groups-rings-fields

[chilly ocean]

What did they say exactly?

[crystal vale]

It says that in book it starts a definition of group with a set such that there is binary operation then follows four axioms. And these axioms are the property of group and empty set satisfy every property of every group because it is a subset of a every set. So empty set is a group

[chilly ocean]

Can you tell me what the axioms are exactly?

[lusty marlin]

If any of the axioms here mention the existence of an identity element, then clearly the empty set does not satisfy all of them

And the definition of a group does require an identity element

[chilly ocean]

The empty set satisfies every statement of the form "for all x in Ø ...", that is what is meant by vacuous truth

But it does not satisfy any statement of the form "exists x in Ø ..."

I think that might be where the confusion lies

[chilly ocean]

If you have some axiom of this form, then the empty set can never be (the underlying set of) a model of your theory

[grave sedge]

You can just project on Z/p[x] which is the same thing and factor the polynomial there (what you're doing here is more or less verifying that the projection is a homomorphism)

[hidden wind]

mhm existentially quantifed statements cannot be vacuously true, but they can be vacuously false

the only group axiom that is not satisfied by the empty set (together with the empty function binary operation), is the identity

[hidden wind]

this is one of the manifestations of the duality “forall” is the same “not exists not”

and vice versa

[hot pebble]

I feel like our professor explained this but i kinda forgot the main idea so i guess i'll ask here

What's the main advantage for talking about field extensions instead of subfields? I remember something about any ring homomorphisms between two fields must be injective but couldn't connect the dots

[hidden wind]

i am very new to thinking fields, but as i understand it, considering field extensions is equivalent to yielding new roots to irreducible non-linear polynomials over the field, which is very interesting

while "taking away" roots is sort of not

somethign along those lines

but it's a good question, with groups we are typically interested in what smaller parts they consist of; with fields we're more interested in what larger structures they can support

[south patrol]

I think that is kind of orthogonal to it rodbet

The thing is that it allows for slightly more flexibility since you don't worry about literally being a subset

But yes every ring hom between fields is injective so it doesn't change much and indeed often it is convenient to view things as subsets

Like say you have a field K and want to embed in some larger field L, it's a bit annoying if you have to construct a map K->L and then replace K by its image in L each time

[crystal vale]

Yes

[hidden wind]

i'm a bit confused how is this different from the situation with groups?

[south patrol]

Well it is how every map of fields is injective

Maybe I have misunderstood the aim of the question

I assumed the point was why do we consider general maps and not set theoretic inclusions

But perhaps yeah it is what you mean lol my bad like why do we care about bigger things than the current field as opposed to smaller things

[hidden wind]

nono i'm the one who's confused here >:S

i should probably just take the time to have a proper look at what you're both saying

[rain grove]

Let $n$ and $a$ be coprime numbers. Show that $n \mid a^{\phi(n)} - 1$.

the results of previous exercises I did were: $\phi(n) = |Z_n^*|$ and $\phi(n)$ is even. And $mx+ny = 1 \iff m$ and $n$ coprime, from this follows that $a+n\mathbb{Z} \in \mathbb{Z}_n$ is invertible then a and $n$ and $a$ are coprime

I need a hint

[cloud walrus]

OHHELLNAH

[hidden wind]

i'm taking a course on commutative algebra and i'm already a bit behind even though the semester started just this week ehehe

[rain grove]

Ohh right, $\phi(n)$ is euler function

[cloud walrus]

OHHELLNAH

[worthy berry]

If you have a k-algebra homomorphism f : A -> k is ker(f) always a maximal ideal?

[chilly ocean]

Assuming A is nonzero, and you are talking about unital associative algebras, I believe so. If A is nonzero, it contains a copy of k, right? So it must be surjective. And a surjective ring homomorphism has codomain a field iff the kernel is a maximal ideal

[arctic trail]

A/ker(f) is a field what does this say about ker(f)?

[lean sail]

Not looking for an answer, but wondering if someone can give me a hint on this one.

[vocal patrol]

When working through my script, I have proved the following:

• every field is an integral domain

• if D is an integral domain so is D[x]

• every finite integral domain is a field

• if F is a field, F[x] can’t be a field

Doesn’t this contradict with one another?

Let’s say D is a finite integral domain then so is D[x]

Every finite integral domain is a field because we can create a function from D onto itself s.t. The function is bijective, hence we have a multiplicative inverse for each element, I.e. all elements that aren’t 0 are units

But there’s no way to find inverses of non constant polynomials

So basically I don’t know which way to think here…

[wraith cargo]

D[x] isn't finite even if D is

[vocal patrol]

Ah okay that makes sense

[vocal patrol]

So that’s why

Then the proof with the bijective function from D onto itself doesn’t hold anymore right?

Where I can prove that all elements in a finite D are units

[wraith cargo]

yeah essentially

[vocal patrol]

Thanks a lot!

[worthy berry]

How do you argue that A/ker(f) is a field? We have that A/ker(f) \cong f(A), but I don't see why f(A) should be k or {0}?

[chilly ocean]

Does A contain a copy of k?

[worthy berry]

As an k-algebra shouldn't k be contained in it merely since we can multiply the identity by scalars from k?

[chilly ocean]

Yes. Can you see how that implies your map is surjective?

[worthy berry]

Is it just cuz for any c in k we have f(c) = cf(1) = c?

[chilly ocean]

Yep

[worthy berry]

Oh cool thanks

[grave sedge]

Fix some of a,b,c,d to be the identity

[lean sail]

Got it, thanks!

[lethal hornet]

if you care, this could be made cleaner; for instance you declare b = yz and d = x'z, then prompty chop off each z from both sides between lines 3 and 4

and also I don't see why we're declaring that set on line 1

[tropic spade]

I'm trying to count the number of conjugacy classes of elements of order p (a prime) in S_n. Is my thinking in this sketch right? Assuming p divides n! an elt of S_n has order p iff it has a cycle decomp as a product of disjoint p cycles. Take k the largest integer such that p^k divides n!. There is one conjugacy class for each partition of n containing a multiple of p as one of its terms and 1s for every other term (by considering the cycle decomp above and the corresponding cycle types). These partitions can be listed as n=(n-p)+p, n=(n-2p)+2p, ... , n=(n-p^(k-1)p)+p^(k-1)p. There are p^(k-1) entries in this list, one for each conjugacy class so our desired value is p^(k-1).

I'm worried I am miscounting

I say sketch because I'm leaning on a bunch of theorems and past exercises for some of this.

[coarse kestrel]

how did you move the y around? I don't see it.

[icy totem]

My lecturer decided not to prove the characterization of UFDs, why do you think so? Because it is too long of a proof?

Im talking about the theorem that states A domain is a UFD iff every irreducible is prime and every ascendent chain of ideals is stationary

[dull ginkgo]

I think it’s every ascending chain of principal ideals

I think there is a metric fuckton of equivalences, like prime ideals having a prime element

[icy totem]

Yea lol I knew the italian name so I made up the english one

[icy totem]

Principal ideals are prime iff they are generated by a prime element

[dull ginkgo]

I think it’s a proof by induction

[icy totem]

On what?

[dull ginkgo]

The prime decomposition of an element

[icy totem]

Ah ok

[dull ginkgo]

assume we have two decompositions, take the difference

Can try to find the proof in Jacobson when I get home in a about 5

[icy totem]

Ok thx, because UFD => the two properties is pretty clear, but the converse is not

I cant understand if she (the lecturer) gave the proof for granted cause it was easy or omitted it because it was too long/advanced lol

[cobalt heath]

It can be both

Too easy but too long if you write it out

[icy totem]

Thx

[lean sail]

Which step?

[lean sail]

So just leave z out altogether?

[crystal vale]

Can you explain more ?

[chilly ocean]

What about it do you want me to explain more?

[knotty badger]

So for example, you want the empty set to be a subset of everything else, and so you want $\forall x, x \in \emptyset \implies x \in A$. But if there actually existed an element in the empty set, this definitely wouldn’t hold for all $A$. Instead the reason this works is that there are no elements in the empty set, and you have vacuous truth in that case

[cloud walrus]

Pseudonium

[knotty badger]

Another way to think about it is - to prove $\forall x \in \emptyset, \dots$ you need to check something works for everything in the empty set. But then there’s no conditions to check, so you get truth for free

[cloud walrus]

Pseudonium

[knotty badger]

Whereas to prove $\exists x \in \emptyset, \dots$, you need to provide an example of an element in the empty set satisfying the condition. But the whole point of the empty set is that it’s empty, so it’s impossible to provide such an example

[cloud walrus]

Pseudonium

[lean sail]

@lethal hornet @grave sedge i just cleaned up the proof from yesterday according to your ideas and feedback. @coarse kestrel i think in this one it's slightly clearer that i didnt move y around. the other proof was very messy, and it did look like that was what happened. here is the new one. please, anyone feel free to provide feedback

[surreal dagger]

I would fix arbitrary x and y first, then define a,b,c,d the way you did and end up with yx=xy for any choice of x,y hence G is abelian.

[lean sail]

are you saying that the proof should explicitly mention that $x$ and $y$ are chosen to be arbitrary first, and the rest of the elements follow?

[cloud walrus]

proofman

[surreal dagger]

I dont want to say it should, but going by the definition of being abelian you have to show that for any choice of x,y xy=yx. So me personally I would fix x,y in the beginning and then work with the axb=cxd implies ab=cd condition

[lean sail]

Okay, and fixing x and y satisfies that?

[surreal dagger]

Well yeah, you fix and arbitrary x,y and then you end up with yx=xy, so G is abelian

[lean sail]

Interesting. Maybe I am a little confused on what fix means. What is the difference between “fix” and “let”?

[surreal dagger]

its the same, tbh it may be that im being pedantic. I think your proof is fine, just me personally I think its cleaner if you explicitly fix x,y in the beginning

[lean sail]

It makes it more clear that x and y are arbitrary, I think is what you’re getting at?

[surreal dagger]

In my opinion, yes

[lean sail]

I think that’s valid and good feedback. I was struggling with the very notion of, how do we know that x and y are arbitrary, when I was writing this.

[surreal dagger]

Great, glad you agree

[crystal vale]

Is existence of an element not a property of set ?

[chilly ocean]

It is

And the empty set does not have that property

[crystal vale]

I understand but I am not 100% sure because I know to be a group it needs to be non-empty, but an empty set follows all properties of all sets

[crystal vale]

But if A has property E(x), then an empty set also has this property?

Is it E(x) is property of an element that belongs to set not of that set

Existence of an element is set property not element property

[knotty badger]

It is not true that an empty set satisfies every property of sets

For example, the empty set doesn’t have any elements

[chilly ocean]

The empty set does not satisfy every property. What it does is satisfy all properties of the form (for all x in \empty set, phi(x))

But there are properties that are not of that form

[crystal vale]

I see

Thank you

[rain grove]

Is there a element of order 5 in GL_n(R)

[chilly ocean]

Depends on the n

[rain grove]

Oh right i meant GL_2(R)

[quiet pelican]

Yes
A rotation by 72 degrees

[rain grove]

ohhh thats amazing

I see now how I have to think about this

[quiet pelican]

What is it?

[rain grove]

So I found an isomorphism from D_8 to O_2 (orthogonal group)

And now I need to find a D_10 isomorphism

But d_8 was easy i just guessed it

[quiet pelican]

An injection?
Or an isomorphism onto a subgroup?
(which are the same)

[quiet pelican]

How do you visualise D_10?

[chilly ocean]

That cannot be since O_2 is infinite but D_8 isn't (I assume you mean the dihedral group by D_8)

[rain grove]

Wait I didn't say that right... i meant there is an isomorphism from D_8 to a subgroup of O_2 generated by matrices

[quiet pelican]

Yes that's right

[rain grove]

And now i need to find a subgroup of O_2 that is isomorphic to D_10.. yeah sry about that

[quiet pelican]

^

[lean sail]

i thought a surjection was onto, no?

[rain grove]

Symmetries of n/2

[grave sedge]

clearly D_5 with the wrong name

[quiet pelican]

So when you visualise it, are you visualising something in R^2?

[rain grove]

Yeah

[quiet pelican]

An isomorphism is still surjective (and I'm using onto to emphasise that)

[quiet pelican]

So can you turn that visualisation into an embedding of D_10 into O_2

[rain grove]

thats so cool. and why is it subgroup of O_2 and not GL_2?

Like it just so happens that these rotations are orthogonal matrices, what proprety in the visualisation or in D_n makes that?

[quiet pelican]

Because preserving the shape of the n-gon you act on is equivalent to preserving lengths and angles everywhere

[tame fog]

can someone help me understand what they mean by relation in this context? (from Basic algebra 1 Jacobson)

[chilly ocean]

They mean that those things are equal to the identity in that group

By relation they mean some equality like xyx^-1 = xy^3

But because of inverses every equality can be written in the form something=1

So they are only writting the something

[tame fog]

ok ty

hmm, I still fail to see the relation though, is xRy iff those three things are equal to 1?

[chilly ocean]

equality is a relation

But yes for each of those things (or even for them all together) you could make a relation that is true iff the equality holds

For example R(x,y) iff xyxy=1 for the last one

I'm not using binary relation notation because there could be more variables

You could have a group where the relation R(x,y,z) corresponding to xyzy=1 holds

[tame fog]

yeah but the 3rd equality refers to two elements so I'm having a hard time seeing how they define if x is in K

[chilly ocean]

x^n is in K, and so is y^2 and xyxy

And K is the smallest normal subgroup containing those elements for all x,y in the free group

[tame fog]

ugh this chapter has me so lost

ok ok got it ty

[narrow zinc]

I was tasked to find an infinite field with a non-zero characteristic. Would Z_2 (X) qualify? Is that even a field?

it should have characteristic 2, right?

[chilly ocean]

I assume by Z_2 you mean the field with two elements. And yes, that is a field

[narrow zinc]

right it is the "breukenveld" of Z_2[X] I found that... I do not know to what "breukenveld" translates

quotiënt body?

[chilly ocean]

Quotient field probably

It's also common to call it field of fractions

[lean sail]

wondering if i can have a hint on this one. we haven't discussed finite groups or order at this point in the book yet. so i am a little confused on where to begin

[chilly ocean]

The finite requirement is unnecessary, I believe

[stone vault]

Try and see if you can organize those elements into groups of 4

[stone vault]

Well if we remove the finite assumption then the number of such elements need not be finite either, no?

[lean sail]

thank you, something else i was thinking about, do i need to consider a particular group?

[stone vault]

It might help to look at examples

[narrow zinc]

right that's the one, I found the wikipedia and turned it to ENglish so that should be the same thing

[lean sail]

cool, thanks for the hints

[rain grove]

Quaternion group is isomorphic to subgroup of GL_2(C) generated by these matrices:

Can this also be visualized somehow?

idk much about quaternions except the definition we had in algebra

[quiet pelican]

I don’t really visualise the quaternion group
Except as acting on 4d real space spanned by i, j, k

(1, i, j, k)

[lean sail]

@stone vault i am struggling hard to find an example of a group where this is actually true. any ideas? so far, i've tried U(14) and U(15)...

did write a nice little python script though.

[rain grove]

D_10 is an example r^5=1

[lean sail]

Thanks I’ll try playing around with that.

[lean sail]

Hopefully a good example like this will inspire a proof.

[spare isle]

We say that the units of a ring form a group, but isn't this false for the trivial ring?

[coral spindle]

The group of units of the trivial ring is the trivial group.

[spare isle]

0 isn't a unit though

[coral spindle]

Yes it is: 0 = 1 is a unit.

Its inverse is itself.

[spare isle]

Ah sorry I was conflating with divisors of 0

[south patrol]

Tbh lol this is something I never thought about so thank you aha

[spare isle]

So would it be right to say that the group of units from a nontrivial ring never form a ring since the 0 element is omitted, or would it still be possible for it to somehow form a ring under a different abelian group (i.e. with a different + operation) with a different additive identity?

[south patrol]

Well "form a ring" is sort of ambiguous

They do not form a subring sure

[spare isle]

Yes it's clear that it doesn't form a subring, but is it possible to form a ring with a different + operation and 0?

[coral spindle]

It would never form a ring, assuming we have the same additive structure, since there would have to be a zero.

[south patrol]

But yes they can certainly be given another operation

[coral spindle]

Yes

But this isn't saying very much

[south patrol]

In some cases anyway

[coral spindle]

Lots of Abelian groups may be rings.

Oh hold on

[spare isle]

I'm just struggling to think of an example where the group of units from a ring can be extended to its own ring

[coral spindle]

you mean multiplicative, my mistake.

No, because 0 must be absorptive

[south patrol]

Oh

[coral spindle]

And there is no absorptive element in a nontrivial group.

[south patrol]

Okay yeah I misunderstood that too lol

Good catch boytjie

[spare isle]

Where did I mean that? Sorry I might have missed something or explained poorly

[coral spindle]

You are saying

given a group, which we will consider 'multiplication'

is there an addition that makes it a ring.

[coral spindle]

.

[spare isle]

ah yes I guess that's what I'm trying to say

[south patrol]

I think we both initially thought like, can the group of units be given a multiplication, rather than can it be viewed as the multiplication of a group

[coral spindle]

Which was just a mistake on my part

[south patrol]

Note that Boytjie's argument works more generally: if you forget addition on a ring, you get a monoid which is never a group (except for the zero ring)

[spare isle]

I never really learned about absorptive behavior which disallows extending the multiplicative group to a ring; are there any resources where I could learn more about this in particular?

[south patrol]

There isn't much to it. It's just that if R is a ring then 0.r = 0 for all r

So if the elements form a group under multiplication, r = 1 for all r (by dividing by 0)

So the ring is trivial

[spare isle]

Ah so basically we have the issue that we are trying to have a group which works both "multiplicatively" and "additively" as their own groups while having different identities?

[dull ginkgo]

It’s like a monoid acting on itself with a different monoid structure sorta

[south patrol]

Hm

No I think the key thing is the interaction between addition and multiplication in a ring

[delicate orchid]

it's a monoid acting on an abelian group

via endomorphisms, I should add

or else you won't get distributivity

[spare isle]

If I'm honest, this is a bit past what I learned in my abstract course 🫠

[delicate orchid]

no idea what we're talking about btw I'm just putting a stop to this babble

[coral spindle]

Fortunately it's irrelevant

[dull ginkgo]

Currying (food)

[spare isle]

Oh wait I think I misunderstood this a bit initially. Basically, rings require some "0" element to yield "0" for all applications of "multiplication," but this would violate the uniqueness of ax = b when viewed as a (nontrivial) group.

[chilly ocean]

Yes, this is consequence of bilinearity of multiplication and + being a group operation: 0x = (0+0)x = 0x+0x therefore 0 = 0x - (0x) = 0x + 0x - (0x) = 0x

[spare isle]

Thank you for the help everyone!

[lean sail]

still working on this problem. i did notice that $D_5$ is an instance of this. i also noticed the following, where r is a 72 degree rotation.

[cloud walrus]

proofman

[chilly ocean]

That is no issue. Infinite numbers can still be multiples of 4

You know about partitions, proofman?

[delicate orchid]

"let p be a prime, then every non-identity element (aka, p-1 elements) of a cyclic group of order p is a generator of that subgroup" is my hint

[lean sail]

yes, i know a little bit about them. i know that if an equivalence relation partitions a set into classes, those classes together equal the set?

[chilly ocean]

Their union equals the set, and they are disjoint

[lean sail]

yes, i recall that now

[lean sail]

i dont want it to seem like im ignoring your hint. where i am at in the book i am reading, they havent introduced generators or subgroups. i happen to roughly recall what they are because i did a chunk of this book some time ago. but i think the author wants us to use some other primitive knowledge we are supposed to have up to this point!

[delicate orchid]

no worries, it's essentially what you noticed - if r^5 = e, then (r^k)^5 = e for 0<k < 5

[lean sail]

cool, good to know, i think i am getting closer

[dull ginkgo]

x^5 = e is equivalent to x^4 = x^-1

Yeah basically that

[delicate orchid]

no it isn't

[dull ginkgo]

Forgot to type -1

[lean sail]

i think the basic idea here still stands though, couldn't i prove something like, if $r^5 = e$, then so do $r^2$, $r^3$, $r^4$? i think what also might be holding me back is... how do i know those are the only ones?

[cloud walrus]

proofman

[dull ginkgo]

If you need a conceptual route to try to quantify “how” it’s divisible by four.

You can consider the collection of elements satisfying x^5 = e, call it S. If you can split this collection up into disjoint subsets all of size four, then we know the whole set is divisible by four in size.

As mentioned, for r in that set, r^2…r^4 are in that set too (giving a collection of size four for each r). But how do you think you can show that they’re distinct from the powers of another u in that set?

[delicate orchid]

wdym the only ones? why would they have to be the only ones

[lean sail]

i think this is more or less what @chilly ocean was getting at

[delicate orchid]

the key insight is that if you have some other x not in <r> with x^5 = 1, then x^2, x^3, x^4 are also distinct elements with the required property

[chilly ocean]

(You also want those elements to not satisfy x=e)

[lean sail]

maybe they dont? as long as the number of elements a multipleof 4?

[delicate orchid]

maybe they don't what

[lean sail]

they may not have to be the only ones

[delicate orchid]

yeah duh

[dull ginkgo]

Did I misread the problem, is it elements of order 5?

[delicate orchid]

annoys me when I split something into two messages and people just do not read the 2nd post

[delicate orchid]

the statement is flat out false if you include the identity

it's 1 mod 4 if you do

[chilly ocean]

It says nonidentity

[dull ginkgo]

Epic

[dull ginkgo]

The major part of note here is that:

Assume x^5 = 1 and x ≠ e, then x^n = e if and only if n is divisible by 5

[delicate orchid]

this is because 5 is prime btw, if we have x^6 = e and x not e, then we could have x^2 = e or x^3 = e, so the "only if" direction fails

[dull ginkgo]

Which prevents shit like if it was x^4 = e, where x^2 would also satisfy

Yeah

[delicate orchid]

u just got trolled buddy

[dull ginkgo]

[lean sail]

im a little lost on whats going on here

[dull ginkgo]

This is to make sure that your set of powers up to 4 has exactly 4 elements

[dull ginkgo]

Well the x, x^2, x^3 and x^4 are all distinct

If they weren’t, then some x^k = e for k < 5 (divide by the lesser power of x)

Can you see why that cannot happen (hint: ||the difference is coprime to 5, and you can use Bezout lemma to show x = e||)

[lean sail]

I’ll think about this, and get back to you.

[dull ginkgo]

Here’s a proof sketch for ya, that you can work out the details

Let S = {x in G : x^5 = e, x ≠ e}

1. for x in S, then x, x^2, x^3, x^4 are all distinct
2. if x and y are in S, then x^a = y^b for some a and b positive (or zero) and less than 5 implies x = y^n or y = x^n for some n positive and less than 5
3. Conclude that for x and y both in S, the sets {x, x^2, x^3, x^4} and {y, y^2, y^3, y^4} are either disjoint or equal
4. show sets of the above form split up / partition S into sets of size 4. Conclude 4 | |S|

Bonus: show this holds for swapping out x^5 = e for x^p = e for any prime p, and showing (p - 1) | |S|

[hidden cairn]

i don't understand why H should not be contained in Z(G) for such g to exist. also dont see why we require H to be abelian.

cant we just take any element g in G - H and see the automorphism formed by conjugation by g is not an inner automorphism of H? simply because g not in H

[knotty badger]

It could be an inner automorphism of H though

[hidden cairn]

this definition implies an inner automorphism of H can only be formed via a conjugation by an element of H, right?

so if i just conjugate by some element g in G - H,

the automorphism can't be the inner automorphism of H

right?

[knotty badger]

Not necessarily

Take some g not in H

You can define an automorphism $\phi_g : H \to H$ sending $h \mapsto g h g^{-1}$

[cloud walrus]

Pseudonium

[knotty badger]

But it could definitely be the case that there’s some $h’ \in H$ such that $g h g^{-1} = h’ h h’^{-1}$ for all $h \in H$

[cloud walrus]

Pseudonium

[knotty badger]

In that case, $\phi_g$ would define an inner automorphism of $H$

[cloud walrus]

Pseudonium

[hidden cairn]

ah right

ok i didnt see it this way

[hidden cairn]

thanks ill try to make sense of the statement one more time now

[lean sail]

appreciate you taking the time to put this together

[lean sail]

i am sorry i just dont see where bezout's lemma comes in, isn't that saying that two elements can be written as a linear combination of their gcd?

[dull ginkgo]

Bezout lemma would state that there exists n, m such that
n(b - a) + m5 = 1
Therefore:
x = x^1 = x^(n(b-a) + m5) = (x^(b-a))^n (x^5)^m = e^n e^m = e which contradicts the assumption that x ≠ e

[grim hawk]

Hey all, if we are working with a general set S that has an identity element and its law of composition is associative, is there a way to prove that the subset of all invertible elements from S is closed under its law of composition? My belief is that there does exist a counterexample but I am struggling to find it.

[south patrol]

Yes this is true

[lethal hornet]

you're describing S to be a monoid, carrying an identity and associativity. if we define U(S) as the set of all invertible elements of S, then we can show that

• U(S) is nonempty, and
• for each x, y in U(S), so too is xy in U(S)

[south patrol]

||the inverse of ab is b^-1 a^-1||

[lethal hornet]

and that's enough to show it's a group because it inherits id and associativity from S

[lean sail]

I get it

[grim hawk]

I think my holdup is the latter of your two points. The fact that it is not empty feels to arise easily from the fact that the identity is its own inverse. However, if we have defined our law of composition * : S x S -> S, then what about its structure tells us in general that the law of composition maintains closure in the subgroup of inverses?

[south patrol]

This

[grim hawk]

Heard. Okay, one more question then. Maybe I'm overthinking this, but does invertibility in this context mean that the inverse is contained within the set or simply that the inverse exists? For instance, if we look at the integers as a set and choose our subset to be Z / {0} then if we choose our binary operation to be standard multiplication, we know that these are all invertible but the inverses do not exist in our U(Z).

My understanding of the inverse axiom is that for a set to be a group there must be an inverse for every element and that element must exist in the set. Is this wrong?

[rocky cloak]

It needs to be in the set yeah

[grave sedge]

"exist" and "exist in the set" are synonims

[dull ginkgo]

How’s it going on the problem

[grave sedge]

The fact that you identify "the integer 2" (which has no inverse) with "the rational number 2" (which has an inverse) is technically an abuse (although a very intuitive and fruitful one)

[stone vault]

I mean, I guess 4\cdot \omega is \omega?

[chilly ocean]

Yes, it is true that 4xAleph_0 = Aleph_0

[arctic trail]

Isnt this false?

Oh wait no youre right

Omega cdot 4 is the one thats not omega

[narrow zinc]

let F be a field with |F| = p^r , with p prime and r a non-zero integer. Would the additive group F, + be isomorfic with the direct sum of Z_p with itself (summed r times?)

[south patrol]

Yes

[narrow zinc]

ah nice

[south patrol]

There are a few ways to see this, but my favourite is this: every characteristic p field is a vector space over F_p. So a finite characteristic p field F is isomorphic to (F_p)^n as an F_p -vector space for some n.

So in particular they are isomorphic as additive groups

[narrow zinc]

yessss

alright I knew I wasn't grasping that from nowhere

I've forgotton so much 😦

yes that makes total sense actually. It was even hinted at in the broader context of the exercise, as it was given that the 'prime-sub-field' = smallest subfield would be Z_p in that case

but seeing it as a vector space of, in this case, r dimensions should be the expected reasoning as well

[south patrol]

It's also how you can prove finite fields have order p^r for some p,r

Nice thing to keep in mind

[narrow zinc]

right that proof came somewhere earlier in the text as well

or was left as exercise too

thx

[narrow zinc]

"Choose a specific representation of F_16 (the field with 16 elements) as Z_2/ (f) , so with f an irreducable polynomial over Z_2, and give in this representation the subfield with 4 elements

so do I choose an appropriate f? from what I understand it should be of the 4th degree?

[quiet pelican]

Irreducible of degree 4 yes

[narrow zinc]

that would be either X⁴+X³+X²+X+1 or X⁴+X³+1 or X⁴+X+1

[quiet pelican]

And yes you pick any f

(That works)

[narrow zinc]

those should be the option for irreducables over Z_2

unless I miscalculated... but then what would be the subfield with 4 elements? Isn't that always 0,1, a, a+1?

or should I 'rebrand' the a? write it in terms of polynomials?

[quiet pelican]

What they’re asking is essentially “what is a in terms of the x”

[quiet pelican]

And yes these are right

[narrow zinc]

so two polynomials that are each other's inverse in Z_2/(f) and the first polynomial squared is the second and the second squared is the first?

so they behave exactly like a and a+1?

right and those will be dependant on the chosen f?

[quiet pelican]

Yes

And yes

[narrow zinc]

ah and because they need to be each others inverses, it suffices to break down f -1 (now reducible) to component parts and test those combinations?

yesss for f= X⁴+X³+X²+X+1 the a-> X^3+X and a+1 -> X²+X appears to work splendidly -> but not for addition

[quiet pelican]

No that doesn’t
Because that “a + 1” isn’t actually a + 1

[narrow zinc]

well no

I mean those seem to fullfill the roles

[quiet pelican]

But for the field structure to work you need “a + 1” to literally be a + 1

[narrow zinc]

hm ah to get your + to work, obv... but I've tested all other candidates?

[narrow zinc]

wait maybe this is a faulty assumption?

every element must have an inverse of course, even the higher order ones... that assumption is ... not great

[narrow zinc]

ah I think it's X³+X² and X³+X²+1 ( provided ones uses that f)

thanks 😉

[quiet pelican]

Ah I’m stupid

I was looking for factorisations of f - 1 in F_2(X), not F_2(X)/(f)

[lean sail]

suppose $x^2$, $x^3$, $x^4$ werent all distinct, why does some $x^k = e$ where $k < 5$?

[cloud walrus]

proofman

[dull ginkgo]

If x^a = x^b, then x^(b - a) = e

[lean sail]

wait, but what if x^a = y^b?

[dull ginkgo]

We know x^5 = y^5 = e

[lean sail]

suppose we have two different elements, but when they are "operated" $a$ and $b$ times...

[cloud walrus]

proofman

[lean sail]

i agree we know $x^5 = e$, im not too sure about $y$

[cloud walrus]

proofman

[dull ginkgo]

y is in our set

So y^5 = e ≠ y

[lean sail]

i mean, i dont think you're wrong, i just don't see how it's in the set

[dull ginkgo]

We are asserting y is in our set

As to show that the sets described are distinct

There is no point assuming it isn’t

[lean sail]

i dont see why not

[dull ginkgo]

We are trying to partition S.

So we are trying to show for x and y in S

That the sets {x, x^2, x^3, x^4} and {y, y^2, y^3, y^4} are equal or disjoint

So we are assuming a priori that they have a nontrivial intersection

I.e some u = x^a = y^b lies in both

Do you see what I’m trying to show?

[lean sail]

i think so... let me see. if we have some x such that x^5 = e, and some y such that y^5 = e, then {x, x^2, x^3, x^4} and {y, y^2, y^3, y^4} are disjoint?

[dull ginkgo]

Or they are equal.

So the only difference between those two cases is that the sets intersect, if they do, then they are equal. That’s what we want to show

[lean sail]

ok, then should we suppose that there are some x and y such that $x^m = y^n$?

[cloud walrus]

proofman

[dull ginkgo]

With the constraint that both n and m are within [1, 4]

[lean sail]

right, we want x^m and y^n to be elements from those two classes

[dull ginkgo]

Yes

[lean sail]

well, actually we dont know that they are classes yet

i dont think we know that the big set S can be partitioned yet

that's more or less what we need to show?

[dull ginkgo]

That’s a pretty smart observation

We are forming an equivalence relation.

To partition the set into equivalence classes yes, that is the goal

[dull ginkgo]

So, would you like a small hint?

[lean sail]

you have given me a lot of hints, i need to start putting this together

i do appreciate it though

[dull ginkgo]

I think it would be more insightful to drop the whole x^5 = e thing

And instead let it be x^p = e

For prime p

Because that’s ultimately what’s behind it all.

[lean sail]

i understand, you want to generalize to p

[dull ginkgo]

The key is Bezout

Toy around with when x^n = y for some n in [1, p - 1], which is coprime to p after all,

[lean sail]

@dull ginkgo i am struggling with the entire thing, i dont see why we even declare the sets: ${x, x^2, x^3, x^4}$ and ${y, y^2, y^3, y^4}$

[cloud walrus]

proofman

[dull ginkgo]

I tried to avoid equivalence relations for your sake but you seem to be comfortable with them

Would you like that approach

[lean sail]

because all these elements when taken to the 5th power, are equal to the identity?

[lean sail]

i do understand that we need an equivalence relation to partition a set S, so i was wondering what the equivalence relation was that we were dealing with here, so far i haven't seen anything like $g \sim h$ or something

[dull ginkgo]

I’m sorry I lead you all over the place on this, let me provide an alternative and likely more intuitive sketch using equivalence relations:

Let G be any group, let p be a prime. Define S = {x : x^p = e ≠ x}. Let x ~ y on S iff x^n = y for some n. That is to say, x ~ y if and only if y is some power of x

Show this is an equivalence relation on S (reflexive, symmetric, and transitive), and show that the equivalence classes are all the same size of p - 1.

Then conclude if G is finite, so is S, and thus S has a cardinality divisible by p - 1

[cloud walrus]

proofman

[lean sail]

it's all good, not your responsibility to solve it for me

[dull ginkgo]

Some people struggle with equivalence relations a lot tbh

Might be easier for ya

Symmetry is the only non-trivial part, but it is due to p being prime and Bezout lemma.

[lean sail]

i didnt read the bottom part of the message, i was just going to ask, does this mean that we just show refl, symm, trans?

[dull ginkgo]

Yep!

[lean sail]

if we do that, then i think there is a theorem that says equivalence relations partition S

[dull ginkgo]

Reflexivity is obvious, transitivity is just substitution… but symmetry we need Bezout to flip it around. We have that the a above cannot be divisible by p

[dull ginkgo]

Partitions ARE an equivalence relation

We say x ~ y if they lie in the same partition set

And the equivalence classes are obviously the partition sets

And if we have an equivalence relation, the equivalence classes form a partition since they can’t overlap

[lean sail]

not to try and split hairs, but from what i understand, a partition is a set of disjoint sets whose union is all of S

[dull ginkgo]

Yes,

In a sense partitions correspond to equivalence relations

[lean sail]

shouldn't this be $x^n = e$?

[cloud walrus]

proofman

[dull ginkgo]

It is y = x^n for some n

That is to say, xRy if and only if y is some power of x,

But we have that x and y are both not the identity

[lean sail]

ah, yeah i am mixing the general case and the particular problem i am trying to solve

[dull ginkgo]

Updated it

[lean sail]

so like for instance, y is related to y^2 since (y)^5 = (y^2)^5 = e?

[dull ginkgo]

Yes!

[lean sail]

so then, y and y^2 belong to the same equivalence class

[dull ginkgo]

Yeppers

y is equivalent to its non-identity powers

Which are powers not divisible by 5

Of which there are 4, since the powers are equal mod p = 5

[lean sail]

right, in the case p = 5

[dull ginkgo]

Yes

The root of the problem is that if we are working with elements who have order p for some prime p, and are NOT the identity, then one being the power of another implies the converse too. That is x is a power of y iff y is a power of x

so that forms an equivalence relation that partitions the set we are trying to describe the order of, and there are only p - 1 non-identity powers of any element in that set

And in general, a good rule of thumb is, if you’re partitioning a set, there’s an equivalence relation involved, and vice versa. Because fundamentally they are two ways of describing the same thing

[dull ginkgo]

This is gonna be once again another ridiculous question

Let’s say we have a free module M over commutative R. We can then consider the tensor algebra T(M), which is a graded ring extending R.

Now, as a graded ring, we have two classical examples of homogenous degree 2 ideals, the exterior ideal generated by elements of the form x^2, and the symmetric ideal generated by elements of the form xy - yx

The quotient rings preserve the tensor powers due to these ideals being disjoint from R, but they also preserve the freeness of these tensor powers

So what if we had some “noncommutative polynomial” in R, that is an element of R[F_n] where F_n is the free monoid in n letters, and considered all “tensor combinations” of these (or image of all the valuations sending the letters to elements of V), I wonder when it also preserves freeness

So like if we had axyx - by or smth and considered the generated ideal of elements of that form, so like when we assume x or y is in V

But it seems x^2 and xy - yx are special in that they admit mappings to the original tensor algebra that you can prove freeness though, so it might just be a quirk of those two being free-polynomials over the integers

[woven plover]

quick question, in any ring with more than one element, 1 and 0 are distinct right? i would try to prove it but my brain has been staring at latex for the past 5 hours and i cant find much on google :p

unital ring* of course

[dull ginkgo]

1x = x for each x, assume 0 = 1, then 1x = x = 0x = 0 for each x

[woven plover]

thanks you put my mind at ease 🙏

[cobalt heath]

The adjoint representation SL(2,C) -> Aut(sl(2,c)) gives rise to a Lie group homomorphism SL(2, C) -> SL(3, C). Denote it by A \mapsto \hat{A}.
Is there a way to compute determinant of \sum_i \hat{A_i}, given information on A_i?

[lean sail]

im getting stuck on the symmetry part. i need to show that if $x^n = y$, then $y^n = x$. i noticed that if $x$ is an element of $S$, then $x^p = e \land x \neq e$ and if $y$ is an element of $S$, then $y^p = e \land y \neq e$. but how do i get to $x^n = y$ and $y^n = x$? im not sure where to go from here...

[cloud walrus]

proofman

[chilly ocean]

That is false if you are using the same n. For example in Z/5Z we have 2*1=2 but 2*2 is not 1

[lean sail]

can you expand a little?

[dull ginkgo]

It’s for some n

[chilly ocean]

What you want to show is that if x,y not identity and there exists some n such that x^n=y then there also exists some k such that y^k=x

[dull ginkgo]

I.e there exists an n such that x^n = y. You want to show there exists an m such that y^m = x

As bequi said

We have the criterion that neither are the identity so for the n where x^n = y that 5 does not divide n, i.e 5 is coprime to n

What can you use then?

[lean sail]

@chilly ocean @dull ginkgo are you both in agreement about what needs to be done? im not sure which messages to be following. i am getting really confused.

[dull ginkgo]

We made the same point yeah

Bequi sent the message before I was done typing lol

[lean sail]

ok, i just wanted to be sure before i started trying to interpret what you were both saying

[lean sail]

i think you are saying we should use bezout's identity, as you mentioned yesterday? but before we get to that, why must it be that $x^n = y$ and $y^m = x$, why not raise both to the $n$? i thought that $n$ was fixed

[cloud walrus]

proofman

[dull ginkgo]

n is not fixed

I said explicitly that n is some number

[glad osprey]

p is fixed, but n must be arbitrary, because we want both y^2 and y^3 to be in the same class as y (if p=5)

[lean sail]

im just trying to think about how this would play out by example

[dull ginkgo]

We have shown you

x^n = y where both x and y are not the identity implies n is not divisible by 5, so has an inverse (m) mod 5. Therefore y^m = x^(nm) = x^(1) = x

[lean sail]

both x and y have an inverse (m) mod 5? im not really sure what that means

[dull ginkgo]

I said n does

[lean sail]

okay, so n times a number m gives you 1 (modulo 5)

[dull ginkgo]

Yes

so y^m = x^nm

And if a = b (mod 5) then x^a = x^b

So x^nm = x^1 = x

[lean sail]

did you just prove the symmetry part?
y^m = x^(nm) = x^(1) = x

[dull ginkgo]

yes.

[lean sail]

ok, i thought that we had to use bezout's identity for this, i was thinking about how we could use that

[dull ginkgo]

Bezout identity is the same thing

Bezout identity literally proves that there’s an inverse mod 5

by Bezout, there exists u, v :
un + 5v = 1, taking this mod 5
un = 1 (mod 5)

So that proves the existence of the inverse

[lean sail]

im not really seeing how this ties in. i understand that n must have an inverse which we call m, and that n * m = 1 (modulo 5), but specifically what i dont understand is, where does the linear combination aspect come in? like, where in the symmetry proof are we saying that there is some linear combination of a GCD?

[dull ginkgo]

are you actually reading my messages before posting.

Linear combination here is to PROVE THERE IS AN INVERSE MOD 5

The GCD is 1

So mn + 5u = 1

Take that mod 5

You have an inverse mod 5

[lean sail]

yes, reading and re-reading, i've probably read each of your messages at least 3 times, it doesnt mean i really understand what you are saying

[dull ginkgo]

Bezout is used to prove that if n is coprime to m, then n has an inverse mod m

Because the one term in the linear combination will vanish modulo m because it’s a multiple of m

Leaving the one scaled term equaling the GCD (1) modulo m

[lean sail]

ok, i mean i've never really heard someone say that before, when i think of bezout, i just think of am + bn = x

[dull ginkgo]

Yes

What happens when you take it modulo n?

[lean sail]

the bn gets dropped

because its like adding a multiple to am

[dull ginkgo]

It is equivalent to the statement that
For integers n, m, there exists an a such that
an = gcd(n,m) mod m

[lean sail]

let me be more clear, i understand the individual things you are saying to at least some extent, but i am struggling to put it all together

[dull ginkgo]

No worries:

It gives us explicitly a way to say if x^n = y ≠ e

then y^m = x for some m

Brb

[glad osprey]

I think if you knew about cyclic subgroups this would be much clearer. We're basically showing that

• if x has order p, then so do x^k for every 1 < k < p
• cyclic subgroups of prime order are either the same or disjoint except at the identity

[lean sail]

using this textbook: subgroups/order are chapter 3, cyclic groups are chapter 4. i am at the very end of chapter 2: groups

[glad osprey]

you haven't learnt about order either?

[lean sail]

no, i just know it exists because i read ahead a little tiny bit

the tools i have currently are: proof techniques, propositional logic, equivalence relations, functions, some elementary number theory, some basics about groups

i also took all the calcs, diff eq, linear algebra, stats, discrete, and proof writing

[shell pilot]

So far I have that $\sigma^i:k\rightarrow(k+i)\text{mod m}$ where k is an element in the m-cycle
$\sigma^i : k\rightarrow k+i, \sigma^i : k+i\rightarrow k+2i, ..., \sigma^i : k+(m-1)i\rightarrow k$ So there are m mappings that are distinct if and only if each k + ai is distinct

[cloud walrus]

Soap_Opera

[glad osprey]

I really dislike when books give exercises before teaching the abstractions that makes them easy to solve. Like, the whole point is to learn the tools that allow you solve problems like this!

[lean sail]

i feel the same way. i dont want to give up on this problem just yet. i am trying to figure it out (with serious help). and at least attempt to follow the spirit of the book.

[glad osprey]

Like when Axler teaches eigenvalues before teaching determinants

[lean sail]

but it is starting to get really frustrating

[lean sail]

yeah, you need to know about characteristic polynomial and all that first, imo

[glad osprey]

So, to show symmetry, you start with $x^n = y$. You know that $n$ and $p$ are coprime, so by Bezout there exist $a, b$ such that $an + bp = 1$. Then $$x = x^1 = x^{an + bp} = (x^n)^a = y^a$$

[cloud walrus]

sheddow

[glad osprey]

This is what Miz was trying to show you. He started with $x^n = y$, then just raise both sides by $m$, ie. the multiplicative inverse of $n$ mod p, so you get $x^nm = y^m$ and $x = y^m$

[cloud walrus]

sheddow

[lean sail]

this is more or less what @dull ginkgo was saying over and over and i was not getting it. laid out like this it is a little easier to follow

but @dull ginkgo was giving hints and trying not to just give it away

[glad osprey]

Sorry if I spoiled it! I assumed you just wanted the solution at this point

[lean sail]

no you didnt, i just realized the way i typed that sounded like i was implying you did

[lean sail]

also, i still have to show transitive yet

[glad osprey]

That should be easier, give it a try and let us know how it goes

[dull ginkgo]

Transitive is easier,

[glad osprey]

This is maybe not the intended solution, but you could prove the more general theorem that if x has order m then x^i also has order m if and only if gcd(m, i) = 1

[cloud lynx]

if we have the field extension Q(sqrt(2),sqrt(3)) of Q(sqrt(2)) then the degree of the field extension is 2 right because both minimal polynomial of sqrt(2) and sqrt(2) have the same degree 2

but what if we have instead of sqrt(3) third root of 3?

then one degree is 2 and the other is 3

so the minimal polynomials havent the same degree. Is the degree of the field extension then 3?

[delicate orchid]

yeah Q(cbrt(3), sqrt(2)) is degree 3 over Q(sqrt(2))

it's not too hard to see that x^3-3 is a minimal polynomial for cbrt(3) over Q(sqrt(2))

[cloud lynx]

so its always the degree of the minimal polynomial which has the highest degree?

or is it too simplyfied

and sqrt 2 has no minimal polynomial in Q(sqrt(2)) right? because X^2-2=(X-sqrt(2))(X+sqrt(2))

[chilly ocean]

It has minimal polynomial: x-sqrt(2) which is of degree 1

[chilly ocean]

I'm not sure what you mean by this.

[cloud lynx]

but isnt the minimal polynomial unique?

[chilly ocean]

It is.

X^2-2 is not irreducible over Q(sqrt(2)) therefore it is not a minimal polynomial

[cloud lynx]

yes i know

i thought X+sqrt(2) would be also a minimal polynomial xD

thank you:)

[chilly ocean]

Let p(x) = x+sqrt(2). What is p(sqrt(2))?

[cloud lynx]

i know:(( ewiofjüwepf

[chilly ocean]

I see

[cloud lynx]

i try to explain it

my english is very bad sry

[chilly ocean]

That's okay

You get better with practice

[cloud lynx]

so Q(sqrt(2))=K

so the degree of Q(sqrt(2),sqrt(3)) over K is 2

the degree of Q(sqrt(2),third root(3)) over K is 3

the degree of Q(sqrt(2),fourth root(3)) over K is 4

and so on

[chilly ocean]

I think that is probably true

[cloud lynx]

and i thought the degree of this kind of field extension is always the degree of the minimal polynomial with the highest degree

you know what i mean?

[chilly ocean]

When you say this kind of field extension you mean Q(sqrt(2), n-th root of 3) over Q(sqrt(2))?

Or do you mean of Q(a,b) over Q(a) for any a,b and are talking about the degrees of the minimal polynomials over Q?

[chilly ocean]

Because this one is false in general

[cloud lynx]

and can we write [Q(sqrt(2)),sqrt(3)):Q(sqrt(2)]=[Q(sqrt(2)):Q(sqrt(2))]*[Q(sqrt(3):Q(sqrt(2)]

[cloud lynx]

more like Q(m-th root of 2, n-th root of 3) over Q(m-th root of 2) where as m<=n

[chilly ocean]

Did you mean to say [Q(sqrt(2),sqrt(3)):Q(sqrt(2))]=[Q(sqrt(2)):Q]*[Q(sqrt(3)):Q]?

[cloud lynx]

would it be right?

no I meant Q(sqrt(2))

[chilly ocean]

I am pretty sure that this is true but I don't know how to prove it. I think it is not easy to prove that

[hidden wind]

this book is already proving great complementary reading to the course in commutative algebra i’m taking

i love john stillwell’s writing

[cloud lynx]

with induction maybe?

[chilly ocean]

I don't know

Also, something to be aware of

Q(sqrt(2), quartic root of 2) : Q has degree 4 and not 8

[cloud lynx]

is it right this way?

[cloud lynx]

ah ok ty^^

[chilly ocean]

no, but it is if I replace the Q(sqrt(2)) in the left side with Q

[cloud lynx]

then mine is right?

yeah it must be right

because i can look at Q(sqrt(2),sqrt(3))/Q(sqrt(2))/Q(sqrt(2))

[chilly ocean]

Well, what you've written is right as it expands to 2 = 1*2

[cloud lynx]

oh brah but it doesnt help ufff

okk tyyy sry for being lost lol

[tardy hedge]

that is really nice. what book is this

[hidden wind]

stillwell's algebraic number theory for beginners

[tardy hedge]

Thanks, im taking alg number theory next year so this looks good

[hidden wind]

it assumes very little specific background aside from some mathematical maturity, and its presentation is the most successfully coupling of the history of the subject together with the actual mathematics i've ever seen

[shell pilot]

If $\sigma$ is an element of a symmetric group such that $\sigma=\alpha_1\alpha_2...\alpha_m$ where $\alpha_i$ are disjoint and the order of $\sigma$ is $a$. Then we can write $\sigma^a=\alpha_1^a\alpha_2^a...\alpha_m^a=e$. Why does it follow that each $\alpha_i^a=e$?

[wraith cargo]

I believe it's because they're all disjoint

[delicate orchid]

I'm presuming you mean \alpha_i^a = e

[shell pilot]

Yes, thank you, I'll edit it

[cloud walrus]

Soap_Opera

[delicate orchid]

no worries, but yeah it's because they're disjoint. It's pretty easy to prove that the support of a permutation is equal to the support of it's inverse

I'll leave the rest of the argument as an exercise for the reader...

[shell pilot]

Ok, so because they are disjoint, it means that no element is moved twice in the entire composition

Therefore it must be moved only once per \alpha_i^a hence alpha_i^a must be e

Well, not moved*

[delicate orchid]

that's the right idea

[shell pilot]

If an element was in multiple \alpha_i then it would not be disjoint cycles

[delicate orchid]

sigma(x) = alpha_i(x) where alpha_i is the unique alpha with x in its support

sigma^a(x) = alpha_i^a(x) = x for all x in the support of alpha_i, so alpha_i^a = e

[barren sierra]

for c.

Should this say "show that the x_1 - a_i are relatively prime" ?

because say $d = 3$ then $p_1 = (x_1 - a_2)(x_1 - x_3)$ and $p_2 = (x_1 - a_1)(x_1 - a_3)$ which are clearly not relatively prime.

[cloud walrus]

Spamakin🎷

[quiet pelican]

It’s not pairwise relatively prime, it’s relatively prime as a set I believe

[barren sierra]

ahhh

[balmy compass]

how do you do something like, given a matrix with char poly f, what are its possible invariant factor decompositions?

just by using that the min poly needs all roots as factors and then the divisibility condition to exhaust cases?

oh and they have to multiply to f

is there a systematic way

[glad osprey]

When talking about the division algorithm for polynomials, ie. f(x) = g(x) q(x) + r(x), the textbook writes that the remainder r(x) either has degree < deg(g), or r(x) = 0. Is this because the zero polynomial has undefined degree?

I kinda thought deg(0) = 0

[void cosmos]

no its undefined

so yeah ig

[barren sierra]

it's not undefined

[void cosmos]

-inf whatever

its conventional

[barren sierra]

you have degree of 0 = -inf

[barren sierra]

well that would break multiplicative properties of degree

[void cosmos]

u should check ur book @glad osprey

as it is conventional

sometime sit is just undefined

in dummit and foote its as spamakan says

[barren sierra]

degree(f(x) * g(x)) = degree(f(x)) + degree(g(x))

that's a property we want

if degree(0) = 0

[glad osprey]

Right, that makes sense

[barren sierra]

then we don't get that property

[glad osprey]

I see

[lapis latch]

if you view degree as a euclidean function then it would not include 0

[barren sierra]

of course also what I wrote kind of assumes we are working over a ring with no zero divisors

[lapis latch]

as per defn of a euclidean function

[glad osprey]

but if we define deg(0) = -inf, then they technically don't need to specify that either r(x) = 0 or deg(r) < deg(g), right?

[barren sierra]

uh

I guess yea

[glad osprey]

Aha, Fraleigh says deg(0) is undefined

that explains it

[barren sierra]

Fraleigh is scared of -inf it seems

[glad osprey]

cowards

[barren sierra]

(although the text I just checked now also leaves it undefined)

[vagrant zinc]

Guys a question in this proposition shouldn't k divide n ?

[south patrol]

No, it's correct

Though remember that "n divides k" means "k is divisible by n"

[vagrant zinc]

I have group Z_18

[south patrol]

Well for intuition note that if you have the trivial group then n = 1 and e^k = e for all k

\mathbb Z_{18}

[vagrant zinc]

|Z_{18}|=18

Ah in this case it is true, because the generator has the same order as the group therefore the order of group n divides the order k

[peak mirage]

R is the ring of nxn upper triangular matrices. what are the R-submodules of R^n?

not sure how to do

[dull ginkgo]

I still haven’t solved a problem from Jacobson and I feel like an idiot lmao

[swift prawn]

I think I had this as an exam question

I guess one good idea is to just start multiplying vectors and seeing what you get

[white oxide]

sorry to interrupt but here do they mean VS under addition

[swift prawn]

Like under function addition?

I’m sorry maybe I don’t know what you mean by under addition

[white oxide]

yea

[swift prawn]

Then yes I think so

If you have two functionals you can add them together and then scale them by constants in F

[white oxide]

ye makes sense

thanks

[vagrant zinc]

$f$ and $g$ is bounded $|f|<N$ and $|g|<M\to |f+g|<|f|+|g|<N+M \to |f+g|<K$

[cloud walrus]

Homology

[vagrant zinc]

If f is continuos => f is bounded

$\mathcal{B}$ is space normed and metric

[cloud walrus]

Homology

[hidden wind]

have good video: https://youtu.be/gHTUhhom9Qc?si=mNoIpzesa8agwq2H

[rain grove]

What is an example of an Ideal that is not a subring?

[grave sedge]

Any ideal basically

[hidden wind]

take for example (3) in Z

it is not closed under multiplication by elements of Z

[grave sedge]

Ok it depends on your definition of subring

[quiet pelican]

Any ideal that isn’t the whole ring won’t contain 1 and therefore won’t be a sub ring
If you don’t need your rings to have 1, then every ideal is a sub ring

[hidden wind]

oop i’m trippinh

[grave sedge]

Not really, you would still need to have an element which works as the identity

[hidden wind]

[grave sedge]

Ok no this is not what i mean

I mean you might not require subrings to have the larger ring's identity

[rain grove]

aha yeah, i got $I \triangleleft K$ and I has an invertible element $\Rightarrow I = K$

[grave sedge]

But still have an identity

[cloud walrus]

OHHELLNAH

[grave sedge]

(which is the case when you have a product of rings RxS, where Rx{0} is an ideal and also a ring with unity but not with the same unity as the larger ring)

[rain grove]

so 3Z is just lacking the identity?

[hidden wind]

i have no idea what prompted me to this example and explanation, both are plainly wrong

[rain grove]

yeah but 3Z is still an example of an ideal that is not a subring

[hidden wind]

mhm it definitely does not have a multiplicative identity, but i don’t trust myself on many more claims rn

[rain grove]

haha its okk and ty for example

[rocky cloak]

This is a very strange thing to do though.

If you require rings to have identity, then it seems identity is part of the structure of a ring. But subrings should inherit their ring structure from their superring.

Anyway, an ideal will be a "subring" in this sense, exactly if it is generated by a central idempotent.

Which happens exactly if your ring is on the form RxS

[warped socket]

Would I be correct in stating that $\mathbb{R}[x]/(x^2 + 1)$ is isomorphic to $\mathbb{R}[x]/(x^2 + x + 1)$, due to both of them being field extensions of degree 2 over $\mathbb{R}$?

[cloud walrus]

ab

[quiet pelican]

Yes
Although the reasoning isn't quite right, or at least isn't how I'd phrase it
The important part there is that the algebraic closure of \bR is of degree 2 over \bR

[warped socket]

ah so moreso using the idea that they're both algebraic closures of R (which in turn is just C)?

[quiet pelican]

Yes

(or at least that one of them is, and they're of the same degree)

[rain grove]

What is an example of a finite group that does not have a subgroup of order n, where n | |G|?

[grave sedge]

A4 and n=6

Iirc its the smallest

Yeah it has to be the smallest because n cannot be a prime power so has to be at least 6 so |G| is at least twice as much

[rain grove]

damn I see, nice

[peak mirage]

sorry, do you remember how you did it? totally blanking

i imagine we want to show there's only some trivial submodules like R^k x {0}^{n-k} or something but not sure how

[dull ginkgo]

Damn that seems like a tricky question

But the ring R of upper triangular matrices is free as a module over the ring it is a matrix ring of, no?

[swift prawn]

I think that's right

I don't really remember how I showed it

I probably took some generic vector of that form (like just fill the entries with * or something) and multiplied it by a generic upper triangular matrix and you can check these are submoduules

but I don't remember how I checked they are the only submodules

[south patrol]

Yes but here it is like different

[south patrol]

Is there a typo here?

I have a feeling it should be this: let R be the ring of n x n upper triangular matrices over a ring A. Then A^n is naturally an R module. What are the R-submodules of A^n?

It's not clear to me how R should act on R^n

Except in the trivial way lol

[peak mirage]

i was thinking just like M(M1, ..., Mn) = (MM1, ..., MMn) but your way makes sense actually

[peak mirage]

in this case how would you proceed?

[rain grove]

Correspondence theorem from my book (just for groups):

Let $N \triangleleft G$: Every subgroup of $G/N$ is of the form $H/N$ for some $H \leq G$ where $N \subseteq H$\

proof: canonic epimorphism: $\pi: G \rightarrow G/N$. \$(\star)$ $H:=\pi^{-1}(H')$ is a subgroup of $G$. In $H$ there are such elements $h\in G$, that the coset $\pi(h) = hN$ is a subset of $H'$. $N\subset H$ because N is the identity of the group H'. Because $\pi$ is surjective: $\pi(\pi^{-1}(H')) = H'$. Written differently: $H' = \pi(H) = H/N$

$(\star)$: $\phi:G\rightarrow G'$ homomorphism of groups: if $H'\leq G'$ then $\phi^{-1}(H')\leq G$

I don't understand the last part where $\pi(\pi^{-1}(H')) = H'$, don't you need $\pi$ to be a bijection for that?

[cloud walrus]

OHHELLNAH

OHHELLNAH

[knotty badger]

$\pi(\pi^{-1}(H)) = H$ works whenever $\pi$ is surjective, actually

[cloud walrus]

Pseudonium

[knotty badger]

Conversely $\pi^{-1}(\pi(H)) = H$ works whenever $\pi$ is injective

[cloud walrus]

Pseudonium

[knotty badger]

Though in this case I don’t think $\pi$ would be injective

[cloud walrus]

Pseudonium

[rain grove]

ohh fk I knew it something like that but couldn't find it online, yeah i tried injectivity and its not

[rain grove]

Why is $d\mathbb{Z}_n := {dx | x\in \mathbb{Z}_n}$ a subgroup of $\mathbb{Z}_n$?

[cloud walrus]

OHHELLNAH

[wraith cargo]

well you usually just check the properties needed for a subset to be a subgroup

[rain grove]

Idk im not sure how

[lean sail]

@glad osprey @dull ginkgo i think this proof might also suffice for what i was working on the other day. FYI still working through the equivalence relation version of the proof. some things came up and i havent had too much time.

[rain grove]

Oh wait so,

unit: 0d = 0
closure: dx+dy = d(x+y)
invertible el: for x = {1...n-1} then dx+d(n - x) = 0
is that good?

[wraith cargo]

yeah

[knotty badger]

It’s also the image of a group hom Z_n -> Z_n

The one which multiplies everything by d

[glad osprey]

What does it mean for x^2 to be distinct?

[hidden cairn]

i proved this by guessing the generators of each cyclic group but is there a better way to know for which n (Z/nZ)^x is cyclic?

[chilly ocean]

There is a characterization of exactly which (Z/nZ)^x are cyclic

But it's not trivial

For just proving for those 3, guessing the generators is probably the easiest way

[chilly ocean]

Namely iff ||n=1, 2, 4 or n=p^k for some odd prime or n=2p^k for some odd prime p||

[hidden cairn]

hmm i see

what does the proof use?

if it's not too advanced i'll search it up

[glad osprey]

Those are the cases where ||the Carmichael function equals the Euler phi function||, right?

[dull ginkgo]

While you’re at it you should try to prove (Z/2^nZ)^x ~= (Z/2Z) x (Z/2^(n-2)Z)

[hidden cairn]

alright ill try my hand at this too

[dull ginkgo]

Actually first is the sub problem:

• what is the order of 3 multiplicatively mod 2^n

The fact that the successor to 2 is prime, and is in fact the ONLY prime with this property, is why the pattern Bequi stated breaks down for n = 2

[chilly ocean]

Yes

[dull ginkgo]

That’s an equivalence pretty much by definition lol

[south patrol]

Well this is essentially true by definition

The Carmichael is the max order of an element and the Euler phi function is the number of elements

[south patrol]

It is funny lol this came up in some topology for me

Very random

[dull ginkgo]

I always wondered why 2 broke down in this pattern and realized ||binomial theorem|| and 2+1 being prime and being the only prime where p + 1 is also prime is why lol

[south patrol]

Oh interesting

Hm why does that make it break down though

[dull ginkgo]

2p^n is cyclic for each prime p EXCEPT 2

[south patrol]

Isn't that equivalent though lol

I don't see how that makes it break down

[dull ginkgo]

I mean the multiplicative group

[south patrol]

Ye I know

[south patrol]

I mean like why do those considerations mean (Z/2^n Z)^x shouldn't be cyclic

Or am I misunderstanding u

[dull ginkgo]

I wrote something small up for it in the past let me find it

[south patrol]

Oh wait no im dumb ye lol

Well it's not the same

It's more like you're saying why 2p^k is cyclic for odd p but not like e.g. 3p^k for p not 3 right

And then it is just Chinese remainder and what you said

[dull ginkgo]

Oh I guess that is a bit easier due to multiplicative group mod 2 being trivial lmao

[south patrol]

Lol

But also like if you have n= p^k q^l then (Z/n)^x = (Z/p^k)^x x (Z/q^l) but p^k(p-1) and q^l(q-1) aren't coprime if p, q > 2 since both are even, so this isn't cyclic

[dull ginkgo]

A n y way

[south patrol]

UwU

[dull ginkgo]

Back to finally doing Jacobson because I skipped that zero divisor determinant problem

It’s too hard

[south patrol]

Oh which one

Injective iff det is not a zero divisor?

[dull ginkgo]

Also realized Leibniz formula is basically writing out a large determinant as a determinant of a matrix power

[dull ginkgo]

Yes

[south patrol]

Wdym

Hm

[hidden cairn]

ok so for this i've done these 2 exercises before so using those results, we have two disjoint cyclic subgroups $<5>$ and $<2^{n-1} + 1>$ and their product's order gives $2^n$, so the result follows.

[cloud walrus]

pink_panther

[hidden cairn]

exercises i mention ^

[dull ginkgo]

Exterior power of an exterior power, determinants are functorial :3

[hidden cairn]

is this a valid proof?

[south patrol]

Hm I thought Leibniz was basically just expanding out the exterior power

[south patrol]

I wouldn't say disjoint but yeah

Two subgroups with trivial intersection

[dull ginkgo]

In a fucked up way I’m pretty sure it’s writing out the determinant of the induced linear map on the second exterior power lo

[south patrol]

What is the induced map? Idk

Maybe I'm silly billy

[dull ginkgo]

think about it

[south patrol]

Well so you have a map V -> W and take top powers and it's nultiplication by the det

What now

[dull ginkgo]

f(x \wedge y) = f(x) \wedge \f(y)

So apply that to the basis vecs

[south patrol]

I know how the exterior algebra works dw

Lol

This is a way to define determinant ue

But why the like exterior power again

Like idk what you meant by that

[dull ginkgo]

Eh I realized actually I think it’s up to sign

No wait

I meant Laplace :c

I got the formulae mixed up, Leibniz is the multilinear form one

[dull ginkgo]

Wait I just had a brain wave

Take matrix A. Assume rdet(A) = 0. I wonder if you can consider the largest sub matrix that doesn’t have a vanishing determinant, and try to use the Laplace formulae to find a vector that will “compute” the determinant to force it to vanish and find a vector in the kernel

I just need to think about how to describe it more clearly

[nimble folio]

Has anyone read Atiyah-Macdonald's book on Commutative Algebra?

It's surprisingly short!

[hidden wind]

i am planning to for this course i just started!

[dull ginkgo]

I am just not well versed enough with determinants to actually be able to solve this damn problem dammit

oh ok

HOW THE FUCK IS THIS ELEMENTARY

that is not obvious at all wtf

[delicate orchid]

elementary doesn't mean obvious

[dull ginkgo]

well i spent literal days on this problem to no avail

which is frankly pathetic but whatever

The only other way I can think of is showing that the exterior powers are free for a free module but that also isn't easy to prove

and in retrospect i shouldn't skip this problem because it's fundamental

and if I do then I definitely shouldn't be doing this textbook lol

[lean sail]

What textbook?

[dull ginkgo]

Jacobson

Basic Algebra I

[dull ginkgo]

Is there an analogue of a PID where we have a module such that every submodule is cyclic

But ISN’T necessarily simple

[rain grove]

why rings have maximal ideals but groups don't have something equivalent, at least it was not mentioned in my book, like a "maximal normal group"

[surreal dagger]

There is such a notion, idk much about group theory, but for example for the Jordan Hölder theorem you care about those

[chilly ocean]

They do! And there's a theorem for groups similar to the one for rings. Namely if I is an ideal of R, R/I is simple iff I is maximal (the simple commutative rings are the fields). Likewise, if H is a normal subgroup of a group G, G/H is simple iff H is maximal.

[rocky cloak]

Fun fact: even though all rings have maximal ideals, not all groups have maximal normal subgroups.

Finitely generated groups do though.

For example Q (under addition) don't have any maximal subgroups

[chilly ocean]

Another fun fact is that if you don't require your rings to have identity, then there are rings without maximal ideals. Like the polynomials over Q with no constant term.

[dull ginkgo]

To add to this: the relation of a subgroup being normal in another is not transitive so you cannot use Zorn’s lemma to assert the existence of maximal ones

[grave sedge]

This is not the issue

Since you can take the poset of normal subgroup ordered wrt inclusion

Which is indeed a poset

The problem is that the union of a chain of normal proper subgroups is a normal subgroup but not necessarily a proper one (i.e. can be the whole group)

With ideals you have that because an ideal is proper iff it does not contain 1

And for finitely generated groups the same thing works (a subgroup is proper iff it does not contain all the finitely many generators)

And the "not containing all the elements of some specific set" being true for the sets in a chain implies it being true for the union relies crucially on the finiteness of the specific set

[lean sail]

i am going back to this. for the relation to be symmetric, shouldn't: $x = y^n$ instead of $x = y^a$?

[cloud walrus]

proofman

[chilly ocean]

The relation isn't that x~y iff x=y^n

The relation is that x~y iff there is some n such that x=y^n

So the n such that x=y^n does not have to be the same as the one such that y=x^n

[dull ginkgo]

I have to be frank with you man, I would not proceed with posting about this problem here unless you have thoroughly reread everything we have posted for you

Because we’ve explicitly mentioned this same thing 3 times

I am trying to give you benefit of the doubt here but please reread what we have already said

[lean sail]

i forgot about the "there exists", that makes sense. thank you.

[lean sail]

this is kind of saying that two elements belong to the same equivalence class?

[chilly ocean]

Yes, if ~ is an equivalence relation, then x and y are in the same ~-equivalence class iff x~y

[dull ginkgo]

Can a lot of the properties of the “property chain” to fields be generalized to modules like Noetherian-Artinian

That is like

PID for rings is that every ideal is principal

I.e what if we have a module such that every submodule is cyclic

Or like ascending/descending principal chain condition which I call principal noetherian/artinian on how ascending or descending cyclic modules stabilize

Or GCD domains where instead we have a module where the binary intersection of cyclic modules is cyclic

Like all of these seem to generalize, is there any use to that?

Actually all of this is boring nvm

cyclic modules are isomorphic to R/J where J is a left ideal

[rain grove]

Hmm I see, I think my book mentions them in a later chapter though idk why not at least introduce them now.

[chilly ocean]

I guess they are not as important to the study of groups as they are to the study of rings

[chilly ocean]

So it's all directly equivalent to properties of the underlying ring?

[dull ginkgo]

Yes afaik

[lean sail]

i think that this should work for the transitive part of the proof. please, anyone feel free to comment. thank you.

for context:

[dull ginkgo]

:3 yep!

Ergo what can you conclude

[lean sail]

i want to say that the subsets that create the partition have $p - 1$ elements in each, but i think i am missing something

[dull ginkgo]

you aren't lol

there's p-1 powers of x

that aren't the identity

and they comprise the equivalence classes

[lean sail]

this i think is intuitive, and then you know each equivalence class "looks the same" - but what proof is there that there are exactly $p - 1$ elements in each one of those classes?

[cloud walrus]

proofman

[dull ginkgo]

[x] = {x^n \neq e , n in N}

but x^n = e iff n = 5m for some m

and x^(pa + b) = x^b

so therefore [x] = {x ... x^(p-1)}

[lean sail]

I will think about this, I think you are getting at, if you go over p, you start repeating

[dull ginkgo]

yeppers

[lean sail]

I guess what I am getting at is, it seems intuitive, but there is nothing in the proof where you see an explicit $p - 1$ “show up”… I am just wondering if there is a way to do that.

[cloud walrus]

proofman

[dull ginkgo]

Can you be a bit specific on what you are trying to prove

I can try to walk you through it

[lean sail]

I just want to say: “therefore, S contains exactly p - 1 elements”

[dull ginkgo]

okie dokie

Trying to show the set {x...x^(p-1)} is the equivalence class or that it has p-1 elements?

[lean sail]

Just show that S has p - 1 elements… I am thinking it is possible to do this by showing that all elements in S are distinct… can the fact that they are coprime to p be used here?

[dull ginkgo]

yep

assume they aren't, then x^a = x^b where a and b are both in 1... p-1

[lean sail]

Sorry by “it” I meant S

[dull ginkgo]

then wlog a > b, so x^(a - b) = e, and a - b is in 1... p-1

thus (a - b) | p. But that can only occur if a - b = 1, thus a = b + 1, so x^(a-b) = x = e, which is a contradiction, thus all the x ... x^(p-1) are distinct

[glad osprey]

S doesn't have p-1 elements, but its equivalence classes does. We're partitioning in S into equivalence classes consisting of p-1 elements (the partitions are cyclic subgroups of order 5 minus the identity). One part of the proof is proving that each equivalence class actually has p-1 distinct elements, another part is proving that the partition actually works (ie. the subgroups are either the same or overlap only at the identity)

[rain grove]

Just making sure but $A_n$ is always a normal subgroup of $S_n$, because it contains all elements of $S_n$ with even signature and $sA_ns^{-1}$, will also have a even signature $(sgn(s) * 1 * sgn(s^{-1}) = 1)$. Can this also be said for any subgroup of $S_n$ that has all elements with even signature?

[cloud walrus]

OHHELLNAH

[delicate orchid]

no

why would it

[rain grove]

Well any such product sAs^{-1} has even signature, now its just the question if it is also contained in A. And I thought maybe those s and s^{-1} cancel eachother so we get something thats also in A

[rain grove]

Whats a counterexample?

[delicate orchid]

{1, (12)(34), (13)(24), (14)(23)} in S_5, conjugate by (15)

[delicate orchid]

yup. It's in A_n

why does it have to be in whatever subgroup of A_n you've picked

[rain grove]

yeah it doesn't, idk I just maybe thought it could be

[rain grove]

Let $X$ be a nonempty subset of the group $G$. Show that the subgroup generated by the set ${gxg^{-1} \mid g \in G, x \in X}$ is the same as the normal subgroup generated by the set $X$ (that is, the normal subgroup containing $ X$ and is contained in every other normal subgroup containing $X$)

[cloud walrus]

OHHELLNAH

[rain grove]

What am i supposed to do in this problem? To me it feels like that set {gxg^{-1} \mid g \in G, x \in X} is the definition of a normal subgroup generated by the set X

[coral spindle]

That is unfortunately not correct

It is more complicated

It won’t even typically be a subgroup

You should just try to prove that any normal subgroup containing X should contain that set, to get one inclusion. As for the opposite, I think you should give it a shot alone when you’ve seen how the first part works

[wild jasper]

Algebra lectures start on monday 😄

[civic cave]

in the context of basic group stuff

is that like

im not sure what |ab| means here

[chilly ocean]

Probably means the order of ab

[civic cave]

order

?

[chilly ocean]

Yes

[civic cave]

i was thinking that but im not sure

hm ur probably right cuz u cant show ab = ba

so

yeah

oh wait that makes a lot of sense, thanks

[dull ginkgo]

Hey chat, for these exercises, do the standard row echelon stuff work BUT you can’t scale a row correct, just remove scaled rows (otherwise the operation isn’t invertible)

[grave sedge]

Yes

[chilly ocean]

You can scale them by -1

[quiet pelican]

(Or, more generally, by any unit)

[dull ginkgo]

I forgot to state that lol

At what point can you determine that you’ve obtained a basis

For instance, the generated module of [0,3] and [0,4] has base [0,1]

Due to Bezout

[grave sedge]

You have a generating set because you're doing row operations

And the set is independent because it's in row echelon form

So it's a basis

[rain grove]

idk how to prove the other inclusion

The first one is easy, every normal subgroup containing X also contains gXg^{-1}. And because its a subgroup and its closed under the operation it contains the subgroup that is generated by that set.

Also... How can I be sure that there exists a normal subgroup that contains X. I mean for that to be true it must be gxg^{-1} in X for all x and for all g. But X was any non empty set of G...

[grave sedge]

You should consider G a normal subgroup of itself for this

[rain grove]

Ohh I think I understand... so if there is some n in this normal subgroup N that is not in X, then it must have been generated by some gxg^{-1}

[rain grove]

Let $G$ be a finite simple group and let $x \in G \setminus {1}$. Show that every element of $G$ is equal to the product of the elements that are conjugate to the element $x$

[cloud walrus]

OHHELLNAH

[rain grove]

I did this: Since G is simple, the smallest normal group containing {x} is the whole group G. Using the result from previous exercise, this means that G is generated by ${gxg^{-1} | g \in G}$

And I think im done here but im not sure... I did not use the finite proprety of G

[cloud walrus]

OHHELLNAH

[quiet pelican]

You don’t need finiteness of G I don’t think

[rain grove]

hmm ok weird

[dull ginkgo]

So for instance

The generated module of the vectors in matrix B is Z^3?

[cobalt heath]

Hmm is row operation fine for finding generating sets

[cobalt heath]

If both row and column operations are fine, this works

[dull ginkgo]

Still a bit confused

Jacobson’s intro is confusing tbh

Like how do we find a basis of the actual module, how do we say it’s a basis or even generates the same module in Z^3???

[dull ginkgo]

I don’t think so actually

I like don’t know if you can just flat out use row echelon here because idk if that would be vectors of a different basis and be different generators idfk

[chilly ocean]

hihi i need help with this problem

part 4 mainly. im trying it but like i can find something by inspection but i dont really know a more general way of doing the same

like from the first couple parts i realize the homomorphism needs to be in the form of phi( (a,b) ) = 3 a x0 + b 2 x1
but then trying to set the kernel to that specfically is hard

because then i need
(10n mod 12) 3 x0 + (6n mod 9) 2 x1 = 0 mod 18
and like thats a bit too many modulos to know what to do with

but i mean i can write a program to find x0 and x1s that work but idk a more theoretical way of doing it

[crystal vale]

How can I show that in a finite Abelian group, for any g in G, g^(ord G) = e. Without using coset stuffs?

Any hint? I am stuck at this problem

Since G is abelian so g->g^(ord G) is a group homomorphism

[tired gyro]

probably easiest to write out the kernel

<(10,6)> = {(0,0), (10,6), (8,3), (6,0), (4,6), (2,3)}

notice that the first component is zero mod 2 and the second component is zero mod 3

[oblique iron]

do you have Lagrange's theorem? That's usually proved using coset stuff though

[tired gyro]

where do you have to send (a,b) so that you get that result