groups-rings-fields

[tidal crag]

I know a conversation about subgroups is happening, and was wondering something since i have been having trouble understanding something around it. I am currently having trouble understanding quotient groups , and specifically finding generators of such. (and also how they actually work)

just a hard topic for me.

[bitter rover]

Ask away

[tidal crag]

For example, using G/H, where H is a subgroup of G, I am trying to figure out what groups are allowed to be quotient groups in general. I am assuming it must be a finite group like Z_5 and D3

But i was also having trouble how to visualize what G does after applying /H to it.

[night tartan]

any group can be a quotient group, just take G = H x Z_2, then G / Z_2 is isomorphic to H

[night tartan]

my groups supervisor explained it like this: G/H is "G with H killed"

[tidal crag]

oh, fair

[night tartan]

you're probably much more familiar with this concept than u might realise

like u've done modular arithmetic right?

[tidal crag]

ye

Is it that easy?

[night tartan]

like there, "mod p" is really just the group Z/pZ

[bitter rover]

Not quite, but modular arithmetic serves as a kind of "easy case" where you can get intuition.

[tidal crag]

I can easily figure it out then

[night tartan]

but you don't think of doing modular arithmetic as
"if i take the coset {..., -6, -1, 4, 9, ...} and add the coset {..., -7, -2, 3, 8, ...}, then i get the coset {..., -8, -3, 2, 7, ...}"

[bitter rover]

Depends on what "that easy" means I suppose

[bitter rover]

Figure what out?

[night tartan]

instead u'd think of it as "4 + 3 = 7, but i consider 7 and 2 to be the same number bcus they differ by 5"

[tidal crag]

All i needed to understand

[bitter rover]

The little engine driving it all is the first isomorphism theorem, once you get there.

[vagrant zinc]

false, $\dim (\mathbb{R})=1$ and $\dim(\mathbb{C})=2$, therefore $\mathbb{R}\ncong \mathbb{C}$

[cloud walrus]

Homology

[bitter rover]

The additive group of the reals and the additive group of the complex numbers are isomorphic in ZFC.

[somber turret]

Assuming vector spaces over reals, yes. You can define them as vector spaces over some other field, like rationals.

[somber turret]

What would be the simplest construction of such isomorphism between R and C?

[bitter rover]

There's no "construction" in the strictest sense. (ℝ,+) ≅ (ℂ,+) implies the existence of a non-measurable set of real numbers, so something close to the axiom of choice is required.

https://www.cambridge.org/core/services/aop-cambridge-core/content/view/AB4402418C7D811572186A23D32D87A4/S1446788700031505a.pdf/div-class-title-a-consequence-of-the-axiom-of-choice-div.pdf

[south rain]

On that topic: I should show that $KN/N \cong K/(K\cap N)$, where $N$ is normal in $G$ and $K$ is just a subgroup. Naturally, I chose the homomorphism $h: K \rightarrow KN \rightarrow KN/N$. Clearly $\ker(h) = K \cap N$. By the first ismorphism theorem I now know that $ K/(K\cap N) \cong \mathrm{Im}(h)$. To show this is onto I took an element $(kn) N \in KN/N$ and then have $(kn) N = kN = h(k)$ and I'm done. Now is this correct? I didn't use that $N$ is normal at all so I'm unsure if I missed something.

[cloud walrus]

dellinger

[south rain]

Oh and $KN = {kn : k \in K, n\in N}$.

[cloud walrus]

dellinger

[dull ginkgo]

Image of R in C under the Q-module isomorphism in the C->R direction?

[south rain]

This then would be one way to show it. Another way would be to use that $ N \trianglelefteq KN$ and $K \cap N \trianglelefteq K$ and then use some epimorphism $h: KN \rightarrow K$, but I couldn't find one that gives me $h(N) = K \cap N$ and $\ker(h) \subset N$.

[cloud walrus]

dellinger

[south rain]

The obvious one first is just $h(kn)= k$ but then $\ker(h) = N$ which is not what I want.

[cloud walrus]

dellinger

[vagrant zinc]

By the fourth theorem of correspondence isomorphism

[south rain]

Ah nevermind, I need $N \trianglelefteq KN$ to get the canonical epimorphism.

[cloud walrus]

dellinger

[tough raven]

In case it is of interest, I have proved that this is true for row matrices: the null space of the 1⨯n matrix (a_1 ... a_n) is spanned by (0 ... 0 a_j 0 ... 0 -a_i 0 ... 0) for 1 ≤ i < j ≤ n.

Also for "almost square" matrices, meaning n⨯(n-1) matrices, provided I can show that the n possible minors obtained by deleting single columns are coprime.

[dull ginkgo]

pretty sure you can also pass through the exterior alg in a fucky way

might be doing the same thing actually

[tough raven]

Oh, that's probably a good way to think about working with all of the minors (of a given size) of a matrix. I'll keep it in mind. Thanks!

[golden turtle]

Why is that a generic element of image(1 tensor phi)? I'm a little confused on what 1 tensor phi is exactly?

[velvet steeple]

Why are we only arranging the e_i vectors in chains?

Are we completely ignoring e.g. f_i?

[dull ginkgo]

Yeah the minors are the components of the induced map on the exterior powers lmao, that's how I use it

biggest PITA is showing the exterior algebra is free

and I wonder how you can extend that to show other quotient algebras of the tensor algebra are free

[tough raven]

1 (⨯) psi is by definition the unique linear map that maps d (⨯) l to 1(d) (⨯) psi(l) = d (⨯) psi(l). By linearity, it must map any tensor ∑_i d_i (⨯) l_i to ∑_i d_i (⨯) psi(l_i). Since any tensor can be written in the form ∑_i d_i (⨯) l_i, that means any element in the range of 1 (⨯) psi can be written in the form ∑_i d_i (⨯) psi(l_i).

[dull ginkgo]

Yo Raghuram

[golden turtle]

Thank you sir, this makes sense

[dull ginkgo]

do you think T(V) / (v (x) v (x) v) would be free?

[tough raven]

I'm not 100% sure about the context, but I think you make a chain for each e_i and each f_i and each g_i and ...

[tough raven]

Is (v (⨯) v (⨯) v) the ideal or vector space generated by {v (⨯) v (⨯) v : v in V}? And what kind of "free"? (It's always free as a vector space if we're working over a field for example.)

[dull ginkgo]

the ideal generated, and free as in each of its powers (or images of the tensor powers) are free as a module over the ring of the base module V

[velvet steeple]

"Where A acts by a Jordan block matrix"

What's this supposed to mean

[rocky cloak]

A restricts to a linear transformation on the subspace, and you can pick a basis so the corresponding matrix is a Jordan block

[velvet steeple]

How do we know we can pick such a basis?

[tough raven]

That's what the page you first posted was about: finding a basis of each direct summand ker(A - l Id)^n with respect to which the matrix of A is a direct sum of Jordan blocks.

[rocky cloak]

You can do it by induction, or through the structure theorem for PIDs, probably in other ways as well

[velvet steeple]

Ok, so we constructed a basis with (A - lambda id)(v_1) = 0 and (A - lambda id)(v_i) = v_(i - 1) for each of the direct summands

How do we know that this property means it will look like a Jordan block?

[tough raven]

A(v_i) = (1) v_{i-1} + (lambda) v_i, so the i^th column consists of a 1 in the (i-1)st row and a lambda in the i^th row - which is exactly how a Jordan block looks.

(And A(v_1) = lambda v_1, so the 1st column is just a lambda in the 1st row.)

[velvet steeple]

Oh.... This is great

Thanks a lot!

[velvet steeple]

So each of these chains is one Jordan block?

But to the same eigenvalue

[velvet steeple]

s is the amount of blocks of size k?

t is the amount of blocks of size k - 1

Etc.?

[tough raven]

Exactly.

[velvet steeple]

Thanks!!

[tacit ginkgo]

Hi I am stuck proving this, i have that both g1,3 and g5,0 have order 2. but that these 2 composed give order 2 when they need to be order 6 to prove that G is isomorphic to D6

[tough raven]

If ℚ embeds in the base ring (i.e., all positive integers are invertible), I believe it ought to be free if V is free. But I don't have a proof yet.

[tacit ginkgo]

am using this theorem
Can someone give me a hand please?

[tough raven]

While it is true that D_n has two elements of order 2 with composition of order n, that is not true for any two elements of order 2. You probably need to try some other 2 elements.

[tacit ginkgo]

Ok thank you

[tough raven]

Drawing ℤ_6 as the vertices of a regular hexagon going (say) clockwise and visualising how g_{1,b} and g_{5,b} act would probably also help understand what is going on.

For polynomials f and g of degree at most m, n respectively, let Res(f, g) denote their resultant (where they are treated as having degree m and n regardless of whether they actually do). Is it true for any scalars a, b, c, d that Res(f((aX+b)/(cX+d)) (cX+d)^m, g((aX+b)/(cX+d)) (cX+d)^n) = (ad-bc)^{mn} Res(f, g)?

[velvet steeple]

From what does this follow?

N_k = Ker(A - lambda id)^k

[tough raven]

Note: for a = d = 1, c = 0 (strictly upper-triangular matrices); b = c = 0, d = 1 (diag(a, 1)); and a = d = 0, b = c = 1 (reflection), it's true according to Wikipedia.

[tacit ginkgo]

I see how that works but am not sure how to get s1 and s0 both order 2 and r order 6

[tough raven]

Compute dim ker(A - lambda)^i for A a Jordan block of size n (answer: 1 for i ≤ n, 0 for i > n) and add up across Jordan blocks.

[velvet steeple]

ah

[tacit ginkgo]

I dont see how i can pick another coefficient for x as only 1 and 5 are coprime to 6 in z6

[tough raven]

Take the reflections that you know work in the visual version and figure out the g_{a,b} that correspond to them. (You can also use the visual to directly guess the isomorphism of groups.)

[tough raven]

That's true, but you can vary the b's. The pair might be of the form g_1b and g_1b', or g_1b and g_5b', or g_5b and g_5b'.

[hidden cairn]

in question 9, how can the kernel have order p^2?

im guessing when H = Z_p^2 x Z_p or the semidirect prod Z_p^2 x Z_p, the complement Z_p is already in the kernel and the if Z_p^2 = <a>, then <a^p> is also in the kernel and that gives another Z_p to the kernel.

is this the correct way to think about this?

[tidal torrent]

this is a dumb question but does the normal subgroup test apply for all subgroups infinite/finite

[sharp ice]

idrk whats referred as the "normal subgroup test" but if its one of these definition, then it works for both
https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Subgroup

[tidal torrent]

its where a subgroup H of G is normal in G if and only if xHx^-1 is a subgroup of H for all x in G

[sharp ice]

then yeah definition 3 on the website. works infinite or finite

[tidal torrent]

since H is assumed as a subgroup this can apply for both finite/infinite

using either the finite subgroup test or otherwise one of two subgroups tests that are infinite

[coral spindle]

Proper? I think you just mean subgroup. If it were proper, then no finite normal subgroups exist.

[tidal torrent]

fixed

what are nilpotent elements of a commutative ring

there arent listed examples in my book

[vagrant zinc]

Guys, this is correct(?)

[coral spindle]

An element r is nilpotent if there exists some natural n such that r^n = 0.

[next obsidian]

No

C is not isomorphic to R^2 as a ring which is kind what you’re trying to do

But I think the easiest way to see they aren’t the same is to count prime ideals

The left has 3, 0 x R x R, R x 0 x R, and R x R x 0

And the right has 2, 0 x C and R x 0

To see that, prove that prime ideals of S x T are of the form p x T or S x q where p is a prime of S, or q is a prime of T

[mystic ether]

How do I show that in a dedekind ring, every ideal has a factorization into prime ideals?

I know that this is often the definition. Definition my textbook is using: a dedekind ring is a ring in which its set of fractional ideals forms a group.

I've already proven that in such a ring, every ideal is finitely generated

so I'm thinking it's probably induction on the number of generators

however I'm struggling to even do the base case (that is, show a principal ideal is prime)

[toxic zephyr]

so for X subset a group G. <X> is all finite words in X. but then isn't that just F(X) but with G's group structure? I heard that <X> is a quotient of F(X) can someone explain that

[mystic ether]

By definition, F(X) has no relations

but G may have relations

for example, let G = Z_8 and X = {2}

<X> is just {2, 4, 6, 8}

but F(X) is the set of all ways to write 2 + 2 + ... + 2

importantly, this addition is formal

it's not the same as the additiion in G

it's just some formal thing (you should think of it like concatenation)

[tardy hedge]

Formal 🙏 🤲 🫡

^remember

[mystic ether]

so F(X) in this case is isomorphic to Z even though <X> is isomorphic to Z_4 (it's order 4 and cyclic)

in order to make F(X) into <X>, I need to quotient out by the normal subgroup generated by the word "2 + 2 + 2 + 2"

this sets 2 + 2 + 2 + 2 = 0, like you want in Z_8

[toxic zephyr]

yeah i know this. but then where is the "quotient" here?

[mystic ether]

here

[toxic zephyr]

oh udgiasdusad sorry

[mystic ether]

np

[tardy hedge]

If you have different relations that make 0 do u just quotient by subgroup generated by all the different ones? How do decide if two expressions that make 0 are different

[toxic zephyr]

so i see this, but like... why specifically 2+2+2+2. i'm guessing because it's equivalent to 0 in the original group, but like what would it be in the general case?

[mystic ether]

in the general case, you can prove any group has a presentation in terms of generators and relations

then take the free group on the generators and quotient by the relations

the proof of this is highly nonconstructive

so there isn't really a good way to just say like "given a group what do I quotient by"

[toxic zephyr]

i see. so just saying <X> is some quotient on F(X) (which kind of "forces" the structure of G into it?) is like the nicest way to say it?

[mystic ether]

but in this case it's basically just that the relation on Z_8 is the relation generated by the word 8 dot 1. So you can look at the intersection of the (normal subgroup generated by 8 dot 1) in Z (Free group on 1 generator) and look at how this normal subgroup intersects 2Z

[tardy hedge]

What do u mean “the relation”, an equation that equal 0?

[mystic ether]

I guess a better way to say it, a free group of a subgroup is a subgroup of a free group

[mystic ether]

a word in the free group

so formally just some list of things, but you should think of it like an equation you're setting equal 0

because you quotient by the normal subgroup it generates, which is just basically quotienting by the least amount of structure to kill exactly that word

[mystic ether]

elaborating on this: if you have a presentation of your bigger group, this induces a presentation on your smaller group

[toxic zephyr]

so then another way to say it is if $\gen{X}\leq G$, and there is some normal subgroup $N$ in $F(X)$ such that

$F(X)/N\cong \gen{X}$?

[cloud walrus]

eigentaylor(got that eigenvalor)

[mystic ether]

yeah

[toxic zephyr]

okay thats really cool

[mystic ether]

yeye

unfortunately a shit ton of stuff about free groups is just straight up independent of ZFC though lol

which is kinda a funny fact

[mystic ether]

ok I still have this question though for anyone who knows anything abt dedekind rings

[toxic zephyr]

oh i guess maybe N is the kernel of the homomorphism from F(X) to G where we map a word w in F(X) to just w (in G) and "compute it" with the group operation. and then N would just be all words in X which reduce down to the identity when computed in G? that would make sense for it being like 2+2+2+2 for <2> in Z8.

[hidden cairn]

can someone help me find the isomorphisms from S to Aut(Sylow-7 subgroup) for these three cases?

[cloud walrus]

pink_panther

[hidden cairn]

not one

[stark mist]

Can I get a hint for this, to get me started? I don't immediately see how the existence of a chain of subgroups showing that a group is solvable imposes constraints on an arbitrary subgroup.

[rotund aurora]

There is a characterization that makes this trivial. ||G is solvable iff G^(n)=1 for some n, where G^(1)=[G, G], G^(n+1)=[G^(n), G^(n)]||

[warm ember]

given an abelian tower of G

think of how to construct an abelian tower of a subgroup

and a quotient group

[lime junco]

i wann amke sure i understand what this question is asking

are they asking me to prove

actually yeah

i have no idea what the bottom part is asking me to show

is it asking me to show that its the union of left cosets of the form xHx^-1 cap H?

[lime junco]

for quotient groups, remember you have a map f: G -> G/H, moreover this map is surjective

if you have a chain in G

what does this look like in G/H?

for subgroups, try intersecting the chain with your subgroup, what do you get?

[dim widget]

no it's asking you to show that double cosets are a union of left cosets (which is obvious) and that you need exactly [H:xHx^{-1} \cap H] many representatives

so any double coset HxH is a disjoint union of y_iH where y_i ranges over a set of size [H:xHx^{-1} \cap H]

for instance if x normalizes H then HxH = xH, but if x does not then you have to work out this decomposition

[lime junco]

ah ok, thanks!

[little shadow]

What does it mean for a homomorphism $\varphi \text{ to factor through a subgroup H? just that } \varphi(h) = 0 \forall h \in H?$

[cloud walrus]

Dompa

[languid trellis]

Yeah, and this implies that G/H is iso to im \varphi

assuming that H is the entire kernel

phi(x) = 0 implies x \in H

[thorn jay]

Subgroup of the domain? Or codomain

Oh its already been answered lol

My wifi is ass

[dull ginkgo]

I'm thinking about doing a small little write up about galois theory and a mildly different way of approaching it maybe as practice for my proof writing skills. Do you think the following would be good, the idea is to keep building up to the fundamental theorem using stronger and stronger conditions.

1. If we have a domain R and a ring K containing R such that K is a finite rank left R-module with a basis of left-units k_i, then End_R(K) is a rank N right K-module with the diagonal basis chi_i(b_j) = \delta_{i,j} b_i.
2. If we impose the condition that both rings are division rings, then the double centralizer of the image of R in End_Z(K) is itself. This gives a correspondence between subdivision rings P of K such that K is a finite-rank P module and finite rank right-K modules of End_Z(K)
3. When we assume K is commutative and thus a field, we have that characters from a monoid are linearly independent.
4. A result of this is that field extensions under this correspondence take the form of the crossed product of the base field and the automorphic stabilizer (the galois group)
5. If F is a field, and we have a finite group of automorphisms, then the fixed field extension has the same degree as the group due to the correspondence.
6. The induced correspondence gives the fundamental theorem of galois theory

Of course with some details filled in

[little shadow]

okay, thank you

[elfin prairie]

what is the motivation for studying structure preserving maps between groups, rings, vector spaces, etc?

and what exactly is meant by preserve here?

[languid trellis]

The motivation is that we can learn things about one group by studying another. What it means to preserve structure is to preserve the operation, let phi :G -> G' be a group hom. then the multiplication should be preserved, phi(g1 * g2) = phi(g1) * phi(g2). Notice that the second multiplication is occuing in G'.

[coral spindle]

The best motivation that I can think of is that maps that don't respect the algebraic structure are just bullshit. I can define some random function between groups, but it won't tell me anything about their relationship without some condition on how it interacts with the operations.

[elfin prairie]

so ig in a loose sense, it tends to reflect how similar say a vector space is to another distinct vector space?

[coral spindle]

There are many ways that a function might have some algebraic meaning, but this is a strong and common way

[languid trellis]

"structure preserving maps" allow us to classify all finite-dimnesional vector spaces, for example

All finite-dim F-vector spaces are just F^n for some n

[elfin prairie]

This must be some big shot result in Linear Algebra

Similar to Cayley's Theorem in some sense

okay I need to clarify smth

[coral spindle]

This is a fundamental result, it is the result that all vector spaces have a basis. It is considerably less trivial than Cayley's theorem.

[elfin wraith]

It’s partly why linear algebra is so nice, vector spaces are just well behaved and easy to classify

[coral spindle]

We have no similar result in group theory

[elfin prairie]

The condition that $\forall a_1, a_2 \in G \phi(a_1 . a_2) = \phi(a_1).\phi(a_2)$ is a sufficient condition for for $\phi$ to be a Group Homomorphism.

[cloud walrus]

James

[coral spindle]

Yes it is, that's right

[thorn jay]

There is a general result called the irredundant basis theorem, which says that the "distance" between irredundant bases must be at most one element

(An irredundant basis is a smallest set of elements which generate the group)

[coral spindle]

I am referring to the fact that there is no classification of groups.

[thorn jay]

Oh okay, apologies

I thought you were referring to the basis thing in particular

[elfin prairie]

It follows from this that $\phi(e) = e'$ and $\phi(g^{-1}) = \phi(g)^{-1}$

[cloud walrus]

James

[coral spindle]

Yes, that's right

[elfin prairie]

Now, this is what led me to infer this statement

I referred to Vector Spaces here but the idea is the same, I suppose

[coral spindle]

Seems like a big jump to me, but if that helps you conceptualise things, more power to you.

[elfin wraith]

It’s not necessarily similarity, but it does tell you that your map sends a vector space to a vector space. It’s entirely possibly to just map out of a vector space in such a way that you lose, say, the abelian group structure on the image, but you’re not entirely wrong

If that map happens to be a bijection then it’s what we call an isomorphism and that is pretty much the idea of equality in algebraic settings

[elfin prairie]

I see

[elfin wraith]

Like it’s not entirely wrong to say the image of the map is similar, it is in some sense, but I don’t think it’s quite the correct takeaway

It just tells you that the map preserves the algebraic structure on the set, that’s it

[elfin prairie]

Is there further classification of these maps based on what kind of map (injective, surjective, bijective) it is?

[hot wadi]

Smallest as in minimal under inclusion or smallest as in achieves the minimum cardinality among all generating sets?

[elfin wraith]

They all have names, a surjective homomorphism is called and epimorphism for example, and different properties but by and large homomorphisms and isomorphisms are the interesting ones

[coral spindle]

What do you mean by that exactly, are you just asking for special names or are you expecting a genuine classification of homomorphisms?

[elfin prairie]

I suppose the latter

[coral spindle]

There is no such thing in general

[elfin wraith]

You can do it for vector spaces though, which is why LA is so nice

[hot wadi]

(If you’re working with rings then not all epimorphisms are surjections)

[elfin wraith]

For example, vector space homomorphism R^2->R^2 are just 2x2 matrices, but you can’t classify all group homomorphisms R^2->R^2

[coral spindle]

I mean, you sort of can if you take for granted a Q-basis of R...

But you certainly can't write these things down

[elfin wraith]

In the categorical setting no, but I was meaning in just the algebraic sense

[elfin prairie]

I can't quite conceptualize this. Can you expand on the reasons as to why we can't do so?

granted I am not as seasoned in Linear Algebra

[coral spindle]

The long and short of it is that R is too big.

[elfin prairie]

I thought so

[thorn jay]

Smallest under inclusion. As in a subset B of G is an irredundant basis if every subset B' of B doesn't generate G but B does

[hot wadi]

Then I don’t understand this theorem

For example

Sn has irredundent basis given by the n-1 adjacent transpositions

And another basis given by a transposition and an n cycle

How are these “distance 1”?

[thorn jay]

It just means that there are also minimal generating sets of cardinality n-2, n-3, ..., 2

Thats what i meant

Sorry

[wild jasper]

Let $n=p_1^{a_1}p_2^{a_2}$ then there integers $u,v$ such that $u p_1^{a_1}+vp_2^{a_2}=1$ how does deviding by $n$ show that a $n$'th root of unity $\zeta_n$ is the product of powers of $\zeta_{p_1^{a_1}}$ and $\zeta_{p_2^{a_2}}$?

[cloud walrus]

mh_le

[hot wadi]

I see, okay. That makes a lot more sense than what I had in mind

[hot wadi]

The idea is that you raise zeta_n to each side

Or, if you want to divide by n first, raise 1 to each side

[wild jasper]

Raise?

[thorn jay]

:3 still a really cool theorem imo

[tardy hedge]

What is the group theory that i need to know for this assignment

2,3,4 specifically

[toxic zephyr]

in class we're proving all finite groups are iso to a subgroup of Sn. we're using that Aut (G) is iso to Sn if G has n elements. I get that all possible bijections from G to G would be Sn, but aren't automorphisms homomorphisms? is it true that a bijection on a group is necessarily a homomorphism?

[coral spindle]

automorphisms are homomorphisms

is it true that a bijection on a group is necessarily a homomorphism?
No.

Exercise: show that the bijections used in the theorem your proved are not homomorphisms (except for the identity map)

[toxic zephyr]

okay turns out AutG was referring to G as the set in the group action

and Aut of any finite set X is definitely just S|X|

[coral spindle]

Understandable that this confused you btw, I think this is poor choice of notation. People more often write Sym(G)

[wild jasper]

@hot wadi I don't understand what you mean by "raise".

[hot wadi]

@wild jasper I mean, if I have two numbers a and b, I can “raise a to the b power” to get a^b

So what I was saying earlier is to raise zeta_n to the [blah] power where [blah] is each side of the equation given

And then use that those two powers of zeta_n are equal to get what you want

[hot wadi]

2 and 4 require very little - lagrange’s theorem and knowing the definitions should be enough. 3 I don’t immediately see how to do, so I’m not sure what is or isn’t needed for it

The fact that 3 is specifically for finitely generated abelian groups seems to imply they would want you to use the structure theorem for finitely generated abelian groups

Which would of course make 3 fairly easy

There could very well be a less machinery-heavy proof though that the prof is hoping you come up with instead

[wild jasper]

@hot wadi thanks 🙂

[dense terrace]

book wants me to show that $xy=x+y+xy,\bR\setminus {-1}$ is abelian group. i did fine for showing associativity and identity, but im struggling with inverse:
\begin{align}
&xx'=e \implies xx'=0\
&\implies x+x'+xx'=0\
&\implies x'+xx'=-x\
&\implies x'(1+x)=-x\
&\implies x'=-x\
\end{align*}
(i got 0 for the identity)
but this doesn't work for x = 1 since we don't have -1
Can anyone give me a hint?

[cloud walrus]

Tristian

[grave sedge]

Your last step is wrong (you should have divided by 1+x)

[dense terrace]

Ahhh

Algebra mistake

Thanks haha

[placid shale]

Hello, I am camping here until I get my active role back, feel free to ask me questions in group or field theory to help me achieve my goal!

[placid shale]

i am struggling to get the meaning of the "=" in

x*y = x+y+xy

[coral spindle]

It is defining the operation *.

[placid shale]

oh yea I though * is mutiplication and that's some kind of a homomorphism

[bitter rover]

Even correcting for the mistake in the last line, this doesn't quite prove what you want.

It proves that if x has an inverse x' under this operation then x' = -x/(1 + x). It doesn't prove that -x/(1 + x) is the inverse of x.

You can either make all those implications bidirectional (and justify that), or you can say: "If x != -1 then -x/(1 + x) is defined and x * -x/(1 + x) = ...", simplifying until you show their "product" (under this operation) is 0.

[stark mist]

Thanks, I think I get the idea but I was struggling for a while to get the intuition behind what happens if the intersections are empty (except with the entire group). In that case, the subgroup is forced to be abelian somehow (edit: by 2nd iso I think?). I guess i need to work thru some examples.

[bitter rover]

Do you know all the isomorphism theorems?

[stark mist]

yep

I see, I should just apply the 2nd iso I guess

well it doesn't really matter whether there's only one intersection

[bitter rover]

They always get numbered differently, but there's one that says if $H,N$ are normal subgroups of $G$ and $N \subseteq H$ then $H/N$ is a normal subgroup of $G/N$ and $$\frac{G/N}{H/N} \cong \frac{G}{H}$$

[cloud walrus]

Cufflink

[dense terrace]

yeah, i did, thank you though

[stark mist]

But I was thinking using the $AB/B \cong A/(A \cap B)$ is more relevant for showing subgroup is solvable (for the part to show that the quotients of the intersections are abelian). I think I'll need the one you mention for quotient groups?

[cloud walrus]

tubelight

[grizzled spindle]

If I take the quotient of a group G twice, so let’s say (G/H)/F, is there a normal subgroup N s.t. (G/H)/F = G/N ?

When I write =, i mean isomorphic ofc

[coral spindle]

Yes

[grizzled spindle]

I’m asking this because this would show that the quotient group of a finitely generated group is finitely generated

[coral spindle]

Hint for finding this normal subgroup: there is a map G -> G/H and G/H -> (G/H)/F, so compose the two and calculate the kernel

[grizzled spindle]

Ohh right yeah that makes sense

[coral spindle]

This will allow you to characterise all normal subgroups of G/H

[bitter rover]

All roads lead to Rome

[tardy hedge]

Thanks for the reply man.

[tender linden]

not sure where to ask this but i am reading this article an I am a bit confused about the wording here: https://www.jstor.org/stable/3072368

what does "restricts to the identity map" mean?

$[a][b]$ denotes an element of $E$ where $a \in \mathcal{T}$ and $b \in \mathcal{O}$ are two group elements

[cloud walrus]

artemetra

[lucid shadow]

So I am recently studying about prime decomposition in number rings.
How should I compute the prime P in Z[ζ_8] such that P^4=(2)?
(My convention is to use ζ_n to represent e^(2πi/n). )

[rocky cloak]

It means that if you restrict the map to T it becomes the identity map on T.

In other words, for t in T we have phi(t) = t.

[tender linden]

why does it say phi([a][b]) = [a_0][b] for some a_0 then? i'm just confused what exactly is different from what they are talking about T and O, for which they just say that it's an identity map

[lucid shadow]

My guess is, T is embedded in E as those elements in the shape of [a][0]

[rocky cloak]

So I guess the elements of E are identified with these pairs [a][b] and the operation is described in terms of some cocycle?

Either way [a][b] is not necessarily in T, so all you're guaranteed is that b is unchanged as that's what it would mean to induce the identity on the quotient.

[rocky cloak]

So I guess you want start with finding the primes that contain 2.

A clever shift in perspective could be to consider the primes in Z[zeta8]/(2) = (Z/2)[X]/(X^4 + 1)

[lucid shadow]

φ([a][0])={a}{0} (Using different brackets for E and E')

[lucid shadow]

I'm listening

[rocky cloak]

Alright, so you're following? Or no?

[lucid shadow]

I can follow

So it is equal to (Z/2)[X]/((X+1)^4)

[rocky cloak]

Indeed

[lucid shadow]

Ah I understand now

[golden turtle]

What is the induced group homomorphism? I tried writing it out following the proof, and like, I just don't understand what they're getting at

we want a homomorphism from ker(dn+1)/Im(dn) to ker(d'n+1)/Im(d'n)

[lucid shadow]

I wonder if this helps

Chain complex or cochain complex doesn't matter much here

Or you may try to use the functoriality of Ker and Im and use category theoretical languages to prove the result

[golden turtle]

We have not done category theory unfortunately

[regal zodiac]

If i consider the following extension Q in Q(a+ib) then is it always true that a-ib is in Q(a+ib)?

[lucid shadow]

Choose one complex cube root of 2, name it as α
Then α's conjugate is not in Q(α)

[rocky cloak]

a - ib = (a^2 + b^2) / (a + ib)

So if a^2 + b^2 is rational that works

[lucid shadow]

yeah and unfortunately in my example it isn't rational

[regal zodiac]

I think the induced homomorphism G_n is just defined by G_n([a]) = [ alpha_n(a)].
Using commutativity of the diagram, it should be easy to check that alpha_n sends ker(dn+1) in ker(d'n+1) and im(dn) to im(d'n)

[regal zodiac]

Yes, thank u so much

Maybe this is what you are looking for @golden turtle

[golden turtle]

Yes, thank you!

[regal zodiac]

Np good luck , not an easy topic

[grizzled spindle]

I was wondering if there was a term for the following properties :
S is a subset of G, for all subset K of S we have that <K> and <K^c> have trivial intersection.
S is a subset of G, for all s in S we have that s is not in <S-{s}>.

Note that the first property is stronger than the second one

When I write K^c, I mean the complement of K in S if that wasn’t clear

[vagrant zinc]

I haven't seen them, honestly.

Hi, could someone recommend me a complimentary book? I'm using the dummit, I know it's good to have a complementary book.

[hot wadi]

I learned and TAed abstract algebra with Silverman’s An integrated approach and I thought it was good

I know there are other good ones out there too, but I don’t know what they are

[lucid shadow]

You write <K> for the subgroup of G generated by K right?

[grizzled spindle]

yeah i think thats standard notation

i just came across these properties when looking at generating sets of groups so i was wondering if there was any work done with these things

its quite tricky, because if <S> is a proper subgroup, adding an element not in <S> to S will not always keep the properties above

[lucid shadow]

I will use "S is a minimal generating set of <S>"

[grizzled spindle]

im not a big fan of calling it minimal, maybe independant is a better word for it

because it is not always minimal in term of the size of S

for example if you take S_n, {(12), (23), ..., (n-1 n)} and {(12), (23...n)} both generate S_n and satisfy the properties

but one has 2 elements and the other has n-1 elements

anyway, both of these properties are strong but they are not well behaved at all from what i can tell

[lucid shadow]

hmm
minimal as in "minimal under the usual "subset" partial order relation" i think
not the least

[grizzled spindle]

right

[vagrant zinc]

[tidal torrent]

how does perfect groups work

isnt a field called perfect if field F has a charcteristic of 0

[hot wadi]

A field is perfect is every polynomial is separable

A polynomial is separable if each of its irreducible factors is squarefree (meaning all roots over the algebraic closure are distinct)

All char 0 fields are perfect (exercise), but so are many others, including all finite fields (exercise)

[warm ember]

every irreducible poly is separable*

[hot wadi]

My Galois theory textbook might’ve had a weird/nonstandard definition of separable

But the pair of defs I gave above correctly characterizes perfection

[tidal torrent]

oh okay

i thought perfect groups have similar properties to those of perfect fields

but all i know is that perfect groups have commutator subgroups of this and that

i dont really understand much about these (they were not mentioned in my book)

[tidal torrent]

my point is how would i find the smallest possible finite perfect group

[bitter rover]

Well, the smallest perfect group is the trivial group.

The smallest non-trivial perfect group is A_5, but there aren't any short proofs I know that don't involve non-trivial theorems like Burnside's theorem.

[rapid junco]

how does one prove that this is a subgroup?

1 \in H' is clear

if a, b \in H' then abH = aHbH = HaHbH = HabH

if a \in H' then a^{-1}H = what?

I dont see this implication at all.

[mighty kiln]

H' is the normalizer of H

That is, H' = {x : xHx^-1 = H}

[rapid junco]

so then

this is what I want

[plucky dome]

ur done already no? if b is an inverse of some other element then its closed under inversion

[rapid junco]

no because you can't conclude that b^{-1}H = Hb^{-1}H

but HcHc^{-1} = Hcc^{-1} right

by definition

if so, then I am done.

I think not though

coset multiplication is defined by this, but im not sure if I am supposed to use coset multiplication here.

[plucky dome]

why not?

H' is nonempty, just suppose you have a elements a,b^-1 in H' and then show with that argument ab^-1 is in H'

[rapid junco]

well b^{-1} is not always in X this is what I am trying to prove

[plucky dome]

then it passes the subgroup test and ur done

[rapid junco]

....

but b^{-1} doesn't nessecarily have this property in the proof above.

this is what I am trying to prove.

[chilly ocean]

Define Torsion submodule M_tor to be the subset of elements x in M such that there exists a in A, a≠0 such that ax = 0.

Does I need ring A to be commutative?

[rapid junco]

Hb^{-1}H = (HbH)^{-1} = (bH)^{-1} = Hb^{-1}

One can do this, but again now im stuck on b^{-1}H

[plucky dome]

but it's in H'

[rapid junco]

but its not

thats what im trying to prove

b is in H'

I am trying to prove that H' is a subgroup.

[chilly ocean]

Actually arki point is good here, you can try that one

Is H subgroup or any subset?

[rapid junco]

H is a subgroup

[chilly ocean]

So you have a problem to show that if a in H' then a^-1 also in H'

[rapid junco]

yes.

[chilly ocean]

Go by definition

[rapid junco]

This for some reason I can't figure out for the life of me

what do you write a^{-1}H as?

[rapid junco]

like this right here works

but H is not normal always.

[chilly ocean]

aH = HaH.

Now you have to show a^-1H = Ha^-1H

Let x in a^-1H so x = a^-1h => a = hx^-1.

Now aH = HaH so for h^-1ah_2 = ah_3

Now it gives you x^-1h_2 = ah_3 => x = h_2h_3^-1a^-1.

Now h_2h_3^-1a^-1 is in Ha^-1H, so x in Ha^-1H

It shows that a^-1H\subset Ha^-1H

[rapid junco]

Okay ill figure out the reverse inclusion

Is there not a simple way to write this with equality of sets?

[chilly ocean]

HaH = {h_1ah_2 | h_1,h_2 in H }, so I took h_1 is h^-1 and h_2 is h_2 so there exists h_3 such that h^-1ah_2 = ah_3

[chilly ocean]

Maybe the normializer idea works here

[rapid junco]

hmm ok

thanks for the help

[rapid junco]

"Now h_2h_3^-1a^-1 is in Ha^-1H, so x in Ha^-1H" why?

this is an element of Ha^{-1}

oh yeah

nvm thats a subset of Ha^{-1}H

okay

[chilly ocean]

Yes

[chilly ocean]

And A has to be an integral domain

[chilly ocean]

I don't get the exact meaning of R-algebra, here internal law of composition refers to bilinear form ?

[rapid junco]

no it just gives your vectors in A a multiplication

Like for example C(K) for K compact is a ring of continuous functions and becomes a C(K)-algebra with multiplcation (fg)(x) = f(x)g(x).

[chilly ocean]

Why do I need K to be compact ?

[chilly ocean]

So it just gives the new operation on A?

[rapid junco]

No K being compact is just a nice space

Take more generally the set of functions from A to B and add a ring structure to it in the natural way.

[chilly ocean]

Okay

So here is a multiplication defined on Mat(R) as ordinary matrix multiplication?

[rapid junco]

Yeah exactly

[chilly ocean]

If I remove the condition that f(s) = 0 for all but finitely many, still Map(S,R) is an R-module ?

I think yes

[grave sedge]

Yes

[chilly ocean]

When I can say A is associative R-algebra?

[alpine island]

so that this is a nicer structer analytically

for non-compact K, you can't get the topology induced by the sup norm

[rocky cloak]

When it's an R-algebra such that
(ab)c = a(bc)

[chilly ocean]

a,b,c from A, right?

[alpine island]

Are non-associative algebras ever used?

[delicate bloom]

does cross product count

[alpine island]

I guess, but the contexts in which cross product is used, nobody knows what an algebra over a field is

[delicate bloom]

yeah I wouldn't count it personally either lol

hopefully someone comes up with a better example than that 🤣

[rocky cloak]

Lie algebras come up a lot.

Other than that, people do study other types of non associative algebras and just study them in general. But it's much more niche

[long swan]

actually i don't think that's a bad example, isn't it the bracket of the Lie algebra of SO(3)?

[rocky cloak]

It's isomorphic at least

[alpine island]

Oh, true. You can think of the Jacobi identity as a generalized associativity though.

[rocky cloak]

Yeah, and there are a few other weak forms of associativity, like Jordan algebras or alternative algebras.

And then you have higher algebra stuff, where things are only associative up to homotopy

[mighty kiln]

Isn't the Jacobi identity like "infinitesimalized" associativity

[alpine island]

Higher!

[alpine island]

Maybe. I really haven’t done lie algebras yet (just no courses on it here I think) so I don’t have a lot of intuition. I just know the def.

[rocky cloak]

I don't think so. Like it's not connecting to associativity in the lie group in a similar way to how the bracket is connected to commutators.

For me the main take way is that [[X, Y], -] is the commutator of [X, -] and [Y, -]

[long swan]

oh interesting

then is it like the induced maps in homology are "truly" associative or something

i guess there's a pretty standard way to force associativity for all these examples

[south patrol]

Yeah

Or more strongly in the homotopy category stuff is truly associative

[long swan]

Ah nice

Tbh I wanna learn more about this after going through the standard homological algebra stuff

[long swan]

Unrelated to what you replied to but I remember seeing that there’s an associator bracket idk how commonly that shows up

Like [x,y,z] = x(yz) - (xy)z

[chilly ocean]

In Q/Z, for any positive integer n there are exactly phi(n) element of order n, correct?

[thorn jay]

Thats often though, its a coping method because complete nonassociativity makes you cry.
When working with loops/quasigroups people often just consider special associativity conditions like alternativity or something

[coral spindle]

Indeed, since the set of elements whose order divides n is iso to Z/nZ

(and is a subgroup, ofc)

[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

[tardy hedge]

FEIN FEIN FEIN FEIN

[chilly ocean]

let M be R-module and M is simple then i have to prove that M is generated by every non-zero element of M.
But if R is not unitary, then there is a possibility that there exist non-zero m such that rm = 0 for all r in R.

does i need to assume R is unitary ring?

maybe i am wrong here, submodule generated by m is {nm + rm | n in Z and r in R}

[coral spindle]

Well no this is incorrect if R is not unital

Remember the definition of a submodule generated by some set is the smallest submodule containing those elements

So by definition <m> must contain m

I can't think of a correct characterisation of this off the top of my head right now

[chilly ocean]

so what is the structure of <m> ?

[chilly ocean]

yes then i can conclude given statement, thank you

[candid dove]

it's some positive integer times m in the sum too

Iirc

[coral spindle]

Yeah I think it's something like this

[candid dove]

Should be there in dummit rings chapter

[coral spindle]

I would imagine {n.m + rm | n in Z, r in R}

But I have not checked this thorougly

[chilly ocean]

isn't?

[coral spindle]

Of course, you can canonically make R into a unital ring and be sane like the rest of the algebraic world

[candid dove]

Nah

[coral spindle]

Oh..................................................................................................... lmao I misread

[candid dove]

Guys if a ring R is commutative and semisimple then it's automatically artin right

Ofc my ring is unital 😛

[coral spindle]

Every semisimple ring is Artinian I thought

Unless you mean something else by Artin

[candid dove]

Oh I need it to be an artin ring cause an artin ring is the product of artin local rings?

[coral spindle]

I think we might be talking past each other

Not sure what you mean by Artin if not Artinian

[candid dove]

I do mean artinian

Probably

Yeah

So a product of fields is always artinian right

[coral spindle]

Yeah, the ideal structure is very easy to describe in that case

[candid dove]

Right

[coral spindle]

There are only finitely many ideals

[candid dove]

Sweet

[coral spindle]

4 in fact

[candid dove]

But I don't understand one thing

Why is there a if further R is artinian line here

[coral spindle]

I fear your textbook may have a special definition of semisimplicity

[candid dove]

Like

By semisimplicity i mean it's like internal direct sum of ideals

Simple

Ideals

Uhh

[coral spindle]

Yeah I don't get it either. Hopefully someone who is more fluent in this stuff can comment

If I speak the name of jagr he shall appear

[candid dove]

Dumb question but like if a ring is commutative and is also left artinian then that makes it artinian right?

Or probably not

Which would be weird

But it would be artinian

[tough raven]

I believe that in some sources semisimple is defined as Jacobson radical = 0, and it's a theorem that for a (unital) ring, "regular module is direct sum of simple modules" is equivalent to "Jacobson radical 0 + Artinian".

[dull ginkgo]

then show commutative artinian rings are products of local rings

[tough raven]

Not 100% sure about that though.

[tough raven]

Wait what

[dull ginkgo]

Yeah pretty much “artinian” as a term means no infinite descending chains. You can slap it on left ideals, right ideals, two sided, subrings, what have you. Just depends on what the chains are and when the property occurs

So yeah left artinian implies artinian for commutative rings

[tough raven]

This definition definitely satisfies left semisimple ⇔ right semisimple (non-obviously) and (left) semisimple ⇒ (left) Artinian (somewhat more obviously - R is fg, so R is a finite direct sum of simple modules which are Artinian, so R is Artinian).

[candid dove]

We also have an equivalence iR is semisimple iff R is left artinian and rad(R)=0

[candid dove]

i already have the result with me

[tough raven]

My very tentative guess is the author originally learnt this with the Artinian hypothesis and didn't want to check all the proofs to see if it was needed anywhere.

[tough raven]

Oh now I get it

[candid dove]

Perhaps

[tough raven]

Theorem (1)A: (i) ⇔ (ii). Theorem (1)B: if R is Artinian, (i) ⇔ (ii) ⇔ (iii).

Read it like this.

[dull ginkgo]

Chinese remainder theorem moment

[tough raven]

Also I wish they said finite product of fields.

[candid dove]

But if R is semisimple that automatically makes it artinian

Oh

Wait

I get it

Okay

Okay

[dull ginkgo]

@tough raven hey so what do you think of this?

Sorry for the ping

[rocky cloak]

I feel like it would have been clearer to have point (iii) be "R is artinian and has no nilpotent elements"

[candid dove]

So in (iii) we can't remove the R is artinian hypothesis can we

[dull ginkgo]

My goal was to kind of break down Galois theory to it’s bare bones and see how the conditions of a Field build up to the correspondence

[rocky cloak]

No you can't remove it, but it's automatic in (i) and (ii) (assuming product means finite product), so kind of a weird way to phrase it

[candid dove]

Okay thanks evrynyan (everyone)

[tough raven]

Re 2., you need to show that the double centraliser of a finite-K-rank subring of End_Z(K) is itself (i.e., that all of them are matched to a P by the correspondence). Re 3., I think you mean that the corresponding ring End_K(L) is a crossed product of L with the Galois group for L/K?

[dull ginkgo]

Yeah sorry, I pretty informally just typed that out in one of my classes

[tough raven]

Otherwise, this looks good mathematically IG?

Pedagogically I would combine 1. and 2., because I think for anyone who hasn't seen any of this it's an unnecessary level of generality (since one doesn't go on to say much about the general setting, just immediately specialises to division rings). But tha depends on your reasons for writing this.

[dull ginkgo]

Fair enough

In retrospect I don’t think the double centralizer part actually depends on the subring being a division ring but eh

[dull ginkgo]

Another thing I wanted to try and find is a left artinian ring R, and a left semisimple module M such that the double centralizer of R’s image in end_Z(M) is NOT the image

[tough raven]

It is in fact true for any free module over any ring, IIRC.

[dull ginkgo]

The simple case does satisfy double centralizer as a very simple corollary of Jacobson density

[tough raven]

Hmm, for any semisimple module M we can write

M = (+)_V V (⨯) Hom(V, M)

[dull ginkgo]

I’m not sure for every ring

Maybe division / left-simple rings?

[tough raven]

End(M) = \Prod_V End_{D_V}(Hom(V, M))

[dull ginkgo]

The problem is those maps

The ones sending a basis to another

*basis vector/element

I don’t think they commute, is what I tried showing

Those maps do generate End_R(M) as a right-R module

[tough raven]

End_{End(M)}(M) = ∏_{V appears in M} End_{D_V}(V)

Which IG is equal to the image of R if M is a finite direct sum of simple modules V, each of which is finite-dimensional over the division ring D_V := End(V)

[tough raven]

So for example this should be true for M and R finite-dimensional over a field.

[dull ginkgo]

Is this true?

[tough raven]

Yep.

[dull ginkgo]

hmmm

[dull ginkgo]

are we talking about division rings here

Because otherwise it doesn’t seem right to me

[dull ginkgo]

Consider M = R^N. Then End_R(M) ~= M_N(R^op)

Which isn’t iso to (R^op)^N

Oh hom(V,M)

[tough raven]

Any semisimple module over any ring.

[tough raven]

up to thi

[rocky cloak]

Since the Jacobson radical annihilates all semi-simple modules you can wlog assume R is semi-simple here.

[dull ginkgo]

Ahh, thanks

I didn’t think about quotienting out the radical

I’ll come back to this after I finish my analysis work

Thank you, all

[chilly ocean]

I don't get what is meant by the diagram of Z-module and Z-morphism here ?

[bitter rover]

ℤ is a ℤ-module

[chilly ocean]

Yes then?

[bitter rover]

There's no ℤ-morphism which makes that diagram commute.

Or are you asking what a diagram is?

https://en.wikipedia.org/wiki/Commutative_diagram

[gusty thistle]

Suppose I have an ideal $(ab,f_1, \dots,f_l)$ of $k[x_1,\dots,x_n]$ where $k$ is some field. I believe there is some sort of condition under which you can say that $(ab,f_1, \dots,f_l) = (a,f_1, \dots,f_l) \cap (b,f_1, \dots,f_l)$. Does anyone remeber what it is? I can't find it.

[cloud walrus]

stable compass needle

[chilly ocean]

No but I don't understand why they take Z-morphism and Z-module ? I don't get the point here

[velvet steeple]

Is there any chance this is wrong?

And it should be N_k relative to N_(k - 2)

The idea should be that we first find a basis of N_k relative to N_(k - 1), then of N_k relative to N_(k - 2) etc. until we are at a basis of N_k

Actually, I guess it is right like this, but what I said is correct too, so f_1, ..., f_t together with B(e_1), ..., B(e_s) form a basis of N_(k - 1) relative to N_(k - 2) but together with e_1, ..., e_s a basis of N_k relative to N_(k - 2)

[tough raven]

I don't think it is wrong. B(e_i) and f_j form a basis of N_{k-1}/N_{k-2} and e_i form a basis of N_k/N_{k-1}; so e_i, B(e_i), f_j form a basis of N_k / N_{k-2}.

[tough raven]

Yep.

[velvet steeple]

Ok, so we just meant different things

Thanks!

[velvet steeple]

And we end this process when we arrive at N_0 = Ker(A - lambda id)^0, right?

I mean my way of thinking about it, finding a basis of N_k relative to N_k - 1, then of N_k relative to N_k - 2 etc.

When we are at N_k relative to {0} we are done

And we get that when we are at relative to N_0, right?

[tough raven]

N_0 = ker(A - l Id)^0 = ker(Id) = 0, in fact, so yes.

[velvet steeple]

Thanks

[chilly ocean]

Anyone?

Why are they taking two different structures of morphism at one time ?

Are they taking identity morphism as Z-morphism?

Got it

[hidden cairn]

i cant see the 6 homomorphisms from Z_4 to Aut(Z_15) = Z_8 in part c

can someone help

[tough raven]

Aut(ℤ_15) = (ℤ/15ℤ)^⨯ = (ℤ/3ℤ⨯ℤ/5ℤ)^⨯ = (ℤ/3ℤ)^⨯ ⨯ (ℤ/5ℤ)^⨯ = ℤ_2 ⨯ ℤ_4, not ℤ_8.

[hidden cairn]

ah

[tough raven]

In fact, there are only four homomorphisms (3 up to isomorphism) from ℤ_4 to ℤ_8.

[hidden cairn]

ah ok ok lemme try again

thanks

[chilly ocean]

I want to prove this one by myself, if there is a way to go step by step then please let me know

[wraith cargo]

it's a diagram chase
for (a) you pick an element b of B' and chase around the diagram to find an element of B mapping to b and you have the case when b is in ker(g') and when b isn't in there

for (b) you have to show that any element in ker(gamma) is 0

also by chasing around the diagram

[chilly ocean]

Let first case, b in ker(g') implies b in Im f' implies f'(a') = b, since \alpha is epimorphism thus \alpha(a ) = a', then...

g(f(a)) = 0

Do I need to construct a new morphism?

[mighty kiln]

If you take a random magma its commutator bracket won't satisfy the Jacobi identity

But conversely if you take any Lie algebra the bracket is the commutator in its universal enveloping algebra

(wait what did I mean by infinitesimalized)

[chilly ocean]

I got this case

Now for the case, y not in ker(g')

g'(y) ≠ 0 and h'(g'(y)) = 0 and gamma is epimorphism therefore \gamma(c) = g'(y)

So it implies that \delta (h(c)) = 0

But delta is monomorphism

h(c) = 0

g(b) = c for some b in B

g'(\beta(b) ) = \gamma(g(b)) = g'(y)

Got it thank you @wraith cargo

[drowsy oxide]

@serene dune @cloud walrus is here too

friendly reminder that not every matrix in O(n) has finite order

[serene dune]

yeah that's crazy

[drowsy oxide]

$\begin{pmatrix} \cos1 & -\sin1 \ \sin1 & \cos1\end{pmatrix}^n=\begin{pmatrix} \cos n & -\sin n \ \sin n & \cos n\end{pmatrix}$

[cloud walrus]

kitty queen

[serene dune]

so not all O(n) is unipotent

idk if it can be classified with classic groups or not

[rocky cloak]

At least SO(2) is just Q/Z x Q^N

For bigger ns I guess the noncommutativity makes it more complicated

[serene dune]

eigenvalues of O(2)\SO(2) is only +1 ?

[rocky cloak]

The eigenvalues of matricies in O(2) \ SO(2) are ±1, if that's what you're asking

[serene dune]

oh i was realy trying to classify the unipotent maps

[serene dune]

$G=\left{e^{i\theta},\middle|,\theta\in\Bbb Q\right}$. Then $G$ is not a closed subset of $S^1$, and therefore it is not compact.

[cloud walrus]

yeshua

[serene dune]

idk if its valueable but a tangent

[rocky cloak]

You mean the torsion elements of O(2)?

[serene dune]

haha, i was trying to reach these sort of statement but i had no information

i have to learn about torsion, is that finite order related ?

[rocky cloak]

Yeah, it just means to have finite order

[serene dune]

right!

[rocky cloak]

unipotent typically means that x-1 is nilpotent

So that's different

[serene dune]

also i was trying to classify the elements outside of O(n) which was looking scary

[rocky cloak]

I think it's probably easier if you don't restrict yourself to O(n)

[serene dune]

wiki says from the characteristics polynomial we can see all the eignevalues should be +1

[serene dune]

that was sort of my naive conclusion

[rocky cloak]

Then I think your looking at the wikipage for unipotent matrices

[serene dune]

https://en.wikipedia.org/wiki/Unipotent

[rocky cloak]

Right, but is that what you wanted to classify?

[serene dune]

i started from there

[rocky cloak]

Okay, I guess I was just confused by the context

[serene dune]

well i wasnt super precise either, i just saw some properties related to it and wanted to classify if thats possible

i once posted this but i thought maybe i was being super dumb

[weary frost]

why isn't the finite order subgroup of SO(2) just Q/Z?

[rocky cloak]

It is

[alpine island]

oh nice pfp by the way

[weary frost]

ohh but why then Q^N?

[rocky cloak]

That's the elements that don't have finite order

All the irrational rotations

[weary frost]

is there an easy to see that it needs to be Q^N?

[rocky cloak]

Don't know if you'd call it easy, but

exp(2pi x): R -> SO(2)

shows SO(2) = R/Z.

R is a vector space over Q, and you can argue by cardinality reasons that R and Q^N have the same dimension. Hence R = Q^N = Q x Q^N

Then R/Z = Q/Z x Q^N

= means "is isomorphic to" everywhere here

[weary frost]

oh that is actually really easy

[serene dune]

is it lie related stuff ?

[rocky cloak]

I mean, the exponential map is there

But otherwise not really

[serene dune]

not being in school is taking a toll on me rn

[weary frost]

if i understand correctly in R = Q^N = Q x Q^N the first Q is generated by the basis vector "1"?

[serene dune]

sumptuous observation 🪷

[rocky cloak]

Yeah that's right

[weary frost]

thx

[elfin wraith]

This is part of an obscenely long problem so I can provide more context if needed but anyway,
\\
Let $R = \frac{k\langle x,y\rangle}{(xy-qyx)}$ for $q \in k^\times$ and $q$ not a root of unity. Let $P$ be a prime ideal of $R$. Ive shown that $y^mx^n \in P$ for some $n,m$ and im now trying to show that one of $x$ or $y$ is in $P$. We have that $y^mx^nR \subseteq P$ since $P$ is an ideal and then i think that we can just pull $x^n$ through $R$ as we can just commute it past any element of $R$ and pick up some like power of $q$ which will still just multiply to be in $R$ so we have that $y^mx^nR = y^mRx^n \subseteq P$
\\
Then since $P$ is prime we know one of $x^n$ and $y^m$ are in $P$ but im not seeing how to get from there to $x$ or $y$ is in $P$. Im guessing its just decomposing it in someway thats probably quite obvious but im not quite seeing it

[cloud walrus]

Nope

[serene dune]

well u can take out countable copies of Q, no ?
but i didnt get the basis argument

[weary frost]

why we take 1 as a basis vector?

or what exactly

[serene dune]

or jagr mentioned vector space, i missed that point

[cloud walrus]

yeshua

[rocky cloak]

Well, by the same argument you did, if x^n is in P, then
xRx^n-1 is in P. Then just induction your way

[elfin wraith]

Oh yeah jesus thats obvious thanks jagr

I will actually ask an intelligent question here at some point, i promise I can actually do algebra

[dull ginkgo]

Better than my babble

[elfin wraith]

With that ive finally finished a page long lemma I need for the actual problem im trying to do, ive got plenty of babble

[bitter meteor]

hi, would anyone be kinda willing to tutor me through group theory using this example question ...?

[wraith nexus]

do vii first after i

[elfin wraith]

Is anyone able to help at all with Weyl algebra problems? I genuinely just dont know how to work with any higher terms than the first, like ive got the problem,

Let $n$ be a positive integer and consider the $n^{\text{th}}$ Weyl algebra $A_n(k)$. Recall that $x^\alpha y^\beta$ with $\alpha,\beta \in \mathbb{N}^n$ is the standard basis for $A_n(k)$. Give formulae in terms of the basis for,
[x_ix^\alpha y^\beta - x^\alpha y^\beta x_i \text{ and } y_i x^\alpha y^\beta - x^\alpha y^\beta y_i.]

[cloud walrus]

Nope

[elfin wraith]

and i just legitimately dont know how to work with this, its just kinda too big, I really cant wrap my head around it.

The end goal is to show that A_n(k) is simple, and this is like easy enough for A_1(k) because you just consider a submodule, take an element of degree n g, then do like 1/(lamba n!) y^n \cdot g = 1 so you generate all of A_1(k), and obviously I get that the idea will be the same for A_n(k) but i just cant wrap my head around the computations in there

[void cosmos]

try asking in #advanced-algebra

[restive birch]

what progress have you made/what part are you on

[bitter meteor]

stuck on the first proof for i) 😢

[void cosmos]

do you know what you have to show

?

[bitter meteor]

just closure, associativity, identity and iunverse?

[serene dune]

first 3 are simple and can be shown in general for every element in mod n

for inverse, try to find out for each element, hopefully u will see a pattern, generalise it for arbitrary element in the group

[hidden cairn]

for part a, i think a=x, b=x^4y works and in part b if we square these we get a^2 = x^2 which generates Z_4 and b^2 = y^2 which generates Z_2. can someone point out the problem here

does <x> x <x^4y> generate G? im very confused

[serene dune]

well i thought u need to figure out order 8, and 4 elements and take their intersection

[gusty tangle]

I'm trying to find the inverse for (i) in sasha's problems, I'm reviewing group theory

[serene dune]

oh what have you thought?
try to word it

[hidden cairn]

order 8 elements: x, xy, xy^2, xy^3.
order 4 elements: y, x^2y, x^4y
Sorry i dont understand what you mean by intersection

like pairing them?

i mean i dont see why we need to take intersection

[serene dune]

oh for some reason i thought u need to express a in terms of x and b in terms of y

[hidden cairn]

ah ic yeah no i need to find the couples a,b which generate G. and from this i see that there are 12 couples but for part b many couples a^2,b^2 also generate H. take for example a=x^2 (generates Z_4) and b = x^4y^2 (generates Z_2)

[gusty tangle]

I think you have to prove the first procedure, the $\vdots$ part is iffy, I think it only works in $\mathbb{Z}_n$ with prime $n$, which is true for $13$

[cloud walrus]

Rain King

[serene dune]

oh yeah u are right, 13 is prime, and if u know some rings that's trivial to figure out

ao u have inverse for all non zero elements, which ticks the final box for group

[gusty tangle]

I'm stumped with 15 here. it's in the first groups chapter of the book so it supposes you don't know many of the strong theorems in group theory, any ideas?

An idea I got: I reformulated the problem to finding a surjective function $f:A^2\rightarrow G$ where without loss of generality $G={e,x_1,x_2,\ldots,x_5}$ with identity $e$ and $x_i$ all different (including different from $e$). $f$ is defined by $f(i,j)=x_i x_j$ and must have the properties

1. $f(i,j)\neq x_i$ and $f(i,j)\neq x_j$. (Otherwise, $x_i x_j = x_i$ then $x_j=e$.)
2. There exists $m$ and $n$ such that $f(m,n)\neq f(n,m)$. (So that $G$ isn't commutative.)

If such a function exists, then not all groups of six elements are commutative.

This reformulation however, doesn't seem to make the problem any less complicated. I guess there must be a counter-example? I think I might not have given it a serious try to prove it must be commutative.

[cloud walrus]

Rain King

[hidden cairn]

have you seen cayley's theorem and group actions?

[gusty tangle]

no, this is on the first groups chapter. maybe it's one of those problems you just can't solve with the available tools

[hidden cairn]

i see then there should be a solution from a simple counting approach i think

[gusty tangle]

yeah but I tried to give a counter example with a cayley table and it just made my brain hurt. maybe you can prove it's commutative

[serene dune]

take possible subgroups (1,x,x^2); (1,y)

[serene dune]

but H has to also satisfy the given constrain, no ?

@gusty tangle u figured it right ?

[gusty tangle]

no, I started some latex notes on the groups chapter, and after that I will give it another try, or maybe tomorrow (I already spent 3+ hours on this exercise)

[serene dune]

oh i was asking coz i dont know if u can use that tool or not, it seemed to be the primitive tool tho

well just write down the possible gorup elements from that, and hopefully u will see the thing

[hidden cairn]

What is the given constraint on H? I thought it is enough to find generators that give Z_4 times Z_2. Since H is isomorphic to that group

[serene dune]

well tbh the problem is a bit dizzing, but i meant this to be the constrain

[hidden cairn]

So a,b should also satisfy a^2b = b^2?

[serene dune]

right

depends from which side u start

[hidden cairn]

Hmm I see yeah I did not pay any attention to that detail

[serene dune]

try to take the elements u have generated and plug in

oh u got it, nonetheless the calculation is a bit dizzy

[hidden cairn]

Yeah the problem makes sense now

Thanks

[gusty tangle]

I don't get how to use this. If I say (1,x,x^2) is a group, then it only has 2 or 1 element depending on whether x^2 is x or 1. If I take x from the original 6 elements group, then (1,x,x^2) might not be a subgroup, if it is, and it's not trivial, then it is (1,x) with x as its own inverse. I get that you can take all five such subgroups, but I don't get what to do with them, it just makes the original group have the elements be all its own inverse, but I still have the problem of proving whether or not there is a counter-example for commutativity or a proof of commutativity. Sorry if I don't get it

[serene dune]

at first prove the existence of such subgroup

then generate all the element of the group

by right and left multiplication

try to compare

[gusty tangle]

This is literally all I get from the first chapter of the book I'm using before it throws to you 14 exercises and then 15 I showed you. Also all example previous to this exercise are commutative.

[serene dune]

right

but u have to consider two ways of multiplication right

to even begin the conversation of non abelian

[gusty tangle]

I will try this. Don't you happen to know a better undergrad abstract algebra book? One that doesn't use categories

[serene dune]

oh boi, books are personal choice imo

i didnt use any book for group theory reading, but dummit foote is a classic

[gusty tangle]

I literally skipped all my group theory classes and somehow passed ring theory kek

[void cosmos]

KEK

[serene dune]

awesome

[void cosmos]

PINTER

ALGEBRA

[serene dune]

i never did math in uni

[void cosmos]

OMG SMAE

same

[serene dune]

but i know a friend who had rings with linear algebra II in their school

[void cosmos]

HUFFMAN KUNZE XXD

[serene dune]

so yeah

wtf is a grassman

[vagrant zinc]

I have never been convinced by any group theory book, I think they explain it in an overwhelming way, things are smoothed out when you get to the rings.

[serene dune]

well understanding happens not in pages

i understood group to somehow degree in order to abstractly classify them
and i mostly used groupwiki and personal scribbles

[serene dune]

i believe ud get two differnt tables, sorry i missed that point

[gusty tangle]

oof, I remember reading this one on my kindle. I think I never found a good pdf scan, I'm looking into one. I loved chapter 1, puts things into context

[chilly ocean]

We can conclude that, g is epimorphism and ker g \subset ker v.

So what the next step? Any idea

[woven egret]

I came across this equation $E^m / \alpha \times E^n / \beta = E^{m+n}$ on a window, but I can’t figure out what it means. The notation leads me to think it’s abstract algebra. Any pointers?

[cloud walrus]

rocketll

[hidden wind]

hmmmm

[mighty kiln]

Why would it be abstract algebra

[woven egret]

Quotient/extension notation, and otherwise wouldn’t make much sense? Shot in the dark tbh

[tough raven]

Maybe alpha bet = 1 🤷

[woven egret]

Managed to find the source through social engineering

[velvet steeple]

In an application, if one quickly wants to find these vectors, one needs to find $v \in \langle e_1, \dots, e_s \rangle \setminus \l \langle \ker(\dots) \cup {v_1, \dots, v_l} \r \rangle$. Is there any general procedure for this?

Without guessing?

[chilly ocean]

I proved all parts but I am stuck at showing f is surjective.

Do I need to use isomorphism theorem?

[grave sedge]

Take an element x in A2', you know π1(i2(x))=0, what can you say about i2(x)?

[chilly ocean]

\pie_1 is surjective and i2(x) is injective

oh wait

[tough raven]

It should follow from the existence of g.

[chilly ocean]

i2(x) in im i1

oh got it, thank you

[chilly ocean]

how g helps me here?

[rocky cloak]

g gives that
pi1 = g pi2

[chilly ocean]

yes commutative

[rocky cloak]

So
pi1 i2 = g pi2 i2 = 0

[chilly ocean]

yes

[wild jasper]

why is the highlighted statement true?

[rocky cloak]

So therefore the image of i2 is inside the image of i1

[tough raven]

IDK. Maybe compute ker(A) by Gaussian elimination, then change to a basis where it has basis the first k vectors (so A has k columns 0 followed by linearly independent vectors). Then you solve for A^2 x = 0 i.e. Ax has all but first k columns 0 (equivalently, ker(B) where B is the last n-k rows and columns of A) and change to a basis where a basis of ker(B) is the next k' columns, etc.

At the end this should leave you with a block-upper triangular matrix with 0 diagonal blocks with the basis being the g_k, ..., f_j, e_i you need to continue

[chilly ocean]

yeah got it, thank you

[velvet steeple]

Thanks! How does one interpret $\langle \langle w_1, \dots, w_n \rangle \cup {v_1, \dots, v_s}\rangle$? It's not the same as $\langle w_1, \dots, w_n, v_1, \dots, v_s \rangle$, right?

[cloud walrus]

Kepe

[hidden wind]

that’s an oddly niche thing to put up on a wall like that

[wild jasper]

Over a perfect field $F$, the splitting field $E$ is Galois over $F$, and since $K\supseteq E$ is radical extension we have that $\textrm{Gal}(E/F)\leq \textrm{Gal}(K/F)$ and since a subgroup of a solvable group is solvable, we are done

[cloud walrus]

mh_le

[rocky cloak]

Not quite.

Gal(K/E) is a subgroup of Gal(K/F), but since E is normal, it's a normal subgroup with quotient group Gal(E/F)

[wild jasper]

Why can we say that the subgroup fixed by E is normal?

[tough raven]

It is.

[velvet steeple]

Oh

[rocky cloak]

Because E is normal.

E being normal implies any automorphism of K leaves E invariant.

[wild jasper]

@rocky cloak what does it mean that a subfield is "normal"?

[rocky cloak]

An extension E/F is normal if any irreducible polynomial over F that has a root in E splits in E.

When I say E is normal, I mean that the extension E/F is normal

And automorphism means F-automorphism

These are all statements about extensions

[wild jasper]

and if E/F is Galois to begin with, why need an extension K?

[tardy hedge]

Whats the difference between prop 29 and exercizs 12

[wild jasper]

Is every splitting field over a field of char = 0 Galois?

[rocky cloak]

Well you wanted to prove that Gal(E/F) was solvable

[rocky cloak]

Yes, every extension in characteristic 0 is seperable

[rocky cloak]

I guess 12 is about products while 29 is about biproducts.

They are the same for R-mod though, except for the extra data of the inclusion maps

[wild jasper]

But if E/F is Galois, why do we need the Galois extension K containing E?, wait is it because Gal(K/F) <= Gal(E/F) ?

[tardy hedge]

Wdym extra data of the inclusion maps ?

[rocky cloak]

... because you wanted to prove that Gal(E/F) was solvable.

Like you know that Gal(K/F) is solvable, and that Gal(E/F) is a quotient group, hence it is solvable.

[wild jasper]

I don't know that Gal(E/F) is a quotient group

[rocky cloak]

Okay, well it is. So now you know I guess

[wild jasper]

can you tell me why?

[rocky cloak]

So say s is an automorphism of K, and let e be in E. Then we know that s must map e to a root of its minimal polynomial. Since E is normal the minimal polynomial has all roots in E, so s(e) is in E.

That means that restricting s to E gives an automorphism of E. So there is a homomorphism Gal(K/F) -> Gal(E/F) given by restriction.

Then you just need to verify that this is surjective. Consider t in Gal(E/F), and consider k in K. We want to extend t to E(k).

Consider t as an automorphism of the polynomial ring E[x]. And let p be the minimal polynomial of k over E, and let q be the minimal polynomial over F. Then p divides q, so t(p) divides t(q) = q.

Since q has a root (namely k) and K is normal, it splits in K. Hence t(p) splits. Let k' be a root of t(p). Then you can extend t to E(k) by maping k to k'. Continuing this with Zorn's lemma you can extend t to all of K, hence the mapping is surjective.

The last part can also be seen as the combined statements of if K/E is algebraic, then any homomorphism from E to an algebraically closed field extends to a homomorphism from K.

And if K/F is normal then any homomorphism from K to its algebraic closure has image K

[rocky cloak]

Well, maybe they're using the notation a little different here. But usually I'd say the biproduct L(+)N consists of projection piL and piN together with inclusions iL and iN such that piL iL is the identity on L, piN iN is the identity on N and iL piL + iN piN is the identity in L(+)N

In practice the proof should be identical either way

[opaque finch]

I have a question that I want to solve.
Factor $x^4 + x + 1 \in \mathbb{F}_2[x]$ into product of irreducibles over $\mathbb{F}_4$.
The facts I understand is $\mathbb{F}_2(\alpha) = \mathbb{F}_4$ where $\alpha$ is the root of $x^2 + x + 1$. But how can I proceed further from here ?

[cloud walrus]

mycroftholmes1703

[rocky cloak]

First step might be to check if it has a root in F4.

And if not you could try to factor it into a product of quadratics

[chilly ocean]

For the first part, we can choose E = A × B such that f(a) = (a,0) and g(a,b) = b

The second part comes from Five Lemma, because we can map A to A by Identity, B to B by identity.

Correct?

[tardy hedge]

I thought E was an extension of B by A

Not an extension of A by B

[chilly ocean]

why? f is injective and g is surjective, yes i wrote g(a,b) = (0,b) it is b

[wild jasper]

@rocky cloak thanks!

[rocky cloak]

Both terminologies are used.

Extension of A by B makes sense, because E contains A, so is an extended version of A (see field extension / extension of rings)

Whereas extension of B by A just comes from matching the order of things in Ext(B, A) (whose notation comes from Hom(B, A))

[opaque finch]

Gotcha

[dull tiger]

Question is: What is the number of different colourings of the sides of a rectangle with k different colours mod $D_4$ (as in the dihedral group). And the solution is $ 1/8(k^4+2k^3+3k^2+2k) $. This is solvable by the non-Burnside Theorem. And i did understand, that the $k^4$ corresponds to all sides have different colouring, $3k^3$ corresponds to 3 sides have different colouring and the remaining one has a different colour and $2k^2$ is for 2 sides share one colour and the other 2 have different colouring. But why 2k?? Following the logic, shouldn't it be just k, cause now all sides have the same colour? And then, those values correspond to the order of the following set: ( {x \in M: ax = x }) for M a set and $a$ in the group that is acting on M. And i get that a is the identity in the first case $(k^4)$ but what happens for the other cases?

[cloud walrus]

Marieeee

[gusty tangle]

hey, I solved it. It was my fault for not reading the examples. In particular there is this one example of a group of 6 elements which is not commutative:

not commutative

[serene dune]

oh yeah that's perfect, now u know dihedral group basically

[gusty tangle]

I got the idea yesterday, that there were some examples of symetry groups which were not commutative. But I went to sleep before looking for an order 6 example

[serene dune]

notice there are 2 subgroups

(1, x, x^2), (1,y)

well this way of generating groups will help u down the way

[serene dune]

now the task is can you claasify all the groups of order 6

you don't have to produce table for all

[toxic zephyr]

doesn't the fact that all elements of a p group have order p^k follow pretty directly from Lagrange?

okay yeah pretty sure it does

whaaa the center of a p group is nontrivial? yo that's sick

[vapid vale]

they dont have to have order p

[toxic zephyr]

meant p^k

for some k

[serene dune]

the converse theorem

cauchy ig !

[vapid vale]

this is a pretty important fact especially if you have not seen class equations

[serene dune]

yeah it can have all sort of center, for example look p,p^,p^3

[vapid dew]

The set $U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}$ forms a group under multiplication modulo 11.
Show that $U$ is cyclic and write down its generators.

[cloud walrus]

Bitwise

[vapid dew]

I got the answers 2,6,7,8 by trial, however the textbook answer just calculates 2 then uses gcd(2^m, 10) = 1 and gets 6,7,8 from there. It never mentioned this before so I don't really understand how that works, could anyone explain?

[vapid vale]

did you show its cyclic?

[vapid dew]

Yes

I showed it by considering 2, which generated all the elements

[vapid vale]

then is that not a generator

[vapid dew]

Yes

it is

I may have phrased it badly, I don't understand this "gcd(2^m, 10) = 1" and how that leads to the other generators

not how they got 2

[vapid vale]

oh i see what youre asking. could you send the excerpt? because that statement is not true

[vapid vale]

oh

[serene dune]

the same question was posted in the North but with 13

[serene dune]

what do u need help with in particular

[vapid dew]

I don't understand where gcd(2^n, 10) = 1 comes from

like how does that give the other generators

[serene dune]

what is your meta goal

[vapid dew]

Sorry?

[serene dune]

ultimately where are u trying to reach

u need to figure out the generators right?

[vapid dew]

Ideally an intuitive reason for why that's the case and a proof, if Im interpreting your question right

[vapid dew]

yep

[serene dune]

well many ways to reach there

[vapid dew]

True, I originally just tried all of them, but I want to understand why this alternative also works

[vapid vale]

the point is that all elements are expressible as 2^n. you want another generator (which we write as the form 2^n) such that the subgroup it generates also has order 10

[serene dune]

observe the order of elements. or take gcd

[rocky cloak]

I think there's a typo here. It should be 2^n where n is such that gcd(n, 10) = 1.

Generally for an element g in a group G, the order of g^n is o(g)/gcd(n, o(g)) where o(g) is the order of g.

So n such that gcd(n, 10) = 1 gives you the other elements of order 10 in the group

[candid dove]

Hey how do I prove this statement on the left ideals of M_n(C)

I mean i don't see how this intersection is not the trivial subspace

[coral spindle]

Remember this is the intersection of null spaces in one ideal

[candid dove]

Yeah

[coral spindle]

So let's choose an ideal

[candid dove]

Okay

[coral spindle]

Let's say n = 2

and let's choose the ideal generated by $\begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}$

[cloud walrus]

$\mathbf{Boytjie}$

[vapid dew]

Yeah I was thinking about that too because if it's 2^n then can't n be 0 to get gcd(1,10) which satisfies their condition but 1 obviously can't be a generator. Is my reasoning correct?

[coral spindle]

OK so we know for sure that the intersection of the null spaces of all the elements of the ideal must, in particular, be a subspace of the null space of this matrix, yes?

Now I claim that in fact this is exactly the intersection

[rocky cloak]

Yes, so it should be gcd(n, 10)

[coral spindle]

Why? Because let's say this matrix I chose is called M. Then everything in the ideal is of the form AM for some matrix A. So in fact if Mv = 0 then also AMv = 0. The null space can only grow.

So we know that it is at most null(M) and at least null(M), so it is null(M).

Clear?

[candid dove]

Right

But your ideal is nice

[coral spindle]

Yes, my ideal is indeed nice.

[candid dove]

What about non nice things

[coral spindle]

The ideal may or may not have trivial intersection, but this was not the question.

[candid dove]

Uh

[rocky cloak]

The great thing is that all the ideals are nice 😉

[coral spindle]

The statement asserts that in fact this is a bijection

[vapid dew]

Alright thanks a lot, this clears it up a little more but I still need to try understand on how gcd(n,10)=1 gives the other elements of order 10

[coral spindle]

Indeed it does not provide a proof, but perhaps you should try showing this

[rocky cloak]

Yes, so I would recommend trying prove that the order of g^n is o(g)/gcd(n, o(g))

[candid dove]

Okay but still I am uh kind of not uh

Uh

Okay for a proper left ideal of M_n(C) why is this intersection non trivial

[rocky cloak]

Think about the converse. Say the intersection is trivial, why is the ideal everything?

Well if v1 is a basis vector it contains a matrix with Av1 nonzero.

Then there is a matrix B that maps Av1 to v1. Composing this with the projecttion onto v1 you get that this projecttion is in your ideal. Same true for all the other basis vectors, then just adding them gives the identity

[candid dove]

No like that converse is clear

[rocky cloak]

Sorry, I meant contrapositive

[candid dove]

Okay

Okay that clears it

One more question

If i replace C by a division ring

Then would it still be true

[rocky cloak]

Yup

Same proof as well

[candid dove]

Thanks somuch

[candid dove]

But before I go does this mean anything deep

You just meant that ideals are precisely the anhilators of subspaces?

Or was it sth else

[vapid dew]

$G$ is cyclic of order $|G|$ with generator $g$. Then $g^n$ is a generator if its order is $|G|$.
The order of $g^n$ is the smallest $m$ such that $(g^n)^m = e$, meaning $g^{nm} = e$. Since $g$ has order $|G|$, $|G|$ divides $nm$. Given $\gcd(n, |G|) = 1$, $m = |G|$ as n and $|G|$ share no factors, so $g^n$ has order $|G|$ and is a generator.

[cloud walrus]

Bitwise

[rocky cloak]

Don't know if it's super deep, but in general for a ring R an ideal is a direct summand of R iff it's equal to Re for an idempotent e.

These matrix rings are semisimple meaning every ideal is a direct summand. So every ideal is Re, and in this case e is a projection onto a subspace (of C^n)

[vapid dew]

I tried doing it this way, is this right?

Thanks alot for the help, I had been stuck on this for a bit and the book being incorrect had just confused me even more

[muted comet]

I dont understand how the elements in a polynomial ring gets constructed. i dont understand polynomial rings at all in fact. can someone help me understand?

[vapid vale]

they are polynomials with coefficients in some base ring

you can add, subtract, and multiply polynomials but you generally can't divide them - hence a ring of polynomials

[muted comet]

define polynomial please

[coral spindle]

Are you asking for a set-theoretic construction here?

This isn't hard

[vapid vale]

for simplicity lets fix one variable X, and call the base ring k. a polynomial is a finite linear combination of k_mX^m + \dots + k_1x^1 + k_0

as polynomials are from grade school math

[muted comet]

i find all constructions ive found to be rather disgusting

[vapid vale]

?

[coral spindle]

OK so you don't want a construction?

So what are you looking for

[muted comet]

yea? im just saying what ive experience

[vapid vale]

are you familiar with polynomials from middle school algebra

[coral spindle]

OK

[vapid vale]

y = x^2 + 2x + 1

[coral spindle]

A polynomial in R is a function $f \colon \mathbb N \to R$ such that, for all but finitely many values $n \in \mathbb N$, we have $f(n) = 0$.

The ring $R[X]$ is then the set of these polynomials. Addition is $$f + g \colon n \mapsto f(n) + g(n)$$, and multiplication is $$fg \colon n \mapsto \sum_{a \in \mathbb N} f(a)\cdot g(n-a).$$

[vapid vale]

did you set your tex bot font

[cloud walrus]

$\mathbf{Boytjie}$

[coral spindle]

unfortunate linebreak there, I fixed it

and yeah I changed it a while ago

[vapid vale]

how to

[coral spindle]

OK Jonas, is this what you're looking for?

[muted comet]

i mean i can accept that but my prof told me the elements in a polynomial ring werent functions

[coral spindle]

Ah I think you need the command ,preamble, that way you can import font packages

[vapid vale]

bleh

[coral spindle]

Right! This is the real question at hand

Indeed

What your prof wants to distinguish is polynomials vs polynomial functions

Let me put it this way

When you write out a polynomial as e.g. 1 + x^2 + 5x^6 or whatever, this looks like it's a function from R to R

But in fact, we don't want to consider polynomials as functions from R to R

That loses a LOT of information

For example, are you aware of the field with two elements, F_2? F_2[x] is an infinite ring, as you can probably see. However there are only finitely many functions F_2 -> F_2

So clearly we are not considering polynomials as functions on F_2 here.

Does this clear things up?

[muted comet]

i mean it clears up why she would say that. but doesnt this mean we have to go back and revise the definition? or do we just live with two different?

[coral spindle]

I feel like I just explained why she would say that. The ring of polynomials and the ring of polynomial functions on R are typically not the same