groups-rings-fields

[slim kayak]

Don't keep us hanging, what's the exercise

[rocky cloak]

A nice little diagram chase

Or just apply ||the snake lemma||

[long geyser]

for the experts it is a chase

for me it is a marathon

[rocky cloak]

Nice long diagram marathon

[slim kayak]

Would be kinda funny to make the ||snake lemma|| moved in front of results like that one

[rocky cloak]

No need to prove it, it was in that movie that one time 😉

[slim kayak]

Lmao

[dull ginkgo]

SN A K E

https://ncatlab.org/nlab/show/salamander+lemma whoever wrote this fuckin nlab article deserves a fucking award for making the first comprehensible nlab article for idiot boys like myself

i spent a long ass time doing some weird homological shit thinking it was whackass and had no ground and apparently it's some weird shit called spectral sequences and looking into it further jumpscared me

[rocky cloak]

Spectral sequences, the mother of all diagram lemmas

[slim kayak]

Are you sure you understood it?
I can write Urs Schreiber an email to add more infinity cats in there to fix it

[dull ginkgo]

NOOOOO

[long geyser]

just finished showing that if the top two rows are exact, then the bottom row is

that was definitely a slog

gave me much needed practice on exact sequences

obviously, I'm not gonna try proving the other half lol

[cobalt heath]

I feel R is a free (abelian) group with a differential d/dt. Is there a way to make this precise?

[south patrol]

is R here the reals?

I don't really know what you mean as you can define the derivation d/dt on any ring

[cobalt heath]

Yeah

[cobalt heath]

Is this as Kahler differentials?

But it can only act on polynomials, right

[cobalt heath]

Basically, I do not understand how analysis differs from algebra precisely in this kind of aspect.

[rocky cloak]

Kahler differentials makes sense on any algebra. It is the universal derivation.

For a ring R and a bimodule M (or if R is commutative you can just use a module) a derivation d is a linear map R -> M, such that
d(rs) = rd(s) + d(r)s

The kahler differential Omega_R/k is then a module equiped with a differential such that every differential factors through it uniquely.

Note that the assumption of linearity depends on the base ring. So for example derivations on the real numbers would be different if you required them to be R-linear than if you just required them to be Z-linear

[rocky cloak]

The difference between analysis and algebra here would be that algebra doesn't see any of the continuity/smoothness.

Like in analysis land you would look at differential forms on a manifold, and in algebra land you would look at kahler differentials on varieties.

If you looked at for example kahler differentials on the ring of smooth functions you would get something much 'bigger' than just the differential forms.

[cobalt heath]

Yeah, I interpret them as "acting on polynomials" which could be imprecise. But still, it's not applying to general smooth.

[rotund aurora]

I don't really know about this but if you have some algebraic variety you can calculate its de Rham cohomology (you have to consider the analytification or whatever) by looking at the algebraic de Rham complex, which is defined by taking derivations of the coordinate ring modulo requiring that differentials of constants be zero. I guess this is saying that to calculate the cohomology you can just look at the algebraic differential forms (although not all differential forms need be algebraic).

Again, I don't actually know about this, I just watched this the other day https://youtu.be/FFH5HSbfcdg?si=xk1jK916cgojCyAh

[cobalt heath]

Yeah the thing is how they have analogue while only analysis got the smoothness.

Where does this come from?

Is it just diff geo envy

[cobalt heath]

Yeah this explains the common properties, but not differences

But yep the fact that algebraic de rahm cohom coincides with singular cohom is indeed surprising

[rotund aurora]

if S=C^infty(M) for some manifold M and you require the differential S-->S to be zero at constant functions, don't you always get that these differentials are of the form f(x) · d/dx, where f(x) in S ?

idk much about differentials on manifolds, hopefully what I said makes sense lol

[rocky cloak]

Like for differential forms df = f'(x)dx and for kahler differentials the same is true for polynomials f.

But if for example M is the real numbers, then the kahler differential d(sin(x)) should be different from cos(x)dx

Like the ring C^infty(R) can't "tell" that cos is the derivative of sin

[chilly ocean]

random sanity check question: any two free modules, over the same ring R and with basis of same cardinalities, are isomorphic right? (same proof as in vector spaces)

[cobalt heath]

Why does d(sin(x)) make sense in the Kahler differentials?

Is it just for being defined over C(M)

[rocky cloak]

The kahler differential is a module together with a map d from the the ring to the module. sin(x) is a smooth function (so in the ring of smooth functions)

[cobalt heath]

Ah. So you cannot say d sin(x) = cos(x) dx because Taylor series does not make sense in algebraic settings

[cobalt heath]

Which stems from the lack of (conventional) limit

[rotund aurora]

it is true in C^infty(R) that all derivations are of the form f(x) · d/dx, I think

[rocky cloak]

Yes a being a derivation is quite a weak condition, just d(fg) = fdg + (df)g
I find it pretty cool that this implies df = f' dx even if it just works for polynomials

[cobalt heath]

Formal derivation >.>

[rocky cloak]

Perhaps if you just mean derivations from C^infty to itself, or require some sort of continuity

But not in general for sure

[cobalt heath]

There is some notion in d(fg) = f dg + df g, but I don’t see it directly

Eh.

[south patrol]

Good talk

[ornate tiger]

why every element here form order 2

do they mean like (0,0,0) (1,0,1) etc ?

or like x,y,z are differtant

[uneven jackal]

there is an element of order one

which is zero

but all others are of order 2 yes

[ornate tiger]

ok thx

[whole moss]

If F is a subfield of K and K/F is a galois extension with K=F(gamma), how would you prove the Fix(Gal(F)) is a subset of F direction for showing that Fix(Gal(F)) = F

i have the idea that it has something to do with automorphisms in Gal(F) uniquely permuting the roots of the minimal polynomial of gamma over F but im not sure how to make things precise

[serene ether]

ı dont want answer ı just know a thing, what is r(a)?

[south patrol]

Radical of an ideal

Often written as $\sqrt{I}$

[cloud walrus]

PD potato

[south patrol]

It is the set of elements x of A such that x^n is in I for some n

[chilly radish]

Hint: show that all the galois conjugates of gamma form a basis

Or rather

Show there is some element a such that sigma(a) for sigma \in Gal(F) is a basis

[serene ether]

thank youu

[chilly radish]

This might be hard to prove without more assumptions on the field

This is the normal basis theorem

You might already know it

[whole moss]

ahh thanks ill give it a shot

nah havent heard of that theorem but ill look it up

its a bit weird becuase the lecturer started off with fields and introduced the galois correspondence theorem in the second lecture, and brushed over separability

[bitter locust]

whats the difference between localization of a ring and using group completion on the monoid of the multiplicative structure already on a ring?

[south patrol]

Well one thing is that group completion inverts everything whilst localisation doesn't

Like you don't want to invert 0 in the monoid for obvious reasons

Moreover the localisation is compatible with the addition etc

[bitter locust]

would it be fair to call localization as "multiplicative group completion that plays nicely"? (by nicely i mean not imverting the 0 and obiding with the addition)

or rather

"partial multiplicative group completion that plays nicely"

tho why we pick just a subset S is kind of a mystery to me

[dull ginkgo]

I internalize the localization as almost a “unitification”

You are essentially taking a multiplicative subset S

And mapping it to a space where that S becomes a set of units

Versus like an ideal being sent to 0 for a quotient

Which is funny because it’s like the opposite, as the units are elements that don’t lie in any proper ideal

Which allows you to take a glance at what happens to the ideal structure under localization

[tribal moss]

One reason not to localize to only some of the non-zerodivisors is if you want to take a quotient of the resulting ring. If you make too many elements into units, the set you want to quotient out afterwards might not be an ideal anymore.

[bitter locust]

so like if i have Z and take S to be the evens 2Z, all elements of 2Z now are units?

[tribal moss]

That actually gives you Q, since every rational can be written as 2m/2n.

[bitter locust]

oh huh

[tribal moss]

But if you take S to be all the powers of 2 (that is, the smallest multiplicative set that contains 2), you get an interesting intermediate ring.

(Correction: you'd get Q if you take S to be 2Z\{0}. If you try to adjoin an inverse of 0, everything always collapses to the zero ring).

[bitter locust]

yeah right 0 isnt really allowed anyway

[tribal moss]

Another interesting case is to localize with respect to all the non-even elements of Z. That gives you the ring of all rationals with odd denominator -- so it won't contain 1/2.

[bitter locust]

oh but odd *odd = even

doesnt S have to be closed under multiplication

oh wait

lol

[slim kayak]

groethendieck, is that you?

[bitter locust]

reincarnation

[bitter locust]

so for this ring we get all fractions with a denominator thats a power of 2?

[tribal moss]

Yes.

[bitter locust]

so 2 is a unit and so is 4 etc

[tribal moss]

(This is usually known as the "dyadic rationals").

[bitter locust]

thats pretty cool

i remember using those before

[slim kayak]

Thats a cool ring, is it used for anything?

[bitter locust]

not really using but i saw those (or their flavor) before in different contexts, like hackenbush, surreal numbers, or some proofs of theorems that divide an interval in half and half

[mighty kiln]

Localization freely generates a group structure indeed

Consider the category C of (R, S) where S is a multiplicative submonoid of a cring R, and the subcategory D of (R, S) where S is a group, I think there's a free-forgetful adjunction here?

[bitter locust]

i dont get adjunctions yet but does that basically mean that localization generates a free group (and forgetful is there because we forget about the additive structure)

[mighty kiln]

You forget the invertibility of S

And freely generate group structure (inverses) for S

[bitter locust]

oh so instead of "enhancing" the original ring R we take a different perspective and instead "enhance" S by using elements from R?

[mighty kiln]

We're modifying R and the S tells us which part to modify

[bitter locust]

ohhhh

ok ty

ig its time for another categorical detour

[limpid ferry]

I am about to learn Ring and Fields, but just want to ask in advance if this term "Fields", and the "Field" in linear algebra, over which a vector space is defined, refer to the exact same thing?

[dull marsh]

Yes, they do

[dull ginkgo]

Ore localization is horrible

[mighty kiln]

How does Ore work

[dull ginkgo]

Trust me you don’t wanna know

[mighty kiln]

Okie

[lethal cipher]

Okay, so I am having a little trouble. Let $I$ be an ideal of a ring $R$ and $\pi: R \to R/I$ be the natural quotient map.
I want to show there is a one-to-one correspondence between ideals of $R$ containing $I$ and ideals of $R/I$, but I am having a lot of trouble.
I do know that since $\pi$ is a surjective homomorphism, that ideals map to ideals, and pre-images of ideals are always ideals.
However, i am having a hard time showing that there is exactly one ideal that corresponds to the ideal in the other ring.

[cloud walrus]

dackid

[chilly radish]

Do you know about the derived category?

Because it's kind of like getting the derived category from the homotopy category

Well it's exactly like that lmao

[mighty kiln]

I read about it a while ago

[mighty kiln]

Show that the two directions form a pair of inverse functions

[mighty kiln]

So do you like

Take an additive cat with one element and "localize"

[rocky cloak]

The basic idea is that when you localize a commutative ring you can write anything as r * s^-1, but in a noncommutative setting this doesn't work because s^-1 * r isn't on that form.

The ore condition just ensures that for r and s there are r' and s' such that s^-1 r = r' s'^-1 and the localization works out nicely

[mighty kiln]

Hmm

And the rest is equation chasing?

[rocky cloak]

I'm not sure what you mean by that. But it's just a condition that makes it easier to explicitly construct the localization

[mighty kiln]

Is it equivalent to the existence of a localization?

[mighty kiln]

In the sense of universal property

[rocky cloak]

No, I think a localization should always exist

The category of rings is complete after all, and it doesn't seem like you should run into any size issues

Like just take the noncommutative polynomial ring R<S^-1> with formal inverses of S and add the appropriate relations. That works, but it's hard to do anything with that construction other than just using the universal property.

[mighty kiln]

Okie, thanks for the help

[chilly radish]

Yea, it's just really badly behaved

[ornate tiger]

hi if i had a question asks to find all right and left cosets where should i start to be able to solve this stuff

if we had this for example

[noble lynx]

I am trying to understand the reason for requiring algebraic indpendence in this. If the xi's are not algebraically independent, is what goes wrong the fact that homomorphisms preserve relations and so you cannot choose the \alpha's arbitrarily?

[true ingot]

Getting back to this now a few days later. The fact that these two are affinely related seems very unintuitive to me geometrically. I guess this is where the fact that we are working with F_3 field comes into play?

[tribal moss]

Yes, and it's also important that those sets are not ordered.
Hint: think of it as ||{(0,0),(0,1),(1,0),(1,1)} and {(0,0),(0,1),(1,2),(1,0)}|| being affinely related, recalling that ||2 is -1||.

[true ingot]

Ahhh when you say it like that... It's just a shear isn't it?

[tribal moss]

Yes.

[true ingot]

Thank you! Thats a cool way of seeing it 🙂

[winged void]

hello i have a question regarding the following because i do not really understand how the following is true

[barren sierra]

which part do you not understand?

[winged void]

well i do not understand the whole thing literaly how is it possible that if tau = sigam tau sigma ^-1 then this equivalent to that on the right side

[barren sierra]

This formula: $\sigma (a_1\ a_2\ a_3\ \cdots\ a_k) \sigma^{-1} = (\sigma(a_1)\ \sigma(a_2)\ \sigma(a_3)\ \cdots\ \sigma(a_k))$?

[winged void]

yes how is that true

[cloud walrus]

Spamakin🎷

[winged void]

it doe not make sense to me

[barren sierra]

ok so I'm going to just assume that tau is a cycle (since disjoint cycles commute + every permutation can be written as a product of disjoint cycles)

[winged void]

and why did you replace tau with (a_1 a_2 a_3 a_4 a_5 ... a_k)

[winged void]

sure i agree on that

but how do you know that tau is a disjoint cycle

and if tau is a disjoint cycle how is that going to help

[barren sierra]

Well if we can prove the formula for a single cycle

we can prove it for arbitrary permutations

for example $\sigma (a_1\ a_2\ a_3\ \cdots\ a_k) (b_1\ b_2\ b_3\ \cdots\ b_k) \sigma^{-1} = \sigma (a_1\ a_2\ a_3\ \cdots\ a_k) \sigma^{-1} \sigma (b_1\ b_2\ b_3\ \cdots\ b_k) \sigma^{-1}$

[cloud walrus]

Spamakin🎷

[winged void]

sure i agree

but then how can we go further from here

[barren sierra]

so without loss of generality we are going to assume that tau is a single cycle

ok well can you compute where a_1 gets mapped to?

Here's a start

Lets say all of these are permutations in $S_n$. For ease of notation let $\tau = (a_1\ a_2\ a_3\ \cdots\ a_k)$. Let $j$ be some integer $1 \leq j \leq n$. So since $\sigma$ is permutation there exists $a_i$ such that $\sigma(i) = j$. There are two cases: either $i = a_l$ for some $l$ or not. Tackle both of these cases separately.

[cloud walrus]

Spamakin🎷

[barren sierra]

I mean the two permutations are equal if and only if they are the exact same functions right, so we can just show that

[winged void]

I see I find though bit difficult

[barren sierra]

well try this

and see how far you get

and when you get stuck tell me your progress (ping me)

[winged void]

sure thanks alot!

[long geyser]

is this true if G is not abelian?

I don't think I used commutativity once when setting up zorn's lemma, so I say yes but then I'm wondering why the question specified abelian

[coral spindle]

I agree with you. The union of a chain of subgroups is itself a subgroup, even if we're non-Abelian. I don't think I'm missing anything. I'm now wondering if this was some kind of gotchya

[long geyser]

maybe the author is instilling in the readers to always check which of the hypotheses they are actually using

[coral spindle]

Just a bit funny to do that while asking questions at that kind of level

[long geyser]

ye lol

[coral spindle]

Perhaps there is an 'easier' argument that requires Abelianity that we haven't noticed

[long geyser]

nah

[coral spindle]

I doubt it too

[bitter locust]

wanted to ask a question here but i just answered it by writing it down

[long geyser]

I have no idea how to do this

plus, the hint is incomprehensible

(these are modules)

[long geyser]

now I understand the hint

(this definition is given 20 pages after the question is asked)

this book is way too comfortable with giving questions that require content further down

so if the exact sequence splits, we know that B is iso to A direct sum C, so what q should do is obvious

and the converse is simply from this, pretend I didn't ask anything

[barren sierra]

tbf I think retraction / section are terms that are introduced in most intro to proof classes (at least they were introduced in mine and ones my friends took as well as the one I TAed for)

however

gun to my head I could not tell you which is which 😵‍💫

[long geyser]

I've seen the term before, that is not the deal

but if you have qi = identity, then it is by definition a retraction

so, what is the hint?

[barren sierra]

oh

yea idk lol

[long geyser]

but the retraction they mean

is different

is the thing

[long geyser]

I don't think this is the same as a set theoretic retraction?

[barren sierra]

it is

the other map is the injection of S into M

[long geyser]

hmm, ok in this case A isn

't a submodule of B

[barren sierra]

since S is a submodule of M it's for all intents and purposes the identity map

[long geyser]

so ig

the hint is saying, try to have A be like a subset of B

[barren sierra]

it sort of is

[long geyser]

which I now get

[barren sierra]

I mean A injects into B

injection, subset," same thing"

[long geyser]

yes

[barren sierra]

this idea of "same thing" in quotes comes up a ton in talking about SES

[long geyser]

idea is you can write b = (b - iqb) + iqb

and it is like B = kerq + im i

and im i = ker p, so p restricted to ker q must be an isomorphism

cool

[slim kayak]

Have you already shown that it splitting is equivalent to B being a direct sum?

[long geyser]

shown that splitting implies direct sum, but the book showed the converse is false

[slim kayak]

Are you sure?

Mod some technicalities the splitting lemma should apply

[slim kayak]

Right, but take some time understanding what exactly fails here

[long geyser]

why?

isn't it because M is like infinitely many A + infinitely many C

idk why that matters, it still fails

[slim kayak]

Because the splitting lemma exists

It being isomorphic + a little bit extra is equivalent

[long geyser]

idk what's going on

brother, idk what the splitting lemma is, and I already solved the problem I asked about

I'm so confused rn

[slim kayak]

Ight

[idle sapphire]

Hi, does anyone have any resources on finite groups of lie type (and the conditions under which they are simple)? I've found some, but I really don't understand what a group of lie type is (and the resources I've found for that aren't helping). Note: I'm not sure if this is the best place to ask this question, seeing as though this is something probably covered in an undergrad math degree

[chilly ocean]

The splitting lemma says that the sequence is isomorphic to the split sequence. It's not enough to require that B is the direct sum. The maps need to be the canonical ones. (i.e. the inclusion and the projection)

Like this. (If you have such a diagram then phi is automatically an isomorphism by the five lemma)

[slim kayak]

Which are said technicalities

[chilly ocean]

I have to prove that a group G with at most three elements is always Abelian

For case 2, is it necessary for me to prove that ea=ae or is it not needed?

Also, is doing this by exhaustion the correct way?

[delicate bloom]

yeah

eg=ge is always true for the identity in a group, I suppose you could prove that as a separate thing altogether

[chilly ocean]

Yeah that's what I thought

So it's not needed here ig

[delicate bloom]

yeah I wouldn't worry about it personally for this

[chilly ocean]

Alr thanks

[delicate bloom]

also I'd probably do it by writing out a multiplication table, might help to see

kind of makes it into like a game of sudoku, but not repeating in a row or column means it's injective (a necessary aspect of being able to invert every element ofc)

[chilly ocean]

Yeah

Multiplication table sounds better, thanks again :p

[true ingot]

Reposting from #linear-algebra : If we denote $Z=Z(GL(n,F))$ we define $PGL(n,F) = GL(n,F)/Z$. At the lectures we said that $PSL(n,F)$ is the image of $SL(n,F)$ under this quotient map and from this it follows $PSL(n,F) = <SL(n,F), Z> /Z$.
How does this last thing follow exactly?

[cloud walrus]

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

[slim kayak]

If you can't nuke it with a certain theorem, then exhaustion is the way to go

[spice whale]

1g = g and g1 = g are usually both axioms i thought

of course you only need one if you know all your inverses are double sided

[dull ginkgo]

Idk how this causes problems for groups lol

For rings it does A LOT

I’m pretty sure if there are two nonequal left inverses there are necessarily countably many

Annoying exercise in Jacobson

[spice whale]

it doesn't

but it's usually included

makes it obvious that a group is a specific type of monoid

[tranquil musk]

Hello, I'm not sure if this is allowed here, but I have a test on basic algebra (groups, rings, fields) and the test is completely open book. So I have the all clear to take in as much material as I want from any textbook. I was wondering if anyone had suggestions on what to take in like books with solutions so I can match it to the questions I get or aything like that

[coral spindle]

We don't know what specific content you are going to be asked about, nor do we know the specific questions you are expected to be able to answer. The best way to prepare for this is going to be deciding for yourself what material to bring in based on the content in your course.

You want to know a secret? That's why your professor is allowing you to take in materials. They are tricking you into revising.

[elfin prairie]

I tried to do induction on n

I can't go from the induction hypothesis to the conclusion

I assumed an arbitrary U(k) has the property

how do I deduce that U(k+1) also has the property

[rocky cloak]

I don't think induction is very helpful here. Try instead to just come up with two examples

(look at U(n) for small n to see if there's a pattern)

[elfin prairie]

but how do I prove it?

[long geyser]

you can name them explicitly

that is the proof

[rocky cloak]

By coming up with two examples and showing that they satisfy x^2 = 1

[elfin prairie]

but that doesn't exhaust all the possibilities

[coral spindle]

You don't need to exhaust possibilities

[rocky cloak]

For example x=1 is always in U(n) and certainly has x^2 = 1

If you can find another you have at 2

[elfin prairie]

hmm okay

[rapid junco]

U(n) corresponds to the circle group

What complex number besides 1 has the property such that its square is 1.

[lusty marlin]

What's the circle group?

[elfin prairie]

you mean nth roots of unity?

that's a different one

[rapid junco]

Sorry I meant U(1)

So it gives an idea on how to proceed for U(n).

[elfin prairie]

$U(n) := {x \in \mathbb{Z^+} : x < n \text{ and } gcd(x, n) =1 }$

[rapid junco]

Oh I thought unitary.

[dull marsh]

And x >= 0

[cloud walrus]

James

[rocky cloak]

Hint is still somewhat relevant

[rapid junco]

James maybe you can use a counting argument

What is the order of the group? 1 is its own inverse and what happens if you assume that every element has an inverse which is not itself?

[elfin prairie]

hmmm

[rocky cloak]

I mean you can use an argument like that, but it feels like it's overcomplicating it quite a bit

[rapid junco]

Ok heres an idea on how to work out the order

Note that 1 and n-1 form a subgroup.

What can you conclude by Lagranges thm?

[elfin prairie]

that isn't introduced yet in the text

lemme skim the wiki

I think I vaguely know about this result

[rapid junco]

Subgroups divide the group in order

And so you have a size 2 subgroup

[elfin prairie]

order is even?

[rapid junco]

Yes

Because 2 divides the group size.

Now 1 is its own inverse, so the set of all elements besides 1 is of odd order. Pair up all elements together and deduce that 1 element must be left over

This works in generality for any group of even order.

So good to know.

[elfin prairie]

so what does this entail?

[long geyser]

if you've noticed this you've already solved the question...

you literally gave away the answer

[rapid junco]

Oh yeah lmfao

I guess so

[elfin prairie]

subgroups aren't even introduced yet

strange

[long geyser]

so, here's how I would approach this

what are the solutions to x^2 = 1 in the integers

[elfin prairie]

+1 and -1

[long geyser]

yeah, so I'm guessing you have seen that operations modulo n give the same result regardless of the representative

so like, if a = b mod n and c = d mod n then ac = bd mod n

[elfin prairie]

yeah

[long geyser]

now, see that if -1 is equivalent to some guy c modulo n, then (-1)^2 = c^2 mod n

and that is essentially the answer

though I guess, you must convince yourself that a guy equivalent to -1 belongs to U(n) in the first place, but I leave this to you

[elfin prairie]

I was thinking about that haha

[long geyser]

well, first think about what can be equivalent to -1 mod n that is a non-negative integer less than n

then check and see if this candidate is relatively prime to n

[elfin prairie]

(n - 1)?

[long geyser]

yes

[elfin prairie]

okay gotcha

now I gotta write down the proof

[long geyser]

[tardy hedge]

d is a common divisor of a,b if (a,b) subset of (d) right?

[next obsidian]

Yeh

if and ONLY IF

[tardy hedge]

Gcd is the smallest ideal that contains (a,b)?

[next obsidian]

No the smallest ideal that contains (a,b) is (a,b)

But actually if that ideal is PRINCIPAL then the generator is a GCD

[tardy hedge]

Oh ok yeah im on the section about euclidean domains so all ideals are principle

[next obsidian]

[tardy hedge]

Im not really sure what im confused about / the question im trying to clear up lol

[tardy hedge]

Oh ok and taking a multiple of the generator and considering the ideal generated by that, it would be a common divisor but not gcd

So d is gcd if and only if (a,b) EQUALS (d)?

[next obsidian]

Yeh

[tardy hedge]

Ty chmonkey

[fresh gate]

how do you prove that the order induced by the positive cone is total?

or in cone terms, a ∈ P ∨ (-a) ∈ P

[rocky cloak]

This is just the definition of positive cone

Like if a-b is not in P, then b-a is, so either a>b or b>a

[fresh gate]

wait now that I think about it

the definition is for a prepositive cone🤦‍♂️ 🤦‍♂️

I really am stupid huh

[tropic obsidian]

not sure how to prove that the kernel is contained in (X_1 - alpha_1, ..., X_n - alpha_n)

[bitter locust]

not sure about containment but i think you could also reason about R[X1, ..., Xn]/Ker f and use the 1st iso theorem

[tropic obsidian]

yeah I guess I am looking for a way to prove it directly

[rocky cloak]

Just by collecting like terms you can see that any polynomial p can be written as
p = q(x1 - alpha1) + r
where r is a polynomial in x2, ..., xn

Repeating for each indeterminate you get that each polynomial is a constant plus something in (x1-alpha1, ...)

[tropic obsidian]

sorry but how can you factor out x1 - alpha1 in the portion of terms not containing x1?

[rocky cloak]

I'm not sure what you mean

[tropic obsidian]

I guess I am just confused why you can write p in that form

[rocky cloak]

For example if you have
p = x1x2 + x3

then

p = x2(x1 - alpha1) + (x2alpha1 + x3)

In general you would right it out like
x1^n pn + x1^n-1 pn-1 + ...
and do normal polynomial division

[tropic obsidian]

ohh thats smart, thanks a ton

[dull ginkgo]

Can we do normal polynomial division in non-integral rings

I guess we need to keep track of the leading coefficient

[rocky cloak]

Yes, you can always divide by monic polynomials

The usual division algorithm only every uses that the leading coefficient is a unit

[dull ginkgo]

I forgot that remainder theorem holds for even noncommutative polynomial rings lol

[limpid ferry]

How comes ℤ_4/⟨(0)⟩ × ℤ_6/⟨(2)⟩ isomorphic to ℤ_4 × ℤ_2 ?
I already understood ℤ_4/⟨(0)⟩ = ℤ_4 and ℤ_6/⟨(2)⟩ is isomorphic to ℤ_2

[coral spindle]

If A is isomorphic to A’ and B is isomorphic to B’ then A x B is isomorphic to A’ x B’. Maybe you should try proving this intuitive fact.

[limpid ferry]

Yeah I should have asked a simpler question like that

[long geyser]

for part 2, I'm thinking of doing 0 -> ker g -> F -> G -> 0 where g and F are borrowed from part (i) and ker g -> F is inclusion
then, since this splits, there is an injective homomorphism G -> F and so G is a submodule of F

and finish with this

I'm still uncomfortable with the free stuff, so just need a check to see if my thought process is good

[vague granite]

To do the second part can’t you just pluck out a group presentation <X|R> and let A be the normal subgroup of the free group on X generated by R and let B be the free group on F(X)?

The fact that it splits tells you the free group is a direct sum of the two either side

But the subgroup of a free group is free and G is a subgroup so in particular G must be free

[long geyser]

I haven't done any group presentations or free groups (only abelian)

at least not in this book

[vague granite]

That’s a real shame because it it’s partly the reason why free groups are a big deal

Every group is the quotient of a free group so the free group is a kind of ‘cover’ that sometimes lets you do nice stuff

[long geyser]

it is a section on modules, not groups...

abelian groups are just a special case

[vague granite]

Oh whoops

In the case of Z-modules that just amounts to being several copies of Z

At least if it’s finitely generated

You can still always construct a group presentation in a fairly elementary fashion (although I don’t remember how you get the relations) perhaps you can study such a construction and try to recreate it in a way compatible with the notation in your textbook

[long geyser]

I don't think I will

I was just asking if my solution had any errors

thanks though

[vague granite]

Actually what I’ve just described was doing part i)

To do part ii) that is exactly what you would want to do

The sub module of a free module is free so since it splits you get that the central one is a direct sum

Boom profit

[dull ginkgo]

Free Abelian groups are so much easier to work with lol

Basically it’s “Abelian group with a basis as a Z module”

Instead of like

Horrors beyond your comprehension

[dull ginkgo]

I have a quick question fam, how does this book describe free Abelian groups

Like is it just the canonical example of Z^S or like any abelian group with a basis as a Z-module

[dull ginkgo]

infinite cyclic groups

JUST SAY Z

[dull ginkgo]

So uhm… here’s an idea right

For a SHORT exact sequence, the idea is A injects into B, B surjects into C

G is at the end right, and we are free to chose our A and B right

… what if your A or B are free themselves?

[long geyser]

am I getting trolled

what do you think my solution is doing?

[dull ginkgo]

Slightly lol

Let me reread it

[limpid ferry]

Hey I see this strange notation and want to ask about it: "the quotient group (Z/2Z)/(Z/2Z) is isomorphic to 1"
I wonder if they mean exact the integer 1 or {1} the trivial subgroup.

[ivory trail]

1 means the trivial group in group theory

sometimes people write 0 for it, too

[chilly radish]

Specifying generators is important

[coral spindle]

The quotient group [...] is isomorphic to 1
So it's a group.

[grizzled crane]

I’m trying to understand the definition of a field extension. A fied E is an extension of F if F is a subset of E and the operations of F are those of E restricted to F.

what does it mean by restricited to F?

[coral spindle]

It means restriction as in function restriction

if f : A x A x ... x A → A is a function

and B is a subset of A

we may restrict f to B x ... x B, which is to say we only consider its behaviour on elements of B

Note that its image may not be contained in B in general once restricted, but this definition is saying that we assume this is the case.

If you have seen the definition of a subgroup, subring, or subspace, this is exactly the same. If you haven't seen any of those things, you may be being a bit ambitious

[grizzled crane]

I was intitally thinking it’s similar to subfield

[coral spindle]

Yes, indeed, a field E extends F iff F is a subfield of E.

The definitions are precisely the same.

[grizzled crane]

the word restricted to F caught me

My guess was operations is restricted as in to the subfield elements.

for example, Q is a subfield of R under usual addition. The operation + is restricited to some elements in Q.

[coral spindle]

Yes this is exactly the same thing.

[grizzled crane]

ie sqrt(2) + pi doesn’t make sense in Q

but ‘+’ extends from Q

[coral spindle]

I don't quite understand the confusion but I hope in any case it's cleared up now

[elfin prairie]

Need some hints for everything except the socks and shoes analogy

[coral spindle]

For the second and third questions, choose a big non-Abelian group and start searching.

You already know infinitely many non-Abelian groups, so choose a big convenient one.

As for the name thing, socks and shoes is when there are two things you have to wear, and this new property needs three things.

[elfin prairie]

I only know the General and Special Linear Groups

[coral spindle]

I sincerely doubt you only know those groups. Have you heard, for example, of the symmetric group?

[elfin prairie]

I have vaguely

permutation group

[coral spindle]

This is a great opportunity to make it non-vague.

[elfin prairie]

okay let me try it out in S_3

found one example!

[chilly ocean]

is this Gallian

[elfin prairie]

yes

[chilly ocean]

nice

[limpid ferry]

can this trivial group "1" be the trivial subgroup of any group?

[coral spindle]

The trivial group embeds in every group.

[limpid ferry]

Given a group G, to prove that the direct product G x {1} is isomorphic to G ({1} being the trivial subgroup of any random group). Can I just define a homomorphism from G to G x {1}, that is bijective?

[coral spindle]

That is what an isomorphism is, yes

[chilly radish]

Mike it seems you lack familiarity with the basics of isomorphism and groups

I suggest you review the first few chapters of any group theory text

[limpid ferry]

I know what that is, I asked that question to know in advance if G x {1} is isomorphic to G, thanks for the advice tho 'cause I am still new anyway

[tardy hedge]

In many definitions of the norm for a euclidean domain, they state that N(ab) > N(b) is an axiom, and that the division algorithm holds. But in dummit and foote, they do not include the N(ab) > N(b) condition. Why is that?

Is that condition somehow implied just by there being a division algorithm?

[rocky cloak]

I don't think it's implied, just not really necessary to work with a euclidean domain.

According to Wikipedia, any euclidean domain can be equipped with a norm satisfying N(ab) > N(b). So I guess it might just be more convenient to assume all norms satisfy this

More specifically if you define N'(a) = inf N(xa) you'll get a new norm with the desired property.

[tardy hedge]

I see, but I wonder what results discussed in these books rely on that fact

probably some results only follow from that condition right?

[rocky cloak]

Well, no results about euclidean domains at least, but there could be results about different norms / when there is a unique norm etc

[tardy hedge]

It seems like the result that units in a euclidean domain satisfy N(u)=N(1) follows from this condition though

[tropic obsidian]

is the geometric intuition for newton's method valid intuition for hensel's lemma? Or since the topology of the p adics is just entirely different so one should not think of the two as the same? Having a hard time understanding why one should expect newton's root finding algorithm to "just work" over the p adics.

[south patrol]

Well the motivatino is the same for both. You want to find a root of a function $f$ and have some initial guess $a_0$. You can write $f(a_0+t) = f(a_0) + f'(a_0)t + \text{higher order terms}$ and if you set this to zero and ignore higher order things you get the formula for Newton-Raphson

[cloud walrus]

PD potato

[south patrol]

Like either way you hope/assume that t is "small" (wrt the relevant norm) and the norms are multiplicative so higher powers are less relevant, so you take the best linear approximation

[rocky cloak]

I feel like this is actually less intuitive for real numbers. At least in the p-addics we know exactly how p^k gets smaller.

So the question isn't "why does this work for p-addics?", but "why does it work for real numbers?"

And there I guess the answer is that it sometimes doesn't even work, so meh.

[limber sequoia]

I know there's only one finite field of size p^n, but does that field split all irreducible polynomials of degree n?

[coral spindle]

Yes

all *irreducible polynomials of degree n

[limber sequoia]

how come there isn't a polynomial that over the extension is reduced to two smaller polynomials?

[coral spindle]

Because they're irreducible...?

[limber sequoia]

i mean in the extension

i.e. why isnt there $f\in F_p[x]$ that is irreducible, and that $F_p[x]/<f>$ doesn't have all the roots of f?

[cloud walrus]

tomer_k

[coral spindle]

There should be an argument here via the fact that the roots of f will be roots of unity of the same degree, but I can't quite seem to think of it

[limber sequoia]

f should divide $x^{p^{degf}}-x$, and that splits

because you can show the finite fields are the sets of all the roots of that polynomial of the relevant size in the algebraic extension of $F_p$

does that make sense?

[cloud walrus]

tomer_k

tomer_k

[rocky cloak]

I guess you can think of it like if x is a root of f, then
x, x^p, x^p^2, ...
gives you all the roots of f

[coral spindle]

Working on that!!!

[rocky cloak]

Yeah, this argument also works

[coral spindle]

Here is a simple argument. If a is a root of f in F_p[x] then so is a^p, a^p^2, etc. So there are at least n distinct roots of f in F_p[a], where n is the smallest natural with a^{p^n} = a.

So what is the order of a? Note p^n -1 | p^d - 1 if and only if n | d. So in fact d=n is the smallest integer such that there is a primitive (p^n-1)st root of unity in \F_{p^d}.

Sometimes I really regret writing in ascii...

[limber sequoia]

wait, i thought $a^p=a$ in $F_p$

[cloud walrus]

tomer_k

[rocky cloak]

Yes, but a is in F_(p^n), not in F_p

[limber sequoia]

right

[coral spindle]

The polynomial x^2 + 1 is irreducible in F_p iff 4 does not divide p-1, so let's say p = 7 and consider the finite field F_49 := F_7/(x^2+1). What's x^7 in this field? x^7 = x.(x^2)^3 = -x.

[limber sequoia]

that makes a lot of sense

[coral spindle]

In fact, let's say x in F_{p^n} and x^p = x, where x is nonzero. Then x^{p-1} = 1. So in fact, it must be that x is in the subfield F_p, since the roots of X^{p-1}-1 are precisely 1, 2, ..., p-1.

Similarly, the fixed points of x |-> x^{p^2} are those in F_{p^2}, and so on.

[limber sequoia]

thank you

[vivid birch]

Is $S_2 \wr S_2$ isomorphic to $D_4$?

[cloud walrus]

Eduude

[rocky cloak]

Yes, there aren't that many groups of order 8

[dull marsh]

What's that symbol though?

[rocky cloak]

Wreath product

[vivid birch]

No one truly knows

[rocky cloak]

More explicitly if D4 = (s, r) with s reflection and r rotation, then (s, sr^2) = (S2)^2 is normal and the action of conjugation by sr is two swap them

[bitter locust]

when taking multiple quotients is it ok to just write (C[x,y]/(y-x^2))/(y-1, x-1) or does it not make much sense? by not making sense i mean, the 1st quotient is iso with C[x], so taking the quotient with (y-1, x-1) in this context is a bit more mysterious, like what even is y here when we have just polynomials in terms of x

[rotund aurora]

yes, it does make sense

it doesn't matter the order in which you quotient ideals out. When you quotient by an ideal after having quotiened by some other ideal, it is understood that you are taking the reductions

[bitter locust]

oh huh i wasnt aware of the order not mattering

C[x,y]/(y-x^2)/(y-1, x-1) - we first pick a function and then evaluate it at a point
C[x,y]/(y-1, x-1)/(y-x^2) - we first pick a point and then evaluate a function there
is this a good way of thinking about it? (at least for this case with these ideals)

tho what would C/(y-x^2) even mean, would that just be C/(0) = C

[rocky cloak]

Yeah, you would just use y and x to refer to their images in the quotient ring. So if you identify the quotient ring with C, then x=y=1

[bitter locust]

i see yeah that makes sense

[rocky cloak]

Also, instead of writing
C[x,y]/(y-x^2)/(y-1,x-1)
you could just write
C[x,y]/(y-x^2, y-1, x-1)

[bitter locust]

thats prob what makes the most sense

also that made me realize one could also write C[x,y]/(y-x^2)/(y-1)/(x-1)

smells like curry

[long obsidian]

Considering that the integers Z are a subring of the complex numbers C

Does it make sense to call the polynomials in 1 complex variable C[x] a Z algebra?

[rocky cloak]

Yes, in fact you can make any ring into a Z-algebra in a unique way

[long obsidian]

So in addition C[x] is an R-algebra where R is the real numbers?

[rocky cloak]

That's right

[tardy hedge]

Anyone know?

[rocky cloak]

Yeah
N(1) = N(u^-1 u) >= N(u) = N(u * 1) >= N(1)

[tardy hedge]

So this property does rely on N(xy) > N(x)?

[rocky cloak]

Yeah, I think so

[tardy hedge]

Yeah i noticed dummit and foote didnt mention this

And that was also the textbook that did not include that condition for a euclidean domain

[winged void]

@barren sierra

[barren sierra]

?

[winged void]

to be fair i do not really understand it

this is what i did until now

but did not get that far

in proving the thing

[barren sierra]

oh oh this question

did you try this?

[winged void]

not really

[barren sierra]

Lets try it

[winged void]

the thing is i do not really have intuition im missing that

[barren sierra]

it's not super intuitive

like what is this saying?

it's saying "if I conjugative tau by another permutation, I get a new permutation of the same cycle type"

[winged void]

yes i mean like what makes a cyclic permutation diffrent from a non cyclic permutation

[barren sierra]

if I were seeing this for the first time and someone asked me "would conjugating maintain cycle type" my instinct would be no

[winged void]

and here when i say the same cycle type then this mean that they have the same lenght

[barren sierra]

oh are you asking why did I reduce tau to a single cycle?

[winged void]

yes

[barren sierra]

rather than consider general tau?

ok suppose we know the result is true for 1 cycle, we immediately get the answer for two cycles

[winged void]

but i mean i thought that this result only true for a cyclic permutation

so for example lets say that we have such a permutation (12)(34)

then this is not going to be true

[barren sierra]

Because $\sigma (a_1\ a_2\ a_3\ \cdots\ a_k) (b_1\ b_2\ b_3\ \cdots\ b_l) \sigma^{-1} = \sigma (a_1\ a_2\ a_3\ \cdots\ a_k) \sigma \sigma^{-1} (b_1\ b_2\ b_3\ \cdots\ b_l) \sigma^{-1}$ so we can now apply the result to each cycle

[cloud walrus]

Spamakin🎷

[barren sierra]

wdym?

[winged void]

this is what i read on wikipedia

gonna send it

[barren sierra]

show me yea

[winged void]

i send you also a friend request because im afraid that the discussion is going to be long to be here

[barren sierra]

we can do it here, no one else is talking here

I don't see how what you sent there makes what we are proving false

all that says is that some permutations are cyclic

[winged void]

i mean then we have two or more cycles right

but a cyclic permutation is a single cycle

[barren sierra]

Lemme restate what I said

I claim that for any two permutations $\tau, \sigma$ that $\sigma \tau sigma^{-1}$ has the same cycle type as $\tau$. To prove this, it \textbf{suffices to prove this} for when $\tau$ is a single cycle.

[cloud walrus]

Spamakin🎷

[barren sierra]

If we can prove it for a single cycle, then by this argument we can prove it for any cycle type

[winged void]

sure i agree

but how are we going to prove it

[barren sierra]

Right so this is what I suggested

so lets try it

Let j be some integer between 1 and n

then there is some i with sigma(i) = j

the claim is that $\sigma (a_1\ a_2\ a_3\ \cdots\ a_k) \sigma^{-1} = (\sigma(a_1)\ \sigma(a_2)\ \sigma(a_3)\ \cdots\ \sigma(a_k))$, that's what we want to show

I am going to show this by evaluating both sides of the equal sign on j. If they agree on all j, then they must be the same permutation

[cloud walrus]

Spamakin🎷

[barren sierra]

so $j = \sigma(i)$ for some $i$. Now we have two cases. Either $i = a_l$ for some $l$ or not. Lets tackle the first case. Without loss of generality lets say $i = a_1$. Then what is $\sigma (a_1\ a_2\ a_3\ \cdots\ a_k) \sigma^{-1}(j)$? Can you compute that for me?

[cloud walrus]

Spamakin🎷

[winged void]

just give me a second

im reading throught it

[barren sierra]

yea take your time

[winged void]

well we know that sigma(i) = j thus this means the following => sigma (a_1 a_2 ... a_k) sigma^- sigma(i) = sigma (a_1 a_2 ... a_k) j

[barren sierra]

you mean i at the end, not j?

sigma^-1(sigma(i)) = i

And then what is (a_1 a_2 ... a_k) i?

[winged void]

yes that is what i mean

[winged void]

well that is what i do not know

[barren sierra]

well there are two cases

either $i = a_l$ for some $l$ (without loss of generality say $i = a_1$) or $i$ is not any of the $a_l$. Lets tackle these cases separately. Lets tackle the first case now.

[cloud walrus]

Spamakin🎷

[barren sierra]

In the first case, what do we get?

[dull ginkgo]

This is why I hate symmetry groups

[barren sierra]

cool man let us work through this problem

[barren sierra]

@winged void you trying this? how is it going?

or are you stuck

[winged void]

Let me think cuz I’m getting error

[dull ginkgo]

ooooooohhhhh

[barren sierra]

don't give away anything Chmiz

[dull ginkgo]

Okay I don’t speak that language but is it about an n-cycle in S_n having cyclic centralizer

[barren sierra]

No

[dull ginkgo]

Not answering it I just was trying to remember if I did this before

[barren sierra]

it's about how if you conjugate a permutation tau by any other permutation, the resulting permutation has the same cycle type as tau

[dull ginkgo]

Oh okie

[barren sierra]

cute (and semi-surprising) fact

[dull ginkgo]

Yeah when I first came across that I had to ponder for like 20 minutes to explain it

[barren sierra]

so you can index the conjugacy classes of S_n by their cycle types, this is useful in representation theory

[barren sierra]

What error are you getting, since this should be almost immediate given that i = a_1?

[winged void]

i will take my time to think about it

now its too late

like almose 12 am

00:00

so i will reread it tommorow

[barren sierra]

ah ok

fair enough

[bitter locust]

thinking about this further, i noticed that if i start at C[x,y] i can pick a function on it like f(x,y) = x + y, i can restrict it to C[x,y]/(y-x^2) to the parabola, and restrict it further to C[x,y]/(y-x^2)/(y-1, x-1) to the point (1, 1)

at first i thought that this was something motivational for introducing sheaves later on, but the restrictions here happen on coordinate rings, rather than on spaces, and that restriction process is done on closed sets (plane -> parabola -> point) rather than open ones

is this related to sheaves in some way or is it just a coincidence

[tardy hedge]

Proofs in dummit and foote showing that certain rings are not euclidean domains do so in reference to a specific norm. For example, they showed Z(sqrt(-5)) is not a ED by showing that its not a PID, but the way it showed this was using a specific norm …

Then it concludes by saying its not a ED for any norm

Im just confused by it using a specific norm, and then that shows its actually not a ED no matter what norm you pick

What if a different norm makes it an ED, with a division algorithm that works etc …

I guess im just not sure how the specific norm function you pick relates to a division algorithm working or not. Because i guess this is implying so far that the division algorithm existing or not doesnt depend on the norm itself, but i guess something more already inherent to the structure of the ring

[rocky cloak]

The crucial part is just that it's not a PID, hence not a ED. The norm they defined is used in showing that it's not a PID, but is not really related to a potential division algorithm beyond that

[tardy hedge]

The proof was assuming the ideal is principal and then showing a contradiction, but the contradiction relied on the norm they picked

[south patrol]

But that's fine

Whether an ideal is principal isn't dependent on the norm or anything

[rocky cloak]

Exactly, they use the norm to show that there are non-principal ideals

[south patrol]

Norms are just a helpful piece of structure in general - and sometimes they also happen to give euclidean functions

I assume the proof they did showed that 2, 3, 1+- sqrt(-5) are prime by using the norm and showed it isn't a UFD?

Or smth

[rocky cloak]

I'm guessing, showing that (2, 1+sq-5) is non-principal because they have no gcd maybe

[tardy hedge]

I mean like, reason why im confused is that if u define some additional structure to the ring (the norm) and ur using THAT to show the ideal isnt principle , like what if the additional info from the norm is whats making the ideal not principle

Cuz the proof used the norm itself to show it cant be principle thats why i dont get it

[south patrol]

I don't really get what you mean

You have extra structure and do something with it

[rocky cloak]

But wether an ideal is principal or not only depends on the ring structure. Imposing extra structure can't change that

Like Z doesn't stop being a cyclic group just because it also is a ring for example

[coral spindle]

It's not cheating to use whatever info you have about a structure to learn things about it. You wouldn't be surprised if, for example, I used the natural ordering on the integers to show that it's an infinite group, right? The norm isn't some extra thing we're tacking on, it's actually a natural thing.

I mean look at the typical norms for EDs that you know. The big one that comes to mind is degrees of a polynomial. That's very natural. We're just noting that a lot of things have this kind of natural notion of size.

[tardy hedge]

Thx for the help so far ill be back with more questions

[tardy hedge]

Ok, yea, cause i was thinking the norm was just some tacked on thing. But idk how the definition of what a norm function is related to the ring. Isnt it just a function that takes ring elements and outputs positive integer or 0?

Please forgive my ignorance. Chatting here and saying dumb questions helps me a lot

[coral spindle]

No, as you know well there are other requirements for a Euclidean norm

If we had any old function it would likely be bullshit

For other kinds of norm we have different requirements

[tardy hedge]

I dont actually know. What are the other requirements? Im pretty sure dummit foote just says its a function that outputs 0 or pos integer

Let me see

It says any function

But then i guess for it to be a ED it says there is SOME norm so that the division algorithm is a thing …

Oh so thats the “euclidean” norm. Ok

Then, im still confused by this example

It shows that ideal cant be principle by contradiction from that random norm it defined

What if that norm was just some bs function that wasnt actually related to ring structure ? How do we even know

[coral spindle]

If it was a random bs function then most of the argument—which depends on the specific definition of N—would simply be false.

[tardy hedge]

I will look more closely at it thanks for help

Omg silly me i think i know where i got wrong

.. k not yet

[kind temple]

is there an example of a non-commutative ring with 1 and a multiplicative subset such that the localization of the ring at that subset doesn’t exist?

i ask because when showing that localizations exist for commutative rings with 1, the only place we use commutativity is when establishing transitivity of the equivalence relation on R x S

i was trying to find a counter example in Mat(n,R), but im having trouble producing a poorly behaved multiplicative set S

[mighty kiln]

Localizations always exist right

In the universal sense

[kind temple]

the universal sense?

[mighty kiln]

Like, you can adjoin inverses of S

[kind temple]

i thought this only works if the ring is commutative though

[mighty kiln]

You generate all elements the same way you generate a free group

[kind temple]

then why not do it that way to begin with

why do all the sources i find assume that the ring is commutative first

[mighty kiln]

In the commutative case you have all elements being rs^-1

So you don't need to look at all "words"

[kind temple]

is this similar to how you can adjoin elements to a polynomial ring

[mighty kiln]

Yes

But polynomial rings adjoin commutative elements

[kind temple]

okay, but before adjoining inverses, there should be a c.e. to this, correct?

[mighty kiln]

What's a c.e.

[kind temple]

counter example

[mighty kiln]

You mean localization in the sense of {rs^-1}?

[kind temple]

yea, something where the construction of S^-1R fails

due to the non-commutativity of R

i.e., the equivalence relation (m1, s1) ~ (m2, s2) <==> there is some t in S such that t(s2m1 - s1m2) = 0 fails to be transitive

[mighty kiln]

Let M be the free monoid on x, y

Localize R[M] by y

Then xy^-1 is an element that isn't in S^-1R

[kind temple]

sorry, what’s R[M]?

[mighty kiln]

The ring you get from adjoining M to R

[kind temple]

i uh. am unfamiliar with these objects and terms tbh

but i believe you, thanks for verifying my suspicion

[mighty kiln]

https://en.m.wikipedia.org/wiki/Monoid_ring

In a sense one literally adds elements of M to R (so that M commutes with R)

[crystal vale]

I think this one is tricky

[ashen heron]

if it were commutative then for any x in X, x not equal to r, x=x.r=r.x. then (x.r).(r.x)=r but this means x=r (contradiction)

[chilly radish]

Localisation still "works" in the noncomm setting, in that it gives you a ring, but it's cery unwieldy

You don't get a nice presentation like in the commutative case really

That's why we need the ore condition

That guarantees a well-behaved localisation, where you can present elements as s^-1r

So generally you will get elements of the form
r_1s_1^-1r_2s_2^-1...r_ns_n^-1
And sums of them

Not very nice

Ore guarantees you a presentation of the form s^-1r or rs^-1 or both, depending on if your ring satisfies the right or left ore condition

The localisation in the sense of the usual universal property exists though, mainly due to abstract nonsense

[crystal vale]

Got it, thank you ❤️
Can I get to know what your thought process is behind this question?

[dull ginkgo]

For the love of all things holy this R better be commutative or I’m raising hell

Actually wait I think you can just do (R[M])[X]/(1 - yX)

because you need two sided units as otherwise it’s very, very scary

[bitter locust]

can i get some opinions on this

[dull ginkgo]

I am trying to jump to the original post but Discord mobile sucks

[bitter locust]

real

god they butchered the search feature too

[dull ginkgo]

I don’t know very much about sheaves or varieties besides fucking around with it without knowing they were an actual thing but

Quotienting out by ideals which are the polynomials vanishing on algebraic sets of like R^n is a major idea in that

For example the parabola (x^2 + 1) as you mentioned

Points on the parabola cause polynomials that are in the principal ideal generated by x^2 + 1 under valuation

and you can use ring isomorphism theorems to kind of go further with that concept and to expand the structure

[bitter locust]

go further in what way?

[dull ginkgo]

Have you ever tried localizing by a prime ideal complement?

Special collections of points are called Varieties, and their ideal is prime.

[bitter locust]

i havent yet, so far i only experimented with different kinds of spaces and ideals

[chilly radish]

the vanishing set of x^2+1 is empty

Over R

[bitter locust]

i didnt even notice that it was x^2 + 1 my brain made it look like y - x^2 lmao

[chilly radish]

y-x^2 will be a parabola

well, the vanishing set of y-x^2

[dull ginkgo]

Just pulled a random one out of my ass

Meant to say C

[chilly radish]

That's still not a parabola

[dull ginkgo]

Fair

[ashen heron]

i mean it says X has more than one element, so there is an r and some x not the same as r. then i just tried contradiction, if it were commutative then x is x.r but x.r is also r.x. now look at the axioms again, x.y = r iff x = y so set x and y to be x.r and r.x...

[dull ginkgo]

@bitter locust here’s a nice exercise chain, skip parts at will

Let R be a commutative ring

1. Show an element of R is a unit if and only if that element lies in no proper ideals

Let R be an integral domain, and let S be a multiplicative subset
2. Characterize ideals in RS^-1 based off ideals in R (similarly to fourth isomorphism theorem)
Hint: ||R injects into RS^-1, and all of the elements in S become units. Use the prior exercise||

Let P be a prime ideal in R.
3. Show that the complement of P in R is a multiplicative subset
4. Describe the maximal ideals in the localization by P’s complement

[bitter locust]

thanks ill give it a shot, i already have an intuitive understanding of that like for example units cannot be in ideals because if they were theyd end up generating the whole ring etc

so it shouldnt be too hard to prove it

[crystal vale]

Yes, thank you

[dull ginkgo]

I have another set after relating to the polynomials lol

[elfin prairie]

I have a little doubt related to cyclic subgroups

If G is a group, then $\langle a \rangle := {a^n : n \in \mathbb{Z} }$

[dull ginkgo]

The generated subgroup of a is the smallest subgroup containing a

[cloud walrus]

James

[dull ginkgo]

all powers of a are in every subgroup containing a because that subgroup is closed under the operation

[coral spindle]

Let's hear the question first

[elfin prairie]

is the group operation multiplication?

[dull ginkgo]

Yes

[coral spindle]

No that's misleading

Remember that the notation we use for a group is arbitrary

[dull ginkgo]

Well the operation I assume he’s calling multiplication

[coral spindle]

We can write it however we like, but usually we write it as if it were multiplication.

[elfin prairie]

is it the same as the group operation,then?

[coral spindle]

But when you write <a> for a generated subgroup, you have to have a group already in mind that a is an element of

Yes, that's right. It's the operation of the group that a is an element of.

[coral spindle]

And I think I've demonstrated why it's misleading to say this :)

[coral spindle]

For example, if we're working in the group Z, then <2> = {n2 | n in Z} is the usual notation for that.

<2> is the subgroup of even integers

[elfin prairie]

can -1 generate Z?

[coral spindle]

Yes, <-1> = Z.

The only subgroup of Z that contains -1 is the whole of Z

[elfin prairie]

but -1 added multiple times with each other only spans the negative integers

[coral spindle]

Ah, but we're not just adding.

We need inverses too, otherwise typically we wouldn't get a group

[elfin prairie]

oh right

[coral spindle]

In general, a better definition of <a, b, ...> is the smallest subgroup containing each element a, b, etc.

So this should explain now why we get inverses in a more straightforward way

[elfin prairie]

I see

[dull ginkgo]

Fair enough!

[hidden wind]

the definition <a> = {a^k : k in Z} doesn’t generalize nicely to generating sets of more than one element

[mighty kiln]

Not quite

X shouldn't commute with all of R[M]

It should be R[M'/(yX=1)], where M' is the free monoid over {x, y, X}

[dull ginkgo]

Oh it shouldn’t good point

[tardy hedge]

oh is it because the field norm in that example is given from (a+bsqrt(-5))(a-bsqrt(-5))?

So its not a bs function, the value of the norm is actually from the ring itself

[rocky cloak]

Yes, and I think even more important here is that it's multiplicative. So it tells you a lot about whether elements are multiples of each other or not

[tardy hedge]

Right, thanks a lot. My question now is that, if a ring is a euclidean domain, would any norm work for the division algorithm?

[tardy hedge]

I understand now that for that example, showing that its not a PID proves that it cant be a euclidean domain for any norm

[rocky cloak]

I guess it depends on your definition of norm

[kind temple]

oh, is it similar to how darnymuckhi described it?

or is it different

[limpid ferry]

"If m divides |G|, G being a finite abelian group, then G has a subgroup of order m". Is this statement a result of Elementary Divisor Decomposition at all?

[chilly radish]

No, because y only has a one sided inverse in his construction

But it's similar

You take the free associative algebra over your ring in however many variables you need and then introduce inverses via quotients

[kind temple]

sorry, i was referring to this

[chilly radish]

Yes

Pretty much

[kind temple]

okay cool

[chilly radish]

This can also be just the 0 ring

Also

Do you happen to know about derived categories

[kind temple]

even if S doesn’t contain 0?!

[kind temple]

no

[chilly radish]

If S contains 0 divisors...

[chilly radish]

Nvm then. If/when you learn about them, compare the way that you obtain the derived category from the homotopy category to Ore localisation. Yakutieli does this explicitly (his book on dg categories)

[kind temple]

will keep this in mind

[dull ginkgo]

horror

[mighty kiln]

Oop I forgot about one-sided inverses

[dull ginkgo]

BIG HORROR

[dull ginkgo]

If there are two nonequal one sided inverses of an element then there is necessarily infinitely many

[tropic obsidian]

is the ideal generated by a single element denoted aR still used for non commutative rings?

[delicate bloom]

yeah but you have to be a little more careful, that'd be a right ideal, which would be generically distinct from the left ideal Ra

[tropic obsidian]

for non commutative do people just use RaR then?

for two sided ideal

[crystal vale]

If the ring has no unity then two sided ideal generated by a look like Ra + aR + na + RaR, where n in Z

[dull ginkgo]

horror

[vivid birch]

Is it true that if I is an ideal in a semisimple ring, and if I^2 = 0 then I=0?

[vapid vale]

yes

use the fact that a semisimplicity implies jacobson radical = 0 and artinian

[rocky cloak]

RaR is used, or (a) sometimes

[vivid birch]

can you elaborate?

[rocky cloak]

The sum of all nilpotent ideals aka the nilradical is contained in the Jacobson radical.

And being semisimple is equivalent to being artinian with 0 Jacobson radical

[bitter locust]

for the last one, is an argument like this fine or is it not general enough?
take any a from R\p
consider a/1 from RS^-1, it has an inverse of the form 1/a, which means that it's a unit, so no element a/1 can be in an ideal, so no element a/1 * 1/s can be in an ideal either
and then one can just check that p and (0) are prime ideals and p is maximal

[dull ginkgo]

p is the maximal ideal

It’s local (the localization)

Hence the name :3

[bitter locust]

yeah given just 3 ideals i can check that its indeed the maximal one

[dull ginkgo]

:3

Yep

[bitter locust]

:3

[dull ginkgo]

But yeah that’s kinda one of the origins of localization

[bitter locust]

i saw a cool vid about it with an example in Spec C[x,y], it showed how the local ring at a point p contains just that point, but also the generic points associated with the space and the curves passing through p, which was quite enlightening

[bitter locust]

also can i ask for this

[dull ginkgo]

In a bit when I get home from work

[bitter locust]

alrighty

[errant wedge]

what would be a more intuitive way to think abt why these r isomorphic? I can see it with certain structures but groups elude me a bit

[prisma ibex]

one way to think about this is if you look at a cyclic group Z/nZ where n=p^{k_1}_1...p^{k_n}_n is a factorization into prime powers, then you have an isomorphism

$\mathbb{Z}/n\mathbb{Z}\simeq\mathbb{Z}/p^{k_1}_1\mathbb{Z}\times\hdots\times\mathbb{Z}/p^{k_n}_n\mathbb{Z}$

[cloud walrus]

nGroupoid

[prisma ibex]

so then if you look at the group of units you have an isomorphism

$(\mathbb{Z}/n\mathbb{Z})^\times\simeq(\mathbb{Z}/p^{k_1}_1\mathbb{Z})^\times\times\hdots\times(\mathbb{Z}/p^{k_n}_n\mathbb{Z})^\times$

[cloud walrus]

nGroupoid

[prisma ibex]

in your example you're looking at the group of units of Z/12Z, in which case you have an isomorphism

[errant wedge]

ohhhh

[prisma ibex]

$(\mathbb{Z}/12\mathbb{Z})^\times\simeq(\mathbb{Z}/2^2\mathbb{Z})^\times\times(\mathbb{Z}/3\mathbb{Z})^\times$

[cloud walrus]

nGroupoid

[prisma ibex]

which makes it clear why this is (Z/2Z)^2

[errant wedge]

Mhm tysm

[prisma ibex]

I should say this first isomorphism comes from the Chinese remainder theorem

[errant wedge]

Mhm, I forgot abt it but ik the theorem, thanks so much :)

[tropic obsidian]

why are pre images of ideals called contractions, instead of extensions? If anything, arent pre images more "complicated"

[next obsidian]

Imagine u have a subring A < B

And took a preimage

It’s like u made it smaller

And contracted ideal into A

[tropic obsidian]

hmm, if you consider a projection instead from R to R/I then wouldnt pre images be bigger?

I guess my intuition is backwards

[rocky cloak]

It would be, but the relationship between ideals in R and R/I is quite simple. Ideals in R/I corresponds exactly to ideals in R containing I.

The relationship between ideals in a ring and a subring is much more complicated and interesting

[tropic obsidian]

I dont know how to show x1^2 + x2^2 + x3^2 is prime

I think I can do case work into different degrees, but is there a general way to do things like this?

[delicate bloom]

I guess x^2+y^2 is more easily seen to be irreducible, and so x^2+y^2+z^2 is irreducible by eisenstein now.

[tropic obsidian]

thanks, I got it using R[x, y, z] is a UFD and just algebra mashing to get a contradiction, since the degrees are small

[south patrol]

I guess this is very nice as it generalises to n not just 3

[vagrant zinc]

name of book ?

[lethal cipher]

Okay, so question.
I know if you have some field R, then R[X] is a PID.
I've more or less accepted this as fact.
But, how would we actually go about showing that? I'm having a bit of trouble piecing this together.

[rotund aurora]

by showing it's a euclidean domain

i.e., by showing that you can do the division algorithm

I guess this is not totally trivial, the key idea here is the ||degree||

[crystal vale]

R is field if and only if R[x] is PID

[errant wedge]

Professor’s notes sorry

[long swan]

direct sum and cartesian product are the same thing for abelian groups right?

$V \oplus W$ versus $V \times W$

[cloud walrus]

chipotle

[dull ginkgo]

For finitely many, yes

[long swan]

gotcha

ty

[dull ginkgo]

For infinitely many, the Cartesian product is as expected

But the direct sum consists of just finite sums :3

[long swan]

yeah that makes sense

[inner harbor]

Guys What's K[X,Y]/K[X]

[lusty marlin]

Are K[X,Y] and K[X] polynomial rings here? If yes, then this quotient doesn't make sense, since K[X] isn't an ideal of K[X,Y].

[inner harbor]

Thats why it doesnt make sense, i thought it was

[warped shadow]

is the exponential surjective from gl(R,n) to GL^+(R,n), i.e. to the space of invertible real matrices with positive determinant?

or do I need gl(n,C)

[rocky cloak]

It is not surjective.

For example the diagonal matrix diag(-1, -2) is not in the image

[warped shadow]

so do I get a map to only positive definite matrices?

well not quite

that's too little I guess

[rocky cloak]

Well -I is certainly there

[warped shadow]

there was as far as I can remember a characterization based on the polar decomposition

[rocky cloak]

So for diagonalizable matrices youd want negative eigenvalues to come in pairs

[warped shadow]

e.g. A = US = U exp(X)

[warped shadow]

in pairs?

[rocky cloak]

Like if you have -1 as an eigenvalue you'd want it to have even multiplicity

So diag(-1, -2) is no good because each negative eigenvalue has odd multiplicity

[warped shadow]

how would I get -I from an exponential

for n even

e.g. diag(-1,-1)

[rocky cloak]

It's the exponential of a rotation matrix

Or of the 90 degree rotation matrix times pi rather

[warped shadow]

I think you mean of a skew-symmetric matrix with theta = pi

exp(-theta J) = rotation by theta

ok, that makes sense

[rocky cloak]

Point being, that every rotation matrix is in the image of the exponential

[warped shadow]

I am asking because I was doing interpolation in SO(n) and SE(n) with the exponential

$A(t) = \exp(t \log(A_1A_0^{-1}))A_0$

[cloud walrus]

criver

[warped shadow]

which gives me the geodesic between A_0 and A_1

if A_0,A_1 are both from O^+(n) or O^-(n), or from E^+(n) or E^-(n)

I was wondering whether the same can be applied for geodesics on GL(n,R)

but I guess not because the exponential is not surjective

of course I could go to GL(n,C) and gl(n,C) (since there the exponential is surjective)

but then I have no guarantee that the curve would not pass through a complex matrix inbetween

basically I wonder why I cannot interpolate within each of the two connected components of GL(n,R)

[rocky cloak]

Well, I don't want to claim that this is the correct answer in this case. But it's not so hard to cook up examples where geodesics don't exist.

Like if you take the plane minus the origin with us usual metric, then there's no geodesic from (0, 1) to (0, -1) for example.

I guess we can imagine something like that is happening for GL(n, R)

[warped shadow]

so you're saying there are "holes" in GL^+(n,R)?

I was kind of imagining GL^- and GL^+ like O^- and O^+, but I guess that mental model was wrong

[rocky cloak]

Not necessarily holes per say. But GL^+ is not compact, so it could be that for example the shortest path between two points is to swing past the zero matrix. But no matter how close you get to the zero matrix you could always swing just a little bit closer, making the path just a little bit shorter

Hence no path is the shortest

In O^+ you don't have this problem. If you keep improving your path you'll eventually converge on the shortest path, because O^+ is compact

[warped shadow]

so I guess something like diag(-1,-2) and diag(1,2) would want to swing through 0?

e.g. you would generally have taken a linear interpolation I guess

but it goes through zero

[rocky cloak]

Yeah, or some other noninvertible matrix, idk

[warped shadow]

how does C solve this, e.g. in the case diag(-1,-2) and diag(1,2)

i guess it would just take the logarithm and get a complex matrix

and then take some funky path

[rocky cloak]

I guess you can take a shortcut through some complex matrices

[warped shadow]

that clarifies it a bit, thank you

I was looking at O(p,q,R) previously also, and there for p,q>=2 the exponential is not surjective, I guess also due to a similar reason

(even if I restrict myself to the connected component O^++(p,q,R) containing the identity)

[young hollow]

Hello,

I’ve posted a precursor question to this in the #linear-algebra section and received response and confirmation for my attempt, which in turn brought me to my next question.

The original question was 2c below:

Redefining addition/multiplication as mod_n allows the integers and positive integers to form a field.

My natural follow-up inquiry is, how can I prove that these are the only ways to redefine addition and multiplication to form a field with the integers or positive integers?

[coral spindle]

So your answer to this is incorrect. That's not a redefinition of addition and multiplication on the integers: you have changed your set from the integers to only (as you might define it) the set {0, ..., n-1}, and not merely the definitions of addition and multiplication. Furthermore typically this will not be a field unless you have a certain restriction on n.

But frankly, 2(c) is a poor question to ask at this stage in time. Yes, there are in fact infinitely many non-isomorphic fields that we can define with underlying set Z, but these will all have very unnatural definitions of + and * that do not respect how we think of the set Z. (This is because there are many fields with countably many elements, so we may pass addition and multiplication through this bijection in such a way that defines a field with underlying set Z.)

Now on the other hand, your confusion did observe that somehow the integers mod n are related to the integers in a special way. You may see this later in your studies, long after linear algebra, but the integers mod n are what are called a 'quotient' of the integers. Indeed these are the only quotients of the integers and they form fields iff n is prime. A quotient of a ring (a ring is like a field but not necessarily having inverses) is a bit like redefining the set without redefining the addition and multiplication, but this is a very simplified explanation.

[rocky cloak]

I guess maybe you can think of it as a multipart question.

Like what happens if you change addition, what happens when you change multiplication what if you change both? Still not an amazing question though

[untold basalt]

Let G be a group of order 357. Prove that the centre of G contains the Sylow 17-subgroup.

I've definetely seen something similar before but I can't remember how it was done

[young hollow]

Thanks so much for your detailed reply! I am somewhat familiar with some ‘higher’ algebra, but I’m trying to learn it more systematically this time. Would it change anything if I redefined the operations as such:

The first attempt that comes to mind is something like:

Define addition as usual, except for any integer i, i+i=0, and likewise for multiplication (i,e., i*i=1, for i ~= 0)

[coral spindle]

This is not a field.

I suggest that you try to see this yourself

[young hollow]

Understood, thank you

Is there any way to prove such a redefinition cannot satisfy the field axioms?

I’ve also attempted that proof to no avail

[coral spindle]

Really? Have you gone through each of the field axioms and checked them in order?

Or are you talking about some other redefinition

[young hollow]

I just mean a proof of the theorem “no redefinition of addition or multiplication exists such that the integers form a field”

[coral spindle]

One cannot prove a theorem that is false

[young hollow]

But by reductio?

[coral spindle]

As I mentioned up here, there are in fact infinitely many ways to do this.

[young hollow]

Perhaps I should wait to learn some of the more advanced machinery

[coral spindle]

I agree

[young hollow]

Thank you for your help!

Perhaps one last meta question, please

Halmos is fairly well regarded

The question does seem out of place. What do you think his intention with it was?

[coral spindle]

I don't know.

[young hollow]

😅

Ok, thanks again

[rocky cloak]

So step 1 would be to show that the 17-sylow subgroup is normal, hint ||use sylow theorems||

Next consider how G acts on the sylow subgroup by conjugation, then the group is in the center if the action is trivial, hint ||The action can be described as a map G -> Aut(C17), what is Aut(C17), which maps are possible?||

[slim kayak]

If I have a finitely generated k-algebra A, and B its integral closure in the fraction field of A, can I conclude that B is a finite A-module?

[rocky cloak]

It's true, but I think it takes some work to prove

Some sources here
https://math.stackexchange.com/a/391220/306319

[slim kayak]

Hm, okay

I am asking this in regards to harstshore II.3.8, showing that if X is of finite type over k, then the normalization of X is finite over X. Maybe there is a more straightforward reason that I am missing

[slim kayak]

oh

[barren sierra]

yea this is why Hartshorne is annoying

[slim kayak]

I am happy that it is at least provided as a theorem for use lol

[next obsidian]

Lmao

[dim widget]

It’s because it’s Japanese

[errant wedge]

Why's it 4C4 and not 6C4 for c_1, ..., c_4?

[vivid birch]

If $K \subseteq F$ are fields and $M$ is an $K$-module, is it true that all the sub-modules of $F\otimes_K M$ as a $F$-module are of the form $1 \otimes N$, for some sub-module $N$ of $M$?

[cloud walrus]

Eduude

[slate fulcrum]

probably a mistake from snew

[lethal cipher]

I'm not too familiar with the terminology in the 2nd paragraph. What does it mean by "I-residue"?

[lusty marlin]

As written in the paragraph itself, the I-residue of X_i being a_i means that X_i - a_i ∈ I

[lethal cipher]

Okay, then followup: why can we say such an a_i exists?

I do understand that there is some polynomial F such that X_i-F is in I, but why can we say that the constant polynomial a_i is such a polynomial?
I understand k=L here, but I'm having a hard time seeing why that tells us this.

[next obsidian]

Just like me

[next obsidian]

It’s literally just the residue class of Xi

Or okay, this is maybe not as instructive

The way k embeds inside of L is by considering the composition k < k[X1,…,Xn] -> k[X1,…,Xn]/I = L

To say that k = L means that this map is surjective, so anything in L lifts to something in k

[ornate tiger]

is there a formula to find the permutations for S_4 or any S_n

[mighty kiln]

The number of permutations is n!

[ornate tiger]

yeah the number of them but what are the elemetns how to know all of them

[mighty kiln]

They are all the permutations on n objects

[ornate tiger]

why they are some with less than 4 objects like not (xyzw)

it be like (xyz)

[mighty kiln]

For the symmetric group over {x, y, z, w}, the notation (xyz) means (xyz)(w)

x → y, y → z, z → x, w → w

[ornate tiger]

is there permutations of the form (xy)(zw) ?

[mighty kiln]

Yes

[ornate tiger]

how did u know that