groups-rings-fields

[cloud solar]

Oh thx

Last question. How do we know (k,p)=1 ?

[rustic crown]

p prime and k was a root of P (mod p) and 0 is not a root as i assumed p didn't divide the constant term of P

[rustic crown]

hewwo ryu

[lethal dune]

Hello

[rustic crown]

[lethal dune]

what have you been up to these days?

[celest furnace]

It should be, sigma is just an auto morphism of H and it doesn’t say anything about the automotohism group acting on itself by conjugation

[rustic crown]

not much new math >< hopefully that changes soon

[celest furnace]

[coral spindle]

I found this turn of phrase in Weibel very pleasing. I suppose that the groups Z/nZ could similarly be called rings ‘by birthright’ :)

[white oxide]

yeah, i guess i'm not seeing how this is "immediate" from these two theorems

[lethal dune]

cool

[white oxide]

oh wait LOL

i forgot that injecting into an abelian group gives an isomorphism with its image, so its abelian

how exactly does this theorem show that if K is abelian over k and K > E > k where E is an intermediate field, then E is abelian over k and K is abelian over E? to show that K is abelian over E, we can't take say F = E in the theorem because then KE = K

because we want to inject into Gal(K/k); hence to use the theorem, one of the factors in the compositum should be K

[tardy hedge]

Ohh the distributive law for rings being required makes it so the addition and multiplication arent just completely separate entities

The distributive law is like what’s necessary for the interpretation of 5*3 being addition of 5 three times?

True?

[coral spindle]

Yes

[tardy hedge]

Cool

The notion of divisibility doesn’t need “inverses” do they?

Cause i know ring of polynomials over a field has a division algorithm defined on it

But, that ring does not contain inverses

[tribal moss]

Division with remainder is different from the kind of division that relies on inverses.

Polynomial division is the former.

[tardy hedge]

Yeah i had no idea about this

I dont think my book has thus far discussed that

Im assuming at some point there is theory that defines the difference?

[tribal moss]

Basically, since polynomial division gives you both a quotient and a remainder, it uses that second output to deal with the fact that not everything is invertible.

[delicate orchid]

you might want to look into euclidean domains for this

[tribal moss]

Division with remainder: Given A and B, the operation produces x and y such that A = xB + y where y is in some appropriate sense "smaller" than B.
Ordinary division: Given A and B, the operation produces x such that A = xB.

[tardy hedge]

I see

For the “ordinary” division, there is not necessarily a notion of size involved?

[tribal moss]

For polynomial division, the "appropriate sense of smaller" is "lower degree".

[tribal moss]

Correct, that's just undoing whatever the multiplication operation does.

[tardy hedge]

I see i see

Can someone give me some insight into how integral domains fits into all of this?

I know something like integral domains are necessarily for some sort of division to take place ?

[coral spindle]

Integral domains have cancellation (i.e. ab = ac implies b = c or a=0) but not necessarily division.

However we can always convert an integral domain into a field by taking the ‘field of fractions’ in a similar way to how we can get Q from Z.

In this way, integral domains are exactly those things that could have division, if you like :)

[tardy hedge]

Ohh thats neat

[tardy hedge]

So i suppose thats the connection

[coral spindle]

If you are feeling motivated, you can try proving this by constructing the field of fractions yourself. It's not super hard per se, but it's a nice exercise where you can get your hands dirty and get creative with defining things.

[white oxide]

@tardy hedge have you seen the field of fractions yet? i remember when i was learning about it fraleigh had a whole entire chapter dedicated to the construction which was pretty funny

[delicate orchid]

good. You should have an entire chapter dedicated to it

localisations are swaggggg (exact)

[white oxide]

could i get a hint for showing that the degree is 20 please? I know that $E = \mathbb{Q}(\sqrt[5]{2}, \xi\sqrt[5]{2}, \xi^2\sqrt[5]{2}, \xi^3\sqrt[5]{2}, \xi^4\sqrt[5]{2})$ where $\xi = \cos(\frac{2\pi}{5}) + i\sin(\frac{2\pi}{5})$, and I tried to use the tower theorem, but I'm having a lot of trouble finding an irreducible polynomial of degree 4 to show that $[\mathbb{Q}(\sqrt[5]{2}, \xi\sqrt[5]{2}, \xi^2\sqrt[5]{2}, \xi^3\sqrt[5]{2})(\xi^4\sqrt[5]{2}) : \mathbb{Q}(\sqrt[5]{2}, \xi\sqrt[5]{2}, \xi^2\sqrt[5]{2}, \xi^3\sqrt[5]{2})] = 4$ (because there are 5 steps in the tower, removing a root from each extension)

[cloud walrus]

okeyokay

[tribal moss]

First note: Your irreducible deg-4 polynomials will have coefficents in Q[root5(2)].
Second note: It's what you get when you divide the known linear factor (x-root5(2)) out from x^5-2.

Note also that you don't need all those generators: root5 and xi·root5 is enough to generate all of E. Namely by dividing those to generators, you get xi itself, and then you can get other xi^n·root5 for free.

[white oxide]

oh right yeah

okay thanks, i'll consider those things

[blissful zinc]

Why do we use "integral domain" for "Anneau intègre" (french)?

Can I also say "integral ring"? Will it be understandable by everyone else? 🤔

[crystal turtle]

Use domain

Because then you'll start talking about (other adjective) + domain, e.g. Euclidean Domain, Principal Ideal Domain, ...

In English, the key part here is domain

[blissful zinc]

Ah, well, teachers always tell me that Anneau is "ring" in English.
Ty for your information

[next obsidian]

It is

[rocky cloak]

It's quite a weird naming convention. Like domain means "doesn't have zero-divisors", but then commutative domains are called integral domains instead of just commutative domains.

[white oxide]

oh, so xi*root5 will be a root of this polynomial which results from dividing x - root5(2) from x^5 - 2

[tribal moss]

It's a root of x^5 - 2, but not a root of x - root5(2) ...

[feral timber]

Can some one confirm this? Resolvent cubic for x^4+ax^2+b must be x^3-2ax^2+(a^2-4b)x, right?

[white oxide]

would this work to show that $\alpha$ is a primitive 9th root of unity? clearly $\alpha \neq 1$. now $\alpha^9 = -\alpha^6 - \alpha^3$; if $\alpha^3 = 1$, then we see that $\alpha^6 \neq 1$, which is a contradiction. since the only subgroups of order 9 of order 1 and 3 and $\alpha \neq 1$, $\alpha^3 \neq 1$, we see that $\alpha$ must be a primitive 9th root of unity

[cloud walrus]

okeyokay

[white oxide]

and there would be 6 possible homomorphisms since there are 6 generators to the 9th roots of unity?

[delicate orchid]

https://tenor.com/view/true-feel-that-thats-true-gif-7410050725709839510

[white oxide]

let's gooooo

wew approved ™

[dire siren]

yes, that's correct;
but this is actually true more generally: a homomorphism is uniquely determined by sigma(alpha) and we see that sigma(alpha) is also a root of that polynomial, so there are at most 6 homomorphisms;
conversely, it's not hard to prove that all those 6 canditates are indeed homomorphisms

I think the hint was given to be able to describe the homomorphisms easier and nicer;
like, they should be the ones that send alpha to alpha, alpha^2, alpha^4, alpha^5, alpha^7 and alpha^8, respectively

[tribal moss]

The intended argument for alpha being primitive ninth root of 1 was probably to note that x^9 - 1 = (x^6+x^3+1)(x^3-1), so the third roots of unity are roots of the right factor, and thus the only ones left to be roots of the left factor are the primitive ones.

[white oxide]

ohh ok

[white oxide]

yeah i think this makes sense

thanks

[fallen trout]

can i get some help with this?

basically i can’t see how zeta -> zeta^3 could generate the other two, and the same for 7 and 9. is it just me or do none of them work as generators??

*the other three i guess, as they do need to generate the identity too

[chilly radish]

It's because you got your exponent laws wrong. It's not (zeta)^(3^3), it's (zeta^3)^3=zeta^9

And the rest follows

[fallen trout]

lol i guess we're at that point in the semester

[fallen trout]

tysm!!! it's embarassing how much i was struggling with that lmao

[chilly radish]

Lmao nw

[round hull]

i think the additive set generated by (z_n)n≥1 works

we also take differences to be in the set

[chilly radish]

Wrong channel

[round hull]

what are they talking about

[chilly radish]

Only God knows

[round hull]

wdym?

i thought G was given

o

no i take
[{z_n}{n\geq 1} \cup {z_m - z_n}{1 \leq m < n}]

[cloud walrus]

quasi_semi_group

[round hull]

ya

the unknown unknown?

never mind that, it's not quite right

[round hull]

i take this

to be my set, and im arguing that this is additive but not atomic

it's contained within G because we have some telescoping behaviour

$z_m - z_n = (z_m - z_{m-1}) + (z_{m-1} - z_{m-2}) + \ldots + (z_{n+1} - z_n)$

[cloud walrus]

quasi_semi_group

[round hull]

then uh

well should i give you the rest of the argument or do you maybe want to think about it

the sets not atomic because of the decreasing nature

so we can construct any term as a sum of something lower, and their difference

$z_n = z_{n+1} + (z_n - z_{n+1})$

[cloud walrus]

quasi_semi_group

[round hull]

both of those summands are in this set

i guess i'm showing that no element is an atom

we're not done btw

we havent shown z_m - z_n isnt an atom

we can

why is z_n - z_{n+1} not an atom then

[round hull]

sorry the set i constructed isn't even additive

i have another argument but it's notationally cumbersome to talk about it

we can take the set
[H := \left{\sum_{s\in S}z_s - \sum_{s'\in S'}z_{s'} > 0 ;\Big|; S,S'\text{ finite subsets of }\mathbb{N}\right} ]
instead

[cloud walrus]

quasi_semi_group

[round hull]

yea i count the empty sum to be 0

but wait maybe i want S, S' parent set to be the part of the index whose differences are monotonically decreasing

ya

actually i think i'm making this overly complicated

no, i let them vary

otherwise H would just be one number

the condition is that the difference of their sums has to be positive

what i want to say is that for any two given finite sets $S,S'\subset \mathbb{N}$ such that $\sum S - \sum S' > 0$, i can raise both sets' indices one by one, until their difference in sums is less

[cloud walrus]

quasi_semi_group

[round hull]

because then i can take the difference between the original difference and the modified difference, which will be positive

and then add that to the modified difference

i know my phrasing is unnecessarily confusing so let me define some notation

[round hull]

because of this

let $S\subset\mathbb{N}$ be a finite set, let $n\in\mathbb{N}$, and define
[ S \uparrow n := {s + n \mid s\in S} ]

[cloud walrus]

quasi_semi_group

[round hull]

so im raising everybody by n

and this is a new finite set in N

[round hull]

i want to show that there is some $n\in\mathbb{N}$ where
[ 0 < \sum (S\uparrow n) - \sum (S'\uparrow n) < \sum S - \sum S' ]

[cloud walrus]

quasi_semi_group

[round hull]

like intuitively the further along we go, eventually the differences should get small

it's notation

i abused $\sum S$ as $\sum_{s\in S}z_s$ sorry

[cloud walrus]

quasi_semi_group

[round hull]

because then
\begin{gather*} \sum S - \sum S' = \left(\sum S + \sum(S'\uparrow n)\right) - \left(\sum S' + \sum(S\uparrow n)\right) \ + \sum (S\uparrow n) - \sum (S'\uparrow n)\end{gather*}
and both are positive

[cloud walrus]

quasi_semi_group

[round hull]

it would if i can show this

also i can't believe you're reading all this so quickly lol

because S is a subset of N

and we want to raise all elements in S by n

ya

so $\sum(S\uparrow n)$ means $\sum_{s\in S}z_{s+n}$ if that makes sense

[cloud walrus]

quasi_semi_group

[stark helm]

If I want to prove that D2n is isomorphic to 𝐷𝑛×ℤ/2ℤ, can I assume that there is a homomorphism phi:D2n->𝐷𝑛×ℤ/2ℤ and define phi(s)=(s,1), phi(r)=(r,1)?

𝐷𝑛×ℤ/2ℤ is external direct product?

[round hull]

we can try but i'm not sure what that would get us

for H

but here we're considering one pair such that that thing is positive

ok so i guess i have an argument?

consider $n_1$ such that $n_1 + \min S > \max S'$

[cloud walrus]

quasi_semi_group

[round hull]

ya

wait what i wasn't done though

i was still thinking about if i should be taking min or max in my arguments

i mean this n1 doesn't work right away

but it pushes $\sum S'\uparrow n_1$ to be really low

[cloud walrus]

quasi_semi_group

[round hull]

and the argument is that this push is bounded below by some constant based on S and S'

and then any such decreasing sequence in R^+ must terminate

i have to go but i'll think about this later

[stark helm]

Can anyone answer this question?

[quartz quiver]

Does anyone happen to know of a resource listing common counterexamples in ring and module theory? (The kind relevant to an undergraduate course)

[mighty kiln]

You should try showing it is indeed an isomorphism

Also it only works for odd n if Dn means dihedral group on n-gon

[stark helm]

Then just proving that phi is isomrorphism here right?

[stark helm]

It is dihedral group, so I assume phi(s)=(s,1), phi(r)=(r,1), do you think it is a correct format?

I am not sure if it is valid to assume homomorphism and define its function on my own

[mighty kiln]

Ye that looks good

But you still have to show it is an isomorphism

i.e. it is a homomorphism and it is bijective

[stark helm]

Here is a another question. If H is a normal subgroup of a finite group G, and G/H contains an element of order n, can I get the conclusion that there must exist an element g such that g is in H with order n?

[lusty marlin]

This is false. For an easy counterexample, consider the direct product of two cyclic groups with relatively prime orders.

[stark helm]

Like U(3) and U(5), then in this case, can we assume that G=U(5) while H=U(3)? but how can I determine elements of U(5)/U(3), where U(5)={1,2,3,4}, U(3)={1,2}, so does it mean there are only elements 1,2 in this quotient group?

[indigo nymph]

you may be interested in this? https://ringtheory.herokuapp.com/

[lusty marlin]

Firstly, U(5) is not a direct product of cyclic groups with relatively prime orders(not counting pairs involving the trivial group). Secondly, the elements of the quotient group G/H are not elements of G. They are cosets of H in G.

[stark helm]

Then is it correct to list U(15)/U(3)? U(15)=U(3) * U(5)

[lusty marlin]

U(3) and U(5) have orders 2 and 4 respectively, which are clearly not relatively prime

Why take examples from unit groups? Consider Z_21. This is the direct product of Z_3 and Z_7. The quotient group Z_21/Z_7 contains an element of order 3, but Z_7 does not. This is a counterexample to the claim that if G/H contains an element of order n, then H contains an element of order n.

[rocky cloak]

When you say "g is in H with order n" do you mean that g^n is in H? If so that's true. But if you mean that g is in H and has order n, then there isn't any immediate relation between H and G/H no.

[stark helm]

For Z21/Z7, does elements in this groups are {1,2,3,4,5,6}, 2*{1,2,3,4,5,6}, 3*{1,2,3,4,5,6}?

[lusty marlin]

You forgot the 0 in each set

Z_7 is the cyclic group of order 7

[stark helm]

Then I don't understand Z21/Z7 contains an element of order 3 means?

If elements are H={0,1,2,3,4,5,6} 2H and 3H

[lusty marlin]

Recall the group operation in quotient groups

[stark helm]

2H={0, 2, 4, 6, 8, 10, 12}, 3H={0, 3, 6. 9, 12, 15, 18}, do you think this is correct? ( I am not sure if I need to write it in mod 21 form)

You refer to me?

[stark helm]

still don't understand this expression. I still feel wonder about how to figure out the order of a coset?

[tribal moss]

That isn't even true. Dn×C2 doesn't have elements of large enough order for it to be D2n.

At least not if n is even.

[velvet steeple]

What would be an example of an operation with no neutral element (since this is a requirement for a group)?

[mighty kiln]

Addition on Z^+

== on {true, false}

[south patrol]

Any order without a minimal element, with the operation being "take the max"

[mighty kiln]

(== is commutative and associative somehow)

[mighty kiln]

[(X is an apple iff X is red) iff X is on the table] is the same as [(X is red iff X is on the table) iff X is an apple]

[velvet steeple]

oh, thanks!

[south patrol]

Any string of objects connected by equals signs is true iff all objects are equal

heh

[mighty kiln]

No this is XNOR

[south patrol]

Are they not the same thing

[mighty kiln]

(A iff B) iff C versus A iff B iff C

The latter means A iff B and B iff C

[south patrol]

Oh sorry ye

I get it

[mighty kiln]

This should return true iff number of false's is even

[late snow]

can someone remind me what this notation means in terms of groups

the weird bracket

I'm looking in our book and having trouble finding where it's introduced

[karmic moat]

group generated by A?

send full context please

[late snow]

ah that could be it

[karmic moat]

yeah

group generated by A

or i guess in this case, subgroup

[late snow]

thanks, was scratching my head at this for a bit trying to remember what that meant

[untold basalt]

Determine the minimal polynomial and the degree over $\mathbb{Q}$ of $1+i\sqrt{3}$

[cloud walrus]

Seagull

[untold basalt]

So what I'm looking for is a monic polynomial with rational coefficients that has 1+isqrt(3) as a root

And the degree will just be the degree of the polynomial

I can't find the polynomial

any ideas?

[south patrol]

well up to shifting this is sqrt(-3)

and hopefully clear enough what the min poly of that is

[tribal moss]

XOR is addition modulo 2, so it is associative. On the other hand (a NXOR b) = (a XOR true XOR b), so it inherits associativity from XOR.

[rocky cloak]

The easiest thing to do is probably just to consider
x = 1 + isqrt(3)
Then manipulate until you get a polynomial for x. So
x - 1 = isqrt(3)
x^2 - 2x + 1 = -3
x^2 - 2x + 4 = 0

[velvet steeple]

#help-18 message would probably also fit here

[late snow]

Is 0 included in G here? Because I don't think I can actually come up with a generator if that is the case.

I know 2 is a generator otherwise, working out which elements would be right now

[dim widget]

No the * means the multiplicative group

1/0 doesn’t exist so it’s not in the group

[late snow]

ah thank you, never saw it defined. will look it up

[dim widget]

it’s the group of all elements of the ring which have inverses

So anything coprime to 11

[rocky cloak]

So at least if A is not C_p^n then it contains CpxCq as a subgroup for two (possibly equal) primes.
Then the sequence with p-1 copies of (1,0) and q-1 copies of (0,1) and one copy of (-1,-1) has value strictly bigger than (p-1)/p + (q-1)/q >= 1.

[velvet steeple]

I assume they used the inverse element of (0 * r) being -(0 * r), right?

Assuming + is defined as normal addition, I guess it would be better to write (0 * r) + (-(0 * r))

Since - is not really defined here

Also they should've stated + is normal addition here?

[rocky cloak]

The usual definition of x-y is x+(-y) yes, where -y is the element such that y+(-y) = 0

[velvet steeple]

And they got that 0 = (0 * r) - (0 * r) from (R, +) being an abelian group, so + having an inverse, right?

[rocky cloak]

Yeah

[velvet steeple]

Thank you

[velvet steeple]

So we have \begin{align*} 0 &= (0 \cdot r) - (0 \cdot r) \qquad \text{inverse exists since $(R, +)$ abelian group} \ &= (0 + 0) \cdot r - (0 \cdot r) \qquad \text{$0 + 0 = 0$?}\ &= 0 \cdot r + 0 \cdot r - 0 \cdot r \qquad \text{inverses cancel}\ &= 0 \cdot r\end{align*}

How is the 0 + 0 = 0 step justified?

Do we take that as an axiom

[dire siren]

that's because 0 is the neutral element in (R,+)

[velvet steeple]

But that's not given? How do you know that + is addition

Maybe it's some other operation

[dire siren]

then how is 0 defined?

In a Ring (R,+,*), the elemnt 0 is defined as the neutral element of (R,+)

[crystal vale]

If x is algebraic over F then deg of x is finite?

[velvet steeple]

That implies there is no operation + that is associative, commutative, has an inverse for all elements, and has a different neutral element than 0

Can we really not find one that has a neutral element of 1, e.g.?

Like what if we define that + as multiplication

That would fulfill everything I listed?

[dire siren]

the problem is that you are thinking of 0 as a number, rather than a symbol

[velvet steeple]

Oh

[velvet steeple]

But when saying we want to prove 0 * r = r * 0 = 0 for all r in R, aren't we thinking of 0 as the number?

[dire siren]

no

[velvet steeple]

Ok, take the ring (A, +, * ) with + being defined as usual addition and * as usual multiplication (that works out fine, right? It's still a ring)

Then our 0 is 1

That'd mean that 0 * r = r * 0 = 0, where you think of the zeros as 1

That'd mean r = 1?

[dire siren]

do you mean \mathbb{R}?

[velvet steeple]

No

I renamed it to A now

[dire siren]

0 is the neutral element in additive notation

1 for multiplicative notation

[velvet steeple]

Yeah

[velvet steeple]

So 0 * r = r * 0 = 0 in that case would actually mean 1 * r = r * 1 = 1, right?

in multiplicaive notation

Since we define + as multiplication

[dire siren]

no, you can't mix them like that

if a property yields for 0, it is not necessarily true for 1 as well

[velvet steeple]

But we defined 0 as our neutral element of +, and now that it's multiplication, we can call that 1?

[rocky cloak]

I'm thinking you wanna do some kind of induction. So if you can show it for Cp. Then if you consider Cp^n, then it has Cp^n-1 as a subgroup. Split the set into elements in Cp^n-1 and others. From the Cp case you know that if you have p or more elements of order p^n, some subset must sum to an element of Cp^n-1. So by grouping appropriately you can reduce to the Cp^n-1 case, hence induction

[rocky cloak]

Yes

The degree of x is just the degree of its minimal polynomial. And polynomials have finite degrees

Yes

So Cp^n / Cp^n-1 = Cp, and elements of order p^n corresponds up elements of order p in the quotient. So if there were more than p such elements you would have sequence with value > p(1/p) = 1

So imagine you have sequence (a1, ..., am), with m>p. If you consider the partial sums
a1, a1+a2, a1+a2+a3, ...
How many are there, and what does that tell you?

Alright, so m>p, and how many possible values could they take?

[velvet steeple]

Can't we?

[rocky cloak]

Indeed, so what must we then conclude

That's right, there must be two that take the same value!

So what do you get if you subtract two such partial sums?

Sure, but like what kind of sum do you get

[sinful kite]

Those are two different symbols (they could be the same element in some cases but that's besides the point).

[rocky cloak]

Maybe we can think of an easy example, say
a1 + a2 and a1 + a2 + a3 + a4 are equal

What's their difference?

[crystal vale]

Okay

[rocky cloak]

Exactly

So the claim was that for any sequence if there is no subsequence that sums to 0, them
sum 1/o(ai) <= 1

o(ai) is always just p, so this is the same as saying such sequences have at most p elements

[velvet steeple]

Yeah, so? We just call our neutral element w.r.t. multiplication 1

[velvet steeple]

Then our theorem saying 0 * z = z * 0 = 0, where 0 is the neutral element w.r.t. + becomes 1 * z = z * 1 = 1, no?

[sinful kite]

no, the theorem is still talking about the identity of the additive group. You can apply the multiplication operator on any element of the ring, and in this case it is being applied to the identity of the additive group.

[rocky cloak]

So not quite, what we did was started with a sequence that has more than p elements, and showed it had a subset that summed to 0

No, that's not really relevant

[velvet steeple]

So + defined as regular addition you mean?

[rocky cloak]

We're already done

Because we've proven what we wanted to prove.

If a sequence doesn't have subsets that sum to 0, there must be at most p elements

[velvet steeple]

We have (A, +, * ) and I meant defining + to mean multiplication, aswell as * .

That'd work, it'd still be a ring

[sinful kite]

yep notationally 0 is always the additive identity, + is the additive operator, . is the mult operator, 1 is the mult identity. You can apply the mult operator on 0 but that doesn't make it 1, like you were suggesting, if I understood right?

[velvet steeple]

Oh, I meant defining + to mean regular multiplication, for which the neutral element is 1

But that 1 is still denoted as 0 then

[sinful kite]

well there are two operators in a ring and they are usually distinguished by + and . because the additive group is abelian.

[velvet steeple]

Yes, but we can define both operations as the same thing

Or,define the second as something else then, it doesn't matter

[sinful kite]

why would you do that?

[velvet steeple]

I'm just saying define the first (+) as multiplication

Then the neutral element of that is 1

But we'd still call it 0, right?

[sinful kite]

why?

[velvet steeple]

(Just to make what I ask in the next line clear)

[velvet steeple]

.

[sinful kite]

I didn't get what you are trying to do.

[velvet steeple]

We have $(A, +, \cdot)$ where $+$ and $\cdot$ are any two operations. \ Let $+$ be regular multiplication and $\cdot$ some arbitrary, fitting operation (that's not multiplication). \ Now, the neutral element of multiplication is $1$, right?\ But we defined our operation $+$ as multiplication, so its neutral element is $1$.

[rocky cloak]

So original claim
If sequence doesn't have a proper subset that sums to 0 then
Sum 1/o(ai) <= 1

In the special case Cp all elements have order p so this is the same as saying
Sum 1/p <= 1
Which is the same as saying there are at most p elements.

What we showed is that if there are more than p elements, then the sequence does have a proper subset that sums to 0. So we can conclude that the only sequence without such subsets have at most p elements, hence
Sum 1/o(ai) <= 1

[velvet steeple]

(A, +, * ) being a ring, I mean

As you said

[sinful kite]

Sure... the + and 0 are notational... they could be whatever.

[velvet steeple]

Yeah, that's what I wanted to know. Thanks!

[sinful kite]

yeah this is where I'm losing you. You don't wanna go to actual numbers since you wanna prove for the general definition of a ring. And, I don't see the benefit either. Like, why do you wanna consider the + to be actual multiplication?

[velvet steeple]

This is not about proving it right now, say we already proved this result now.

Think of this as trying to find an error in it to gain a deeper understanding, lol

[rocky cloak]

So if you take a sequence in Cp^n that sums to 0 and reduce modulo Cp^n-1 you get a sequence in Cp. All the elements in Cp^n-1 just become 0, so we can ignore them. The remaining elements can be grouped in subsets of size at most p, that sum to 0 (modulo Cp^n-1).
And since p/p^n <= 1/p^n-1 we can replace any such subset by its sum without reducing the value.

[velvet steeple]

So in this particular case, why does the theorem 0 * r = r * 0 = 0 still hold?

If we plug in our actual number neutral element of +, 1, we have a false statement

1 * r = r * 1 = 1

[sinful kite]

oh so you are making a concrete example? ok, but then why not just keep + as +? I guess your example is fine so long as you can distinguish between your additive 1 from your multiplicative 1... 😄

[velvet steeple]

Yeah

[sinful kite]

The statement isn't false because 1 is your 0, remember? 😄

[velvet steeple]

I'm defining + as multiplication because the theorem 0 * r = r * 0 = 0 only applies for the neutral element of +

[velvet steeple]

r is a number, right?

So we can switch to numbers from notation, no?

0 * r = r * 0 = 0 is notation

If we plug in our actual neutral element of + into 0, then we only have actual numbers

[rocky cloak]

That's a typo, supposed to say Cp^n

[sinful kite]

yep your 0 is now 1 like you said, so where's the problem?

[velvet steeple]

Yeah, then 1 * r = r * 1 = 1, where all of this is actual numbers

1 is an actual number

not notation

Right?

[sinful kite]

yep so where's the problem?

[velvet steeple]

Well r is an actual number too.
That means r = 1?

But r is just some random element in our set A

so why does it need to be 1

[sinful kite]

It's all poorly defined :D, like what is your multiplicative operator now? Your + is . ok but what is your .?

[delicate orchid]

$0_A = 1_{\mathbb{Z}}$

[cloud walrus]

Wew The Lads Tbh

[velvet steeple]

Yes

[delicate orchid]

ok what's the problem

[velvet steeple]

We have the theorem $0_A \cdot r = r \cdot 0_A = 0_A$ for all $r \in A$, right?

[delicate orchid]

yup

[velvet steeple]

So let's switch all this notational 0 to actual numbers, 1, as you stated above

Then we have 1 * r = r * 1 = 1

[delicate orchid]

yup

[velvet steeple]

Yeah. But r can be any arbitrary number in A

we are saying it has to be 1 with this?

[delicate orchid]

why on earth would we be saying that

[velvet steeple]

1 * r = r * 1 = 1

implies r = 1

no?

[delicate orchid]

right this is going to be one of those conversations isn't it

no

[velvet steeple]

Oh

So there's the mistake

[delicate orchid]

because we're not working in Z

[tardy hedge]

R subset of integral domain D. Show that if R is a ring under operation of D, it is a subring of D.

The main point here is that we need to show the identity of R is the same as the identity in D?

[delicate orchid]

I think we just need to start actually labelling things properly

[rocky cloak]

I can give an example maybe.
Say we're in C9 and we have a sequence
(1, 2, 3, 1, 4, 7)
Then we reduce mod 3 we get
(1, 2, 0, 1, 1, 1)
This we can group as {1, 2} and {1, 1, 1} for example, which corresponds to {1, 2} and {1, 4, 7}
Now o(1) = o(2) = o(4) = o(7) = 9, so
1/o(1) + 1/o(2) = 2/9 < 1/3 = 1/o(3)

So if we replace the original sequence by
(3, 3, 12)
Then the value has not decreased, so if we're looking for a maximal value sequence we might as well consider this one

[velvet steeple]

Why are we not working in Z? In 1 * r = r * 1 = 1, 1 is in Z, r is in Z

no?

Or which part is wrong

[rocky cloak]

Now maybe this was a bad example because 7+2 = 0 mod 9, but the point was about how you reduce the sequence to one in Cp^n-1

[delicate orchid]

1 is the additive identity in your ring. You cannot cancel the additive identity - that's essentially division by 0
hence you cannot conclude that r = 1

if we restrict r to be like, in A intersect Z

[sinful kite]

for the additive group inverses need to exist... it's all screwed up 😛

[delicate orchid]

I'm taking * to be multiplication in A as well, by the way

[delicate orchid]

hence why I'm assuming A is a larger ring that contains Z

Q, for instance

anyway this is an incredibly unnatural question to ask and will never come up again after this conversation

[dire siren]

any operations, but don't forget about the distributivity of multiplication over addition

[velvet steeple]

Oh, it's equivalent to the additive identity, right

[delicate orchid]

no it just is

not equivalent. It just is

[velvet steeple]

yeah

equal to

[south patrol]

i think this notation is slightly bleh too

[tardy hedge]

Im not sure if this is really easy or im missing something

God damn its bumpin in here today

[tardy hedge]

Can i just say element of R times identity in R = element of R times identity in D

Its an integral domain so use cancellation law

And the identities are the same

So R is a subring of D

[velvet steeple]

Alright thank you, now my original question was about

And that 0 + 0 = 0 step is justified by 0 being the neutral element

But why is it justified by that?

[delicate orchid]

wait why was this entire discussion about rings relevant to this at all?

[delicate orchid]

what is the definition of a neutral element

[velvet steeple]

It isn't, this was a tangent since after looking very close to the definition, I found this interesting, lol

[delicate orchid]

fairs

[velvet steeple]

Oh... a * e = e * a = a

And we think of one 0 as being a and the other as e

with + being *

[delicate orchid]

well no, both are 0s
0 is the netural element here

there is no e

but yes that's the idea

[languid axle]

Okay so getting the basic definitions down

A field is a set that is a group for both addition and multiplication plus is communitive plus is distributive

[tardy hedge]

Is this good

[languid axle]

And a ring is a set that is a group over addition but only needs to have an identity element for multiplication?

[tardy hedge]

Well its a group over addition

So it has an identity for additiin

And inverses

[languid axle]

Yeah but im saying the only thing it satisfied for multiplication is an identity element

[tardy hedge]

Yeah

Some elements may have multiplicative inverses and some may not

[delicate orchid]

what's S here

you wrote x,y in S

ah ok cool, it's like the lattices then

you mentioned the chain idea for monoids, I think that should generalise to G here (G is a semigroup)

[sinful kite]

ring may not have mult identity

[delicate orchid]

I'll go read that convo then come back

I definitely agree with quasi's approach

[sinful kite]

If you are already given that it is a ring under operation of D and it is a subset of D, is there anything left to prove?

[delicate orchid]

wait hold on yeah

[tardy hedge]

Just to show that the mult identity in R is same as in D

In my book it states defn of ring has mult identit

And it says a subring needs to have same mult identity

As the bigger ring

Q5

[delicate orchid]

the sequence is strictly decreasing, so for all n z_n cannot be written as the sum of z_{m_i} with each m_i > n. But, since z_n-z_{n+1} is in G, we always have the decomposition of z_n = z_{n+1}+(z_n-z_{n+1}). But this all holds true for z_{n+1} as well, so inductively we can always write z_n as the sum of two other elements in G - hence none of the z_n can be atomic. Hence the subsemigroup of G generated by the z_n and the z_n-z_{n+1} cannot be atomic

I think this works

[languid axle]

Then what does a ring need to have for mult?

[delicate orchid]

rings usually have a multiplicative identity but this differs based on authors - check your course notes

[round hull]

i got pinged

[delicate orchid]

up here boss

[languid axle]

But for it to exist it doesn't have to have it?

[round hull]

i don't think the last part is true

or there's still some work

well we discussed this too

[delicate orchid]

I've proven that they're not atomic in G

[round hull]

but what about their differences

[delicate orchid]

good point, put those in there as well

so yeah we've both ended up with the same H

[sinful kite]

ok my bad, maybe books define it differently. but mult assoc and distributivity is mandatory... identity maybe depending on book i guess

[round hull]

like why isn't z_n - z_{n+1} atomic

[delicate orchid]

who cares

it's atomic if and only if every element can be written as the sum of atoms

we've found elements that can't, so it isn't atomic

[languid axle]

Okay and it needs to be closed too I assume?

So the only thing it doesn't need is an identity?

[round hull]

why can't z_n be written as a sum of differences

[delicate orchid]

the entire point is that it can be?

[round hull]

i mean purely differences

[delicate orchid]

count the number of z_ns that appear in both sides

ah wait yes, we need that each of the z_is are different because it's strictly decreasing

[round hull]

but this really is possible for an actual sequence

like we can have z_1 = (z_2 - z_3) + (z_2 - z_3) + ... + (z_2 - z_3) by coincidence, for instance

and it could be that z_2 - z_3 are atoms (a priori)

[delicate orchid]

true - I'm thinking of these as free generators when they're not

but if we assume this is true for every z_i is there a contradiction

remember we only need a single z_i

[round hull]

ya idk

i want to say that any argument needs to rely on the archimedean nature of R_{\geq0}

[delicate orchid]

unfortunately I don't think so, take z_i = 1/2^i

[round hull]

[delicate orchid]

ok so lets look at these z_n-z_n+1s

they form another sequence in G with y_k = z_k-z_{k+1}

[tardy hedge]

For a given ring , a subset that is also ring need not have the same multiplicative identity as the original ring right?

And in that case it is not considered a subring?

[sinful kite]

yep closure too. i guess there are other optional things too like commutativity, division ring etc

[delicate orchid]

if this is strictly decreasing does my argument for z_n hold, y_k-y_{k+1} = z_k-z_{k+2} = (z_k-z_{k+1})+(z_{k+1}-z_{k+2}) so their differences are in G again
so my argument showing that each of the z_n holds for these y_n - I think? I can't actually see where I've used strictly decreasing
now we can inductively do this on the differences of the y_n, and then on the differences of those differences, etc.
So if any of the y_k are atomic, this implies that this induction stops at some point

[languid axle]

Yeah I'm just trying to get the most plain bagel simple definition rn since I have a good bit of familiarity with groups and fields and none whatsoever with rings

Just filling that little gap

I'm not in a formal abstract alg course RN it just happens a part of my discrete math class is dedicated to introductory study of groups

[sinful kite]

well I'm new to the subject too but afaik, this is from D&F and just 1) is the basic definition

[languid axle]

So like we've done a bit on regular groups, abelian groups, and group isomorphisms

And then now we are finishing the semester off with cryptography

[white oxide]

i don't follow

isn't that just a subring

in any case, if the ring is unital, then the subring must have a unit element as well

[round hull]

why is the next one strictly decreasing

[delicate orchid]

I don't see why it has to be

[round hull]

maybe you can choose a sequence whose differences strictly decreases

[delicate orchid]

that would imply data on the derviatives of the original sequence that we just don't have

my argument would require each higher derivative to be strictly negative for all n if I've indeed used the fact that the original sequence is strictly decreasing anywhere, which I must have done

so it's essentially a complete bust

[round hull]

it would be nice if someone gave a counterexample for non-archimedean uh

linearly ordered cancellative monoids

[delicate orchid]

I don't know what "archimedean" means

[round hull]

all nontrivial (sub)semigroups are unbounded

[delicate orchid]

oh I see

[rocky cloak]

Is it clear that archimeanness is relevant though?

[delicate orchid]

not to me personally

this seems almost lattice theoretic

and it's the fact that the set is bounded from below is the important part

I feel like there's something obvious i'm missing

[round hull]

never mind, it's complicated to extend subtraction in general

[tardy hedge]

Maybe the defn in my book is weird

[round hull]

i don't even know if there's a way to extend the order from M (cancellative) to K(M) (grothendieck completion)

[tardy hedge]

Like for example, it says that even tho the subset of additive identity (0 element) is a subset of every ring, its not a subring cause it doesnt have the same multiplicative identity …

So it says the zero ring is not a subrint

Subring

[round hull]

in particular m > n doesn't imply m - n is canonically inside M

[delicate orchid]

problem with taking grothendieck completions is that in a group we don't really have this notion of atoms

[round hull]

ik

[delicate orchid]

I feel like your H from earlier is the right answer though

assume towards a contradiction that all of the differences y_k := z_k-z_k+1 can be written as a sum of atoms

[tardy hedge]

Wikipedia article about subrings say same thing

[delicate orchid]

If your subring does not have the same identity we do not have an embedding of that subring back into the larger ring

ring morphisms preserve the identity

being able to embed back inside seems like a property we want subobjects to have!

[tardy hedge]

What is my book talking about then

It says this whole thing about this being in contrast to subgroups

[delicate orchid]

a subgroup automatically has the same identity by closure and inverses

let H < G h in H, then hh^-1 = e in H

[tardy hedge]

True

[velvet steeple]

Btw, my lecture notes go on to say that if $1$ is our neutral element, then $0 = 1$ and so from $r \cdot 0 = 0 \cdot r = 0$, it follows that $r = r \cdot 1 = r \cdot 0 = 0$ for $r \in R$ and so $R = {0}$.

[tardy hedge]

So for rings we need to state it has identity because rings dont always have inverses

[velvet steeple]

Isn't this very similar to what we discussed?

[tardy hedge]

@delicate orchid whats the purpose of the question i posted then

[delicate orchid]

correct, you can think of it as the multiplicative monoid of the subring has to be a submonoid of the multi. monoid of the larger ring

[velvet steeple]

But how do they follow that r = r * 1?

[delicate orchid]

huh, maybe you don't need full inverses and instead just need cancelabiltiy

[velvet steeple]

Isn't it basically what you said, division by zero

[tardy hedge]

Subset R of integral domain D is a ring. Show that R is a subring of D

[delicate orchid]

this is quite literally the definition

[tardy hedge]

That was the question

[delicate orchid]

95% of your problems can be answered by just using the definition of these things, kepe

[velvet steeple]

But 0 = 1

[delicate orchid]

ok?

[velvet steeple]

Oh wait

The problem we faced before was r * 1 = 1 * r

That was different

[delicate orchid]

so for integral domains we have that ab = ac if and only if a is 0 or b = c

[velvet steeple]

we couldn't have divided by 1, but here, we aren't doing that

[tardy hedge]

Yeah

Like it says R is already a ring. In our book, that means it has identity

So i was thinking the “work” of the question is showing its the same identity as D

[velvet steeple]

Well, where did you get this definition from? Is it literally a definition that in a ring, 1 * a = a?

[delicate orchid]

once again! that's the definition of a neutral element!

[tardy hedge]

The defn of subring for my book says it needs to have the same identity element as the bigger ring

[velvet steeple]

Oh, of a neutral element

[delicate orchid]

it is

[tardy hedge]

It is?

It is the work ?

[velvet steeple]

Thank you Wew

[delicate orchid]

I'm thinking we do some 1_Dr = r = 1_Rr kind of deal for r \neq 0

yeah there we go

[delicate orchid]

no worries, but try to work through these yourself a little bit more in the future lol

[tardy hedge]

@delicate orchid https://math.stackexchange.com/questions/426962/ring-and-subring-with-different-identities

I guess this is what i was talking about with “different identities”

[loud merlin]

Hi, I think I'm having a brain fart. Suppose we have a short exact sequence of $A[G]$-modules
$$ 0 \to M^\prime \to M \to M^{\prime \prime}$$

how can I show that the induced sequence obtained by taking the $G$-invariant is exact? I believe injectivity is trivial but can't show exactness at $M$

[cloud walrus]

Louis

[delicate orchid]

yes, and it can happen in non-integral domains, I don't really get why you're bringing this up lol. I guess some authors use a different (objectively incorrect, why would you make subrings not be subobjects categorically, that's disgusting) definition

[delicate orchid]

you can think of taking G-invariants as multiplying by 1/|G|\sum_{g in G} g iirc

that might help

[rocky cloak]

So suppose m in M^G maps to 0, then it's in M' and also in M^G hence in M'^G

[tardy hedge]

Ok yeah so the point of the hw question was to show that for integral domains any subset that is a ring must be same identity

I think i get it now anyway thanks

[delicate orchid]

oh yeah duh 0 is fixed

alternatively, this is the first homology lolololol

[loud merlin]

So that part doesn't need the exactness of the first sequence?

[loud merlin]

Yeah, the worst is that I am able to prove it for H^0 (by viewing it as a Hom) but eh.. I'm tired.

[rocky cloak]

What do you mean?

[delicate orchid]

you can view H^1 as a hom functor as well

[loud merlin]

Uh yeah I meant H^1

[delicate orchid]

ah, ok

wait

[loud merlin]

?

(And thx)

[delicate orchid]

taking G invariants is naturally isomorphic to Hom_A[G](A, -)? With A being the trivial A[G]-module

what's the natural isomorphism uhhh

[loud merlin]

That was indeed H^0 lol.

H^0 = G invariants

[rocky cloak]

a map A -> M is uniquely determined by where it maps 1

[delicate orchid]

yeah but what makes G-invariants special here

oh right

you send 1 to a G-invariant m and that preserves the (trivial) G-action

hence it's a homomorphism of A[G]-modules

[loud merlin]

(I think you can say rep we know what you're talking about)

[delicate orchid]

I'll say character if you're not careful

and Hom_{A[G]}(A, -) exact if and only if A is uhhh

I'm gonna get the wrong duel

pro...jective?

[rocky cloak]

Projective is indeed correct

[delicate orchid]

oh shit we don't know that A[G] is semisimple the argument isn't as simple as I thought

no, wait - it is

[rocky cloak]

What argument are we talking about here?

[delicate orchid]

that A is a projective A[G]-module

[rocky cloak]

It's not

[delicate orchid]

WHAT

[rocky cloak]

Usually not anyway

If it was then all group cohomology would be trivial

[delicate orchid]

b-b-b-but... \chi_Tr^0 is a constituent of \phi\overline{\phi} for every brauer character

anyway

oh right yeah fields aren't algebraically closed

hmmmm

oh we just need LEFT exactness

I didn't notice!

yeah we chill then

[rocky cloak]

Like M' being the kernel of M -> M'' is using exactness

[loud merlin]

Yeah you're right, sorry

Having a bad day and not being able to solve such a simple thing is something I don't recommend

[delicate orchid]

I wouldn't call this simple!

But yeah I've been there... a lot...

[loud merlin]

The proof is easy and left to the reader - Silverman (and my prof but I'm pretty sure he just copied this part of Silverman for my galois cohomology class)

[languid axle]

This proof is easy and left to the reader (Riemann's Hypothesis)

[pliant rivet]

How does one go about finding a polynomial with rational coefficients such that one of the below is a zero? I completely see how to do it when the number is a sum of no more than 1 irrational number or a root of such a sum, but I have no clue for either of the below

ie, I know and understand the procedure for something like 1+i or 3+sqrt(17) or roots of those

[coral spindle]

Take powers and check for relations

So for example, take a few powers of the first number and see what you get.

Once you learn the tower law you can figure out an upper bound for the number of powers you need.

[pliant rivet]

Oh, that's very helpful, thank you. I think I have the first one

[dire siren]

@pliant rivet denote the number by x and try to manipulate the relation;
for example, for the first one you have x=sqrt(2)+sqrt(3), so x^2=5+2sqrt(6), so x^2-5=2sqrt(6), and square this to get (x^2-5)^2=24, so x^4-10x^2+1=0

[velvet steeple]

an injection, not a bijection ofc, sry*

[chilly radish]

Also a bijecton for nonzero a

What's the inverse?

And no, x and y here are interchangeable so obviously there's no difference

[velvet steeple]

Yep, nonzero a, sry

[karmic moat]

where does the name "graded" come from for "graded rings?"

i understand the definition, just curious if there's some motivation behind calling it "graded"

[coral spindle]

Well it has grades; it has parts of various ranks

[karmic moat]

oh

wait yeah lol

i forgot about that definition of grade

oops

[delicate orchid]

gradiant

[crystal turtle]

It received a poor grade on a previous exam

[rustic crown]

should have studied instead of procrastinating like det ><

[crystal turtle]

Ryx should be studying more >.<

[rustic crown]

det too ><

but det always super tired when det get back home

[karmic moat]

was there a curve

[crystal turtle]

Nope!

[tardy hedge]

No way … is all this proposition saying is that subtituting a value into a polynomial yields a unique value!?

If so, that is so funny

Anyone?

[rustic crown]

no, not just a unique value.

it says substituing defines a map R[x] --> S

and this is a ring homomorphism. and is uniquely defined by extending R --> S by sending x to a given element s in s

when you're constructing ring homomorphisms from strange rings, it's annoying to check it preservers addition and multiplication, the more exotic your algebraic structures get, this verification would be annoying

its nicer to think in terms of basic maps and composing ring maps to get more such

[crystal turtle]

Another way to state it: maps R[x] --> S is correspond bijectively with pairs of a map R --> S and an element of S, and they all look like this "substitution" homomorphism

(something something R[x] represents Hom(R, -) x U(-) : CRing --> Set, blah blah obligatory degenerate category theorist comment)

[rustic crown]

(U(-) : R-alg --> Set )

[crystal turtle]

Me forget about algebras a lot

[rustic crown]

[tardy hedge]

I think your guys’s explanation went a bit over my head

[tardy hedge]

Oh ok i think I understand what ur saying here

[crystal turtle]

(this was not meant for you to understand)

[tardy hedge]

I know haha

Just meant the parts i am supposed to understand im a bit weary on

[rustic crown]

me realize me should have said,
U(-) forget about algebras a lot?

or do U(-) forget about everything?

[crystal turtle]

[delicate orchid]

Sludgebabble but correct

[crystal turtle]

Fuck you

[tribal moss]

the sequence is strictly decreasing, so for all n z_n cannot be written as the sum of z_{m_i} with each m_i > n.
That's doesn't sound true. Consider e.g. z_n = (golden ratio)^-n. Then we always have z_n = z_{n+1} + z_{n+2}.

[delicate orchid]

You’re right sorry, I meant m_i < n

I’m chopping off everything before z_n so I don’t have to worry about it - the differences of the z_m_i can still make an appearance however

Tbh I’ve forgotten most of the details about this problem and I didn’t know how to solve it anyway

[tribal moss]

I'm trying to find a clear statement of your argument, but there's been a lot of other things going on in the chat, so I'm not quite sure what has been superseded by later corrections ...

[delicate orchid]

Let’s just go from the top

[tribal moss]

Ah, okay. I hoped you had a solution -- I've gotten nowhere with it all day ...

[delicate orchid]

Nope, I have no clue sorry

Google was no help either

I feel like the way to go is an explicit construction and then showing that specific subsemigroup isn’t atomic

An existence proof just doesn’t seem to be feasible

It’s a strictly decreasing sequence that’s bounded below, so the differences between terms must go to 0 as n goes to infinity

So the sequence of differences has a strictly decreasing subsequence

More importantly, each difference is contained in a strictly decreasing subsequence. So can we reframe onto this subsequence and induct?

[tribal moss]

I've been toying with the idea of splitting into cases. By omitting some of the z's, I think we can reduce to instances where either all the deltas are rationally independent, or there's a finite number of them that they're all rational combinations of.
In the first case the deltas are atoms, and it feels possible to arrange things so none of the z's are rational combinations of them, so then we're done.
In the second my gut feeling is we might be able to prove the semigroup generated by the deltas is atomless. But the details elude me.

[delicate orchid]

I’m not seeing how rational independence implies atomicness

They’d definitely be atomic among themselves but who knows what dwells in G

[tribal moss]

If we look at one delta, it is the smallest number in the semigroup whose decomposition into rational combinations of the deltas (unique because I assume independence) involves that delta itself.

So it cannot be the sum of two numbers whose decompositions don't.

[delicate orchid]

Yeah, by independence. I got you now

Like vectors almost

[tribal moss]

Exactly vectors over Q, in fact -- I'm considering the semigroup as a subset of the subspace generated by the deltas, defined by "the vectors whose coordinates are nonnegative integers, except 0 is not there".

[delicate orchid]

It’s actually closer to characters in the semiring of reps but you get me

[delicate orchid]

I was just thinking this earlier!Except I was thinking about the (“positive chamber”? I guess) of the R-vector space generated by the z_ns, your formulation is much more to the point

[tribal moss]

The troublesome case is when there are relations between them, because then the map from the nice integral half-lattice is not an injection anymore, and then it becomes hard to keep track of which ones are atoms and what the integer span of those atoms is.

[delicate orchid]

As long as one of them doesn’t appear in a relator then we’re still ok I think

[tribal moss]

Hm, might be. But is there any particular reason we might hope for that?

[delicate orchid]

Because I believe it to be so with all my heart

[delicate orchid]

Does this help

Apologises if my quips aren’t very astute I’m quite distracted while posting atm

[tribal moss]

Almost certainly we'll need to use that. By skipping some of the z's we can even arrange it such that each delta is smaller than half the previous one -- or however fast we want them to go to zero.

[delicate orchid]

I’m also very interested to know what evil mastermind set this problem up

This can’t have been for homework right?

Putnam
Right so there’s some bullshit reframing we’re all missing

Yeah that’s what I meant by bullshit reframing

It’s the putnam, they’re designed to be as annoyingly tough as possible

However this does allow us to rethink our approach

[tribal moss]

I'll give up for now and go to bed. Ping me if you figure it out. :-)

[delicate orchid]

Do not expect a ping!

[tribal moss]

Sure, keep it to yourself, then.

[delicate orchid]

I appreciate the confidence

I for one do. It’s just the only reasonable answer

wait didn’t I show that the z_ns cannot be written as a sum of the other ones or was that proof bogus

I have extremely poor short term memory you’ll need to help me out here

Right yeah. And we don’t know why z_n-z_n+1 can’t be written as a sum of the other geezers

Ok I just completely give up

I tried iterating H down to smaller subsemigroups by taking a decomposition of a difference d and the removing those factors from the generating set but that then means some of the z are now possibly atomic

There’s no structure to work with here and I’m tired

Find the solutions and post them here

[tardy hedge]

Can i have help for if u is a unit in R and a nilpotent, then u-a is a unit

[cobalt heath]

What have you tried?

[delicate orchid]

there's a ring theory way of proving this and a nonsense way of proving this which one you want

[round hull]

this approach doesn't work

the positive differences need not be in G

on the bright side

if this is a putnam at least the solution is elementary

[tardy hedge]

Ring theory

The book said as a hint start with u = 1

If its a^2=0 then (1-a)(1+a) = 1

Is it just continuing some pattern like that?

[delicate orchid]

right so your book is hinting at the BULLSHIT method

[tardy hedge]

😦

Bad book

[delicate orchid]

but yes, that's a good idea

[round hull]

how can a unit be nilpotent

oh

[delicate orchid]

see if you can find the inverse when a^3 = 0

[round hull]

that's confusing

[delicate orchid]

trivial ring

[round hull]

i meant for their question

[delicate orchid]

trivial ring

[tardy hedge]

(1+a)(1+a^2)

[delicate orchid]

I don't think so - we'll need negatives somwhere

to cancel out the a's and the a^2s

[tardy hedge]

(1-a)(1+a)(1+a^2) is (1-a^2)(1+a^2)and then its (1-a^4) = 1 no?

[round hull]

ya

extrapolate from there

[delicate orchid]

oh sorry yeah

I was doing 1+a not 1-a

this is correct

wait, I said a^3 = 0 not a^4

[round hull]

same thing

[delicate orchid]

I never understood why this proof was the go to one for textbooks. The maximal ideal argument is cleaner

[tardy hedge]

The inverse of 1-a when a^n = 0 is (1+a)(1+a^2)(1+a^4)…(1+a^(n/2))

[delicate orchid]

and for n odd?

[round hull]

it relies on zorn's

[delicate orchid]

for what, the existence of maximal ideals?

[round hull]

ya

[tardy hedge]

Is it just the same thing but do it till (1+a^((n+1)/2))

[delicate orchid]

a ring theory without maximal ideals is not one I want to be in

[round hull]

yes

actually i think n/2 is already quite large

the ceil of log_2 n probably also works

[delicate orchid]

I don't buy that for a second
we'd then have two different units with equal inverses

[round hull]

sorry i misread

[tardy hedge]

So for the general u-a case, is it just a similar kind of thing

Modified version of it

[delicate orchid]

you can turn u-a into u(1-u^-1a)

[round hull]

actually we can just take the infinite product

since a^m eventually vanishes

[delicate orchid]

I'll tell you how this goes canonically once it won't spoil the entire thing

[tardy hedge]

If u show (1-u^-1a) is a unit then multiplication of units is still a unit?

[delicate orchid]

yur

[tardy hedge]

For a^2=0 the inverse of (1-u^-1a) is (1+u^-1a)

Then Etc its same kind of thing

Probably

[round hull]

exactly

because u^{-1}a is also nilpotent

[tardy hedge]

Yeah

Thanks

[round hull]

yus

wait wew mods tbh wanted to say the canonical version

[tardy hedge]

Ok go

[delicate orchid]

take the series expansion of 1/(u-x) via the binomial formula

since x is nilpotent this terminates in finitely many terms and is thus a polynomial

[round hull]

geometric?

[delicate orchid]

even better actually

good observation

$\sum_{i=1}^{\infty} ua^i = \sum_{i=1}^{n} ua^i = \frac{u}{1-a}$

[cloud walrus]

Wew The Lads Tbh

[round hull]

n-1?

[delicate orchid]

excuse the poor latex there

[delicate orchid]

sure, point is it's finite

[round hull]

ngl i was 40% expecting "assume u + a is not a unit. then u + a is contained in some maximal ideal M such that R/M..."

[delicate orchid]

nah, that's the correct way of doing it though

not only do I accept the axiom of choice. I force it upon those around me

[round hull]

thus giving them no choice

[south patrol]

Lol I thought the geometric series is the standard way

[sinful holly]

Just those four elements? Well, it's a ring without identity. It's just the eM_2(F_2) where e is the fourth matrix on your list. That's a ring without identity and e e_{1,2} = e_{1,2} and e_{1,2} e= 0.

[long obsidian]

How important is it for a free object F on a set X of a concrete category to have the map of sets X->F be injective?

I'm reading hungerford and he doesn't seem to say it's injective but I remember it to be injective. Wiki says it's the canonical injection

[dreamy schooner]

All rings of order 4 are commutative. As a general result, all rings with order equal to a squared prime are commutative: Ring of order p2 is commutative? is that true

[sinful holly]

Rings of order 4 with identity are, yes. Your ring above doesn't have identity, but some people consider that a ring.

[dreamy schooner]

all right that was causing some confusions thanks

[broken quartz]

guys my lecturer's (hes a post doc) mentor taught us a lesson today

hes actually da goat

ic now why my guy is a fields medalist

[dreamy schooner]

is that true ?

[mighty kiln]

n has this property for Sm iff its smallest prime divisor is greater than m I believe

[dreamy schooner]

so its false
could you provide me with a hint how to disprove it

[ebon pine]

If x^p is an isomorphism from G to G (ie an automorphism on G) then is it necessary than G must be abelian

This holds for p=2, 3 easily but I couldn't figure it out for any higher terms

Couldn't find counterexample either

[chilly ocean]

What's G_1G_2 in group theory

[round hull]

only a little related but there's some result that says that if x ↦ x^n is a homomorphism for some consecutive n, then G is abelian

[round hull]

quaternions Q_8 has exp(Q_8) = 4. so x ↦ x^5 is the identity map

[ebon pine]

Bruh, I forgot about them. Thanks

[ebon pine]

Yeah if that holds for 3 consecutive integers then G is abelian

[mighty kiln]

Do you know the criterion for two elements of Sm to be conjugates

That turns it into a combinatorial-ish problem

[dreamy schooner]

iff they have the same cycle type
so every permutation fails
as an example if n=9 then the permutation (1,2,3) fails
like wise for every other n right ?

[mighty kiln]

👍

[dreamy schooner]

but now i realise even without knowing that i coulld have answered this
just by realising that two conjugates have the same classes
and the class of the identity element contains just the identity element

[dreamy schooner]

thanks

[cobalt heath]

This appeared in my HW (not the problem itself):
Suppose I is an ideal that for all a \in I, 1 + a is left-invertible (A (1 + a) = A). Then, is 1 + a right-invertible as well ((1 + a) A = A) ?

I get that not all left-invertible elts are right-invertible, but I don't see the examples for the ones where the entire ideal has the property.

[crystal vale]

A?

[cobalt heath]

Ah, A is a ring

[crystal vale]

Can you please share question pic

[cobalt heath]

What, no

I am not asking for the answers of my homework

[crystal vale]

I am not getting the question properly therefore I asked

[cobalt heath]

Ugh, I mean all these are typical notations

Well I might have had to specify that I is an ideal of A, but that's it

Ah well, left-ideal, to be specific

[rustic crown]

take a look at the jacobson radical, there are many equivalent definitions

rad(A) is defined as the intersection of all maximal-left ideals. but this happens to be a two-sided ideal

and has the nice description that
rad(A)
= {a in A : 1 - xay is invertible for all x, y in A}
= {a in A : 1 - xa has a left-inverse for all x in A}

[cobalt heath]

Yep, and.. that was the homework

Oh

I should study towards "1 - xay is invertible" part then

Thank you!!

[rustic crown]

[rocky cloak]

So the trick here is to consider the left inverse of 1+a and show that it also has a left inverse

And the way you would to that is ||to show that the left inverse also can be written as 1 + b for some b in I||

[cobalt heath]

Oh thanks, that is more straightforward

Somehow I came to dislike simple algebraic manipulations..

[dire siren]

btw, I'm pretty sure you also need u and a to commute

[rustic crown]

you really need commutativity. else consider R = Mat_{2x2}(k)
and

u = [1 1]
    [0 1]
a = [0 0]
    [1 0]

u+a isn't invertible.

[rustic crown]

another way it is usually defined is as the intersection of all ann(M) where M varies over simple left-A-modules. to see this is the same,
notice that for each non-zero m in M, ann(m) is a maximal left-ideal, and to see that the intersection is two-sided, check that ann(M) is actually two-sided.
the nice thing about the description
rad(A) = {a in A : 1 - xay is invertible for all x, y in A}
is that it shows this is a purely symmetric notion, all other had some "left" in it. in particular intersection of all maximal right ideals is also this.

[icy bear]

why are conjugacy classes defined like they are

I kiiinda get it, but not really

like a mix of right and left and right cosets, which are analogous to modulo

[coral spindle]

Conjugacy classes are orbits under the conjugation action

And orbits are a very natural definition

[icy bear]

what's an orbit ;-;

[delicate orchid]

lots of group invariants either only care about things up to conjugacy, or are themselves are all conjugate

in a sense, two elements being conjugate basically means we can consider them as "the same" in a weaker sense than straight up equality

[icy bear]

well yeah it's an equivalence class

[delicate orchid]

like two matrices are similar if and only if they're conjugate in GL(n, R)

[icy bear]

oh yeah matrix similarity

actually with that it makes sense now

[delicate orchid]

yeah it's a good example

[icy bear]

it's doing x, doing a and undoing x

it's like a but in a different context

[delicate orchid]

it just is that

[icy bear]

just like matrix similarity is doing the same transformation but in a different coordinate system and then coming back

that makes sense

[delicate orchid]

yeah, the vibes are the same for general groups

just no coordinates obvs

you'll start seeing a lot more use out of conjugacy when you study Symmetric groups/the Cayley embedding and group actions

[icy bear]

hm

[teal vessel]

Today in "trying things that you would know don't work if you just remembered things":
Trying to show that SL(2,F3) is not isomorphic to S4. "well all symmetric groups contain all smaller symmetric groups, so I'll just check for S3"

reminder text: all non-abelian groups of order 6 are isomorphic to S3

[velvet steeple]

To prove the binomial theorem, my lecture notes let $R$ be a ring with a multiplicative neutral element an $ab = ba$. then, they define $mr \coloneqq 0$ if $m = 0$ and $mr \coloneqq r + \dots + r$ with $m$ summands if $m \geq 1$. Why is this necessary?

[delicate orchid]

m is the characteristic of the ring

it isn't needed to prove the binomial theorem

[mighty kiln]

Integers aren't ring elements, so mr has to be separately defined

mr is shorthand for repeated addition of m ring elements

[delicate orchid]

oh right good point, yeah - which part are you actually asking about

[velvet steeple]

What do you mean it's not necessary, they go on to prove it right after this

[velvet steeple]

Why are integers not ring elements?

[teal vessel]

is the ring a ring of integers?

[mighty kiln]

Say Z/2Z

[velvet steeple]

No, it's just some ring R

[teal vessel]

then you can't be assured that the integers are ring elements in that ring, so we have to define it separately to explicitly allow access

[velvet steeple]

Oh, you are saying that'd be a ring but wouldn't include all integers

[mighty kiln]

Ye

[velvet steeple]

Then why not say "ring R with Z \subseteq R"

[teal vessel]

because what if you want a non-real ring?

[mighty kiln]

(Though you can map the integers into any ring R by letting the integer m be the multiplicative identity added to itself m times, and this yields the same result)

[delicate orchid]

the binomial theorem holds automatically from commutivity and distrubutivity, it does not depend on characteristic

[mighty kiln]

Where do you see characteristic

[mighty kiln]

There's no mention of characteristic, they prolly just misread

[velvet steeple]

a ring having complex numbers?

[teal vessel]

or a ring not having numbers at all

remember a ring is a set with two operations we call addition and multiplication (with particular restrictions). They needn't be specifically traditional integral addition and multiplication

[tardy hedge]

Oh yeah the question did say R was a commutative ring

[dire siren]

@velvet steeple just think about a ring of matrices, say 2x2 matrices with real entries
does this ring contain integers? No. Yet it is convenient to define nA as A+A+...+A (n times).

the same idea for a general ring

[velvet steeple]

Oh

So they define their multiplication operation with this

[dire siren]

without that notation we would have to write stuff like
(a+b)^3=a^3+a^2b+a^2b+a^2b+ab^2+ab^2+ab^2+b^3 (assuming ab=ba), which is not convenient

[velvet steeple]

Oh

[teal vessel]

more abstract. They define a collected notion of quantity with this, in other words, instead of Δ+Δ+Δ (where triangle is not a number in any respect), we can simply say "3Δ" to represent collecting all those instances into a single collection to save space and recognize the sameness of the objects. The multiplication could be that triangle times pentagon is now nonagon, or something (being very loose with this), but defining a coefficient here is distinct from the ring multiplication.

[velvet steeple]

Thanks! But now what does this have to do with "then you can't be assured that the integers are ring elements in that ring, so we have to define it separately to explicitly allow access"

[chilly radish]

nr where n is an integer is defined to just be repeated addition, it's not the same as saying n is an element of the ring and multiplying them together

If you know what a module is, you can think of this as being a Z-module, and this is the definition of the scalar multiplication

[teal vessel]

because the choose function and the exponents are integers in order to make for easy collection of terms and show the underlying structure. a and b are not necessarily numbers, but they exhibit the same "behavior" using this form of collected addition.

[velvet steeple]

Oh, yep

[teal vessel]

3a^2 in this ring would be (a∙a)+(a∙a)+(a∙a) if we were to be perfectly explicit with our notation and only use ring elements. This is, however, not convenient, so we find it nicer to count the number of multiplications and additions using integers to keep track. These integers serve no purpose in the ring itself, and are simply there to make it sensible to us as we seek to understand the structure.

[velvet steeple]

Oh.

[velvet steeple]

Wait, but we haven't defined this with exponents yet, then

[teal vessel]

a ring is a group

it's baked in with the group definition

[velvet steeple]

By saying mr := 0 if m = 0 and mr:= r + ... + r m times if m >= 1 we only define the "collection"

But not exponents yet

right?

We have to do that

Saying a^n*a := a^(n + 1) and a := a^1 maybe?

[teal vessel]

technically, yes, but in group theory, it is already defined clearly that x^n is the collection of n copies of x which have been joined with the group operation

[velvet steeple]

So it's defined for exponents but not for factors?

[dire siren]

well, the ring with multiplication is not a group, but a monoid (assuming the ring has identity)

[teal vessel]

more commonly, yeah. We define things like $D_8 \coloneqq \langle s,r | s^2=r^4=1, sr=r^{-1}s \rangle$ with the explicit assumption that these exponents apply repeated operation

[cloud walrus]

GoldenPhoenix

[teal vessel]

point being that exponents are ubiquitous, but coefficients are not always as obvious, so it's better to be explicit

[velvet steeple]

Alright, thanks.

Yeah, my lecture notes don't define it for exponents

Only for coefficients

[teal vessel]

if you play MtG, it's reminder text XD

[dire siren]

in general, say you have an associative operation denoted multiplicatively, then x^n is defined with respect to this operation to be x*x*...*x (here x appear n times);
if the operation is denoted additively, then x^n is replaced by nx

[teal vessel]

I'm really stumped on showing these two groups being non-isomorphic tbh

[delicate orchid]

different subgroups of order 8

you are doing SL(2, 3) and S_4 right

[teal vessel]

yeah but for some reason my brain really doesn't want to let me find them

I know the SL subgroup is isomorphic to the quaternion group of order 8, but for some reason the S4 subgroup is escaping me

wait a second....

[delicate orchid]

hint: a square has FOUR corners

[teal vessel]

D8

dangit

that took too long

[delicate orchid]

there's two non-abelian groups of order 8 so there's not much choice here

[teal vessel]

that would have been nice to know

probably not all that hard to prove tbh

[delicate orchid]

it's true for all primes

[teal vessel]

...8 isn't prime?

[delicate orchid]

2 is though

[teal vessel]

I'll get there eventually, I'm sure

[delicate orchid]

the structure is weird for p = 2 though but in general it looks like the special upper triangular 3x3 matrices over F_p and some weirdo C_{p^2} \ltimes C_p

which I think you can realise as a specific subgroup of Aff_1(F_p)

this is for groups of order p^3

[teal vessel]

oh, dumb thing to notice: SL(n,F2)=GL(n,F2) for all n

I think, pretty sure

certainly is for n=2,3

[rocky cloak]

Not so many choices for what the determinant can be 🙂

[teal vessel]

yeh

[white oxide]

can somebody explain to me how the extension being degree 2 implies that i is in Q(a)?

[rustic crown]

what will [Q(i) : Q(a) n Q(i)] be?

[white oxide]

it can't be one since otherwise Q(a) n Q(i) = Q(i) which is impossible since a is real?

[rocky cloak]

In the sequence p-1 of the elements should be (1,0), q-1 of them should be (0,1) and one of them should be (-1, -1)

[white oxide]

oh wait and that would imply that Q(a) is contained in Q(i)

[rustic crown]

other wya

[white oxide]

ah right

[rustic crown]

bigger n smaller = smaller

[white oxide]

ye i always get confused

about intersections and inclusions

[white oxide]

oh so then, [Q(i) : Q(a) n Q(i)][Q(a) n Q(i): Q] = [Q(i): Q], which is two, but [Q(i) : Q(a) n Q(i)] = 2 which we have shown must hold, then [Q(a) n Q(i): Q] = 1

[rustic crown]

ye sure, (i was thinking the other way but same thing lol. if Q(a)nQ(i) has degree 2, then Q(a)nQ(i)=Q(i) which is bad).

[white oxide]

yee ok makes sense

thanks a lot!

[tardy hedge]

Hey guys

A set A with a horizontal line above

What does it mean

[white oxide]

the complement?

[tardy hedge]

Q12

Is it complement? Ok thanks

[rustic crown]

why not just write (A u B) \ (A n B) so much easier to read >.<

[delicate orchid]

A nice way I visualised this problem is by noticing that Q(a) \cap R = Q(a), but clearly but Q(i) \cap R is just Q, so Q(i) \cap Q(a) = Q(i) \cap Q(a) \cap R = Q(i) \cap R \cap Q(a) = Q \cap Q(a) = Q

which I think is just what det was saying but not in galois language

so I could actually understand it

[white oxide]

wew cmon ur galois theory can't be that shit right

[delicate orchid]

I have never done a formal course

[white oxide]

that's a lot of caps, translating into latex currently

[delicate orchid]

I understand we're using the tower property but it's not that intuitive to me

[rustic crown]

why so

[delicate orchid]

ok let me elaborate - the tower property is VERY intuitive