groups-rings-fields

[fading basin]

Well done

[crystal vale]

Yes, thank you

[ebon pine]

I am stuck in this question - Let R be a unital commutative ring then prove that there is exactly one integral domain R such that the map f(a) = a^6 is a ring homomorphism.

Since f maps 1 to 1 and f(ab) = f(a) f(b) the only thing that remains is f(a + b) = f(a) + f(b)

ie a^6 + b^6 = (a+b)^6

[dull ginkgo]

are there two different integral domains or

Or is it an endomorphism

[ebon pine]

No f is an endomorphism

[dull ginkgo]

uhhh I haven’t done this exercise before but my immediate inclination is to try forcing characteristic

[ebon pine]

As in showing n = 0 for for natural number?

[dull ginkgo]

(1 + 1)^6 = 2^6 = 2 for the map of Z into this ring

thus (2^6 - 2) = 2(2^5 - 1) = 0

it’s an integral domain

[ebon pine]

Yeah so 2 = 0 or 2^5 = 1

For first I think the only option is R must be Z/2Z

But not sure what to do with the other case

[dull ginkgo]

(2^5 - 1) means it is of characteristic 31

If it’s of characteristic 2 then (x + y)^2 = x^2 + y^2

(x + y)^6 = ((x + y)^2)^3 = (x^2 + y^2)^3

Actually

There’s an easier way

[rocky cloak]

You can just continue, like if 3^6 = 3, then 3(3^5 - 1) = 0

[dull ginkgo]

Yeah

[south patrol]

ig the first like 5 words of the question are redundant

lol

[dull ginkgo]

I thought he meant it was a function between integral domains

[south patrol]

I'm pretty sure there are it more than one integral domain but maybe I am silly

don't ||Z/2Z and Z/3Z|| both work

[rocky cloak]

||2^6 = 1 in Z/3Z||

[dull ginkgo]

I’m pretty sure the remaining power might be of issue

[south patrol]

oh yeah sorry

i am silly LOL

[ebon pine]

Well that's how it was given in the question...

[south patrol]

Sure

[dull ginkgo]

More formally I guess

Let D be that integral domain with that “psuedo-Febronius” endomorphism

Then composition of the map of Z into D with that endomorphism gives some kernel.

Assume (n + m)^6 - n^6 - m^6 is always in this kernel. By induction: n^6 - n = n(n^5 - 1) is in the kernel for EACH n

Because D is integral, thus so must be the image

Thus that implies the kernel is prime

so n is in the kernel or n^5 - 1 is in the kernel

However, n is coprime to n^5 - 1

[rocky cloak]

I guess a naive way to solve it is just that ||(1 + x)^6 = 1 + x^6 is a degree 6 equation, but has every element as a root. So there are at most 6 elements, i.e. you're in a finite field.||

||Then just check the 3 cases||

[dull ginkgo]

Hey chat I am stupid there is a much easier way to do this than that

[dull ginkgo]

wait you asserted it’s an integral domain prior. It has prime characteristic

[rocky cloak]

I guess 1 + (-1) = 0 shows it to be of characteristic 2, no matter if integral or not

[ebon pine]

Wait are you asking if it has a prime characteristic or are you telling?

Lemme just post the question to avoid confusion

[dull ginkgo]

So if we map Z into D, then apply the endomorphism we have a kernel, but (p) is already maximal so the kernel must be a sub ideal of (p). However we also know that n^6 - n for each n must also be in this kernel. thus (q^5 - q) = q(q^5 - 1) must be in this kernel. This implies p divides q^5 - 1 for each prime q ≠ p). Clearly p divides 2^5 - 1 = 31 which is prime. Thus p = 31

[ebon pine]

This is the exact question

[rocky cloak]

Hmm, if one replaces 'integral domain' with 'connected commutative ring', how many are there then?

[dull ginkgo]

What do you mean by connected

[rocky cloak]

Not the product of two rings / only idempotents are 0 and 1

/ Spectrum is connected / whatever equivalent definition

[dull ginkgo]

Wait this is wrong too, shit

I mean another route is to assume the characteristic of D is p,

Then 0 = (n + (p - n))^6 = n^6 + (p-n)^6

Chose n = 1, then -1 = (p - 1)^6

[ebon pine]

Btw does characteristic p for given integral domain R imply that R is isomorphic to Z/pZ?

[south patrol]

Well, every field is an integral domain, so, ...

[rocky cloak]

No, so you're still only half way

[ebon pine]

Half way as in you need R to be a field?

Are there any other integral domains of characteristic p?

[rocky cloak]

No, I mean for your problem. R has characteristic 2, but you have to pin down that it's Z/2 and not any other characteristic 2 domain

[rocky cloak]

Polynomial ring over Z/p for example, or finite field of order p^n, or bunch of other stuff

[ebon pine]

Ah okay okay

[dull ginkgo]

Z/p^nZ

[dull ginkgo]

If you have a ring in general with unity, you can always map Z into it

[ebon pine]

Is Z/p^nZ an integral domain?

[dull ginkgo]

No wait

Yeah it isn’t brain fart

[rocky cloak]

I guess F2[x1, ..., xn]/(x1, ..., xn)^2 is an infinite family of examples. Oh well.

[dull ginkgo]

I can’t find what I did wrong here

Because it doesn’t hold for Z/31Z

Oh wait

(Z/pZ)/(nZ/pZ) = Z/nZ

Quotient of Z/pZ satisfying (n + 1)^6 = n^6 + 1 mod p

0 = (-1)^6 + (1)^6 = 2* 1 so yeah 2 | n

gtg

[ebon pine]

Yeah but integral domain + characteristic 31 doesn't imply R = Z/31Z right

It's weird like rest questions were easy and then this is off the charts 😭

I wonder if we are missing something obvious

[delicate orchid]

what we doin

ok that's what we doin

[topaz solar]

They’re cookin but they’re overcookin

[topaz solar]

Perhaps

[delicate orchid]

R = Z/7Z :pack:

[summer path]

,ti 233993116418441216

[cloud walrus]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

[summer path]

[rocky cloak]

Aren't we looking for overcomplicated solutions on purpose here?

[ebon pine]

Definitely not, at least not for me. I just want to find a method to solve it.

So far we have characteristic is either 2 or 31

[delicate orchid]

char 31

[rocky cloak]

So from this comment you know it's either characteristic 2 or 31, and since 3^6 = 3 it can't be 31, so you know it's characteristic 2

[ebon pine]

Oh, I completely missed that my bad

[rocky cloak]

Then I don't think anyone's really addressed the rest, except here I guess

[south patrol]

Eep

[rocky cloak]

But that solution is kinda boring

[ebon pine]

If it works it's okay I guess cause there's another part to this which asks about f(a) = a^15 being a ring homomorphism

But yeah it's more or less brute forcing it

[rocky cloak]

I guess it's more fun if you write out the polynomial a bit more, like
||(1 + x)^6 = 1 + x^6 means
1 + x^2 + x^4 + x^6 = 1 + x^6, so
x^2(1+x)^2 = 0
hence x = 0 or x=1||

[delicate orchid]

this is nifty

jagr can you help me with my group theory homework

[rocky cloak]

You have group theory homework?

[delicate orchid]

yes I just need to do one little calculation

[rocky cloak]

Damn, calculation. That sounds hard

[delicate orchid]

All I have is calculation

[rocky cloak]

And you can't even throw it into Wolfram alpha

[rocky cloak]

Alright, so whip it out

[ebon pine]

Wait in which ring do we have (1+x)^6 = 1 + x^2 + x^4 + x^6? Am I missing something

[rocky cloak]

Characteristic 2 rings

[ebon pine]

Ohh, Alrighty

Thanks

[delicate orchid]

I just need to locate for $S \in \text{Syl}_2(\text{Spin}_7(q))$ the $\chi \in R(S)$ such that $\chi(z) = \chi(z')$ for all involutions $z, z' \in S$ and $\chi(x) = \chi(y)$ for all $x \in y^{\text{Spin}_7(q)}$...

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

this is due in 1 hour... should be easy...

maybe I should just try and find Irr(S) lol

[rocky cloak]

R(S) being characters, and involutionsbeing what?

[delicate orchid]

involutions are order 2 elements

[rocky cloak]

I see

[delicate orchid]

the second condition is nearly all that you need but there's a cheeky central involution in Spin_7(q) that we want

q odd btw

[languid trellis]

spain without the a

[delicate orchid]

this guy gets it

and then we just ocnsider the "spain without the i" of whatever mfs we find

[languid trellis]

i am in spain without the spin

[celest cairn]

While I don’t know much about inverse limits, I’ve heard you can represent towers of extensions in terms of them.
How can we represent the tower $T = \mathbb{Q}(\sqrt{2}) \subseteq \mathbb{Q}(\sqrt[4]{2}, \sqrt{\omega}) \subseteq \mathbb{Q}(\sqrt[8]{2}, \omega)$ in terms of inverse limits, where $\omega = e^{\pi i / 4}$?

[cloud walrus]

Sapphire

[delicate orchid]

now how do I explain this without using the words "indexing category"

so this would be the inverse limit over the inclusion maps a_n -> a_n+1 with a_n = Q(2^(1/2^(n+1)), omega^(n/2))

since this is a finite sequence of inclusions you just get the inverse limit being isomorphic to the last term

but if you went on forever you'd get something new

[celest cairn]

Okay thanks! I made a criterion and decided inverse limits could play a role in it, so I wanted to get a better understanding of them.

[delicate orchid]

in that case I recommend reading a proper source about them

unironically the wikipedia page is quite good

[winged void]

I was thinking that since ( \sigma^{280} = e ), and the prime factorization of 280 is ( 2^3 \cdot 5 \cdot 7 ), I could construct a cycle of length 8, then 5, and then 7 because ( 8 + 5 + 7 = 20 ). Then, the sign of ( \sigma ) will be ( (-1)^{17} = -1 ), so it must be odd.

is this enough

[cloud walrus]

Mootje

[winged void]

is this enough

and i did just like a random cycle of lenght 8, 5 and 7

is that enough

[pliant forge]

Not sure if this is enough information, but suppose you have a R module homo f: A -> B that is 1-1, and another homo g: M -> B such that im g ⊂ im f (≅ A). Then can one conclude that there is a map h: M -> A such that g = fh?

[fading basin]

Yes

[coral spindle]

what did you just call me

[pliant forge]

homo

[rocky cloak]

Just compose with the isomorphism imf -> A

[pliant forge]

ahhh

[coral spindle]

You can see the existence of an injective homomorphism f : A → B as actually saying that A is a submodule of B

You're kinda describing how it lands inside B

With that understanding this is saying almost nothing, right – it's like saying that if the image of g is in A then A contains the image of g lmao

[pliant forge]

thanks

[rotund aurora]

what are examples of this

[south patrol]

They are ubiquitous in many areas, for example anything relating to hom alg (the (co)homology groups/rings of smth for example)

[next obsidian]

k[x] over k[x]

[barren sierra]

yea polynomials are the most basic example of this (and probably the motivation for such a definition if I had to historically guess)

[mighty kiln]

Singular homology is a graded module over singular cohomology

[dull ginkgo]

Tensor algebra over a vector space

Likewise, exterior algebra

[mighty kiln]

Those be graded rings

[dull ginkgo]

Never said it wasn’t a module

technically the underlying vector space is a module lmao

OH

MISREAD

[crystal vale]

Let R be a commutative ring with 1 and Spec(R) be the set of all prime ideals of R, called the prime spectrum of R. For an ideal I in R, let V(I) = { P in Spec(R) | I is a subset of P }

I need to prove Spec(R) is non-empty, hint?

And V(I) = Spec(R/I).

I am not sure about how prime ideals in R/I look ?

If P is in V(I) then it's structure is different from prime ideals in R/I, right?

[barren sierra]

wdym Spec(R) is empty

because in general it is not empty

[crystal vale]

No sorry, it is not empty

[cobalt heath]

Woh you already doing this stuff?

[crystal vale]

If R is a non-zero ring then (a) is ideal and then there exists a maximal ideal which contains it so that the maximal ideal is prime ideal, hence Spec(R) is non-empty, right?

[cobalt heath]

Ye

[crystal vale]

Okay thank you

[crystal vale]

I don't understand

[cobalt heath]

Spec to me is quite advanced stuff, so I was surprised that you are learning these now.

[crystal vale]

No, they introduce Spec in a problem set so I don't know much about it

[cobalt heath]

Still.

[crystal vale]

Any hint for V(I) = Spec(R/I)

[cobalt heath]

Oh, that will take some manipulations

[winged void]

Can someone please check if my explanation is good ? (:

[rocky cloak]

Correspondence theorem

[rocky cloak]

What exactly are you explaining?

[winged void]

Just a second

Oh I see now I posted only the answer without the question

The answer was this

And the question

Is find an element that belong to S_20 with order = 280 and find its sgn

So that’s how I solved it

But not so sure if that’s enough

[rocky cloak]

Looks good to me

[winged void]

Thanks a lot. I was afraid because my proof is so short that it’s wrong

[crystal vale]

Yes but I am not sure about the correspondence theorem, set of all ideal of R/I is bijective to set of all ideals of R containing I, right?

[rocky cloak]

Yes, and also R/J = (R/I)/J when J contains I

[crystal vale]

So When J is prime in R containing I then R/J is an integral domain and it implies that (R/I)/J is an integral domain which implies that J is the prime ideal in R/I, right?

[chilly ocean]

[crystal vale]

4Z is maximal in 2Z?

Then 4Z is not prime ideal in 2Z but there is a question such that a 2 sided maximal ideal is prime ideal in any ring.

[rocky cloak]

Your rings should have 1

[crystal vale]

Yes, but they state that any ring

[rocky cloak]

Any ring has 1

[flat tree]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[cobalt heath]

(This channel is for groups, rings and fields)

[crystal vale]

They don't specify that R has 1

[rocky cloak]

If it doesn't have 1, it ain't no ring, is all I'm sayin

[tawny magnet]

How can I find the exact number of pairs of permutations that generate Sn for small n? For example S5?

I understand Dixon and Babai each have results when n is large but is there a nice way to calculate for small n without bash?

[narrow wagon]

@tawny magnet

what do you mean generate Sn?

powers of some element in Sn will generate the group Sn?

[tranquil musk]

If I have a group say z/Z9 = {0,1,2,3,4,5,6,7,8}

and then another group G = {0,5}

Does this mean that in G if i go 5+0 it is just 5 and then say 5+5 = 0? since it rotates through?

[coral spindle]

The first group you mentioned has no bearing on the second group.

Every group defines its own operation and does not have to be dependent on anything else.

As it happens, no matter what the operation is, we know that G will be isomorphic to Z/2Z, but you have made no specification as to which element is the identity.

It is also worth pointing out that if you meant for G to be a subgroup of Z/9Z, you have gone wrong. It is not.

[tranquil musk]

So the first group is defined by addition, I need to specify that in the question, each number has a BAR ontop of it so I think this means modulo right?

that means it would be a subgroup then right?

[coral spindle]

What is "it"?

[mighty kiln]

{0, 5} is not a subgroup cuz 5+5=1

[coral spindle]

As I said once again, {0, 5} is not a subgroup of Z/9Z.

It is only a subset and nothing else.

[tranquil musk]

Well my question asks to show that its a subrgroup

[coral spindle]

Show us the whole question.

[winged void]

I have small question regard the following notation what I know that $c_n \times c_m $is the direct product but what does $c_{nm}$ means

[cloud walrus]

Mootje

[coral spindle]

The same thing as C_n and C_m but with mn instead of n or m.

[mighty kiln]

nm is n times m

[winged void]

sure but i mean is it like the product of two elements

i do not get it that much

can you give like a direct example

[coral spindle]

n and m are integers.

They are whole numbers

n = 5, m = 6 is an example, then nm = 30.

[winged void]

sure agree but i assume that C_nm is a group right

[coral spindle]

Are you asking what the notation C_k means?

[winged void]

i guess its a cyclic group

[mighty kiln]

C_5 × C_6 = C_30

[coral spindle]

Yes.

Whatever you are reading explains this, I have no doubt.

[winged void]

but i mean im not really seeing how those are isomorphic

[winged void]

they did

explain the direct product really good

[mighty kiln]

Does it state the proof

[coral spindle]

They are not always isomorphic.

[mighty kiln]

This is true iff gcd(m, n) = 1

[winged void]

yes

[coral spindle]

You should see this in the things you're reading. There will be a condition.

[winged void]

yes i know

[mighty kiln]

Have you read it

[winged void]

yes i read it and i understand the proof

[tranquil musk]

Consider the group G := (Z/Z8,+) = {0,1,2,3,4,5,6,7} (These numbers all have bars ontop of them but I dont know how to insert this into discord). Show that the set H:= {0,4} (once again bars ontop of these numbers) is a subgroup of G.

[winged void]

really good

[fading basin]

i think it represents the space of real matrices with $m$ rows and $n$ columns iirc

[cloud walrus]

Nats

[winged void]

the problem that im facing is that how does the groupC_nm look like

[coral spindle]

So indeed you didn't post the actual question, because although {0, 4} is a subgroup of Z/8Z, this says nothing about {0, 5} being a subgroup of Z/9Z.

[winged void]

you can use overline for bar

[mighty kiln]

No

[coral spindle]

Do you have any questions about this or is that all?

[tranquil musk]

hahah damn it got the numbers wrong

[fading basin]

okay

[winged void]

im not really sure if its easy to think of an example like simple example where i can see it

[mighty kiln]

C_30 is the group {0, 1, …, 29} under + mod 30

[coral spindle]

Who got the numbers wrong?

[winged void]

use as well Dollar sign for a good notation as tip (:

[coral spindle]

Did you ask GPT something? Do not ask GPT about maths, it is uniformly horribly wrong.

[tranquil musk]

So then in H, 0+4 would be 3 right?

[fading basin]

what are you doing

[coral spindle]

Why do you think this

[winged void]

agree on that but its really good in real analysis though

[coral spindle]

You are saying that 0 + 4 = 3 mod 8?

Why.

[mighty kiln]

Press [x]

[tranquil musk]

Im asking how it works becaus I dont necessarily understand the notation

[winged void]

no joke it helps in analysis

well it does not give the answer but if you ask ik correctly it will give a good intuition

[tranquil musk]

4+4 = 3 right because it goes to 0 then 1 then 2 then 3? or am I wrong here

i made a typo earlier

[narrow wagon]

What are you finding?

[winged void]

the example that you gave how is that releated to the direct group

[narrow wagon]

4+ 4 mod ?

[tranquil musk]

4+4 in the group H I defined earlier

[coral spindle]

I'm going to leave this to you Dubs

[narrow wagon]

Can you show the orginal problem?

[mighty kiln]

One case of the theorem is that C_5 × C_6 is isomorphic to C_30

Do you understand every word in this statement

[winged void]

to be fair not really

[tranquil musk]

this is the original question here

Consider the group G := (Z/Z8,+) = {0,1,2,3,4,5,6,7} (These numbers all have bars ontop of them but I dont know how to insert this into discord). Show that the set H:= {0,4} (once again bars ontop of these numbers) is a subgroup of G.

[winged void]

let me explain

what i understand

[narrow wagon]

Why is it typed so bad?

[tranquil musk]

I typed it out of a book

[narrow wagon]

you want show {0,4} is a subgroup of Z/ Z_8?

[tranquil musk]

It is asking me to show that {0,4} is a subgroup Z/Z8

[winged void]

well i know that the direct product is the followin so lets say the direct product of C_5 * C_6 = (c_1, c_2) where c_1 belongs to C_5 and c_2 belongs to C_6 and im not really sure what is the operation here but lets assume that in C_5 its modulo 5 and C_6 modulo 6 as an example then if you take two arbitrair elements say (c_1, c_ 2) and (a_1,a_2) then the following is true --> (c_1+a_1 mod 5 , c_2+a_2 mod 6)

[narrow wagon]

What do you think is the operation is in this set?

[coral spindle]

You mean Z/8Z, not Z/Z8. Z/Z_8 is not meaningful.

[narrow wagon]

I really mean Z8, the subgroup of integers

[tranquil musk]

addition but is it addition modulo 8, Im not too familiar with modulo stuff

am i right?

[narrow wagon]

I can give you some intuion

think about it as clock,

[tranquil musk]

Yes

[narrow wagon]

in clock we count in sort of count modulo 12

[winged void]

well you do not really to think about it as modulo. I faced also lots of problem regarding that. think about the clock you are wearing its alos modulo

but its modulo 12

only change 0 with 12 to make it more convenient

[narrow wagon]

if it’s 13:00 you will say it’s 1:00 clock right?

[mighty kiln]

Yes, this is correct

[winged void]

do not make it that difficult

[narrow wagon]

essentially think about wrapping the integers around 12 corners.

[winged void]

sure how is that then releated to C_30 i do not really see it

[tranquil musk]

So in H, what is 4+4 equal to? is it 0?

[narrow wagon]

if you taking modulo 12

[mighty kiln]

(c1, c2) + (a1, a2) = (c1+a1, c2+a2), more precisely

[narrow wagon]

4+4=8 =0 mod 8

[mighty kiln]

Do you know what an isomorphism is

[winged void]

i know its homorphism but injective and surjuctive

but i do not really have that big intuition

for it

[mighty kiln]

Right, a bijection that is a homomorphism

Does the proof state which bijection it is

[winged void]

what do you mean exactly ?

well there is only one type of bijection except if im missing something

[narrow wagon]

@tranquil musk for example. if you want to find 5+3 in mod 8, just add them and find remainder upon division by 8

[mighty kiln]

A bijection is a function that is injective and surjective

[tranquil musk]

Ok thankyou. And then from there to prove its a subgroup is quite easy I believe. The second part is to find all the left cosets of H in G

[mighty kiln]

That is, the elements of the domain and codomain are paired up exactly

[winged void]

that is right

i agree on that

no doubt

[narrow wagon]

what have you tried?

[mighty kiln]

When we say two things are bijective / isomorphic, that means there is some bijection / isomorphism between them

[winged void]

well you can use lagrange theorem to know how many elements might be in each coset that will make it easier to find all left cosets because each left coset has the same size

[winged void]

sure

[mighty kiln]

Does the proof state which isomorphism it is?

[winged void]

well not really

[narrow wagon]

isomorphic is really a way to relabel elements in a group

[tranquil musk]

I havent attempted the second part yet, I need to read up on what a left coset is. But to prove its a subgroup Ill just show that 0+4 and 4+0 is in H and then the other properties of a subgroup should be easy to prove I believe

[mighty kiln]

Is it actually a proof

[winged void]

or use the criterion of a subgroup that is easier

its two in on e

[mighty kiln]

Surely you would state the function and then prove it is bijective and is a homomorphism

[winged void]

sure agree

[narrow wagon]

what about 4+4? is it all there?

[winged void]

but until now the point that i do not really get C_30 does not has anything to do with direct product

right

[tranquil musk]

4+4 = 0mod8 right

[mighty kiln]

One isomorphism you can find from C_30 to C_5 × C_6 is f(x) = (x mod 5, x mod 6)

So e.g. f(13) = (3, 1)

[narrow wagon]

4+4+4=?

[mighty kiln]

13 is an element of C_30, and (3, 1) is an element of C_5 × C_6

[winged void]

i see

so C_30 does not have the same operation as C_5 * C_6

am i right

C_30 is the cycle (1 2 3 4 5 .... 30 )

[mighty kiln]

Indeed they are defined differently

But they turn out to be isomorphic

[tranquil musk]

4+4+4 = 4 + 0 = 4mod8?

[narrow wagon]

right

so its sort of cyclic right?

do you know cyclic groups?

[tranquil musk]

Yeah I think so

so its easy to show that its closed under any operation then just show that 0 is the identity and 4 is the inverse operation and associativity should be easy

[narrow wagon]

H={0,4}?

[tranquil musk]

yes

[narrow wagon]

you don’t need to show associativity

these elements belong to a set with is already a group, so it’s elements enjoy associativity already

have you learned subgroup test?

[tranquil musk]

I dont believe so

[narrow wagon]

To show any non empty subset H is a subgroup, you need show ab^-1 is in H whenever a,b in H

[tranquil musk]

Ahh ok easy

Thanks bro I really appreciate it

[grizzled crane]

ookie np

[winged void]

Sure I will draw a table if I’m home

To get mor intuition

Around it

[winged void]

That’s the subgroup criterion

You can also prove ab is inside

And for every a there is an inverse a^-1

[grizzled crane]

yes, all of these are implied in this one step

set b=a, you get e in H

set a=e, you have inverse

so on

[smoky ivy]

does $G/K \cong G/H$ and $H \subseteq K$ imply $H=K$?

[cloud walrus]

lewis

[dull ginkgo]

First isomorphism theorem applied to the composition of the map G onto G/K into G/H

That kernel necessarily is K, but there’s a composition-inclusion law for kernels

[south patrol]

No

[dull ginkgo]

Oh?

[smoky ivy]

whats a counterexample

[dull ginkgo]

Probably some free group schenanigans

[south patrol]

For example take G = Z (+) Z (+) ... and let H be the first Z and K the first two

[south patrol]

Almost yeah

[south patrol]

Then both quotients are iso to G

[smoky ivy]

ah dang, right

lol

[south patrol]

The issue is that you are only requiring an iso rather than anything to do with how the subgroups fit inside G

If G is finite, then this is true just by counting

[smoky ivy]

yeah

actually, let me ask this in another context

[dull ginkgo]

Hmm I thought $\mathrm{Ker}(a) \subseteq \mathrm{Ker}(b \circ a)$

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[smoky ivy]

in this proof

[dull ginkgo]

So I guess if you could impose something on the composition

[south patrol]

Yea

[smoky ivy]

how do i follow that ker \phi \otimes \phi^\prime = G

[dull ginkgo]

I’ve actually used this lemma a lot in some proofs

Very nice, also nice for chain complex proofs

e.g salamander lemma

[placid tide]

sorry to derail the conversation here, but I was wondering if this channel was just for discussion, of if I could ask a question, because a hmwk problem I have that I was wondering is how you can number R^N with R?

[south patrol]

You can ask though it's not quite the right channel

[placid tide]

oh sorry, what would that be?

[south patrol]

I guess that would go in #proofs-and-logic

[placid tide]

alright thanks

[smoky ivy]

dont quite see how it follows from just C \otimes C^\prime = (B \otimes B^\prime)/G

[south patrol]

hm where do they say this

oh like by the "clearly"?

[smoky ivy]

oh, no, that direction is clear

like

G is a subset of the kernel

[south patrol]

of what

oh

do you mean like

[smoky ivy]

of \phi \otimes \phi^\prime

[south patrol]

"why is it nothing bigger than G"

[smoky ivy]

yes

[south patrol]

and to check, are these like abelian groups

[smoky ivy]

yea

[south patrol]

tensored over Z

okay sure

[smoky ivy]

yeah

lol

surjectivity follows from the first isomorphism theorem

[south patrol]

so what's the issue with their proof

[smoky ivy]

im confused on how the kernel is a subset of G

they only show G \subseteq ker

[south patrol]

Ah okay. Well the point is that the induced map (B (x) B')/G --> C (x) C' is an iso

in particular, it has trivial kernel

so if (φ (x) φ')(y) = 0

then y maps to zero in (B (x) B' )/ G

i.e. y is in G

[smoky ivy]

oh

okay yeah, thank you

[south patrol]

np

you can also view like

"f: A -> B is a surjection with kernel K" is exactly the same as saying that the induced A/K -> B is an iso

and more generally like

[smoky ivy]

yup

that's just first iso theorem

[south patrol]

if L \subset K \subset A then like

kernel of the induced A/L -> B is K/L

[smoky ivy]

i see

tyty

[crystal vale]

If M_n (Q) is a set of all matrices of n × n over Q then it is a local ring, right? Is it independent of n?

[rocky cloak]

Local ring usually means it has a unique maximal left ideal (or equivalently a unique maximal right ideal). The matrix ring does not satisfy this. It's a simple ring though.

[crystal vale]

The set of all 2-sided ideals in R is bijective to the set of all 2 sided in M_n(R) if R is commutative and it has unity, right?

[crystal vale]

Q is a simple ring, right?

[dull ginkgo]

Q is a field so yes

[rocky cloak]

Yes, that's correct. Hence it's a simple ring

[dull ginkgo]

The bijection I don't think needs commutivity?

Just left/right/two-sided correspond

[crystal vale]

No it states that its not necessary onto for left/right ideals

[dull ginkgo]

huh

[rocky cloak]

It's just the two-sided ideals that gives you a correspondence, but you don't need the commutativity

[crystal vale]

If it is a local ring then a set of all non-units is a maximal ideal but there is only one maximal ideal (0) and the set of all non-units is not equal to (0), right?

[rocky cloak]

Thats right

[crystal vale]

Okay, thank you

[rotund aurora]

yes but the next trivialest example would be like k[x]^2. But there seems to be many ways to grade this. Also, in the definition of graded ring you only consider nonnegative degrees

like say R is graded

is there a more natural grading on R^n than any other?

[rocky cloak]

I guess the most natural would just be to consider it as a direct sum of graded modules.

One example could be that if R is a ring you can consider R[x]/x^2 with R in degree 0, x in degree 1. Then a graded module is exactly a complex of R-modules.

[delicate orchid]

are we coming up with examples of graded rings

[rocky cloak]

Think so

Or graded modules maybe

[rotund aurora]

are you saying that you wouldn't usually grade the entire module R^n?

[rocky cloak]

No, I'm saying R is a graded module, so you just consider R^n as the n-fold direct sum of R with itself

So put (R0)^n in degree 0, then (R1)^n in degree 1, etc

[hollow mica]

Wait I was thinking about this some more, how does giving X some zero eigenvalues accomplish this? In the image I linked, isn't it jus that some of the x_i become 0, nothing more?

And btw the matrix X I chose just has each x_i along the diagonal

[dim widget]

The x_i are the eigenvalues of X yes

But making k of the eigenvalues 0 makes the elementary symmetric polynomials of degrees > n-k vanish

In particular if n-k of the eigenvalues are zero, then we get the equality e_k(x_1,…,x_k) = 1/k \sum blah from the case that the number of variables is the degree of the polynomial

[hollow mica]

Am I correct in thinking that if we want to prove the case of n variables x_1, ..., x_n and you want p_{n-1}, we would work with the n+1 by n+1 matrix with the x_i along the diagonal and a single 0?

[delicate orchid]

homology is a classic

group algebra is good as well to generate a G-graded ring for arbitary G

[dim widget]

It’s just nxn but yes

[hollow mica]

But there's n+1 eigenvalues

[dim widget]

Im not sure what you’re trying to prove

[hollow mica]

Sorry

[dim widget]

So there are two integers

[hollow mica]

the equation at the bottom of this image, for k = n-1

[dim widget]

n is the degree of the p_n and m is the number of variables

When n = m we can prove the identity

[rotund aurora]

But I'm still put off by the condition that the sum of the graded parts of the module goes from n=-infinity to +infinity, when in the case of rings you only consider nonnegative degrees. Why is that change in the definition needed?

[dim widget]

When m > n then an equality of homogenous polynomials of degree n can be checked after setting all tuples of m-n variables equal to 0

Which you can prove by induction

Basically, no term involves more than n variables, and equality of polynomials is checking the equality of terms

[hollow mica]

What do you mean by "setting all tuples of m-n variables equal to 0"

Is this supposed to correspond to the operation you described of setting some eigenvalues to 0?

[dim widget]

Like for any size m-n subset of {1,…,m} set those variables equal to 0 and check if the resulting polynomials are equal

[hollow mica]

Are you saying:
For every size m-n subset of {x_1, ..., x_m}, consider the matrix X with eigenvalues x_1, ..., x_m EXCEPT making the x_i chosen by the subset 0 , and then look at the char polynomial p_X ?

That's a total of m choose n polynomials

[rocky cloak]

You could allow the ring to be supported in negative degrees as well. But then it becomes a little more complicated, since the positive part doesn't become an ideal.

As for the modules, you want to be able to shift the degree arbitrarily up and down, so then you have to allow it to be negative.

[hollow mica]

Oh I see and your claim is all of those polys are equal

[dim widget]

No

[dim widget]

We’re trying to prove this equality of two homogenous polynomials of the same degree

My point is that if n > k the equality holds if and only if it holds for the nC(n-k) polynomials obtained from the left and right sides by setting n-k x_i = 0 on each side

[winged void]

im studying rightnow the chinese remainder theorem

what does actually C_1 ^{-1} mod m_1 means ? I know its the inverse

so is it like the number when divided by m_1

it gives me the identity

[delicate orchid]

it’s a ring element such that c1c1^-1 = 1 mod m_1

Not really much else to be said

[winged void]

i did not yet study rings

well we solve it using the euclidean algorithm

right now

i do not get a good intuition about it

like why when we say for example $b1 = 35^{-1} mod 3 $ here b1 must equal to 2 because 70 mod 3. why idk

i do not really get what you mean here so for example b1 = 7 mod 10

am i right

and we see that 7*3 = 21 and 21 mod 10 = 1

indeed

but from here did you get the 37 = 21 = 1 + 210!

how did you get that ?

[dim widget]

For instance if $m_1 = 10$ and $c_1 = 7$ then $c_1^{-1} = 7^3 = 3$ because $37 = 21 = 1 + 210$

[cloud walrus]

missa inte chansen

[dim widget]

It was just some problem either with this app

[winged void]

sure but what is actually the diffrence between the extended euclidean algorithm and the euclidean algorithm

[dull ginkgo]

You compute the Bezout coefficients simultaneously

Alternatively, one kinda does it in $\mathrm{SL}(2,\mathbb{Z})$

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[hollow mica]

Fixing variables x_1, ..., x_n, how can I show that the polynomials f_k = (x_1)^k + ... + (x_n)^k do not generate the ring of symmetric polynomials? Everything takes in place over Z (i.e. the polynomial ring is Z[x_1, ..., x_n]).

I know that if Z is replaced with Q then the opposite is true

[dull ginkgo]

Which part

[random pasture]

im on part a rn

my intuition says we just consider s=0

but I'm not sure how to use the results of that

[dull ginkgo]

well, if ra = rb, what can we say about a - b?

[random pasture]

if it's injective we just have a=b, right?

[topaz solar]

If it’s injective, and s=0, what does that say immediately

What do rs be

[random pasture]

0 right

[topaz solar]

Yeah

So if rs = 0…

That’s one direction of part a

[random pasture]

yeah i think i figured it out

[random pasture]

i proved that (1) implies (2) and (2) implies (1) for the proof

how do I approach part b

[topaz solar]

Nice

[topaz solar]

See the hint

And use part a

How does s -> rs work

[vocal fern]

how can i approach this?

[coral spindle]

Part a: by being creative. The best hint I can come up with is to prompt you to think about when it is that fg = 0.
Part b: picture the functions and think about finding an h that isn't invertible. Think about what continuity requires of such an h.

[south patrol]

||Writing fg for the pointwise product pains me||

[random pasture]

how do we do spoilers?

[topaz solar]

Surround your text by ||

Like ||this||

[random pasture]

||i think it's like any set of functions where f(x)=0 for every value except a certain range where it will be nonzero, and since it's over R, you can pick an infinite number of such ranges. ex. f(x) = 1 for x in [0,1], and f(x) = 0 otherwise||

[hollow mica]

why is it that any field extension of K in which p splits, it always holds that L is inside (can be embedded in) this field?

[south patrol]

This isn't immediate from the definition - it is a theorem. Note that if M is an extension of K in which p splits then M contains a splitting field - consider the field generated over K by the roots of K

Then a theorem says any two splitting fields are isomorphic (in a way that fixes K)

[hollow mica]

Ohh, I see

[south patrol]

Good question though

But yeah basically the idea is any two splitting fields are generated over K by the roots of p and you can send the roots in one field to those in the other

You have to be more careful but yeah

[hollow mica]

I see

[south patrol]

This theorem should be in any good field theory or galois theory text and you can induct on the degree of the poly

[hollow mica]

How does one use the galois correspondence to find the intermediate fields of a field extension?

Like say we are given the extension of Q(sqrt(2), sqrt(3)) / Q, and I want to find the intermediate fields

let G be its galois group

Galois correspondence says that the set of subgroups of G is in 1-1 correspondence with the set of intermediate fields

• where a subgroup H is sent to the largest intermediate field K^H that all of its automorphisms fix, and going backwards an intermediate field K is sent to the subgroup H_K of automorphisms that fix K

[coral spindle]

If you can compute a basis for the field, and generators for a particular subgroup as linear transformations on that basis, then the fixed field is the intersection of all the 1-eigenspaces of those generators. There are effective computational ways to do so.

This is just the approach that comes off the top of my head. I wouldn't be surprised if computational algebraists have nifty approaches that work faster.

[hollow mica]

I see

If F is a field, and p is an irreducible polynomial in F[x], how do I show that (p) is a prime ideal ?

Suppose not, then there exists f, g such that p = fg and neither f nor g are divisible by p

Since p is irreducible, either f or g is a unit, suppose it is f

What's supposed to go wrong

[hollow mica]

Oh, I guess since F[x] is a PID, prime elements will coincide with irreducible elements, but I was looking for an approach more along the lines of above

[topaz solar]

What’s a unit in K[X]

[primal tusk]

how would you go about proving that there exists a homomorphism/isomorphism between O(2, Z_4) and D_4 ? You could write the tables out and compare them but is there an "easier" way?

[hollow mica]

ohh. thanks

[topaz solar]

Also consider division to get to a monic irreducible to be nice

[hollow mica]

How does that help

p = fg -> f is a unit -> f is a constant -> g is in (p) -> contradiction

that's how I understood

[chilly radish]

It guarantees uniqueness

[topaz solar]

Well how is that a contradiction

[chilly radish]

Rather than uniqueness up to a constant

[hollow mica]

f, g are by assumption not divisible by p ?

[topaz solar]

Ah right

Reading comprehension moment

[hollow mica]

Ok I still dont see how uniqueness matters here

But the proof works

[topaz solar]

As I said, to be nicer

9 times out of 10, monics good

[split vector]

hey, can someone help me understand how come my textbook (abstract algebra, thomas w. judson) defines the characteristic as in the first image, defining it with nr = 0 for all r in the ring R, whereas wikipedia defines it in second image with n(1) = 0 i.e. just using the multiplicative identity for the definition

i noticed that my textbook's definition is also listed on the wikipedia page, but i still dont quite understand the reason for the discrepancy between the two definitions

[void cosmos]

if n(1) = 0 then n(r) = n(1+1+1...) = 0

if n(r) = 0 for all r then n(1) = 0 (assuming whatever has to be assumed unital bla bla )

[split vector]

ah thats really simple but that clears it up for me lol i didnt think of that

thank you :)

[void cosmos]

yw 😄

yeah if you dont like it you can like say

n(r) = n(1)(r)

@T

whatevers

that was a good question

[hollow mica]

Oh, so like
n(r) = n(1r) = 1r + ... + 1r = (1 + ... + 1)r = 0

[void cosmos]

no just n(r) = n(1)(r) = 0

but the thing is

if you have a ring with unity

there is a homeomorpihsm from Z to that ring

so

what is the kernel of that homeomorpihism

[hollow mica]

Oh this uses the fact that (ab)r = a(b(r)), where a, b are positive integers

[void cosmos]

yeah

[hollow mica]

But I guess that's supposed to be obvious

[void cosmos]

over rings multiplication is not necessairly commutative

so you have to be careful

[hollow mica]

(n) where n is the characteristic of R

[void cosmos]

yeah exactly

[hollow mica]

oh were you just telling me that

[hollow mica]

or does that relate to this

[void cosmos]

yeah

[hollow mica]

oh idr see

[void cosmos]

no i meant thats why like

when you asked why every element is 1+1+1 ..

[hollow mica]

oh

I see

[void cosmos]

not always true ofc 😄

[hollow mica]

in a ring of characteristic n, the only elements that look like 1+1+1+... are im(homo from Z) which is exactly 1, 1+1, ..., 1+1+... where the last guy has n 1s

[void cosmos]

yeah Z/nZ you mean

last guy has n-1 1s

[hollow mica]

well I said n so I could include 0

I guess when the char is 0 this isnt exactly right

[void cosmos]

yeah its kinda like why it"s called 0 cuz like

ur modding out by nothing so u get Z which has 1+1+1 forever okay

kinda like that

whatevers

[rapid junco]

What does it mean by a commutative ring containing Z_p for some prime p?

just that we have a Z_p-algebra structure on R? i.e. an injective hom into R?

[hollow mica]

It probably means Z_p can be embedded inside your ring R, i.e. there exists an injective hom from Z_p into R

[rapid junco]

I am trying to prove if R is a ring containing Z_p for some prime p. Then $x \mapsto x^p$ is a ring hom.

[cloud walrus]

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

[rapid junco]

multiplicativity is trivial

(x + y)^p

[hollow mica]

expand using binomial theorem

[rapid junco]

yeah

it follows by freshmans dream but looking at proof again

what do we define in a ring to be (p choose k)?

since division is not always here.

[void cosmos]

wow what a coincidence

[hollow mica]

in (p choose k), p and k are both integers

[void cosmos]

if your ring is commutative then the homormophism from Z --> A works

[hollow mica]

binomial theorem in a ring R says that for elements x, y in R and an integer n, (x+y)^n = sum ((n choose i) * x^n y^{n-i}) from i=0 to n

[rapid junco]

p!/(p - k)!k! is always an element of our ring?

do we take the division to be inverses in Z_p?

[void cosmos]

you wont divide

this is always an integer

it's pretty cool

[hollow mica]

you know how in an arbitrary ring R, we can define for a positive integer n the product nr = r + ... + r

[void cosmos]

the number of ways of choosing n things k ways is always an integer

[rapid junco]

yes

[hollow mica]

the same thing is going on in the statement of the binomial theorem

[rapid junco]

oh I see

[hollow mica]

(n choose i) * x^n y^{n-i} is meant to represent the sum of (n choose i) copies of x^n y^{n-i}

[rapid junco]

So p! is (1 + 1 + 1 + 1)(1 + 1 + 1)(1 + 1)

[hollow mica]

if you expand out (x+y)^n out by hand this is excatly what you want

no no no

[hollow mica]

remember in (n choose i), n and i are both integers

[rapid junco]

yes

[hollow mica]

you just compute the binomial coefficient normally

[dull ginkgo]

p choose i for any i greater than 0 and less than p is divisible by p

[rapid junco]

yeah because the denominator isnt divisible by p

[hollow mica]

I think you misunderstand what binomial theorem is saying @rapid junco

[dull ginkgo]

$\binom{p}{n} = \frac{p!}{(p - n)!n!}$ and the denominator never gets to $p$ lo

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[rapid junco]

Ohhh I see now

Heres why I was confused

[hollow mica]

it tells you that multiplying n copies of x+y is the same as adding (p choose i) multiples of x^i y^(n-i) for each i

[rapid junco]

I was like oh ok Z_p embeds in R

but then why can't and element of Z_p have its inverse being in R by chance, not in Z_p.

But this follows by uniquenss of inverses

So then the division is well defined

[hollow mica]

inverses shouldn't matter here

[rapid junco]

well it should

[hollow mica]

that's what I'm trying to explain

[rapid junco]

no?

[hollow mica]

you shouldn't have to think about them

why would they matter?

[rapid junco]

so p choose i is not an element of our ring

rather its a notion of addition

[hollow mica]

p choose i

[rapid junco]

p choose i times

[hollow mica]

is always an integer

integers

[rapid junco]

I see

okay

[dull ginkgo]

Also this is called the Frobenius Endomorphism just in case you want to look into it

[dull ginkgo]

Bro I just saw you real time get very active i think lol

[hollow mica]

yee

[dull ginkgo]

congrations you doned it

[rapid junco]

ah so all terms where 0 < i < p exist make it so p divides the number.

so it must be a multiple of p.

say kp for some k.

[hollow mica]

Think of (p choose i) * x^i y^(n-i) as nr, where n = (p choose i) and x^i y^(n-i) as r @rapid junco . This is defined as adding (p choose i) copies of x^i y^(n-i)

ok I used n twice there

[rapid junco]

yes I see now

so if p choose i is kp for some k in Z

Then we say kp x^i x^(p - i) is zero?

How

[dull ginkgo]

There is actually a map of Z into every (unital) ring

[rapid junco]

is this ring of characteristic p?

[hollow mica]

yea

[rapid junco]

Not always

[hollow mica]

prove it

[rapid junco]

Z_2 embeds into Z_6

[hollow mica]

well

I assumed homo are unital

so 1 has to be sent to 1

maybe it's true without this assumption

not sure

[rapid junco]

how is Z_2 mapping 0 to 0 and 1 to 1 not an embedding into Z_6

[dull ginkgo]

I don’t like to argue with people’s religions (that rings without unity are remotely interesting)

[hollow mica]

if f: Z_2 -> Z_6 is your function then f(1+1) = f(0) = 0 and f(1) + f(1) = 1 + 1 = 2 @rapid junco

[rapid junco]

ah

wait what is the canonical embedding then

[hollow mica]

who is supposed to canonically embed into who

[rapid junco]

Z_2 into Z_6

oh there is none

[hollow mica]

Usually "canonical embedding" means it is known ahead of time that there is an embedding lol

[rapid junco]

so if there is an embedding of Z_p into R then R is char P.

[dull ginkgo]

I have to impose great restraint here because it’s an advanced channel

[dull ginkgo]

not particularly

I don’t think?

Unless it’s a domain

[rapid junco]

what did you mean here then

[hollow mica]

I think so

[dull ginkgo]

i know the kernel of Z into your ring is like the upper bound of the “additive order of each element” if it’s nontrivial

E.g it’s prime ring

[rapid junco]

f(0) = f(p) = 0 = p f(1) = p * 1 = p.

[dull ginkgo]

Since you can still factor out your 1 from every element

Once again, if Z just doesn’t flat out embed into R

[rapid junco]

this is fine right?

[hollow mica]

If f: Z_p -> R is our embedding, f(1 + .. + 1) = 1 + ... + 1. If there are less than p ones in the sum then the sum is nonzero by injectivity, and if there are exactly p ones then the sum equals f(0) = 0 @dull ginkgo

Doesn't this show that
Z_p embeds into R <=> char(R) = p

[dull ginkgo]

I was going off of a weird def of char(R)

Literally it’s just the n corresponding to the kernel of Z into R

[hollow mica]

That defn is equivalent no?

[rapid junco]

Characteristic is the kernel of the n corresponding to the principal ideal

yes equivalent

[dull ginkgo]

Well that last one I said is

[hollow mica]

Oh, you were using something else before

[rapid junco]

But yeah on the other hand if char(R) = p the by definition we have an embedding on Z_p into R.

[hollow mica]

Idk what you're trying to do here

[rapid junco]

0 in Z_p is p

f(0) = f(p)

but f(0) = 0

so f(p) = 0

but f(p) is pf(1)

[dull ginkgo]

Random question

What were we originally trying to prove here lmao

[rapid junco]

but f(1) = 1.

(x + y)^p = x^p + y^p in char p comm rings

[hollow mica]

So what have you shown @rapid junco

Or like, what was the point of doing that computation

Here you're basically there

I'll just tell you

(x+y)^p = x^p + (p choose 1) x^(p-1)y + ... + (p choose p-1) xy^(p-1) + y^p

[rapid junco]

yeah once you have char p its trivial

[dull ginkgo]

Yeppers

[rapid junco]

then pk for some p is pk times 1

[hollow mica]

every term on the RHS except the first on the last has coefficient divisibile by p

[rapid junco]

but this vanishes since R is Char p

[hollow mica]

Oh above you were trying to show that the characteristic is p? For that you can just do. 1 + .. + 1 = f(1) + ... + f(1) = f(1 + ... + 1)

[dull ginkgo]

It gets a bit weird because “p” isn’t “really” an element of this ring, it’s really f(p) where f is the map from Z into R.

which honestly I actually sometimes when doing stuff like that use rho(p) in order to represent that “prime” map

Just so I don’t screw myself up sometimes

[hollow mica]

(here f is the embedding f: Z_p -> R

If there are p 1s in the equation, then this quantity equals 0, and if there are less than p 1s, then this quanitity is nonzero by injectivity of f.
These are exactly the conditions required for char(R) = p

[dull ginkgo]

On the topic of binomial theorem another thing you can try is proving the nilradical (set of nilpotent elements) is an ideal for your commutative ring

[rapid junco]

intersection of all prime ideals so intersection is an ideal no?

[hollow mica]

You can "do" computations like this for intuition, but you really shouldn't refer to the element p in Z_p in an actual proof, unless you are thinking of Z_p as a quotient ring and know what's going on underneath

[rapid junco]

yeah its pk(1)

[dull ginkgo]

Proving that it me the intersection of all prime ideals is a bit tricky if you don’t shortcut through localization lo

[rapid junco]

which is an element of the ring

also

if I give you a surjective ring hom

can it be the case that we nessecarily have two different sets mapping to the same ideal

say I is an ideal and you look at its preimage

[dull ginkgo]

hm

Preimage of that ideal will be an ideal btw

[rapid junco]

yese

but could you have two ideals that have the same image

or is this preimage unique

[dull ginkgo]

i mean probably, let me think

[rapid junco]

Okay wait if P is a prime ideal and you look at the preimage of P then are we good?

[dull ginkgo]

Wait

[rapid junco]

then preimage of prime is prime.

[chilly ocean]

,rotate

[rapid junco]

I am trying to prove b btw

[hollow mica]

If P is a prime ideal in R containing ker(phi), show that

1. phi(P) is prime
2. phi^-1 sends phi(P) to P

[void cosmos]

ig all you have to do here is prove that it does indeed send prime ideals the kernel to prime ideals of S

[rapid junco]

Oh image of prime ideal is prime?

[void cosmos]

no the question is if you have phi a surjective ring homo

is the preimage of a prime ideal in S a prime ideal containing the kernel?

[hollow mica]

if your homo is surjective yes

[rapid junco]

0 is contained in all prime ideals

[rapid junco]

It's further. How do you not have two ideals mapping to he same ideal

[void cosmos]

yeah but that's later

isnt this c)

and ur asking for b

[rapid junco]

Injectiveness is tribial

You need to also proof for well-defined Ness, that if I give you two prime ideals, then their image of the map is the same i.e. the pre-image of them corresponds to the same prime ideal

Injectivity is a part of the well defined Ness of the map to my knowledge

Am I wrong here?

[topaz solar]

Injective should be easy, since if p, q in Y both map to J, then phi(J) = p, phi(J) = q so uhhh

[topaz solar]

Now here’s the kicker, is this your midterm?

@rapid junco

[rapid junco]

Not mine

From 2 months ago

[topaz solar]

A midterm now would be odd timing wise, but just to be sure

Why are you doing this now

[rapid junco]

final next week

[topaz solar]

Acceptable

[rapid junco]

I still dont understand how phi is injective

[topaz solar]

So have you shown phi inverse is well defined?

[rapid junco]

b) preimage of prime is prime

GOOD

[topaz solar]

As in, have you shown the preimage of prime is prime

[rapid junco]

yes

but how do i Know if A and B are two ideals that map to P then A = B.

[topaz solar]

Ok, suppose f(p) = f(q) = J

[rapid junco]

i.e. when I take the preimage

how do I know it corresponds to only 1 ideal

[topaz solar]

$\phi^{-1} p = \phi^{-1}q= J$

[cloud walrus]

Sage Sharp

[topaz solar]

When is x\in \phi^{-1}?

[rapid junco]

im not at injectivness yet

I need to show still if P = Q then f(P) = f(Q)

or in other words phi^-1(P) = phi^-1(Q)

the preimage of a prime ideal is unique.

i.e. there are not prime ideals A not equal B such that phi(A) = phi(B).

[topaz solar]

That’s well definedness

[rapid junco]

yes, and I have no clue

[topaz solar]

If f is a function and x=y then f(x) = f(y)

[rapid junco]

I know the defn of well definedness

I am stuck on how to show it

i.e. the preimage of a prime ideal is a unique prime ideal

not two seperate ones

or am I messing this up dealy

preimage of P is all x \in R such that f(x) \in P.

Oh Im dumb

OK so well definedness is easy then

[topaz solar]

Yes

[rapid junco]

Now suppose f(P) = f(Q). Then phi^{-1}(P) = phi^{-1}(Q).

Now suppose they are not equal. Choose x such that WLOG x \in \phi^{-1}(P) such that x is not in \phi^{-1}(Q).

Choose an element in phi^{-1}(Q).

say a

Then xa is contained in \phi^{-1}(Q) (ideal)

[topaz solar]

I think you might be over complicating matters to an extent

Pick x in f(P)

x is in f(Q) = f(P)

so phi(x) in Q

But phi(x) in P

[rapid junco]

phi(x) is in P?

oh yeah

[topaz solar]

That’s what it means to be in the preimage dawg

[rapid junco]

why does this imply P = Q. Sorry im slow

oh

bruh

if x \in F(P). then phi(x) \in P. Then \phi(x) \in Q.

Symmetry implies equality

[topaz solar]

phi(x) = P iff it’s in Q

But what do we know about phi

[rapid junco]

and finally surjectivity of f is a result of the well definedness of phi

phi is surjective

[topaz solar]

Yep, so phi(f(P)) = P

[topaz solar]

How so

Why is every prime ideal containing the kernel a preimage

[rapid junco]

if you have an element in P in X. Then phi(P) is prime in S. So phi^{-1} of phi(P) is what we want.

[topaz solar]

Why is phi(P) prime

Second, how do you know there’s no prime P < Q so that f(phi(P)) = Q

[rapid junco]

hmm idk

if x, y \in S such that xy \in phi(P). Then xy = phi(a) for some a \in P.

yeah honestly no clue does this require tech

or am I missing something easy again

[topaz solar]

Perhaps easy

[west nova]

is it saying G' is isomorphic to some subgroup H of G? and what is the relation between H and G by that unusual arrow?

[topaz solar]

Just saying it’s isomorphic to a subgroup yes

[west nova]

i see thanks

[winged void]

I have small question that I was thinking about lately. It’s about literally about the inverse of a function f(x) why is it always reflection on the line y=x but is the inverse not the following like f(x)*f(x)^-1 = identity but what is then the identity element for a function

Maybe it’s a stupid question

[topaz solar]

Compose functions not multiply pointwise

f(f^-1(x)) = x

[winged void]

Sure agree

But does this mean the identity =x

Or not really

[topaz solar]

Yes

[winged void]

But why is it the identity can you give some intuition about it

Please

[topaz solar]

Id(x) = x

So f(Id(x)) = f(x)

Same with Id on outside

So it’s the identity

[rocky cloak]

$\hookrightarrow$ means inclusion

[cloud walrus]

jagr2808

[crystal vale]

Give an example of a non-trivial group that is not of prime order and is not the internal direct product of two non-trivial subgroups.

Does Z/(4Z) work ?

[south patrol]

Yes perfect

Indeed Z/p^k works for any prime p and k >= 2

You could also just do something like Z if infinite groups are allowed, otherwise the above is your only option for the abelian case

[crystal vale]

Okay, thank you

[fading field]

yar

[hidden wind]

i keep stumbling over such elementary things just trying to learn basic group theory 😭

(well actually so far i’ve only really stumbled over cardinality arguments, which probably just means i should do more of those)

these kinds of things are no fun

[dull ginkgo]

Is it possible to derive a lot of Galois theory from independence of characters?

I do wonder if there’s some use for a field F, the group ring/algebra F[Aut(F)]

I just got brain blasted

I have an idea. Baby’s first thought

Each automorphism of F lifts to F[Aut(F)], and we also have valuation linear maps from F[Aut(F)] to F

Finite subgroups are special in this regard because it kinda just shuffles the subring F^S[S]

[dull ginkgo]

WAIT KERNELS OF THE VALUATIONS

I CAN USE CHINESE REMAINDER THEOREM SOMETIMES

watch out chat I am no longer cooking I am fucking BAKING

Actually each of these valuations are into a field so the kernels are all maximal and thus coprime unless they’re literally the same fucking map

[feral timber]

If F_q^* is the finite field of order q excluding 0, then the number of characters on it are (q-1). Why?

Can I say that this is because it is an abelian group, so all irreducible reps have degree 1, and each correspond to a character, and the sum of degrees must be the order of the group, hence number of characters is (q-1)?

[rocky cloak]

Sure, that argument works.

You could also say that the number of characters is the number of conjugacy classes, and since it's abelian each element has it's own conjugacy class

[south patrol]

You can also just do it by construction. Fix a generator of F_q^x and you can send it to any (q-1)st root of unity to get a 1D rep

Then ig you need to argue all characters are 1D which follows from being abelian

[feral timber]

Thanks

[heavy verge]

Can someone please explain what my professor is saying in question 2 with his solution. The hand waveyness of it seems to be confusing me, I dont see how this is an application of the third isomorphism theorem nor do I understand how its acceptable to apply \phi to the top and bottom of the quotient? Can someone help me make this rigorous?

[celest furnace]

Yes and I think Artin took this approach because he was “appalled that Galois theory depends so much on the primitive element theorem” or something like that

He prob has a book on it

[dull ginkgo]

I refuse to open Artin algebra

I’ll just keep trying stuff out using Chinese remainder theorem later

[celest furnace]

Why

[dull ginkgo]

i don't wanna sad face

Though with dedekind independence of characters I wonder if you can prove some stuff here

[grizzled crane]

can a polynomial have a zero in a field F but not reducible over F?

I feel like no, because if the polynomial f(x) has a zero at ‘a’ then (x-a) divides f(x), now by division algorithm, I’m guaranteed to get the quotient q(x) with in F such that f(x)= (x-a)q(x)

[coral spindle]

x-1 is irreducible in all fields :)

If you restrict to degree at least 2, you’re good.

[grizzled crane]

I mean, given any polynomial f(x), if we know f(x) has a zero in F, can we conclude f(x) is reducible over F, not (x-a)

[coral spindle]

You’re not seeing my point, I think

[grizzled crane]

i think. I didn’t get it 😦

[coral spindle]

Does x-1 have a zero in Q?

And note of course, it is irreducible.

[grizzled crane]

It’s irreducible and has zero when x=1

[coral spindle]

So...

[grizzled crane]

i should ask, can all polynomial

[coral spindle]

OK

[fading field]

wait now i don’t know what you’re asking (i’m risking sounding like the comedy sketch that children do, now you’re modifying the first part but now i’m asking “can any polynomial what?” but i don’t get what you’re trying to ask now)

[coral spindle]

I think that goes both ways

[grizzled crane]

if f has zero in F then f is irreducible over F

is this statement true

[fading field]

no

[grizzled crane]

that’s what i was asking

[coral spindle]

...

[grizzled crane]

yes, i think that’s one example right?

[coral spindle]

That might be why I mentioned it!

If you restrict f to be degree at least 2, you’re good.

[fading field]

if f has degree 2 or more and f has a root a, then f has to be reducible for the reason you said earlier, because (x-a) divides f

if f has degree 1 it’s always irreducible (why?)

[grizzled crane]

the only way you could express (x-1) is (x-1) times some unit

[fading field]

yeah that’s right

wait

no

your thing is close and i think you meant the right thing

[grizzled crane]

(x-1)= (-1)(-x+1), irreducible over Z, since -1 is a unit.

anyhow i try to express, it will lead to something like this.

[fading field]

yeah. if you’re over a field you can do this with any nonzero element

[grizzled crane]

yes

[fading field]

i only mention this because you seemed like you cared about the case where the underlying ring was a field

[south patrol]

Idk what you mean by the "since" there lol

[fading field]

in fact, there are degree 1 polynomials that are reducible if the underlying ring is NOT a field

[grizzled crane]

I think 2x+4 is reducible over Z[x]

it’s degree 1, but 2 isn’t a unit in Z

[south patrol]

Ye

[grizzled crane]

Given a field F, if f is reducible over F then f has a zero in F.

is it true?

in a ring Z[x], 2x^2+ 2 is reducible, but it doesn’t have zero in it

what about fields in generals ?

please note, i’m asking about the convese

[celest furnace]

False

Take the product of any two irreducible quadratics

[grizzled crane]

right

[coral spindle]

You say "Given a field F..." but you are looking at Z[x].

[grizzled crane]

No, i was showing an example i had

[grizzled crane]

^

[fading field]

Z is not a field