#groups-rings-fields

406252 messages · Page 483 of 407

oblique river
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start with the identity permutation

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then swap 1 with whatever should be in the first spot

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then swap 2 with whatever should be in the 2nd spot

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etc

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I guess that's a bit simplistic

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but it's the general idea

shy bluff
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I mean I looked at S_4 and it seems to me that it's ker(s) is made up of all elements of S_n that have an even number of transpositions right?

oblique river
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yes

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but you don't need transpositions to describe A_4

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you can also describe it as

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the permutations whose corresponding matrix has determinant 1

shy bluff
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Yes

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But then how do you describe when it has determinant 1?

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Yes

olive mirage
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except that you use permutation parity in the definition of a determinant, typically

oblique river
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what do you mean

olive mirage
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(or, if you use THAT definition, in the proof of existence)

oblique river
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zeta, we went over that

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yeah

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also liria what do you mean "how do you describe"

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what you said is how you describe them

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that is the definition

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yes, there are other descriptions

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but I don't think the one we have now requires any additional description

bleak abyss
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The new meta: define the determinant of a complex matrix to be the product of eigenvalues and use that to define sign

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So to compute it you have to triangularize the matrix

olive mirage
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I love it!

bleak abyss
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Definitely not a bad strategy by any means -Axler

shy bluff
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Well

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I want [S_n : A_n]

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I know what |S_n| is

oblique river
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oh

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yeah that's easy

shy bluff
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So I need |A_n| rigght?

oblique river
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and you don't need |A_n|

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sorry by "easy" I just mean that the solution is short

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not that it's necessarily easy to come across

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[G : H] = |G / H|

olive mirage
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I would say you can define the permutation parity by an explicit homomorphism from S_n to {+1, - 1}, which makes it index 2 almost by definition.

shy bluff
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The index is the number of distinct left or right cosets

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Yea that

oblique river
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zeta, that's exactly what we've done in this problem haha

olive mirage
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sorry, I should have read back 😛

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but if you want to avoid determinants you can use that Product of (i-j)/(pi(i)-pi(j)) if you like

oblique river
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yes but that doesn't really help solve the current problem

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liria has a fixed definition of A_n

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and is looking to find the index of A_n in S_n

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using that definition

olive mirage
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ahh, so the explicit definition is "even number of transpositions"?

oblique river
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no

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maybe you should just scroll up........

shy bluff
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Ok so the index is defined to be the number of distinct left or right cosets

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In this case the number of cosets of A_n

oblique river
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yes

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but just using that definition as it's written probably won't be helpful

shy bluff
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oh

oblique river
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because, like you said earlier, we don't really know what A_n looks like

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let's think about how A_n is given to us in this problem

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what theorems do you think about when you think of kernels

shy bluff
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rank-nullity

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Uh how it can only be a bijection of if the kernel is trivial

oblique river
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that's a theorem in linear algebra

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the second theorem is true but not relevant here

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since S_n isn't in bijection with {1, -1} unless n = 2

shy bluff
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Yea

oblique river
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there's a theorem that's kind of like rank-nullity but for groups

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well, not really "kind of"

shy bluff
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Yea I thought that there was something like that

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Uh

oblique river
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what other theorems do you know about kernels

shy bluff
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If f : G-> H is a homomorphism then ker(f) <= G and im(f) <= H ?

oblique river
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yep, that is true and relevant and we're getting there

shy bluff
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Ker(f) is a normal subgroup of G?

oblique river
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getting warmer!

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what do we do with normal subgroups?

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(meaning, what can we do with normal subgroups that we can't do with non-normal subgroups?)

shy bluff
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Uh they're commutative?

oblique river
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no

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also that's false

shy bluff
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Or like not commutative

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But like aH =Ha

oblique river
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sure, that's the definition of normal

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that's not what we do with normal subgroups

shy bluff
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Oh

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What do we do with normal subgroups?

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Direct products?

oblique river
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what do you mean

shy bluff
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aHa^{-1} = H?

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I dunno

oblique river
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what can you do with a group and a normal subgroup

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that you can't do with a non-normal subgroup

shy bluff
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I have no idea

oblique river
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surely you've talked about quotient groups in your class...?

shy bluff
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Oh I think that part of it is that the homework that's handed out is kinda wonky and sometimes includes the next week's lecturse

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I think we're covering that next?

oblique river
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oh

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so "the first isomorphism theorem" means nothing to you then

shy bluff
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We covered uh definition of a group, elementary properties, homomorphisms, isomorphism,s subgroups, coset,s lagrangge's theorem, and normal subgroups, and the last thing that we did was direct products?

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Nope we have not gotten there yet

oblique river
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oh

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that's the theorem I wanted you to name earlier 😅

shy bluff
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Oh

oblique river
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I assumed you had done that because that tool will solve this problem in like

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15 seconds

shy bluff
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Oh

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Well

oblique river
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so here is the gist of it, tell me if any of this sounds familiar

shy bluff
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Oke

oblique river
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if s: G --> H is a homomorphism

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then the cosets of ker(s) are in bijection with the image of s

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the idea being that if two elements are in the same coset of ker(s)

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then they must map to the same thing under the homomorphism

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if we call K = ker(s)

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and if aK = bK

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then a = k^(-1)*b*k; for some k, k' in K

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and therefore s(a) = s(b) since s(k) = s(k') = id

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in particular all of this means that |G / ker(s)| = |im(s)|

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the left hand side is what we want to compute in this problem

bleak abyss
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That's weird that they give this problem without first iso

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Maybe the idea was to show that multiplication by a given transposition gives a bijection A_n to complement?

shy bluff
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That last bit I get

oblique river
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since |G / ker(s)| = [G : ker(s)]

shy bluff
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Ok wait that last bit sounds familiar

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I think that the prof mentioned something similar to that at the end of the latest video?

oblique river
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and the right hand side is {1, -1} since in this case s is surjective

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(1 is clearly in the image of s, just verify that -1 is also in the image by finding an explicit permutation with s(sigma) = -1)

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and therefore

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|im(s)| = 2

shy bluff
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But uh yea I have definitely seen |G/ker(s)| = |im(s)|

oblique river
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oh

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yeah

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well that's the theorem you need

shy bluff
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Rest of this is somewhat going over my head 😅

oblique river
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sorry, I was just trying to justify to you why that statement is true

shy bluff
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No it's ok

oblique river
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maybe you used a different proof of it

shy bluff
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I'm pretty sure that I've read about the 1st isomorphism theorem

oblique river
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than what I tried to explain

olive mirage
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If you define A to be the even permutations and O to be the odd permutations, and you let t be the identity matrix with the first two rows swapped, then you can verify that the function A \to O : x \mapsto tx is a bijection

shy bluff
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But that it just hasnt' been formally covered in class yet maybe

olive mirage
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that would give you that the index is 2 in a sneaky way without needing any theorems

shy bluff
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Wait what do you mean zeta?

olive mirage
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so, if you can write down a bijection between two sets, they are the same size

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Do you agree this is doable without assuming anything we're not supposed to Bananas?

shy bluff
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yes

oblique river
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that would show you that |A_n| = |S_n \ A_n|

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that's a set-minus on the right

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not a quotient-set

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and you need to argue from there that that implies that |A_n| has index 2

olive mirage
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and that certainly tells you the index is two if you know that cosets have the same size.

shy bluff
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but how do you know that the cosets have the same size?

oblique river
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that's one way to do it

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that should have been something you proved about cosets

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when you defined them

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if not... your prof really didn't give you the tools you needed to solve this problem...

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also I gotta go

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sorry

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gl with this

shy bluff
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wait

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By "know that the cosets have the same size"

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Do you mean the cosets of |A_n| and the cosets of |S_n \ A_n|?

olive mirage
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S_n \ A_n is a coset of A_n, that is equivalent to what you are proving (that the index is 2)

shy bluff
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A coset of A_n is a set that's smaller than A_n such that you take one arbitrary element of S_n and multiply all of A_n by it right?

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Err take the product of that element and all of A_n

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Right?

uncut girder
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Theres only 2 cosets of A_n

olive mirage
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it's not smaller than A_n, it is exactly the same size as A_n

uncut girder
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And they both have the same size

olive mirage
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so here, G= S_n and H = A_n, in your definition.

shy bluff
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Yes

uncut girder
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but how do you know that the cosets have the same size?
Left multiplication by an element of the group is a permutation on the elements of the group. Thus, it sends a subgroup to a coset necessarily of the same size

shy bluff
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I see

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And how do we know that S_n \ A_n is a coset of A_n?

uncut girder
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The cosets partition the group, do you know this?

shy bluff
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Yes

uncut girder
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So why is S_n\A_n a coset of A_n? You tell me

shy bluff
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S_n is the union of all left (or right?) cosets of A_n

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... Right?

uncut girder
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Yeah

shy bluff
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I don't really ese how removing A_n gives me a coset though?

uncut girder
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How many elements does S_n have

shy bluff
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n!

uncut girder
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How many does A_n have?

shy bluff
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We don't know?

uncut girder
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We actually do. Why is it n!/2 ?

shy bluff
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|Im(S)| = 2

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It's like rank nullity somehow I think

uncut girder
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?

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What is S

shy bluff
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Err s

uncut girder
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Sign ? Ok

shy bluff
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Yea

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But we haven't learned first isomorphism theorem?

uncut girder
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Rip

shy bluff
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Oh

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Is that what you were trying gto get at?

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I'll just email my prof about it 😅

uncut girder
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I was trying to get at |A_n| = n!/2,
There there are only two cosets, A_n and its complement

shy bluff
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Wait but how do you know that?

uncut girder
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Know what

shy bluff
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How do you know that there are only two cosets

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And that they're A_n and its complement

uncut girder
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Ok take for granted |A_n| =n!/2, for now.

Why does that tell you there are two cosets? I dont wanna give you the answer

shy bluff
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Lagrange's theorem?

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Because n! = |S_n|

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Wait

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Not Lagrange's theorem

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Sorry

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but Laggrange's theorem tells us that |A_n| has to divide |S_n|

uncut girder
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Yes, you should understand why that formula works tho

shy bluff
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Um well the index thing is by definition

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By the formula do you mean lagrange's theorem?

uncut girder
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Nah I mean [G:H] = |G|/|H|

shy bluff
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Thats... A definition?

uncut girder
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Yeah regardless you should understand why it counts the number of cosets

shy bluff
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It's bc you can partition it up right?

uncut girder
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Yes

shy bluff
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Ah ok nevermind I went through the prof's videos aggain

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There's a slide that says |Im f| = [G: ker f]

stone fulcrum
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What's the question?

shy bluff
oblique river
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yeah liria that should answer it for you

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there are only two options for the image: {1} or {1, -1}

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so you need to just need to determine whether or not -1 is in the image

shy bluff
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Yea

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I mean -1 is in the image

oblique river
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great

mild laurel
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It's free in the sense that there are no relations between the generators

lime skiff
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just making sure I clearly understand

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if K is a field

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then the units in K[x] are just K{0} correct?

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I mean invertible elements

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because otherwise I'd need some form of polynomial with negative exponent

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which isn't in K[x] right?

upper pivot
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yeah its just K

lime skiff
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so in particular

upper pivot
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you can generalize this to a general ring A[x] btw

lime skiff
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if I have a polynomial f of degree p

shy bluff
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For this here, <S> indicates "the group generated by all elements of S" right? like say that S = {1, 2, 3} or something then <S> would be {1^n, 2^n, 3^n}?

📎 unknown.png
lime skiff
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say in R[x]

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if I can show f = gh, where g,h\notin K

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then I've shown f is reducible right?

upper pivot
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yeah

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assuming you meant K[x]

lime skiff
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oaky good

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yeah

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well I mean, that would be true for a general ring R anyway right

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or A, if you're French speaking

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🙂

upper pivot
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well the general statement is this btw

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$f(x)=a_0+a_1x^1+\dots a_nx^n$ is an unit in $A[x]\iff$ $a_0$ is an unit in $A$ and $a_1\dots a_n$ are nilpotent in $A$

cloud walrusBOT
lime skiff
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ah I see that makes sense

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but the only nilpotent elements in a field of char 0 is just 0 right?

upper pivot
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yep

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(in any field not even char needed)

lime skiff
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oh yeah ofc

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but like

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take a ring with char 4 (say)

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then a unit isn't necessarily just the ring, correct?

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err

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just in the ring

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as in, the base ring

upper pivot
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yep

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cause 2 is nilpotent

lime skiff
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okay, thanks

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yeah that's what I had in mind

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oh that makes sense

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I'll have something like

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2x * 2x

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= 4x^2 = 0

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which give me what I want

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or something like that

upper pivot
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yeah 1+2x is an unit for example

lime skiff
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makes a lot of sense

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thanks, I didn't know this more general statement

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(1+2x)(1-2x) = 1 + 4x^2 = 1

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so it's a unit

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good

upper pivot
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yep

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theres some nice general statements like this in atiyah macdonald ch 1 problem 2 and 4, if interested

lime skiff
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yeah this is the next thing I'll do once I'm done with this exam

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I've browsed through AM but haven't done the exercises yet

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I was just reviewing some ring theory first

upper pivot
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yeah its a nice book

lime skiff
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I don't know if you know but people have been typing up solutions to the problems

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I was re-texing the exercises

upper pivot
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i do that too!

lime skiff
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and people started giving solutions to the problems

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oh very nice

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hi yami chan tooru

upper pivot
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yeah its a good way to motivate yourself to do all the problems i think

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hey yamin

lime skiff
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yeah

uncut girder
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Isnt Kuri the one with the AM solutions? thonkzoom

lime skiff
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especially since I'd like to be able to start doing AG this summer

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yeah I'm the one that started typing the solutions

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but I haven't done the exercises yet

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well, not typing the solutions

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typing the exercises

uncut girder
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Hmmmmm

lime skiff
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people type the solutions in the doc

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I don't see them

upper pivot
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oh i just typed them in my texstudios

uncut girder
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Just gathering solutions for your HWs

upper pivot
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rather than overleaf

lime skiff
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I wish

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there's no CA class here

upper pivot
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rip same tho lol

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ASU has 0 math classes higher up rip

lime skiff
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rip

uncut girder
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3rd sem grad alg is commutative Algebra (and homological algebra) at my school

shy bluff
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Uh

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How does this work?

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Arent' g_1 and h_1 in two different groups?

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How do we gog and take (g_1h_1) if they're in different groups..?

upper pivot
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they are in the same group

uncut girder
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No g1, h1 are in G1

lime skiff
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check the notation

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it's a bit confusing

shy bluff
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Oh

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Uh

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Then how does this work?

lime skiff
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well you create a new group out of two other groups

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take Z_2 and Z_3 to get some intuition if you want

shy bluff
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Sure

lime skiff
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you'll have something like this:

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(0,0), (0,1), (0,2), (1,0), (1,1), (1,2)

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do you understand what's happening here?

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for instance, what's (1,1)(1,2)?

shy bluff
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(1, 2)?

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err under multiplication (1, 2)?

lime skiff
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yeah that would be true under mux

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what about addition?

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(1,1)+(1,2)

shy bluff
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(2, 0)?

lime skiff
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well

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close

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you correctly noted that 1+2 mod 3 = 0

shy bluff
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Yea

lime skiff
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what's 1 + 1 mod ... =

shy bluff
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Oh

lime skiff
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is it mod 3 or mod 2?

shy bluff
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That first one is mod 2?

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Because it's (Z_2 x Z_3)?

lime skiff
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well is it Z_2 x Z_3 or Z_3 x Z_3?

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right

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exactly

shy bluff
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Oh

lime skiff
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it's really natural I think?

shy bluff
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So it'd actually be (0, 0)

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Oh ok

lime skiff
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think of addition in R^3

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R^3 = R x R x R

shy bluff
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Yea prof just notated it funny I think

lime skiff
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it's the same thing

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which reminds me

shy bluff
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I was very very confused because he didn't provide any further explanation on the notation so I kept thinknig that (g_1, g_2) x (h_1, h_2) where g_1 and g_2 are in the same group

lime skiff
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I was in an algebraic topology class

shy bluff
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And h_1 and h_2 in the other group

lime skiff
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and the prof called the direct product "the abelianization of the free product"

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smh

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don't become algebraists kids

shy bluff
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The only word that I know in that sentence is "abelian"

lime skiff
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don't worry it was just a quirk of my prof

upper pivot
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i mean hes not wrong

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lol

lime skiff
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no no it's totally right

shy bluff
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this is my first foray into uh a real math class I guess

lime skiff
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but imagine actually doing that lmaooo

shy bluff
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Level of difficulty is uh an order of magnitude above that which I'm used to 😔

lime skiff
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"apply the ab functor to get H_1"

upper pivot
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hehe

lime skiff
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It's ok Liria

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don't hesitate to ask

upper pivot
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yeah dw we were all there once

shy bluff
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Yea this discord is great and people here are very patient with my uh slowness 😔

lime skiff
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can you guys check my understanding

shy bluff
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Everything being online is not being conducive to my learning sadly

lime skiff
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I have the polynomial x^4 - 2x^3 + 2x^2 + x + 4

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if I go mod 2

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I only get x^4 + x = x(x^3 + 1)

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so this means the polynomial is reducible right?

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am I allowed to do that

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the polynomial is clearly in Z[x]

upper pivot
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yeah should be if im remembering correctly

lime skiff
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oh wait I think I got it wrong

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it's

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if irreducible in Z_p[x] => irreducible in Z[x]

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aaaaa

uncut girder
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Yeah

lime skiff
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not reducible in Z_p[x] => reducible in Z[x]

upper pivot
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oops

lime skiff
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what a pain

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it might be irreducible in another localization

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aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

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err

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another Z_p

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my bad

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I mean I can probably go a more direct route

uncut girder
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You know, the degree of the irreducible factors when you reduce mod p, (if p doesnt divide the discriminant), tell you about the existence of certain cycle types in the galois group of the polynomial over Q

lime skiff
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and try to factorize it

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really yami?

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I haven't done galois theory so far

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well not much at least

upper pivot
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same my galois is minimal

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i just read lang and didnt do the exercises rip

lime skiff
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oof

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Iktf

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I'm reading Milne's field and galois theory rn

upper pivot
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oh yeah ill prolly read those at some point

lime skiff
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I need it for arithmetic geometry aa

upper pivot
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yep i went to it for AG lel

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cause hartshorne chapter 1 is

lime skiff
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hahaha

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yeah

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I agree

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I have Liu and Vakil for AG

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though I might do exercises in Hartshorne

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not yet sure

upper pivot
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yeah i am prolly gonna use hartshorne only for exercises

golden pasture
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same

uncut girder
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Yes, if you have a degree 4 irred polynomial p(x), and reduce mod 5, and say 5 doesnt divide discriminant of p(x), and say p(x) factors into a product of irred quadratic mod 5, then you know there exists a (2,2)-cycle in your galois group of p(x) over Q

lime skiff
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that's sick

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I should learn galois theory

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this semester was spent learning algebraic topology and some elliptic curves

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so it's not wasted

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but I should really learn more algebra

upper pivot
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that sounds nice

lime skiff
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yeah it was pleasant

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wish we had done some cohomology though

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I only know De Rham from differential geometry

upper pivot
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oh yeah im in an AT "class"

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but i dont think we will do cohomology rip

lime skiff
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we went through most of Hatcher 1-2

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like first two chapters

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the prof is a symplectic geometer so he was big on fixed point theorems and stuff

uncut girder
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I learned galois theory by doing hws weeks later than they were due opencry

lime skiff
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oh this one's opencry

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got it

upper pivot
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brauwer fixed point is extremly sexy tho

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with homology

lime skiff
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yeah I agree

uncut girder
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Holomogy

olive mirage
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(co)homology is awesome. I really like Harder's Lectures on Algebraic Geometry I (which does not actually contain much Algebraic Geometry) as a really wideranging dive into the topic.

olive mirage
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do you mean group theory? Are you talking about a presentation?

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a presentation is a list of generators and relations, that describe a group

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e.g. <r,s | r^4=s^2=rsrs=e>

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I don't thin of generators as having an order. Can you give us a sense of where you're coming at this from, so we can tailor our responses accordingly?

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ahh, yeah, so in the case of a rubiks cube there are a small number of moves you can think of as generating the group

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but the thing that makes it complicated is that, unlike in linear algebra with a basis, those generators are not at all independent

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roughly speaking independence in this context would mean that different sequences of moves put the cube into different states

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exactly

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so a presentation is the set of generators (that part is easy) plus a set of relations that capture all the ways in which two different sets of moves give the same output

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(and that is the part that is messy)

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and when I say complicated, in the case of the Rubik's cube group, it is something like this:

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perfect, yes

chilly ocean
#

sry to interrupt, zetamath do you have any problem sets for NT similiar to the algebra one?

olive mirage
#

wHat kind of number theory?

chilly ocean
#

any I guess

#

I dont know much, so from basic one

scarlet estuary
#

@chilly ocean have you seen the group axioms?

#

Then that's what a group is

olive mirage
#

gotcha, let me dig up my homework set from last time I taught number theory. I will say, though, that intro number theory problems are an enormous pain to come up with, so I have never been super happy with the set of problems I've come up with.

scarlet estuary
#

Any structure with a binary operation that satisfies those axioms is a group

chilly ocean
#

wait so you made up your own problems for the class???

#

btw what other NT classes did you teach if I may ask?

scarlet estuary
#

Formally, a group is an ordered pair (S, •) where S is the set of all elements in the group and • is its binary operation satisfying the group axioms

olive mirage
#

I basically always make my own problems for courses I teach, though of course a lot of them are inspired by problems I collect from elsewhere. I've also taught crytpography a number of times. I don't think I have any problems laying around on algebraic or analytic number theory, or arithmetic geometry.

scarlet estuary
#

Generators give an alternate way to "construct" groups

#

But it still ultimately becomes this definition

#

Of a set of elements and a binary operation

olive mirage
chilly ocean
#

wow thats so nice of you, thank you. Were those courses made for undergrad students?

olive mirage
#

In this case, yes. The algebra problems I shared before were for grad students.

chilly ocean
#

That's intersting, I thought it would be rather other way around

olive mirage
#

The structure I have seen most is that there is a year of algebra in undergrad, and then a year in grad school, taht goes over similar topics in more depth. One place I was at actually had two years of graduate algebra.

scarlet estuary
#

IIRC the Rubiks cube group is generated by the 6 clockwise rotations

#

What do you mean?

#

It means we can represent any series of rotations (i.e. any arrangement of the Rubik's cube) as just a series of clockwise and counterclockwise rotations

#

I'm not really sure what you're asking, beyond that

#

A presentation of a group includes its generators and relations between its generators

#

I'm not sure off-hand what the presentation of the Rubik's cube is

#

It's likely very sophisticated

#

44 relators

#

Well, there are alternate ways to use groups, but looking at its presentation is one way, yes.

#

I'm not sure the Rubik's cube group is much more than a novelty though

golden pasture
#

it is kinda hard to play with like presentations tho

scarlet estuary
#

Uh

#

I'm still confused what you're asking honestly

golden pasture
#

eh so say your problem is solving a rubiks cube, how would you go from [state of rubiks cube] to [some sequence of generators]

#

going would already involve solving it

scarlet estuary
#

I guess? In the same way that when you draw a free body diagram in physics, you stop focusing on the rest of the problem and work through the physical geometry calculations

#

An element of the group

#

Idk what you mean by "unique"

golden pasture
#

Well yes but there can be different ways of writing it out as a some sequence of generators

scarlet estuary
#

But yes

#

And ° is function composition here

#

Or well, rotation composition

#

Ie do one rotation, then the next

#

(or one series of moves, then the next; whatever)

golden pasture
#

well if you think about it the moves of a rubiks cube is kind of just some permutations, and all the moves is a subset of some permutation group of the rubiks cube

scarlet estuary
#

Yes, that's the group operation

#

The group operation is just composing elements

#

Each element represents some series of moves

#

The group operation says to do the first series of moves, then the next

#

Which gets you a new element although

#

Oftentimes this new element "reduces" to another

#

Like 3 counterclockwise rotations of a given face

#

Is the same as a clockwise rotation of that face

golden pasture
#

yes

scarlet estuary
#

Yes

lime skiff
#

ah the rubik's cube was the last part of our algebra lectures

soft elm
#

[G:H n K]=[G:H][H:H n K]

soft elm
#

actually that wont work consider the intersection of the cosets of H and K consider aH n bK now this is equal to zH n zK for z in aH and bK both as zH=aH and zK=bK now since the number of such zH is finite and the no. of the cosets zK is finite their intersection is finite too.

#

im pretty sure i was answering to a question here

#

or i wasnt

olive mirage
#

there is a version of the diamond isomorphism theorem that implies this trivially, but the traditional version assumes one of the groups is normal

#

so if you talked about the second isomorphism theorem you might go hunt around there

supple rampart
#

@soft elmhehe

I got that done friend

#

So deleted the question

#

But wht is zeta math talking about

olive mirage
#

oh I was talking about (your?) question about proving that if H and K are finite index so is H \cap K

fading sparrow
#

do they perhaps mean K integral over A

olive mirage
#

by "A integral" they mean "A an integral domain" in which case B being integral over A means that B sits isomorophically inside the algebraic closure of the field of fractions of A.

fading sparrow
#

ah

#

thanks

olive mirage
#

(I should have sounded less confident when I said that, I did not do the exercise, but that is my reading of it)

sinful zephyr
#

Hi I'm trying to find the intermediate fields of Q \subset Q(zeta_7). I know the Galois group is Z_6 so I should be able to find intermediate fields equal to Z_2 and Z_3, but idk how to / what to try.

mild laurel
#

Well, the fundamental theorem of galois theory tells you that the intermediate fields are the fixed fields of those subgroups right

sinful zephyr
#

ooooo

sinful zephyr
#

so one of the subgroups is {id, phi_4} that sends z_7 to z_7^4, and that permutes the roots

0 1 2 3 4 5 6 -> 0 4 1 5 2 6 3

so one of my intermediate fields equal to Z_2 is all elements of the form

a + b(z7 + z7^2 + z7^4) + c(z7^3 + z7^5 z7^6)

mild laurel
#

Wait I'm not sure that's the subgroup because (phi_4)^2 isn't the identity right

sinful zephyr
#

fff

#

{phi_1, phi_6}

#

so that should give me the field Q(z^1 + z^6, z^2 + z^5, z^3 + z^4)

#

where z = zeta_7

mild laurel
#

Yeah

#

There should be an easier way to describe this field but this works

#

I think it's just Q( cos(2pi/7)) but don't quote me on that

sinful zephyr
#

thank you thank you

#

then I should find a way to describe the other one, by looking at {phi_1, phi_2, phi_4}

#

and it's the thing I found earlier

scarlet estuary
#

elements of a group are... elements of the group

#

everything "in" a group is an element of it

#

oh wait

#

i thought by "or" you meant like

#

"which one are they"

#

lmao

#

ok so when we use the term "word"

#

we're usually referring to like

#

some specific combination of elements (united by the group operation)

#

so a given "element" might have multiple different corresponding "words"

#

for example, in the dihedral group $D_4 = \langle a, b \mid a^4 = b^2 = e, ab = ba^{-1}\rangle$

cloud walrusBOT
scarlet estuary
#

$e = a^4 = a^8 = b^2 = a^4b^2 = a^8b^8 = $ lots of other things

cloud walrusBOT
scarlet estuary
#

all of these are the same element, the identity

#

but in a sense, they're different "words"

#

anyway, the generators of $D_4$ are $a, b$

cloud walrusBOT
scarlet estuary
#

the elements are:

#

$e, a, a^2, a^3, b, ba, ba^2, ba^3$

cloud walrusBOT
scarlet estuary
#

and again, each element is gonna have multiple different words corresponding to it

#

(infinitely many, in the case of D_4)

#

does that clear things up?

#

so the answer is neither, "word" and "generator" are distinct (though related) notions

#

that said, if youre familiar with the idea of "equivalence class"

#

you could think of elements as equivalence classes of words

#

if you don't know what that means, don't worry about it

#

well

#

when you talk about free groups

#

you're building a new group off your "base" group

#

the elements of the free group are the words of the "base" group

#

up to group axiom simplifications

#

and in that case, yes, the elements of the "base" group are the generators of the free group.

#

so:

#

words of base group = elements of free group (up to group axiom simplification)
elements of base group = generators of free group

#

here the "base group" is just the group you built your free group off of

#

its not a common term, i just made it up

#

perhaps "base set" would be

#

a better term

knotty mason
#

it's usually called the generating set

scarlet estuary
#

since the underlying set is not necessarily a group

#

yeah

knotty mason
#

and i don't think it should have group structure

scarlet estuary
#

well it can

#

this arises when working with a lot of geometric notions

#

eg hyperbolic geometry

#

anyway uh

#

So let's say for a free group, every element of a set is a word, but not all words are elements of a set because not all words are necessary reduced

#

i guess i'd adapt this to

#

for a free group, every element of the free group is a word of the generating set

#

they all are except for when simplifications are "obvious" due to the group axioms alone

#

that is to say, $aa^{-1} = e$

cloud walrusBOT
scarlet estuary
#

this follows just from group axioms

#

and holds in any group

#

so $abcdd^{-1}$ would be the same element as $abc$

cloud walrusBOT
scarlet estuary
#

but besides that, they all are

#

its equal to e, the identity

#

so yes, but no one would write aa^-1

#

"equal" means "literally the same thing"

#

so you could write e as aa^-1

#

and it wouldnt change anything

#

besides perhaps annoying some people

#

they are literally the same object

knotty mason
#

it's a word but not a reduced word

scarlet estuary
#

well, in english grammar, A and a act differently

#

whereas in group theory, there is no difference between e and aa^-1

#

they are the same thing

#

i mean, i'd generally say that "A" and "a" are distinct things in English, since the sentence "I am a cat" is grammatically correct while "I Am A cAt" is not

#

but if you're saying uppercase = lowercase

#

then yes

#

they're the same

#

well, in group theory, we care very much about the "grammar" of groups (i.e. how they behave under the group operation)

#

much more than about how we write the individual elements

#

[this notion is captured by isomorphism]

#

but yeah i see what you mean

#

sure, yeah

#

that works

acoustic pine
#

If someone was having a touch of trouble while going through their first "modern algebra" course, but generally could grasps the concepts decently well, what texts would yall recommend to reinforce learning?

#

I was looking at Dummit and Foote, but its quite pricey

sharp sonnet
#

you can download the pdf for free on libgen

#

a cayley graph is a graph and everyone will understand you if you talk about edges and vertices/nodes of it

acoustic pine
#

@sharp sonnet I prefer paper copies of texts sadly. More than preference, its almost a requirement 😦

chilly ocean
#

you can print them

acoustic pine
#

:/

chilly ocean
#

sometimes i do that

sharp sonnet
#

hmm

acoustic pine
#

lotta paper

sharp sonnet
#

there are alternatives

#

artin's book is nice, but it covers less

chilly ocean
#

artin is great

acoustic pine
#

I've heard " A Book of abstract algebra" is p good

sharp sonnet
#

herstein's book topics in algebra is popular, but im not familiar

#

in general, algebra is well understood and i think most books do a good job

#

if you just want to reinforce learning, download multiple books

#

print the exercise pages

#

and do the exercises

fading sparrow
#

I did the case A integral

#

but now I'm lost

woeful flint
#

Is this a correct way to prove that if an ideal $I$ is maximal then $R/I$ is a field?:

Let $I$ be maximal and suppose $R/I$ is not a field. Take $J/I\subset R/I$ a non-trivial ideal of $R/I$. There exists a homomorphism $\phi :R\to R/I$ and so $\phi^{-1}(J/I) = J^\prime$ is an ideal of $R$. But $0\in J/I$ so $I\ subset J^\prime$ and $I \neq J^\prime$ as $J/I$ is a non-trivial ideal of $R/I$. So $I \subsetneq J^\prime \subsetneq R$ contradicting the fact that $I$ is maximal.

cloud walrusBOT
knotty mason
#

why don't you consider principal ideals in the ring R/I

#

I realized that this only works if you are dealing with commutative rings so perhaps ignore my comment

fading sparrow
#

so I figured that if ab=0 then f(a)=0 or f(b)=0, but not necessarily both...
what do I do then?

upper pivot
#

hmm i know 2 ways to do this but it doesnt follow the hint necessarily

#

@fading sparrow

#

do you know the analogous statement for fields?

#

that if E is an algebraic extension of F then embeddings of F into algebraicly closed fields can be extended to E

#

you can either 1. ||p much use the same proof changing little things to adapt to rings||

#

or 2. ||use some clever mods and localizations to deduce it from the field statements||

fading sparrow
#

so this is what I did so far

#

@upper pivot

upper pivot
#

well you say there is an inclusion B to Fbar

#

but thats what needs to be proved basically

#

so what is your proof for that

#

@fading sparrow

fading sparrow
#

uh

#

every $b\in B$ is the root of a polynomial with coefficients in A

cloud walrusBOT
fading sparrow
#

and the roots of those polynomials is contained in $\overline F$

cloud walrusBOT
upper pivot
#

the place where this will fail is that this isnt necessarily a homo

fading sparrow
#

hmm

#

I don't see the problem

#

if $b,b'$ are roots of $p(t)$ and $p'(t)$, then the product is a root of $(pp')(t)$

cloud walrusBOT
upper pivot
#

well not the product of the poly but i get what u mean

#

the idea is that how do you know you picked the right root i guess

#

for everthing to be consistent

#

you could do this if your extension was finitely generated, but that is not necessarily the case

fading sparrow
#

which part are you talking about right now?

upper pivot
#

just a respond to what you said

#

hmm so you are defining f(b) to be a root of the poly of b in Fbar right

fading sparrow
#

yes

upper pivot
#

i am asking how do you know which roots to pick, cause it needs to remain consistent

fading sparrow
#

I see

upper pivot
#

yeah its a bit wierd to think about

#

anyhow one of my hints will be helpful, if you scroll up

fading sparrow
#

thanks

#

but what do I do when A is not integral?

#

it's not obvious that f maps all zero divisors to zero

upper pivot
#

you mean when A is not an integral domain?

#

well consider the image of A

#

(in the field that is)

#

anyhow try looking up the proof for the fields statement, thats the important part

fading sparrow
#

right

slate brook
#

channel free?

chilly ocean
#

nope, $5.99

slate brook
#

seems free, ok

#

i'm trying to show that the middle term != 1 but that requires me figuring out what the lhs is

#

is this approach doable? or should i consider stuff like, a non rational cube root can not be expressed as a linear comb. of square roots

wind steeple
#

your approach is right but why computing F(cbrt(2)) ?

slate brook
#

i thought that if

#

is not 1, then they're not the same, hence cbrt2 not in F?

wind steeple
#

yes but that's not the good direction to take imo

#

you can look a sub extensions of F

slate brook
#

i.e. some field L such that
F(cbrt2) subset L subset F?

slate brook
#

i suppose this will work

#

since F contains Q but not Q(cbrt2)

wind steeple
#

It was that I was thinking

slate brook
#

nice, thanks

slate brook
#

i'm here for your algebra support once again for god knows how many times in the past 24 hrs

#

am i missing anything?

charred bison
#

$(R, +, \cdot)$

cloud walrusBOT
charred bison
#

when we write rings like this and say "additive group of the ring" etc. it doesnt actually refer to addition and multiplication but general terms for any operations, right?

mild laurel
#

any operation that satisfies the axioms yes

charred bison
#

damn i was confused for a while because "addition" and "multiplication" was used everywhere. got it, thanks

olive mirage
#

there was a textbook I saw that, "to avoid confusion" used an upward pointing triangle and a downward pointing triangle and it was awful

supple rampart
#

Can anybody tell me a few words on inner automorphism...specifically about its usefulness

olive mirage
#

Well, for any mathematical object, one of the best ways to understand it is to understand its symmetries, which for most objects form a group

#

(indeed, that happens almost by definition)

#

so, one strangely ends up understanding groups by undrstanding their automorphism groups

#

the inner automorphisms are a subgroup of the full automorphism group, and unlike the full automorphism group they are readily computable

#

in some sense, the inner automorphisms are the boring ones

#

so the outer automorphisms are the interesting ones, and frequently tell you when soething interesting happens

#

so for example, S_n has no non-trivial outer automorphisms except in the exact case that n=6

supple rampart
#

Wht should be my first reaction to this theorem
G/Z(G) isomorphic to I(G)=group of all inner automorphisms of G
By reaction..i mean
Things like
"Well thats obvious from definition or the way things are defined..." things like this

olive mirage
#

I don't think it is obvious, but I think it's proof is a really beautiful illustration of the first isomorphism theorem

supple rampart
#

Anything besides its beautiful proof ??

olive mirage
#

I think to me the most interesting part is taht it shows that all non-abelian simple groups include directly into their automorphism groups

supple rampart
#

Hmm..

dawn kiln
#

has anyone been able to get a good intuition for why there is one in the n=6 case

scarlet estuary
#

sorry, in what context?

#

ah, nontrivial outer automorphisms of S_n

dawn kiln
#

ya

scarlet estuary
#

i'd assume you want a better intuition than "the construction works"

#

but i'm afraid i'm not aware of anything that isn't just a proof sketch

dawn kiln
#

fair

#

tbf a lot of properties of S_n just seem to be because "that's how the algebra ends up"

#

like simplicity of A_5 or wtev

scarlet estuary
#

yeah, honestly im not sure theres a great intuitive way to reason about the internal structure of S_n

#

it doesnt help that |S_n| is big

#

which makes "concise examples" tough

mild laurel
#

was a nice read, only tangentially related

scarlet estuary
#

but im not sure it helps intuition for why S_6, specifically, is special

#

i mean, it should be fairly clear that the construction has problems in some other cases, but i cant think of a nice way to show that no construction works in any other case

#

other than just doing the algebra as you say

olive mirage
#

oh, it's because there are the same number of 3 cycles and 3-3 cycles

#

so the outer automorphism interchanges them

#

but generally, automorphisms preserve conjugacy classes, and the conjugacy classes in S_n are HUGE, becuase they correspond to cycle types

#

and certainly you can only exchange them if there are two conjugacy classes that represent elements of the same order in the group and have the same number of elements in them

#

and that is a backbreaking restriction

#

so I think the "right" intuitive view is that accidents like that shouldn't happen, but as so often happens with groups, it happens once

shy bluff
golden pasture
#

eh not really afaik it is easily verified

chilly ocean
fading sparrow
#

How do I read the hint?

#

if a is not completely contained in any of the ideals, then a is not contained in the union?

mild laurel
#

yes

fading sparrow
#

ok

fading sparrow
#

I was thinking of trying to show that the ideals $\mathfrak{b}$ are maximal in the ring $\bigcup_{\mathfrak{b}\in\varphi^{-1}(\mathfrak{p})}\mathfrak{b}$

#

ok TeX doesn't know mathfrac...

oblique river
#

it's mathfrak

fading sparrow
#

ups

oblique river
#

$\mathfrak{p}$

cloud walrusBOT
oblique river
#

frak is short for fraktur

fading sparrow
#

right

cloud walrusBOT
fading sparrow
#

do u have a hint for me?

oblique river
#

sorry, no, I just happened to open discord right as you made that comment

#

I'm abotu to go to bed, sry

fading sparrow
#

oh, good night then

golden pasture
#

hm it's atiyah macdonald's propostion 1.11

#

but the idea is you go by induction and construct an element that is in a but not in the union of ideals

left monolith
#

Hi

#

I was reading an abstract algebra proof and had a notation doubt

#

I don't know what does he mean when he writes (R)ϕ or (a1)ϕ

#

Any ideas?

sour plume
#

Looks like they're just writing functions to the right of their arguments

#

So (a)phi rather than phi(a)

#

Weird thing to do tho

left monolith
#

Oh

#

That would make sense

#

I was thinking about generating sets

#

Very weird thing to do

#

Thanks very much!

left monolith
#

Hi again 🤦🏽‍♂️

#

I don't understand why this reasoning proves the claim

#

Any ideas?

sour plume
#

Except for the statement about "Kernel of Homomorphism on Cyclic Group" not having a proof on its page, this seems like a fine proof

#

It ends with the general statement that we have now represented the subgroup H as the kernel of a group morphism going from a cyclic group to another group

#

And it seems to be a general statement that such kernels are again cyclic

left monolith
#

We wanted to see that H was cyclic, and we end up stating N is cyclic. At what stage we assume H=N or N cyclic => H cyclic?

sour plume
#

Oh, I see, they mistyped; N should be H

#

'cause if you look into the proof, "N" was never defined

left monolith
#

I see

#

So this row
∀x∈G:qH(x)=xN
should also be corrected to
∀x∈G:qH(x)=xH

#

Right?

sour plume
#

yuss

#

I did a sneaky edit, so now it should be fine

left monolith
#

Nice

#

Thanks again :))

charred bison
upper pivot
#

shows its closed under addition

#

btw this is more commonly refered to as Ann(U), the annihalator of U

charred bison
#

oh right, got it. yeah the notation was a bit weird to me someone else also told me. alright, thanks

kindred lichen
#

That is the subgroup test, A subset H of a group G is a subgroup of G if and only if for all elements a and b in H, ab^-1 is in H. In ring theory, the group operation is + thus you get a+(-b)=a-b

kindred mist
#

So I get that a module being Noetherian just means we have the ACC (ascending chain condition) on submodules by why would a Noetherian module necessarily have a maximal proper submodule? (i.e. why would the set of proper submodules have a maximal element by assuming the module is Noetherian)?

upper pivot
#

ACC is equivalent to saying every set of submodule has a maximal element

#

(try proving this)

hot lake
#

maybe you're confusing a maximal element with a greatest element

kindred mist
#

uhhh... if I didn't then there'd exist an infinite sequence M1, ... of distinct submodules. I'll try think of the other direction

#

is this first part of what I said correct ^^^? If so then that's what I need

upper pivot
#

yeah its correct

#

also zef dont the two mean the same thing unless im missing something?

hot lake
#

I went to wikipedia just to make sure

#

greatest element would be global maximum

kindred mist
#

I just need a maximal submodule, not maximum

#

in the same way that (p) prime ideals in Z are maximal but not maximum

hot lake
#

yeah

kindred mist
#

that way if M2 is such a submodule, M/M2 is simple which is what I was looking for

upper pivot
#

ah i see

kindred mist
#

Thank you by the way

upper pivot
#

oh are you trying to make a composition series btw?

#

np

#

yeah composition series are p cool, although you need more than noetherian to show its existence

kindred mist
#

wha really? hmm I thought noether. was good enough Ok I need to look into this now ty 😄

#

yes though I want a comp. series

#

The other direction of that theorem I think I might have: assume ACC does not hold, then we have an infinite chain M1, ... i.e. with no maximum

upper pivot
#

yeah should work

kindred mist
#

well actually btw I just wanted a finite chain that starts and ends wherever where each consecutive quotient is simple

#

do I recall correctly that a composition series would be "maximal" in the sense of starting at the whole module and ending at zero?

upper pivot
#

yeah

kindred mist
#

ok so noetherian should be good enough not for comp. series but just to have a finite sequence Ma, Mb, ... , M_whatever such that each consecutive quotient is simple (which isn't necessarily maximal)

upper pivot
#

hmm i dont think so

#

so you want a $M_1 \subset M_2 \subset \dots$ with $M_{i+1}/M_i$ simple

cloud walrusBOT
upper pivot
#

if im understanding right

woven delta
#

what do you mean simple?

kindred mist
#

ya but it can be length 2 also

#

no proper nonzero submodules

upper pivot
#

no non-trivial submodules

woven delta
#

oh ok

upper pivot
#

i think you need DCC to garuntee this exists

woven delta
#

can you prove DCC from this?

kindred mist
#

pretty sure no

#

DCC => ACC but not the other way around iirc

woven delta
#

lol

upper pivot
#

not in a general module

kindred mist
#

on modules at least ok nvm idk 😄

upper pivot
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this is only true for rings

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artin rings are noetherian rings of dim 0

kindred mist
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wait if we can step back to the other part though

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if ACC is equivalent to: every set of submodules has a maximal element, can't we just take the set of all proper submodules, call M2 a maximal element of that, then M/M2 should be simple, no?

upper pivot
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yeah we could do that

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but then you have a descending chain

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M subset M2 subset dots

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and this doesnt need to be finite

kindred mist
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yes, makes sense

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I didnt understand the dimension zero thing though, could you elaborate on that?

upper pivot
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oh yeah thats just a fancy way of saying

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all prime ideals are maximal

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(dimension is basically the supremum of the lenght of prime ideal chains in a ring)

kindred mist
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hmm that's neat hopefully this will show up in the Hungerford book Im reading it sounds really interesting

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I always liked studying the structure theorems in rings/modules

upper pivot
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yeah theres even stronger structure theorems for artin rings

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(ill not spoil em its nice)

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and particularly in regards to composition series, theres some satisfying theorems about them which i think you will enjoy once you read about it

kindred mist
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whee ok ty

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Ill look forward to that

brisk granite
golden pasture
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$$\frac{\mathbb Z[x]}{(p,f(x))}\cong\frac{\frac{\mathbb Z}{p\mathbb Z}[x]}{f(x)}$$

cloud walrusBOT
mild laurel
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@brisk granite

brisk granite
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oh, I see

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so f has to be irreducible mod p?

mild laurel
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yep

brisk granite
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I don't fully see why those two things are isomorphic tho

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should I try to make a map or is there a better way?

lime skiff
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I'm trying to find the splitting field of x^3-3

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is it just Q(3^(1/3), alpha, conjugate of alpha)

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where \alpha is the complex root of x^3 - 3?

fierce perch
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@brisk granite the natural map Z[x]/(p, f(x)) into Z/p[x]/(f) is surjective. What is the kernel?

brisk granite
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uh, what is the natural map?

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ok, so, I found a nice homomorphism from Z/pZ [x] to Z[x]/(p,f(x)) with a kernel of (f) so I think Im good

left monolith
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Talking about rings, this looks like true, but is there any where I can find a proof, don't know how to look for it really..

cloud walrusBOT
mild laurel
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How have you defined <X>?

left monolith
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So, given a ring R and a non-empty subset, <X> is an ideal

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I defined <X> as the intersection of all ideals I_i which contain X

golden pasture
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<X> would also be the ideal generated by X then?

left monolith
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Exactly that

golden pasture
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<X> intersect (anything containing <X>) is <X>

mild laurel
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Wait I'm confused, if you've defined <X> in that way, then this is the definition

golden pasture
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i think this is the kind if things that has like a few equivalent ways

left monolith
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Yeah, I just copied it from my classnotes

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I see the idea behind but I'd like a rigurous proof

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And can't find it

golden pasture
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The ideal generated by X is:
RX(rx for all r in R, x in X)
The intersection of all ideal containing X

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Show that if a ideal contains X, it must contain <X>

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for whatever definition of <X> you have

left monolith
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I'll give it a try that way

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Thanks

random ravine
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Is anyone able to give a hint for 2a

wind steeple
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you have a theorem for eigen values of self-adjoint endomorphisms

random ravine
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i'm still lost

wind steeple
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what do you know about reduction of self-adjoint endomorphisms
?

left monolith
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Sorry for asking so many questions in such short time, I hope I'm not breaking any rule or anything, I'm just studying for my exams next week and have questions 😭
Do ideals obey a distributive rule with the ring's operations?
Formally:
Let (R,+,·) be a ring and A1,...,An,B ideals of ring R.
Is it true that (A1+A2+...+An)·B = A1·B + A2·B + ... + An·B ?

olive mirage
#

Try to prove mutual inclusion

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it will be helpful for breaking down the definitions involved

left monolith
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So get an x, suppose it belongs to the ideal in the left hand side and prove it then belongs to the right hand side.

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And the same the other way, right?

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Previously proving that both sides are ideals, which is pretty trivial

brisk granite
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so, showing (A1 + A2)B is in A1B+ A2B is easy. For the other inclusion, note A1 is in A1 + A2 and hence A1B is in (A1 + A2)B. Similarly, for A2B. Hence, A1B + A2B is in (A1 + A2)B

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induction will get you to the case with n ideals

left monolith
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I hate induction 😭 never manage to get where I want

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Thanks for your answer, I'll give it a try

solemn rain
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@left monolith suppose the statement is true at n = k

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show that its true at n=k+1

left monolith
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Ok! Got it!

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Any clues for the first inclusion prove?

solemn rain
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i got lost what exactly

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do you want to prove

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@left monolith

left monolith
#

Sorry for asking so many questions in such short time, I hope I'm not breaking any rule or anything, I'm just studying for my exams next week and have questions 😭
Do ideals obey a distributive rule with the ring's operations?
Formally:
Let (R,+,·) be a ring and A1,...,An,B ideals of ring R.
Is it true that (A1+A2+...+An)·B = A1·B + A2·B + ... + An·B ?
@left monolith

This is my claim

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Proving by reciprocal inclusion is a good idea

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I got the inclusion from right to left

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I'm missing the left to right and have no clues on how to start it

solemn rain
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okay so you want to show

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for your induction

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(A_1+A_2)B = A_1B + A_2B

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right?

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@left monolith

oblique river
#

@left monolith your "proof by induction" isn't really using induction at all...

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in an induction proof, you prove P(k) => P(k+1) (or P(k-1) => P(k) depending on how you want to index it)

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where in your proof have you ever assumed P(k-1)?

solemn rain
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wass are you done?

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i have a problem

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too 😄

left monolith
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I fixed it after mo2men told me @oblique river and didnt sent the correction

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My problem now lies on the first inclusion

solemn rain
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(A_1+A_2)B = A_1B + A_2B
[8:43 PM]

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right?

left monolith
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right?
@solemn rain
Yes, and I use induction for the inclusion from right to left

solemn rain
#

what?

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u are now using induction and now trying to prove base case

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after u do that

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u assume the statement is true for k and show its true for k+1

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got it?

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@left monolith

left monolith
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Yes, induction part is SOLVED

solemn rain
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okay so your stuck with the base case?

left monolith
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I used induction (and didnt share my final sheet) to prove that:
(A1+A2+...+An)·B ⊃ A1·B + A2·B + ... + An·B
Now I'm missing the proof that
(A1+A2+...+An)·B ⊂ A1·B + A2·B + ... + An·B
And those to proofs together, proof my initial claim:
(A1+A2+...+An)·B = A1·B + A2·B + ... + An·B

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So I'm missing the left to right inclusion

solemn rain
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you knwo what those operations eman right?

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mean*?

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between the ideals

left monolith
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Yes

solemn rain
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okaay so

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let x be in (A_1+A_2+....A_n)B

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x = (a_1+a_2+....a_n)b for some b in B

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is (A_1+A_2+....A_n)B an ideal in R?

left monolith
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is (A_1+A_2+....A_n)B an ideal in R?
@solemn rain
Yes, because sum of ideals is ideal and product of ideals is again ideal

solemn rain
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okay so can we distribute (a_1+a_2+....a_n)b

left monolith
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x = (a_1+a_2+....a_n)b for some b in B
@solemn rain
And a_k in A_k?

solemn rain
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yea

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sorry

left monolith
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okay so can we distribute (a_1+a_2+....a_n)b
@solemn rain
But distributing means using what I'm trying to proof isn't it?

solemn rain
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x is an element in a ring

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(a_1+a_2+...a_n)b is an element in a ring

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multiplication distributes over addition in rings no?

left monolith
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Oh yes

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Got ya

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And distributing I get a_1b + a_2b + ... + a_nb

solemn rain
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yea

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which is an element of ?