groups-rings-fields

[rustic crown]

last one is false

the usual way to think would be to lift(extend?) the action of Z to an action of Q, now they have different dimensions as Q-vector spaces.

[lethal dune]

as a Q vs they have diff dim, but does it mean they are non isomorphic?

[rustic crown]

but if you wanna remove this language, then consider (1, 0) and (0, 1) on the right, if there was an isomorphism, then you can pull these back to rationals a and b.
||But any two rationals are Z-linearly dependent, which will prove (1, 0) and (0, 1) are dependent||

[next obsidian]

Yes, but this requires a little work

[hidden haven]

Wasn't what I meant by the last one

I didn't want to say which ones my hint solves

[rustic crown]

(oh oops)

[rustic crown]

so it's like if we had an isomorphism between Q and Q x Q as Z-modules, then we can lift it to an isomorphism of Q-modules. But Q-iso would imply the dim are same.

[next obsidian]

You have to use the fact that Q = Frac(Z) for this to work

[lethal dune]

ye (took no work at all)

[shell brook]

is there a nice way to find the inverse of a polynomial in Z_n[x]

sry this is a vague question i guess

but say i give you a polynomial and tell you it has an inverse

is this just super annoying

or maybe just check if its a unit at all

[rustic crown]

can we say that all coefficients other than constant term will be divisible by every prime factor of n?

[next obsidian]

So like

There’s a criterion for a polynomial being invertible

Constant is a unit, all others are nilpotent

Under those assumptions you can manually compute the inverse

So yes, you can

[trim grove]

no one tagged me

[shell brook]

what is nilpotent

[rustic crown]

some power is 0

[next obsidian]

r^n = 0 for some n

[shell brook]

ah ok

interesting

i realized that in my specific example i was just fucking up something w a system of equations but very cool

[trim grove]

like (example)?

[hidden haven]

What do you mean

Do you want a hint on how to prove that

[trim grove]

i just want to know any subgroup of Q , other then Z , {e}, itself

[hidden haven]

ℤ is a subgroup of ℚ

Multiples of ℤ

[lethal dune]

p-adic numbers?

[rustic crown]

so moldi which options does that solve... i'm confused 🙈

[hidden haven]

The numbers of the form m/2^n

[hidden haven]

They don't sit inside ℚ

[trim grove]

i missed this one

[hidden haven]

3, 4

[rustic crown]

so what about this... given a subset P of all prime numbers, consider the subgroup generated by 1/p^n for p in P and n in N

[hidden haven]

That's what this is

Oh wait

All primes

[rustic crown]

(i meant for the first one)

[hidden haven]

Isn't that just ℚ

[rustic crown]

we can recover that subset P from the subgroup

[hidden haven]

Oh

P is any subset

Right should work

[rustic crown]

right

[hidden haven]

Nice

[lethal dune]

we can decompose every number in ℚ in the form a1/p1^k1 + a2/p2^k2+..?

[rustic crown]

yea but primes are given to us

so no

else do partial fractions decomposition

[lethal dune]

no my statement was independent of yours

[trim grove]

is this work for first one?

[hidden haven]

Yes

[rustic crown]

since there are infinitely many primes, there are uncountably many such subsets P. which gives you uncountably many subgroups of Q

[rustic crown]

so say you wanted 5/4 * 9 * 7
since 4 * 9, 4 * 7 and 9 * 7 are coprime, we can find
x * 4 * 9 + y * 4 * 7 + z * 9 * 7 = 5

[lethal dune]

yes

I was thinking about the finitely generated one

(kinda not sure)

like one of gens doesn't necessarily have to be 1/pⁿ, could be 2/pⁿ, not sure if that'll make a diff

though the set not being dense automatically makes them cyclic

but I'm not fully convinced

[trim grove]

i am not getting it who this is uncontable?

[hidden haven]

Prove that adding a generator to a non dense subgroup can't make it dense

Or like sum of 2 cyclic subgroups is cyclic lul

[lethal dune]

that approach is clear, just wondering about a purely number theoretic approach

[rustic crown]

purely nt would be just show that subgroup generated by the rational is the cyclic one generated by their gcd

[hidden haven]

This is all number theory though

[lethal dune]

ye

[hidden haven]

GCD in ℚ

Ik what you mean

[rustic crown]

🙈

[hidden haven]

But GCD in a field

[lethal dune]

you know we were taught that shit in grade 7

gcd in ℚ

[rustic crown]

GCD in field of fractions of a UFD 🙈

[hidden haven]

[rustic crown]

gcd of num/lcm of denom

[lethal dune]

even there was one question asked in UPSC about it

ffs

[trim grove]

[lethal dune]

translation: find the lcm of 4/3, 8/9, 3/5

[hidden haven]

What language is that

Bengali?

[trim grove]

W.B

[lethal dune]

bengali

[hidden haven]

[lethal dune]

that's pseudomath

[rustic crown]

[hidden haven]

Least common integer multiple

As long as they clarify

[lethal dune]

no that literally mean least common multiple

[trim grove]

looks like this is not form UPSC

[lethal dune]

idk where it was from but one of those

[gritty sparrow]

i never liked lcm's in general

i feel like they should all be 0

[lethal dune]

[hidden haven]

But that's not even defined though

[trim grove]

because as per i know they ask subjective questions

[lethal dune]

pseudo mathematiks

idk shit about upsc

[paper flint]

Classic "find the next number 1,4,9,25,..."

[rustic crown]

16

[lethal dune]

no it's iota

[paper flint]

Sorry the correct answer is 18961

[rustic crown]

[trim grove]

see carefully it should be iota raise to power iota

[hidden haven]

Not well defined

[lethal dune]

see it's i

[rustic crown]

is there a formal way to define these next term question?

[trim grove]

loks like prime square so i will go with 49

[hidden haven]

It's obvious if your IQ is high enough

[lethal dune]

ye like newton diff =0

maybe

[hidden haven]

The fact that you even ask

[rustic crown]

🙈

[paper flint]

No, it's a meme

[hidden haven]

1 is not prime

[trim grove]

sometimes yes sometimes no

depend upon question asked

[rustic crown]

it can be fibonacci squared idk

[hidden haven]

[lethal dune]

[hidden haven]

What kinda math you been studying

[rustic crown]

[hidden haven]

That doesn't have fundamental theo arith

[coarse forge]

OEIS time

[hidden haven]

Uniqueness up to multiplication by the special prime

[lethal dune]

classic your mom sequence

[hidden haven]

Bruh

[gritty sparrow]

well a formalization might be something like find the minimal length first order formula that defines a function that works for the given values or something.

[hidden haven]

[rustic crown]

is there a way to prove stuff with that definition?

[gritty sparrow]

probably not easily

[hidden haven]

Absolutely not

[lethal dune]

(abstract nonsense)

[trim grove]

its look like just author knows what will happen next in missing sequence question

[gritty sparrow]

ie you would upper bound by the polynomial trick then do the rest by case analysis

[paper flint]

You could define a first order formula ascribing any value to the missing term though, right?

Maybe it won't be minimal length though

[lethal dune]

only take the values you are given?

[hidden haven]

He is ordering all formulas

[paper flint]

Oh

[trim grove]

BTW Thankyou all for helping

[hidden haven]

Where the bking emoji at

[gritty sparrow]

what is bking?

[rustic crown]

bleak cat king i suppose

there was probably bking emoji at some point, don't remember

https://cdn.discordapp.com/emojis/913315700993577031.webp?size=160&quality=lossless

[hidden haven]

It was so bad though even though and it were made from literally the same cat photo

[desert dome]

I am working on (b). Any hint on how to think of such an example?

[hidden haven]

Add a root of a polynomial but don't add all the roots

And then in L do add all the roots to make it normal

If you still have a hard time you first gotta see that the polynomial has to be degree at least 3 (otherwise adding even a single root splits the polynomial)

[desert dome]

I saw this https://math.stackexchange.com/questions/2078494/show-that-mathbbqa-mathbbq-is-normal-where-a-is-a-root-of-the-irred and I thought it is some general theorem.

will try that!

[hidden haven]

Yeah so sometimes, adding a single root automatically adds all the roots

You have to avoid that

And there are simple enough ways to do it for a polynomial over Q

Like you have to choose a polynomial

A hint that works for some easy choices is that R is a subfield of C

You can also use this hint to come up with sort of a stupid example lul

With the fact that C is algebraically closed

[desert dome]

Thanks! will try that now

[sturdy marsh]

[delicate orchid]

The great bleakening must come to an end

[summer cradle]

when you define in formal way you give the answer (a general theorem says the next term of any series is )

[summer cradle]

by choice axiom

there can be multiple functions

for infinite series U{x}

[desert dome]

I guess I can let F=Q, E = Q(a) where a = sqrt(1+sqrt(3)) and L = algebraic closure of Q.
f(x) = x^4 - 2x^2 - 2 has a as a root but doesn't split completely

[trim grove]

any hint?

[hidden haven]

For what

[trim grove]

for first one

[hidden haven]

Oh the 17 thing is wrong

[trim grove]

i think 14 will work 14 = 7 x 2 and (sqrt(13)-1)(sqrt 13 +1)?

[hidden haven]

Ye that sounds fine, you'd have to show that all those factors are irreducible though

[trim grove]

i will

[trim grove]

can you please help we in 1st one?

[hidden haven]

Seems true but

[trim grove]

but

[hidden haven]

[trim grove]

okk i will try

[trim grove]

every time i have to google what these emojis says

[hidden haven]

This emotion has not been invented yet

[rigid depot]

Hello, I have a question. Let $B$ be an integral domain. Then $B$ is noetherian of Krull dimension $\leq 1 \iff$ for all $b \neq 0$ the quotient ring $B/bB$ has finite length. I have already shown the implication $\implies$. Assume $B/bB$ has finite length for all $b \neq 0$. Then $B/bB$ is an artin ring, hence any prime ideal is maximal. Since the prime ideals of $B/bB$ map bijectively to the prime ideals in $B$ which contain $bB$, it follows that $B$ has dimension $\leq 1$. But how do I show it is noetherian?

[cloud walrus]

shu

[next obsidian]

A ring is Noetherian if all the prime ideals are finitely generated, a proof can be found in eg Matsumura

Now, let P be a non-zero prime ideal, and take b in P nonzero, then P/b is finitely generated, say by the images of x1 through xn

It follows that P is generated by x1 through xn and b, so that P is finitely generated

This shows all of the height 1 primes are finitely generated, and all that’s left is the 0 ideal which is finitely generated

@rigid depot

Actually, I think you don’t even need to use the fact that you can check Noetherianness just on primes

Just replace P with any non-zero ideal, and run the same proof and you get that any non-zero ideal is finitely generated

Then all that’s left is the 0 ideal which is finitely generated

Cool problem

[rigid depot]

Why is P/b finitely generated?

[next obsidian]

B/bB is Artinian

[rigid depot]

Oh my bad

Thank you! 👍

[next obsidian]

I guess you can say something more general

A ring R is Noetherian iff R/r is Noetherian for all r in R

You just run the same proof for the <= direction

And the => is immediate

[rigid depot]

Is that really true? I only know that R noetherian => every finitely generated R-algebra is noetherian, in particular R/r is noetherian

How would you show the converse?

[next obsidian]

Just do the same proof

Grab a non-zero ideal

Grab an r in I

Look at I/r

Lift up generators, and then add in r

It’s exactly the same thing I did in the proof I outlined above

[rigid depot]

Cool, thanks

[next obsidian]

Oh I guess I should say non-zero r here

If you let r be 0 then this is trivial lol

[rigid depot]

Yeah lol

[chilly ocean]

Why wouldn't just saying this work, that the mapping argument with prime ideals pulling back to B from B/bB gives you the ascending chain condition since B/bB are all finite length by hypothesis

[foggy merlin]

wouldn't one have to prove that bB is noetherian anyways if that line of reasoning is used?

[next obsidian]

What do you mean that bB is Noetherian?

And no, it works fine because if you had an infinite ascending chain
I_0 < I_1 < … then WLOG you can assume I_0 is nonzero

Then for some b in I_0, append (b) to the front of this

Mod out by (b) and you have your infinite ascending chain in B/b

[foggy merlin]

yeah that makes sense

[trim grove]

some one would like to help me in first option?

[hidden haven]

If it factors into linear factors, then that factorization would also be a factorization over Q[sqrt 13], so it must be the same as the factorization you found before up to units, just show that then it doesn't work out (I think it won't but idk)

The problem is that it could factor as a quadratic and a constant

Oh wait it is monic

It can't fact as a quadratic and a constant because then that constant would have to be a unit making the factorisation trivial

So it should be fine

[trim grove]

this is the same method i tried to apply , but get confused so much that i leave the question

[prisma shuttle]

hey guys i am a bit confused about the concept of an ideal in a ring

is an ideal a subgroup of the ring or a subring?

what does it mean for something to be a subgroup of a ring (like it is only additive and it has no multiplication operation?)

[hot lake]

it's neither a subgroup nor a subring

[delicate orchid]

an ideal is just any subset that satisfies the properties of an ideal. It's useful to think of them as analogous to "normal subgroups" from group theory. Ideals aren't subrings in general though unless you're talking about rings without identities

[hot lake]

a subgroup of a ring would be a subgroup of the additive group of the ring

[delicate orchid]

^

beat me to it lol

[prisma shuttle]

ah ok thx for the clarification

then what does this part of the book mean

is that incorrect?

[hot lake]

an ideal is a subgroup of R but with additional properties

so all ideals are subgroups but not all subgroups are ideals

[prisma shuttle]

oh ok i see

so an ideal is a subgroup of R

but its like more than that

[delicate orchid]

that's a really weird way of defining it

[hot lake]

it is a subgroup I of R such that forall r in R, rI is contained in I

it's a pretty concise way of defining ideals

[delicate orchid]

I'd just say an ideal is a subset $I \subseteq R$ such that for all $i_1, i_2 \in I$ and $r \in R$ we have $i_1+i_2 \in I$ and $ri_1 \in I$

[prisma shuttle]

ok thx so much

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

I prefer this definition actually because I think you can then show that I is an additive subgroup just from these two properties, which means less checking

[misty harness]

An ideal of a ring is **always **a subring of that ring. Moreover, ideals aren't defined for groups, only for rings.

[next obsidian]

Ideals are not a subring

Unless you don’t require rings to have a 1

In which case, perish

[misty harness]

So the only ideal that is also a subring is the ring itself?

[next obsidian]

Yes

If you want to treat ideals and rings similarly you treat them as modules

Which is a very fruitful perspective

[chilly ocean]

im having trouble figuring out how to approach this problem, I went to the help channel and was recommended to come over here for help.

[next obsidian]

You can characterize maps out of a cyclic group in terms of the order

Note that if it’s generated by g, that where you send g determines the entire map

i.e. phi(g^n) = phi(g)^n

So you just have to figure out how many choices of what phi(g) is are valid

This is true for maps from G to any group H even, not just for maps G -> G

[chilly ocean]

okay so

[next obsidian]

The criterion is given purely in terms of the order of the thing you set phi(g) equal to, and I’ll let you try and figure out what it is exactly

[chilly ocean]

i would first find the generators?

[next obsidian]

No

You just assume that into existence

It’s generated by g, okay done

Let H be any group, and G be a cyclic group of order n, generated by g

Try to find out when phi(g^n) = h^n is a valid homomorphism for h in H

It’s purely in terms of the order of h

And n

[tribal moss]

Surely you don't mean "in terms of the order of h and n"?

[next obsidian]

Maybe try to figure out if you can say something about |h| relating to n if there existed a phi:G -> H such that phi(g) = h

(Order of h), and n

n and the order of h

[chilly ocean]

im so lost man

[next obsidian]

Idk is that less ambiguous?

[chilly ocean]

whats the best book for abstract alg

[tribal moss]

Oh! It looks like n has two different meanings here. You meant a fresh variable instead of n when you wrote phi(g^n)=h^n, right?

[chilly ocean]

One big reason we like to talk about ideals rather than "subrings" is cause people often carry two definitions of what a ring is that are at odds. Some define a ring with identity necessarily, and some don't. This makes "subring" ambiguous. But the definition of an ideal is unambiguous, and also ofc it plays the role of normal subgroups in the context of rings etc etc

[next obsidian]

Oh, yeah

I meant n the order of G

And I guess phi(g^k) = h^k

[tribal moss]

for all k.

[next obsidian]

Yeah

[chilly ocean]

I'm not sure if i understood anything you wrote tbh

[next obsidian]

Okay, I tried to generalize this to get at a really general statement

[tribal moss]

So let's take it a small bit of the time.

[next obsidian]

Let’s just do the case you have where G is order 30

First note that if you had a map phi:G -> G

[chilly ocean]

yes

[next obsidian]

That you’re forced to set phi(g^k) = phi(g)^k

And since g generates G, all elements are of the form g^k for some k

So phi(g) determines what you have to send everything to

Now there’s 30 elements in G

e,g,g^2,…,g^29

[chilly ocean]

yes

[tribal moss]

("Let G be a cyclic group of order 30" implies you're being promised that there is a g that generates G; you don't need to "find" it before you can use it).

[next obsidian]

You have to determine when the map given by phi(g) = g^m is valid, as in it actually is a group homomorphism

Aka phi(g^k) = g^mk

[next obsidian]

I wrote it as phi(g) = g^m because this is the only choice you’re making

Once you pick where g goes, everything else is forced to go to a power of g^m

[chilly ocean]

i thought homomorphism were when phi(xy) = phi(x)phi(y)

[next obsidian]

Yes

[chilly ocean]

our phi function doesnt act the same since we have

[next obsidian]

So you have to determine for which m the map phi(g^k) = g^mk actually satisfies that rule

[tribal moss]

Note that if you set both x and y to be g in phi(xy)=phi(x)phi(y) you get phi(g^2)=phi(g)^2.

And if you then set x to g and y to g^2, you get phi(g^3)=phi(g)^3.

And so forth. That's what Chmonkey has summed up as phi(g^k)=phi(g)^k.

[chilly ocean]

ohhhh okay it makes sense

[next obsidian]

Sorry, I didn’t realize that part was unclear

[chilly ocean]

where does the m take play here

[next obsidian]

m is any integer

[tribal moss]

Since every element of G is of the form g^m (that's what we're assuming about G), it means that the value of phi(g) is all we need to know about phi to reconstruct all the other values of phi.

[next obsidian]

You’re picking an element of G to send g to

Elements of G look like g^m for some m

If you like, you only have to let m be from 0 to 29

[next obsidian]

Because of this

[chilly ocean]

yeah

[next obsidian]

Anyway, it might seem trivial that under this definition of the map that phi(xy) = phi(x)phi(y)

Like, if you have phi(g^kg^j)

This is phi(g^k+j) = g^m(k+j)

But phi(g^k)phi(g^j) = g^mk g^mj = g^m(k+j)

So wtf???

Actually so this shows it is a group homomorphism

The actual problem is if the map is well defined

note that g^k can be equal to g^j even if k ≠ j

An example is that g^30 = e

So you need to know that then, phi(g^30) = phi(e)

Note that if you tried to like, send g to 1 inside of Z, then this would be asserting that 30 = 0 inside of Z

Which is false

Maybe that last point is more confusing than anything else

[tribal moss]

I think it might be a bit confusing that you're solving a problem that doesn't actually arise in Par's exercise.

It's still work that needs to be done, of course.

[next obsidian]

Yeah, but you have to address it still

Does the issue of well-definedness make sense?

I can try to illustrate an example again to show what could go wrong

[tribal moss]

Let's try a concrete example.

[chilly ocean]

theres too many variables bro

im like lost

[next obsidian]

Do you know what it means for a map to be well-defined?

[tribal moss]

For example, we have proved that if there is a homomorphism phi: G -> G such that phi(g)=g^5, then it must be that phi(g^k)=phi^5k for all k.

[chilly ocean]

my notes look nothing like what you talked about

[tribal moss]

But is there actually such a phi or not?

[chilly ocean]

oh okay

well defined means one element from g could be mapped to something in h

right?

[next obsidian]

Or well

To not be well-defined

Means that one thing might potentially be mapping to more than one thing

[tribal moss]

And what we need to check is that, for example g^17 = g^47, so it had better the case that g^(5·17)=g^(5·47) too; otherwise phi could not have all of the properties we have assumed about it, at the same time.

[next obsidian]

So the issue here is the map phi we defined

Is that phi(g^k) = g^mk

But k that we took there can vary

Like if k = 1

Or k = 31

[tribal moss]

I think "is this phi well-defined" is a confusing way to word it. We should be saying "have we actually defined a phi at all"?

[next obsidian]

g^1 = g^31 = g^61 = …

So when we want to send that element to g^mk we need to know that for k = 1,31,61,… that those are actually the same thing

[chilly ocean]

alrighty

[next obsidian]

Oh here’s a super explicit example to show how this can fail

You’ve seen Z/nZ?

[chilly ocean]

yes

[next obsidian]

Do you know the elements as like

[k]

Or something

[chilly ocean]

we didnt go over any of the mod stuff

tbh

[next obsidian]

Oh okay well how would you write down the elements in Z/nZ

If you have to list them off

[chilly ocean]

honestly no clue man

i think its time for me to read the book one more time

[next obsidian]

Okay let’s just list them off as k mod n

Where k mod n = j mod n if k - j is divisible by n

[chilly ocean]

this subject is too hard for me

[next obsidian]

So suppose we wanted to make a “function” f:Z/nZ -> Z

Which said I map k mod n to k

This isn’t well-defined

Because look, 0 mod n maps to 0

And n mod n maps to n

But those are the same thing!

0 mod n = n mod n

So it’s like saying, if I set x = 0 mod n

Then I need f(x) = 0

But also f(x) = n

[chilly ocean]

ohhh okay

[next obsidian]

So going back to the thing with G the cyclic group of order 30

I want to make the map phi(g^k) = g^mk

But g^k = g^(30+ k)

So I need g^mk = g^m(30 + k)

It’s the same thing like with Z/nZ

Does that make sense?

[chilly ocean]

yeah kind of

i just lack knowledge bro

imma just read the book

and hope for the best

[next obsidian]

Alright

[chilly ocean]

nothing makes sense

like

idk why

[oak grove]

what knowledge do you think you lack

specifically

is it just a confidence gap

[chilly ocean]

no no

its like literally

i know what a group is

and its properties

i know a subgroup and its properties

[oak grove]

you know generators

[chilly ocean]

i know generators kinda

[oak grove]

and you said what a homomorphism is

and an isomorphism

and some properties of each

[chilly ocean]

yeah i know definitions

thats it

im a dictionary rn LOL

anyways imma take a break

thanks for helping out everyone

[prisma shuttle]

hey guys

for this probelm

why can't you just do $a^m = a^n\implies a^m(a^{n-m}-1)=0\implies a^{n=m}=1$

[cloud walrus]

JustKeepRunning

[foggy merlin]

because $(a^{n-m}-1)$ might not be zero, it might just be a zero divisor, take the element 2 in the ring Z/6Z as example for a

[cloud walrus]

Dr. J. Stockfish

[chilly ocean]

2 is a zero divisor tho so that can't be a anyway
I think the argument you (justkeeprunning) are trying is pretty close to what the author does anyway

[foggy merlin]

Well it's trying to show that a is either a zero divisor or a unit. So 2 can definitely be a.

But yes the argument is essentially the same

If one writes literally one additional line to be more precise

[rustic minnow]

Im wondering what to do for c)?

how is it related to irreducible polynomials?

[south patrol]

it's a field iff <x^2 +c> maximal

Consider what happens if x^2 + c is not irreducible

[rustic minnow]

do we end up with a denominator of zero in that case? So we need x^2 + c to be irreducible?

[tribal moss]

You'd end up with zero divisors -- most unfieldlike!

[rustic minnow]

right

Im still confused as to why the notation is <x^2 +c> though

is it for a ring?

[tribal moss]

That's the usual notation for a principal ideal, isn't it?

[rustic minnow]

oh right, you are correct.

so c=1 and we get no roots in Z_3

[cloud walrus]

dumbo

[tribal moss]

State the equality in terms of individual elements of the (shared) domain of f and g.

Sure, but it sounded like you wanted a more detailed phrasing.

f(a) and g(a) are each an element of G×H, so suppose f(a)=(g1,h1) and g(a)=(g1,h2). Then g1 = pi1((g1,h2)) = pi1(f(a)) = pi1(g(a)) = pi1((g2,h2)) = g2, and h1 = pi2(f(a)) = pi2(g(a)) = h2, and therefore f(a) = (g1,h1) = (g2,h2) = g(a). Since a was arbitrary this shows that f = g.

[prisma shuttle]

in the def of a product of ideals presented here

https://commalg.subwiki.org/wiki/Product_of_ideals

why is the third definition equivalent to the first definition?

shouldn't the third be $\sum_{i=1}^n \sum_{j=1}^n a_ib_j$ in order to generate all the elements of the first definition?

[cloud walrus]

JustKeepRunning

[delicate bloom]

no it shouldn't because that's more restrictive, your sum factors as (a_1+...+a_n)*(b_1+...+b_n) which is only an element of the form ab for a in I and b in J, but there are more elements that we get by adding elements of this form together that can't be factored

the additive subgroup generated by elements of the form ab means we take elements of that form, and then let ourselves add them in every possible way. Elements of the form ab are not in genearl going to be closed under addition

[trim grove]

how the definition of irreduicibility varies over I.D, UFD and fields ?

[rustic crown]

it doesn't vary

an element a in R is called irreducible if any decomposition of a = b * c satisfies that exactly one of b or c is a unit.

over a general integral domain you can't say much more... but over UFDs these are same as non-zero prime elements.

fields have no irreds because 0 = 0 * 0 and for a unit u, u = 1 * u.

[lethal dune]

for maximal ideal do we need the ring to be commutative or that it has an identity? (like R/M is a field so commutative)

[chilly ocean]

but all rings are commutative

[rustic crown]

if you're quotienting then your ideal would be two-sided, i wonder if that makes the difference

[lethal dune]

no like M is maximal <=> R/M is field, is it true even if R is not commutative

[rustic crown]

(but usually i've only seen these definitions for commutative rings, for non-commutative they explicitly say maximal left ideal and stuff)

oh that should false lol

the ring M(n, k) of n x n matrices over k is simple

[lethal dune]

on the same not I've seen ppl requiring 1 to be in S for a subring S or R, i.e. ideals are almost never a subring

[rustic crown]

yea the usual notion of subobject is like the inclusion S --> R should be a ring homomorphism

[lethal dune]

(even rings may not even be associative, so much variation fuck)

[rustic crown]

so not just that S needs to have a 1, that 1 needs to be same as that of R

(most people have associative rings with unity lol, only non-associative thing i've seen are lie algebras)

[lethal dune]

I don't even know what I'm reading anymore

every author seems to have their different definition of a ring,subring stuff

[rustic crown]

also in general for a commutative ring R, the (two-sided) ideals of M(n, R) are precisely M(n, I) for ideals I of R

[rustic crown]

yea lol, but the more general you go, the less nice theorems you get

[lethal dune]

ye

[rustic crown]

also this reminds me....

I had a question about lie algebras which i was trying yesterday

so are epimorphisms of lie algebras surjective or not?

[lethal dune]

(idk epi generally means surjective, at least what I have encountered so far)

[rustic crown]

the situation of lie algebras feel pretty similar to that of groups, so i'm expecting it to be true... but the proof for groups was kinda whacky, it first says that it's enough to show for inclusion maps H --> G, and if H was normal then we're done because G --> G/H the quotient map and the zero map both agree on H, so they need to agree on G, which means G = H. This told us that H has at least 3 cosets, say H, Hu, Hv. We can define a permutation f of G which swaps hu and hv (for each h) and leaves everything else fixed.
Then the proof constructs two maps G --> Perm(G) such that they agree on H but don't agree on all of G. one map is the usual left multiplication thing. other one conjugates this by f.

I tried to copy something similar, but wasn't able to succeed

for a lie algebra L and a subalgebra K i need to find some object Z and two maps L --> Z which agree on K but not on L.
looking at the thing for groups, I hoped that L --> gl(L) the adjoint representation would do the job...

for the conjugation thing then i required something in GL(L). but to make this agree on K i would require some properties, but the obvious ones were not working out nicely

the map f in GL(L) would need to satisfy [x, f(y)] = f([x, y)] for each x in K and y in L. how should i find such a map? (but not for every x in L and y in L)

one condition where this would hold is when f : L --> L is a map of lie algebras and f fixes K. but it doesn't feel like something this would work, idk

when will you wake up >.<

[mystic radish]

So, I'm pretty new to abstract algebra and I'm studying the Fraleigh book on my own.
What is exactly the difference between an orbit and a cycle?
Fraleigh defines a cycle to be "at most one orbit containing more than one element."
But AN orbit is also ONE orbit.

[trim grove]

Finally ,i think i got it

[desert dome]

Any suggestion for this problem?

[rustic crown]

you wanna prove that minimal polynomial looks like x^(p^n) - a for some a in k

so start with minimal polynomial of alpha, say it's f in k[x]. what can you say about f given that K/k is purely inseparable?

[desert dome]

it has repeated roots. its derivative = 0

[rustic crown]

well just a little thing... alpha could have been an element of k to begin with, so we can't say that for sure

but you're right, if we assume that f was inseparable, then f' = 0

so f(x) = g(x^p)

notice that eventually we wanna show that f is a polynomial in x^(p^n)

we've already kinda seen that it's a polynomial of x^p

[desert dome]

I dont quite get this

[rustic crown]

oh try thinking about

start with any such polynomial say a_0 + a_1x + ... + a_n x^n, what does f' = 0 tell us?

[desert dome]

the coefficients in f' are 0 so either original coefficient is 0 or a_i * i is zero

so a_i * i are multiples of p

[rustic crown]

yep, so a_i * i is 0 in the field k, if a_i was non-zero this would force i = 0

so i is a multiple of p

[desert dome]

does i have to be a multiple of p?

[rustic crown]

yep, that's the only way an integer in the field k would be 0

(recall that p was the characteristic)

[desert dome]

a_i * i has to be multiple of p, is it possible that none of them is multiple of p, but their product is

[rustic crown]

a_i * i has to be multiple of p
this statement the way it is written is incorrect

because the coefficients a_i are in k, so a_i * i is an element of k... what does it mean for it to be a multiple of p?

p = 0 in k

[desert dome]

ahhh

if a_i * i is a multiple of p, it is 0

so i has to be multiple of p

i see

[rustic crown]

yep, and we had that directly from the derivative condition

a_i * i = 0

[desert dome]

right right

[rustic crown]

now if a_i isn't zero then by field stuff we can say i must be zero

also maybe i should say that we're confusing an integer n in Z with the element 1 + 1 + ... + 1 (n times) in k so it could feel a little weird

anyway, so i = 0 in the field k. This can only happen if i is a multiple of p in Z

[desert dome]

yes

to make i = 0, the power of x has to be p^n?

[rustic crown]

nah... just any multiple of p would also make it 0

all we can say that is the polynomial now looks like
a_0 + a_p x^p + a_{2p} x^{2p} + ...

which is what i wrote f(x) = g(x^p)

[desert dome]

why dont we have a_{p-1} x^p-1?

[rustic crown]

oh because we proved that if a_i is non-zero then i is a multiple of p. So if i isn't a multiple of p then a_i must better be 0.

[desert dome]

ohhh I see.

[rustic crown]

only way the derivative is gonna die is because of those p from the powers of x

this a pretty standard thing... issokie, you might be seeing it for the first time

so if f is an inseparable irreducible polynomial, then we get that f(x) = g(x^p)

[desert dome]

yes

[rustic crown]

we'll repeatedly use this now

let's go back to the problem

so we had alpha in K, with irreducible polynomial f in k[x]

if f was insep, then f(x) = g(x^p)

notice that this g is also irred, because if it were reducible then so would be f

now iterate the reasoning

if g was insep, then g(x) = h(x^p)

with h irred...

since the degree goes down, we'll eventually hit a separable polynomial!!

f(x) = g(x^p) = h(x^(p^2)) = .... = f_{sep}(x^(p^n))

so alpha^p^n satisfies f_sep right

which means alpha^p^n is separable /k

[desert dome]

right

[rustic crown]

pure inseparability tells us it better be in k

[rustic crown]

(this n, may even be 0, in which case f was separable to begin with, that is alpha itself was in k)

that pretty neat right... it basically tells you that inseparability occurs in clumps of p^n

[desert dome]

because otherwise, these alpha^p, alpha^p^2 ... can be distinct roots so it will be separable?

[rustic crown]

alpha^p may not be a root

all we know is that alpha^p^n is a root of f_sep

and f_sep has no repeated roots

which by definition means that alpha^p^n is separable over k

it satisfies a polynomial with no repeated roots

(f_sep is actually irreducible, so it's also the minimal polynomial of alpha^p^n)

[desert dome]

sorry I still dont get it. if a^p^n in k, then f_sep probably looks like x - a^p^n

I dont understand why a^p^n in k...

[rustic crown]

take a look at the definition of K/k being purely inseparable

[desert dome]

ahhhh I see.

I was being stupid

thank you so much!

[rustic crown]

nah, that's probably my fault, i should have written pure here >.<

[desert dome]

this is too hard for me... I feel so stupid. Thank you for helping me out

[rustic crown]

you'll be good with some practice

inseparability felt very weird the first time i read it

[desert dome]

yeah, I still dont quite get this concept. I'll probably understand better after this problem

[rustic crown]

yee, good luck!

i had to work through these things a couple of times before finally understanding it nicely

[rustic crown]

Hello Ryu

[desert dome]

ahhhh could i ask another question about the problem above? I didn't see where we use f is inseparable to derive f(x) = g(x^p)?

[rustic crown]

we used that its derivative is 0

wouldn't happen without insep

insep + irred => f' = 0

[desert dome]

right!!

[rustic crown]

insep only tells us that (f, f') is not 1

irred says that (f, f') is either 1 or f

[desert dome]

I was also confused about a_i i = 0 again.. is it because field don't have zero divisor so if a_i != 0, i = 0? (which means that i is multiple of p)

[rustic crown]

right

[desert dome]

then is i an integer or element of k

[lethal dune]

hi

[rustic crown]

like when we write an integer n inside a ring R, we're basically using the fact that there is a unique map Z --> R

so if you really wanna be explicit, i lives on the left

the map would be Z --> k here

and its image is 0

there are two ways to think about n * r inside the ring R

Rings are abelian groups, so n * r already makes sense

it's r + r + ... r (n times)

this is same as thinking (1 + 1 + ... + 1) * r

and 1 + 1 ... 1 is nothing but image of n under Z --> R

[desert dome]

yea but if i is in k, I can just use k don't have zero divisor. but if it is a_i + ... + a_i for i times, I am confused on why i has to be 0 again

[rustic crown]

oh that's what we're saying right

i is an integer

but i * a_i = phi(i) * a_i

where phi : Z --> k is the unique map

[desert dome]

right

[rustic crown]

this says phi(i) = 0 by no-zero divisors

so i is in the kernel of that map

which is pZ

[desert dome]

make sense

[rustic crown]

but in practice people confuse i and phi(i) all the time, it's usually clear from the context

[desert dome]

so in general I don't have to care too much if it is integer or element in k right

it's actually the same

[rustic crown]

yep

[rustic crown]

this is the equation which translates the two

on the left i * means action of the integer i on the abelian group k.

on the right it means multiplication of phi(i) and a_i inside of k

[desert dome]

i * a_i = phi(i) * a_i is because of this reasoning

[rustic crown]

yep

[desert dome]

I think I get it now. Thank you again

[rustic crown]

[lethal dune]

so have u decided where to apply next?

[prisma shuttle]

ok i see

but why do they like have the same indices $i$ like i thought that restricts it even more does it not?

[cloud walrus]

JustKeepRunning

[delicate bloom]

why do you say that

[prisma shuttle]

idk maybe i just don't get the notation but it seems like each a_i corresponds to a b_i, and for example, if a_ib_i is used, then a_ib_j cannot be used in the sum for any i\neq j

[rustic minnow]

I thought for the first question I should check possible remainders for polynomials when divided by x^3 and create a multiplication table to see if there are any zero divisors. i. e 2 elements ab = 0 or ba= 0.

would the possible remainders be 0,1,x,x+1,x^2, x^2+1 ?

[next obsidian]

You aren’t dividing by x^3

You’re setting x^3 equal to 0

The elements are gonna be of the form ax^2 + bx + c

[rustic crown]

aren't they the same things?

[next obsidian]

I mean…

Brain power 🧠

Maybe

It’s just hella more complicated to go about dividing

Also I’m not sure

What do you get as a remainder from x^4 + 1?

I feel like you don’t just get 1

No, yeah no way right?

[rustic crown]

x^3 * x + 1

quotient = x, remainder = 1

[next obsidian]

I’m not gonna lie

I don’t remember how division with remainder works

I’ll just say yeh

I guess so

¯_(ツ)_/¯

[rustic crown]

a = bq + r >.<

[next obsidian]

[rustic minnow]

so I need to consider possible remainders of ax^2 + bx + c?

[rustic crown]

(no, they are the possible remainders where a, b, c vary over {0, 1})

[rustic crown]

you basically missed the quadratic remainders which also had some linear terms

[rustic minnow]

yes, I mean over the field Z_2 .

[worthy haven]

For any sufficiently nice ring R (I think a domain is sufficient?) you can do a division algorithm on polynomials with unit leading coefficient; given a(x) and b(x), there exists q(x) and r(x) so a(x) = b(x)q(x) + r(x), where deg(r) < deg(b). Because in a field, any polynomial is monic you can do this to whatever, so thinking of Z_2[x]/(x^3) is the same as thinking of the remainders when polynomials are divided by x^3, but I agree it's way more cumbersome

[next obsidian]

I think you can do division by a monic polynomial in any ring

I had to do it once

And I don’t think it was assumed to be a domain

But idk

[rustic crown]

you probably won't have uniqueness >.<

[next obsidian]

¯_(ツ)_/¯

[worthy haven]

oh yeah maybe

[rustic crown]

no wait, uniqueness also works idk

[sharp sonnet]

leading coefficient needs to be a unit

[worthy haven]

oh oops yeah that's what I meant, not monic

Thank you

[rustic minnow]

@rustic crown so, x^2 + x +1, and I need to consider x^2 + x, x^2 +1

[rustic crown]

right x^2+x and x^2+x+1 were missing in the earlier list

[rustic minnow]

0,1,x,x+1,x^2, x^2+1 , x^2+x+1, x^2+x, x^2 +1

and the rest of the questions follow very easily after I have constructed the multiplication table

[rustic crown]

you wrote x^2+1 twice

[sharp sonnet]

(i also want to note that you are actually dealing with equivalence classes, you just computed "canonical" representatives that you now work with)

[rustic minnow]

oh, yeah haha

@sharp sonnet canonical as in typical?

[sharp sonnet]

as in "obvious choice"

[rustic minnow]

yeah

[sharp sonnet]

smallest degree in this case

[rustic minnow]

aha)

[sharp sonnet]

(i only mention this because in polynomial rings with multiple variables, the choice isnt necessarily as obvious and you need to be aware that you are actually dealing with equivalence classes of polynomials, not polynomials)

[rustic crown]

I need some help >.<

[spare quest]

i need help my teacher said prove to me if there is integer solutions for a, b and c and n > 2 for the equation : a^n + b^n = c^n

[chilly ocean]

#❓how-to-get-help

[rustic crown]

(0,0,0)

[spare quest]

also theres another problem

[ember field]

y isnt the trivial ring initial in the cateogry of rings?

[thorny fiber]

I don't think it's easy enough in order to post a help on these channels to be honest :s

.<

[spare quest]

if you have an odd number you multiply it by 3 and add 1 and if its even you divide by 2 is there a starting number that doesn't end with the pattern 4,2,1

[rustic crown]

because we require 1 to map to 1.
so if 0 --> R is a ring map, then as 1 = 0 on the left, we would get 1 = 0 in R which means R = 0

[chilly ocean]

so how much do you know about elliptic curves?

[rustic crown]

(it is final tho)

[spare quest]

i am andrew wiles

i know loads about modular and elliptical curves

[chilly ocean]

going senile? didn't you write a paper on this a while ago?

[spare quest]

ahhh yes i remember now

thankyou good sir

[lethal dune]

[rustic crown]

[lethal dune]

what did I just witness

[next obsidian]

VERY FUNNY

[rustic crown]

i'm going to bed, should be up in 9hours or so. let me know if there is a simpler approach which i'm completely missing >.<

[pallid ore]

How can i prove that a fuction X is epimorph. I mean, i know that i need to prove that its a morph and its surjective. Am i right?

[chilly ocean]

uh

[chilly ocean]

can you be more specific?

is this about groups? vector spaces? rings? what?

[pallid ore]

ah, i forgot to mention, its rings

[chilly ocean]

okey

then not all epimorphisms are surjective

however every surjective morphism is a epimorphism so if it is surjective you can just do that

[pallid ore]

okey perfect, then, how can i prove that a fuction is 2 rings its surjective?

[chilly ocean]

rings are not surjective

surjective is a property of a function

not of a ring

[pallid ore]

exactly, i wrote it bad sry, can your explain it

[chilly ocean]

f surjective means that for every element y in the codomain there's an element x in the domain such that f(x)=y

[pallid ore]

oh, LOL, i forgot that "theorem". Thank you for the time spent with me

[chilly ocean]

it's not a theorem it's a definition

you might want to learn basic set theory better before doing abstract algebra

[worthy haven]

for odd n, take a = 1, b = -1 and c = 0

[worthy haven]

0 doesn't end with 4, 2, 1

since 0 just goes 0, 0, 0, 0, 0, 0, ...

[pallid ore]

i don't need it tbh, its my last "abstract subject". Sooo. Just need to study this and move on. And don't have time too ahah. But thanks for spending time explaining

[delicate orchid]

ahah.

[chilly ocean]

why are you laughing

[terse crystal]

,,, it’s way too basic perhaps you better figure it out yourself…

[chilly ocean]

[delicate orchid]

aba^-1b^-1

Note that ab = ba => b = a^-1ba
=> a^-1b = ba^-1

So a^-1 and b^-1 commute

You should be able to see where this is going

As for the other direction, start by multiplying both sides by b on the right

Something most magical will happen

[sly crescent]

What is the smallest simple group with an order divisible by 37620?

[gusty cave]

what does an upside down capital pi mean?

oh wait is it disjoint union

[next obsidian]

I don’t know, technically you could use the classification if you can understand how it works. But I want to know, why on earth do you want to know this?

[delicate orchid]

I can give you an upper bound of A_37620

[delicate bloom]

idk, cyclic group generated by an element of order 37620

I guess it's not simple 😭

[sly crescent]

I can give you an upper bound of A₁₉.

[tribal moss]

The https://en.wikipedia.org/wiki/O'Nan_group is smaller than A_19.

[wicked zephyr]

of course not, since the only simple cyclic groups are Z/pZ

for p prime

[tribal moss]

and the orders of the simple groups of Lie type all have fairly specific factorizations that seem to make it difficult to fit one in below O'N.

[next obsidian]

I still want to know why you want to know this

Where could this possibly be showing up

[sly crescent]

I want a simple group with subgroups isomorphic to both PSL₂(11) and PSL₂(19).

[gritty sparrow]

It should be equal to the number of elements a of Z/mZ st na=0 mod m i think. So given a prime factorization of both m and n you can probably get an explicit formula out of this

[tribal moss]

gcd(n,m) I'd guess.

[gritty sparrow]

I’m getting an answer like product over i of (k_i+1) where k_i is the power of the prime p_i in m which is in the prime factorization of n

[chilly ocean]

[tribal moss]

Is k_i the prime power itself or just the exponent?

[next obsidian]

It’s actually isomorphic to Z/gcd(m,n)Z

As an abelian group

[gritty sparrow]

It was the power of the prime, i realise I made a mistake and the +1 is not supposed to be there. That makes it the same as the gcd lol

[tribal moss]

ah, then it looks like we all agree.

[next obsidian]

The stupidest proof I know of is that Ext^i_Z(G,H) ≈ Tor^Z_i(G,H) non-canonically for finite abelian groups G and H

And then computing it via tensor product is easy

I think this is circular tho, lol

[chilly ocean]

[next obsidian]

I think you get the result by just manually computing Hom, Tensor, and the first derived functors for Z/nZ and Z/mZ

Then apply the classification

Then because everything is an additive functor and by direct sum being direct product it just gets pooped out

[final oasis]

I am trying to work out the Legendre symbol $\left( \frac{2}{p} \right)$

[cloud walrus]

bchaotic

[final oasis]

here we treat it as a homomorphism from F_p to {1,-1} , taking a to a^{(p-1)/2}

[rustic crown]

there are two nice approaches for this, one is to go from Z to Z[i] because 2 ~ (1+i)^2

second one mimics the proof of euler's theorem and gives you some sort of summation which is easy calculate for a = 2.

[rustic crown]

(in the first one you could also use F_{p^2} if you liked finite fields more)

[barren sierra]

I can already feel this class is gonna kill me

[broken stirrup]

Hi everyone hope you are having a good day

Can you recommend a book or source that provides lots of examples in abstract algebra?

especially in ring theory

[weak oriole]

I'm not, but thank you

Have you tried dummit foote?

[next obsidian]

https://ringtheory.herokuapp.com

[broken stirrup]

yep

[broken stirrup]

thanks

[median pawn]

https://math.stackexchange.com/questions/4358866/if-g-subset-gl-n-bbb-c-is-a-closed-subgroup-gamma0-vert-gamma

The problem uses some #point-set-topology and #multivariable-calculus too, but it seemed like something I'd call #groups-rings-fields

[desert dome]

is F_p the prime subfield?

[delicate bloom]

Q and all the F_p are prime subfields

it just depends on what the characteristic of the field is

the basic way to think of it is, in every field you have 1, so now you can add this to itself as much as you like and as long as you don't get 0 you can divide these as well, these are the elements of the prime subfield

if it's characteristic 0, you get Q, if it's characteristic p, you get F_p

[desert dome]

Thanks! I am not entirely sure about the definition of F_p. Does F_p equal Z/pZ?

[delicate bloom]

yeah

[desert dome]

[delicate bloom]

keep in mind F_{p^n} is not Z/p^nZ for n>1

just a thing to be aware about

[desert dome]

why is that?

[delicate bloom]

I guess simplest reason is Z/p^2Z has an element p which satisfies p^2=0

so it can't be a field with zero divisors

[desert dome]

I see. thank you

[hidden haven]

[delicate bloom]

you're welcome 😌

[rigid depot]

Context: $K_1/K$ is a separable field extension and before we constructed a $K-$embedding $\rho: K^h \to K_1$. where $K^h$ is a subfield of $K_1$.
Can someone explain how do you extend $\rho$ to an automorphism of $K^s$ (separable closure of K)? I only know that such an extension is possible for algebraically closed codomains

[cloud walrus]

shu

[rustic crown]

extend to the algebraic closure then restrict to the separable closure?

[cloud walrus]

det

det

[rustic crown]

the map K^s --> K bar will factor through K^s because it's image will be separable inside of K bar, so contained in K^s

[rigid depot]

oh that makes sense

thank you ^^

[rustic crown]

hello Carla_

[worthy haven]

What's the motivation for the definition of a split short exact sequence? here's the definition that I have

and like, I kind of get that diagram commuting; it's saying that (after identifying M with M' oplus M''), the image of M' under f loooks exactly like the copy of M' in M' oplus M'', and the copy of M'' in M' oplus M'' looks like the image of M under g

but there are lots of commutative diagrams you could draw

why this one?

[chilly ocean]

i can't say too much right now but it's basically just that the middle term is built out of the first and last ones

in a nice way

[rigid depot]

by this definition, the five lemma immediately implies that alpha is actually an isomorphism. i've never seen this definition though

[chilly ocean]

i'd personally wave around my hands and say something like "the SES is the same as the one you trivially get from M' and M'' "

[rustic minnow]

Would the image again be the set of positive reals? Im not sure how I can have a kernel for x^n, n>0 as its impossible for elements to be mapped to zero.

[keen sparrow]

The group is with multiplication. What is the identity element here?

[hidden haven]

The commutative diagram means that the top row and the bottom row are basically the same short exact sequence. That diagram is just an isomorphism of short exact sequences.
When you work with vector spaces, it turns out that in an any short exact sequence, the middle thing looks like a direct sum of the other 2, and the first map is the inclusion of the first factor and the second one is the projection onto the second factor. This is not always the case with modules. When this does happen, we call the SES split. The diagram just puts this "looking like" thing into formal language using isomorphisms of sequences

I guess you kinda already said that

But maybe this helps idk

The direct sum is like the trivial way to fill in the middle thing in an SES

[rustic minnow]

@keen sparrow The multiplicative identity would equal to 1

[keen sparrow]

Right, so what is the kernel?

[hidden haven]

A slightly more technical reason for caring about split SESs is that if you apply a functor to a random SES, it need not give an SES, but a split SES always a split SES if you apply a functor (assuming that the functor is additive)

@worthy haven

[hidden haven]

This is a very important property when you do anything homology related

[worthy haven]

It is very helpful to know that that always happens with vector spaces

[worthy haven]

also this is very interesting; I only know an epsilon worth of category theory but this seems like a very helpful property

[worthy haven]

We haven't proven the five lemma. This was one of the first definitions in the course

[keen sparrow]

@rustic minnow is it clear now what the kernel should be?

[rustic minnow]

no, its not as I cannot choose 0

[keen sparrow]

The identity is not 0

It’s 1

Also zero is not in the group anyway

[hidden haven]

The five lemma here will just tell you that α is an isomorphism as soon as the diagram commutes. Since it's anyway assumed to be an isomorphism in the definition itself, it's not necessary to use the five lemma

You can read the five lemma statement it's very simple

[rustic minnow]

@keen sparrow so I choose 1 as it is the inverse image of 1

[hidden haven]

And also tells you things about SESs which you should know

[worthy haven]

yeah I looked up the five lemma, the proof and the statement both don't look too difficult. I think we're gonna do them at some point

[hidden haven]

Namely that these weird isomorphisms of SES really do capture sameness

Well it doesn't entirely

[keen sparrow]

@rustic minnow right, in the positive real the only number whose power is 1 is 1

[hidden haven]

The point is that you say that 2 SES are isomorphic if each corresponding pair of modules is isomorphic, and the maps in between also look the same. This second thing gets captured by saying that the squares commute, and to see that you just gotta think about it yourself for a couple minutes lul (The commutativity of the squares says that elements that correspond under the isomorphism map to corresponding elements when you apply the respective map)

[rustic minnow]

@keen sparrow right thanks. For the last part of automorphism to I need to show the surjective property? I have already showed that the property of homomorphism holds.

[keen sparrow]

Right

[rustic minnow]

so x^(1/n) as (x^1/n) ^n = x ?

for n>0

[keen sparrow]

Positive Real numbers have positive nth roots,yes

[rustic minnow]

so this property holds an we can conclude that the group is an isomorphism onto itself

[keen sparrow]

Right

[rustic minnow]

And if we have homomorphisms which are both into and onto itself then we can say that we have an isomorphism as it is defined as a homomorphism with a mapping which is bijective?

[keen sparrow]

Yes, that’s right

[rustic minnow]

@keen sparrow Thanks. So, one final question. If I was dealing with the group of positive rational numbers (Q_>0, *) instead I could not say that the isomorphism is onto itself as the positive nth roots might not exist (i.e irrational numbers) ?

[warm holly]

If we have H normal sub group of G, g in G and h in H do we have the following equality for right class of H: hgH=gH? As proof I would write hgH=Hhg=Hg=gH but I’m not sure if it is true in general

[rustic crown]

yep that's true

[next obsidian]

:D

Your proof is good

[rustic crown]

if you've seen quotient groups, hgH = (hH) * (gH) = (eH) * (gH) = gH

[warm holly]

Kk thank you I was juste told that it might not hold in general so I was a bit confused

[next obsidian]

I think they meant for H a general subgroup

[warm holly]

But on the other hand I have a proof xd

[next obsidian]

ie not normal

[rustic crown]

for general it's false

[warm holly]

Ye if H is not normal I’m agree

But it is told H normal

[rustic crown]

(your english is kinda cute >.<)

[chilly ocean]

🇫🇷

[warm holly]

frenglish

[next obsidian]

Soit R un Anneau Locale Noethèriennes

Or something like that, idk

[keen sparrow]

@rustic minnow If you were dealing with the positive rationals, then you wouldn't be able to guarantee nth root. So, the homomorphism wouldn't be an isomorphism because, for example, f_2 then 2 wouldn't be in the image since the square root of 2 is not rational

[rustic minnow]

yep, thats what I thought.

its the next part

although will the function be defined as: $ f_n : \mathbb{R}^* \longrightarrow \mathbb{R}{>0}$ and $ f_n : \mathbb{Q}{>0} \longrightarrow \mathbf{R}_{>0}$ in these different groups?

[keen sparrow]

For the second one, you might as well make the image Q since power of rationals are rationals

[rustic minnow]

good point

I K.o'ed latex

[chilly ocean]

How do I know how many isomorphisms exist between 2 finite cycle groups each having an order of 30?

I know there are 30 homomorphisms

Cause gcd(30,30)

[next obsidian]

you have to figure out which ones are bijective

which ends up amounting to counting the number of elements with order 30

[chilly ocean]

so the elements of the set are {0,...,29}

and i count the number of elements inside that set with order 30?

and that will be my answer ?

[woven delta]

Yeah you are essentially picking a generator to map 1 to

[chilly ocean]

but

idk how to do that tbh

This is what I have so far

[bleak abyss]

Yeah, so the idea is that you can map 1 to x

And then your homomorphism is gonna map n to nx

So when is "multiplication by x" invertible mod 30?

[chilly ocean]

hmm

so am i basically finding an element g where g^k = e?

where does the mod30 take play in all of this

[chilly ocean]

where did this come from?

[bleak abyss]

I mean G is a cyclic group of order 30, I'm thinking of it as Z/30Z

[chilly ocean]

can we assume that

for all cyclic groups?

say the order was like 20

[bleak abyss]

Z/20Z

Cyclic groups of order n are all isomorphic

[chilly ocean]

got it

so how do i find the number of isomorphisms

[chilly ocean]

i know this

[bleak abyss]

So the basic idea here is, and in fact I can say it without resorting to thinking of it as Z/30Z so I'll do that

A map between finite sets is injective if and only if it's surjective, right?

[chilly ocean]

yeah

[bleak abyss]

So if we have a map phi:G->G where G is cyclic of order 30

Well take a generate g of G

The image of phi is what?

[chilly ocean]

g again

phi(g)

[bleak abyss]

Well, I meant by that to describe to me the image of G under phi

But this is step 1 in doing that

We know phi(g) is in the image

[chilly ocean]

yes

[bleak abyss]

So let's say h is in G

Since g is generates G, how can we write h?

[chilly ocean]

so h is an element of G

phi(h) ?

[bleak abyss]

Yes, but try to write h in terms of g

[chilly ocean]

hmmm

g/20g?

im kinda confused

[bleak abyss]

h = g^k for some k

[chilly ocean]

oh okay

since its cyclic i would just use the def

okay

h = g^k

[bleak abyss]

So phi(h) = phi(g^k) = phi(g)^k

So the image of phi is actually going to be generated by phi(g)

[chilly ocean]

yes

[bleak abyss]

So if we want phi to be surjective

Fix a generator g of G. Then we can map phi(g) to any element which generates G

And that's all

[chilly ocean]

so i can choose any generator g

from G

[bleak abyss]

Well, fix one in advance

So given G, take g

Now what's a hom phi:G->G? You can set phi(g) = h for any h, and then you've determined phi

When is phi an isomorphism? It's precisely when h is also a generator of G

So the number of isomorphisms G->G are the number of elements which generate G

[chilly ocean]

damn man im so lost

idk why

[bleak abyss]

Hmm

[maiden heath]

Can anyone help explain what an action actually does? And how the first and second definition in the attached image relate?

My understanding is an action on a group takes elements of the group and maps each of them to a permutation in the set omega, is that correct?

[chilly ocean]

Honestly I think the easiest way to think about group action especially at first is a function f: G x Omega -> Omega, (g,x) maps to gx. This map has to satisfy 1x -> x and g(hx)=(gh)x. So it's just multiplying elements of Omega by group elements from the left side

[maiden heath]

I think I understand. Cheers

[gusty cave]

on the part about u, why does the order of x1 and y1 switch around?

omg wait im stupid lol

i was pondering this question for like 15 minutes and the second i posted it here the reason popped up into my brain : |

thanks brain

[chilly ocean]

[wise igloo]

is this good enough for me to learn group theory from? it seems short but I wanted someone to skim through it and lmk if it's sufficient https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.maths.gla.ac.uk/~mwemyss/teaching/3alg1-7.pdf&ved=2ahUKEwj0kLvoqLr1AhVBSTABHclpC5QQFnoECBAQAQ&usg=AOvVaw30jXL--dtOBp7PCSGOPIWn

[coarse forge]

it looks fine if you're just looking to study some introductory group theory

[wise igloo]

okay ty

[cloud walrus]

Hausdorff

Hausdorff

Hausdorff

[median pawn]

If the supremum is finite and equal to M, then I can possibly consider f(X)^M?

Otherwise I do not yet have ideas on how to solve this

[cloud walrus]

Hausdorff

[chilly ocean]

so would the answer be 30?

all of em are isomorhpic

[bleak abyss]

So I'll explain the idea by identifying G = Z/30Z

So if I have a homomorphism Z/30Z -> Z/30Z

Then it's determined by where the number 1 goes, correct?

[keen sparrow]

@median pawn the coefficients are element of the finite group Z_n

[median pawn]

Agreed

[keen sparrow]

So, does that give you a bound for the nilpotent index?

[cloud walrus]

Hausdorff

[bleak abyss]

Well, if you just want a homomorphism, then you can send 1 to anything. If we want it to be a bijection... well let's say we've sent 1 to x. Then our homomorphism is just multiplication by x. When does it have an inverse? It has an inverse if x has a multiplicative inverse mod 30