groups-rings-fields

[golden turtle]

but I don't see why that is

[vast stump]

you can show each b can be represented as phi(a) + mu(c) (i.e. mu(c) is in B, and considering b - mu(c), applying phi yields c - c = 0, so b - mu(c) is in phi kernel, or psi image, so there is psi(a) = b - mu(c))
and this representation is unique (assume they’re not, i.e. psi(a) + mu(c) = psi(a’) + mu(c’))
and psi(A) intersect C’ is trivial
hence mu(C) direct sum psi(A) = B

[tardy hedge]

Why can we do this R quotient on both sides ?

This is about abelian groups can be embedded into a divisible group

I guess its just that, if R is a subgroup of F and F embeds into D, then R is a subgroup of D. So u can quotient by R on both and get that inclusion still there

Right?

[rocky cloak]

Yup

[tardy hedge]

Ty

[tardy hedge]

If a direct sum / product of modules is projective or injective, we must have every summand be proj/injective?

Iff?

[dull ginkgo]

Is this a binary sum/product?

[coral spindle]

Well a direct summand of a projective module must equally be a direct summand of a free module.

As for injective modules, you can actually do this pretty straightforwardly from the definition

You just need to project down onto one of the summands

[dull ginkgo]

Would the general argument just have to be broken down into products for injectives, coproducts/sums for projectives?

[coral spindle]

Seems likely, yeah

[dull ginkgo]

I might try to see if you can just extrapolate it immediately from the whole "factoring through" parts of both universal properties

[dull ginkgo]

well I guess it would be dual lol

I'll try drawing it out in a bit perchance

[rocky cloak]

A summand of a projective/injective is always projective/injective.

The direct sum of projective modules is always projective and the direct product of injective modules is always injective.

Fun fact, a ring is Noetherian iff direct sums of injectives are always injective.

[tardy hedge]

Thanks ppl

[rocky cloak]

Direct products of projectives being projective can also be characterized by the ring, but is more complicated (some kind of condition on finitely generated ideals and stuff I don't remember)

[dull ginkgo]

Oooooh, is that last fact by considering ideals as submodules of the ring as a module over itself?

and would a ring be Artinian iff direct products of projectives are projective?

[rocky cloak]

It's R being left perfect and every finitely generated right ideal being finitely presented.

For R commutative this is equivalent to being artinian.

So this result is more like a nice dual in that case

[rocky cloak]

True for commutative ring, but not in general

[dull ginkgo]

I definitely need to learn more module theory, most of this stuff is new to me

I only know the basic defs by osmosis

[jolly jewel]

Is anyone able to help me out on this?
On 1 attempt I picked d, it was wrong
On 1 attempt I picked c, it was wrong

[delicate bloom]

sounds like you're guessing, can you explain why

[rocky cloak]

Did you try picking both c and d?

[weary frost]

d) is true actually

so is c)

[jolly jewel]

That's what I'm thinking but for some reason I got 0/5 for that

[weary frost]

pick c) and d)

[jolly jewel]

c makes sense as well

[rocky cloak]

It's not unlikely that you get 5 points for correct and 0/5 for not correct

[jolly jewel]

What if a^2 has an odd order?

[weary frost]

uh

[rocky cloak]

Is anything different in that case?

[jolly jewel]

what if a has order of 3, what would a^2 be? and would it divide?

[rocky cloak]

3 divides 3

[jolly jewel]

ig it would be 3 as well

[weary frost]

the order of a^2 is still 3 yea

[jolly jewel]

I see, so c is true

d is true by laws of powers

[rustic rapids]

sorry to interrupt, but just want a quick sanity check for this one. It suffices to show the seq {A^k}N is dependent right? since M2x2 has dimension n^2, do i just pick N > n^2 and im done?

[weary frost]

I can also tell you b) is false because it lacks inverses

a) is false for many non-finite groups

[jolly jewel]

I absolutely agree, because a^-1 wouldn't be in it

[weary frost]

e) is false, take Z_2

[toxic zephyr]

if A is an abelian normal subgroup of G then conjugation is a group action of G/A on A. does this not work if A is not abelian? it seems like the map might not be well defined

[weary frost]

that makes sense, at some point they have to be linearly dependent

[wraith cargo]

You need A abelian here

Think of it like this

G has a natural action on A by conjugation

If A is abelian

Then multiplying elements of g by elements of a still gives the same action

So we're actually working on the level of cosets

Hence the action reduces to an action of the quotient

Basically choose coset a representative of gA for example, let's say ga
Then we have that (ga)b(ga)^-1 = g(aba^-1)g^-1 (now since a and b commute we have that) this is equal to gbg^-1

This is just to illustrate that the action is well defined

You should go and prove it more rigorously yourself

[opaque finch]

Yes. Apologies for missing this part.

[trail cave]

quick question, when my professor says two isomorphic groups have the same algebraic properties, he means that anything proven of one binary operation holds for the other, right?

[vagrant zinc]

Ye

[cloud walrus]

Homology

[trail cave]

ty

[tardy hedge]

On my module theory midterm, i dont know why i couldnt get this question:

R an integral domain. I is an ideal of R thats not a principal ideal. I is not free

I guess as an R module

[next obsidian]

If it’s free it has to have rank 1 so the isomorphism R -> I gives you a generator

You can see its rank 1 because when you take field of fractions you get an injection of Frac(I) -> Frac(R) and the latter is rank 1

So the rank of Frac(I) = 1, but if I is free of rank n, Frac(I) is rank n

[tardy hedge]

Hm

[dim widget]

here is a simpler argument

if a, b are two generators then they satisfy the relation ba - ab = 0

thus the module is not free

[tardy hedge]

Oh cool thanks

Tbf i havent thought about field of fractions in a while

[next obsidian]

Why do a and b have to be generators here? You could just grab x and y in I and then observe xy - yx = 0

[dim widget]

so that it's dispositive

[sweet pendant]

For (2) I have defined a linear transformation map $T_{a} := R \rightarrow R$ such that $r \rightarrow ra$. I have shown that this is indeed a linear transformation. Then I have also shown that $ker(T_{a})={0}$ so it is injective. Then I used the hint provided and currently I have $rank(T_{a}) = dim(R)$. What can I conclude from this to show that $R$ is a field?

[cloud walrus]

mss

[next obsidian]

You want x and y to be basis elements if I isn’t rank 1 right?

Or I guess a and b

[dim widget]

I want to assume there is some R^n \cong I and then take two basis vectors

yeah

[next obsidian]

Right, but all I was pointing out is it seems like you were saying I isn’t principal so you take a,b to be generators as an ideal

[dim widget]

just part of a set of generators is all I mean

two random elements of a free module can have relations but two generators ought not to

[next obsidian]

Well I disagree still

Generator isn’t a basis elements

[chilly radish]

They'll in particular be a part of a set of generators

[next obsidian]

Z is generated by 1 and 2

[dim widget]

okay but this is just quibbling over terminology

[next obsidian]

It just gets confusing when you’re dealing with an ideal and there’s already a notion of generators there

[dim widget]

it's the same notion as module generators

[next obsidian]

And in particular if you’re taking a non-principal one and taking two generators

[dim widget]

so it's not that confusing

but I'm taking two elements of a minimal set of generators

[next obsidian]

Like it’s not clear that you’re actually saying they are independent coming from some isomorphism with R^n

[dim widget]

is I guess the key part

[chilly radish]

You can actually phrase this as a direct proof if you phrase it as proving any map R^n->I is not an isomorphism for n>1

Well not "direct" in the constructivist sense but still

[dim widget]

Yeah perhaps I should've framed it as there is no injection from R^2 \to R

[next obsidian]

Anyway I also framed my proof the way I did because it generalizes to show you can’t have larger rank free modules inside of smaller rank ones

The fact this proof works when the target is R is sort of an accident I think

[dim widget]

yeah both arguments are fine.

[proven mango]

If I want to show that there is no isomorphism between two groups, specifically the dihedral group D4 and Quarternion group Q8, I know that we can do so by finding a group property that they don't share.

Currently, I am kind of stuck since I checked the orders which is 8 and both groups are not abelian.

[dim widget]

have you thought about counting the number of elements of a fixed order in each?

eg how many elements of order 2, how many of order 4, etc.

[velvet steeple]

My text only mentioned the centralizer of G being a normal subgroup but that's not because this is not the case, right?

[dim widget]

if you take S=G it's just G

which is, to be fair, a normal subgroup

[velvet steeple]

..only if we assume G is a normal subgroup, no?

[dim widget]

subgroup of what?

all you've mentioned is G

[velvet steeple]

Ok what I said doesn't make sense here

[dim widget]

ok

agreed

[velvet steeple]

So basically what you're saying is that G is a normal subgroup of G

Right?

[dim widget]

yep

[velvet steeple]

Yeah that's trivially true

Alright

Thanks!

[velvet steeple]

When referring to the multiplicative group of a field, why do we need to say that we only consider nonzero elements?

The determinant is a homomorphism of the multiplicative group of square matrices to the multiplicative group of a field

It can take on zero, so how come we still have that requirement?

[weary frost]

0 has no multiplicative inverse, wouldn't form a group

[velvet steeple]

Oh true..

[weary frost]

btw determinant 0 matrices have no inverse either lol

so those aren't in a multiplicative group of matrices

[velvet steeple]

Oh, so we get rid of them with that requirement

Thanks!

[proven mango]

Yeah I found the number of elements of each order to be not the same

[hidden cairn]

what are x bar and h bar referring to here?

[nimble folio]

Havn't had much luck in the linear algebra channel

Let $A = \left(\begin{matrix} 5 & 1 & 0 & 0 \ 0 & 5 & 0 & 0 \ 0 & 0 & 4 & 0 \ 0 & 0 & 0 & 4 \end{matrix}\right)$\hspace{1pt}. I can easily compute that $c_A(x) = (x-5)^2(x-4)^2$. However, since $A$ is in Jordan canonical form, can I by inspection say that $m_A(x) = (x-5)^2(x-4)$?

[cloud walrus]

clubsoda14

[warm ember]

yes

becasue the eigenvalues

[hidden cairn]

i think yes because minimal polynomial has the same roots as the characteristic polynomial but is the smallest degree polynomial st p(T) = 0

there's a better explanation of why the powers in the minimal polynomial are taken as the dimension of the greatest jordan block for each eigenvalue

but i dont remember it off the top of my head

[trail cave]

In these questions (I believe Z is (Z,+)) how do I determine the amount of onto homomorphisms in 1 and ensure my map is a homomorphism in 3? I know I can map 1^n to some element b^n to determine a homomorphism since these are cyclic

[nimble folio]

i wasnt sure if it was biconditional

[hidden cairn]

it is iirc

[nimble folio]

good enough for me

ty

[hidden cairn]

you might wanna check FIS linear algebra on jordan canonical forms

[nimble folio]

what is that?

[hidden cairn]

which is ch 7 iirc

[hidden cairn]

linear algebra book written by friedberg et al

[nimble folio]

Found a pdf and cant control f

WHATEVER

i take it as fact

[vagrant zinc]

$\varphi(x)=4x \mod (6)$

[cloud walrus]

Homology

[leaden mantle]

hai

anyone really keen on group theory
tryna prove this isomorphism without using a function which all the answers use
(R,+) and (R+, x)
the answers just use e^x
which yeah like works
but it feels dirty to use lol

[trail cave]

can you explain please

[vagrant zinc]

Test if it meets the homomorphism condition

[trail cave]

I believe it does

[charred iris]

I'm pretty sure the condition for a map to be a homomorphism practically forces you to have an exponential

other maps probably exist, but they're going to be very not nice

[nimble folio]

e^x is the "simplest" one, see https://math.stackexchange.com/questions/302257/is-ex-the-only-isomorphism-between-the-groups-mathbbr-and-mathbb

[charred iris]

(similarly to how a homomorphism (R,+) to (R,+) is basically going to be cx unless you go really horrid and use the axiom of choice)

[leaden mantle]

can you show a homomorphism exists without saying what it is?

like a more rudimentary proof

[leaden mantle]

what is cx btw

[leaden mantle]

right,,,

[charred iris]

I meant the map that takes x to c*x, I should have clarified

[leaden mantle]

oh okay

thank you

[trail cave]

Would anyone help with 4/5 here with z being the only commutative nonidentity element in 4? I understand the geometric intuition but I don't know how to prove this

[hidden cairn]

for 4, it is enough to show z commutes with r and s. for the uniqueness, assume another r^t commutes with all other elements and show t = k

s can not be a candidate for a second commutative nonidentity element as it does not commute with r to begin with

[lime junco]

recall D2n is generated by two element, r and f. it is easy to show f does not commute, based on the laws for r^if = f r^2k - i. For an element of the form r^i, you can easly show the only one that satisfies this is r^k, by the previous equality. Now you need only to consider elemnents of the form r^i s. But if we pick this element, we get that r^i s r =/= r r^i s, so we conclude that r^k is the only element which satisfies this

[trail cave]

Could you clarify the equality r^if = f r^2k - i ? I'm having trouble parsing it

[trail cave]

My trouble is with showing z commutes with r and s. z obviously commutes with r but why with s?

oh wait, I've got it

Thank you guys

[lime junco]

this defines D^2n

r is rotation, f is flip

rf = fr^n - 1 in D2n, basically if i flip then rotate

then its the same as i rotate n - 1 times and then flip

try drawing it out with a triangle

[trail cave]

Yes I understand now

thank you

[dreamy edge]

Are there any good resources for getting a basic level understanding of the groups of lie type in the classification of finite groups? Mainly I just want to know what they look like in terms of the multiplication tables and/or cycle graphs for specific examples.

[vapid vale]

the multiplication table will not be illuminating at all

[dreamy edge]

Ok

Something else then

[vapid vale]

you should know a bit about the theory of reductive groups, but i dont really know a good resource for it offhand

this is not elementary though – i think you really need some basic algebraic geometry first

[dreamy edge]

I appreciate the effort, but surely there exists a simplified overview for someone who understands basic group theory, I don't need to have a deep understanding of it.

Or any understanding for that matter

[vapid vale]

there really doesnt

perhaps someone has some rather low level explanation that gives an intuition if you're familiar with matrix groups and the like, but im not aware of one

the groups are generally extremely large

[dreamy edge]

There is a way, it just depends on how much you want to 'lie' in the explanation.

Well thanks anyways

[vapid vale]

i think some classes of them may be explainable with the simplicity you desire but i dont know that those expositions currently exist

[dreamy edge]

On a side note, what would be some good books after "abstract algebra: theory and applications"?

[vapid vale]

i think the best way to get a feel for it is find a professor that knows the topic well and is willing to chat to you about it

[vapid vale]

what did you like most in that book

[dreamy edge]

I don't really have anything specific, everything was generally explained in a way that was easy to understand which I appreciated, it's also the first math book I've actually read for personal interest not just a class.

I just want to study more abstract stuff but dont know what because theres just so much I dont know

[vapid vale]

if that book is a general overview of abstract algebra, there is no specific next step to take. galois theory, commutative algebra, representation theory, are all interesting topics

[bitter meteor]

is it enough to disprove that two groups are not isomorphic because one group is cyclic and the other isnt?

[vapid vale]

you can find a lot of these things in a graduate level abstract algebra text

[vapid vale]

yes

[bitter meteor]

cause isomorphisms preserve cyclic properties right

ok awesome ^_^

[dreamy edge]

Is there anything you would recommend? What topic is your favourite?

[vapid vale]

i like representation theory - i think steinberg's representation theory of finite groups is interesting and a pretty easy read

[dreamy edge]

Thanks, I ll look into it.

[static mauve]

Im working out a question where it asks:

"How many of the 19,683 possible binary operations on a set of 3 elements give a group structure?"

I know that a binary operation on a group structure yields all of the elements of the set for any given row/column of table that has all of the operators outputs...

I came up with 2 when working out the possibilities, and I think I can swap any given rows and columns and that makes a "new one." With my method, that works out to like (3!)^2, but the answer in the back of the book says 3.

Trying to understand where Im going wrong.

[bitter rover]

How much abstract algebra are you allowed to use?

[tough raven]

It's not true that you can independently apply any permutation to the rows and any permutation to the columns, because (for example) the left and right identities are equal in a group, and if you apply different permutations to the rows and to the columns you might break that.

Moreover, not all pairs of permutations of the rows and columns give rise to different multiplication tables. Here's an example in your case: for any correct multiplication table for three elements, there should exist two elements such that if you swap those two rows and those two columns, the multiplication table is unchanged.

[bitter rover]

There's one group of order 3 up to isomorphism. It's cyclic and isomorphic to {0, 1, 2} with addition mod 3.

If you have a set {a,b,c} then you can send any of those to 1, and that determines a group structure. So there are 3 ways you can map {a,b,c} to a group of order 3.

Basically, once you decide which element is playing the role of 1, all of the other elements' roles are determined completely.

[static mauve]

Ah ok, that actually clears up just about everything! Thanks for your thoughts 🙂

[somber sleet]

Can $\mathbb{R}$ be seen as a $\mathbb{C}$ vector space?

[cloud walrus]

damn_guuurl

[somber sleet]

If yes, is there a way to prove it without the continuum hypothesis?

[serene dune]

scalar multiplication can take a vector outside of the space

[wraith nexus]

he means some operation

R x C to R

that agrees with vector space axioms

not the obvious one from complex mult

[serene dune]

then im unaware, sorry

[sharp sonnet]

not with the addition from R

i dont see how the continuum hypothesis would be related to this

you can take your favorite bijection from C to R and move all the operations along i guess?

[somber sleet]

That is the problem, how do you prove that $(\mathbb{R},+) \cong (\mathbb{C},+)$, how do you find this bijection?

[cloud walrus]

damn_guuurl

[somber sleet]

I saw that one can find this bijection by seeing both R and C as Q vector spaces and with the help of the continuum hypothesis these are isomorphic

[sharp sonnet]

thats the axiom of choice

[somber sleet]

fair

[sharp sonnet]

and that works, yes

you need choice to get a basis for every vector space

[somber sleet]

then is there an easier way, where one does not need the axiom of choice?

[sharp sonnet]

R and C have the same cardinality, so they are the same as Q vector spaces

this way you can move addition along the bijection

and Q-linear scalar mulitplication

[somber sleet]

yes exactly, this is what I found too

[sharp sonnet]

then C-linear could probably follow?

actually hmm

you can show that this bijection cant be continuous

so it probably breaks here as well

🤷

[golden turtle]

I've been following along the hint pretty easily up until, deduce that beta(b)=psi'(a') for some a' in A'

[somber sleet]

mmh I need to find a solution, I'm really curious if there is some way

[golden turtle]

I'm not sure how to show that is the case, but I assume that it uses alpha being surjective since of the maps involved that's the only information that we have

[somber sleet]

How far did you come?

[golden turtle]

So far I am up until that step completely

I think

[rocky cloak]

So alpha being surjective means that you can find a such that alpha(a) = a'

[somber sleet]

I mean, $\varphi$ is surjective, which means that if $c \in ker \gamma = {x \in C \mid \gamma(x)=0}$, then there must be a $b$ by surjectivity such that $\varphi(b)= c$

[cloud walrus]

damn_guuurl

[serene dune]

a map from C \to R
a +ib \maps to b

[golden turtle]

yes I have this already

[rocky cloak]

Or I guess you mean you're struggling with deducing this step? Use exactness

[golden turtle]

Yes deducing that step

Im(psi)=ker(phi)

Im(psi')=ker(phi')

[somber sleet]

oh sorry then

[golden turtle]

Oh I guess if phi'(beta(b))=0 then beta(b) is in the kernel of phi'

and thus must be in the image of psi'

ty

[somber sleet]

yes of course!

[sharp sonnet]

what i outlined actually just works

take your favorite bijection (any works) f: R -> C

define addition in R via x + y : f^(-1)(f(x) + f(y)) and multiplication via zx := f^(-1)(zf(x))

[golden turtle]

So the next part then, I can easily see that there is an a in A with alpha(a)=a' since alpha is surjective

[sharp sonnet]

the rest should go through, but exercise for the reader

[golden turtle]

but I'm not sure how to show that there is an a satisfying that and beta(psi(a))=beta(b)

[rocky cloak]

From looking around it seems that all sets being measurable implies that R and C are not isomorphic. So some choice is needed

[rocky cloak]

It's not really "there is an a", it's just "a satisfies this". You've already chosen a and b

[golden turtle]

so by surjectivity of alpha, there exists a in A with alpha(a)=a' and I then need to show using that speciifc a that the other things also hold

and that just comes from the map being commutative I think

for all x in A, psi'(alpha(x))=beta(psi(x))

wow

I think I got it

[golden turtle]

So Prop36 says that a module Q is injective if and only if for all left ideals I of R and any R module homomorphism g:I to Q can be exteneded to an R module homomorphism G:R->Q. Also, if R is a PID then Q is injective <=> rQ=Q for all r!=0 in R.

So when I'm looking at this, I'm thinking that Z_m is a PID and so it suffices to show that each Zd and Zm term in the sequence is injective by showing that it satisfies rQ=Q for all r in R

But I'm not sure how to do that, and I'm not entirely sure that I'm applying the theorem in the right way

[rocky cloak]

Z/m is not a PID, because it's not a domain.

But you can think about in which ways an ideal I can map to Z/m

[rocky cloak]

Another approach:
If R is any ring, then Hom_Ab(R, Q/Z) is an injective R-module. This is somewhat easy to see using Hom-tensor adjunction and that I is injective iff Hom( -, I) is exact.

Then you can note that Hom(Z/m, Q/Z) = Z/m

[hallow bough]

Hello! I need help with an exercise. I need to calculate the number of fields $L$ such that $\mathbb{Q} \subset L \subset \mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{3}, \sqrt[3]{5})$. For now, I'm trying to determine the Galois group for $ \mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{3}, \sqrt[3]{5}, \omega) / \mathbb{Q} $, where $ \omega $ is a primitive cubic root of unity. I believe I can then determine the subgroup related to $ \mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{3}, \sqrt[3]{5}) $. At the moment, I'm confused about the Galois group of the splitting field mentioned above. Is there another way to think about solving this exercise?

[cloud walrus]

gcp00_13776

[void cosmos]

why are u adding the cubic root of unity

[rocky cloak]

To make it Galois

[void cosmos]

oh yeah right

mb

[rocky cloak]

So you can look at the Galois group of
Q(cb2, cb3, cb5, w) / Q(w)
which isn't too hard to see should be (C3)^3

And Q(w)/Q has Galois group C2 so it will be a semidirect product of those.

[opaque finch]

I think your description provides the best possible way using Galois correspondence.

[golden turtle]

I’m trying to show this is an injective resolution using the blue prop

But I get stuck trying to show the homomorphism extends

Not sure what to do from here

[weary frost]

you basically are trying to prove injectivity of the modules?

[rocky cloak]

So an ideal in Z/m looks like (n) for some n that divides m.

So a homomorphism (n) -> Z/m is just given by maping n somewhere. What are the possible homomorphisms?

[tulip scaffold]

Hi, how do I prove that x^4n + x^n + 1 is irreducible in the finite field F_2, where n = 3^k for some positive integer k?

I am confused how to do it

Help would be greatly appreciated

[rocky cloak]

There might be a shorter argument, but:
||You can show that x^4 + x + 1 is irreducible just by brute force||

||Let z be a root of x^4 + x + 1, then F2(z) is a field of order 2^4 = 16, so the multiplicative group is C15. The minimal polynomial of z has degree 4, so the order of z is at least 4, so must be 5 or 15. It's not 5 since z^5 = z^2 + z. So it's 15.||

||That means z can't have a cube root, so x^3 - z is irreducible.||

||Calling a root of x^3 - z for z1, we have that z1 is a root of x^12 + x^3 + 1 and that the extension F2(z1) = F2(z)(z1) has degree 4*3 = 12. So this must be the minimal polynomial of z1, hence irreducible.||

Picking up a nifty lemma:
||Now let us consider a field F of order m such that 3^k divides m - 1 but 3^k+1 doesn't. Consider y an element that doesn't have a cube root. Then the multiplicative order of y is a multiple of 3^k. Consider y' a root of x^3 - y. Then the multiplicative order of y' is 3*o(y). Now F(y') has order m^3.||

||m^3 - 1 = (m - 1)(m^2 + m + 1). Since m is 1 mod 3, m^2 + m + 1 is a multiple of 3. But since x^2 + x + 1 has no solutions mod 9, it's not a multiple of 9. This means that 3^k+1 divides m^3 - 1, but 3^k+2 doesn't. Since y' has order a multiple of 3^k+1, it can't have a cube root.||

Turning back to the original problem. ||Let z2 be a root of x^3 - z1. By the above argument x^4*3^2 + x^3^2 + 1 is the minimal polynomial of z2. Continuing by induction x^4n + x^n + 1 is irreducible for all n = 3^k||

[bitter rover]

That's intense haha

[velvet steeple]

The semidirect product preserves the law of composition of the group, right?

Lang writes the following:

(thanks)

Once, we're considering A and N as multiplicative, and once with law of composition being the composition of mappings

For G = AN = NA we need the law of composition to be composition of mappings

Or am I mistaken somewhere

[next obsidian]

Sure, but under the given maps composition of T_a,0 and T_a’,0 is T_aa’

And for the other, T_1,b and T_1,b’ is T_1,b+b’

So up to these identifications it’s the same

[velvet steeple]

True, yeah

[velvet steeple]

But still, really we should consider A and N to have a law of composition exactly like you stated, right? Even if in the end the result is the same

[next obsidian]

Also, idk if you’re French or lang just uses that phrase, but “law of composition” isn’t a thing anyone really says in English, we just call it the group operation

[velvet steeple]

Ah

Yeah Lang uses it

[next obsidian]

I mean it doesn’t really matter. You’re implicitly working through an isomorphism. It’s like when you take a standard open in a manifold M, technically that open U consists of points in M, but for proofs you just say it’s R^n (or like, the unit ball in R^n)

Like if M is S^2 or whatever, technically points in U are points in R^3 with norm 1

But when you do proofs you just say x and y are coordinates and prove stuff using calculus in R^2 and then apply it to an open in S^2

[velvet steeple]

Alright, thanks. So it's really a 'simplification'

[next obsidian]

I guess so, idk if I would say it’s a simplification more than a tacit identification

You know two things are isomorphic and are identifying the “actual thing” with the isomorphic thing

[velvet steeple]

Yeah that's better

[next obsidian]

Because it’s easier

[velvet steeple]

Yep

Thank you!

[next obsidian]

[velvet steeple]

Out of $H$ being a subgroup of $K$ being a subgroup of $G$ with $H$ normal in $K$ implies $K \subseteq N_H$, the normalizer of $H$, it directly follows that the normalizer of $H$ is the largest subgroup of $G$ in which $H$ is normal, right?

Seems like the answer is no

[fast coral]

why not?

[next obsidian]

The normalizer of H in G is basically defined to be the largest subgroup H is normal in

[velvet steeple]

Because it seems proofwiki takes a whole lot more to prove this

[velvet steeple]

I define $N_S \coloneqq {x \in G \mid xSx^{-1} = S}$ as the normalizer of $S \subseteq G$

Is this it?

[velvet steeple]

From this, the last statement obviously follows

[fast coral]

if you’re given “H normal in K implies K is subgroup of normalizer” then it pretty much directly follows doesn’t it

[velvet steeple]

Yeah that's what I'm thinking

But

[velvet steeple]

So my proof wouldn't be enough?

[fast coral]

ah okay. they showed H is normal in its normalizer, yes sure that would/could be necessary

[fast coral]

that combined with this ^ would complete the proof for the normalizer being the largest […]

[velvet steeple]

Obviously H is normal in its normalizer. Take any element x in the normalizer. Then by definition xH = Hx

QED...?

[fast coral]

yes pretty much

[velvet steeple]

This whole proof is like one sentence

[fast coral]

it is fairly trivial you are correct

[velvet steeple]

Proofwiki still takes much longer, perhaps just because of different definitions?

[fast coral]

proofwiki is a bit interesting yes, sometimes their proofs are a bit longer or shorter than one might expect

[velvet steeple]

But yeah, someone on MSE also answered this

Their definition is a bit different than what we used but well

It's also a line

[golden turtle]

(n) onto anything in Zm ?

[velvet steeple]

Thank you!

[fast coral]

welcome 👍

[golden turtle]

I've shown that it is an injective resolution

Now I want to compute the Ext_Z_m^(n)(A, Z_d)

Yes I had this but I’m

This is the definition we had in class

I'm wondering if when they say, "in terms of the dual group Hom_Z_m(A, Z_m)"

if it means the same thing as "derived from the functor Hom_R(_,D) or if I'm potentially not looking at the right definition for this part of problem?

[graceful dome]

a) (1)(2 3)(4 5 6)(7 8 9 10).......
b) (1 2 3 4 5 6 7 8 9 10 11......)
c) (1 3 5 7 11 13......)(2)(4 6)(8 9 10)(12 14 15 16).....

are these correct?

d) (1 2 2^2 2^3 2^4....)(3 3^2 3^3 3^4 3^5....)....(p p^2 p^3 p^4 p^5...)....

[toxic sapphire]

slightly silly question but for an unspecified group in Dummit and Foote, is it assumed to have addition as a group operation or multiplication?

I'm struggling to find an answer in the text

here is the exercise I'm doing

[next obsidian]

Your question is not really well posed

The convention is arbitrary, you can denote the operation any way you want

That being said, for normal groups you will usually call it multiplication and say g^-1 and call the identity 1

Consistent with the notion of “multiplication” we’re used to working with

[toxic sapphire]

oh right of course

[next obsidian]

And then with abelian groups you often switch to + and 0

To be more consistent with Z as an abelian group

[toxic sapphire]

if there's only one operation it doesn't matter what you call it

[next obsidian]

Yeh

[toxic sapphire]

thanks

[toxic sapphire]

hm is it arbitrary in this case also?

should I consider (1,1) to be the identity or (0,0)?

I guess it must be (1,1) since otherwise the identity doesn't appear

[next obsidian]

They wrote down 1 somewhere

In a group you don’t have a 0 and a 1

You just hav an identity and here they wrote a 1 somewhere so it’s clear they mean multiplicative notation

[leaden mantle]

from the fact that (0,1) (the interval) has the same cardinality of the reals, could i say that R+ and R has the same cardinality

it feels wrong

[next obsidian]

Yeah

Why does it feel wrong

You chopped infinity in 2, it’s the same infinity

[leaden mantle]

hmmm i guess so

what about like

R+ but take out 1

it would have the same cardinality right

as R

[kind temple]

yes

[leaden mantle]

hmmm

i see

thank you

[rocky cloak]

Well let's think of an example. Say m=6 and n=2.

Is there a homomorphism from 2Z/6 to Z/6 maping 2 to 1?

[golden turtle]

I ended up figuring this out I think earlier, ty for the reply, no there isn't

I went with

[hollow topaz]

hello everyone! why did the author write this proof, when he already proved that the left inverse is the right inverse?

[rocky cloak]

In 1.1.15 they start with a matrix that has a left inverse and a right inverse.

In 1.2.20 it has a left inverse or a right inverse.

So 1.2.20 is proving something more.

[hollow topaz]

oops, i somehow missed that. thank you! i'm still confused though, why did he need to introduce the elementary matrices before he could prove 1.2.20, when 1.1.15 followed from just the definitions?

[rocky cloak]

Well, 1.1.15 is actually true for any associative operation. Whereas 1.2.20 is more specific to square matrices

I don't know if that answers "why" for you, but to prove 1.2.20 you really need to use something about matrices

[hollow topaz]

ah thank you, i guess what im confused about is why 1.2.20 is specific to square matrices. is it possible for there to be a left inverse but not a right inverse?

[rocky cloak]

In general it is possible yes.

Let's for example think about functions from N to N under composition.

Consider f(x) = x+1, and g(x) = x-1 for x>0 and g(0) = 0.

Then gf is the identity, but there is no function h such that fh is the identity.

[hollow topaz]

youve really cleared it up for me, thanks!

[muted comet]

How do i formalize the notion of a division ring? specifically i have to talk about the "multiplicative" part of the ring. i tried this:

A ring $(R,+,\cdot)$ with identity 1. Is a division ring iff all the following are true:
$$\exists 1 \in (R,\cdot) \ \exists 0 \in (R,+)(0 \neq 1)$$
$$\forall x \in (R,\cdot) \ \exists x^{-1} \in (R,\cdot) (xx^{-1} = x^{-1}x = 1)$$

[cloud walrus]

Jonas!

[muted comet]

saying stuff like (R,+) doesnt seem to make much sense to me. but i cant think of something else rn

[grave sedge]

Does "(R{0},•) is a group" work for you?

[rocky cloak]

I mean, you can just say, for example, 1 is in R.

R is the set that contains the elements after all

[muted comet]

so you would change the first line to, for example: $$\exists 1 \in R\ \exists 0 \in R(0 \neq 1)?$$

[cloud walrus]

Jonas!

[rocky cloak]

Sure. And in the second line I would add the condition x =/= 0

[muted comet]

ok. But i dont see why you would add that. thats just what the condition on top says. these are formally an and statement?.

[rocky cloak]

Well as written now you're claiming that 0 should have an inverse

[muted comet]

oh i get it. thanks

[dull ginkgo]

Galois theory is boring

Did introduce me to crossed products of rings and groups

So I guess I’ll give it that

[tardy hedge]

Really?

[topaz heart]

that's a controversial opinion

[dull ginkgo]

Not much to it unless you look deeper at the topological side

[atomic forge]

tbh I’ve always thought that way as well, that a first course in galois theory by itself is not especially interesting, but it’s a super useful tool (and the topological/etale versions are very pretty and pleasing)

like I felt most courses overemphasised it (I guess practice is important, but most people who end up doing number theory or algebra can get that practice elsewhere, and people who go in other fields don’t really need it) while something like rep theory or artin-wedderburn theory has more content and is more interesting to me at more or less the same level

[tardy hedge]

I see

I am taking it next term lol

[dull ginkgo]

If [A : B] = N and is a Galois extension then End_A(B) ~= B cross Gal_A(B)

If b_n is the basis of B as an A module, then the maps chi_n(b_m) = \delta_{n,m} b_m

Should serve as another basis too of that module

[dull ginkgo]

Dedekind independence of characters asserts Gal_A(B) is linearly independent as characters over B so they are linearly independent in End_A(B) as a B vector space

Invariant Basis Number means its a basis of End_A(B)

[languid trellis]

Z, considered as a Z-module over itself seems to satisfy the ascending chain condition, but consider the set { (2), (3)}, the ideals generated by 2,3 respectively. Neither is a subset of the other as 2 and 3 are coprime, so how can this set have a maximal element by inclusion?

Are we to say that both are maximal elements? In which case the use of "a" in "a maximal element" makes little sense.

[rocky cloak]

They are indeed both maximal, which is way we say "a maximal element" as opposed to "one maximal element" or "a unique maximal element"

(definition of maximal is just that no other element is larger)

[languid trellis]

Okay my english evidently needs work

Thanks jagr

[dull ginkgo]

Now find an analogous one for Artinian (descending chain condition) modules :3

[orchid onyx]

Hey,
I've got a question on lattice reduction, that I already asked in the help channels, but I got a DM, telling me that this may be a more appropriate channel. :)

Suppose I have a base b1, in which I have the biggest (with the biggest norm) vector b1Max und the smallest (with the smallest norm) b1Min.
The ratio b1Max/b1Min is what I want to optimize with an reduction algorithm.
A high ratio would be bad and a ratio near one would be great. My goal is to have my basis kind of "symmetric" (ratio about 1) but I can't seem to find any existing algorithm doing this.
LLL and BKZ seem to be reducing the base with focus on shortening most of the vectors and making them more orthogonal, but they do not seem to focus on reducing this ratio..?
I found this ( https://eprint.iacr.org/2016/847) paper which I actually do not really understand yet. But reading the abstract suggests, that their algorithm does exactly, what I want.

My questions are:

• is it a common task to reduce the defined ratio of the biggest and smallest vector in the base?
• if so, this ratio will most likely have a name?
• are there any algorithms for reducing a base with focus of bringing this ratio closer to 1?

[languid trellis]

im not getting distracted mining for iron when i got diamonds here

[dull ginkgo]

bruh

[hidden wind]

you’re boring >>>:S

grrr

[south patrol]

Theory of inseparable extensions is interesting though

[uncut girder]

False

The reality is that you are boring

[uncut girder]

???

[dull ginkgo]

that’s the Jacobson part, exponent one inseparable extensions lol

[south patrol]

Only for exponent one yeah

There has been much more work though e.g. https://people.maths.ox.ac.uk/brantner/FundamentalTheoremOfPurelyInseparableGaloisTheory.pdf which is fancy

[tropic spade]

For the c=2 case here, how are they concluding one of a,b is even and the other is odd and what is the "quick check" that shows s=1, t=all that garbage?

Ignore my bad scribble in the middle

[dull ginkgo]

HIGHLIGHTER

[next obsidian]

Look at the equation below the line with (*)

[tropic spade]

i need color coding

Hence highlighter

[next obsidian]

When c = 2, if a and b are both even then a + bsqrt(-19) you can factor a 2 out of the numerator

[tropic spade]

Yeah the even case I get but why not both odd?

[next obsidian]

Yknow I was gonna type my argument after that

But it wasn’t right

[tropic spade]

It's okay lmao

I'm wondering if they're waving away a reduction mod 2 trick (like the c=3 part) or some garbage after that. Or just rewriting (a+bsqrt(-19))/2 in the case a,b are odd to rule it out.

[next obsidian]

Yeah idk

[tropic spade]

Actually I think I see why both can't be odd. I think it's just that if you have ((2k+1)+(2m+1)sqrt(-19))/2, you can rewrite it as an element of R with some algebra.

[next obsidian]

I mean so it comes down to whether or not

(1 + 1sqrt(-19))/2 is

And actually wait -19 is 1 mod 4

So I think you can always do this yeah

[tropic spade]

Okay I see what they mean by s and t. I thought these had to use the earlier equations but this is a totally separate case and what they claim is obvious by substitution.

[grizzled spindle]

how many generators of Sn can i have without having one of them being generated by the others

i know its at least n-1 from transpositions

[void cosmos]

you can have 2

generators

of S_n

[grizzled spindle]

i know but that wasnt my question

im looking for most generators with none of them being useless

[void cosmos]

(1j)

{(1j) | 1<j<=n}

generate S_n

[grizzled spindle]

yes but could there be a generating set with n elements

[void cosmos]

indeed (ij) = (1i)(1j)(1i)

[grizzled spindle]

and none of them being useless

[void cosmos]

how many elements are there

[grizzled spindle]

i knwo these standard generating sets

[grizzled spindle]

exactly n-1

[void cosmos]

ig to do this you would need to like

take one of ur n-1 set of generators

[grizzled spindle]

and check that if theyre all useful (none is obtained from the others), it generates S_n?

or for counter example, find one for which this is not true

maybe theres a more general statement that could solve this?

[void cosmos]

i dont think we can do that

yeah i tried it out for like small ns and its impossible but idk how to prove it

[rotund aurora]

Since there is no useless element, by forming certain products you will obtain many distinct elements. Maybe you can say something non-trivial if you do that

[void cosmos]

like over counting?

some kind of overcounting argument?

[grizzled spindle]

sounds like an angle

maybe doable

[vagrant zinc]

Any idea why they are not isomorphic ?

[cloud walrus]

Homology

[delicate bloom]

play around with trying to make an isomorphism between them and show us what you come up with

[graceful dome]

Shouldnt this be H/G and [H : G] for the left cosets of a right group action H on G

They even say so earlier in the book

[thorn jay]

An isomorphism of field extensions over Q must fix Q (why?) and therefore sqrt(2) and sqrt(3) must be roots of the same polynomials over Q (why?)

That naturally creates a contradiction as sqrt(2) does not satisfy the equation x^2 - 3 = 0

[rotund aurora]

@grizzled spindle There is a stupid argument that proves your integer is <=n log log n+O(n). If a_1,...,a_m is a minimal generating set of S_n, then <a_1> subset <a_1,a_2> subset... subset <a_1,...,a_m>=S_n where the inclusions are proper. If d_i denotes each index of the "tower" then d_1...d_m=n!, where d_i>1 (by your condition, although this is weaker I think). m is minimized when all the d_i are primes, but you can calculate the p-adic valuation of n!, which is <=n/(p-1). Then, m<=sum_p n/(p-1) where p ranges over primes p<=n. This is n log log n+O(n) (by Merten's theorem)

[grizzled spindle]

damn i dont have the background for this type of argument but that is a good bound

thx for taking the time to work something out

[rotund aurora]

I have no reason to believe this is true, but it would be nice if the actual number is <n

[grizzled spindle]

well if its less than n, it has to be n-1

[rotund aurora]

yeah

maybe a simpler problem is when all the a_i of the minimal generating set have the same order

[vagrant zinc]

Aaah mmm

[rotund aurora]

maybe n-1 is not actually the answer, since no set of <n-1 transpositions generates S_n. I don't know if every set of n-1 (distinct) transpositions generates S_n

well definitely not every such set, but I meant a set of transpositions with no "useless" elements. Maybe it is true idk

[grizzled spindle]

if you take the transpositions (12), (23), ..., (n-1 n), any removal will leave 2 consecutive numbers unable to interact with each other right

it will split the permutations over n elements into permutations over k elements and n-k elements

[grizzled spindle]

well the transpositions need to act transitively on n elements if they have no useless element i think

[rotund aurora]

ah yeah

[grizzled spindle]

and reordering the set {1, ..., n} should give you back the standard generators

[rotund aurora]

if that's true, then that's S_n

[grizzled spindle]

i think an induction argument works for this

[thorn jay]

I hope you can fill in the details from there

[nimble folio]

I'm having trouble showing $V_\mathbb{C} \cong \mathbb{C} \otimes_\mathbb{R} V$. Recall that $V_\mathbb{C} = V \oplus V$ is a $\mathbb{C}$-vector space with multiplication defined as:

\begin{equation*}
(a+bi)(v_1,v_2) = (av_1-bv_2,av_2+bv_1).
\end{equation*}

I want to solve this using the following universal property: let $\iota_V :V \rightarrow \mathbb{C} \otimes_\mathbb{R} V$ be defined by $v \mapsto 1 \otimes v$; given $t \in Hom_\mathbb{R} (V,V \oplus V)$, there exists a unique $T \in Hom_\mathbb{C}(\mathbb{C} \otimes_\mathbb{R} V, V_\mathbb{C})$ so that $t = T \circ \iota_V$. My goal is to find an inverse for T.

I'm confused about two things. 1) How is $T$ even defined in the universal property? 2) How does complex multiplication behave the same in these two spaces? Clearly:

\begin{equation*}
z_1(z_2 \otimes v) = (z_1 z_2) \otimes v,
\end{equation*}

but I don't see how this relates at all to:

\begin{equation*}
(a+bi)(v_1,v_2) = (av_1-bv_2,av_2+bv_1).
\end{equation*}

[cloud walrus]

clubsoda14

[vagrant zinc]

$\mathbb{C}\cong \mathbb{R}^{2}$ \
$(x,y)\cdot (u,w)=(xu-yw,xw+yu)$

[cloud walrus]

Homology

[nimble folio]

Right

That is the first equation in what I wrote up

[vagrant zinc]

From which bibliography are you studying ?

[rocky cloak]

One thing you can notice is that C(x)V = 1(x)V (+) i(x)V.

Then you can try to multiply (a+bi) times 1(x)v1 + i(x)v2 and see what you get

[hollow topaz]

hello! what does the author mean when he says the set of n by n matrices forms a "space with dimension n^2"?

[rocky cloak]

The set of nxn matrices is a vector space, and it's dimension is n^2.

The dimension of a vector space is the size of a basis for that space.

[hollow topaz]

is it fine if i dont understand that? lol

[rocky cloak]

Idk, depends what you're doing I guess.

Have you not done any linear algebra before this?

[hollow topaz]

nope. should i? im at chapter 1, and its about matrices though.

[rocky cloak]

Hard for me to say. I have no idea what you're reading, or why you're reading it.

Like it's common to have linear algebra as a prerequisit for abstract algebra, but you can make a book without this prerequisit.

I also don't know if the fact that R^nxn is a vector space becomes relevant in the next paragraph or not

[hollow topaz]

oh, makes sense. thanks!

[heady shuttle]

If $G$ is the quotient group of $\operatorname{PSL}(2, \mathbf Z)$, find the number of supgroups of $G$ of index $m$ for $m \in {1, 2, 3, 4, 5, 6}$

[cloud walrus]

imo2025

[heady shuttle]

For 2 to 6 I got ||1, 1, 1, 0, 1||

Not sure it's right

[wild jasper]

Show that if $\varphi: \mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$ then $\varphi$ is forced to fix $\mathbb{Q}$ and $\varphi(\sqrt{2})=\sqrt{3}$, and then show that this is a contradiction.

[cloud walrus]

mh_le

[dense root]

anyone able to explain this to me please

[wild jasper]

what is the question?

[dense root]

no question, just the concept

on dividing groups/rings

[wild jasper]

you are not "dividing" group or rings, but forming quotient groups and rings

[dense root]

yeah that

quotient groups

how do they work or form

[wild jasper]

Z/2Z is the same as Z_2, so what is that?

[wild jasper]

And note that Z_p={0,1,…,p-1}

[delicate bloom]

I don't agree that phi(sqrt(2)) must be sent to sqrt(3), but maybe you have a different argument in mind than I do

||I just see that phi(sqrt(2)) = a+b sqrt(3) for some rational a, b. Then squaring both sides gives us 2 = a^2+3b^2 + 2ab sqrt(3) so you can now solve for sqrt(3) in terms of rational numbers - that's the contradiction I have in mind. ||

[wild jasper]

Do you agree that phi must fix Q?

[wild jasper]

Your argument works

[delicate bloom]

that's how I reason that phi(sqrt(2)) squared becomes 2 in my argument

there are plenty of elements that are not fixed in either field other than sqrt(2) and sqrt(3)

[thorn jay]

||i think simpeler is showing that, as phi fixes Q, it must fix polynomials, so a polynomial which has as a root sqrt(2) must then also have as a root sqrt(3)||

[delicate bloom]

"must" - why?

how do you know sqrt(2) can not be written as a+b sqrt(3)

[thorn jay]

Oh right i see now

Lol my bad

[delicate bloom]

for instance, Q(golden ratio) to Q(sqrt(5))

yeah you're good

[thorn jay]

I mean i guess you could say Q(sqrt(2)) and Q(sqrt(3)) are splitting fields of x^2 - 2 and x^2 - 3 and do something with that lol

[delicate bloom]

ok say it

[severe linden]

Is it correct that the answer is $$\mathrm{ann}_{\mathbb{Z}/(n)}(\overline{m}) := { \overline{k} \in \mathbb{Z}/(n) \mid \overline{km} = 0}$$?

[cloud walrus]

Yuese

[rustic crown]

its correct, but notice that this is a subgroup of a cyclic group, so is again cyclic. so we can write it in a more explicit way :3

[wild jasper]

||phi(a+bsqrt(2)) =a+b phi(sqrt(2)) = a+b sqrt(3)||

[heady shuttle]

Anyone have any ideas?

[frail summit]

Hey, in D&F they say that if K is the splitting field of f over F and F' is an extension of F, then KF' is the splitting field of f over F'. This makes sense, but I'm not sure how to go about proving it. Any hints?

[wild jasper]

Let K/F be a splitting field of some polynomial f over F, then since F is contained in F', which is contained in KF', then f is a polynomial over F', but then KF' must be the splitting field of f over F' as well

[frail summit]

That shows that it splits in KF', but why can there be no smaller subfield where it splits?

[wild jasper]

because KF' is the smallest field contaning K and F'

[frail summit]

And why do u need both K and F' for it to split?

[wild jasper]

Because because KF' is an extension of F' and K is an extension of F

[frail summit]

I don't see the your argument

[frail summit]

What does this have to do with needing K and F' to split?

[wild jasper]

The splitting field of a polynomial over F' must be an extension of F'

[frail summit]

True

But you haven't argued why one needs all of K over F' for it to split

[wild jasper]

We know that f splits in K, so the splitting field L of f over F' must contain K (because splitting fields are unique) and because L must contain F' as well, we have have L=KF'

[frail summit]

Why would L need to contain K? How does uniqueness give you this?

[wild jasper]

that means that f does not split over any field not isomorphic to K

[frail summit]

Yeah, it means that L must contain an isomorphic copy of K, but not necessarily K itself then...

So why would L need to contain K?

[wild jasper]

well it needs to contain an isomorphic copy

it's pretty much the same thing

[frail summit]

It's not at all the same thing

[wild jasper]

Let K' be the isomorphic copy of K then see if you can convince yoursels that K'F' is isomorphic to KF'

[frail summit]

If you want to argue that way then you would need to show that all extensions isomorphic to your splitting field are also splitting fields

[wild jasper]

Indeed, and that's why I told you, it follows from the uniqueness of splitting fields (up to iso), which is a theorem in DF

[frail summit]

That theorem gives you one direction: splitting fields are isomorphic

Not that isomorphic fields to the splitting field are also splitting fields

[wild jasper]

you have the other direction, because KF' is a splitting field of f

Since KF' is a splitting field, we have that KF' is isomorphic to L, call this iso phi, then phi(K) =K' is an isomorphic copy of K in L. but note that we must have that F' is contained in L, hence we must have phi(F')=F' hence im phi = L = K'F'

[frail summit]

You still haven't shown that an isomorphic copy is also a splitting field

But nevermind. I found a solution

[wild jasper]

L is a splitting field KF' is a splitting field. Hence they are isomorphic, hence L=K'F' where K is iso to K'

[frail summit]

Hm. Ok, I might buy that then. Although why do you conclude that phi(F')=F'?

Wait, now I'm confused. You started off with assuming that KF' is a splitting field, so I presume you're working with the base field being F'. Then we take an extension F of F' and claim that KF is the splitting field of f over F

[wild jasper]

no

Over F'

[frail summit]

Well then it doesn't make any sense to say that KF' is a splitting field when that's what you want to show...

[wild jasper]

We know that K is a splitting field of f over F. F is contained in F' and from this we can conclude that KF' is a splitting field of f over F'

[frail summit]

breh

Ok, nevermind

[south rain]

i'm currently trying to prove that if i have two cyclic groups $G$ and $H$, that $G \times H$ is cyclic $\Leftrightarrow$ orders of $G$ and $H$ are coprime, ie gcd$(|G|, |H|) = 1$. Now since every cyclic group is isomorphic to some $\mathbb{Z}_n$ it suffices to show it for $G = \mathbb{Z}_n$ and $H = \mathbb{Z}_m$ . Therefore I assumed the product is not cyclic. Then every element in the product has order < $nm$. Now I should get that lcm$(m, n) < nm$, which somehow makes sense but I can't quite exactly see how I get this. Is every product group generated by order elements of order $m, n$ of the order lcm($m,n$), and if so why?

[cloud walrus]

dellinger

[south rain]

Nevermind I got it.

[delicate bloom]

this is wrong because you have not argued why phi(sqrt(2)) = sqrt(3)

read the discussion I had with the other person at that time to see why

[cloud walrus]

mh_le

[wild jasper]

I see my mistake. Thanks

[toxic zephyr]

can someone tell me if i'm understanding free groups correctly here? trying to explain it in my own words

[coral spindle]

You seem to have the right idea

[bitter rover]

It can be helpful to specifically talk about the inclusion map.

[toxic zephyr]

and if |X|=|Y| then F(X) is iso to F(Y) right? and woul that mean that understanding the "structure" of a free group just depends on how many elements are in the generating sets?

[bitter rover]

You also don't have to stipulate the stuff about inverses and whatnot, because otherwise the map wouldn't be a homomorphism at all.

[next obsidian]

Yeah

It don’t matter what your alphabet is, when you work formally

[toxic zephyr]

that makes sense

[bitter rover]

Yes, that's sort of the spirit of "free". The set X is like the least structure we can put on anything at all: it's a collection of distinct objects, with nothing to be said about them except how many there are.

[toxic zephyr]

one other question, say we have like X={a,b,c} and we have the word abc. then we can define an equivalence class for all the words obtained from insertions and deletions on abc (like aba^-1 ac). is the free group the set of equivalence classes or is there another structure that is the equivalence classes (like after we mod out by something)

[chilly radish]

The free group is the set of these equivalence classes, but you can always find a unique reduced representative

i.e. one without any xx^-1

So you can just think of the elements as reduced words

[toxic zephyr]

okay yeah thats what my friend was saying. so like basically we just cancel out all the inverses we can and thats the obvious choice for the equivalence class representative

[next obsidian]

Yuh

[toxic zephyr]

okay this is pretty cool

[next obsidian]

You know what else is cool?

Chmoker

But they removed him. Killed him even

[toxic zephyr]

im wondering if this counterexample works for b and c. like {2} is a LI set in Z (which is rank 1, right?) but <2> is not Z, and further we cannot extend it to a basis because if we add any element <2,n> then 2(-n)+n(2)=0 shows its not LI.
and then for d, does the example <2,3> is a generating set (since they're coprime) work? we can't just take 2 or 3, that wouldn't generate Z. and then the last part of d, can i say that F is guaranteed to have a basis?

[coral spindle]

C.f.: Given a set X and a field k, the set k[X] of functions v : X -> k for which all but finitely many values of x give f(x) = 0 is a vector space over k. Identify the elements of X with the indicator function for that element.

For any k-vector space V and function f : X -> V, there is a unique linear map phi : k[X] -> V with phi(x) = f(x) for every x in X.

[coral spindle]

Your counterexample for (b) works. I am reading (d) now

Oh, also for (c) ofc.

F is guaranteed to have a basis, but why does this imply m <= n? You don't seem to have justified this.

The rest is fine.

[toxic zephyr]

yeah still thinking about it

[toxic zephyr]

tysm, it's so helpful to know i'm on the right track

[coral spindle]

An equivalent question is saying that a rank n FAb cannot be generated by fewer than m elements.

Whoops, fewer than n*

[coral steeple]

If G1, G2 are simple, and K is a normal subgroup of G1 times G2, with pi(K)=Gi (i.e., the projections of K are equal to the Gi), does it follow that K=G1 times G2? I think so, but I have no idea how to show this.

I'm guessing the condition that the Gi are simple isn't necessary here.

[coral spindle]

I think this is false for G1 = G2 = Z/pZ and K being the diagonal.

[toxic zephyr]

im kind of struggling on this last part. if F=<x1,...,xn> and we have a basis {g1,...,gm} then we basically want to show that we can't have m>n. but im not sure how to get a contradiction here (if thats the way to go). like i wrote each gi as a combination of the x's, and then took a linear combination of the g's, but now i'm not sure how to show something goes wrong.

[coral spindle]

OK quick question: do you have as given that all bases have the same cardinality?

[toxic zephyr]

yeah

[coral steeple]

Thanks 👍

[coral spindle]

So try to transform a generating set in some way to produce a basis. This is the best approach that I can think of quickly

[coral spindle]

Btw I think this is true if both Gi are non-Abelian

[coral spindle]

So we know that we cannot do this simply by removing elements, so perhaps there is a way of substituting then removing

The usual way I would want to show this is unfortunately too high-powered...

[coral steeple]

I'm trying to characterise the normal subgroups of G=G1 times G2, so I'm guessing that this means it's not worthwhile looking at cases based on whether pi(K) is Gi or 1

[coral spindle]

There's some theorem, I forget what it's called. But we can just argue directly.

So suppose (a, b) is in K. Then assuming G1 is non-Abelian, we may conjugate by some element to get (c, b) in K also. Overall we get (ac^-1, 1) in K. Now agan arguing by normality we conclude that G1 x {1} is a subgroup of K. Repeat with the other side and we get that G1 x {1} and {1} x G2 are subgroups, whence K = G1 x G2.

We can guarantee such an element (a, b) where a,b are non-identities using these surjections you posit

[coral steeple]

How are you using non-abelianness?

[coral spindle]

All conjugacy classes are nontrivial (except the identity's of course)

So c is not a

So we get an element of the form (g, 1) where g is nontrivial.

[coral steeple]

Ah, I didn't know that

[coral spindle]

Well imagine we have a group where all the conjugacy classes are trivial

I.e., the conjugacy class of g is {g}

So what is aga^-1 for some a? it must be g

so aga^-1 = g

so ag = ga.

[coral steeple]

Ok, I think I see

[coral spindle]

You probably know this already from studying the class formula

[coral steeple]

We don't know that either

[coral spindle]

Class equation, whatever you call it

[coral steeple]

We don't know much other than definitions and the first isomorphism theorem

Anyway, that's very helpful. Thank you

[toxic zephyr]

gonna come back to that other problem later. if i have a homomorphism from F to Zn, then the kernel has index n right?

trying to explain that if g1 is in a basis for F a free abelian group, then <ng1> is a subgroup of index n.

[coral spindle]

The index is equal to the order of the image by Lagrange. In general there may be homomorphisms into Z/n whose image is not Z/n, so no. However the index must divide n.

[toxic zephyr]

but if its surjective then its true?

[heavy minnow]

yes (im assuming by Zn you mean Z/nZ)

[coral spindle]

By Lagrange, yes.

[toxic zephyr]

my map is basically c1g1+...+cmgm to (c1+nZ). so we have f(ig1)=i+nZ is surjective since g1 is LI

[coral spindle]

You don't need to be specific, this comes completely for free abstractly

The preimages of elements are cosets of the kernel

There are n images, so there are n cosets of the kernel.

[toxic zephyr]

aha

so then this idea does work, and <ng> is a subgroup of F with index n for any n in Z and g in any basis of F?

[coral spindle]

There are no subgroups of a free Abelian group of finite order

[toxic zephyr]

sorry meant index

[coral spindle]

OK I see you changed that

Let me answer this with a different thing

Z^2 is a free Abelian group of rank 2

What's the index of the subgroup {(2n, 2m) | n, m in Z}?

[toxic zephyr]

2 right?

[coral spindle]

Try finding an appropriate group G and surjection Z^2 -> G with kernel the given subgroup

Well, finding a group G of order 2 with these properties would prove that

However if no such thing exists then we know this isn't true

[toxic zephyr]

map (a,b) to (a+2Z,b+2Z)

[coral spindle]

OK, what's the kernel of this map? I will let you assume without proof that this is a homomorphism

[toxic zephyr]

2Zx2Z

[coral spindle]

Yeah, great OK. So what's the group G?

[toxic zephyr]

Z2 x Z2?

[coral spindle]

Yeah, so we know this subgroup is of index 4

Now is this what you meant when you wrote <ng> where g is in a basis?

I realised too late that you may not have meant what I thought you did...

[toxic zephyr]

i suppose in this case of Z^2, then (taking the basis (1,0) and (0,1)) we would have <(n,0)> and <(0,n)>

[coral spindle]

Ah. OK, fortunately we can still use the same method

[toxic zephyr]

map (a,b) to a+nZ or b+nZ

[coral spindle]

So let's consider the subgroup <(2, 0)> instead. This is equal to {(2n, 0) | n in Z}, yes?

[toxic zephyr]

yeah

[coral spindle]

OK, so what's the group G here and what's the kernel?

[toxic zephyr]

so we have that 2Zx0 is the kernel of a map to just Z_2

[coral spindle]

Hm is that true?

Where does (0, 15) get sent by your map?

[toxic zephyr]

the map (a,b) to a+2Z

oh

[coral spindle]

Yeah

The kernel is 2Z x Z.

[toxic zephyr]

you're right

[coral spindle]

The subgroup <(2, 0)> is of countable index in Z^2

An appropriate surjection exists onto Z/2 x Z

[toxic zephyr]

so <(n,0)> is not index n, it's index infinity? or i think i have the wrong map.
(a,b) maps to (a+2Z,b) i think might work. so its a map to Z/2 x Z like you're saying

i might be confusing myself

[coral spindle]

Yeah this is correct

It's worth saying that it's of countable index rather than merely infinity because Lagrange is a statement about cardinals -- we can still distinguish subgroups of index different cardinals

[toxic zephyr]

ah true

[coral spindle]

It's a nitpick though

The funny thing about free Abelian groups is they always embed in a Q-vector space

So you can actually safely imagine a vector space and remember that our object looks 'discrete'

But these torsion bits are alien to vector spaces.

Well. Vector spaces in fields of positive characteristic...

[toxic zephyr]

the exercise is to show a nonzero free abelian group has a subgroup of index n for all n positive integer. i think i still want to try to find a surjective homomorphism from F (or Z^m) to Z_n. but it seems i'm not managing to find a good map.

[toxic zephyr]

interesting

[coral spindle]

Well ok I'll give you a hint

We looked at the example <(2, 0), (0, 2)> being of index 4 earlier

because it's the kernel of a surjection onto Z/2 x Z/2

Could you adapt the map for Z/n x Z/m, where n and m are possibly distinct? What happens if I set m = 1?

[toxic zephyr]

yeah i was trying to think about that. is Z/1 just {0} or Z? i would think its the former

[coral spindle]

Well ok this is maybe worth getting into the weeds with

Z/1 is just notation for Z/1Z = Z/Z

So to count the number of elements of Z/1, we need to count the number of cosets of Z in the group Z

I think hopefully it should be clear what the size should be

[toxic zephyr]

just 1 right

[coral spindle]

Yeah, that's right

(And there is only one group, up to isomorphism, with that size)

[toxic zephyr]

the trivial group

[coral spindle]

Exactly

[coral spindle]

Is this clear btw? It occured to me that you may not actually have seen quotient groups.

[toxic zephyr]

oh so would it just be (a,b) to (a+nZ,b+Z)? but then, like... isn't b+Z just 0? or i suppose maybe there is a slightly more nuanced distinction? like whats the difference between Z x Z to Z/n x Z/Z and Z x Z to Z/n x 0 or even just Z/n

[toxic zephyr]

yeah this is clear, it makes sense

[coral spindle]

Yes, b+Z = Z is the identity of the group Z/1

There is no group-theoretic difference between Z/n x Z/1 and Z/n; they are isomorphic as groups.

What I showed you was a trick to get you to see an approach

You can simplify now if you like. What's the kernel of this map, since we know its index now?

[toxic zephyr]

oh... <(2,0),(0,1)>?

[coral spindle]

Yeah that's right

[toxic zephyr]

omg okay i see it now

[coral spindle]

So I think you can generalise this

Yeah nice

[toxic zephyr]

so then the map c1g1+...+cmgm to (c1+nZ) was right, i just didn't have the right kernel?

[coral spindle]

That's correct

[toxic zephyr]

i see. does this still work with an infinite basis? or do we not consider that

[coral spindle]

It's funny how just that dummy Z/1Z term makes all the (mental) difference

[toxic zephyr]

yeah totally

[coral spindle]

We can certainly have infinite bases, and yes it works

You just need to be a bit aware of how generation notation works for infinite sets

Just like in vector spaces, we don't have a notion of infinite sum, so we have to only consider finite sums.

Notation for a free Abelian group on a set X that I am used to is Z[X] or sometimes Z^(X)

(This mirrors notation for free modules over a ring)

[toxic zephyr]

so then the answer isn't <ng1,g2,g3,...,gm> i have to allow for an infinite basis like <ng1,g2,g3,...> (for {g1,g2,g3,...} a basis for F)

[coral spindle]

Yeah that's right

[toxic zephyr]

okay yeah that's what im used to in linear algebra. span of an infinite set is all finite linear combinations

so then i guess to define the map properly, we have to define it on each basis element. f(g1)=1+nZ and f(gi)=1+Z or 0 if i>1. writing it in that sum form is a little sus

[toxic zephyr]

could this also work for an uncountable basis? like can we have Z^|R| or something lol

[coral spindle]

Yes, the cardinal is immaterial here

You just pick one element and double it and leave the rest alone

[toxic zephyr]

that's kind of sick

[coral spindle]

That's algebra baybee

[toxic zephyr]

this is actually kinda hype. i was lowkey dreading this group theory course.

[coral spindle]

Nooooo group theory is the fuckin best

[toxic zephyr]

i just have one more conceptual question. so in this theorem can we say that the cardinality of the basis of F is the same as the cardinality of the set X described in part iv? like we could consider the free group generated by the real numbers R and that would sort of induce a free abelian group of rank |R|. and F would be isomorphic to Z^|R|?

[toxic zephyr]

i was in an undergrad research project on group theory for a year and i kind of hated it, but i think that's probably because i just didn't know enough group theory. things are starting to fall into place and make sense in retrospect

[coral spindle]

Z^|R|
watch out, this is not the notation for the free Abelian group of rank |R|

This will typically have many more elements. I'm not sure in this case

[toxic zephyr]

oh i see

[coral spindle]

This is, put simply, allowing infinite sums in this notation

[toxic zephyr]

so maybe like... Z^|R| (but finite support)? idk if there's a super nice notation to specify that the elements have "finite support" (if considered as functions)

[coral spindle]

I think (ii) has some subtlety because one needs to be careful as to what one means by an infinite internal direct sum. But yes this all holds

[coral spindle]

Yeah it's $\mathbb Z^{(\mathbb R)}$ as I've seen it

[cloud walrus]

$\mathbf{Boytjie}$

[toxic zephyr]

oh that's nice

[coral spindle]

E.g. in Rotman's intro to homological algebra iirc

[toxic zephyr]

@coral spindle thank you so much, I feel much more confident in these concepts. I really appreciate your help

[coral spindle]

No worries Taylor

I'm sure everyone else here would be glad to help you out if you have more questions, and I'll stick around for a bit longer too

[toxic zephyr]

can a homomorphism from G (abelian) to itself be surjective but not injective? what about free abelian

[void cosmos]

its true for free abelian

yeah its true for any f.g mod

nakayamas lemma

not this

lmao

[coral spindle]

Think about this in relation to functions of sets f:X -> X and maybe think about what would happen if we looked at the free Abelian group generated by X, and the lift of f we get to those groups

[coral steeple]

Sorry to bother you again about this. If I understand correctly, your proof says that if G1 is not abelian, then we can define a function f from G1 to itself mapping a to a different element of the conjugacy class of a. We need to show that a |-> af(a) is a surjection (in your notation, we need to show that one can obtain every possible (g,1) with g in G1 by applying the process you describe). I'm not sure how this would work. It looks sort of like the statement that left-translation is a bijection, but I'm not seeing it.

[coral spindle]

We need to show that a |-> af(a) is a surjection
No, we don't.

We obtain a single element (g, 1) in K

Just one works

It doesn't matter which one

The argument then continues by seeing that we can just look at G1 now

Looking at a subgroup generated by <(g, 1)> is the same as looking at a subgroup generated by <g> in G1

Now in particular since we have normality, we know that G1-conjugates of <g> cover G1

and equally G1 x {1} -conjugates of <(g, 1)> cover G1 x {1}.

[coral steeple]

Ahhh ok I was missing this

[coral spindle]

We did not need to appeal to this function being a surjection, and it will never be one in fact (since the identity is never hit).

[coral steeple]

Is this just a consequence of the definition of normal subgroup or is there some work involved in this statement?

[coral spindle]

No this is precisely what it means to be a simple group

We can look at the smallest normal subgroup containing <g>

It must be the whole group, so we know that conjugates of <g> cover G1.

[coral steeple]

I see, clever.

[coral spindle]

Or well... products of conjugates. It would get messier actually...

The point is that the smallest normal subgroup of G1 containing <g> is G1, so the smallest normal subgroup of G1 x G2 containing <(g, 1)> must, in particular, contain G1 x {1}.

[coral steeple]

I wondered whether I hadn't done enough work on this before asking for hints but seeing the whole proof tells me there wasn't any hope of me finding it independently. Thanks again!

[restive birch]

okay yall might have talked about this before but i dont remember seeing it
show that if $1 \unlhd M_1 \unlhd M_2 ... \unlhd M_r = G$ and $1 \unlhd N_1 \unlhd G$ are two composition series of G, then $r = 2$ and either $M_1 \cong N_1$ or $M_1 \cong G/N_1$

[cloud walrus]

henryduke

[restive birch]

grad student and i tried to do this for an hour

got nowhere

[sharp ice]

look up the jordan holder theorem

[toxic zephyr]

was thinking of a way to do this. where if m>n, then we can define a homomorphism f from F to F such that f(gi)=xi for i≤n, and then f(gi)=0 for i>n. this would define a surjective but not injective homomorphism from F to F. I was wondering if that would be a sufficient contradiction.

[toxic zephyr]

what is a lift?

[restive birch]

this problem is trying to prove a special case of it

[nimble folio]

When can we ignore that arbitrary elements of tensor products are finite sums?

Like, is there a certain condition where I can prove a property (like linearity) with pure tensors instead?

[toxic zephyr]

injective but not surjective seems possible. like Z to Z via f(n)=2n. but surjective but not injective seems impossible

[nimble folio]

i guess i should be more specfic, I have homomorphisms $T \in Hom(V_\mathbb{C},\mathbb{C}\otimes_\mathbb{R}V)$ and $S \in Hom(V_\mathbb{C}, \mathbb{C}\otimes_\mathbb{R}V)$. Would I need to show that $S \circ T = id_{V_\mathbb{C}}$ and $T \circ S = id_{\mathbb{C}\otimes_\mathbb{R}V}$ with finite sums of tensor products or can I use pure tensors?

[cloud walrus]

clubsoda14

[tough raven]

Let A be a "generic" m⨯n matrix, i.e., A = (a_ij)_{i,j=1,1}^{m,n} with entries in the polynomial ring R := ℤ[a_ij : i=1,...,m, j=1,...,n]. Is the null space of A (over R):

• 0 for m ≥ n
• for m < n, spanned by v_S for every set S of m+1 columns of A, where v_S is a n⨯1 column matrix such that (v_S)_i = 0 for i not in S and is the m⨯m minor of A corresponding to the columns S\{i} for i in S with an appropriate sign.

For example, for A = (a b c), v_{1,2} = (b -a 0)^T, v_{2,3} = (0 c -b)^T and v_{1,3} = (c 0 -a)^T.

[tardy hedge]

I was just reading this comment, but I was wondering, Why do you say phi is a linear map when k[X] isnt a vector space, is it?

wait yes it is

Lmao

[tardy hedge]

really? why?

Why does surjective but not injective seem impossible?

[toxic zephyr]

i mean, how would i define a surjective homomorphism from Z to Z that isn't injective? it's uniquely defined by where i map 1. to be surjective n has to be 1 or -1. then it's not injective.

[tardy hedge]

Are you saying Z because we are only talking abou free groups or something ?

[toxic zephyr]

yeah

free abelian

for a generic group, a surjective but not injective homomorphism might totally be possible, but im not sure

[tardy hedge]

yea true

[next obsidian]

Take an infinite direct sum of Z, this is a free abelian group and the map which sends the n-th coordinate to the n-1st coordinate is surjective but not injective

[tardy hedge]

weird

[toxic zephyr]

the way i'm thinking about it is that unlike vector spaces where we can divide scalars, any non-unit multiple of a basis element will no longer function as a basis element. this give injectivity (due to the independence) and then we lose surjectivity. so surjectivity is more finicky than injectivity. i guess since it's easier to lose generation than losing independence (due to the general lack of divisibility by scalars).

[toxic zephyr]

but i suppose for an infinite rank group, it makes sense we can similarly use the shift operator like we can in infinite dimensional vector spaces

in the problem i'm working with, it's finitely generated

[next obsidian]

For any finitely generated abelian group a surjective endomorphism is injective

[toxic zephyr]

aha

but not the other way around

[next obsidian]

Such a group is Noetherian as a module

And if the kernel is nonzero

When you iterate the map you get growing kernels

Contradicting Noetherian

[tardy hedge]

does that mean an ascending chain condition on its submodules or smth?

I havent learned much noetherian stuff yet

[next obsidian]

For an Artinian module an injective map is surjective

[next obsidian]

Yes

[next obsidian]

Because if it wasn’t the successive images of iterates will keep shrinking

[tardy hedge]

Whats an artinian module>

?

Idk what artin rings are

[next obsidian]

Descending chain condition

[tardy hedge]

Oh really?

Its like opposite of noetherian?

[next obsidian]

Yeah but for commutative rings it implies Noetherian

[tardy hedge]

Why are noetherian rings useful? Somehow i have never learned them yet

[next obsidian]

Specifically it’s a dimension 0 Noetherian ring

[next obsidian]

Nice properties

You can have Artinian modules that aren’t Noetherian though

[toxic zephyr]

so i basically tackled this problem like it was a module problem. by defining a homomorphism from the basis {g1,...,gm} to a generating set {x1,...,xn}. supposing that m>n gives a hom that is surjective but not injective (f(gi)=xi where xi is just 0 for i>n). but uh... i don't think i'm supposed to use that stuff for this group theory problem.

like we haven't really talked about that. i don't even know if we proved that we can always define a hom based on where it takes a basis.

[tardy hedge]

oh because there is no ascending chain? dimension of a ring is like the longest chain of ideals or smth?

[next obsidian]

Prime ideals

And you index starting at 0

There’s sometimes a few notions of dimension, this one is called the krull dimension

[toxic zephyr]

am i overcomplicating the problem or is what i'm doing a good solution

(trying to do the second part)

trying to do a direct proof was really nasty

[next obsidian]

Can you localize to become over Q?

[toxic zephyr]

idk what that means ;-;

[tardy hedge]

fr

neither do i

lol

[next obsidian]

Then no

[tardy hedge]

what does that mean

[next obsidian]

Add in fractions formally to make it a vector space

[tardy hedge]

why is it localize?

[next obsidian]

AG

[toxic zephyr]

wait if B={g1,...,gm} is a basis for F, then can i say F is isomorphic to the free group generated by B? and then i can still define that homomorphism by doing the function h(gi)=xi (and 0 if i>n) from S to F and extending it to a group hom from F(B) to F

[next obsidian]

Yeah

[toxic zephyr]

okay okay. and then i guess how would i argue that the group hom (phi, lets say), which is clearly surjective (since the x's generate) but not injective provides a contradiction?

i'm just not sure how to take the free module-lingo to free abelian group lingo

[tardy hedge]

What are we doing again

I legit forgot lol

[toxic zephyr]

here. i'm supposing that we have a basis for F with more elements than a generating set

trying to find a contradiction from that

to show that rank(F)<=n (the size of the generating set)

if the basis B={g1,..,gm} has more elements than the generating set {x1,...,xn}, then the map from B to F via h(gi)=xi (where we let xi=0 for i>n) can be extended to a group homomorphism from F to F (using the inclusion map?) which is surjective because every element in the generating set has a preimage but not injective because we have a nontrivial kernel element g_n+1

so then... why exactly is that not gucci?

i don't think we've really covered by this point that endomorphisms on free abelian groups cannot be surjective and not injective. it seems like that's something i would have to prove separately, which makes me think i'm overcomplicating it lol

not sure if there's something more elementary i'm missing from this logic. like maybe this idea provides a window into a direct proof that i'm not seeing

[next obsidian]

I think if you are willing to analyze matrices hard enough you can probably get it

An endomorphism of a rank n free module is an n x n matrix

And you can maybe try to mimick linear algebra in terms of the rank of the matrix and stuff

But idk how easy that is

[toxic zephyr]

i would love to use a matrix, but i feel like that's going even farther from the "spirit" of the problem... this is chapter 2 of hungerford

[next obsidian]

¯_(ツ)_/¯

[toxic zephyr]

😭

[hidden cairn]

why is a --> a^2, b --> ab not counted as one of the elements in Aut(H) of order 2? (H is Z_15)

or a --> ab^4, b --> b^4

btw <a> = Z_3, <b> = Z_5

[hidden cairn]

ah nvm i see why

a^2b does not even have the same order as a to begin with

[vagrant zinc]

Dummit?

[hidden cairn]

yes

[toxic zephyr]

WAIT. first iso theorem. h is surjective so F (the image) is isomorphic to F/ker(h). ker(h) is nontrivial. can we use this to say it's a contradicted? like can we guarantee that F/ker(f) is "smaller" than F?

F is iso to Z^m, since <g(n+1),...,gm> is contained in the kernel, we can say F/ker(h) is iso to something "smaller than or equal to" Z^n?

[golden turtle]

How does it follow that imaage(1 tensor psi) subseteq ker(1 tensor phi)?

I'm a bit confused by what computation they're even performing

[lone niche]

Above the red line you drew, is a generic element of the image(1 tensor psi). And thus that equation shows that (1 tensor phi) sends any element of the image to zero.

[cloud walrus]

Nope

[elfin wraith]

Well I have spotted the issue and its that I forgot the rings in this course on non com algebra are not commutative

[south rain]

This might be a pretty stupid question but why do we choose the subgroup condition as $a \circ b^{-1} \in H$ instead of $a \circ b \in H$ where $H$ is the subgroup of $G$? Is it just for convenience in other proofs? These definitions are of course equivalent right?

[cloud walrus]

dellinger

[weary frost]

yes (edit: no)

since b is in H so is b^(-1)

so that means a * (b^(-1))^(-1) = a*b iis in H

[somber turret]

but is it in G? We need H to be a subgroup of G.

[weary frost]

oh right there is something missing

in my statement

[somber turret]

Actually I'm confused too

[south rain]

what do you mean?

[somber turret]

$a\circ b^{-1}\in H$ asserts the closedness of the operations

[cloud walrus]

luke1337

[south rain]

$b^{-1}$ is always in $G$ if $b \in G$

[cloud walrus]

dellinger

[weary frost]

ah okay so what we do is g*(g^(-1)) = e is in H, so e is in H

[somber turret]

right

[weary frost]

then e*(g^(-1)) = g^(-1) is in H

[somber turret]

I think G is irrelevant actually

[weary frost]

ok so actually the condition with the inverse is stronger

[south rain]

i think thats already enough, at least thats what I got too.

[somber turret]

In this context, $b\in H\implies b^{-1}\in H$ follows from the subgroup condition. If the subgroup condition did not hold true, we cannot assert$b\in H\implies b^{-1}\in H$ and thus we cannot say H is a group (let alone subgroup of G)

[cloud walrus]

luke1337

[south rain]

Ah yea, that makes sense

If we know its a subgroup it follows immediately

[somber turret]

ye

[south rain]

Ah, got it, thanks.

[weary frost]

yea you actually have a counter example with the natural numbers as a subset of Z

[somber turret]

(I think the existence of identity should be included here as well, but I guess it was omitted for brevity?) actually maybe follows too

[south rain]

Its basically in the proof of this condition. The one direction is easy to see but the other is more work.

[south rain]

Identity in $H$?

[cloud walrus]

dellinger

[somber turret]

If we set $a = b$, $a\circ a^{-1}=e\in H$

[cloud walrus]

luke1337

[south rain]

Yea that follows from choosing a = b.

Yea

[bitter rover]

We need the subgroup to be closed under both the group operation and inverses. Saying $a \circ b \in H$ for $a,b \in H$ doesn't guarantee that. For example, the set $\left{2, 4, 6, \ldots, 2k,\ldots\right}$ is closed under addition but certainly doesn't have any inverses. It doesn't even have the additive identity $0$!

\\

However, $a \circ b^{-1} \in H$ for $a,b \in H$ if and only if $H$ is closed under both the operation and inverses. (And being closed under those is enough to guarantee the identity is in $H$, since $e = a \circ a^{-1} \in H$.)

\\

So it becomes a more compact way of saying that $H$ is a subgroup. Rather than having to prove $H$ is closed under the operation and inverses separately, we can prove $a \circ b^{-1} \in H$ in one go and conclude $H$ is a subgroup.

[cloud walrus]

Cufflink
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[stone sky]

That’s how i do it, if i can show that the subgroup contains an inverse with an operation in one motion, it would be a quicker check. Separately verifying for the inverse is also valid, but takes an extra step.

[south rain]

Thanks, that's clear now!