[vast verge]

Like you repeat it infinitely

[rocky cloak]

It might be a good idea to identify which elements definitely map to themselves, then compare them to other elements with the x>y condition.

For example you know that 0 and 1 map to themselves

So you can distinguish between things <0, things > 1 and things between 0 and 1

[vast verge]

Well I think every integer maps to themselves but the properties of ring isomorphisms

[rocky cloak]

Right, so that means that for example pi would have to map to some real between 3 and 4

[elfin wraith]

I think you’re over thinking this

[rocky cloak]

Any other numbers that definitely map to themselves?

[vast verge]

Oh okay because $3 < \pi < 4 \implies \phi(3) < \phi(\pi) < \phi(4) \implies 3 < \phi(\pi) < 4$

[cloud walrus]

Tropical Greens

[glad osprey]

How do you know phi(3) = 3?

[vast verge]

I feel like the next logical thing to say is the rationals although I can't explain how

[vast verge]

Integers map to themselves under the ring automorphism on the reals right?

[rocky cloak]

Well let's think about it.

Could it be that for example 1/2 maps to 1/4 or something?

What would that entail if so?

[glad osprey]

Yes, but it's best if you spell out exactly why

[rocky cloak]

Maybe just figure out the idea first before spelling out all the details

[vast verge]

It's because for any integer $n$ we have $\phi(n) = \phi(n \cdot 1) = n \cdot \phi(1) = n \cdot 1 = n$ since $\phi(1) = 1$

[cloud walrus]

Tropical Greens

[glad osprey]

Nice 👍 didn't mean to be pedantic, just wasn't sure if you understood why

[narrow temple]

Im not entirely sure what is meant with data in the second part of the question.
My answer would be something like: A R-algebra hom C -> A is the same as a R-Algebra Hom R[X] -> A such that (x^2+1) is the kernel, which corresponds to elements r in A s.t. r^2=-1.
So the data I need to have is the existence of elements r in A such that r^2=-1 in A?

[vast verge]

Yeah I understand

[narrow temple]

thanks jagr

[vast verge]

No? Because only squares can map to squares

Actually 1/2 is a square number in the reals

So maybe

[rocky cloak]

Well 1/2 is the square of 1/sqrt(2)

But a different thing you can think about. How would you think about 1/2 ring theoretically? (So not using division, only addition multiplication and integers)

[vast verge]

1/2 = 1/4 + 1/4 or 1/2 = 2 * 1/4

[rocky cloak]

Yeah, so multiplication by 2 might be relevant here

What is special about 1/2 when it comes to multiplication by 2?

[vast verge]

It's the multiplicative inverse of 2

I think I'm starting to see what you're getting at

[rocky cloak]

Right, 2 * 1/2 = 1

So that's a ring theoretic property of 1/2

[vast verge]

You're using the fact that 1/2 and 2 are multiplicative inverses to make it so that phi(1) is not 1

[rocky cloak]

Yeah, if phi(1/2) = 1/4, then
phi(1) = phi(2 * 1/2) = 2*phi(1/2) = 1/2 =/= 1

So it's not possible for phi(1/2) to be 1/4 for example

[vast verge]

So the next step would be to do the same thing but for arbitrary rational numbers

i.e. any rational number must be fixed under the automorphism

[rocky cloak]

Yes, then after that you can move onto numbers like pi

[vast verge]

Hmm okay

[vast verge]

Here's what I did for the rational numbers

[rocky cloak]

Right, so phi(m1/n1) doesn't even need to be rational apriori.

But you know that m1/n1 is the unique x such that
n1 x = m1

[vast verge]

Now I know that the rational numbers are fixed, I can do the same thing with the real numbers right?

[rocky cloak]

Well, think about what we did with pi before.

You know phi(pi) is between 3 and 4.

What other bounds can you give for phi(pi)?

[vast verge]

phi(pi) is contained in every open interval with rational endpoints that also contains pi