groups-rings-fields

[boreal inlet]

Ok that's fair

I can't factor the map otherwise

[next obsidian]

Idk what you mean

🗿

[boreal inlet]

What I mean is I can't write phi as some f • pi

Where pi is the surjection to R/I

[next obsidian]

Yeah

It literally isn’t possible

Cuz you pickup a kernel

Unless I is 0 I guess lol

[boreal inlet]

Yeah then I guess there's no way to actually do it except using First Iso

[next obsidian]

Well no

You can do the other thing I said

You can induce maps on quotients in both directions and conclude they’re inverse

[boreal inlet]

How do you induce without the universal property

[next obsidian]

Right so

I was gonna say showing you can induce such maps is how you prove first iso

But it “isn’t the first isomorphism theorem”

Chmonkey

[boreal inlet]

Ok see I get how R/I -> S/J should look like, but I don't see how it can be well defined

It is, but it's not like trivial

[next obsidian]

Yeah

[boreal inlet]

Unless you already use first iso

Then yes trivial

https://cdn.discordapp.com/emojis/756386972662235286.png?quality=lossless&size=48

[next obsidian]

Well to prove first iso involves showing maps descend

The machine that shows it’s well defined is used to prove first iso

[stark helm]

so because S->S/J is surjective mapping, so I think we can have the ring homomorphism between R and S/J, while the kernel is I, then we can use first iso to argue that R/I is isomorphic to S/J?

[boreal inlet]

Yeah so technically we are not doing anything new

[next obsidian]

Yes

[boreal inlet]

Yes ^^

[hollow mica]

is it true if I have a group G, normal subgroup H, and a subgroup K such that {k + H | k ∈ K} forms a partition for G (i.e. the elements of K together form representatives for the cosets in G/H), then G/H is isomorphic to K ?

I'm pretty sure that k -> k + H is an isomorphism here?

[stark helm]

i am actually considering since we have isomorphism phi: R->S with phi(I)=J, can I just argue that there is an isomorphism from phi^-1: S->R with phi^-1(J)=I? And say S->R->R/I and argue that this map has ker=J, so S/J isomorphic to R/I?

[boreal inlet]

Surjectivity is not obvious. Though, I don't have a counterexample rn so it possibly is true

[stark helm]

I mean argue it by using phi^-1, but I want to confirm if it is essentially. the same?

[boreal inlet]

Like for example let's pick an element from G/H. It should be of form g + H, where g \in G. Without knowing what's exactly is there in K i don't know if we can prove surjectivity

[hollow mica]

Given a coset a + H, we know that a will be in some set k + H (since sets of this form are assumed to partition G), and ofc a ∈ k + H implies that a + H = k + H, so k -> k + H = a + H

[boreal inlet]

Ohh right right my bad

[crystal vale]

Is there any relation between H and K what if K is a subset of H

[boreal inlet]

Set of cosets is et of equivalence classes

[crystal vale]

Yes

I don't understand how this makes partition of G ?

[boreal inlet]

The fact that K generates all partitions is a proof K cannot be a subset of H

We assume it's the set of all partitions using the equivalence relation

[hollow mica]

This is part of the hypothesis to my statement

[crystal vale]

Okay

[boreal inlet]

So are we like trying to prove the uniqueness of the group structure of G/H

Also one I thing I'm not really sold about is, what is the group that K is a subgroup of

I don't think it's a subgroup of G

It simply cannot be unless like H is trivial

Simply stating K is a group is enough

[hollow mica]

IDK I just formulated that statement after observing that R[x]/(x) = {r + (x) | r ∈ R} so I wanted to conclude right after that R[x]/(x) = R

this is for rings but it's like the same thing

[crystal vale]

Then how do different groups operate?

[boreal inlet]

Yeah but the addition of R[x] is quite different than R

[crystal vale]

Yes

[hollow mica]

what hwats wrong with my original statement again

[boreal inlet]

"Subgroup K"

Just say it's a group with some operation (it'll be the coset operation after you show its isomorphic)

Because like

For the homomorphism part you need it

Bijection you have shown

[hollow mica]

If K isn't a subgroup how do I make sense of k + H for k ∈ K ?

[crystal vale]

I don't understand if it has different operation then how { k + H | k belongs to K } makes partition of G ?

[boreal inlet]

Let's say K is indeed a subgroup of G

Then G/H is a subgroup of G right, at least upto isomorphism

Is this true always

[crystal vale]

Yes, I think so if K is a subgroup of G then maybe it makes sense

[boreal inlet]

Let's look at the isomorphism R/Z \cong S^1

[hollow mica]

I think if I add the original requirement that each k + H is distinct then I'm good

like k1 + H = k2 + H implies k1 = k2

that was sorta implicit in my original formulation

[boreal inlet]

I don't see how S^1 (or if you want, z \in C such that |z| = 1) can be embedded in R

[crystal vale]

How?

[boreal inlet]

we started with a subgroup K of G and then we show G/H is isomorphic to K

[hollow mica]

Ohhhh I see

lemme think

[boreal inlet]

The thing you're saying is fine because everywhere wjat you're doing in the ring context is addition, even though the additions are different, some amount of properties are preserved because universal properties are there

So yeah it does look like the exact same operation

But it's just not

[crystal vale]

Oh, I thought in general how G/H is a subgroup of G

[boreal inlet]

k + H is the just the equivalence class related to the operation of G

Doesn't mean this + is same as the other plus

[hollow mica]

Mmmm

[crystal vale]

But if K has a different structure than elements of G then how k+ H works?

[boreal inlet]

it's just the set of all elements y in G such that there exists an element h_1 in H with the property y - h_1 = k

[hollow mica]

In my R[x]/(x) = R case, I had G = R[x], H = (x), and K = R, but this was something special cuz R embeds naturally in R[x], so k + H where k ∈ K makes sense

[boreal inlet]

That's what it is

[boreal inlet]

Correct

[crystal vale]

So K must be a subgroup of G ?

[boreal inlet]

It doesn't need to be at all

If it is

Then nice

If it is not, property still holds

It definitely doesn't need to be a requirement

[crystal vale]

But then operation is different so how does it operate?

[boreal inlet]

As long as K is a group with its own operation

[boreal inlet]

It depends on K?

Like the case I gave, unit circle operates on complex multiplication

[crystal vale]

What if G is a group with multiplication and K is a group under composition?

[boreal inlet]

It can be

The isomorphism only requires the operations preserving some structure between them

[hollow mica]

I think Notknow's problem, and lowkey mine too, is how do I make sense of k + H when k is not in G

cuz you're saying you can drop the requirement that K is a subgroup of G

[crystal vale]

Then what do they look like { k + H | k belongs to K} ?

[boreal inlet]

k + H is nothing but a notation

[hollow mica]

LMAO if k + H is a formal symbol then of course {k + H} is iso to K

[boreal inlet]

If you name it like say $ it doesn't change its properties

[hollow mica]

but if k + H is a formal symbol how do I make sense of the assumption that {k + H} partitions G

[crystal vale]

Yeah notation but there is operation between elements of K and H so what will be the operation here?

[boreal inlet]

Any Equivalence relation always generates a family of sets which satisfies every property of a partition.

[hollow mica]

I know that but im still lost

[crystal vale]

Then I need to define the first $ operation in how they operate with k and H

[boreal inlet]

Pick two equivalence classes let's say C1 and C2.

C1 is the set of all such element g of G such that for each such g there exists an element h_1 in H with the property g * (h_1)^-1 = c_1

Similarly define C2.

This definitions come from the equivalence relation (x is related to y iff x(y)^-1 belongs to H)

This is how the cosets are constructed anyways

Now the operation of C1 and C2 is defined as usual

That is, C1 ∆ C2 = set of all such elements g of G such that for each such g there exists element h in H with the property g * (h)^-1 = c_1 * c_2

Now the set of all equivalence classes is a group under ∆

[hollow mica]

Are you defining what k + H means for k ∈ K ?

[boreal inlet]

(* is the operation of G)

Yes, while I say a general group G, not something with the addition you guys are mistaking it for

[hollow mica]

and C1 and C2 are two cosets in G/K ?

[boreal inlet]

Yes

Which have their own operation ∆

[hollow mica]

I don't get why you need two cosets

[boreal inlet]

....for a group you need binary operation.

[hollow mica]

• should accept two inputs, an element of K and a coset

oh

[boreal inlet]

Of course you need two cosets to describe an operation.

[hollow mica]

you're defining a group structure on G/H then?

[boreal inlet]

Yeah.

This is true for any general group G and a normal subgroup H

[hollow mica]

that's just the normal quotient group yea

so now what's the operation + that takes in an element of K and a coset from G/H ?

[boreal inlet]

goodness how i blame the common algebra books for making everything as + now

[hollow mica]

no I just thought you were defining k + H

[boreal inlet]

I am defining k + H yes

In this context it's an equivalence relation

It's not the element addition to a set always

It definitely can be addition, sometimes (like in cases of vector spaces i think), but not always

[hollow mica]

Ok so I know what the quotient group structure is on G/H , but I still don't understand your definition of k + H for k ∈ K, where K is an arbitrary group

[crystal vale]

No I know + is just notation

[boreal inlet]

I don't mean to say this in a bad way but I think I genuinely can't convey what I want to mean for my lack of presentation

If someone else understands please help

[crystal vale]

So always can I find an operation between K and subgroup H which makes sense

[crystal vale]

Is it something like that K acts on set of all subsets of G by +

[minor wraith]

what does a priori mean in a math proofs context?

and a posteriori

[boreal inlet]

This is probably okay to say, I'll have to check later and tell

[cobalt heath]

What is K and G here?

[cobalt heath]

Sorry, I don’t know without the context.

[vivid tiger]

a fortiori.

a priori as in "beforehand"

or "on priors"

"a posteriori" I've never used.

[cobalt heath]

Wish I knew English better.. anyway there you go, kakaka

[vivid tiger]

it's latin

[boreal inlet]

G is a group and K is a subset of G such that {k * H | k \in K} forms a left coset space of G. If K is a group, it is identically the quotient group G/H, as K can only be a group if H is normal

[cobalt heath]

Hmm, so + could be a confusing notation here

[vivid tiger]

a fortiori means "as a special case of something stronger".

[summer path]

You could also just use Google to find some examples

I mean I think it's clear in context

[boreal inlet]

It's not could be, it's precisely the problem

[crystal vale]

So K is a subset of G, right ?

Then it makes sense

[tough raven]

Since no-one seems to have mentioned semidirect products:
I assume you mean that every left H-coset contains exactly one element of K.
Then yes, the quotient map restricted to K is an isomorphism. Even more is true: such a K exists iff G is a semidirect product of H and K. (And no, this is not always possible: the standard (and smallest) example is G = Z/4Z, H = Z/2Z.)

[chilly radish]

Le extension problem!

[tough raven]

Indeed.

[boreal inlet]

Wait, how do you construct the semidirect product here?

[tough raven]

If H is a normal subgroup of G and K is a subgroup with the mentioned property, then the construction “has been done for you”.

[dull ginkgo]

Is semi direct product a way to try to generalize the notion of a split exact sequence from abelian groups to all groups

[delicate orchid]

it's a useful construction by itself, but yes that is one of the consequences

[dull ginkgo]

The only mention of it in Jacobson is a single problem and then one about wreath products

[tough raven]

Specifically, let K act on H by conjugation (since H is normal); call this action σ.
Then if we map H × K to G by (h, k) -> hk, we have
h1 k1 h2 k2 = h1 k1h2k1^{-1} k1 k2
= (h1 σ(k1)(h2)) (k1 k2).
So the map is a group homomorphism from the semidirect product H \rtimes K to G.

This much is true for any normal subgroup H and subgroup K. The assumed property of K is precisely that this map is bijective (and hence a group isomorphism).

[stark helm]

for the correspondence theorem for rings, i am confused by J->J/I can be injection map? Because it seems that if we have a map between a ring to a ring, then map is injection iff ker={0}. But J->J/I given a->a+I for a in J, if a is in I, then the kernel will have elements a not 0, can someone help me figure out where I made mistakes?

[viscid pewter]

essentially, the ideals in R/I are bijective with the ideals of R that contain I

with the same structure

for example, the ideals in Z_12 that contain <6> are in bijection with the ideals in Z_12/<6>

the key is that I < J < R

[delicate orchid]

the canonical map induces a bijection between the sets {J is an ideal of R containing I} and {ideals of R/I}

it says nothing about a bijection between J/I and J

[rocky cloak]

I wouldn't say that it generalizes the concept from abelian groups, but semidirect products do correspond to split exact sequences.

Like if you have an exact sequence

1 -> N -> G -> G/N -> 1

Then G is a semidirect product iff G -> G/N splits.

And then worth noting that the splitting lemma doesn't hold for nonabelian groups, so just because G->G/N splits does not mean N -> G splits. In fact N -> G splits iff G is the direct product.

[stark helm]

I am wondering what will the ideals in Z12/(6) look like?

[delicate orchid]

so lets try and use the corrispondence theorem to see

ideals of Z12 that contain (6) are (6), (3), (2), and (1)
so mapping these through to Z12/(6) we get the ideals {0+(6),6+(6)}, {0+(6),3+(6),6+(6),9+(6)}, {0+(6),2+(6), ..., 10+(6)}, {0+(6), 1+(6), ..., 12+(6)}
which simplifes down to just {0+(6)}, {0+(6), 3+(6)}, {0+(6), 2+(6), 4+(6)}, and the whole ring. And these are all of the ideals of Z12/(6)

alternatively one can use the isomorphism between Z12/(6) and Z6 and then these ideals are (0), (3), (2), and (1)

[crystal vale]

In the ring, the right inverse does not imply the left inverse, right?

[delicate orchid]

they can be different if your ring is non-commutative

in fact the existence of one does not imply the existence of the other

[crystal vale]

Yes I want an example, I first thought about matrix but I think in matrices it will work that if the left inverse exists then the right inverse also exists, right?

[delicate orchid]

take the group of sequences on Z, then consider the endomorphism sending a sequence a_n -> b_n with b_0 = 0, b_n = a_{n-1}. This only has a left inverse, sending a_n -> b_n with b_n = a_{n+1}

[delicate orchid]

yeah, although my example can be quickly changed around to give you a linear operator on an infinite dimensional vector space that is only invertible on one side

[crystal vale]

What is the operation here ?

[delicate orchid]

ok fine, change it to Z

[crystal vale]

You take a set of all endomorphisms to be a ring?

[delicate orchid]

the set of endomorphisms of any abelian group is always a ring

[crystal vale]

Yes

[delicate orchid]

then why are you asking

[crystal vale]

I just want to confirm that

[rocky cloak]

If they exist they are equal. But one can exist without the other.

[delicate orchid]

really? huh

are you positive?

[dire siren]

~~Ring of functions from R (real numbers) to R, with addition and composition
f has a right inverse <=> f is surjective
f has a left inverse <=> f is injective

[or change the domain and codomain to positive integers to avoid the axiom of choice]~~

[delicate orchid]

ah no I see it now

xy = 1, zx = 1 => z = zxy = y

[rocky cloak]

Composition doesn't distribute over addition here sadly 😢

[stark helm]

we know there is also a ring map phi': R->S/J and we can get there is an isomorphism phi^bar: R/I->im(phi'). But we want to prove R/I and S/J is isomrophism, so we have to prove image(phi')=S/J? I am somewhat confused about how to argue the image(phi')?

[barren sierra]

Can you be more explicit about how you want to use this map phi'?

First of all what is the definition of phi'? Second of all what theorem are you trying to apply with phi'?

[stark helm]

second: I want to use first iso thm. first: since I have the ring map phi:R->S, and S->S/J is natural surjective ring map, so I get phi':R->S/J

[barren sierra]

Can u be explicit as to what this "natural surjective ring map" is

[stark helm]

surjective ring homomorphism

[barren sierra]

And what the conditions of the first isomorphism theorem are

[barren sierra]

Please give me an actual definition

I give you x in R, what is phi'(x)?

[stark helm]

phi'(x)=phi(x)+J? Or do I still need to add something?

[barren sierra]

That's correct

[barren sierra]

And you want to apply this theorem. I believe that you know this map is a surjective homomorphism but what other part of the first isomorphism theorem do you need to satisfy to apply the result?

[stark helm]

so I have phi':R->S/J given by x->phi(x)+J, and ker(phi')=I, therefore I apply first iso to get phi(bar):R/I->im(phi') given by x+I->phi(r)+J

[barren sierra]

You have to prove maybe why ker(phi') is I (very easy, one-two sentences but actually write that out instead of just declaring it)

Otherwise yea seems good to me

See how actually making things explicit and writing out definitions / theorems helped?

[stark helm]

I am actually confused about the image {phi(r)+J}=S/J?

[barren sierra]

?

[stark helm]

I mean because it writes isomorphism R/I->S/J

[stark helm]

but we only prove isomorphism: R/I to image(phi')

[barren sierra]

What is the domain and codomain of phi'

[stark helm]

domain: all x in R, codomain: phi(x)+J? or domain:R, codomain:S/J?

[barren sierra]

What is the domain and codomain of phi?

[crystal vale]

Can there exist a distinct right inverse? Like here in Z_8 2•1= 2 and 2•5 = 2 so there is particular two distinct elements which give me same element 2 so is there any example where distinct right inverse exist ?

Is the right inverse unique? Or if the left inverse exists then it will be unique.

[barren sierra]

What map did you compose with phi to get phi'?

[rocky cloak]

If an element has both left and right inverse, then any left/right inverse is a two sided inverse, and two sided inverse are unique.

If you don't have a left inverse you may have several right inverses

[stark helm]

map: R->S given by x->phi(x) and map: S->S/J given by phi(x)->phi(x)+J, and get phi':R->S/J given by x->phi(x)+J

[crystal vale]

Yes, so is there any example?

[barren sierra]

Ok so phi' is a map from R -> S/J?

What is the domain and codomain of the map you got from the first isomorphism theorem?

[stark helm]

I think it should be x and phi(x)+J respectively?

[barren sierra]

well x is an element

Not a set

And phi(x) + J is an element of S/J

Not S/J itself

So that can't possibly be right

[crystal vale]

Is the set of all endomorphisms is the ring? Because I think they defined multiplication operation as composition

[rocky cloak]

So Wew already gave you an example.
Take the set of all sequences of integers. And let R be the ring of all endomorphisms with composition as multiplication.

Consider the map that sends
(a1, a2, a3, ...)
to
(a2, a3, a4, ...)

Then the map sending (b1, b2, ...) to
(0, b1, b2, ...)
is a right inverse while the one sending it to
(b1, b1, b2, b3, ...)
is another. There are infinitely many in fact.

[stark helm]

Then based on the first iso thm, we get R/I isomorphic to im(phi'), is it correct?

[barren sierra]

But what is im(phi')...

That's the whole crux of the question

[stark helm]

This is the stuffs that I feel confused now

[stark helm]

I only know that this set contains elements phi(x)+J

[barren sierra]

What did you write here? Please read what you wrote here

[stark helm]

so is it adequate to demonstrate S/J=im(phi'), only because codomain of phi' is S/J?

I mean codomain is equivalent to image?

[barren sierra]

You have that + phi' is surjective

so you know phi' is a surjective map from R to S/J

[crystal vale]

So here composition holds distribute property?

[barren sierra]

such that x -> phi(x) + J

So then the map you get from first isomorphism theorem has what domain, codomain, and how are elements mapped?

[rocky cloak]

Yeah, homomorphisms preserve addition.

[crystal vale]

Yes

[crystal vale]

Got it, thank you

[stark helm]

first of all, I think because phi' is surjective map, so the whole image should be S/J. But I still have one question. do you mean because we have S->S/J natural surjection, so R->S/J surjection?

[barren sierra]

phi is an isomorphism remember?

[crystal vale]

How can I prove that Q is the smallest subfield of R?

[dull marsh]

Show that Q is a subfield and that every subfield of R contains it

[dull ginkgo]

Q is the smallest subfield in general of char 0

[crystal vale]

Like I can generate Z then take inverses of elements of Z then it makes Q? Like fraction field of integers?

[dull ginkgo]

Do you want a clue to the way I understand it?

Yeah

Any of the prime fields are generated by 1

[crystal vale]

Yes

[crystal vale]

Yes, but what does it mean generated by 1

[dull ginkgo]

Hmmm

[dull ginkgo]

You can construct it by induction within any field basically

[crystal vale]

Induction?

[dull ginkgo]

Recursion

[dull marsh]

Just think of all the elements you can get starting from 1 with the use of the field operations (addition, subtraction, multiplication and division)

[dull ginkgo]

Brain fart lo

but yeah

That’s the prime field, either it’s Q or Z/pZ depending on the characteristic, I.e if 1 + 1 + 1… ever equals 0

You can always map N into your field by adding 1 to itself over and over again

If it is injective, then you have characteristic 0, and you can then append the negatives to embed Z in your field

[crystal vale]

Yes

[dull ginkgo]

Because you have a field, you can then map Q into it injectively via the inverses and quotients

That means that every field of characteristic 0 either contains or is Q

I.e Q is the “smallest” field of characteristic 0, which includes R, Q, your p-adic fields, etc

[crystal vale]

Okay thank you

[dull ginkgo]

For characteristics p, you can show that some 1 + 1 … 1 = 0 for a prime p, then you can inject Z/pZ into the field which is itself a field

[dull marsh]

Isn't that the definition of characteristic p though?

That p * 1 = 0

[crystal vale]

Yes Z-> F by n-> n•1 and kernel is pZ

[dire siren]

yes, I was silly
but I guess that can be fixed if instead of regular functions we consider the endomorphisms of (R,+)

and I probably need to use Hamel basis to get a counterexample

[stark helm]

I see, I ignored this one.

[dire siren]

the example with sequences can be incorporated in this: we take a Hamel basis B and we extract a countable subset, say A={b_1, b_2, ...}
we let f(x)=x for x in B-A and f(b_1)=b_2, f(b_2)=b_3, ...

hopefully this works; f is injective, but not surjective; so it has right inverse, but not left inverse

[tough raven]

Personally, I consider them to be generated by the empty set as well.

[dull ginkgo]

banned

pean in the ass

[tough raven]

Not really.

[dull ginkgo]

it was an afront to god (a joke on the name peano axioms)

[vivid tiger]

ℤ in any subfield via 1+1+1.... from this, ℚ

[tough raven]

Oh, that was intentional! Not bad.

[stark helm]

i am confused about why we can map from Z to F( I am considering if this map is adequate enough to argue this proof)? I mean if I change Z into other domain R and do the following steps in this picture, what will happen? it seems to me that it will not have an obvious result, but we can confirm that the intersection of all subfield of F must be embedded in the image X, so the map from Z to the image(X) will show the result of all intersection of subfield of F that is isomorphic to Q and Zp. Do you think this is the reason why this proof is adequate?

[rocky cloak]

What do you mean by change from Z to another domain? Like what would be the map you considered?

[stark helm]

I mean define x: R->F, R is a specific domain that I imagine, I just want to know if this argument is adequate enough?(Why picking domain Z and the result of its image isomorphic to Zp or Q can prove this statement)

Is it just because the intersection of all fields should be embedded in image X, so the proof showing image X isomorphic to either Q or Zp is adequate?

[stark helm]

and the possibiliity of mapping from a domain R(not Z) to F and its image will be different ( not isomorphic to Zp or Q)

[rocky cloak]

Well the map from Z makes sense no matter what the field is, so the image is contained in all subfields.

If R is just any domain then there's no guarantee that you can construct a map at all.

[dull ginkgo]

Hur dur recursion theorem gives a map from N

[stark helm]

so if there exist a map that we can construct, then it is valid right?

if the map helps us get the result of imX isomorphic to something

[rocky cloak]

I mean, vacuously that's true I guess.

[vivid tiger]

Z via 1+1+1+1...

if char = p then Zp else Q

(cyclic and countable, so these are your only options)

[south patrol]

p adics

[dull ginkgo]

,rotate

[dull ginkgo]

I think this is just explicitly wrong

I’ve done it thrice and got incorrect answers

[cobalt heath]

Which part are you getting incorrect answers here?

[dull ginkgo]

all of it

The verification of group action

[cobalt heath]

Hm

[dull ginkgo]

I tried looking up a solution but the solution posed the definition entirely differently

Even the def of a semi direct product differently

[cobalt heath]

Is this semidirect product?

Thats different notation than I've seen

Yeah it does seem to be using different conventions mostly

[dull ginkgo]

Wreath

[cobalt heath]

Ah, wreath product

[dull ginkgo]

Wait

L

[cobalt heath]

How is it defined, may I ask?

[dull ginkgo]

,rotate

[cobalt heath]

Ahhh

[dull ginkgo]

But if you try 11 as it’s posed

The composition fails

[cobalt heath]

Hmm

[dull ginkgo]

If you try it you’ll see

Idk I still think it’s wrong

[cobalt heath]

Hmm, let me see

[rocky cloak]

Yeah, something doesn't seem right here. Seems like it should be

(f, h) (t, s) = (f(hs)t, hs)

for it to work out.

[cobalt heath]

So (f1, h1) (f2, h2) = (f1 (h1 * f2), h1 h2) part is simply from definition of semidirect product

Then,

Ah, now I see where the problem is, h is somehow omitted in left part of the action

[dull ginkgo]

Yeah

[rocky cloak]

Just from the fact that
(f, h) = (f, 1)(1, h)
It should be clear that how f acts has to depend on h in some way

[dull ginkgo]

f(hs) or f(h^-1 s)?

[rocky cloak]

I think f(hs) is the thing that makes it work, but you should check both

[cobalt heath]

What is the textbook? Might be in the errata

[dull ginkgo]

Jacobson

Basic algebra I

[cobalt heath]

Huh, seems like there is no errata..

[dull ginkgo]

hausdorffT1

Wrong channel homie

[long obsidian]

Say f:A->B is a ring hom.

Say p is a prime ideal of B. Is it possible that im f intersected with p is non empty and im f intersected with p complement is nonempty?

[dull ginkgo]

If gA = B is also in the partition set, does that imply each partition set is bijective to the next

Like a coset

[crystal turtle]

Yes. The ideal P = <2,x> is prime (maximal) in Z[x]. Let f : Z --> Z[x] be the inclusion at constant polynomials. Then P satisfies what you want.

Most exampes will satisfy this.

[rocky cloak]

It's actually impossible for that not to be the case. As the image will contain 1 which is not in p, and it will contain 0 which is in p.

[dull ginkgo]

I can’t tell if each set in the partition is bijective or not lmao

[crystal turtle]

Oh I think I read it backwards as im f complement intersect p; that can be empty (say if im f = B) but yeah what jagr said

[dull ginkgo]

Nvm it’s only bijective if it’s transitive on the partition

[toxic zephyr]

so in a PID, gcds exist and are exactly the generator of the ideal generated by the elements?

that is consistent with Bezouts lemma, that
gcd(a,b)=d=ax+by for some integers x,y and d is the least positive integer that is an integer linear combination of a and b.

so is this how we find the generators of an ideal generated by multiple elements: just find the gcd?

[ivory trail]

https://en.wikipedia.org/wiki/GCD_domain look at the chain on this page

[tough raven]

I just realised that G = G×G/D, where D = {(g,g)} is the diagonal subgroup and we map (g,h) to gh^{-1} to define a map from G×G to G, and a (G×G)-invariant measure on G is the same as a two-sided Haar measure on G.
So this should yield a criterion for unimodularity (possibly the same as the usual one).

[crystal vale]

In the formal series R[x] where R is an arbitrary integral ring with identity element 1. It states that a_0 + a_1 x + a_2 x^2 +....... is a unit in R[x] if and only if a_0 is a unit in R.

I think this statement is wrong, right? Because it states that if a_0 is a unit in R then a_0 + a_1 x +...... is a unit in R[x] so if I take R is a set of real numbers then 1 +x + x^2 +3 x^3 +...... how it is unit in R[x] ?

[next obsidian]

The set of power series is denied R[[x]]

[next obsidian]

This has inverse 1-x

This is the geometric series

[crystal vale]

I wrote 1+x + x^2 + 3 x^3 +........

[next obsidian]

Okay?

Its inverse is 1-x

Proving that it’s a unit

[crystal vale]

How it's inverse is 1-x ?

[pliant forge]

3x^3

[dire siren]

just calculate (1-x)(1+x+x^2+...), which is equal to (1+x+x^2+...)(1-x)

[pliant forge]

Let R be a finite dimensional k-algebra and let M be finite dimensional R-module Let $$rad(M,N) = {g \in Hom_R(M,N) : hg \in rad(End_R(M)) \forall h \in Hom_R(N,M)}$$ where $rad(End_R(M))$ is the radical of the ring $End_R(M)$. I am trying to show that $g \in rad(M,N) \iff hg \text{ nilpotent } \forall h \in Hom_R(N, M)$. I understand that when g is in $rad(M,N)$, then hg is in the radical of the endo, which is nilpotent by the fact that the endo is artinian, but I cant figure out why the other implication works. Is it immediate that the radical is the set of all nilpotent elements (non-invertible maps) in the endo?

[cloud walrus]

Eso

[crystal vale]

?

[dire siren]

if the result is 1, this means that 1+x+x^2+... is the inverse of 1-x, right?

[delicate orchid]

as chmonkey said, write out the geometric series of 1/(1-x) and you'll see that it's 1+x+x^2+...
hence (1-x)(1+x+x^2+...) = 1

[crystal vale]

But my series is not 1+ x + x^2 +x^3+.......

My series is 1 + x+ x^2 + 3 x^3+.....

[dire siren]

what's the pattern then?

[crystal vale]

Got it, thank you ❤️

[crystal vale]

It's random polynomial

[delicate orchid]

ah ok so you can't be looking for an explicit inverse

[crystal vale]

Yes

[delicate orchid]

good LUCK

[dull ginkgo]

You’d have to construct it likely via recursion and that isn’t fun

[dull ginkgo]

In general, a unit in a ring is an element that lies outside of all proper ideals

We have a map from R[[X]] -> R injective sending a series to its constant coeff.

Perchance consider that implication

[delicate orchid]

surjective

[dull ginkgo]

Yeah sorry lol

I originally wrote epi but I changed it to the wrong word

[delicate orchid]

and this is true for polynomial rings as well, so you still need to use something specific about R[[x]] to show what the maximal ideals are

[south patrol]

Epi for rings is diff to surj anyway ig

[delicate orchid]

yeah but no-one cares about

[south patrol]

Real...

[delicate orchid]

uhh which one was it

Q -> Z?

[south patrol]

Well like all localisations

[dull ginkgo]

Z -> Q is binorphic

Biomorphic

[delicate orchid]

that's the one

[south patrol]

I think it's important to bw precise for that reason

[delicate orchid]

I just say surjective because that's almost always what I mean

all categories are concrete etc. etc.

[dull ginkgo]

I think poly ring unitality requires all other coeffs to be nikpotent

[south patrol]

I also would say surj lol

[south patrol]

Ye

Being a unit implies ur a unit in R/p[x] for all m

[delicate orchid]

it does

[pliant forge]

radical as in the jacobson radical*, intersection of all maximal ideals etc.

[south patrol]

Then you know units in poly ring over a domain are constants

Gg

[delicate orchid]

that proof is lame cause I don't understand it. Simply raise the mf to a high enough power

you'll end up with just some constant term, which is a unit by assumption (if I'm following the conversation properly)

[south patrol]

I don't think that works right

like if b is nilpotent then for all n (a+b)^n will still have terms like a^(n-1)b

Idk

[south patrol]

I don't understand this

You may be doing the wrong direction

[delicate orchid]

no that's a fair point

I think I am yeah

[south patrol]

But dw

[delicate orchid]

oh well I literally never have to care about this in real life

[dull ginkgo]

So all that’s needed for the power series case is just the constant is unital?

[delicate orchid]

yur

[crystal vale]

If I want to prove that Laurent series are an integral domain under an integral ring.

Let two non zero Laurent series such that p= a_n X^n + a_n+1 X^(n + 1) +...... where a_n ≠ 0 and another polynomial q= b_m X^m +....... where b_m ≠0 then pq has coefficient a_0× b_0 x^(n+m) and a_0 × b_0≠0 so pq≠0.

I think the same we can prove for Laurent polynomials.

[delicate orchid]

(p, x) is maximal in R[[x]]

with (p) maximal in R

[delicate orchid]

yeah this works, although I presume you mean a_nb_m instead of a_0b_0 at the end?

[crystal vale]

Yes

[delicate orchid]

I saw the sticker

[crystal vale]

😂

How can I prove that a Laurent series a_nX^n +..... with a_n ≠ 0 is a unit in R<X> if and only if a_n is a unit in R ?

[dull ginkgo]

I still wish to at some point prove Hilbert basis theorem without using ultrafilter lemma or leading coefficient mappings

I want to know how the structure of R[X]’s ideals depends on R’s

I know there’s like a “laying over” property on R[X] for primes but

In order to use that for Hilbert basis you need ultrafilter or an equivalent to show every ideal is in a prime ideal

My idea was to assume R is noetherian, and we have a strictly increasing sequence of ideals J_n

J_n cannot be strictly contained in (X), as R[X]/(X) is iso to R violating the noetherian condition

So eventually J_n must intersect R nontrivially

However I don’t think I can do much wirh this. The intersection of J_n, J_m (n < m) with R need not preserve strict inclusion

Let $R$ be Noetherian. Let $K_n$ be a sequence chain of ideals in $R[X]$. The auxiliary sequence of ideals $A_n = K_n \cap R$ must stabilize.

[cloud walrus]

Mizalign #1 simp

[dull ginkgo]

$R[X] = R + (X), R \cap (X) = (0)$

[cloud walrus]

Mizalign #1 simp

[dull ginkgo]

Makes me wonder, assuming A_n must stabilize after a certain point…

I need to find a way to show $B_n = K_n \cap (X)$ does too

[cloud walrus]

Mizalign #1 simp

[dull ginkgo]

Key thing of note is that if $B_n = (X)$ after a certain $n$ then it must stabilize because the ideals can be mapped to $R$ via $R[X]/(X)$

[cloud walrus]

Mizalign #1 simp

[dull ginkgo]

So the remaining case is $K_n \cap (X) \subset (X)$

[cloud walrus]

Mizalign #1 simp

[crystal vale]

A positive integer is square free if it has form n=p_1× p_2×.....× p_k and all these p_i are prime and are different for all i=1,2,....k.

Right?

[dull ginkgo]

if $K \subset (X)$ then shouldn’t $\frac{K}{x}$ be an ideal?

[cloud walrus]

Mizalign #1 simp

[dull ginkgo]

Idk how to prove that off the top of my head

Or introduction it

Actually I don’t think it’s true

p(x)xH = xH forall p(x) doesn’t necessarily imply p(x)H = H

Though it does if aH = H

Let xH = K, K an ideal of R[x]
a(x)K = K
a(x)xH = xH
By cancellativity of x in R[X]?
a(x)H = H?

Another question: assume $A \subset B$ are rings, and we have a principal ideal $(b) \subset B$. If $A \cap (b) = (0)$ and $A + (b) = B$, then does the at imply $B \cong A[X]$?

[cloud walrus]

Mizalign #1 simp

[crystal vale]

So if n is square free then I want to prove that Z_n has no non-trivial nilpotent elements.

So if there is a nilpotent element in Z_n, say a≠0 and a<n that means a^k=0 mod n for some positive k>1. Thus n divides a^k.

So if n= p_1 × p_2 ×......× p_m. Then p_i divides a , thus p_1 × p_2 ×.....×p_m divides a means n divides a. But a<n so it is not possible.
Hence there are no non - trivial nilpotent elements in Z_n.

Is it correct?

[south patrol]

and it's easier to prove this than the corresponding result for the poly case

which is funny

[dull ginkgo]

Poly case is by induction

[south patrol]

Well you don't need induction if you use the proof I showed tbf

As in, one direction is easy (unit + nilpotent is always a unit)

and the other like, if $f \in R[x]$ is a unit, then for each prime ideal $\mathfrak p \subset R$, $f$ becomes a unit $\bar{f}$ in $(R/\mathfrak p)[x]$ and by the corresponding result for domains it follows that $\bar{f}$ is constant i.e. all non-constant coeffs of $f$ lie in $\mathfrak p$. But $\mathfrak p$ was arbitrary, so all non-constant coefficients are nilpotent

[cloud walrus]

Süßkartoffel

[south patrol]

and the result for domains is easy since then degree is a multiplicative thing

[hollow mica]

in the polynomial ring of a commutative ring with unit, a polynomial is a unit if and only if its leading coefficient is a unit and the rest of the coefficients are nilpotent

Does a similar result hold in the non-commutative case?

[delicate orchid]

Always funny when we get two questions in a row on the same topic from different people. Makes me think u guys are in the same class

[rocky cloak]

I suppose it depends how you define polynomial ring over a noncommutative ring

[delicate orchid]

It would have to be the free algebra right

[rocky cloak]

Well, how do you define 'algebra' over a noncommutative ring

Like if algebra just means ring with map from your base ring, then the free algebra looks pretty horrible.

[delicate orchid]

A ring S with a map R -> End(S)? Or are you talking about left/right actions

Yeah I didn’t say it would be nice we’re doing ring theory here nothing is nice

[rocky cloak]

Well, you want some compatibility between the ring structure and module structure, no?

[delicate orchid]

True

Although I think what the asker is after is just elements of the form a_0+a_1x+… with a_0 in R.

Product given by convolution - who knows if this is a ring or not

But if it isn’t that answers the question as well

[delicate orchid]

My intuition is telling me that associativity breaks

But maybe not

[rocky cloak]

That should be a ring. And then the constant coefficient would have to be a unit for the polynomial to be a unit. So then if you multiply by the inverse of the constant coefficients, you can look at things with constant coefficients 1, and then I think the proof from the commutative case more or less goes through.

Though just because b is nilpotent and a is a unit, I don't think you can conclude ab is nilpotent

[crystal vale]

I want to prove that in Z_n if n is the power of prime number. Then a set of non -units elements of Z_n is an additive subgroup of Z_n.

So if n= p^k, for some positive integer k.
Let S be a set of all non-units elements of Z_n. So 0 belongs to S.

And if a belong to S that means gcd( a, n) ≠1.
So let a and b belong to S so gcd(a, n)≠ 1 and gcd(b,n)≠ 1. So it follows that p divides a and b. Thus p also divides a-b so gcd(a-b, n) ≠ 1 so a-b belongs to S. Hence S is an additive subgroup of Z_n under addition.

Is it correct?

[south patrol]

yes

you have even found an explicit formula for the non-units

namely, subgroup generated by p

Since you know a bit of commutative algebra, another way to phrase this is that the ring Z/p^k has exactly one maximal ideal, namely pZ/p^k, e.g. by the correspondence theorem (since (p) is the only maximal ideal of Z containing (p^k))

and for local rings, the maximal ideal is precisely the set of non-units

Of course overkill here lol

but perhaps worth highlighting if you want to study non-units more generally since they will come up

[hollow mica]

what breaks down in the ring structure of R[x] when R is not commutative?

[rocky cloak]

Nothing breaks down, or I'm not sure what you mean

It's just clear what the definition should be

[hollow mica]

Oh I thought you were hinting that the repeating the standard construction of a polynomial ring in the non-comm case doesn’t work

[rocky cloak]

Well what is the 'standard' construction

The usual construction uses that R is commutative

[south patrol]

I think one thing is that usually you want like, a map R[x] -> S to be a map R -> S equipped with a choice of element of s

but then if S is non-commutative, that changes since s would need to commute with all of S

[hollow mica]

I’m asking where that is

[rocky cloak]

Like should
rx = xr or not?

Whether you think so depends on how you define the polynomial ring

[hollow mica]

Let R be a non comm ring. Define addition and multiplication of polynomials in R[x] as you usually would if R was comm. Is R[x] not a non-comm ring?

[rocky cloak]

You keep saying 'as you usually would'. So then it depends what you would usually do

[hollow mica]

like

[south patrol]

gotem

[delicate orchid]

jagr stop being a little nerd

[south patrol]

But no i mean you can view them as finitely supported sequeencesof elements with $(a_i) \cdot (b_j) = ( \sum_{i+j=n} a_i b_j)_{n \in \mathbf N}$

[cloud walrus]

Süßkartoffel

[delicate orchid]

Actually don’t we love you for it

[south patrol]

i think

[delicate orchid]

But I said all this earlier RAAAGGGGH

[south patrol]

and then just take x = (0,1,0,...)

[hollow mica]

the sum of p = (sum a_i x^i) and q = (sum b_i x^i) is (sum (a_i + b_i) x^i), and their product is (sum a_i b_j x^(i+j))

[south patrol]

and that gives you a construction with x in the centre

[rustic crown]

you just had to say "yes" to this

[south patrol]

lol

[hollow mica]

yes

I kind of wrote it poorly above but this is what I meant

i don’t see where we use the commutativity of the base ring, besides showing the polynomial ring is commutative

[south patrol]

you don't use it

[rocky cloak]

Yeah, so that's a ring.

And it should be clear that the constant coefficient has to be a unit for it to be a unit

[hollow mica]

so what’s wrong

[rocky cloak]

The other coefficients are more complicated though

[hollow mica]

you’re saying
p(X) is a unit implies that the constant term of p(X) is a unit and that the reverse direction is harder?

[south patrol]

reverse direction isn't true

[hollow mica]

Like formulating a sufficient condition is harder is what I mean

[crystal vale]

.

[hollow mica]

Mhm

Since there’s no evaluation universal property everything normal gets messed up

[rocky cloak]

So for example if R is the 2x2 matrix ring
n = [0, 0; 0, 1]
m = [1, 0; 0, 0]
a = [0, -1; 1, 0]

Then (a + nx) and (-a + mx) are inverses of each other, without m or n being nilpotent.

[dull ginkgo]

Sup homies

[rocky cloak]

I think maybe something like
a0 + a1x + a2x^2 + ...
is a unit iff
a0^-1 ai is nilpotent, but I haven't checked the details on that

[dull ginkgo]

[crystal vale]

But if it is a formal power series then it is true, right?

[crystal vale]

Okay thank you

[tough raven]

Yes?

[crystal vale]

No, sorry actually there is discussion about the use of commutative property in the construction of Ring R[x], so I remember that you answered this question so therefore I just mentioned it .

[tough raven]

I see.

[crystal vale]

But i want to prove it's reverse part , if a set of non -units elements of Z_n makes an additive subgroup of Z_n, then n is the power of a prime.

I assumed if n has two prime divisors then it will contradict but how ?

[rocky cloak]

Are you familiar with Chinese remainder theorem?

[south patrol]

ye

[south patrol]

real

to be fair you don't quite need Chinese remainder either

you use the fact that like ||union of incomparable subgroups is never a subgroup||

[crystal vale]

Yes

[dire siren]

is this even true?
In Z_12, the set {0,4,8} is a set of non-units, and it is an additive subgroup

[crystal vale]

Set of all non-units

{0,2,3,4,6,8,9,10}

[dire siren]

I see, but you said "a set" initially

[crystal vale]

My mistake

[crystal vale]

What is an incomparable subgroup?

[south patrol]

That's like asking what an isomorphic group is :)

I mean that neither is contained in the other

[south patrol]

Could also just do 0

[dire siren]

yeah, I noticed shortly after I wrote that

[crystal vale]

Okay

[crystal vale]

Then i think you want to say that the set of units in Z_n is an additive subgroup of Z_n, right?

[crystal vale]

If n has at least two distinct prime divisors

[south patrol]

No

It evidently isn't since 0 isn't in it

[crystal vale]

Then how do I use that fact here?

[south patrol]

If you're a nonunit then ur in pZ_n for some p|n

And the union of those is precisely the non units

[dire siren]

@crystal vale your idea of assuming that n has two distinct prime divisors works

[crystal vale]

It is true for all n here we did not use n has two distinct prime divisors?

[crystal vale]

How?

[dire siren]

say the primes are p and q
then p and q are non-units, and apply Bezout

[crystal vale]

Yes

[dire siren]

you get that 1 is in that subgroup, which is a contradiction

[crystal vale]

So here n=p_1 × p_2 or n haz at least prime divisors p_1 and p_2 ?

[dire siren]

at least two

[crystal vale]

Got it thank you

[south patrol]

Oh yeah that is easier lol

[crystal vale]

Actually I thought about what if p-q is a unit but it is not true in general

[crystal vale]

I want to understand your way

[crystal vale]

For its reverse if I let n is not square free then there is a prime p which divides n and power of p is more than 1 , say k, so I take a= n/p which is <n and then a^k=0 mod n. So it has a non - trivial nilpotent element.

Is it correct?

[tough raven]

BTW, I am now pretty sure that this (F was a division ring BTW) is false because it implies something I know to be false.
If anyone is interested, ping me and I'll (later) try to flesh out the contradiction into an explicit counterexample.

[rocky cloak]

So what does that tell you about Z/mn when m and n are relatively prime?

[tough raven]

(Rough sketch: it would imply linear independence of characters valued in division rings, but conjugations (which are characters of the group of units of the division ring) by linearly dependent (over the centre) elements are linearly dependent.)

[crystal vale]

Groups of units of Z/mn is isomorphic to the direct product of groups of units of Z/m and group of units of Z/n

[rocky cloak]

Yes, but not just the group of units

Z/mn is isomorphic to Z/n x Z/m

Do you see two nonunits in Z/n x Z/m that sum to a unit?

[frigid sparrow]

I have a question about separability over different fields, as far as i understand the polynomial X^p - t is separable over F_p(t) since there is an extension (its splitting field) where it only has p’th root of t as the single root and since it’s irreducible we say it’s not separable - (my definition: a polynomial is said to be separable if its irreducible factors don’t have multiple roots) however over the splitting field the polynomial already splits and has irreducible factors X- ^p √t and they obviously only have one root so the polynomial consists of separable factors and is separable..?
is the point here that we look at whether a polynomial has multiple roots not over which the field is defined but rather its extensions or even the algebraic closure..?

[rocky cloak]

I can't quite parse you're last question, but with the definition you're using it matters which field you consider the polynomial over, yes.

[marsh quarry]

i’ve seen people say that (a) is a “multiple” of (b) when (a) \subseteq (b).

i can imagine this terminology being motivated by ideals in the integers. but in general, is it true that if (a) \subseteq (b) then a = cb for some c \in (b)?

say we’re working in a commutative ring

[coral spindle]

(b) = {br | r in R} in a commutative ring.

So since a in (a)...

[marsh quarry]

ahaha ok i see

a = br for some r since a is in (b)

[random pasture]

Is this just:
H, {(1 3),(1 2 3)}, {(2 3),(1 3 2)}, {(1 2 3),(1 3 2)}, {(1 3 2),(1 2 3)}

[delicate orchid]

those last two are the same set

and they aren't disjoint

we know there are |G|/|H| = 6/2 = 3 cosets

[white wraith]

I think it's only the first 3

At least that's what I got when I worked it out, how did you get the last one?

[random pasture]

nah i fucked something up

you right

[crystal vale]

(0,1) and (1,0) ?

[coral steeple]

I am asked to prove that an isomorphism phi: F[x] -> F[x] which fixes the constants preserves irreducibility. Isn't enough that phi is an isomorphism? If f is irreducible and f=pq, then phi(f)=phi(p)phi(q). By definition one of p,q is a unit and one is not, and since phi is an iso, the same is true for phi(p), phi(q), ie phi(f) is irreducible.

[next obsidian]

Yeah

Meme problem

[coral steeple]

How odd

THanks

[old hollow]

If I have a field F and an automorphism φ of F[x], why is it true that φ(c) = c for all constants c?

[crystal vale]

If the field F has characteristic p then x^p=x for all x belongs to F?

[crystal vale]

How many automorphisms of F exist ?

If F is a finite field and has characteristic p then x-> x^p is an automorphism. But I am not sure where F is finite I used?

[crystal vale]

for onto, right?

[celest furnace]

Injectivity is clear. Onto is NOT clear and NOT true for infinite fields of characteristic p in general

However, injective maps from two sets of the same finite order are surjective

This is where it’s used

[rocky cloak]

If phi is an automorphism of F-algebras this is true by definition. If phi is just an automorphism of rings, then this is not true.

[crystal vale]

Yes, but how injective is clear? If x^p = y^p then (xy^(-1))^p =1 then how does it imply that xy^(-1) =1 ?

[celest furnace]

Once you’ve verified that x|->x^p is a homomorphism you can just work with kernel is 0

What are the solutions to x^p=0?

[crystal vale]

Yes got it

[crystal vale]

0 because the kernel can be only {0} or F

[celest furnace]

This is true for integral domains too

[crystal vale]

And if the kernel is F then 1^p=0 means 1=0 but I assumed that the characteristic of F is prime not 1.

[celest furnace]

A product of terms is 0 in an integral domain if one of the product terms is 0

[crystal vale]

Yes

[celest furnace]

That works too

[crystal vale]

And x^p = x is it true?

[celest furnace]

For F_p in particular yes but not for any other extension of it ( at least finite)

[crystal vale]

F_p notation mean? Z_p

And if I want an infinite field which has finite characteristics then if I take the fraction field of Z_2[x] then will it work ?

[chilly ocean]

Yes

Another example is the algebraic closure of F_2

[crystal vale]

Okay thank you

[old hollow]

Ok then let’s say phi fixes the constant polynomials, ie phi(a) = a for all constants a. Then what is phi(x)? Is there enough information to determine phi(x)?

[rocky cloak]

No, phi(x) can still be pretty much anything

[old hollow]

Really? Hm

[rocky cloak]

Or if you want it to be an automorphism it would have to be something of the form
ax + b

But that's still a lot of choices

[old hollow]

This is the problem I was given

[old hollow]

Oh yeah

Yeah how do I show it’s of the form ax + b

I’ve been stuck on this

I feel like it should be simple lol

[chilly ocean]

What happens if its not of that form?

[old hollow]

Well, if phi(x) = x^2, then i figured applying phi iteratively gives you higher and higher degrees

But I don’t see a contradiction…

I mean, I think the x —> ax+b part should be straightforward, no?

[rocky cloak]

if phi(x) = x^2 what is phi(f) for arbitrary f?

[old hollow]

f(x^2)

[rocky cloak]

Indeed

And if this is an automorphism, then there must be an f such that phi(f) = x

[old hollow]

I mean, could I not make some sort of argument about cosets? Like, the degree 1 polys map to degree 1 polys, and degree 2 maps to degree 2, etc? I don’t like this contradiction business

[old hollow]

Oh interesting

Hm

[crystal vale]

Actually the same question I asked a month ago , it shows that automorphisms preserve degree

Exact question

[cobalt heath]

Do you recall what is automorphism?

[old hollow]

Bijection, right

[cobalt heath]

You also need homomorphism, but right.

[old hollow]

Oh sht

[cobalt heath]

Ah, I guess this was already explained. But for this to be bijective, you need f where phi(f) = x.

Or alternatively, a homomorphism phi^(-1) shall exist

Which sends x to f, then phi sends f to x; I think you can derive from here than f is of degree 1 without contradiction

[old hollow]

How do you get it without contradiction

?

[cobalt heath]

Put phi(x) = g(x), phi^(-1) (x) = h(x)

Then, you have h(g(x)) = x

[dire siren]

to add to what was mentioned: x^p=x in a field can only happen for at most p elements; this is because the polynomial X^p-X must have at most p (its degree) roots

if the field has characteristic p, then 0,1,2,...,p-1 are the only elements for which x^p=x

[crystal vale]

Okay thank you

[crystal vale]

I want to prove that if R is a ring with 1 such that non - units in R form a subgroup of (R,+), then char(R) is either 0 or else a power of a prime.

If n has at least two distinct prime divisors say, p_1 and p_2.

Then p_1 is non-unit because if it is then char(R) will be less than n. Similarly p_2 is non -unit in R. But we know that there exists integers and t such that p_1 s + p_2 t=1 which implies that a set of all non -units is not a subgroup of (R, +).

Is it correct?

And if R has no unity then in R[x] what will be the coefficients of x or x just the element of R[x].

And I want to show that if R ≠ (0), the variable x is not a zero divisor in R[x].

If it is then there is non-zero polynomial f such that f × x =0 but since f is non -zero polynomial so there exist a_k x^k term in f(x) with a_k ≠ 0.
f × x has non-zero term a_k x^(k+1) so it can not be equal to 0.

Is it correct?

[crystal vale]

I want to discuss ring theory, does anyone want?

[rocky cloak]

Yes

[crystal vale]

Okay, variable x is not a zero divisor in R[x] proof is it correct?

[rocky cloak]

Looks good

[rocky cloak]

Discuss in what way?

[crystal vale]

I have doubts and want to solve that

[rocky cloak]

What kind of doubts?

[crystal vale]

1. If I define 0 as zero- divisor then if ab is zero divisor then is ba zero divisor?

2. Is there any specific form in M_n (Z) such that there are two zero divisors whose sum is not a zero divisor ?

3. Let R= M_n (Z) and N be the subset of all strictly upper triangular matrices then x^n =0 , for all x belongs to N.

4. If I take R= S_3 and define addition as S_3 operation and multiplication a•b = (1) then is it ring. If yes then what will be char(R).

and many more....

[rocky cloak]

1. No it's possible for ab to be a zero divisor and ba to be a unit for example.

2. Unsure what you mean by form, but if you take a diagonal matrix with 1s and 0s along the diagonal you can find to 0 divisors that sum to the identity.

3. Yes

4. Addition isn't even commutative, so not a ring.

[crystal vale]

Hint for 1.

I thought about it but I don't know how to show it. Maybe there are many calculations for 2.

I don't know how to prove 3.

I think I get it 2. Thanks

[rocky cloak]

1. I think you've seen an example of this when asking about one sided inverses earlier.

2. There are indeed lots of examples.

3. You can use linear algebra, think about the characteristic polynomial.

[crystal vale]

Okay, thank you

[crystal vale]

In the ring of endomorphisms of groups of sequence, additive identity is (a_1, a_2,.....) -> (0,0,0,....) , right?

[rocky cloak]

That's right

[crystal vale]

So if I take
a is (a_1, a_2, ....) -> (a_2, a_3, .....)
b is (b_1, b_2,.....) - > (0,b_1, b_2,.....)

Then ab is unity but ba is not unity and ba ≠ 0.

So if I take c is (c_1, c_2,.....) -> (1,0,0,.....)
Then bac =0, right ?

[rocky cloak]

Yeah, well c defined like that is not a homomorphism, but you basically have it

[dull ginkgo]

For $G$ a group, $G^{S}$ for any set $S$ has the structure of a group, where $f_1(s)f_2(s) = (f_1f_2)(s)$

[cloud walrus]

Mizalign #1 simp

[dull ginkgo]

if I can get this right, then $S = \mathbb{N}$ in this case?

[crystal vale]

Then what will be c ?

[cloud walrus]

Mizalign #1 simp

[crystal vale]

Is c (c_1, c_2,....) -> (c_1,0,0,....) work ?

[dull ginkgo]

Any non-unit in a ring must be contained in an proper ideal

Therefore your statement that non-units form a subgroup of (R,+) might be equivalent to the union of all ideals is an ideal, implying there is a single maximal ideal. i.e local

[south patrol]

lol this is how i proved it

[crystal vale]

Can you please give me example where R is a ring and S is a subring of R and both have distinct unity but they have the same char.

[south patrol]

Well to me a subring necessary has the same unit - isn't that the standard convention

[crystal vale]

No

{0,2,4} is subring of Z_6 and has unity 4

[south patrol]

okay this is just a different definition then

hm

[crystal vale]

Yes

[south patrol]

well you can do Z/2 as a subring (in your sense) of Z/2 x Z/2

as in like {(0,0), (1,0)}

[crystal vale]

Yes

[south patrol]

then they have distinct units but the same characteristic

[crystal vale]

But they have distinct unity right

Yes thank you

[south patrol]

np

[crystal vale]

And if S= Z × R, R is a ring.
If I define (m ,x)(n, y) = (mn, my + nx + xy).

R is a subring of S. There is a statement that any ring R can be realised as a subring of a ring S with unity.

If R has no unity then is this statement true?

[cloud solar]

All functions f:F4->F4* are polynomial?

[coral spindle]

All functions on finite fields are polynomial

You can construct them quite easily if you know the idea behind Lagrange interpolation

[delicate orchid]

oh yeah? watch this.... sin(2 mod 5)....

[coral spindle]

Yeah what is it nerd

Tell me the value

[delicate orchid]

good question

gimme 2 mins

chat I have divided by zero

it's 4+O(x^5)

[coral spindle]

So true

[cloud solar]

What about the degree of a polynomial P in F4[X] with P(x)=f(x) for every x in F4

I am trying to prove there is such polynomial P irreductible

[dire siren]

the degree doesn't have to be fixed

x^4=x, so the polynomial P(x^4) also works, for example

[crystal vale]

Let R= M_n (Z) and N be the subset of all strictly upper triangular matrices then I want to prove that N is non- commutative if n>=3.

So if I have A and B elements of N then let AB= C.

So C[1,3] = Summation A[1,j] × B[ j, 3] , j runs over 1 to n. And we know A[1,1] = 0 and B[ j,3 ] =0 for all j>=3. Hence C[1,3] = A[1,2 ] × B [ 3,2].

So we can construct a matrix where A[1,2] × B[ 3, 2 ] ≠ B[1,2 ] × A[ 3, 2].

Is it correct?

[dire siren]

yes

[vivid tiger]

okay, that's pretty neat

[sterile condor]

What’s a polynomial ring?

[delicate orchid]

A ring of polynomials

[vivid tiger]

well, usually it's the ring of polynomials over a ring.

[rustic crown]

well they didn't say how many variables >.<

[vivid tiger]

okay, it's a group monoid ring where the group is free abelian and written multiplicatively

[rustic crown]

monoid ring!!

.<

[delicate orchid]

that's R[x, x^-1]

[vivid tiger]

... whoops

[delicate orchid]

hate to be the ermm aktually-er but it is the truth...

[vivid tiger]

dude this is mathematics, the subject is founded on "ermm aktually"

[delicate orchid]

I'm trying to make your definition even more unwieldy

[vivid tiger]

it's less unwieldy.

because now it's correct.

[delicate orchid]

it's the category algebra R[(N, \leq)] I think

[vivid tiger]

now that's more unwieldy

congratulations

(now pls explain)

[delicate orchid]

or maybe we just take the nerve of X_1 -> X_2

my head hurts

no (N, \leq) should work

[delicate orchid]

(N, \leq) looks like 0 -> 1 -> 2 -> ... with a unique map in each Mor(x, y). So composing two arrows is like adding two numbers. The analog for "x" in "R[x]" is the morphism "0 -> 1" here

again - I think

no! it doesn't work! gaaah!

[vivid tiger]

okay but we want many vars

[delicate orchid]

we need a single object with one endomorphism for each natural number, and composition defined obviously

[vivid tiger]

that's easy to do though

just harder to type in ascii

[delicate orchid]

yeah, I just had to actually arrive at that construction

[vivid tiger]

it's just like with groups

[delicate orchid]

yes it's the delooping uhh

monoidoid...

lol

[vivid tiger]

...is there a "monoidoid"?

because there's a groupoid

[delicate orchid]

they're more commonly called "categories"

[vivid tiger]

...oh you literally thought the same thing

[delicate orchid]

a monad is a monoid in the monoidoid of endofunctors

if you want more examples of this the general process is called "horizontal categorification"

abelian categories are "ringoids" for instance

[chilly ocean]

A monad is a monoid in the monoidoid of monoidoid morphisms from a monoidoid to itself

[rotund aurora]

is this poetry or math

[chilly ocean]

This is croqueta3385

[rotund aurora]

🤘 😔

[south patrol]

the poly ring over R is just the ring corepresenting the forgetful functor RAlg -> Set

[delicate orchid]

u could just say "the free R-algebra" home dawg

[dusk whale]

out of curiosity is it common for a first course in group theory to be egregious

[delicate orchid]

in what way

[dusk whale]

idk if its the concepts themselves or the professor teaching it that's making it difficult

cus i heard people say algebra is easier than analysis but im finding it to be the opposite

[vivid tiger]

it would be strange to cover the egregious theorem

[dusk whale]

never heard of that

[rustic crown]

found this
https://grossack.site/2022/08/21/monoidal-monoidoidoids.html

[chilly ocean]

im a fan of chris grossack

[rustic crown]

(= (strict) 2-categories)

[delicate orchid]

oh goody! the worst thing I've ever seen

[next obsidian]

Lmfao

[vocal fern]

what does it mean by "induces an action in the usual way"?

[rotund aurora]

it's written in the screenshot

but how is this an action, it's not associative

[vocal fern]

wait but how does sigma define an action on Z^N^3? isn't it defined for N^3? sorry im very confused by actions

[rotund aurora]

it's just permuting the coordinates

like the permutation (1,2,3)-->(2,1,3) would translate to (x,y,z)-->(y,x,z), and so on

but well they are actually not doing that, I have no idea why they take sigma^{-1} instead of sigma

[vocal fern]

so when its saying like $\sigma \cdot p$, does it mean $(\sigma \cdot p)(x_0, x_1, x_2) = p(x_{\sigma^{-1}(0)}, x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)})$?

[cloud walrus]

board 2

[delicate orchid]

it's a right/left action thing

[delicate orchid]

yes that's the action on N^3, and you can think of Z^{N^3} as functions f(x1, x2, x3) -> Z with xi natural numbers, so you're just permuting the inputs of these functions

which is an action on Z^{N^3}

[vocal fern]

i see, thank you so much!

[random pasture]

Do i just prove identity and compatibility?

[delicate orchid]

have you tried setting (h,k) to the identity

[random pasture]

yeah i got it i think

setting h,k = 0

gives 0 + f(a) = f(a) for the identity

the other part i denoted (h1,k1) and (h2,k2)

and tried to show ((h1,k1)*((h2,k2)*f(a))) = (h1+h2,k1+k2)*f(a)

[delicate orchid]

(h1,k1).((h2, k2).f)(a) = (h1, k1).(h2+f(a-k2)) = h1+h2+f(a-k1-k2)
to make it even clearer:
let g(a) = h2+f(a-k2)
then (h1, k1).g(a) = h1+g(a-k1) = h1+h2+f(a-k1-k2)

I'll let you do the other part

[random pasture]

oh i see

i just expanded wrong i think

for part c of the same problem, how do i get those values from ((h,1).f(a)) = h+ f(-1+a)=f(a)?

is it because f(-1+a) = f(a)+k for some k in {0,1,2}?

[delicate orchid]

are they using 1 to mean 1 here or the identity

right nvm they wrote 0, so they mean 1

[random pasture]

yeah

[delicate orchid]

this is a bit more interesting

f : Z/3 -> Z/12
(h,1).f(a) = h+f(a-1) = f(a)
it's hard because f is just a set map

[delicate orchid]

could you explain how you derived this

[random pasture]

f is a function mapping from Z/12 to Z/3 right?

or did i read the notation wrong

[delicate orchid]

has to be the other way around otherwise how are we adding an element of Z/12 (h) to an element of Z/3 (f(-k+a))

actually it just tells us lol

[random pasture]

oh

wait so why is it 0,4,8

[delicate orchid]

I can see how they have a fixed point though

set f(a) = 4a,
(4,1).f(a) = 4+f(a-1) = 4+4a-4 = 4a = f(a)

but this works equally as well for f(a) = xa
then (x, 1).f(a) = x+f(a-1) = x+xa-x = f(a)

so I'm confused

[dull ginkgo]

high quality Polish image

[random pasture]

f(a) can be any function from z/12 to z/3

so it could be any set mapping

[dull ginkgo]

[random pasture]

but it also means that f(a) ∈ {0,1,2}, so f(a-1) ∈ {0,1,2}

[delicate orchid]

you got it backwards again

[random pasture]

oh it's the other way around?

oh fuck

[delicate orchid]

ok time to do this purely combinatorially and not care about functions

think of f(a) as a sequence of length 3 with values in {0,...,12} and our action (h,k) sends (a_0, a_1, a_2) to (a_{0-k}+h, a_{1-k}+h, a_{2-k}+h) where the indices are taken mod 3

this induces a homomorphism from Z/12xZ/3 into S_{1728}

it's clear that the generator of Z/3 is sent to (1,13,25)(2,14,26)...(12,24,36) up to conjugacy

actually, is that clear

[vivid tiger]

functions don't exist, they can't hurt you
(but morphisms are very real)

[delicate orchid]

no because it fixes diagonal elements

oh lord it's not S_{36}

ohhhhh lorrddddd here we GOOOO

yeah so this is like an action on base-12 numbers

[random pasture]

i mean ultimately it's proving that h=f(a)-f(a-1)={0,4,8}

[delicate orchid]

ok fine I'll actually answer the question instead of being massively distracted

so we permute the coordinates counter-clockwise one and then add 0,4 or 8, obviously (0, 1) fixes the tuple (0,0,0), corrisponding to the trivial homomorphism Z/3 -> Z/12
(4,1) fixes (0,8,4) and (8,1) fixes (0,4,8)

if (h,1) fixed some (a_0, a_1, a_2) we'd need a_1+h = a_0, a_2+h = a_1, a_0+h = a_2
so a_1+h = a_2+2h = a_0+3h = a_0, so 3h = 0

implying h in {0,4,8}

@random pasture done!

[delicate orchid]

time to return to getting massively distracted

[random pasture]

lmao

so how do i apply burnsides to this

[delicate orchid]

actually I'll translate this back into functions for you

if (h,1) fixes f, then f(1)+h = f(0), f(2)+h = f(1), f(0)+h = f(2)
so f(1)+h = f(2)+2h = f(0)+3h = f(0), so 3h = 0

[delicate orchid]

using the same thing you can show that (h, 2) has a fixed point if and only if h in {0,4,8}
(h,0) doesn't fix anything, (0,0) obviously fixes everything

[random pasture]

ohh i se

z/3 makes this a lot clearer to see

[delicate orchid]

yeah I just needed to stop thinking about f as a function lol

now the number of fixed points (h,1), (h,2) have I'm not too sure, I think there's 12 each

[random pasture]

so you just get f(x) = f(x) + 3h i can't believe i missed that lmao

[delicate orchid]

think how I feel

I legitimately wrote out ``the generator of $\bZ/3$ gets sent to $(1, 144, 12)(2, 288, 24)...(11, 1584, 132)(13, 145, 156)(14, 289, 168)... = \prod_{(a_0, a_1, a_2) \neq a_0(1,1,1)} (a_0+12a_1+144a_2, a_1+12a_2+144a_0, a_2+12a_0+144a_1)$"

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

like what am I smoking

[random pasture]

i was so lost after that notation

[delicate orchid]

you and me both

I just wrote out f as a tuple (f(0), f(1), f(2)) cause it was easier for me to see the action on it

and then I didn't look at the action on it

[random pasture]

wait so is |X/G| just 1/36 * (Fix((h,0)) + Fix((h,1)) + Fix((h,2)))

[south patrol]

Essence of Burnside

[delicate orchid]

|X| is much bigger than 36

it's 1728

it isn't, it's like

[random pasture]

oh it's |X| not |G|

i mixed them up

[delicate orchid]

no no

omg no it is |G|

jesus I'm skill issuing all over the place today

yes you're completely correct

now just to find all of the fixed points of (h, 1) and (h, 2)

[random pasture]

(h,1) is just 12^3 * 3

from part b right

and (h,2) would be the same?

[vivid tiger]

I just realized that the lemma that is not burnsides feels pretty obvious

|G| |X/G| = sum of fixed points of g

[delicate orchid]

so for (4,1) the fixed points look like (n, n+8, n+4) for any n in Z/12

so there should only be 12

[random pasture]

but that would give 12+36+36

so 84/36 as the answer?

[delicate orchid]

no obviously not

I'll leave it up to you to compute the other fixed point sets

[delicate orchid]

(h, 1), (h, 2) look very similar to this
(0, 1), (0, 2) should be obvious
(0,0) is obvious
and everything else doesn't fix anything as we previously came up with

[random pasture]

wait but for (h,0) it'll be 121, (h,1) is 12 3 and (h,2) is 12*3

[delicate orchid]

(h,0) doesn't fix anything

f(a)+h = f(a) => h = 0

[random pasture]

oh identity doesn't count as fixing

mb

[delicate orchid]

what

no it just doesn't fix anything

[random pasture]

this was part b that i already did, which is why I was confused

[delicate orchid]

yes this agrees with me

[random pasture]

wait so just to make sure

|X/G| = 2?

[delicate orchid]

lemme figure it out one sec

|(0,0)| = 1728
|(4,1)| = |(8,1)| = |(4,2)| = |(8,2)| = 12
|(0,1)| = |(0,2)| = 12
1728+12*6 = 1800
1800/36 = 50

[south patrol]

Wew btw

Have you seen the character-theoretic proof of the 5/8 bound

like for commuting elements

[delicate orchid]

I recall seeing it at some point

[south patrol]

Ah okay

Thought you might enjoy it if not lol

[delicate orchid]

do I remember the details? no lol

[south patrol]

Oh I mean me neither lol

I think it is a simple dimension bound tho

[delicate orchid]

oh so that's how it goes

ahhhh

yes ok I've rederived it

or have I

no I haven't

@south patrol if you can find it somewhere I'd love to reread it

[south patrol]

I can't remember the trick off the top of my head tho

Okie I shall

I think will sawin posted it on stack exchange

[delicate orchid]

it'll be the character degrees squared adding up to |G| I have a feeling

[south patrol]

Over maybeoverflow

Yes that's what I mean, exactly

Idk how you link it to the probability tho

Presumably because centralisers correspond to conj classes

[delicate orchid]

hmm

[south patrol]

Probability g commutes w h is probability C_G(g) = C_G(h) and then do a weighted average over all conj classes

Presumably gives u smth in terms of characters

Idk

Here's another dumb question

How interesting is the character table of a group as an invariant

Do people study it like this at all

I guess it is just equivalent to saying character rings are iso lol nvm

[delicate orchid]

I'm not familiar with invariant theory so could you be a little bit more precise

[south patrol]

I just mean like call two groups the same if they have the same char table

How interesting are those classes

[delicate orchid]

they're more or less isomorphism classes lol

[south patrol]

But I imagine it is just this

But isomorphic as rings, not hopf algebras

[delicate orchid]

on abelian groups they are isomorphism classes

and the only... - ok I'm gonna ignore you bringing up hopf algebras - counter example I know is D_8 and Q_8

[south patrol]

Is that just by classification thm

[delicate orchid]

yur

[south patrol]

Same lol

[delicate orchid]

not true

there's two groups of order 16 as well I think

SD_16 and something maybe

one mo

yeah @south patrol

[south patrol]

ah okay

[delicate orchid]

Q_16 and SD_16 have the same character table

[south patrol]

hm

[delicate orchid]

now the question becomes, find an example which isn't a 2-group

anyway D_8 and Q_8 are interesting because it shows you can't find the splitting field for a group just from the character table

all of D_8s reps can be realised in Q but Q_8s are over Q(i)

[crystal vale]

Okay, thank you

[south patrol]

Lmfao wew

I tried recreating it

and just ended up finding a non-character-theoretic proof

but i think it is the standard proof?

idk

[delicate orchid]

if you're fucking around with centraliser sizes, that sounds like the standard proof

eitherway still really impressive

[south patrol]

like probability two elements commute is $P = \frac{1}{|G|} \sum_{g \in G} |C_G(g)|/|G| = \frac{1}{|G|} \sum_{g \in G} \frac{1}{|g^G|}= \frac{|Z(G)|}{|G|} + \frac{1}{|G|} \sum_{g \in G \setminus Z(G)} \frac{1}{|g^G|} \le \frac{|Z(G)|}{|G|} + \frac{|G| - |Z(G)|}{|G|} \cdot \frac{1}{2} = \frac{1}{2} + \frac{1}{2} \frac{|Z(G)|}{|G|} \le 1/2 + 1/4\cdot 1/2 = 5/8$

[delicate orchid]

I should be able to do the character proof

[south patrol]

cannot find an error bruh

i think

[delicate orchid]

2nd summation

[south patrol]

Oh