groups-rings-fields

[coral spindle]

In the integers mod m anything that isn't a zero divisor is indeed a unit.

[vocal patrol]

And equally gcd(a,m)!=1 implying it’s a zerodivisor since they’re not relatively prime

Ah okay thanks

[coral spindle]

This can be seen from the fact that any finite integral domain is a field.

[vocal patrol]

Yes ok makes sense then

[coral spindle]

Exercise: prove this.

[vocal patrol]

I just prove that for every element there must exist an inverse

For the neutral element it’s trivial

[dull marsh]

More generally we have the same property for elements of an arbitrary finite unital ring

[vocal patrol]

For the rest I guess I’d need to make use of the fact that in finite algebras a^n = a^m with m>n

So a^n - a^m = 0

But how do I transform this into an inverse of a because with minus I can’t use exponent laws

[rocky cloak]

a^n - a^m = a^m(a^n-m - 1)

So if this is 0, either a is a zero divisors or a^n-m = 1

[vocal patrol]

Yea that makes a lot of sense

Tysm

[crystal vale]

@barren sierra

[crystal turtle]

Please don't randomly ping members asking for help.

[rocky cloak]

I don't know the full problem, but A is a subgroup of AB, so if A has order p^n, then AB has order a multiple of that

[untold basalt]

How can I find the minimal polynomial of $\sqrt{3+2\sqrt{3}$ over $\mathbf{Q}$? The hint is that $3+2\sqrt{3}$ is the square of something, but I can't see what.

[cloud walrus]

Mike Hawk

How can I find the minimal polynomial of $\sqrt{3+2\sqrt{3}$ over $\mathbf{Q}$? The hint is that $3+2\sqrt{3}$ is the square of something, but I can't see what.
```Compilation error:```! Missing } inserted.
<inserted text> 
                }
l.52 ... minimal polynomial of $\sqrt{3+2\sqrt{3}$
                                                   over $\mathbf{Q}$? The hi...
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.```

[stark helm]

Suppose there is an ideal I and J, and I want to demonstrate an example to show that IJ may not be the same as {ab| a is in I, b is in J}, I want to suppose I is Z3, J is Z4, then we a1=1, a2=2, b1=2, b2=3 such that 12+23=8 in IJ, but it can't happen for {ab|a in I, b in J}, can someone help me check if it is correct?

[chilly ocean]

In that case they are the same. Also 8 isn't in IJ.

[stark helm]

why 8 is not in IJ?

[chilly ocean]

how would it be

3Z×4Z=12Z

[stark helm]

pick a1=1, a2=2, b1=2, b2=3, then we have a1b1+a2b2=12=12+2*3=8, why it is not correct?

[chilly ocean]

2 isn't in any of the ideals tho

[coral spindle]

Neither is 1, for that matter

[stark helm]

i am wondering that if we say a1 is in I, and I let I be Z3, which contains 0,1,2. Why 2 isn't in I?

[chilly ocean]

3Z is not 0,1,2. It's all multiples of 3

[stark helm]

but I don't refer to 3Z

[stark helm]

I refer to Z_3

[coral spindle]

Are you confusing the ideal 3Z of Z with the ring Z/3Z?

[chilly ocean]

That isn't an ideal

[stark helm]

Since I am considering that Z_3 is a subring of Z, is that true?

[coral spindle]

Firstly, you should not think of the elements of Z_3 — better written Z/3Z — as 0, 1, 2. They are really equivalence classes of integers, but this is besides the point. The more important point is that Z_3 is neither a subring nor an ideal of Z.

[chilly ocean]

It's not a subring of Z.

[coral spindle]

It's useful to note that there is only one ideal that is also a unital subring of any unital ring :)

[stark helm]

I am thinking that Z/3Z must satisfy abelian under addtion( have identity, inverse, commutative and associative and closure under addition), and it still satisfy closure under multiplication, with associative and distributive.

[coral spindle]

We all know Z/3Z is a ring.

But it is not a subring of Z.

Think about this: what is 1 + 1 + 1 in Z? Now what is 1 + 1 + 1 in Z_3?

Remember that to be a subring, the operations must be the same.

[stark helm]

one is 3, but the other is 0, so the same operation must generate the same result?

[coral spindle]

Last time I checked.

[stark helm]

Then Z/3Z will not be the subset of Z?

[coral spindle]

Z/3Z is not a subring of Z.

[stark helm]

i mean for subset

[coral spindle]

Because you seem to like thinking of Z_3 as {0, 1, 2}, it could be seen as a subset, but this does not matter in the least.

As it happens, the construction of Z/3Z makes it not a subset of Z.

Again, this does not matter.

[stark helm]

because I am considering thinking it by definition that a non-empty subset that satisfy closure under addition and multiplication

[chilly ocean]

they are completely different elements though

One are numbers modulo 3, the others are intergers

Elements of Z/3Z are equivalence classes, they aren't really a subset of Z

[coral spindle]

We've already talked about how Z_3 is not a subring of Z. There's not more to say.

[stark helm]

already understood, really appreciate that

[formal mulch]

How can i prove if there exists isomorphism or not? would appreciate any hints

[slim kayak]

nana :(

[formal mulch]

i remember how you helped me with first one and i solved it, thank you so much

[slim kayak]

Well, depending on how your solution looked like it would be very suggestive of how you might construct an isomorphism here

[formal mulch]

do you mean solution of first question?

[slim kayak]

yup

[formal mulch]

i just showed that this basis defines any function from F(A) at every point, do you think it's enough?

[slim kayak]

No

[formal mulch]

what else?

[slim kayak]

Well, what is the free abelian group with basis A?

[formal mulch]

if about definition, then set from the task condition, isn't it?

[slim kayak]

Huh? The task condition?

is that really the definition you were given? Oh well

[formal mulch]

yes, about constructing here

and if about definition, there is first line

[slim kayak]

that doesnt look right, did you copy in the definition correctly?

[formal mulch]

first line?

[slim kayak]

nvm

[formal mulch]

[slim kayak]

Okay. Do you see how the basis you constructed in the first part of the exercises might be related to the definition of a free abelian group

?

[formal mulch]

wait, you don't agree with my proof of basis?

[slim kayak]

wdym

[formal mulch]

this

[slim kayak]

It isnt an isomorphism so idk what you mean by it being enough

[formal mulch]

yes, but i thought you was asking about first question, it's about proving basis

[slim kayak]

So you have a basis for F(A), now consider some other abelian group with basis in A and you want to show an isomorphism between them

[formal mulch]

yes

[slim kayak]

How would you go about defining an isomorphism between a pair of vector spaces of dimension n?

[formal mulch]

oh, with coordinates of vectors?

[slim kayak]

i guess

[crystal turtle]

How do the "coordinates" connect with a basis?

^^ for nana

[formal mulch]

i misread the question

[rocky cloak]

To find the minimal polynomial, the easiest way is just to give the element a name (like x), then start solving.

x^2 = 3 + 2sqrt(3)
(x^2 - 3)^2 = 12

So there's a polynomial with it as a root. And since 3+2sqrt(3) doesn't have a square root in Q(sqrt(3)) you know the minimal polynomial should have degree 2*2 = 4, so that must be it.

[stark helm]

Suppose there is an ideal I and J, and I want to demonstrate an example to show that IJ may not be the same as {ab| a is in I, b is in J}. So for this question, is it true to let the whole ring be Z/6Z and let I be A/6Z={2,4,0} and B/6Z={0,3}?

[barren sierra]

What are A and B?

[stark helm]

A is 2Z and B is 3Z?

[barren sierra]

And can you give an explicit example of why IJ and the other set are different with this example

[formal mulch]

as linear combination of basis vectors with the same coefficients?

[stark helm]

Since I list a1=2 a2=4, b1=3, b2=3, so we have a1b1+a2b2=23+43=18 in IJ, while in {ab| a in I, b in J}, 18 is not in it since the maximum number should be 3*4, I don't know if my constructiion of ideal is correct?

[barren sierra]

All the arithmetic is done mod 6...

So 18 = 0

Also type \* to type *

[barren sierra]

So here I = <2> and J = <3>

The ideals

Right?

Can you show that IJ ≠ { ab | a ∈ I, b ∈ J}

[rocky cloak]

A hint could be that ||this will never work with principal ideals||

||Try maybe looking at R[x, y] or some other ring which has non-principal ideals||

[barren sierra]

Or show that these are the same in your example

And conclude that you need a different example

[stark helm]

Could you explain what does R[x,y] mean? I mean a brief example might help

[stark helm]

like <x,y> in R[x,y] if we need to find R[x,y]?

[rocky cloak]

R[x, y] is the polynomial ring in two variable with coefficients in R

And indeed (x, y) is an ideal that needs 2 generators

[stark helm]

I am actully wondering if (x,2) can be an ideal in R[x,y]?

If there are less than 2 variable

[rocky cloak]

So x = 2(x/2), so the x is redundant then

[stark helm]

In that case, x is redundant even if in R[x]?

[rocky cloak]

Yes, the ideal (x, 2) is principal whether you're in R[x] or R[x, y], that's right

[stark helm]

Do you think it will work for <x,2> and <x,3> in Z[x]?

[rocky cloak]

Yes, that should probably work.

[boreal inlet]

I'm trying to find an counterexample, if that exists. I am given an L/K, a field extension, where alpha is algebraic over K.

I have proved the fact if this extension was of odd degree, K(alpha) = K(alpha^2)

I'm trying to check if the converse is true or not

That is, I have K(alpha) = K(alpha^2), where alpha is algebraic over K

[coral spindle]

Hint: find a minimal polynomial for alpha in the extension K(alpha) : K(alpha^2)

You will not always be able to guarantee it is minimal, but it will be a start.

[boreal inlet]

Isn't it just x^2 - alpha^2

[coral spindle]

That's a great idea

So what does this polynomial tell you about the degree of K(alpha) : K(alpha^2)

[boreal inlet]

It's 2 or less than 2, that is 1

[coral spindle]

Yeah exactly. So either K(alpha) = K(alpha^2), or the degree is 2

[boreal inlet]

Oh ny god

That explains it

[coral spindle]

So you're looking at the degree of the extension K(alpha) : K right :)

I'll stop now since it seems you've got it

[boreal inlet]

yes

K(alpha) : K = [K(alpha) : K(alpha^2)] [K(alpha^2) : K]

oh wait.

If degree is 2, then extension to K(alpha) cannot be odd at all

[boreal inlet]

But how is this conclusive

x^2 - alpha^2 isn't in K[x]

[coral spindle]

Ah sorry, I got confused as to what we were proving. This is just a reproof of what you said you'd already proved. My mistake.

[boreal inlet]

oof that's fine, it happens to best of us 😭

Hmmm..

[
\bQ(2^{\frac{2}{3}}) = \bQ(2^{\frac{1}{3}})
]

[cloud walrus]

Riku

[boreal inlet]

Is this correct

[coral spindle]

What you write is true

[boreal inlet]

I think this is a counterexample then

Because the degree is odd

[boreal inlet]

I think I also need to write a proof for this though... I do know how Q(2^1/3) looks

Oh nvm

2^(4/3) = 2 * 2^(1/3)

And 2^(6/3) = 4

Yeah

....... i am an idiot

I had to find an even extension lmao

wait lemme see

[boreal inlet]

This did not age well

[cloud walrus]

Riku

[boreal inlet]

Even this doesn't work

The latter is just Q(√6)

[rocky cloak]

You mean you're wondering if K(alpha) = K(alpha^2) implies the degree of K(alpha) is odd?

If so consider ||the case where alpha is a 3rd root of unity.||

[boreal inlet]

Q(zeta_3) right?

That has a minimal polynomial x^2 +x + 1, and is a degree 2 extension

Oh right.

(zeta_3)^2 is another root of this polynomial

And Q((zeta_3)^2) = Q(zeta_3)

[rocky cloak]

In fact this will work for any element with minimal polynomial x^2 + ax + b, with a nonzero

[boreal inlet]

Even after looking at like everything I didn't look at the most obvious example

Thanks a lot

[stark helm]

Can someone explain how to show that it is a ring( don't know how to show there exist an identity 0?)

[daring nova]

identity 0 ? This is pointwise addition, it's a well known group
Do you mean identity 1 ?
This one is the meat of the exercise yes

(well, and distributivity and associativity)

[stark helm]

I actually feel confused about if pointwise addition can imply that it is a group?

[stark helm]

in that case, if I can tell that the functions are 1. distributive, 2. associative, 3. closure under multiplication, then it is a ring right?

[daring nova]

Assuming there's an identity and all the required properties for the underlying group, yes

[stark helm]

Do you refer to identity 1?

assume that there is an identity 1 in functions f?

[daring nova]

Yes
Though ofc the identity is not the constant 1 here

[daring nova]

Prove that there's a neutral element for this binary operation ?

[stark helm]

I still have no idea of this part, can you breify introduce how to get started?

[rocky cloak]

Let's say f is the neutral element. That would mean

Sum[xy = n] f(x)g(y) = g(n)

What is the easiest way to choose f to make the left hand side equal to the right hand side here.

[fringe heath]

hey, could i get some help with (3) please?
(2): isomorphic to Z/2 x Z/2

when i try to think of the symmetries, labelling 1,2,3,4 from the top corner i get:
r: (1234), r^2 = (13)(24), r^3 = (4321), r^4 = e
then for reflections you can either reflect down the middle or diagonals to get
(12)(34), (14)(23), (24), (13) [just like you'd expect, since its essentially D4)
then
now, the set does not in fact contain r, r^3 or the transposition reflections but im really unsure why?
in fact, i don't see what difference it makes to have a chessboard and a square 😵‍💫

[crystal vale]

Any hint for c part

[coral spindle]

in fact, i don't see what difference it makes to have a chessboard and a square 😵‍💫
symmetries of a chessboard must also preserve the colour pattern

I'm unsure what exactly the question is going for too, tbh though. They really ought to have made it clearer.

[fringe heath]

oh right right rihgt
so i can only reflect diagonals and rotate to preserve order of squares thing thing
but that makes it more confusing to me because that gets rid of the rotations i didnt want but surely gets rid of the reflections that i want

[chilly ocean]

it would work if you chose the points to be midpoints of chessboard edges

[fringe heath]

oh it does

thats good

thank you

[rocky cloak]

If you're redefining both addition and multiplication, then the only thing you're keeping is the integers as a set. As a set the integers are countable, so the question becomes, does there exist a countable field

[stark helm]

I might choose f is 1, but feel confused about why writting f(x)g(y)=g(n)

[rocky cloak]

f is a function, so I'm not sure what you mean by "choose f is 1".

If you mean that f(x) = 1 for all x, then this doesn't work since the left hand side will become

Sum[d divides n] g(d)

Which is not equal to the right hand side

[stark helm]

I think I may not understand what does this question mean, because I feel cnfused about why when f(x)=1, the left hand side becomes sum(d divides n) g(d), i mean why it becomes g(d) on the left side?

and also, what will be the format of f(x) here, like constant or with variable x?

[rocky cloak]

So you where given the definition

(f*g)(n) = sum[xy = n] f(x)g(y)

Now if f is a neutral element, f*g will just equal g, so the left hand side in the equation above just says g(n).

Now, xy=n is just saying that x and y are divisors of n. So summing over all such x and y is summing over all divisors.

Now think about a simple case, say n=2. Then we have
(f*g)(2) = f(2)g(1) + f(1)g(2)

We need this to equal g(2). What could be natural choices for values f(1) and f(2) that makes that work? Can you generalize from that?

[stark helm]

f(2)=0, f(1)=1? so f(1)=1 while f(x) =0 when x is not 1

I think it should be the way to generalize f*g(n)

[stark helm]

Then when I do this assumtion and prove that we can have neutral element for this binary operation( pointwise add and product), then we can have the group with unity? And what other thing we have to do in order to say that R is a ring?

I think I still don't know what kind of condition we can make to say R is a ring as we have shown that we have neutral element for binary operation?

Probably, I still need to know what does pointwise addition and product imply?

[rocky cloak]

Just checking the ring axioms. So you gotta be an abelian group under addition, and have a multiplication such that it is associative, distributes over addition and has a neutral element. That's it

[stark helm]

I am actually wondering what does abelian group under addition really mean, does it refer to for every a and b in R, we can conclude that a+b=b+a( or we need to satisfy that it is closure, associative, commutative and existence of identity and inverse under addition)

[rocky cloak]

It just means that if you consider R with the operation of addition, then that is an abelian group.

So you need a+b = b+a, but also the existence of 0 and -a such that a + (-a) = 0, and all the other group axioms.

[stark helm]

Then go back to the question I posted before, can I just say that group must be abelian under addition as we are given that f is pointwise addition and product?

[crystal vale]

Yes I got it thank you

[stark helm]

and since we have proven that we have an neutral element, then the question is done.( all underlying properties must be satisfied because of pointwise addition and product)?

[rocky cloak]

So the fact that pointwise addition makes it an abelian group is fairly immediate, since we verify the conditions pointwise.

But checking that multiplication is associative and distributes over addition is really something that should be checked.

[teal vessel]

a homomorphism f being f(xy)=f(x)f(y) doesn't necessarily imply surjectivity, right? so when talking about this homomorphism, it is mapping to a subgroup of H, which may be all of H, but needn't be?

[daring nova]

yes

[coral spindle]

a homomorphism f being f(xy)=f(x)f(y) ...
I hope you are aware that this holds for all homomorphisms of groups, this is not some kind of special property as repeating it seems to imply

[daring nova]

but that's always been
For functions in general, you say f : R -> R, x -> x², even though it's not surjective

[teal vessel]

Yeah, it was a clarification of definition, not an extra condition

[coral shale]

The 0 map 0: G -> G is a homomorphism (0 being the identity)

[stark helm]

One more question is why we need to prove the existence of neutral element here? in a ring, it does not require unity?

[rocky cloak]

Many authors include 1 in their definition of ring, but either way it's exercise (ii) in your picture.

[stark helm]

I am kind of confused if this solution solve my second question

[static glen]

what is a product vector space?

[viral mountain]

nvm, wrong channel, and also the message sent before i could complete it lol (was trying to go onto the next line)

[viral mountain]

meant to ask in #linear-algebra , and i meant product "of" vector spaces, my bad

[wooden fulcrum]

Hello, could anyone here please help me understand what the abelianization of a group "looks like"?

[formal ermine]

wdym

[wooden fulcrum]

I'm trying to find some intuitive understanding about how the resulting classes look

like how the group falls apart and into what with the factorization

[daring nova]

ab D(G) = a D(G) b D(G) (by virtue of being a normal group, this is multiplicaiton in the quotient group)
contains a [a^-1, b] b D(G)
= a a^-1 b a b^-1 b D(G) = ba D(G)

But this isn't visual

[wooden fulcrum]

ahh that is an explanation for why any 2 classes commute right

[coral spindle]

There's no really nice way of seeing what an Abelianisation of a group is in general. The most visual example is how the free group F(2) on two generators becomes Z^2 after Abelianisation. If you can picture that tree becoming a grid, there you go.

The following is less visual but helps in terms of working with the actual result. The point is that G is some kind of combination of D(G) and Ab(G). We can say this succinctly by stating there is an extension 1 → D(G) → G → Ab(G) → 1, but this will typically be nonsplit so it is hard to use this to say much about the group in general.

[wooden fulcrum]

if I have lets say 10 different elements with there being a "non commutative pair" for each of them within the group, it puts them all in one class, right?

or more like a non commutative path I guess?

between each 2

[rustic crown]

again not very visual, but you can think diagramatically via its universal property.
if you have a group map f : G --> H, then notice that it sends the commutator [a, b] to the commutator [f(a), f(b)]. If f : G --> G is conjugation by some g, then this shows that [G, G] is a normal subgroup.

And so, if f : G --> A maps to an abelian group, then every [a, b] dies. So f factors through G/[G,G].

[wooden fulcrum]

thanks for the explanation, I haven't gotten to learn much about extensions yet but I think I get it the first example, I will read this reply more times over time haha

[coral spindle]

Just think of it as saying that G is some combination of D(G) and Ab(G), that's the point.

It's really saying very little indeed.

The derived subgroup allows us to cut a group into layers like this

If you somehow believe that Abelian groups are very elementary then cutting a group into Abelian segments appears fairly useful

[wooden fulcrum]

interesting thanks, I'm not entirely sure how to do the diagramatically part but it makes sense nonetheless I think

[coral spindle]

^ this is a diagram

[wooden fulcrum]

I think I grasp a bit more now, so is this part also correct?

[coral spindle]

We need to wait for a GGT person to come along and comment on that

[wooden fulcrum]

I know but idk what universal properties are really so I just tried to understand it the standard way xd

[rustic crown]

universal properties tell you how to give a map out of or into your object

in our case it says that the data of a map G --> A to an abelian group is same as the data of the map Ab(G) --> A

[wooden fulcrum]

hmm this seems .ike a very useful and nice way to put things

[rustic crown]

so if you're mapping to abelian groups, you might as well replace G with Ab(G)

[wooden fulcrum]

ok sec I will draw ze diagram

[rustic crown]

yee diagrams are

[coral spindle]

try using https://tikzcd.yichuanshen.de/

[wooden fulcrum]

woah

[wooden fulcrum]

so what you're saying is any G->A can be written as f o v_[G,G] for precisely one f ?

[rustic crown]

yep

in that sense G/[G, G] is the best possible abelian approximation to G

[cobalt heath]

It took too long for me to realize that this means
im \cong coim

I doubt I can pursue math at this point, missing this basic fact

[wooden fulcrum]

so is it true that G/[G,G] puts together into classes the subsets in which there was a path of non-commutativity between any 2 members?

like if I were to make a graph of the set and use "doesn't commute" as an edge then the classes of G/[G,G] would be the connected components?

hmm wait no that doesn't work with [G,G] because 1 communtes with everything right, so maybe the other classes? or is this complete nonsense lol

[rustic crown]

"commutes with" isn't the best relation :p

[wooden fulcrum]

[cobalt heath]

@rustic crown how can I improve from this kind of state?

[rustic crown]

too long depends on when you saw the definition of "coim"

[cobalt heath]

Yeah, I first encountered it in my graduate algebra course

[rustic crown]

hmm

i first saw then when i was reading about abelian categories

because you add coim --> im is iso as an axiom

aka "first-iso-theorem"

[cobalt heath]

Ah, interesting.

I learned it in module theory part without mentioning abelian category..

[hidden kite]

If anyone is free and willing to help, could you help me see if you think my proof is alright?

Such as in terms of phrasing

It isn't too concise, is it?

[limber sequoia]

That qed symbol

[hidden kite]

Yes, handdrawn by

[limber sequoia]

How did you add it?

So cute

Sorry I'm not parsing the proof what are the I?

i

[hidden kite]

Just the natural numbers that are part of conventional cycle notation

[hidden kite]

I just yoinked 's code

[cobalt heath]

What is a (i_1, ..., i_r), Do you mean conjugation?

[hidden kite]

No I was using cycle notation

Isn't this conventional

[cobalt heath]

Oh wait I mean

[hidden kite]

Ahhh

[cobalt heath]

$\alpha (i_1, ..., i_n)$

[cloud walrus]

Absta

[hidden kite]

Its just composition of functions

Namely

(\alpha \circ (i_1,i_2,\dots,i_r))

[cloud walrus]

grass

[hidden kite]

Oh nooo

I mistyped

[cobalt heath]

Then I don't see how this holds.

[hidden kite]

there's supposed to be an a^{-1} at the back 🤦‍♂️

[cobalt heath]

Exactly.

[hidden kite]

Yep sorry about that!

Other than that, is it ok?

[cobalt heath]

Ah one more, I don't get why you took this beta

[hidden kite]

I wanted it to look cleaner

[cloud walrus]

Absta

[hidden kite]

Ahh... yes

How did I not notice...

[cobalt heath]

Other than that, it's all good imo!

It's okay, typo is common even in published papers

[hidden kite]

I see, thank you for your help! I apologise for the confusion 😅

[cobalt heath]

Nah, it's fine! No problem at all~

[hidden kite]

Okie me back with another simple algebra problem done

If anyone is free and wants to, could you help me see if my proof is alright?

Please feel free to leave any suggestions, for instance, if you think that something could have been phrased better.

[alpine plank]

yep looks good

ig you should mention the product of disjoint cycles result too though

[hidden kite]

I see, thank you!

ig you should mention the product of disjoint cycles result too though
Hm why though? I don't think I used the fact that any permutation can be decomposed into a product of disjoint cycles, did I?

[rocky cloak]

Your proof only shows how to write a cycle as the product of the given transpositions

[hidden kite]

Ah that's true

Forgot about that for a sec oops

[still pulsar]

btw for <V> = S_n take arbitrary permutation and proceed with bubble sort to get id. g=g^{-1} for transposition, so take them in reverse order to get inverse.
(k k+1) = (1 k)(1 k+1)(1 k), so <U> contains V.

[lavish sigil]

Let $V_\mathbb{R}$ be the real vector space of sequences of real numbers, and let $V_\mathbb{Q}$ be the vector space over $\mathbb{Q}$ of sequences of rational numbers. Is it true that $V_\mathbb{R}$ is isomorphic to the tensor product $V_\mathbb{Q}\otimes_\mathbb{Q}\mathbb{R}$?

[cloud walrus]

gustavn64

[lavish sigil]

This was used in an argument for the fact that the dimension $\dim_\mathbb{R}(V_\mathbb{R})$ is uncountable, as follows: By a cardinality argument, $\dim_\mathbb{Q}(V_\mathbb{Q})$ is uncountable, and we have $\dim_\mathbb{Q}(V_\mathbb{Q})=\dim_\mathbb{R}(V_\mathbb{Q}\otimes_\mathbb{Q}\mathbb{R})=\dim_\mathbb{R}(V_\mathbb{R})$. But I'm unsure about the last part.

[cloud walrus]

gustavn64

[south patrol]

Yes

[limber sequoia]

Grass can you share the code?

The special qed code

[rocky cloak]

Consider (ri) a sequence of linearly independent real numbers put into a sequence (for example pi^i)

If it was the case that VQ(x)R equaled VR, then we could find sequences sj and real numbers tj such that

Sum sj(i)tj = ri

Which would mean each ri is in the span of the tj. But there are only finitely many ts and infinitely many rs, so that's impossible.

Hence VR is bigger.

[vale locust]

can I just say $\bZ_{p^m}\oplus \bZ_{p^n}$ and then use the five lemma to prove that any other group would that can be put there is iso to A?

[cloud walrus]

DarQ

[vale locust]

I prolly need to construct the morphism between them tho hmm

[rocky cloak]

The dimension argument still works though, since dim(VR) >= dim(VQ(x)R)

[rocky cloak]

The 5-lemma only works if you already have a morphism.

For example it's not true that Z/4 = Z/2 + Z/2 even though it fits into a sequence

0 -> Z/2 -> Z/4 -> Z/2 -> 0

[vale locust]

oh hmm

I see thx

this looks like the splitting lemma

[cloud solar]

A finite ring that has only trivial idempotents is local?

[coral spindle]

I don't think Z/6Z is local

Oh wait, it does indeed have nontrivial idempotents

I was thinking nilpotents!

[rocky cloak]

That's right, you can even replace finite with artinian.

[cloud solar]

Yes thanks!

[stark helm]

Is there any example of a quadrilateral Q with symmetry group sigma(Q), which contains the family of all orthogonal transformation isomorphic to Z2?

[coral spindle]

Yes

Wait

The example I thought of was in fact... wrong

But there is an example that works

[rocky cloak]

Like a ||trapezoid||?

[coral spindle]

Spoooillleeeerrsssssss

[rocky cloak]

A ||kite|| is also an option I guess

[slim kayak]

Whats wrong with straightup checking if they define an ideal

or, where does it go wrong?

[coral spindle]

How does the argument work here? It's been on my mind. I was trying to see if it followed from Artin–Wedderburn but I couldn't seem to argue that the nilrad was zero

[rocky cloak]

Why do you need the nilradical to be zero?

You just do AW, see that R/J is a field, boom boom

[coral spindle]

If the ring isn't semisimple then how does AW apply?

[rocky cloak]

If you mod out the radical it becomes semisimple

So R is an artinian ring with trivial idempotents -> R/J is semisimple, and since idempotents lift modulo J, R/J has trivial idempotents. -> R/J is a division ring -> R is local

[coral spindle]

Riiight right right

That's very nice

Thanks jagr

[tiny jolt]

Working through a hint to solve this problem that says to prove that for $\alpha\in\mathcal{O}_K$, $\alpha(1-\zeta_p)$ has trace is a multiple of p. I'm convinced that the trace of $1-\zeta_p$ is $p$, but I haven't gotten any further than that. Any hints?

[cloud walrus]

pramana

[south patrol]

Lol is this Napkin

[tiny jolt]

Yea

[woeful sage]

exact sequences?? darQ prepping for your AT arc??

[silent delta]

I was looking for a proof that all the subgroups of Z x Z are generated by multiples of the canonical basis, and I came across this answer on Stack Exchange. I am confused on the step N = Zy_1 + (N \cap Z(0,1)), which is shown in purple font. I think what is shown is that N is a subset of it, and I don't understand why we have an equality. Any help would be appreciated

[warm wyvern]

That's hatcher lel

I'm already doing AT

I would never do algebra otherwise

[warm wyvern]

Anywho, this is kicking my ass lol

all I can gather is that i(1) has rank p^n and if j(x)=1 then x has rank a multiple of p^m

And x and i(1) generate A

So A has at most two generators

I feel like A has to be
[A=\bZ_{p^i}\oplus \bZ_{p^j}]
Where $i+j = n+m$

[cloud walrus]

DarQtmrw

[chilly ocean]

Can you explain why at most dim R+1

[warm wyvern]

ig $A$ is a quotient of
[\bZ_{p^n}\oplus \bZ_{kp^m}]?

[cloud walrus]

DarQtmrw

[warm wyvern]

Where k is some positive integer

And the quotinting set has to have index k, right?

[rocky cloak]

The ideals in the chain must have different dimensions, the possible dimensions of an ideal is
0, 1, 2, ..., dim R

So there can't be more than dimR + 1 different ideals in the chain.

[rocky cloak]

So they show that N is a subset of it. But it should be clear that it is a subset of N, since when you intersect a set with N, what you get is a subset of N.

Since both contain each other, they're equal.

[vale locust]

anyone?

[woeful sage]

based

[formal ermine]

unbased

[cobalt heath]

But you have to deal with noncomm algebra in AT

[still pulsar]

Yup! It's true

[elder wave]

When computing pi_1 maybe but apart from that?

[cobalt heath]

Yeah, pi_1. Can you avoid it?

[elder wave]

I sure can

[warm wyvern]

yea, but like

how do I prove that?

[elder wave]

Pretty much all the spaces I work with are simply connected

[cobalt heath]

I.. ugh.

[cobalt heath]

How do you distinguish between simply connected spaces?

[elder wave]

Higher homology/homotopy

[cobalt heath]

Is the higher homotopy practical to deal with?

[elder wave]

Not really but at least it’s abelian

It also carries a lot of information especially if you restrict to cw complexes

So it makes sense that it’s hard to compute

[south patrol]

Your brain on rational homotopy theory

[limber sequoia]

What's the deal with exact sequences actually

Lang defines them but it did not appear in class

[static glen]

Often when you are dealing with short exact sequences there are a lot of "2 out of 3" theorems for instance if $1\to U\to G\to H\to 1$ is a short exact sequence of groups and $U$ andf $H$ are solvable then so is $G$ and often $U$ and $H$ are easier to understand than $G$

[cloud walrus]

Max

[slim kayak]

You can define quite a lot of things using them and use them to make sense of the notions such as (left-/right-)exact functors and (co)homology

[static glen]

that too, they work pretty nice with functors

[slim kayak]

Are you taking a generic abstract algebra class? Those often dont interact too much with material in which this formalism helps significantly

[limber sequoia]

It's basically a group theory class

But the prof said that he'll do some category theory if time allows, so I think he likes abstract stuff

So far we've investigated normal subgroups and the homomorph connection on week 3

[slim kayak]

Yeah, exact sequences are pretty far off then, at least doing useful things with them that go beyond funny ways of reformulating stuff

[limber sequoia]

Ok cool

[slim kayak]

Its the statement that if N and G/N are solveable, G is solveable.

[static glen]

Well okay just another way to formulate it since short exact sequence is equivalent to $U\leq G$ and $H\cong G/U$

[cloud walrus]

Max

[limber sequoia]

U is injected into G, i.e. U is a subgroup, and then U is the kernel of G->H, i.e. U is normal and G/U is H?

[static glen]

yes

[limber sequoia]

Ah cool way to write it then

[static glen]

Thats also true if you replace Groups by rings (or by C*-algebras if you are an analyst)

then you have to replace normal subgroup by ideal (or closed two-sided ideal when C*-algebra) ofc

[limber sequoia]

In class we called this the universal property

Might be something a bit different

That you can reconstruct homomorphs through G/ker

[static glen]

do you mean that a homomorphism $\varphi :G\to H$ descends to $\varphi:G/N\to H$ when $N\subset ker\varphi$?

[limber sequoia]

Yeah

[static glen]

That is indeed a universal property of the quotient

[cloud walrus]

Max

[static glen]

I know it as the Homomorphism theorem

[limber sequoia]

"descends"?

I get it I think

[static glen]

That's you how this phenomenon is called

terminology

[limber sequoia]

Because the quotient is smaller?

[static glen]

Probably comes from this picture

so that the diagram commutes

[limber sequoia]

I like how random things (like whether G/N is drawn below or above) is more common spread than actual terminology lol

Anyway tyty it's clearer for me now

[static glen]

this is common also push-forward comes from the thinking that one abstract object is placed before another

[cloud solar]

If K is a finite field with 12n+1 elements with n non zero show there is a in K s.t. the polynomial f=x^3-a is irreductible in K[X]

What I did:

K* cyclic group of order 12n by cauchy there is z in K*{1} s.t. z^3=1

If we suppose that for every a in K the polynomial f=x^3-a is reductible in K[X] this means that for every a in K there is x in K s.t. f(x)=0

This means that for every a in K, there is x in K s.t. x^3=-a and that means that for every a in K, there is x in K s.t. x^3=a so the function g:K->K, g(x)=x^3 is surjective but because K is finite g must be injective but this is false because g(1)=g(z) (z from cauchy)

Is that ok?

Because idk why they put that 12

[final dragon]

yo guys how do i proove that some mirroring (idk how to say it) is a isomorfism of a ring

[barren sierra]

Can you give a definition of such a mirroring

[final dragon]

its fi:(R,+,*)->(R,some other operation,some other) , fi(a)=a-1,R is a R1NG

the other operations are deffiened as the first one is a operation b = a+b+1 and second operation is a operation b = ab+a+b

and i have to proove that fi is isomorfism of ring

[rocky cloak]

An isomorphism of rings is a bijection such that

f(a + b) = f(a) new+ f(b)
and
f(a * b) = f(a) new* f(b)

Where new+ and new* are the addition and multiplication in the second ring

[final dragon]

yea ty

i just did it

np

i just saw these rules in the book

[rocky cloak]

That's right. And you're correct that it would work perfectly well with 3n+1, so I don't know why they used 12. Maybe to throw you off.

[low wyvern]

It was mentioned when talking about the matrix exponential if that helps

[rocky cloak]

I guess it would come up in a first course on Lie theory.

I'd say at least a first course in algebra, like groups and rings. Then some basic differential topology, like knowing what a manifold is.

And a good amount of linear algebra, like knowing about the matrix exponential at least.

It doesn't necessarily require that much, just depends on the structure of the course

[low wyvern]

Ye, that is the thing, lots of things can be 'speedup' for physicists but I don't want to do that. I want to learn it like a mathematician would.

In the correct order at the correct time and in the correct detail

[rocky cloak]

I don't know that everything has a correct order or a correct amount of detail necessarily

But I guess some things definitely have a wrong order

[low wyvern]

Like , we are told 'we won't go into that' lmao and I hate it

[rocky cloak]

Yeah, physicists are good at shoving things under rugs

[low wyvern]

I quote from a first year lecturer in electromag when talking about the RLC diff eq, 'we could solve it, but we're physicists and we already know the answer' 🤣

[boreal inlet]

How do I show that the action of SL(2,R) on the upper half Argand plane is transitive?

I cannot find one element that generates everything

[cloud walrus]

Riku

[boreal inlet]

This is the action btw

[prisma ibex]

take z_1=x_1+y_1i and z_2=x_2+y_2i, then an element g of SL_2(R) such that g.z_1=z_2 is given by g=g_2Ag_1 where uhhh

$g_2=\begin{pmatrix} \sqrt{y_2} & 0\ 0 & 1/\sqrt{y_2} \end{pmatrix} \quad A=\begin{pmatrix} 1 & \frac{x_2}{y_2}-\frac{x_1}{y_1}\ 0 & 1 \end{pmatrix} \quad g_1=\begin{pmatrix} 1/\sqrt{y_1} & 0\ 0 & \sqrt{y_1} \end{pmatrix}$

[cloud walrus]

nGroupoid

[prisma ibex]

also a nice way to think about this is that the upper half plane H is the quotient SL_2(R)/SO(2) and SL_2(R) obviously acts transitively on this

[boreal inlet]

Holy shit, I'd have never thought about this

Thanks

[prisma ibex]

btw the way you can come up with this is using the Iwasawa decomposition

SL_2(R)=KAN

this works much more generally for any semisimple Lie group for instance

[rose prism]

you can alternatively prove iwasawa using this

[boreal inlet]

I think the point that upper half plane is even isomorphic to this comes from the fact that this action is transitive

[prisma ibex]

Yup

To be fair I’m thinking kinda backwards to how one typically proves this but I’m just saying how I came up with this

[rose prism]

theres a vaguely geometric argument for why SL(2,R) (and in fact AN := upper triangular matrices) acts transitively on the unit tangent bundle of H

in particular you can guess the formulas from that perspective

and a (much easier) geometric argument indicates that the stabilizer should be SO(2)

a hint for how to see this is

think about how the diagonal matrices act on (i,i) in the unit tangent bundle T^1 H

you will get a curve

and for any other (z,w) you can try to find a matrix in N which bends that curve so it passes through (z,w)

something like that

[boreal inlet]

Wait, this isn't giving me z_2 from z_1

I think i messed up the order of multiplication because

the way you gave g_2, A and g_1 doesn't give me it

I'll check for the correct order

[ebon pumice]

Hello, anybody can help me🥹 This is a question of the final exam. When I was writing it, I thought that this Q is the field of fractions of Z/5Z. Then my professor told me that "Since Z/5Z is a field, its field of fractions is isomorphic to Z/5Z" I got it. But why the Q is not the field of fractions of it ??? I constructed Q by definition. And even if it not, it still a field. Q[square root of 2] still satifies the condition.

I am so confusing🥲

[boreal inlet]

I think the field you're looking for is

(\mathbb{F}_5[\alpha])

[summer path]

\mathbb

[cloud walrus]

Riku

[boreal inlet]

Yes, I thought it would work here 😭

[ebon pumice]

this is not a field

[boreal inlet]

Wait, x^2 - 2 isn't irreducible in F_5[x]?

[ebon pumice]

I know the answer is this. But I dont know why I am wrong🥲

[cloud solar]

But what if K has 12n-1 elements? The result remains true? (This is b) )

[boreal inlet]

I mean Q is field of fractions, yes, but it just has the exact same structure as Z/5Z

And yes, Q[√2] is exactly what is the field you need here, in fact it's probably the smallest field satisfying the conditions

[rocky cloak]

Maybe try checking the field with 11 elements

[ebon pumice]

why??Maybe Q has these elements??

And the professor said:

[boreal inlet]

What is 2^-1 in F_5?

It's 3.

[ebon pumice]

3

[boreal inlet]

Hence 1/2 in F_5 is actually just 1 * 3 = 3

In fact, each of thise fractions will be reduced to an element of F_5

[ebon pumice]

oh I got it

[boreal inlet]

Oof that is harsh 😭😭😭

[ebon pumice]

I dont know why he said I look for a field with 25 elements inside Q. I did this ???🥲

[ebon pumice]

I think so....Thats why I ask here

[cloud solar]

It works

[boreal inlet]

If you're extending the field, the new field is a vector space over the previous one. As alpha^2 = 2, the degree of extension is 2 [in fact, you do write a + b√2, which clearly shows how the basis is {1, √2}].

Now for finite dimensional vector spaces over finite fields, the cardinality equals the cardinality of field raised to the dimension. [try to prove this result. There's a simple set bijection]

So, 5^2 = 25

[ebon pumice]

I understand. Thanks!!🫶

[boreal inlet]

You're welcome 🤗

[hidden kite]

It's det's code, you should ask him about it

[warped shadow]

What do they mean by

T is a representation of the semigroup (R_+, +)
https://en.m.wikipedia.org/wiki/C0-semigroup#Formal_definition

Do they just mean that T and o are "consistent" with the group operation, i.e. T(a+b) = T(a) o T(b)

What about the T(0) = I part?

[rustic crown]

it's snow's code :giggwe:

[hidden kite]

Hewoo detto

[rustic crown]

hewwo gwassu

[cloud solar]

Idk how to answer the question

[rocky cloak]

Think about the answer you gave. You said x |-> x^3 couldn't be surjective, because 12n was divisible by 3. Is 12n-2 divisible by 3? What does that tell you about wheter or not the map is surjective?

[cobalt heath]

I just heard about cyclotomic cosets and cyclotomic polynomials over F_q. Could someone give a brief explanation about them in algebraic perspective?

What I can find is just the coding theory ones.

[cloud solar]

Oh the map is always bijective

Cuz 3 and 12n-2 are coprime

And extenting the function to K

We get x^3 is bijection

[rocky cloak]

Exactly

[cloud solar]

And this is equivlent to for every a in K, f=x^3+a is reductible in K[X]

[boreal inlet]

[
\begin{bmatrix}
1 & -\sqrt{\frac{x^2 + y^2}{y}} \
\sqrt{\frac{y}{x^2 + y^2}} & 0
\end{bmatrix} \cdot (x + iy) = i
]

[cloud walrus]

Riku

[boreal inlet]

This simplifies things

This also gives me the set of matrices that fix i

And we can use this to show the action is transitive, rest follows

[velvet shard]

Let $n,m\in\mathbb{N}$ and $\gcd(n,m)=d$. How to show that $$\mathbb{Q}(\zeta_n)\cap\mathbb{Q}(\zeta_m)=\mathbb{Q}(\zeta_d)$$

[cloud walrus]

Ham Sandwich

[velvet shard]

I have shown that RHS is contained in LHS. But have not idea how to show the converse

[rustic crown]

use that there are a, b in Z such that an+bm=d

[velvet shard]

I have tried that but I still have no idea

Oh I have shown that $$\Bbb{Q}(\zeta_n,\zeta_m)=\Bbb{Q}(\zeta_l)$$ where $l=\operatorname{lcm}(n,m)$

[cloud walrus]

Ham Sandwich

[wraith cargo]

Show that the degree of the extension is 1

[velvet shard]

Since $$\phi(l)=[\Bbb{Q}(\zeta_l):\Bbb{Q}]=[\Bbb{Q}(\zeta_n,\zeta_m):\Bbb{Q}(\zeta_n)][\Bbb{Q}(\zeta_n):\Bbb{Q}]=[\Bbb{Q}(\zeta_m):\Bbb{Q}(\zeta_n)\cap\Bbb{Q}(\zeta_m)]\phi(n),$$
we know that $$[\Bbb{Q}(\zeta_m):\Bbb{Q}(\zeta_n)\cap\Bbb{Q}(\zeta_m)]=\frac{\phi(l)}{\phi(n)}.$$
On the other hand, $$\phi(m)=[\Bbb{Q}(\zeta_m):\Bbb{Q}]=[\Bbb{Q}(\zeta_m):\Bbb{Q}(\zeta_d)][\Bbb{Q}(\zeta_d):\Bbb{Q}]=[\Bbb{Q}(\zeta_m):\Bbb{Q}(\zeta_d)]\phi(d).$$ So, $[\Bbb{Q}(\zeta_m):\Bbb{Q}(\zeta_d)]=\frac{\phi(m)}{\phi(d)}$.
Therefore, $[\Bbb{Q}(\zeta_n)\cap\Bbb{Q}(\zeta_m):\Bbb{Q}(\zeta_d)]=\frac{\phi(m)\phi(n)}{\phi(l)\phi(d)}=1$.

[cloud walrus]

Ham Sandwich

[velvet shard]

Is this correct?

thx for help

[wraith cargo]

Sounds good!

[cloud solar]

Let K be a finite field with q elements and d>=2 s.t. d divides q-1 and let f be a polynomial in K[X] with deg(f)=d. Show that there is a in K s.t. the polynomial f-a does not have any roots in K

If I suppose that for every a in K, there is z in K s.t. f(z)=a that means the polynomial function f:K->K is surjective

And because K is finite that means f bijective

Now I am trying to figure out how to find a contradiction using the fact d divides q-1

Any hint?

[teal vessel]

fun news, guys: I've finally gotten to chapter 3 in my journey through every exercise in D&F's Abstract Algebra. I will now be able to engage with quotient groups like you guys have so frustratingly been suggesting the entire time I've been doing these examples.

[wraith cargo]

Why are you doing every exercise in dummit and Foote?

[teal vessel]

because I can, and I'm teaching myself, and by the end of this I might actually understand what's going on.

I was a bit lazy with my selection of exercises in my last couple textbook readthroughs (linear algebra and axiomatic set theory), so I've forgotten and misunderstood a sizeable chunk of stuff.

[slim kayak]

You might want to consider just going through a subselection, for example by finding some course with lectures notes and psets availible online

[teal vessel]

nah

I set out to do every example, so I'm doing every example. I got a documentation of my progress and everything

[cloud solar]

.

[wraith cargo]

Consider the elements of order d in the group of units of K

[outer valve]

Omg me too 🥹

[cloud solar]

The subgroup <b^((q-1)/d)> where b generates K* ?

[wild solar]

Could anyone help me with this exercise? I have the final exam tomorrow and I would really like to understand this... I know the universal property of localization of rings, but I don't know how to prove this.

[delicate orchid]

well we have to show that M_S exists first, which is just writing down what the localisation of a module is. The choice of j_M should be fairly clear from that point. The hint I'd give for the second paragraph is that since each element of S acts via a bijection on n, we can take their inverses (almost like writing a/b as ab^-1...)

b) is the standard proof "object satisfies universal property => object is unique up to isomorphism"

[wild solar]

um, acts by bijection meaning n -> sn is always a bijection for any s?

[delicate orchid]

yeah

[wild solar]

oki

sorry I dont get the part about taking inverses

[slim kayak]

The localization of modules fulfills that property, I wouldnt think too much about "using it"

[delicate orchid]

it's used to construct the map f bar

like what's a natural choice for f bar(m/s) to be, considering we want to extend the map f

[wild solar]

f(m)/s?

[cloud solar]

??

[wild solar]

wait no N is an A-module

[delicate orchid]

yeah so "/s" doesn't make sense but you're basically there

[slim kayak]

Actually, I am not seeing how the elements of s inducing bijections would get us inverses of s here

[delicate orchid]

it's not direct but the idea is that you associate each s to an automorphism m_s of N and then send fbar(m/s) = m_s^(-1)(f(m))

[slim kayak]

like n -> sn being a bijection doesnt imply that its inverse has form n -> tn

[delicate orchid]

yeah it's a bit more annoying than when you take localisation of rings, in which case you can just write "f(s)^-1" iirc

[wild solar]

that doesnt seem obvious😭

I hope questions like this wouldn't be asked on the exam

[delicate orchid]

there may be a simpler way I'm missing but that's just what first came to mind

[wild solar]

ummm wait so how do I show uniqueness

[wooden fulcrum]

hello

[delicate orchid]

It's the standard thingy for universal properties: assume there is another localisation M'_S, then this localisation satisfies the universal property as well, then the two universal properties induce two maps M_S -> M'_S and M'_S -> M_S - which you then use uniqueness to conclude they must be inverses. I'll leave you to fill in the details/draw the triangles

[wooden fulcrum]

with a surjective group homomorphism f, it's not generally true that f(x) in f(H) implies x in H right?

[coral spindle]

It does not. Find a counterexample.

[wild solar]

I meant in part (a)

[slim kayak]

I think they meant it as in showing that the thing they defined fulfills the universal prop

[wild solar]

yis

[delicate orchid]

oh

[slim kayak]

I mean, is there any other choice for the map here?

[delicate orchid]

I forgot how this part goes lol

I hope the composition being f is clear but yeah uniqueness

hmm

[slim kayak]

Maybe go back and define f_bar "canonically"

[rocky cloak]

f(m) = f(s m/s) = sf(m/s)

Since multiplication by s is a bijection f(m/s) is uniquely determined

[delicate orchid]

so we need some g(m/s) such that g(m/1) = f(m), then g(m/1) = g(m/1 * s/s) = g(ms/s) = sg(m/s) = ...

wow i see how it is

[slim kayak]

it must send this here, and and the stuff from S must be send to invertible thingies

[wooden fulcrum]

thanks

[slim kayak]

oh right module homomorphism were a thing lmao

[wooden fulcrum]

I spent like the last 2 hours trying to come up with how to prove it holds because it seemed like the correct step to do in the proof lol

[cloud solar]

@rocky cloak can you help me with this one please?

[wild solar]

yesyes, thank you!!

[rocky cloak]

Maybe, the answer isn't obvious from a first glance, so I would have to think about it

[narrow wagon]

For an element $a \in \mathbb{Z}_n$, for a to be a generator, why does gcd(a,n)=1?

[cloud walrus]

Dubs

[slim kayak]

Bezout

You need an element that is 1 mod n

[coral spindle]

If a is a generator, there is some m such that m.a = 1 mod n, right? Go from there.

[delicate orchid]

a fate worse than death

[narrow wagon]

if (a,n)=1, then by bezouts $ax_0 + ny_0= 1$

[cloud walrus]

Dubs

[coral spindle]

You really want the converse to Bézout

gcd(a,b) divides every integral linear combination of a and b

You should etch this into the wrinkles of your brain

[narrow wagon]

I have a way in my mind

Essentially wanted to compare with others

Can I say my idea?

[slim kayak]

No, you need to get your ideas sharing license first

[narrow wagon]

$ax \equiv b \pmod{m}$

[cloud walrus]

Dubs

[narrow wagon]

This is solvable iff, (a,b) divides m

or b= (a,b)k

ie the smallest i can get is gcd(a,b)

My gut says, if it can reach 1 mod m, then i can multiply both sides and reach anything

[slim kayak]

Yeah. You have to generate 1 mod m somehow anyways, and from there you get anything else. Now you have to convince yourself that this can only happens if they are coprime

[rocky cloak]

My google-fu found a result called Hermites criterion, which says that a polynomial is a bijection iff

f(x)^t mod (x^q - x) has degree less than q-2 whenever t <= q-2.

If d divides q-1 you can pick t so that f(x)^t has degree q-1.

So maybe you can look up the proof of Hermites condition...

[narrow wagon]

1= (a,b)k, since k and gcd can only be integer value

It forces us to conclude k=1

ie gcd(a,b)=1

is it correct?

The reason why i am asking this is, my professor doesn’t talk about modular at all

he found it so obvious to say (a,m)=1 for a generator

so i was wondering how people see it

[slim kayak]

i thought you wanted to show gcd(a,m)=1

[narrow wagon]

i mean b=1

it b=1

[slim kayak]

Also, maybe think of the orders of elements and the substructure they then generate

[narrow wagon]

it means (a,m) divides 1

This same reasoning goes for why relatively prime numbers are not zero divisors right?

in ring Zn

[cloud solar]

Oh wow

I think I found a particular solution

For this problem

What I did:

q-1=dr with 1<=r<=q-1

f=a_d x^d +...+a_1 x + a_0

f^r=(a_d)^r x^(q-1) +...= sum k=0 to (q-1) b_k x^k (notation)

sum when a in K of f^r(a)=-(a_d)^r≠0

I used the fact that sum a^k = 0 for k<=q-2

And sum a^(q-1)=-1

if f is surjective cuz K finite we have f bijective

so sum a in K f^r(a)= sum a in K a^r

so we have from here r must be q-1

so d=1

But d>=2

So contradiction

It is ok?

[mighty kiln]

[severe linden]

Is this the case iff k=n?

With the k-element subsets they mean subsets of {1, ..., k}, and by k-tuples they mean tuples (of any length) with all elements in {1, ..., k}

[cloud solar]

Let (M,•) be a monoid and f:M->M with f(f(x)•f(y))=x•y for every x,y in M. Show there is a in M with a^3=e and ax=xa for every x in M and a homomorphism g:M->M with g(a)=a and g(g(x))=x for every x in M s.t. f(x)=a•g(x) for every x in M

[rose prism]

are you sure about that

[severe linden]

Oof, major interpretation mistake. Let me redo the exercise

[severe linden]

Redoing I would find that the answer for a) is 0 < k < n and for b) k > 0

[surreal sluice]

I don't understand why p is an irreducible here assuming $p=\pi \pi'$ where $\pi'$ is a unit.

[cloud walrus]

J

[surreal sluice]

Oh wait it is clear, I missed the part where it said in $\mathbb{Z}[i]$!

[final dragon]

how do i proove that something is a monomorfism

[slim kayak]

for groups/ring/etc?

[final dragon]

ring

[slim kayak]

Formally it means being left invertible, which is equivalent to being injective

[final dragon]

the thing im trying to proove is that if R is a ring that has a finite number of elements and there exist an element a from R that is not the left devisor of zero proove that the f:R->R deffined as f(r)=ar,r from R is injective

[slim kayak]

Can you see why that should be true intuitively?

[final dragon]

no

idk what that even means

[slim kayak]

Which bit

[final dragon]

that i have to proove its injective

[slim kayak]

ar = as -> r=s

[final dragon]

so the ring only has 1 element

?

[slim kayak]

no?

a function f is injective if f(x)=f(y) implies x=y

[final dragon]

ohh

and cause the a is not the left devisor

i can reduce it

[slim kayak]

or equivalently x =/= y implies f(x) =/= f(y), altho you should know that if you are working with rings

wdym reduce it

[final dragon]

i mean like devide

[slim kayak]

Why would you be able to do that

[final dragon]

cause its not the left devisor of zero

[slim kayak]

Element can be non-invertible and also not be zero divisors

Like, consider Z. 3 definitely isnt a zero divisor, but you can invert it either

[final dragon]

then how did you do this

[slim kayak]

i just stated injectivity

[final dragon]

oh ok

and i have to proove that

1 exists in R

[slim kayak]

1 as in the identity?

or unit, the multiplicative identity I mean

[final dragon]

1 as in neutral element for multiplication

[slim kayak]

Did you define them that way?

[final dragon]

and i have to proove that if and element of R is not zero devisor its invertible

[slim kayak]

Rings having always having a 1?

[final dragon]

no

if ring has a 1 its R1NG

[slim kayak]

thats vile

[final dragon]

what does vile mean

[slim kayak]

Also then this is not the correct way. And this (reply) is wrong for the integers

non-zero divisors of a ring are invertible if the ring is finite, but wrong in general

[final dragon]

it doesnt say that its a ring containing integers

[fossil hound]

You don't have to. What you have to prove is that the map is njective?

Which is just the definition of a non zero divisor

[final dragon]

no it says here i have to in the text

its 3 part problem

[slim kayak]

I am using the integers to show that your statement is wrong. Non-zero divisors dont have to be invertible

[final dragon]

yea but it says that i have to proove that in this ring they are

[fossil hound]

Ah okay. Well it's an injection of sets with the same cardinality...

[slim kayak]

his rings dont have 1's, I think the equivalence fails there

[final dragon]

so its not correct

[fossil hound]

You don't need 1 for that, it's injective either way.

[final dragon]

why doesnt it have 1

[fossil hound]

But you do need 1 for it to be a unit

[final dragon]

what is a unit

[fossil hound]

Invertible element

[final dragon]

oh

[slim kayak]

whats the argument for injectivity then?

[fossil hound]

If a is a non zero divisor and you have ax=bx then (a-b)x=0 so a=b so injective, don't need 1 for it

[slim kayak]

you are right, I mixed up the identities lmao

[final dragon]

and how do i proove that 1 exists in this ring

[fossil hound]

You don't

[slim kayak]

is your exercise saying it has a 1

[final dragon]

yes

[fossil hound]

?????

then you dont need to prove it

[final dragon]

it says proove it has a unit element

[fossil hound]

Oh

[final dragon]

is 1 and unit element not same

[slim kayak]

So yeah, finiteness comes into play here

[final dragon]

i am bad at english

[fossil hound]

Yes

Sorry

Not to that

[fossil hound]

meant that

It's an injective map of finite sets with the same cardinality

Sooo..?

[slim kayak]

For any r in R applying f(r)=ar successively gives you an orbit

[fossil hound]

Point is that the map is bijective

[final dragon]

what is an orbi

orbit

[fossil hound]

That's the only thing you need

[final dragon]

wait it is?

[fossil hound]

Yes

Cause it's an injective map of finite sets of the same cardinality

[final dragon]

yea it is

and so what if its bijective?

[slim kayak]

the collection {r,ar,aar, aaar, ... } in this case

[final dragon]

ok and how does that give me the unit element

[slim kayak]

the ring is finite

[final dragon]

ok so

[fossil hound]

[delicate orchid]

so this set must also be finite....

therefore at some point we have r = a^nr right?

[final dragon]

and then a^n=1

yo is unit = 1

[fossil hound]

😢

[final dragon]

idk im serbian

what does unit mean like what are its properties

[fossil hound]

unit is an invertible element

[final dragon]

well then it doesnt have to be 1

[fossil hound]

A unit doesn't have to be 1

That's true

[final dragon]

but it says here in the text that i have to proove that R has a 1

[fossil hound]

You still need to show that a^n is the multiplicative identity for all elements of the ring tho

[fossil hound]

Yes

A unit doesn't have to be 1

[slim kayak]

"1" = "multiplicaitve identity"

[fossil hound]

But what does this have to do with anything?

[delicate orchid]

we just did

[fossil hound]

Yeah but just having xy=x for one element x does not directly show it's also the multiplicative identity for every other element

I mean sure its easy

But i guess he needs to show that as well

[delicate orchid]

yeah there's a little bit more to the argument, the "n" in the exponent could change if you picked a different r but the point is there will exist some n

[fossil hound]

Sure there will but you need to show that r^n = r^m then

[slim kayak]

just take the product

[fossil hound]

Like you need to show it's the multiplicative identity for all elements of the ring

[final dragon]

why does the n depend on r

[delicate orchid]

set r = a

[final dragon]

oh yea

you are corrcet

[fossil hound]

I mean that will work, but in general that's not true

[final dragon]

you guys are very smart

amazing

[delicate orchid]

if you keep taking the product it will eventually work

[fossil hound]

Why would it?

[slim kayak]

in general i wouldnt give rngs a second look tbh

[delicate orchid]

because the exponent is larger than 1 for almost all elements of the ring, so eventually you'll end up with a^n with n > |R|

[fossil hound]

Yes but this doesn't show its the identity for all elements

[delicate orchid]

what

if a^mr = 1

then a^mra^mr = 1

so on

[fossil hound]

What?

You need to prove 1 exists

[delicate orchid]

sorry I missed an r

[slim kayak]

and an r on the other side

instead of 1

[fossil hound]

So while you might have r^n*a = a it doesn't follow that r^nb=b, it follows as a is a non zero divisor

Cause you will get r^n*a*b = a*b and as a is a non zero divisor you get r^n*b = b

That's the only reason

[slim kayak]

you build a "partial inverse" for any r in R. Then you take their product. This product is guranteed to fix any element in R

[fossil hound]

Why?

[delicate orchid]

oh yeah is R commutative or

[slim kayak]

product of the exponents I mean, sorry

[fossil hound]

There is no reason for it to fix it at all

Yeah doesn't matter

Like

[slim kayak]

If a^nr=r, then a^2n r = r, can we agree on that

[fossil hound]

Right

[slim kayak]

and if a^n r = r and a^m s = s, then a^nm r = r and a^nm s = s

since a commutes with itself

[fossil hound]

Oh I thought you meant the sum

of the exponents

nvm then

Yeah that also works

Like just taking the products of them

Cuz that wouldnt work

[slim kayak]

yeah this was an exercise on our communication skills 😭

[final dragon]

and what about this if an element of R is not the zero devisor its invertible

[slim kayak]

I am getting sullied for speaking the truth, is this really what mathcord has come to smh

[delicate orchid]

for "invertible" to even make sense we need a 1

[final dragon]

it has

we just proove it

i mean you guys

[delicate orchid]

was that not the problem lol

[final dragon]

its multiple

questions in one

[fossil hound]

Again cause the map is bijective

x->ax is bijective

So if you showed that 1 exists it needs to have a preimage

[slim kayak]

or in reverse order, since for some n, a^n=1

[final dragon]

so are only a and 1 invertible

[chilly ocean]

Silas is a very bad communicator, not you're fault

[fossil hound]

no every non zero divisor is invertible

[slim kayak]

How did you escape discussy 😬

also thanks ❤️

[chilly ocean]

only to bully sials

silas

hes mean to me irl so gotta be mean here 😈

he sucks

[final dragon]

yo chill

what does it mean to be nielpotent

and idempotent

[slim kayak]

x^2=x

I mean, did you try to google it first lol

[fossil hound]

do you even need to ask that question?

[final dragon]

if in ring R exist non zero nilpotent element proove that there exist a from R so that a^2=0

[slim kayak]

W hating, gotta respect the grind

[final dragon]

yea lol

[fossil hound]

Did you even try it first?

It's not a difficult exercise

[final dragon]

no i go try it

[slim kayak]

this might be weirdest question ive seen in a while

[fossil hound]

If you just want the answer you won't learn anything

[slim kayak]

how did they even define nilpotents

[final dragon]

yea correct

[chilly ocean]

🫡

[fossil hound]

🙏

go away nine

[slim kayak]

I enjoy your company nine, stay as long as you like

[chilly ocean]

♥️

[chilly ocean]

https://tenor.com/view/spongebob-squarepants-spongebob-evil-laugh-laugh-laughing-gif-5059320

be careful, he might say kill yourself again

[final dragon]

does sombody have like a guide to rings groups

i dont know anything its so hard

any like suggestions

[slim kayak]

have you taken linear algebra I and II yet?

[final dragon]

its in next semestar

we only do ring fields and groups this semestar

[slim kayak]

Yeah this explains stuff

[fossil hound]

This is so weird

[final dragon]

wdym

[fossil hound]

Who starts with rings first

[final dragon]

we do

[slim kayak]

not even rings, RNGS?!

[fossil hound]

And then not even rings with 1

[slim kayak]

not even commutative

[delicate orchid]

that's beyond bizzare

[slim kayak]

what kinda place is this

[final dragon]

bro what is happening

[fossil hound]

i think your professor hates you

[final dragon]

yea he does

we did some arithmetic first like module and all that

and then we do rings groups and fileds in that order

[slim kayak]

Sono Chi no Sadame starts playing

[final dragon]

first we did rings and then some isomorf stuff

and then he said there are 3 therorems very important

for isomorf

and i dont know them

[fossil hound]

Okay you gotta be kidding me right?

[final dragon]

bro im dead ass

why is that

did you do different

[fossil hound]

You did modules before vector spaces?

[final dragon]

vectors are next year

in linear algebra

[slim kayak]

okay to put things into perspective

[fossil hound]

yes lets start with modules before cause why not

[slim kayak]

this is like learning calc before trigonometry - its very weird

[fossil hound]

[final dragon]

why is it important to start with vectors

[delicate orchid]

that's not what they meant

[final dragon]

wait acttually

[delicate orchid]

because vector spaces are the nicest algebraic objects

[final dragon]

bro these rings are so hard

and the mf doesnt even care

[fossil hound]

What then?

[delicate orchid]

so it makes sense to introduce them first to get students used to thinking abstractly

[delicate orchid]

arthimetic was the first thing they did in the module

which is still hard to believe

[final dragon]

yea

they said that

[fossil hound]

Oh

[delicate orchid]

they said that and then didn't introduce vector spaces

[slim kayak]

rings and groups were introduced in LA in my case, even with some results and exercises on them

[final dragon]

we need to learn to think abstract so we can learn rings

[fossil hound]

I completely misread that

[slim kayak]

But doing abstract ring theory is unhinged

[final dragon]

i mean we dont even have

calculus

we start off with real analysis

first year

and abstract algebra

bro i just got out of highschool leave me alone

[fossil hound]

Before LA?

💀

[final dragon]

yea

[fossil hound]

You sure you didn't pick the wrong courses broski?

[final dragon]

no bro

i can send you my subjects

linear algebra is second year

what is so wierd guys

how did you start

[fossil hound]

linear algebra

[delicate orchid]

literally first semester we did linear algebra and didn't do ring theory until a year later

[slim kayak]

^ (silas) and real analysis, standard fare

no wew why

[delicate orchid]

we did group theory 2nd sem

[final dragon]

you see

[delicate orchid]

I don't speak serbian

[final dragon]

we have real analysis 1 and abstract algebra

and second semestar you see linear algebra

[delicate orchid]

ok yeah that is pretty clearly "analnza"

[final dragon]

yea

[slim kayak]

can you tell us what the less recognizeable words mean

[fossil hound]

Maybe thats why I don't know any Serbian mathematician

[final dragon]

osnove matematike we do like axiomatic theory of sets

and osnovi programiranja we do programing c laungauge

[chilly radish]

Abstract algebra before linear algebra is completely fine

It might just be a bit harder to grasp but not by much really

Also not that uncommon

[final dragon]

my type of guy right here

[chilly radish]

Historically abstract algebra came first anyways

[final dragon]

but still i dont know shit in abstract algebra

real analysis is easier than this shit

[chilly radish]

It'll come with time

And practice

[slim kayak]

abstract algebra with non-unital rings tho?

like youd at least start a bit watered down

[fossil hound]

Makes no sense though

[final dragon]

what does it mean non unital

[fossil hound]

I don't think there is any benefit to it