groups-rings-fields

[little ocean]

What is an upper chunk?

[thorn jay]

All the ideals containing I

[little ocean]

Yeah

Again, my bad for not knowing all these English words tho

Isn't this Zorn's lemma, then?

[thorn jay]

No

[rocky cloak]

So if you have a chain of ideals
J1 < J2 < ... in R/I
what does the correspondence theorem give you?

[south patrol]

I have never heard of upper chunk personally and study maths in UK lol don't worry

[thorn jay]

Not official terminology but i always have this picture in mind

[little ocean]

Eh?

[rocky cloak]

I mean it's just normal English.
chunk = piece of something
upper = near the top

[thorn jay]

Con A is the ideal lattice of the ring A (in this case)

Anyhow, better to focus at the task at hand

[south patrol]

Yee

Just how it was said at the time made me think it was some mathematical terminology lol

[little ocean]

For every J in R/I there exists a unique K < R that contains I such that J = K/I?

[thorn jay]

I mean i couldve said like "principal up-set" or smt but thats even less helpful

[little ocean]

But we know that the chain of all those respective K's will stabilise, given the noetherian nature of R?

Cool

Now is the proof for artenian the same?

[thorn jay]

Sadge

[thorn jay]

Yes

[little ocean]

There's other questions too tho, if you guys still wanna help me out ;-;

[thorn jay]

Sure!

[little ocean]

The next exercise wants me to show that every artinian domain is a field

How do I even begin?

I suppose that I need to show that every element has a multiplicative inverse?

[thorn jay]

This is all commutative yes?

[little ocean]

No

[thorn jay]

Oh man

[little ocean]

Yeah

I can't be bothered to figure that out tbh

[thorn jay]

Left artinian then?

[little ocean]

Uhm

There was no specification on left or right artinian

[thorn jay]

How do you guys define domain?

[little ocean]

I believe just ring with no zero divisors

[rocky cloak]

It's only true in the commutative case

[little ocean]

Nvm

[rocky cloak]

Unless field means skewfield I guess

[thorn jay]

No that would be called delingsring in dutch at least

[little ocean]

"A commutative ring R != 0 without zero divisors is a domain"

[thorn jay]

Dunno abt flemish

[thorn jay]

Ah okay

[little ocean]

Yeah it's the same

[little ocean]

Altho it's explicitly stated in my syllabus that delingsring is more a Ditch word than a Flemish one

[thorn jay]

Lol

Just like you people actually call it veld instead of lichaam

[little ocean]

Indeed, no one does that

[little ocean]

Ok but any tips?

[rocky cloak]

I guess a hint could be to think about the ideals
||(x) > (x^2) > (x^3) > ...||

[marble hinge]

Is there a mistake in the problem statement here? If psi is just an injection, then S1 can have two elements, S2 can have three elements and then A(S1) will have two elements and A(S2) six elements, so obviously no isomorphism…

[thorn jay]

Huh yeah

[rocky cloak]

Presumably the author uses 1-1 to mean bijection

[delicate orchid]

I'm assuming one to one means bijective

[thorn jay]

Not a fan that they dont use injective surjective or bijective

[rocky cloak]

Some authors do this

[marble hinge]

I checked his definitions and no

[thorn jay]

But in the latter sentence they say "one to one onto"

Rather than one to one into

[delicate orchid]

well then the problem is obviously nonsense

try proving the true statement instead

[rocky cloak]

Maybe they forgot what definition they used midway

[thorn jay]

Sym(1) = Sym(1000000)

[thorn jay]

Execution

[delicate orchid]

I'll map my car one to one into your living room window in a minute

[marble hinge]

And bijection he calls “one-to-one correspondence between X and Y”

[thorn jay]

Its just a stupid typo then

[rocky cloak]

Then I'll put my money on this problem being a consequence of this terrible convention

[thorn jay]

Into and onto are very close on the keyboard

[marble hinge]

Yeah, seems so

[little ocean]

Does this not just show that x is nilpotent, so the only case in which this is true is if x=0?

[marble hinge]

I also quite like Bourbaki’s terms

Instead of this into/onto/shmonto business

[thorn jay]

Cant have a nilpotent element in a domain

[delicate orchid]

the humble 0

[little ocean]

Well yes, only 0

[thorn jay]

0 DOESNT EXIST

[rocky cloak]

Well you have this chain no matter what x is

[thorn jay]

THE FEDS LIED TO YOU

[rocky cloak]

But the artinian condition tells you it stabilizes

[little ocean]

Yes okay, but this means that x^{n+1} \in (x^n) for some n

[thorn jay]

Thats always true

[little ocean]

No... Right?

[thorn jay]

The other way around, you get that x^n in (x^n+1) for some n

[rocky cloak]

Yeah x^n in (x^n+1)

[little ocean]

Wait what

[thorn jay]

x^n+1 = x^n * x in (x^n)

[little ocean]

Bruh, I was thinking about R[x^n] for some reason

Sry mb

[thorn jay]

Why

Lol

[little ocean]

Because who tf uses x as an element in R??? It's mad confusing

Technically it shouldn't matter

But still

[thorn jay]

Cuz x looks pretty

[little ocean]

It really doesn'

t

[thorn jay]

Acquired taste

[delicate orchid]

$\mathcal{X}$

[cloud walrus]

DM Modmail For A New Nickname

[little ocean]

Ew

[thorn jay]

Goated

[little ocean]

Ok so I know then that x^n is in (x^{n+1})

[thorn jay]

Something something character variety

[thorn jay]

And what does that give you? What are the elements of (x^n+1)

[rapid cave]

,, \mathfrak{X}

[cloud walrus]

ExpertEsquieESQUIE

[little ocean]

Isn't (x^n+1) just x^{n+1}R?

[thorn jay]

Yes

[terse crystal]

(a)=(b) gives you b=ua for some unit u

[little ocean]

Right

[rocky cloak]

(in a domain at least)

[terse crystal]

Yes, domain, forgot

[delicate orchid]

oh my science

[little ocean]

And now what?

[terse crystal]

a=x^n+1, b=x^n, plug in

[little ocean]

Damn

Right

And since x was chosen random, x must be invertible

Cool

Now there's another exercise 😅

It says: given an artenian ring R such that the intersection of all prime ideals is trivial. Show that there exists a finite amount of prime ideals P_1, ..., P_n such that their intersection is trivial

[twilit wraith]

Now im trying to think of a non domain where two principal ideals are generated by non associate elements

[thorn jay]

I'm sure the Z/nZ will give you plenty of examples

all ideals are principal

[twilit wraith]

Yeah

[thorn jay]

can't be fucked to do it though so go wild

[twilit wraith]

(2) and (4) in Z6 i think

[thorn jay]

4 = -2

[twilit wraith]

Ah shit

[little ocean]

Genius ooc quote

[south patrol]

What on earth

[twilit wraith]

Its 7 am and I didnt sleep at all so maybe this isnt the best time for me to be pondering this

Uhhh

[south patrol]

Jk

[delicate orchid]

(3) and (6) in Z/Z9... no wait! (4) and (8) in Z/12Z... no wait! (5) and (10) in Z/15Z... no wai

[twilit wraith]

Fuck my chudslop equivalent up to renaming life

Do i even know any non domains that arent Z/nZ

[delicate orchid]

... any matrix ring?

[twilit wraith]

Oh probably

[little ocean]

Polynomial rings?

[twilit wraith]

Or definitely

[south patrol]

0

[thorn jay]

eeughh noncommutativeee

[south patrol]

Ok ig n = 1

[thorn jay]

that mf has only a single element

[twilit wraith]

https://tenor.com/view/speed-frustrated-speed-over-it-speed-crash-out-speed-crashing-out-crash-out-gif-12382450018351543960

[thorn jay]

oh wait we're asking for non domains

[south patrol]

Empty ring

[little ocean]

Honestly the 0 ring is based asf

[south patrol]

(Ragebait)

[thorn jay]

that mf can't even map into any other ring

useless ahh

[twilit wraith]

"How did you know I secretly work for the IRA and am trying to get you to pay federal taxes?"

[little ocean]

It can, just only to the 0 subring in that other ring

[delicate orchid]

requiring the multiplicative identity to be preserved is hold over from the dark ages of model theory

[thorn jay]

k[e]/(e^2)

[twilit wraith]

What the hell is k or e

[south patrol]

0 subring 🥀

[thorn jay]

NONUNITAL CHUD ALERT

[thorn jay]

k is field

e is infinitessimal element

[south patrol]

If you work with rings with identity aka are sane then what wew said is correct

[thorn jay]

(hence the quotient by e^2)

[twilit wraith]

Wait a damn minute

[delicate orchid]

I don't like 0 being a subobject because the initial object should not have any subobjects

[thorn jay]

the biggest UA texts use nonunital rings I hate my people 💔

[twilit wraith]

Is k[e]/(e^2) not just k[e]

[south patrol]

With nonunital rings 0 is initial rather than Z

[little ocean]

Wym bro 😭 any object is a subobject of itself??

[twilit wraith]

Or did I assume ur talking about dual numbers

[delicate orchid]

I'm actually gonna slime u out lil bro

[thorn jay]

yeah?

[thorn jay]

0 is not the object itself?

[delicate orchid]

but all rings are unital

[south patrol]

By definition yes

[delicate orchid]

but ring homomorphisms need not preserve the multiplicative identity

[south patrol]

Why must you hurt me like this

[delicate orchid]

it's called a full subcategory chud you might want to google it

[fading acorn]

https://tenor.com/view/cooked-gordon-ramsay-memes-cooking-noob-gif-1145964353964846098

[south patrol]

?? Lol

[little ocean]

What is even "chud"?

Never heard of it

[thorn jay]

great job on angering both algebraists and model theorists

[south patrol]

https://tenor.com/view/chud-chudjak-wojak-soyjak-marcvin5-gif-13903021270190271130

[delicate orchid]

https://tenor.com/view/ishowspeed-try-not-to-laugh-gif-7682731162751353849

[delicate orchid]

look in the mirror boss man

[little ocean]

Ok anyways imma do my homework because my algebra exam is literally monday

Have fun you guys

[delicate orchid]

exams in the glamorous 26 💔

[supple ice]

good luck

[little ocean]

Could've just stopped at "exams 🥀 "

[twilit wraith]

Foidchasing receipt paper touching Wikipedia ignoring chud vs based agarthan positive canthal tilt having Google gemini observing alpha male

[delicate orchid]

my gonal angle isn't even crystallographic lil bro I mog you six ways to sunday

[twilit wraith]

Six seven ways

Get it

[delicate orchid]

no

[tardy hedge]

Why is it bumpin in here

[twilit wraith]

Reference to the hit six seven meme

[twilit wraith]

This message lowkey cleared ts out

[thorn jay]

"Google gemini observing alpha male"

[terse crystal]

intersection of one prime >= intersection of two primes >= intersection of three primes>=… I think. I know nothing about fundamental staffs, any issue in set theory occurs this way?

Oh set of finite intersection of prime ideals

Having a minimal

[thorn jay]

Consider the set S of all finite intersections of primes.

Then use Zorn

(First prove it satisfies the conditions of Zorn's lemma)

[south patrol]

Wouldn't this work?

I guess more precisely you can consider the set of finite intersections of primes which then has a minimal element by being Artinian and then this is kinda obviously 0 by the hypothesis

[thorn jay]

Yes

[terse crystal]

Yeah. It’s just that I forgot why chains being stable <-> any subset having a minimal/maximal are equivalent in the first place. Former is always indexed by N, latter indexed by any set

[thorn jay]

Well that is exactly what Zorn's lemma is for

[south patrol]

Shouldn't need Zorn at all for this

At least w what I said

[south patrol]

It's clear that if every nonempty collection of ideals has a minimal element, then every descending sequence stabilises. Conversely, if a nonempty collection does not have a minimal element, then you can pick any element, pick one smaller than that, etc etc to create a non-stabilising descending sequence

[terse crystal]

Oh

[thorn jay]

DCC <=> every subset has minimal element doesnt use zorn?

[terse crystal]

Yeah, memory back

[south patrol]

Ye by the above

Still uses like dependent choice ig

[thorn jay]

The picking an element feels icky but yeah i see

You dont need the full power of Zorn

[south patrol]

Ye my point was not rly about set theory but more just that the style of argument isn't rly Zorny

[thorn jay]

I suppose its because every chain stabilizes rather than just has its bound in the set

[rocky cloak]

I guess the obvious thing is just to freely construct an example, e.g.
k[x, y, z, w]/(y = xz, x = yw)

[thorn jay]

Of course k[X \cup Y] where X and Y are algebraic sets such that X is not contained in Y and vice versa

[crystal vale]

how do i show rank of free abelian group is well define?

and free product of Z/2Z and Z/2Z is not free group?

[delicate orchid]

why would it be? it has two non-trivial relators

el classico PSL(2, Z) is the free product of Z/2Z and Z/3Z

[crystal vale]

what do you mean by it has two non-trivial relators?

[delicate orchid]

the free product is <a,b |a^2 = b^2 = 1> = <a,b | a^2 = 1, b^2 = 1>

[crystal vale]

yes

[delicate orchid]

so how is that free when it has relators lol

[south patrol]

One way is to go from Z to Q. Like if you have two isomorhisc free abelian groups then this gives you a linear map between Q-vector spaces of the same ranks, and now you use that dimension is well-defined

[delicate orchid]

now prove that Q-dimension is well defined

but yeah tensoring with Q is clever

[crystal vale]

i know for vector space rank is well define

[crystal vale]

ah, I just know free object in terms of category language but i don't know much about free group just universal property

[delicate orchid]

a group is free if and only if it has no relators and it's a good exercise to show that this is equivalent to the categorical definition

[crystal vale]

and if basis exists then its cardinality is unique, i think we don't need aoc here( existence preassumed)

[crystal vale]

okay, relators mean, there is some relations between generators?

[delicate orchid]

have you not seen presentations before?

[crystal vale]

i seen but i am not sure about terminology

[delicate orchid]

like, writing a group as <X|Y> where X is some set of generators and Y the set of relations

[crystal vale]

yeah

[delicate orchid]

I swap between relations and relators because I'm silly so my apologies in advance - they mean the same thing

[crystal vale]

np

[delicate orchid]

yeah so like, if you want to prove that, say G = <x | x^5 = 1> isn't a free group, you can map {x} into any group H whose order is coprime to 5, but because x^5 = 1 in G, by Lagrange there cannot be a homomorphism extending your map {x} -> H

you can do this same game with any relation in your G, so free groups cannot have any relations at all

[knotty badger]

there are only silly examples where a group with relations is still free

like <a, b, c| c = 1>

[delicate orchid]

yeah ok when I say "any relations" I mean any set of relations which can't be turned into the empty set by applying Tietze transformations

but that's a lot of words

[knotty badger]

oh i haven't heard of tietze transformations before

[delicate orchid]

they're exactly what you think they are LOL

[knotty badger]

i mean i've only heard of, like, tietze extension theorem

[delicate orchid]

"u can add a generator along with the relation that generator = some relation, and any relation that is logicially deductible from other relations can be removed"

you have to phrase the latter by uhhh

the relation to be removed has to be writable as some word in the free group generated by the other relations? it's a tad complicated

[marble hinge]

Just today I read this:

Somewhat unusual definition/approach, at least not the same that I’ve seen in Wikipedia

[crystal vale]

i am following this definition, it is from Hungerford, so initially we have function a: GROUP -> SET between categories( not functor) with some property, here they denote g(G) as G, so i think you assumed a as forgetful map but what about if there exists other such map g, and how do i choose X, here you choosed X as {x}

[delicate orchid]

this is pretty much how I learnt it icl

[delicate orchid]

X is the generating set for your group

why the hell are they doing it with a CONCRETE CATEGORY and not just saying "it's the adjoint to the forgetful functor, so there's a bijection of hom sets"??? they're going to this generality and then not taking advantage of it? That's terrible

[crystal vale]

okay, can you tell me your definition of free group?

[delicate orchid]

I think this definition is fine but like. Not for first learning about free groups

the free group on a set X is the set of all words in X, together with formal inverses, with the group operation being concatenation and cancellation of words

[crystal vale]

and with some universal property?

[delicate orchid]

well no, you prove that this group has that universal property

[crystal vale]

oh

[delicate orchid]

hence the two definitions are equivalent up to unique iso

[crystal vale]

so universal property is if there is a f: X -> G then there exists unique homo g: F -> G which extends f. where F is a free group on X

[delicate orchid]

yur

[crystal vale]

i think i can see why it is true

[noble nexus]

I thought this was the standard approach

[delicate orchid]

Hom_Set(X, U(G)) is isomorphic to Hom_Grp(F(X), G) where U is the functor sending a group to its underlying set and F(X) is the free group on X

[noble nexus]

Free groups should def be taught using universal properties

[delicate orchid]

I think you should introduce both the universal property and the concrete object

[noble nexus]

yeah of course

[delicate orchid]

same goes for tensor products imo

[knotty badger]

yeah it's important to know that there actually exists an object satisfying the universal property

[marble hinge]

This is what Wiki says:

[noble nexus]

The universal property is also a lot less confusing when you compare it to the universal property from linear algebra that students have been using for their entire math career without knowing its a universal property

[crystal vale]

but then there assumed forgetful mapping

[thorn jay]

This is how any free object is defined

But i suppose introductory textbooks tend to aim for a construction first approach

[delicate orchid]

using the grothendieck completion of the free monoid is brave

[noble nexus]

Tbh it's a lot easier to prove things about the free monoid

so I kinda like that approach

But honestly trying to do things with the free group completely formally is imo more confusing than it's worth

[crystal vale]

where you guys learnt this?

[delicate orchid]

fourth year of uni personally

[thorn jay]

Youre gonna end up with using the universal property under the hood anyways lmao

[noble nexus]

When I first did free groups my prof did everything super formally using equivalence classes of sequences that are trivial almost everywhere and it's just kinda painful

better to treat the construction somewhat informally imo

and focus on the universal property

[thorn jay]

A Course in Universal Algebra was my personal first proper experience with free objects (as explicitly mentioned)

[knotty badger]

free groups in first year, but their universal property only really in 4th year

[crystal vale]

oh universal algebra is a course

[thorn jay]

No thats a book, by Sankappanavar and Burris

[crystal vale]

okay thank you

[noble nexus]

The classical way to do this is to count the number of homomorphisms from the group to Z/2Z

using the universal property

[crystal vale]

btw what i need if i want to learn universal algebra?

[crystal vale]

why?

[noble nexus]

By the way the universal property you should think about is as being very similar to bases for a vector space

If you have a basis for V, then to define a linear map from V to W it's enough to specify what happens to the basis

[crystal vale]

yes

[noble nexus]

And conversely, every linear map is determined by what it does to the basis

for free groups it's exactly the same

[crystal vale]

i see

[noble nexus]

so to count the homomorphisms from F(S) to a group G, it's the same as counting the number of functions from S to G

[crystal vale]

yes

[thorn jay]

Some interest in more logic stuff, and I'd recommend some group and ring theory

[crystal vale]

but here we have to basis set so how does counting of functions from S to G helps?

wait

if there two basis with cardinaility x and y then 2^x = 2^y but how do i say x = y

[marble hinge]

Consider logarithms 😅

[crystal vale]

no, because x and y are not numbers

[marble hinge]

Consider categorical logarithms!

[thorn jay]

It is an interesting question if G is a free group on an infinite set

[crystal vale]

what?

[marble hinge]

Sorry, I’m just kidding

[thorn jay]

But in the case that G is free on a finite set X, then Hom(G, Z/2Z) is finite so it cannot be free on an infinite set

And from there the cardinality of a basis is unique by taking log_2

[knotty badger]

This is essentially representations of functors

[amber burrow]

im trying to understand the free group construction thing and i am a bit confused

so dummit and foote uses the analogy from linear algebra "Any set map from the basis S to another vector space V extends uniquely to a linear transformation from F(S) to V"

I'm a bit confused how this actually works tho, in linear algebra

dont you have to specify a basis in both the domain and codomain vector spaces?

[kind temple]

nope

just the domain

[amber burrow]

how?

i mean the linear transformation is determined by where you send the basis of the domain

but if you have a given vector space codomain, what is the unique way of doing that

[rocky cloak]

If you have a basis for V, b1, .., bn, then any element of V is a linear combination
a1b1 + ... + anbn

So a linear function T has
T(a1b1 + ...) = a1T(b1) + ... + anT(bn)

I.e. it is determined by T(b1), T(b2), ...

[amber burrow]

oh ok i think i was thinking that for any set S and vector space V, there should only be one map phi

but no, the map phi, S, and V together determine a unique F(S) upto iso

[thorn jay]

no

[rocky cloak]

I think you might still be confused.

The point is that if S is a basis for V, then V ~= F(S). Because a linear map out of V is uniquely determined by a function out of S

[amber burrow]

i think its supposed tob e into V

"Theorem 17. Let G be a group, S a set and q; : S � G a set map. Then there is a unique
group homomorphism cJ> : F(S) � G such that the following diagram commutes:"

[kind temple]

try some concrete examples.

say V is the vector space of polynomials of a single variable with real coefficients. R^3 is freely generated by S = {e1,e2,e3}.

any function f : S —> V can be extended to a linear function R^3 —> V. test this out with a few functions f.

[rocky cloak]

Yes, the point is that any function from S to a vector space extends to a linear map from F(S). And it does so in a unique way

[rocky cloak]

I guess correcting what you said here:

F(S) only depends on S, but the function S -> V uniquely determines a linear map F(S) -> V

[knotty badger]

You can think of $F(S)$ as consisting of functions $S \to \mathbb{K}$ with finite support

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

If you have a map $f : S \to V$, then given such a function $g : S \to \mathbb{K}$, you can form the weighted sum $\sum_{s \in S} g(s) \times f(s)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

This is “extending by linearity”

Each element $s \in S$ corresponds to the “indicator function” $\delta_s : S \to \mathbb{K}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

Defined by $\delta_s(t) = \begin{cases} 1 \text{ if } s = t \ 0 \text{ otherwise } \end{cases}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

Then you have a formula $g = \sum_{s \in S} g(s) \times \delta_s$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

And so any linear map $L$ applied to $g$ must satisfy $L(g) = \sum_{s \in S} g(s) \times L(\delta_s)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

A function $f : S \to V$ can then be thought of as just specifying the values of the linear map on these indicator functions, by $L(\delta_s) := f(s)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

Then you extend by linearity!

[knotty badger]

If you’ve done distribution theory then this is the analog of $f(x) = \int dy f(y) \delta(x - y)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

Does that make sense @amber burrow ?

[tulip otter]

Its probably obvious and i am just dumb, but i am not sure why g is the unique homomorphism that satisfies this property, I mean the projection maps are surjective but not injective so there arent left inverses for these maps in order to compose this left inverse with both sides of g_i=p_i o g

[terse crystal]

p_i • g=p_i • g’ for all i implies that for any x in G’, i th component of g(x) equals i th component of g’(x) for any i. Which by definition of Cartesian product implies g(x)=g’(x)

[tulip otter]

ah lol, i was so fixated on composing with an inverse of one these maps that i forgot about everything else

[tulip otter]

tysm

have a great day/night

[tardy hedge]

Does every graded ring have a unique homogenous maximal ideal?

For the rings that do have a unique homogeneous maximal ideal, is every prime ideal contained in it? Or is it more like every graded prime ideal is contained in it

Bruns herzog defines something called *local

I dont think the ring itself needs to be local

[modest dune]

are you discounting the irrelevant ideal?

[tardy hedge]

I dont think so

Oh ok so i thinj what i said is true then

The definitions arent straight in my head tbh

[modest dune]

i think u would usually be more concerned with the prime homogenous ideals than the prime ideals

[tardy hedge]

Reason I asked is to verify if dim R = dim R_m where m is the unique homogeneous maximal ideal (also called irrelevant ideal …?)

[cloud solar]

I have a question. If G is a finite abelian group with at least 2 elements and Spec(G) = Spec(Aut(G)), then who is G up to isomorphism? Any ideas?

G must have an even order (if not x^-1 has order 2 in Aut(G))

Btw Spec is the spectrum of the group, the set of all element orders

I am trying somehow to show G is a 2-group but i dont really know how

Spec(G) is just the set of all divisors of exp(G)

[karmic moat]

no. e.g., by projective nullstellensatz, points in P^n (over an alg closed field) correspond to homogeneous maximal ideals of k[x0, ..., xn]

[cloud solar]

So for every such divisor there must be an automorphism with that order

And such groups exist. For example Z/4Z x Z/2Z

[tardy hedge]

Which graded rings do then?

[karmic moat]

oh i see you're including the irrelevant ideal

there's a notion of graded local rings

iirc if you have a graded ring A with A_0 local, then A itself is graded local

[tardy hedge]

Irrelevant ideal is the ideal generated by elements of degree “greater than 0” , so like, that only makes sense if its like Z or Z^n graded right

For the “greater than 0” part

[karmic moat]

i don't really think about gradings other than Z^n so i'm not sure

[rocky cloak]

I mean, if you just give the ring the trivial grading all ideals become homogeneous, and it's of course not true that every ring has a unique maximal ideal.

If you require the degree 0 part to be a field and you're non-negatively graded, then the irrelevant ideal will be the unique maximal homogeneous ideal

[tardy hedge]

I wont care either then. Lol

[karmic moat]

if G is your additive abelian group giving a grading on R, you can probably define an "irrelevant ideal" analogously by assuming that everything below R_g (for some g in G) is 0

well i guess "below R_g" is ambiguous

probably need some ordering on G

maybe someone else can answer better

[rocky cloak]

Consider for example k[x] with the usual grading. Then (x) is this maximal ideal, and (1+x) is prime (not homogeneous)

[tardy hedge]

Is dim R = dim R_m?

[karmic moat]

you need noetherian-ness

well, i think you need noetherian

[tardy hedge]

Yea this is nice cuz i always see these hypotheses and i never knew what is used where

[rocky cloak]

Still not true in general right?

I'm not sure how R being graded plays into it...

[tardy hedge]

I had been studying combinatorial algebra and when checking cohen macaulay property they do so by considering R_m, yet the dimension they use is dim R

The ring in question is nonegative graded

[rocky cloak]

dim R should equal sup dim R_m over all maximal ideals m though

[karmic moat]

yeah

[tardy hedge]

Ok well basically ive only been considering polynomial rings k[x1, … xn] and the irrelevant ideal is (x1, … xn) , any proper ideal is contained in that so dim R = dim Rm

Maybe i said something wrong idk i wasnt confident w it lol

(x-2) is not in (x) lol nevermind then

[rocky cloak]

It is true that the localization of the polynomial ring at (x1, ..., xn) has the same dimension though

And I guess more generally that any maximal ideal of the polynomial ring has height n

[tardy hedge]

Why is that? I should probably study more dimension theory

[tardy hedge]

Some krull height theorem or something?

[rocky cloak]

I mean there's an automorphism taking any maximal ideal to any other I guess (for k alg closed that is). And more broadly just the like explicitly describing the prime ideals.

Otherwise I guess there's the alggeo picture, k^n does not consist of components of different dimension

[rocky cloak]

You can use that to establish that (x1, ..., xn) has height n sure

[cloud solar]

Any idea?

[rocky cloak]

I misread the definition of spec, nvm

[wicked patio]

With multiplicity?

Otherwise I don't see why Z2×Z2 doesn't work

[rocky cloak]

Aut(Z/2 x Z/2) = S3

Which has element of order 3

[wicked patio]

oh

[rocky cloak]

I think Z/2 x Z/4 might be an example though

[cloud solar]

Yes that is one example i found too

[modest dune]

okay I Have been staring at this for way too long, probably because I have pretty little experience with tensor products of modules. I am finding them hard to work with. Currently I am looking at the claim that (A ⊗ l) ∩ K(A) = A.

This makes intuitive sense to me, but I don't get how to work with the tensor product in a nice way to show it. I assume I want to look at the elements like a/s ⊗ 1 and show that if that is equal to some linear combination inA ⊗ l then s is actually a unit in A. But I do not have a clue how to do that. I do think it is clear that all the elments in the linear combation have the second component in k instead l though.

[terse crystal]

Z/2Z times Z/2Z times Z/3Z also works

Did you obtain anything? Like a range of p or the range of power for each p, I haven’t though. I tried to deduce power of odd p<=1 and there is no p>=5 both haven’t succeeded yet…

[modest dune]

I also thought about trying to create a bilinear map to something to try to learn about the tensor product. Alternatively, I guess I want to try to find some combination of terms that equals 0 so I can assert that the left hand side is 0? I guess if I have a value of k on the right side I can just move it to the left?

I guess the trick is that if I can rule out that the right hand side has any element of b_2, ... then I can move over the k factor from rhs to left and have 1 on the rhs of the tensor, then equate the two lhs and be done

[terse crystal]

x in the intersection, meaning x otimes 1=x otimes b1=Σ ai otimes bi for some ai in A. But you already have K(A) otimes l is a field extension over K(A) with basis 1 otimes bi, it forces x=a_1 in A, and a_i=0 for i>=2

[modest dune]

oh wait I guess once I have that the tensor product with k is just A right?

[terse crystal]

What tensor with what

[modest dune]

that that part shows that A tensor l intersect K(A) is A tensor k

and A tensor k is A

I think I have all the steps I just cant convince myself I actually showed the statement

[terse crystal]

If you can show this, sure. Though I don’t know how you did. I simply used the fact that K(A) is a field over k, thus flat, tensor_{k} with injection k->l is still injective, thus K(A) otimes l has 1 otimes bi as a basis, thus compare coefficients x=a_1 in A

[modest dune]

ok so when you write x otimes 1=x otimes b1=Σ ai otimes bi for some ai in A

you have something slightly different, i had

f/g otimes 1=x otimes b1=Σ ai otimes bi for some ai in A

so I still was wanting to show that the representation on the left was actually f/g in A. But I think I see how to finish that now

can you explain what you meant by flat?

i think "thus flat, tensor_{k} with injection k->l is still injective" is the part that I am not understanding

nvm I think I found something to read about that

[terse crystal]

exactly the same thing can be said to A: k is a field, any k-module is free (thus flat), so A otimes_{k} applied to the injection k->l is still injective

[terse crystal]

How do we calculate order of elements of Aut(A_p) in general? Where A_p is a p-finite abelian group. A_p is direct sum of Z/p^ei Z where 1<=e1<=…<=en. I know Aut(A_p) is the group of {n by n integer matrices M such that p^(ei-ej) | M_ij, and M mod p is in GL(n, F_p)} quotient {matrices N such that p^ei | N_ij} But how to calculate their order? In fact a special case, how to calculate order of elements of GL(n, F_p)?

[terse crystal]

Seems like it’s not an elementary problem. I thought there were existing results

[tribal moss]

In fact a special case, how to calculate order of elements of GL(n, F_p)?
You could always do something like

k = |G|
for each prime factor d of |G|:
  while d|k and M^(k/d) == I:
    k = k/d

This assumes you can factor |G|, but then at least should be faster than the naive "compute successive powers of M until you hit I".

[terse crystal]

I see

[tribal moss]

Now I wonder if we can start with a smaller upper bound for the element order than |G|, though. |G| itself is surely too large since GL(n,F_p) is not abelian.

[terse crystal]

I heard someone mentioned a paper titled polynomial time theory of matrix groups might be relevant

[tribal moss]

Hmm, |GL(n,p)| is (p^n-1)(p^(n-1)-1)···(p^2-1)(p-1)·p^(n(n-1)/2), and most of the final factors of p cannot appear in the order of M.
Namely, move temporarily to an appropriate algebraic extension of F_p where the charpoly of M splits. We can then write M in Jordan normal form. Each Jordan block λI+N becomes I when we raise it to k=(p^n-1)(p^(n-1)-1)···(p^2-1)(p-1)·p -- the binomial expansion of (λI+N)^k becomes λ^k·I+N^k because p|k; the multiplicative order of λ divides one of the (p^i-1) factors; and N^k disappears because even p^n-1 alone is >= n.

In fact, we can also omit the p^i-1 factors where i<n/2 because p^i-1 always divides p^2i-1.

Or better yet: instead of the product of p^n-1 down to p-1 we can take their lcm, since only one of them is relevant for each Jordan block.

[terse crystal]

I see. Thanks a lot. Should have thought using Jordan. I think it can be done over a finite extension of F_p right

[tribal moss]

Yeah.

[terse crystal]

Thanks

[tribal moss]

Most of the p become p^m, except the final factor of p itself which was just there to be the characteristic rather than the size of the field.

[terse crystal]

I see

[little ocean]

I have a question, is there an 'easy' method to find all normal groups, the center and the commutator group of a given group G?

And with easy method I mean something simpler than looking at all elements and checking the definition for center, or looking at all subgroups and checking the definition for normal subgroup

[candid patrol]

No

[little ocean]

Sadge

[cloud solar]

Omg

Ok so it is not a p-group like I thought

[rocky cloak]

The first isomorphism theorem can be helpful sometimes. For example the commutator subgroup is the intersection of kernels of all maps from G to an abelian group.

Identifying such homomorphisms is sometimes less work than checking all elements.

Similarly normal subgroups are exactly kernels of homomorphisms, so you could for example find all normal subgroups of index n by finding all surjective homomorphisms from G onto group of order n (this is not too hard if you know all groups of order n)

[little ocean]

Uhm

Right

I also have another question

What is $\mathds{C} / \mathds{R}$?

[cloud walrus]

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

[little ocean]

Nop

$\usepackage{dsfont}
\mathds{C} / \mathds{R}
$

$\usepackage{dsfont} \mathds{C} / \mathds{R}$

[cloud walrus]

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

[little ocean]

$\mathbb{C} / \mathbb{R} $

$\mathbb{C} / \mathbb{R}$

[cloud walrus]

Nico

[little ocean]

Yay

[little ocean]

What is this? Is this just (isomorphic to) R?

[noble nexus]

Yeah

It's just R^2/R

[little ocean]

And what is $(\mathbb{C},+) / (\mathbb{R},+)$?

[cloud walrus]

Nico

[little ocean]

I meant $[(\mathbb{C},+) / (\mathbb{R},+)]$

[cloud walrus]

Nico

[noble nexus]

That's what I thought you were asking

[little ocean]

Oh no

[noble nexus]

what did you mean by C/R if not the additive groups

[tribal moss]

Real vector spaces, perhaps.

[little ocean]

Yes I did

But now I mean the index of R in C

What is the index of R in C?

Is it like

[noble nexus]

ah, it'd be uncountably infinite

[little ocean]

Omega

[noble nexus]

as a group

[little ocean]

Or smth

How big is R?

Is it omega?

[noble nexus]

omega isn't a cardinality but it's larger

[little ocean]

What is omega, then?

I thought it was also some infinite size

Or is omega the first ordinal number after aleph null?

Idk how to explain

[noble nexus]

it's an ordinal, aleph null is a cardinal not an ordinal

[little ocean]

Yeah but I mean

After aleph null amount of numbers, the number 'after that' would be the omega-th number

Or something

[noble nexus]

but really referring to specific cardinals besides "countable" and "uncountable" is not really something you do outside of set theory

[little ocean]

Ah

[noble nexus]

no, aleph 1 would be the first cardinal after

[little ocean]

Oh

Nvm then

[little ocean]

So I should just answer this as 'the amount of all real numbers'?

[noble nexus]

Yeah it would just be the cardinality of the real numbers

[little ocean]

Cool

Thx

[noble nexus]

As a group though, if you mean index as a field extension then its 2

idk which kind of a course this is for

[little ocean]

Excuse me?

[little ocean]

Introductory abstract algebra

[noble nexus]

or as vector spaces it's index 2

[little ocean]

No, not as a vector space

It's as groups

[noble nexus]

alright then yeah uncountable, because the index is just the cardinality of the quotient

[little ocean]

Right

Thanks!

[tulip otter]

https://arxiv.org/pdf/math/0605185 I havent read the whole paper but this paper aims to find Aut(G) for any finite abelian group G

[terse crystal]

Thank you. Someone recommended exactly it to me before. This is from where I know Aut(G) is group of matrices of the said form.

[tulip otter]

ohhh i see, actually i was searching for some results about Aut(G) in general because of a question asked in this channel a few days ago that i found interesting, and i stumbled on this paper in the process

[terse crystal]

It seems all the ways lead to it.

[tulip otter]

sotrue

[spiral tendon]

ngl we are taking vector spaces soon

and i see a lot of questions that ask

prove that (an equation) is a vector subspace of R

lie

like

A = (x,y,z <= R^3 : 2x+3y-z=0 ) , B=R^3 , we need to prove A is a vector subspace of B (B is a vector space over R)

we are taking it in a week but im kinda into finishing it rn

[thorn jay]

Boy do i have the perfect channel for you

[tribal moss]

Yeah, vector space exercises go in #linear-algebra.

(But you seem to have neglected to ask a question about that exercise).

[rocky cloak]

You can do it Gibbons! Solve that exercise

[thorn jay]

show that the kernel of every linear transformation is a subspace, and then prove that A is the kernel of some linear transformation and you're done!

[tribal moss]

I think it's more instructive to do it directly the first few times, and then notice that the argument generalizes.

(For one thing, recognizing that expression as a linear transformation is about the same amount of work as showing the subspace properties directly).

[thorn jay]

I mean atp just go all the way into the generality of 0-regular varieties where 0 generates a trivial subalgebra

[thorn jay]

yeah lol I was jesting

too many of these kinds of exercises and it becomes a slog though

probably not how you spell it

[cloud solar]

I feel like for the problem I asked this suggests to look on Zsigmondy primes in the order of Aut(G)? The formula for the order is quite hard

I wonder if Z/2Z x Z/4Z and Z/2Z x Z/6Z are the only solutions

[tacit condor]

hey just wondering what does Z/nZ mean

[candid patrol]

It reminds me something

Given a finite group H, there are only finitely many isomorphism classes of finite groups G whose automorphism group is isomorphic to H.

Pretty cool

[cloud solar]

Residue classes modulo n additive group

[tacit condor]

can you give an example

like Z/2Z

[cloud solar]

Is a set of 2 sets: the set of integers that have remainder 0 when we divide them by 2 + the set of integers that have remainder 1 when we divide them by 2

You may saw this written like Z_2, but this notation Z/2Z tells us this is a factor group. 2Z is a normal subgroup of Z, and the group Z/2Z is just a group of cosets of 2Z in Z. This sets called residue classes are equivlance classes and the form a partition of Z

In my problem I think as the order of G has more distinct primes then somehow Aut(G) creates new primes in his order but I don’t have any idea how to prove that. Even that if we assume G is a p-group (2-group cuz it’s even), it’s still very hard to manage the orders of the automorphisms

[valid fox]

Does anyone have a source which lists subgroup lattices for some small groups?

[noble nexus]

https://people.maths.bris.ac.uk/~matyd/GroupNames/

this one is great

[valid fox]

Brilliant, thank you

[tacit condor]

ok i think this gave me a better idea about them

thank you

[little ocean]

What is a generalised quaternion group?

[frosty mauve]

Q4n

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

[tardy hedge]

How do u get n <= ht(x1, ... xn)?

oh thats obvious part

(0) < (x1) < (x1,x2) ...

[tardy hedge]

ok i finally understand why dim Rm = dim R for R a face ring and m = (x1, ... xn)

i find bruns herzog a challenging text there are always so many details that i dont understand without really going into it

[candid patrol]

A generalisation of the quaternion group

😇

[river heart]

IIRC it's the group that has presentation <i,x | i^4=1, x^n=1, i^2x=xi^2>.

[little ocean]

Cool

Also completely other, unrelated question, if R is a ring, when is R[x] a domain?

[rocky cloak]

Maybe you can try this yourself. How can two things in R[x] multiply to 0?

[little ocean]

If the coefficients of every power of x is 0

[fading acorn]

here's a nice exercise for it

(assuming A is a comm ring w/ 1)

[little ocean]

Omg

I literally have that same exercise somewhere written down

[fading acorn]

Lol

[little ocean]

Yh

Were you the guy that spoke Dutch?

[fading acorn]

no?

[little ocean]

Oh, sorry mb

I don't like algebra :/

I'm no good at it

[rocky cloak]

In case you don't want to do the whole exercise:

If R is not a domain, could R[x] be?

[little ocean]

I have "if R is a domain, then the units in R[x] are the units in R"

[rocky cloak]

That's true sure

But what does it mean for R to not be a domain?

[little ocean]

And stronger: $(R[x])^{\times} = (R)^{\times} \Leftrightarrow R$ has no nonzero nilpotents

[cloud walrus]

Nico

[little ocean]

There are a and b in R, both nonzero, such that ab = 0

[rocky cloak]

So if there are such a and b, could R[x] be a domain?

[little ocean]

Well, no

Because R[x] contains R

[rocky cloak]

Boom boom

Now if R[x] isn't a domain, what would that mean?

[little ocean]

There are f and g in R[x] such that fg = 0

again, with f and g both nonzero

[rocky cloak]

And what do elements look like in R[x]

[little ocean]

Polynomials

[rocky cloak]

So you can write
f = ...

[little ocean]

$f = \sum a_ix^i; g = \sum b_ix^i$

[cloud walrus]

Nico

[rocky cloak]

And then their product will be what?

[little ocean]

$fg = \sum (\sum a_ib_jx^{i+j}$

[cloud walrus]

Nico

[rocky cloak]

And what does it mean for this to be 0?

[little ocean]

But fg = 0 so a_ib_j=0 for all x^{i+j}

But a_i and b_j are in R

So R is not a domain

[rocky cloak]

Yeah.

So it's a little more complicated since i + j can be equal for different i and j, but you can look at the leading coefficient for example

[little ocean]

Yeah, but writing down the product of 2 sums with sigma notation kinda sucks

[rocky cloak]

Yeah, in this case it's much easier without using sigmanotation

[little ocean]

$fg = \sum_{i=0}^{max(n,m} \left( \sum_{j=0}^i c_jx^j\right)$

[cloud walrus]

Nico

[little ocean]

So anyways

I have another question

When is the criterion of Eisenstein actually useful?

[rocky cloak]

When you want to prove a polynomial is irreducible

[little ocean]

Yeah ik, but

Like

So?

It's only in Q[x]

[quiet pelican]

It generalises to any sufficiently nice ring in place of Q (where I can’t remember the correct “sufficiently nice” adjectives)

[little ocean]

What is p² in this notation?

[rocky cloak]

It generalizes from Z[x] to any integral domain. Then going from Z to Q should work for UFDs / integrally closed rings

[quiet pelican]

Square of the ideal

[little ocean]

As in a subspace of R x R?

I'm confused, I've not heard about the square of an ideal before

[rocky cloak]

It's just the product of the ideal with itself.

I * J = < ij : i in I, j in J >

[little ocean]

Ah

[knotty badger]

Angle brackets means generated

[little ocean]

And what can you do with the universal property of free modules?

That also looks kinda niche and unhandy to me

[thorn jay]

Its the defining property of a free module, first of all

[knotty badger]

They help a lot for defining maps out of free modules

[thorn jay]

So you can eg prove it is injective

Projective*

[fading acorn]

kan extension

[thorn jay]

It looks unhandy at first but the more you work with modules the less you start to work with the elements explicitly, and the more you begin thinking in morphisms and diagrams

The basis definition of a free module then becomes annoying and clunky to work with as opposed to the universal property

The universal property is, in fact, a restatement of the existence of a basis, but in terms of morphisms

At least, it's definitely not niche lol because it is one of the most important things in algebra

Having an object with the "free" property

[little ocean]

Sure

[thorn jay]

One thing that follows from the universal property: for every module M there exists some free module F such that there is a surjection F -> M. M is finitely generated iff there exists such F where F is free on a finite set

This is nice because you can sometimes reduce problems down to proving it for (finitely generated) free modules, and prove that it holds for quotients

[tardy hedge]

How does that first statement follow from uni property

[thorn jay]

Take the underlying set of M

[tardy hedge]

I didnt know uni property can construct free modules for you?

[thorn jay]

The there is a set function f : M -> M which must extend to a homomorphism f* : F(M) -> M which is surjective as f is surjective too

[knotty badger]

the universal property tells you what a free module "does"

[south patrol]

Let me pedantically write U and F for the forgetful and free functors between sets and modules. Then the identity map U(M) -> U(M) gives you a map FU(M) -> M, and moreover this is surjective, e.g. as we have a factorisation U(M) -> UFU(M) -> U(M)

or just by looking by hand

[thorn jay]

No because you already know that a free module exists on every set

[tardy hedge]

Yeah i guess i forget how we know thay

That*

[thorn jay]

Because theyre precisely the direct sums of R

[tardy hedge]

Yeah

[thorn jay]

You can prove this a multitude of different ways, the cleanest are using either Yoneda or the Hom-set isomorphism

[tardy hedge]

I think dummit and foote used some indicator function thing

[thorn jay]

Uh yeah that works too

Its more elementary lol

[tardy hedge]

Yea i guess thats like explicitly creating the module

With elemets

Elements

[thorn jay]

I.e. notice that Hom_R(R^(X), M) ≈ \prod_x in X Hom_R(R, M) ≈ \prod_x in X M ≈ Hom_Set(X, M)

These are all natural isomorphisms of set-valued functors

[thorn jay]

Yis

[tardy hedge]

I dont think i understand what happened in the second and third =

[thorn jay]

The first is the definition of a coproduct

[knotty badger]

triangular identities

[thorn jay]

The third is just some set theory

[knotty badger]

for this you do need that the forgetful functor is faithful though

[thorn jay]

The set of functions from X to Y is naturally (in Y) isomorphic to the product of Y "X times"

[knotty badger]

you have surjectivity of the map UFU(M) -> U(M), and you use faithfulness to argue that the original map FU(M) -> M must have been epi

[knotty badger]

power/cotensor

[thorn jay]

Yurrr

Cartesian closed category

[knotty badger]

mm this is a little different to that

[south patrol]

well kiand said surjective which i assumed meant set-theoretically

but this is also true ye

[knotty badger]

yeah that's fair

actually i think right adjoint being faithful is equivalent to the counit being epi

and right adjoint being full is equivalent to the counit being split mono

[tardy hedge]

Erm is there an example of a ring with no prime ideals?

Haha

Dum W

Q

no

[thorn jay]

trivial ring

else, the top ideal R is finitely generated, so the ideal lattice has maximal elements

and maximal elements in the ideal lattice are maximal ideals, which are prime

[tardy hedge]

Lattice ok

[thorn jay]

replace lattice with poset and same thing

you can try to prove it yourself using Zorn

but this is a standard result about compact elements of posets, and it so happens that finitely generated ideals are exactly the compact elements in the ideal poset

well, not like anyone knows order theory nowadays -_- lol

[rocky cloak]

Depends how strongly you believe the axiom of choice 😉

[tardy hedge]

If Big Math gives me the OK then im OK 👌

[rocky cloak]

AoC is equivalent to every nonzero ring having a maximal ideal. The existence of prime ideals is strictly weaker, but still independent of ZF

[balmy python]

im trying to do a problem where i show φ(mn)=φ(m)φ(n) iff (m,n)=1. The solution uses the isomorphism between cyclic groups to imply that their multiplicative groups are the size, how is this possible if the isomorphism is a group isomorphism and not a ring one

[rocky cloak]

It is an isomorphism of rings though

[rapid cave]

its also a ring homomorphism by CRT

[balmy python]

ah havent learnt about that yet

😭

i think its soon

[rocky cloak]

I guess you could argue using the automorphism group instead, but then you need to argue that
Aut(Cn x Cm) = Aut(Cn) x Aut(Cm)

[supple ice]

the key map comes from the Chinese Remainder Theorem which is a ring isomorphism it respects multiplication and 1 any ring isomorphism sends units to units so it induces an isomorphism between the unit groups and therefore their sizes multiply

[thorn jay]

notice (n) \cap (m) = (nm) in general, and (n) + (m) = Z iff n and m are coprime. Then as the variety of rings is congruence-permutable just use A Course in Universal Algebra Chapter II Theorem 7.5 to conclude that Z/(nm) = Z/(n) x Z/(m)

[balmy python]

😭😭😭

[tulip otter]

you can always cheat and use φ(n)=n\prod (1-1/p) sotrue

[balmy python]

thats what i gotta prove after

which i havr now done

[thorn jay]

lmaooo

[balmy python]

ngl these are how my exams are sometimes

[tardy hedge]

Y is bro giggling

[tulip otter]

prove it before that and in the next part say "see the solution of the previous part"

[thorn jay]

why tf did I press enter

[rapid cave]

because you pressed enter

[balmy python]

https://tenor.com/view/chris-pratt-mind-blown-parks-and-rec-surprised-shookt-gif-9771169

[tardy hedge]

Wat

[thorn jay]

To give a serious answer:
Consider the map from Z/(nm) -> Z/(n) x Z/(m):
a + (nm) -> <a + (n), a + (m)>
Show that this map is well-defined, and in show that it is injective whenever n and m are coprime. Then by comparing sizes conclude that it must be surjective

yes, this is slightly different from CRT, as it uses finiteness rather than 1 in (n) + (m)

[balmy python]

yeahh

thats what i was trying to do but it jhst got long and i couldnt be bothered

[thorn jay]

shouldn't take more than like 2/3s a page

[terse crystal]

I also have an approach. φ(m)φ(n)=φ(d)φ(l), and mn=dl, where d=gcd(m,n), l=lcm(m,n). So you can just show φ(d)φ(l)=φ(dl) for some d|l iff d=1, pretty straightforward

(Smith form tells us Z/mZ times Z/nZ is isomorphic to Z/dZ times Z/lZ)

[marble hinge]

Pinter strikes again with his left/right inconsistencies:

He uses right cosets, and f . g = f(g(x)), so this exercise just doesn’t work and h is not a homomorphism. The fix is either switch to f . g = g(f(x)), or to left cosets and xH |-> (ax)H. @rocky cloak - another typo? 🙂

[knotty badger]

why is this called "a sharper cayley theorem"

[marble hinge]

Not sure, it looks like he ends up proving something weaker actually, but I haven’t got to the end yet 🙂

[quiet pelican]

If you prove something like “the kernel is the core of H”, then it’s stronger
Otherwise, it’s neither weaker nor stronger, it’s just closely related

[marble hinge]

The rest is this, if anyone is curious:

[quiet pelican]

Yeah 3 proves what I was saying

4 then makes it explicit that this is a sharper Cayley

(That set in 3 is called the core of H)

[marble hinge]

So it is stronger because for a normal Cayley we can just take H = 1, but this also proves more isomorphisms?

I was somewhat confused initially, and superficially thought that it’s weaker because it shows some isomorphism only when there are conditions on H, (and normal Cayley shows unconditional isomorphism), but now I read it more carefully and see why it’s “sharper”

[quiet pelican]

Yes

[thorn jay]

Plainly, its sharper because there always is a case that implies cayley

(H = 1)

[tulip otter]

i have a (probably silly) question, why are all the coefficients of f_1 in A?

[agile burrow]

By definition, each of the coefficients of f(X) is divisible by cont(f)

So you can factor out cont(f) and still have a polynomial f_1 with coefficients in A

[tulip otter]

sure, but i mean the coefficients of f are in Frac(A). So why would the coefficients necessarily become elements of A after factoring out cont(f)

[agile burrow]

Oh I see. If the coefficients belong to Frac(A) and are not in A, then ord_p(f) will be (negative) the largest power of p dividing one of the coefficients. So once you factor that out, none of the coefficients have denominator divisible by p. Once you repeat for all irreducibles, all of the denominators are in fact units of A hence the coefficients lie in A itself

[marble hinge]

Yes, I figured that two messages above 😉

[tulip otter]

ohhhh right i see, tysm for the insight. have a great day/night!

[tulip otter]

so this means any polynomial f(X) which has a factorization in K[X] also has a factorization in A[X] right? K is the quotient field of A

because c_gc_h in A and g_1,h_1 in A[X] so one can rewrite f(X) as g_2(X)h_1(X) where g_2(X)=c_gc_hg_1(X)

[modest dune]

so this means any polynomial f(X) which has a factorization in K[X] also has a factorization in A[X] right? K is the quotient field of A
your statement here is too general

[tulip otter]

is it wrong tho

I mean too general implicitly means that its wrong ig

but where does it go wrong

ah right well its implicitly assumed

[modest dune]

i think im getting my terms mixed up

[tulip otter]

like its one of the hypothesese of gauss's lemma, and this is a corollary of gauss's lemma

[modest dune]

integrally closed is the name

[tulip otter]

yea its a factorial ring, or a unique factorization domain

[modest dune]

oh wait thats about the zeros being in the domain

ignore me

[tulip otter]

does this mean that every element has a factorization into irreducible elements which is unique up to a unit?

[tulip otter]

ohh i see

[modest dune]

yeah i was getting it mixed up with the zeros bering part of the domain instead of just the factorization being valid on the base ring

[tulip otter]

so it does hold for factorial rings

[tulip otter]

btw whats the difference between algebraic and integral closure

[modest dune]

i am not a good person to ask clearly since i just misread what you originally posted 😆

it looks like integral closures only care about monic polynomials?

[tulip otter]

at least you knew the term

[tulip otter]

i see

[modest dune]

i would search up what it actually means cuz im confused

[tardy hedge]

here comes Mr. Peace

[thorn jay]

yes, given a ring homomorphism f : A -> B (or in particular a subring A < B), an element b is integral over A if it is the root of a monic polynomial with coefficients in A

uhh I think it's iff A[b] is a finite A-module

One direction of that is fairly easy to see (as it will be generated by a finite subset of { 1, b, b^2, b^3, ... }

for the other uhhh I'm pretty sure you can use lattice theory for that? If you've got an algebraic closure operator and X = C(S) for some finite S, then for any T such that C(T) = X, there is a finite subset T0 of T with C(T0) = X

yeah because X will be a compact element and you can just take the cover X = \bigvee_{t in T} C(t)

[tulip otter]

of course you will know what the details mean @thorn jay

i didnt read the details and i will probably have to look up some terms in order to have a chance to understand the argument

but i sent the proof of lang anyways

[modest dune]

oh integral closure is relative to a field i think that is why i was getting vonfused by the terms

so you can say like Z is integrally closed in Q

[thorn jay]

for INT 1 => INT 2 is super weird

If X^n + a_n-1 X^n-1 + ... + a_0 = 0, then X^n (and thus all higher powers) are an element of the A-module generated by { X^n-1, X^n-2, ..., 1 }

so you're done

[thorn jay]

uhhh

it outsources the technical details of the argument

[modest dune]

it could at least tell you where the forward reference is to

instead of yeah we will prove that later

[thorn jay]

istg 😭

[tulip otter]

huh

[thorn jay]

Bosch's Algebraic geometry and Commutative algebra has this proof but like actually told properly

[tulip otter]

its just that i jumped to chapter 10 in order to check this

nothing related to this mentioned in chapter 4 (the chapter i am reading rn)

[thorn jay]

I still don't really think I "get" what integral extensions are

I guess they're generalisations of algebraic numbers and algebraic extensions of fields

[modest dune]

well i think your original question is answered by the fact that you do have a ufd assumption since youre using gauss' lemma and ufds are integrally closed in their fraction field

[tulip otter]

ohh so integral closure is related to whats going on i see

tho the proofs doesnt use integral closure as far as i can see, at least not explicitly

[modest dune]

apparently k[x,y]/(x^2−y^3) is a good bad ring to play around with

and Z(sqrt(5))

[thorn jay]

bad ring!

bad ring!

[tulip otter]

so in general given a ring A which is integrally closed in its quotient field K, if f(X) in A[X] has a factorization in K[X] then it also has a factorization in A[X]?

[modest dune]

x^2 + x - 1 = 0 seems to be a ring that factors in K(Z(sqrt(5)) but not in Z(sqrt(5))

i dont know if that requires UFD or if integrally closed alone is enough

cause integrally closed only tells us about the roots of monic polynomials

[velvet hull]

hell yeah determinant trick

[modest dune]

but clearly your text is only providing it for ufd rn

[velvet hull]

🔥 trick

[tulip otter]

well the answer to my question is yes and it follows from the corollary in the image (since A is a UFD), but the example of Z[sqrt(5)] and K(Z[sqrt(5)]) shows that it need not hold for integrally closed rings over their quotient fields

[modest dune]

Z(sqrt(5)) is not integrally closed, it just shows that it does not hold for rings that are not even integrally closed

we would need an example ring that is integrally closed but is not a ufd

i think i have some if you want to play around

k[w,x,y,z]/(wz - xy)

Z(sqrt(-5)) should be integrally closed...

[tulip otter]

wait how, i mean x^2-x+1 has coefficients in Z[sqrt(5)] and its roots are in K(Z[sqrt(5)]) no?

[modest dune]

integrally closed should mean that if the root is in the fraction field, its in itself

[tulip otter]

ah wait right, i was looking at the definition of an integral element over a ring lol

mb

[modest dune]

Z(sqrt(-5)) is apparently integrally closed though

oh i misread something else

i was wondering why it would be different than Z(sqrt(5))

[tulip otter]

x^2+sqrt(-5)x-3 doesnt have roots in Z[sqrt(-5)]

[tulip otter]

ohh i see

[tulip otter]

yea thats a nice trick

(idk the trick bc i didnt even read the page)

[modest dune]

now i am confused

[velvet hull]

the cool thing about the determinant trick is that it works as long as your ring of scalars is unital and commutative

because that's all you need for the adjugate matrix to exist

[tulip otter]

its always exercises labelled easy the wrong exercises

[modest dune]

your example doesnt work because the root is not in the fraction field

i think this text takes integrally closed to mean integrally closed over K()

[tulip otter]

ah wait right i misread the roots from wolfram lol

i misread a sqrt(7) as a 7

[modest dune]

yeah

well i verified the trick that breaks sqrt(5) doesnt work for -5

[tulip otter]

ohhh i see, I think I might check the proof soon. Tho its still a few chapters away lol, but I might jump to it and see it idk

[modest dune]

2x^2 + 2x + 3 should factor over the fraction field but not over the ring

if i did a bit of mathing correct

so integrally closed over the fraction field is not strong enough, need ufd

after i fix my typos at least

[tulip otter]

this one doesnt

[modest dune]

i corrected the sign

i made the same mistake when typing into wolfram alpha before typing into discord

[tulip otter]

the first one you gave also works, ie 2x^2+x-3 because its roots are 1 and -3/2

[modest dune]

that one better factor over Z then lol

[thorn jay]

the adjugate matrix is so scary

like why do you exist

[tulip otter]

wait right lmaoo

what

rip

[karmic moat]

That one humble paragraph in atiyah macdonald

[thorn jay]

if I close my eyes it won't hurt me

[tulip otter]

wait so this is wrong?

[modest dune]

its true for ufd

[tulip otter]

I mean Z is a ufd

[modest dune]

(x-1)(2x-3)

[tulip otter]

ah yea nvm i am stupid rn

forget what i said

alright so the conclusion is that it has to do with ufd but not with integral closure

[modest dune]

yeah

[tulip otter]

i see, tysm and sorry if i confused you midway

[modest dune]

i think i confused myself

by thinking that integral closure could generalize it

i guess that it probably would, for monic polynomials only...

[tulip otter]

well that was a fun thing to explore imo

ohh wait we forgot something important ig

[tulip otter]

which is that integral elements should be roots of monic polynomials, so this doesnt count?

[modest dune]

i guess it may still not work

because the factor might not be degree 1

not sure if you could say anything about it if it factors an irreducible quadratic in the fraction field

[tulip otter]

i see

that was fun

tysm again and have a great day/night

[crystal vale]

Okay so far i didn't use the fact E is finite Galois extension of F, so maybe I need this for define a functor Field_E -> O_G^op such that it will be inverse for given functor

[crystal vale]

So here can I extend any field morphism K -> L ( Field_E ^ F ) to automorphism E -> E ?

[rocky cloak]

You mainly need that for the being a bijection on objects part

[crystal vale]

I am trying to show they are both isomorphic

I am trying to define a functor Field_E -> O_G^op

So any K will go to { \phi \in G | \phi(k) = k for all k in K }

But to map a morphism f: K -> L i need some more things

Any hint?

[rocky cloak]

If E is normal you can extend maps K -> E to automorphisms of E

[crystal vale]

Sorry I don't know much field theory, what does it mean here E is normal?

[rocky cloak]

That polynomials with a root has all roots in E.

Something is Galois iff it is normal and seperable

[crystal vale]

I see

[rocky cloak]

I guess a more direct approach could be to notice that if you compose the embedding
K -> E with an automorphism of E, then it is fixed iff the automorphism fixes K. So the set of embeddings K -> E is your G-set

From here there is a fairly natural way to turn embeddings L -> E into embeddings K -> E given a map K -> L

[crystal vale]

embedding means injective mapping, right?

[rocky cloak]

Yeah. All homomorphisms here are injective, so I guess you can just say homomorphism

[crystal vale]

yes

[crystal vale]

yes, composition but i don't see the use of G set here

[rocky cloak]

What do you mean? G is the automorphisms of E

[crystal vale]

i mean yes we can turn L -> E into K -> E given a map K -> L, by composition

[crystal vale]

but i don't know its importance here

[rocky cloak]

I'm not sure I understand.

You wanted a functor from intermediate fields to this category of G-sets, and this is it

I guess you still have to prove these G-sets are transitive. But this is at least a nice description of the functor

[crystal vale]

but that category O_G^op how isomorphic to this category of G-sets?

[knotty badger]

Hm this might be related to what’s called a grothendieck construction

[crystal vale]

[knotty badger]

You’re essentially taking the category of elements of a functor G -> Set

To get the orbit category

[crystal vale]

elements of functor?

[delicate orchid]

ok no no no no no we're not doing this

[knotty badger]

This is analogous to how you can take a function X -> {0, 1} and get a subset out of it

[knotty badger]

Any set-valued functor has an associated category of elements

[delicate orchid]

actually it flat out spells it out in the image you posted lol

the correspondence H -> G/H yields a functor from O_G^op -> G-set

then, as jagr said, once you have that all of the G-sets arising from the functor he constructed are transitive, you can take the preimage under the functor O_G^op -> G-Set and you get back into O_G^op

[crystal vale]

man, it is interesting i don't know what's going on but still interested

[fading acorn]

i.e., brainrot

[knotty badger]

Welcome to category theory

[crystal vale]

how k(x) is isomorphic to subfield of k((x)), formal power series?

i mean how do i map 1/ g(x) ?

does for each such function there exists power series which converges to g?

[empty kernel]

Let L be a field and K a subfield of L. Prove that char(K)=char(L)

Obviously we know that $char(K) \leq char(L)$ but now I have to prove that $char(L) \leq char(K)$ to prove the equality but im stuck on that part

[cloud walrus]

you_are_me

[empty kernel]

i could use a little push in the right direction

[quiet pelican]

Prove something stronger
Char(K) divides Char(L)

Then ||what do you know about characteristics of fields?||

[rocky cloak]

I mean the objects are G-sets G/H, and the morphisms are G-equivariant maps....

[empty kernel]

oh ok i found the theorem abt char(F) having to be prime or 0
thanks

[limber tapir]

I think K((x)) are formal laurent series, which is the fraction field of the ring of formal power series k[[x]]

[empty kernel]

how does opne go about proving an element is transcendent?

[tribal moss]

That will depend essentially on what you already know about the element.

[empty kernel]

lets say pi

[tribal moss]

The proof there is a corollary of the Lindemann-Weierstrass theorem (which I don't know how to prove).

[empty kernel]

soo the proof is beyond the average algebra student capabilities?

[tribal moss]

Yes.

[knotty badger]

I think even proving pi is irrational is tricky

[empty kernel]

what abt e then?

[knotty badger]

It’s actually fairly easy to prove e is irrational

[empty kernel]

the proof abt irrationality of e is in our textboook

[empty kernel]

but prove that it is transcendent?

[knotty badger]

Fairly difficult but doable compared to pi I believe

[tribal moss]

https://en.wikipedia.org/wiki/Lindemann–Weierstrass_theorem contains a proof (or at least something that claims to be a proof, I haven't read it in detail).

[empty kernel]

yeah if its beyond the scope of an algerba course then i wont rlly bother with it

i already have enough stuff im confused abt lol

[tribal moss]

It's at least as much analysis as it's algebra.

[empty kernel]

im not much better at analysis lmao maybe i will take a look at it after the exams

[crystal vale]

yes

[limber tapir]

So... Polynomials live naturally inside the ring of formal power series, so it should make sense that fractions of polynomials live in the fraction field of the ring of formal power series

[delicate orchid]

yes this is true

cause localisation is exact :trollface:

[crystal vale]

[ Q( 2^1/4, √3 ) : Q ] = 8 because we can write [ Q( 2^1/4, √3 ) : Q ] = [Q(2^1/4, √3) : Q(2^1/4) ] [ Q(2^1/4) : Q ].

So it is enough to show x^2 - 3√2 is irreducible over Q(√2) = Q[√2], which i showed explicitly, because if a+b√2 is a root then we have a^2 + 2b^2 = 0 and 2ab = 3, so it gives already a = b = 0

I mean no solution can exist simultaneously

[quiet pelican]

That’s enough yeah

[next obsidian]

This isn’t quite enough

[delicate orchid]

subobjects are monomorphisms and exact functors preserve monomorphisms trollface

[next obsidian]

Yeah but when you localize at the complement of 0 from the polynomial ring you don’t know this adds enough denominators to actually give the fraction field of the overring

You have to localize the target even further and use that since the target is an ID it embeds into any localization of itself

[delicate orchid]

I'm claiming that a localisation of a subring is a subring of the localisation. Nothing more

[next obsidian]

Yeah yeah but you need to do a trivial amount more to get it to extend to fraction fields

It’s not worth even saying whatever

I just didn’t want someone to walk away thinking that if S = A\{0} then S^-1B = Frac(B)

Whateverrrrrrrr

[delicate orchid]

sure I can concede that. Perhaps I need a stronger tone indicator than ":trollface:"

[next obsidian]

I don’t even care dude just leave it alone bro I never even cared that much to begin with bro like seriously I’m done, I’m done with this I got over it ages ago bro

[delicate orchid]

Yeah well I literally never even cared chat chat I don't care chat no chat that's fine I don't care I literally don't care chat I don't I don't care chat

[thorn jay]

insert wojack sticker

[karmic moat]

chmonkey more like dont even caremonkey

nonchmonkeyalant

[next obsidian]

More like amanono

[crystal vale]

Corollary 1.26, let denote L be the set of all such elements which are algebraic over F, then proposition 1.23, F(L) is algebraic over F, i.e., F(L) \subset L, so L = F(L), hence L is a field, right?

[velvet hull]

For an alternative proof, if alpha and beta are algebraic over F then F(alpha, beta) contains all their sums and products but is contained in L. So L is closed under field operations

[crystal vale]

yes

x^4 -24x^2 + 4 is irreducible over Q, right?

and \sqrt5 + \sqrt 7 is a root

therefore Q(\sqrt 5 + \sqrt 7) = Q(\sqrt 5, \sqrt7)

[lethal cipher]

I could use a little help figuring out part 1.
Here's what I know: if we take the coefficients $mod p,$ $(Z/pZ)[x]$, we see that $f=x^k (x^{n-k}+\cdot+a_k)$ (modulo p).

However, my issue is whether or not this can actually tell us there exists a factorization in Z[x]

[cloud walrus]

dackid

[lethal cipher]

It also says a factor of degree at least k, but this'll pull out a factor of degree exactly k