groups-rings-fields

[half nebula]

So I just have to find a way to eliminate the possibilities of them having orders 1 and 3. I already know if it has order 1, it's the identity element so it can't be that.

[oblique river]

no

[half nebula]

Oh.

[oblique river]

that's not what the problem asks

it doesn't say "show everything has order 9"

[half nebula]

Oh wait okay.

[oblique river]

it says that a^9 = e for all a

[half nebula]

Oooh yeah because if an element has order 3 then that has to mean it raised to the power of 9 is still e.

So is everything else still right? Like every element in a group of order 27 has to have an order that divides 27 right?

[oblique river]

there's a theorem that says that's the case

[half nebula]

Okay cool ^^

But it can't be 27 since it's non cyclic.

Also I swear I've lost 4 freaking pencils today, no idea how.

[oblique river]

rip

[golden pasture]

could be super trivial but like anyone has an example of some subgroup $H$ of $G$ and some element $k$ such that $Hk\subsetneq kH$?

[cloud walrus]

ariana:

[solemn rain]

any non normal subgroup?

or am i missing something

[golden pasture]

nonnormal we have $Hk\not\subset kH$

[cloud walrus]

ariana:

[upbeat juniper]

I was thinking of the group of bijections on the positive reals with composition

let H be the subgroup generated by the map x -> 2x and k be the squaring map

[golden pasture]

hm something along that line seems to work nice

[shy bluff]

Ah @next obsidian apparently the question I showed you yesterday, the one about the centre of an action, that was actually supposed to be the kernel of an action

[next obsidian]

Okay, that makes more sense.

[shy bluff]

Wait question

What is the kernel of the conjuggation action?

Conjugation takes in 2 things doesnt' it?

[wicked bobcat]

I believe the kernel of the conjugation action is all elements of G such that the conjugate of the element fixes the element

and if G is abelian, then the kernel is G itself

Think of conjugation as taking in k instead of k and g

I believe, i havent dont group theory in some time

[shy bluff]

Hrm

So treat g like ti's fixed?

[wicked bobcat]

I believe the kernel is the set of all g that fix k, yes

[shy bluff]

oke

[wicked bobcat]

So consider it going through all the g, picking out the ones that fix all the k that would be the kernel

[shy bluff]

Oh so the kernel of the conjugation action is all $h \in G$ such that $ghg^{-1} = h$?

[cloud walrus]

Liria ^(;,;)^:

[wicked bobcat]

Replace the first h with g and i believe thats it

[shy bluff]

Oh ok

I see then

I think

Err wait no what I meant to say was

If $\phi$ is the conjugation action on $G$ and $h \in Ker(\phi)$ then $h$ is all $h \in G$ such that $ghg^{-1} = h$ for all $g \in G$

[cloud walrus]

Liria ^(;,;)^:

[shy bluff]

And then you just gog and right multiply both sides by $g$ to obtain $gh = hg$ and thus it's the centre

[cloud walrus]

Liria ^(;,;)^:

[wicked bobcat]

The kernel is a set of elements of the group doing the action G, not on the set being acted upon

[shy bluff]

Pardon?

[wicked bobcat]

As in, the kernel is $g \in G$ such that $ghg^{-1} = h$ for all $h \in G$

oh hang on

[cloud walrus]

danii:

[shy bluff]

Oh

I swapped them

I see

[wicked bobcat]

Yeah, the kernel contains the elements of the group actually doing the action

it's a bit confusing here because G is acting on itself but you get the idea

[shy bluff]

I think I see it

[wicked bobcat]

The definition of the kernel should be in any group theory textbook/you could probably find an online pdf which explains it better than i could do

[next obsidian]

g is in the kernel of conjugation if and only if gkg^{-1} = k for all k in G

If you view an action as a map from G into the symmetric group on the set you’re acting on it’s the usual notion of kernel

[half nebula]

Hey is this channel still being used?

If not, could I have some help getting this started? It's been a hot minute since I've done Iso and Homomorphisms.

[wicked bobcat]

to prove a, just show that (a,b)->a preserves the group operation. as in (denoting group operation by . ) p((a.c,b.d))=p((a,b)).p((c,d))

then it should be easy to show that it is onto

from that you should be able to get the kernel immediately

[half nebula]

Wouldn't that be ab=ac though?

[wicked bobcat]

oh, mistyped it

there

[half nebula]

Oh and that still works because x=(a,b) and y=(c,d) so xy is (ac,bd)

Okay I get it ^^

[wicked bobcat]

Yeah!

Glad to help :)

[half nebula]

And as for onto, that's just that it's surjective or something right?

[wicked bobcat]

Yep

For the kernel, remember that it's defined as all the elements that go to zero in this case

[half nebula]

So like when a goes to 0 since that's what's outputting?

[wicked bobcat]

actually wait, this is a group homomorphism

The kernel is defined as all the pairs (g,h) such that p(g,h)=1 (identity in G)

[solemn rain]

do you mean 1

do you mean the kernel is all ordered pairs (a,b) ssuch that phi(a,b) = 1 ?

[wicked bobcat]

Sorry, yes i believe that is it i must be getting multiple things confused

[solemn rain]

oh yea cool

hahaah same

[half nebula]

Wait so I just say that ker(phi) is all ordered pairs (a,b) in GxH such that phi(a,b)=1? Like that's all I'd need?

[wicked bobcat]

Well, you need to get what the pairs (a,b) in GxH such that phi(a,b)=1 are

cause Find

[half nebula]

Well that would just be pairs where a=e right because 1 is the identity element.

Right?

[wicked bobcat]

It would be (e,h) for all h in H yep

[half nebula]

Okay cool ^^

[wicked bobcat]

which is isomorphic to H

[half nebula]

But don't I also need to account for the other way? When it's (h,e)?

[wicked bobcat]

Well it's an direct product/ordered pairs, h is not an element of G so (h,e) doesnt exist in GxH

[half nebula]

Sorry for so many questions but I just want to make sure I get this all as right as I can for my final next week. For a, I said that since H is a subgroup in G that means h in H is also in G. Since G is an Abelian group that means ah=ha and since h is all elements in H that means aH = Ha. Is that correct?

[solemn rain]

yea

[half nebula]

Okay cool ^^

[solemn rain]

cool

subgroups of abelian groups are all normal

but the converse is not true

cool af

[half nebula]

Oh is that what that proves?

[solemn rain]

yea

a normal subgroup is a subgroup that satisfies

what u just proved holds for abelian groups no?

do u know about normal subgrous

[half nebula]

Yeah it's where, in this examples, aha^-1 is in H right?

[solemn rain]

yea

aH = Ha <--> aHa^-1 is in H

[half nebula]

Oh cool ^^

[solemn rain]

<--> normal

[half nebula]

I feel kinda dumb for asking this but for the subgroup made by 11 in Zx_24, do I add or multiply? Like is the second element 22 or is it 121 which turns into 1 (121 (mod 24)=1)

[solemn rain]

whats Z_2

Z_24

the cyclic group of 24 elements

or Z/24Z

[half nebula]

This

[solemn rain]

yea it depends on notation for me

but yea

so here this means U(24)

if this makes sense to you

U(24) = { a |(a,24) = 1 }

define on this multiplication mod 24

so yea its multiplication

mod 24

[half nebula]

Okay cool.

Thank you ^^

[solemn rain]

np 😄

[half nebula]

So the subgroup of 11 in U(24) is 0, 1, and 11 right?

[solemn rain]

subgroup generated by 11

u mean?

[half nebula]

Yes.

Actually 0 isn't in it I think. I'm unsure though.

[solemn rain]

then just do the operation

powers of 11

where operation is mod 24

[half nebula]

Because it starts with 11, then 11^2=121 which is 1, then 11^3=1331 which is 11.

[elder valley]

0 isn't in U(24)

[half nebula]

Okay cool. So just 1 and 11.

[elder valley]

so can't be in <11>

[half nebula]

So now for the coset part, I just multiply them until it loops back around.

Like this but instead of addition, multiplication

Or well actually yeah because 1 times the set is still just the set 1 and 11 and 11 times it is also still just 1 and 11.

Right?

[stone fulcrum]

The × means "multiplication" in this sense

@half nebula
Sorry to ping 40 mins in advance

[half nebula]

Lol you're good. I already knew that.

Oh wait you meant for the coset part?

[stone fulcrum]

$\mathbb{Z}^{\times}_{24}$

[cloud walrus]

Kaynex:

[stone fulcrum]

The × means "take the elements that don't have an inverse, and get rid of them. Whatever's left is the group" and in this context implies multiplication

@half nebula
Again, sorry for late ping

[half nebula]

I thought it was units that aren't divisible into 24.

Like 5 is only divisible by 1 and itself and it doesn't go into 24 so it's a member of Zx_24

But if it was Zx_25, then like 2 would be in it since 2 doesn't go into 25.

[next obsidian]

It's equal to the elements that are coprime to the number

But that is exactly the same as those which have an inverse under multiplication by Bezout's lemma

[elder valley]

i think his point was that $R^\times$ is defined for any ring in the way that kaynex described. and the fact that in $Z_n$ it's just the numbers which are coprime to n is a theorem

[cloud walrus]

Auvera:

[half nebula]

Idk what coprime means or a ring means tbh. Our professor's notes just say that it's numbers that aren't divisible into the bottom number.

[stone fulcrum]

Fair! Then go with that

[shy bluff]

Coprime means that their only common divisor is 1

[stone fulcrum]

It does mean "multiplication" here

[half nebula]

That you times it.

I guess ><

[elder valley]

i never thought about it but does the "co" in coprime come from some duality statement in a categorical sense?

[shy bluff]

So given this

With the "usual action" being this

What exactly is the stabilizer?

Like the stabilizer of x is all in g in G such that the action of g on x does nothing, if that's what I'm understanding yes?

[latent anvil]

@elder valley probably not

I think coprime as a term is older than category theory

[half nebula]

Either way though I'm ready for my final that will more than likely be in no way whatsoever similar to the review because our lasts two test were the exact same way, where the review only covers the first half of the chapters and the actual test is the second half.
Also thank you all for helping me out this semester ^^

[latent anvil]

I've always assumed it meant "co" as in "together"

[elder valley]

i figured, wishful thinking xD

[latent anvil]

like, jointly prime

[next obsidian]

@shy bluff yeah that’s the definition

[shy bluff]

But how does that work here?

[next obsidian]

Idk what it is in that case tho

¯_(ツ)_/¯

[shy bluff]

:x

Guess I'll email my prof 😔

[next obsidian]

I mean

Do you mean you don’t get what it is as a set?

Like you can’t compute it?

[shy bluff]

hrm

So the usual action on functions that we've defined works on f: X -> Y

But b maps from 2 inputs to 1 output ,not from 1 inptu to 1 output

[next obsidian]

No

The domain is still one thing

Like the pair (a,b) is ones object

[shy bluff]

Oh

[next obsidian]

I think like g(a,b) would be (ga,gb) there

[shy bluff]

Oh

[next obsidian]

So you can just let it act on each like index

[shy bluff]

So then g^{-1}(a, b) = (g^{-1}a, g^{-1}b)

[next obsidian]

Yeah

[shy bluff]

And then b of that whole thing is the dot product of that

Ok

I see

ok nvm then I think that this problem is doable

[next obsidian]

Yeah do

So*

It says diagonal action on the product

So to me that’s what I described

In general for a set S, the “diagonal” thing on S x S

Is just do the same thing on both

So the diagonal of S is as a set literally just the set of (a,a) in S x S where a in S

[shy bluff]

Oh that's what they menat by diaggonal action on

Err so that's how you're suppoesd to use that bit

And what about the trivial action on R?

[next obsidian]

Uhh

That’s like

The do nothing action

[shy bluff]

So g * x = x for all x in R?

[next obsidian]

Every group acts on every set by doing nothing

Yeh

[shy bluff]

Oke

[shy bluff]

Wait question, I get to like $g \cdot (j^Tg^{-T}g^{-1}k)$

[cloud walrus]

Liria ^(;,;)^:

[shy bluff]

Because the action of $g$ on R is trivial, we have that this just becomes $j^Tg^{-T}g^{-1}k$

[cloud walrus]

Liria ^(;,;)^:

[shy bluff]

OH wait nvm I see how this works now

[shy bluff]

How are you supposed to go about showing this? I see that |G| = n!, and that you can use orbit-stabilizer theorem but I dont' think that you're allowed to use that here

[next obsidian]

Wtf is P_sigma?

Does that like jumble up the coordinates the way an element of S_n would do it?

Like if you have (12) the first coordinate to second, second to first?

Because if so then it’s just combinatorics

The size of a stabilizer is fixed for all elements in an orbit by orbit stabilizer, or you can show it by hand

SO WLOG you can make it go into blocks so like

Lambda1 occurs m_1 times in a row

So it looks like (lambda_1,lambda_1,...,lambda_1,lambda_2,...,lambda_2,...,lambda_k,...,lambda_k) where each “block” has size m_i

Then an element which fixes it needs to permute only those elements in each block

You don’t even need to put it into blocks like this, but that’s the easiest way to see what’s going on

@shy bluff

[shy bluff]

P_sigma is just permutation matrix

Uh

Yea that's what I was thinking

But I do'nt really know how to show it lol

[next obsidian]

That’s the proof tho

You can put it into that form

And you can do combinatorics from there

[shy bluff]

huh

[next obsidian]

Anything which permutes the elements in the blocks will fix it

And vice versa

So like you have m_1! ways to permute the first block

m_2! ways to permute the second

...

m_k! ways to permute the last

and then you multiply them because anything which fixes it is necessarily a product of stuff that permutes each block

It’s like a cycle decomposition almost

[shy bluff]

That's what it is yea

and thats' why each one is a factorial

[next obsidian]

Yup

The only thing you have to justify is putting it in that form

Now you don’t haaaave to

But writing it would be a pain in the ass without it

But you can justify that anything in the same orbit has the same sized stabilizer

Either by orbit stabilizer

Or explicitly constructing a bijection

And from there clearly there exists an element of S_n which puts it in that form

Man, I love group theory

[shy bluff]

I wish I understood enough to love it :x

[latent anvil]

It'll come in time

[next obsidian]

Tbf me and Shamrock learned from the same group theory obsessed 17 year old

So

Biases

[shy bluff]

17 year old?

[next obsidian]

But a lot of algebra people seem to like groups a lot

[shy bluff]

wow

Idk I'm just curious what pmath is like and so far it seems to be "go very fast and ggive no examples"

this is my uh first real pmath class I guess

[next obsidian]

It depends on the person

And where you’re at

If you’re at Harvard you’ll get a different experience than like, idk some small state school in Vermont m

[shy bluff]

Mmm yea I think that everyone else in my class came from the advanced math streams at my school

while I came from engineering

[stone fulcrum]

Group actions are pretty bae. They are used as easy roads to get some of the more advanced group theorems

[shy bluff]

I see

[next obsidian]

Group actions took my second time around to get it

[shy bluff]

I see

[next obsidian]

But it’s amazing because in reality a group action is a map from G to S_X

[stone fulcrum]

The stabilizer of common actions usually define useful subgroups

[next obsidian]

The symmetric group on X

So if X is finite you get a map into S_n where n = |X|

And being able to get these maps for free is really useful

[shy bluff]

Yea

That bit I get and it's cool

[next obsidian]

And yeah, I think Kaynex makes a good point

I think you had an assignment to show the kernel of conjugation is th center

[shy bluff]

I think that I'm just having a lot of trouble expressing it mathematically?

Yea

I got that one

Once someone explained to me what I was actually conjuggatingg

[next obsidian]

It comes with time for sure, you’ll see more and more how useful things are described in terms of maps and things acting on another thing

Like a lot of topological group stuff is about group actions, and stuff you’re familiar with are secretly group actions and then you can apply group theory to it

And group theory is nice because groups are really nice, they’re simple and the theory is well-developed

[latent anvil]

it's really important to keep examples in mind imo

[next obsidian]

^^^

[shy bluff]

Yea I'm starting to see that a lot of stuff is group actions/etc but jsut not called such

[latent anvil]

Like, you should be able to list all the groups of order < 12

[next obsidian]

Algebra is born of examples. really all math is tbh

[latent anvil]

And most of the groips of order 12

[shy bluff]

Uh

Nope I ggot nothing

[next obsidian]

I just like algebra a lot so I have a lot of example to keep in mind

[shy bluff]

Dihedral groups?

And symmetric groups?

Those are the ones that come to mind

[next obsidian]

Yeah, so you know semi direct products now right?

[shy bluff]

Kinda yea

[next obsidian]

If you have time and like it and want to try something, try to classify groups of order 8

[latent anvil]

It sounds dumb but I think it's useful to literally just go order by order

[next obsidian]

You’ll learn a lot

[latent anvil]

Like start at 1

[shy bluff]

oke

[latent anvil]

What groups of order 1 are there?

[shy bluff]

Trivial group?

[latent anvil]

Yup

[shy bluff]

Nothing else right

[latent anvil]

and that's it, right?

yeah

[shy bluff]

Ok yea that's what I thought

[latent anvil]

What about 2?

[shy bluff]

Z/2Z?

[latent anvil]

Sure

Can there be any others?

[shy bluff]

I dont' think that dihedral ggroups work at n = 2?

[next obsidian]

It does, but it ends up being the same.

[shy bluff]

Oh

[woven delta]

Well it works formally

[shy bluff]

I guess that S_2 is technically still a group?

[next obsidian]

Think about groups of prime order

[shy bluff]

Everything is just isomorohpic to Z/2Z though if of order 2 thouggh

[next obsidian]

Can you say anything useful about them?

[shy bluff]

I think

[woven delta]

Yeah that's correct

[shy bluff]

Groups of prime order... they only have trivial subgroup and itself?

[next obsidian]

Sure

And that’s characteristic of what kind of group?

[shy bluff]

Oh this is the thing that we don't have time to cover in our course I think

Simple group!

[next obsidian]

That’s true indeed

[woven delta]

No, that's if no nontrivial normal

[next obsidian]

But that’s for normal

[shy bluff]

oh

[next obsidian]

So let’s think about elements

The identity

Ignore that

It’s in everything

We can say something crucial about any non-identity element tho

[woven delta]

By Lagranges theorem, we know the order of an element has to divide the order of the group

Therefore if your group is prime order...

[shy bluff]

Every element has the same order, and that order is either 1 or p? And I take it that it's p because they're uh cyclic?

[woven delta]

Yeah

[next obsidian]

Well it’s backwards

Because order is p

It’s cyclic

So groups of prime order must be Z/pZ (up to iso)

So you’ve classified all groups of prime order

[shy bluff]

oh wait question what does "up to" mean

[next obsidian]

🥳

It just means like

They’re all isomorphic to Z/pZ

Because you can as a set express it a gajillion ways

You can think of it as modding out by the relation G = H iff G is isomorphic to H

[shy bluff]

hrm

[next obsidian]

But idk if that will help haha

Basically by virtue of the fact it has an element of order equal to the group

You can make an iso to Z/pZ

[shy bluff]

Why don't we just say that "groups of prime order must be isomorphic to Z/pZ" instead of "up to isomorphism"

[woven delta]

That is the same thing

[shy bluff]

oh ok

[next obsidian]

I just want to say

“We have one group, up to isomorphism”

Becaus ethen I can also say “we have two groups, up to isomorphism”

[woven delta]

Yeah

[next obsidian]

Instead of like, any groups of ___ category are iso to either G, or H, and G and H are not isomorphic

That’s a mouthful

[woven delta]

Well you can also just say "a group H of order n is iso to G_1 or G_2"

Oh lol

Sure

[next obsidian]

But you’d have to specify G_1 and G_2 aren’t iso

So it gets tedious

[woven delta]

just say xor

[next obsidian]

Unless you mean exclusive or

Lmfao

Anyway, you classified all groups of prime order so you hit off2,3,5,7

So the only ones left are 4,6,8

[latent anvil]

I said < 12

[next obsidian]

Shhhhh

[woven delta]

Lol

[latent anvil]

so also 11 is done and 10 to go

[next obsidian]

Sure

And 12

Oh, <

[shy bluff]

4, 6, 8, 10 we can def say permutation + dihedral groups right

[woven delta]

Well 9 and 12 are a bit hard

Oh lol

[latent anvil]

liria, what's the permutation group of order 4?

[next obsidian]

4 wont have any symmetric group

[latent anvil]

or 8? or 10?

[next obsidian]

But you don’t want to list examples

[latent anvil]

what's the order of the nth symmetric group?

[shy bluff]

Oh wait

Whoops

[next obsidian]

You need to show for any group of order 4, that it’s isomorphic to either G or H

[latent anvil]

no this is good!

[shy bluff]

Yea it's order n not the other way around

Lol

[latent anvil]

this is why we do examples

it's order n!, not n

[shy bluff]

Yea

Dihedral group though, there is one of order 8, 10, 12?

[next obsidian]

Have you classified groups of order p^2?

[latent anvil]

Yup, any even number

but lets go order by order

not group by group

what groups can you think of of order 4?

[shy bluff]

D_4, Z/4Z, Z/2Z x Z/2Z?

[latent anvil]

okay so just to clarify

[cloud walrus]

shamrock:

shamrock:

[latent anvil]

there's two different conventions so I want to check first

[shy bluff]

The former

[latent anvil]

okay, cool

Those are three groups of order 4

[shy bluff]

[next obsidian]

So two of those are isomorphic actually.

[latent anvil]

But I claim you've only listed 2 groups up to isomorphism

[next obsidian]

Can you try and guess which one?

Which two*

[latent anvil]

so two of those are isomorphic but they aren't isomorphic to the third

[shy bluff]

D_4 and Z/4Z are the same up to isomorphism?

[next obsidian]

Let’s think about it

Let’s think about elements

[shy bluff]

Sure

[next obsidian]

Let’s list the orders

In Z/4Z you have 1 element of order 1

[shy bluff]

D_4 = {e, r, s, s^2}, Z/4Z = {e, 1, 2, 3}, Z/2ZxZ/2Z= {(e, e), (1, e), (e, 1), (1, 1)}

[latent anvil]

that D4 you listed has 5 elements

[next obsidian]

Uh, D4 has 5 elements under that

Cool

[latent anvil]

and your Z/2Z×Z/2Z has 25

[shy bluff]

Is that right?

[next obsidian]

2 = e in Z/2Z

[latent anvil]

no

There should be 4 elements in each

[next obsidian]

You listed off Z/3Z x Z/3Z

[shy bluff]

whoops

[next obsidian]

They edited

[shy bluff]

ok now it's edited to be right

[next obsidian]

Okay perfect

[latent anvil]

no worries!

being wrong is how you end up being right

[next obsidian]

So in Z/4Z what are the orders of each element?

[dull ingot]

at most 4

[latent anvil]

also, your D4 is wrong

what's the presentation of D_2n?

[next obsidian]

Oh it should be rs instead of s^2?

[shy bluff]

Order of e is uh 1, order of 1 (under addition) is 4?

[latent anvil]

Yes but I didnt want to say that

[shy bluff]

Oh

Oh rigght it should be rs not s^2 yea

[dull ingot]

What is e? Is it the identity element?

[next obsidian]

Yeah

[shy bluff]

Yea

[dull ingot]

thx

[next obsidian]

So we can actually speed this up a bit I think

Let’s go with orders of elements of D4

|e| = 1 right?

[shy bluff]

Yep

[next obsidian]

It’s just, gotta be

[shy bluff]

|r| = 2

[next obsidian]

What about s?

[shy bluff]

And |s| = n

[next obsidian]

Which is in this case?

[shy bluff]

Um 4

[next obsidian]

So you’re done!

It has to be cyclic

[latent anvil]

no

that's not right

can you write down the presentation of D4?

[next obsidian]

Oh shit yeah

[latent anvil]

Alex think about D6

And the order of the rotation

[next obsidian]

I get mixed up by the D_2n thinf

[latent anvil]

yeah

[shy bluff]

Wait is D_4 not the square

[next obsidian]

Yeah

[latent anvil]

Nope

it's not the square

[next obsidian]

It’s a... line segment?

[latent anvil]

That's D8

So the geometry is a little weird

Liria, do you know what I mean by "presentation"?

[shy bluff]

no

[latent anvil]

Hmm okay, how do you define Dn?

[chilly ocean]

Deez nuts

Ha got em

[shy bluff]

Uhhh a dihedral symmetry is the set of rotations and reflections about a n-sided polygon?

[next obsidian]

That’s if you go by the convention that D_n has order 2n

[latent anvil]

sure, so that definition doesn't really work for a 2-sided polygon

[chilly ocean]

Had an order of 2 nuts

Lmao got em

[latent anvil]

it's too degenerate

[shy bluff]

Wait give me like 5 mintues I gotta go grab the laundry

[next obsidian]

No worries

[latent anvil]

Did you know that world-renowned writer Stephen King was once hit by a car? Just something to consider

[chilly ocean]

I got something in the mail today. Do u know what it is?

[latent anvil]

Oh yeah you asked about knot theory and Operads

Earlier

So basically, lots of knot invariants can be thought of as coming from hopf algebras

If a hopf algebra is good enough then the braid group will act on its category of modules

and this can be used to produce braid invariants, and then knot invariants

Specifically the "good enough" condition implies that the category of modules is a braided monoidal category, since the braid group always acts on braided monoidal categories

my project is trying to generalize this kind of process from braids to singular braids (where they can have self intersections)

[shy bluff]

I would like to understand these sentences but alas they mean nothing to mt

[latent anvil]

It turns out that braided monoidal categories are exactly the algebras over a certain operad, whcih is built out of braid groups

And that's why you get an action

So we're trying to generalize this and build an operad out of the singular braid monoids

Then look at the categories it has as algebras

And use this to produce singular knot invariants

The operad and knot theory stuff is actually pretty independent

@next obsidian

@shy bluff sorry this was in response to a question from earlier

Not your thing

[shy bluff]

Naw I recognize that

I would however like to understand what those words mean

[latent anvil]

I'm doing an REU this summer

research thingy for undergrads

I didn't understand most of these words three weeks ago

[shy bluff]

wow you're in undergrad too

[latent anvil]

Magician and I are classmates

[shy bluff]

I see

[next obsidian]

So Liria idk if you want to continue with classifying groups or do your hw haha

But you can classify groups of order p^2 where p is prime

You can show they’re either isomorphic to Z/p^2Z or to Z/pZ x Z/pZ

So in particular they are all abelian

That gets you groups of order 4

Groups of order 6 aren’t too bad

Then there are 5 groups of order 8

If you’ve covered semi direct products and Sylow’s theorem it’s doable to classify them

If you haven’t touched Sylow’s then groups of order 8 might be hard

[shy bluff]

Oh yes we learned that

But we did'nt calle it sylow's theorem

Or rather we did'nt ggive a name to it

Sorry I fell asleep

Or rather we learned some of those htheorems

[woven delta]

That's not sylows theorem, it's a corrolary

[olive mirage]

wait... how does Sylow's theorem help you classify groups of order 8?

[raw moth]

Groups of order 8 is okay

It goes along these lines:

Element of order 8 iff C8

Elements of order only 1 or 2 iff C_2^3

Otherwise let r be element of order 4

[olive mirage]

Wait, are we determining the isomorphism type, or proving we have a complete list?

[raw moth]

Both

[olive mirage]

mostly I was just thinking Sylow's theorem doesn't say anything about groups of order 8

[raw moth]

Suppose all elements besides r^2 are order 4; then must be Q8

Otherwise let s be element of order 2 other than r^2 and consider G as the union of <r> and s<r>

Which you can then show to be either D8 or C4xC2

Ofc there are details in each step

So the overall proof is not short

But each step is really quite short

[olive mirage]

yeah I like DF's writeup of this

though it is weird that 2^3 and p^3 are different

[raw moth]

Hmm, yes and no

[olive mirage]

Though I just found a nice write up by Keith Conrad that is pretty clean, as his writeups always are.

[raw moth]

The difference seems to me to stem from the fact that

[next obsidian]

Oh yeah, Sylow doesn't do shit for groups of order 8

I was thinking it might help for 6

And it will for 12 probably

[raw moth]

If you have all elements of order 2, you can easily conclude that it's abelian

[olive mirage]

(With 6 you can just use Cauchy)

[next obsidian]

Yeah, maybe, idk

[olive mirage]

yeah, the groups are fundamentally different

[next obsidian]

I just didn't want to promise you can classify them w/o too much trouble

[olive mirage]

since there is a group of order 27 where every element has order 3

[raw moth]

For 6 considering as cosets of order 3

[next obsidian]

and if it requires Sylow if she doesn't know Sylow

[raw moth]

Yeah Heisenberg group on Z/3Z

[next obsidian]

Some1 teach me where to learn about wreath product

I want to know them for no reason other than the name

[raw moth]

It's a little strange but understandable that 2 is special for this

[olive mirage]

yeah, I was just saying the p=2 case is fundamentally different. Though weirdly still 5 groups in each case

[raw moth]

Hmm

That is strange

[olive mirage]

Wreath products are just a particular form of semidirect product

[next obsidian]

Wait really

They sound cool tho

and special

[olive mirage]

so, discover it for yourself: Describe the Sylow 2 subgroup of symmetric groups, one at a time.

and you will naturally discover the idea of a wreath product

[next obsidian]

[next obsidian]

@latent anvil Dude you know that thing about like

Surjective for free module

of same rank is injective

[latent anvil]

yes

[next obsidian]

There's something kinda like it in Matsumura

If M is a finite A-module and f a surjective endomorphism of M

then it's an automorphism

It just reminded me of that other thing

[latent anvil]

Oh cool

I wonder if you can do this the same way

Wlog A is local

then linear algebra(???)

Idk

[next obsidian]

You use Nakayama

Consider M an A[x] module via xm = f(m)

then by assumption on f being surj, XM = M

So there is a y in A[x] such that (1 + xy)M = 0

now assume that u is in ker f

then 0 = (1 + xy)u = u + yf(u) = u

So u = 0, f injective 🙂

[latent anvil]

That's pretty cool

And makes sense

[undone orbit]

so, discover it for yourself: Describe the Sylow 2 subgroup of symmetric groups, one at a time.
@olive mirage wreath products are so cool! I've never tried to rediscover them this way but it seems fun

Another good way to understand how they work is from the lamplighter group examples imo

also if I remember correctly you can define wreaths in terms of group extensions with a universal property but I'm not quite sure which that would be

[lament dawn]

does anyone know wtf this notation d = (m,n) means? My textbook randomly started using it without any explanation whatsoever. This is in the section about cyclic groups

[woven delta]

It means gcd

@lament dawn

[lament dawn]

!thanks

oops

thanks!

and this? Sorry for asking so many notation questions, but the book doesnt really explain it and it is really hard to look up lol

the angle braces I mean

Is it a group generated by that single element or something?

[mild laurel]

Yes, dummit and Foote defines this earlier

[lament dawn]

must have missed it, thanks

also pretty neat you can recognize the tb from two propositions lol

[mild laurel]

I just taught this section like Wednesday so I remember reading it

[thorn delta]

I want to check if I understand what these vertical isomorphisms look like:

For the left one, is it like g maps to h if their images are the same under f and the inclusion respectively?

For the middle, its the identity.

For the far right, it looks like g maps to h if their preimages are the same under g and the canonical map respectively?

oops, forgot to mention, H -> G is the inclusion and G -> G/H is the canonical map here

[next obsidian]

@mild laurel you teach sections?

Also isn't class over?

kxrider, I think that's the case yeah

[thorn delta]

hm alright, thanks

[mild laurel]

@next obsidian me and max both are teaching a summer online camp for high schoolers

[next obsidian]

lol wut

like the AT thing?

or a different summer online math camp

[mild laurel]

Yeah that

I'm teaching some number theory classes

[next obsidian]

Wait so is this like official?

Are you associated to a uni / getting paid / whatever

or is it just like, ppl on this server were like "teach me math pl0x"

[mild laurel]

No, it was started by a Harvard undergrad

If you Google Alec sun summer camp you'll find it

It's a "donate whatever you want" model so

[next obsidian]

Oh, huh

[latent anvil]

It's p cool

I wish I could be involved

I'm helping hsers in max's class on here I guess

But I didn't want the time commitment with my REU

[uncut girder]

Btw @mild laurel is there anyway I could get involved even tho I'm like 3 weeks late

[next obsidian]

in a local ring with maximal ideal m, is being equivalent to 1 mod m equivalent to being a unit??

That can't possibly be true right?

[latent anvil]

No

Take a field

[next obsidian]

ughhhh, wtf is this step in this proof

[latent anvil]

It's being equivalent to a unit mod m I think?

[next obsidian]

no lol

[latent anvil]

No maybe not

[next obsidian]

take a field

[latent anvil]

Take a field?

For my thing?

[next obsidian]

that's the counterexample

yeah

[latent anvil]

no?

[next obsidian]

wut

[latent anvil]

Equivalent mod m means equality

[next obsidian]

Oh

[latent anvil]

In a field

[next obsidian]

Wait

I missed "unit"

ughhhhh

This is so dumb

I have some $a_i$ right, finite

[cloud walrus]

Mathemagician:

[next obsidian]

and have an expression of the form $a_i = \sum a_j c_{ji}$

[cloud walrus]

Mathemagician:

[next obsidian]

we also know that $a_i \notin (a_1,\dots,a_{i - 1},a_{i + 1},\dots,a_n)$

[cloud walrus]

Mathemagician:

[next obsidian]

and from this we somehow derived that $1 - c_{ii} \in m$

[cloud walrus]

Mathemagician:

[next obsidian]

and that $c_{ij} \in m$ for $i \neq j$

[cloud walrus]

Mathemagician:

[latent anvil]

I think I need more context?

Are we assuming the ideal generated by the ai is proper?

err, no I'm still confused

[next obsidian]

not necessarily I thinkj

I could try and transcribe the entire proof lol

Or like, take a photo

That's probably better

[latent anvil]

You get that (1-c)ai in (a1,...\hat{ai},...,an)

Right?

From that expression

[next obsidian]

Uhhhh I think so?

oh...

Sure

[latent anvil]

Hat meaning delete

[next obsidian]

one maximal ideal

...

[latent anvil]

But I don't get how you get that 1-c in m

because that ideal might not be proper

and it might not be prime

[next obsidian]

Maybe I can show it is?

[latent anvil]

If it's always prime I'm good with this

[next obsidian]

I mean you get that (1 - c)a_i in m

if the ideal is proper

[latent anvil]

Right

[next obsidian]

local ring

[latent anvil]

But you get 1-c in m if it's prime

I don't see how you get that without extra assumptions

[next obsidian]

like if 1 - c is prime?

[latent anvil]

If (a1,...,\hat{ai},...,an) is prime for each i

then I understand the implication

[next obsidian]

Umm, the a_i are basis elements if that helps

probably should've mentioned that

[latent anvil]

No they aren't?

They're elements of A

[next obsidian]

Oh yeah

lmao nvm

the u_i are

lol

[latent anvil]

Hmm this seems weird

[next obsidian]

It makes more sense what c_{ij} being in m

that just says it isn't a unit

But even that is hard to see how that gets implied, maybe we have a way to write it as a zero divisor?

[eager willow]

Is there a name for subgroups that do not contain any nontrivial conjugacy classes? They're kind of like anti-normal, in a way. One property of such a subgroup $H < G$ is that $G$ acting on $G / H$ by left translation is faithful.

[cloud walrus]

datorangeguy:

[next obsidian]

I've never heard a name associated to it, but I'd scour groupprops

[eager willow]

never seen that before, thanks for the tip

[next obsidian]

If u like group theory, that site is your friend

[latent anvil]

hmm yeah Alex my brain isn't working on this right now

Watching avatar with my dad

[next obsidian]

Fair

[latent anvil]

Sorry I didn't mean "I'm not working on this right now", but "I'm having trouble thinking about it"

[next obsidian]

So brendan I looked at stacks project

and it makes more sense

So the reason c_{ij} is in m is that it can't be a unit

[bleak abyss]

Wait what's the problem we're talking about?

[next obsidian]

because if it is, move all the other stuff to the other side

So you're expressing c_{ij}a_j = linear combo of other a_i not equal to a_j

Then multiply by the inverse of c_{ij}

Then you get a_j as a linear combo of the other a_i, a contradiction

Conversely, if (1 - c_ii) is a unit

then you already noted (1 - c_ii)a_i in (a_1,..., hat{a_i},..., a_n)

multiply by the inverse of (1 - c_ii)

then a_i is in that ideal, a contradiction

[latent anvil]

Ahh I think I get it

1-ci isn't a unit by that

[next obsidian]

yeg

and same for the c_{ij}

[latent anvil]

And so ci can't be in the Jacobson radical?

Wait

1-ci is in m

Yes

Good

[next obsidian]

Nah so (1 - c_i) in m

lol

Separate thing, if you have a basis set, and another set of equal cardinality, and the like "change of basis matrix" is invertible, that implies your other set is also a basis right?

Yeah, I think so. The inverse matrix to that shows how to express your basis set in terms of the other in a way that undoes the original. So using that you can show a linear dependence among the other set induces a linear dependence among your known basis

Bruh, how does this process terminate uggggh

Oh cuz countable, yeet

Woooooooah, this is cool, but also makes sense

Let $l(M)$ be the length of a module, then if the following sequence is exact,
$$0\to M_1\to M_2\to\dots\to M_n\to 0$$
and all $M_i$ have finite length, then $\sum (-1)^il(M_i) = 0$

[cloud walrus]

Mathemagician:

[eager willow]

@next obsidian it seems like theres a stronger version of what I was looking for called 'malnormal' subgroups, for which all conjugates by a non-member intersect trivially. Can't find a name for exactly what I was looking for, though.

[next obsidian]

huh, that's weird

[next obsidian]

I went back over my proof that commutative rings have the Invariant Basis Number property, dang it is super slick

[latent anvil]

lol

I thought you just like, tensored with A/m?

[next obsidian]

Maybe that’s a cool way to do it

But I showed that if R/I has IBN for any I then R does too

From there you just mod out by a maximal ideal

So like, yeah the tensor part is the same as modding out, but moreso the part where you can lift IBN from a quotient up is the cool part, and then you just go to vector space land

[fading crag]

So I am a computer science person with little math background

what pre-req subjects should I know in order to take a stab at abstract algebra? I really want to learn group theory eventually

[golden pasture]

the ibn proof ik off is basically pure lin alg lol

i rmb seeing an alternative in AM but didn’t bother using it xd

[bleak abyss]

@fading crag some linear algebra might help but in principle you don't fully need it

[fading crag]

Ok

[bleak abyss]

There's a book called Artin which basically starts from 0 with the linear algebra

[fading crag]

For abstract algebra or for linear algebra?

[bleak abyss]

It does both

[fading crag]

Found it! Sweet.

[bleak abyss]

So if you haven't had LA or you had but it wasn't proof-based

Then that's the book to read

[fading crag]

https://www.amazon.com/Algebra-2nd-Michael-Artin/dp/0132413779 this guy right?

[bleak abyss]

Yup

[fading crag]

Cool. Do you know how much of a jump it is from AA to group theory? I have a little group theory book I tried to work through, but got stuck where I didn't know the LA concepts

[latent anvil]

Group theory is part of abstract algebra

It's pretty much the first subject introduced in most algebra books

[bleak abyss]

I think he meant LA given the next sentence

And depends on what you're doing in group theory

Group actions involving matrix groups might actually ask for some linear algebra knowledge

[golden pasture]

la really only becomes useful in like modules and grp actions

[latent anvil]

[next obsidian]

@golden pasture yeah Matsumura has you do both proofs lol. The linear algebra one is about rank M = largest degree of non-zero minor

Oof, I pinged some random person 😦

What does it intuitively mean to be a module of finite presentation? I feel like I have a handle on what being finite is, so for things like “here’s a SES of modules, under ____ assumptions prove that N is finitely generated” it seems pretty clear how you could go about proving that

But in the case where you replace that with proving N is of finite presentation I suddenly feel super lost. Does anyone have a good way to think about modules of finite presentation?

[latent anvil]

How concretely do you want a way to think about it?

And also do you have a feel for what it means for (nonabelian) groups?

[next obsidian]

Nope nope

I just currently am rocking with “finitely generated and the relations among generators is finitely generated” which basically means nothing to me

[bleak abyss]

So, when you think of a presentation of M

You've got a map from a free module R^S -> M where S is the generating set

[next obsidian]

Yeh

[bleak abyss]

And then giving me a generating set for the kernel

[next obsidian]

yeh

[bleak abyss]

So you have an SES 0->ker(phi)->R^S->M->0

So the question is whether the kernel is finitely generated

[next obsidian]

You mean finitely generated?

I mean I get that

My issue is the bit about the kernel being finitely generated seems too far away from M that I have no intuition why sticking something in the middle of a SES with two finitely presented modules implies it too is finitely presented

[bleak abyss]

Well you can I think take it a bit further in fact

Let's replace R^S with F

[next obsidian]

Some random ass free module?

[bleak abyss]

I'm just renaming so I don't need to use the ^ key

Well we can find another free module F' which surjects onto ker(phi)

[next obsidian]

Oh okay

Yeah

[bleak abyss]

So I think being finitely presented is the same thing as an SES F'->F->M->0 where F and F' are finite rank free modules

[next obsidian]

Yeah it is

That’s literally the definition I’m working with lol

[bleak abyss]

Wait I'm not sure what the problem is then

[next obsidian]

Intuition

I don’t have issue with the definition

Like being finitely generated is easy to think about

[bleak abyss]

I mean I think it's just saying that you can recover all the info of the module by setting finitely many words equal to 0 right?

[next obsidian]

Boom I can just grab finite m_i so it’s <m_1,...,m_n>

I suppose, but it’s too far away from the module itself that I don’t see how I can think about it. Like I’m trying to show that if you have
0 -> L -> M -> N -> 0 exact and L and N are finitely presented then so too is M

[bleak abyss]

Like if I have the presentation <a,b|2a = 0>, well also I know 4a = 0

But that just results from knowing 2a = 0

[next obsidian]

But the difference between finite generation and presentation is so unclear to me that I don’t see how only fitting L,M,N in that SES can imply stuff about how M is presented

[bleak abyss]

I feel like that'll be snake lemma

[next obsidian]

Idk

[bleak abyss]

Don't quote me on it but that feels snake lemma ish to me

[next obsidian]

I wrote a diagram out

But it doesn’t seem exact enough

Like the vertical maps don’t seem exact

But maybe I’m going about it wrong

[bleak abyss]

Oh okay I did something similar and uh, it might be sneaky lol

[next obsidian]

Did u just now solve it

[bleak abyss]

No but I'm thinking of a problem I did once which was something like

If F is finitely generated, M is finitely presented, and we have F->M->0, then the kernel is finitely generated

[next obsidian]

Oh I solved that

It was actually done in Matsumura too

But that one I just did with elements

Nothing fancy

[bleak abyss]

Oh I did it with snake lemma and the diagram was odd

[next obsidian]

Wtf

[bleak abyss]

Like not one you'd expect

[next obsidian]

Whack okay

[bleak abyss]

Like you start a new row with M at the end and link by the identity

[next obsidian]

Wait that is the diagram I made

Wait shit can you just do snake lemma on that???

[bleak abyss]

Yeah

[next obsidian]

Bruh I’m boutta off myself lmao

[bleak abyss]

What did you think you needed for snake lemma?

[next obsidian]

I forgot I can just shove the kernels above

And then put maps there

That make it commute

Oh wait

No, how do you know the maps are surjective, the ones going down?

Or errr

[bleak abyss]

I don't use that

The only thing I use is that identity has 0 kernel and cokernel

[next obsidian]

Oh shit yeah

[bleak abyss]

Because then coker(g) = coker(f)

[next obsidian]

Damn yeah I haven’t ever used it like that

I’ve only ever done it when I have the huge commutative diagram

But that makes sense

And I used it a lot 😭

I’m just thinking about how I probably could’ve simplified some proofs

[bleak abyss]

Yeah and tbh I think the characterizations of finitely presented modules from earlier are basically how you think of them

Rather than some kinda like

idk M is purple iff it's finitely presented

Not purple but you get the idea I'm not sure how much intuition there is to be had

[next obsidian]

I guess

It’s just that being finitely generated is so intuitive

you just get your hands on a finite generating set and then you can go to town

Figuring out how to leverage that the kernel is finitely generated is way harder

[latent anvil]

The thing I was going to say was about finitely many relations in a presentation

Like I feel like I have a sense of what a group presentation is

Finitely many relations means you can actually write it down

[next obsidian]

I guess, but I don’t see how I’d ever use that

[woven delta]

Computable set of relations

[next obsidian]

Computable might as well mean nothing to me lol. I have no idea what that would imply

[woven delta]

Like there are some infinite sets of relations that you can still get your hands on

[bleak abyss]

I think his problem is more

Why does knowing shit about the relations tell me something about the module?

[next obsidian]

^^^^

Like I’ve used it before for something with PIDs or something

But i still have no handle on how to relate it

[bleak abyss]

I mean so for Noetherian rings finitely generated = finitely presented lol

And tbh in general I'm not sure how much else there really is to being finitely presented

Like it just means I can write it down on a page

[next obsidian]

Yeah it seems like Matsumura wants you to get comfortable with them

But like damn I don’t see the point and have very little experience with them

Everyone and their grandma has tons of experience with finitely generated modules

[woven delta]

Oh I was more talking about sham's comment

Sorry

[latent anvil]

That's true liquid

I was thinning about the infinite braid group recently

which has an infinite presentation but is easy to write down

Also magician you know what computable means

It's literally "can you write a computer program which prints out all the elements"

Or more precisely "for each element, there's some N such that after N steps the program will print it out"

[vestal snow]

Can someone help me out with understanding tensor products, exactness, and flatness? I feel like I know what these things mean, but I don't have an intuition for solving problems with them. I'm using Atiyah MacDonald (chapter 2), if that's relevant.

I feel like I either lack the prereqs or I'm missing the bigger picture

[woven delta]

Sham that's computably enumerable

[latent anvil]

oh sorry you're right

computer program to check if it's an element in finite time

I'm used to decidable vs recursively enumerable

[next obsidian]

@vestal snow I’ll be honest, the way I understood it is by just doing a lot of exercises with them

I think tensor products and flatness are pretty famous for being unintuitive and confusing until something just clicks

Concretely, for the tensor product I’d recommend trying to never ever peer into it directly, and get everything via its universal property. To make a map out of it, define a bilinear map from M x N, to show something is iso to it show it satisfies the universal property.

To show some element a (x) b is non-zero come up with some bilinear map from M x N such that (a,b) isn’t mapped to 0, etc

[bright knoll]

what is abstract algebra ?

does it have calculus ?

i dont even see this on my colleges class list lol

[scarlet estuary]

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

your school might call it "modern algebra" or similar

or use terms for subfields like "group theory" or "ring theory" or "galois theory"

or, it might not offer classes in algebra at all, such as if it's a community college

as for whether it has calculus:

basically every mathematical subject can involve "calculus" in some form

but algebra tends to be more removed from calculus than subjects such as analysis

(in fact, analysis is, in a way, generalized calculus)

[solid drum]

Don’t forget C^* algebras, sigma algebras etc and their importance in analysis 🙂

[next obsidian]

I don’t think people really consider a sigma algebra as very algebraic

Like you certainly can put an algebraic structure on them and make sense of that, but all uses I saw didn’t use any algebra, to derive results about them

[vestal snow]

@next obsidian thanks for the reply. One of the main challenges I'm facing when doing the exercises are the details.
I feel like if I think of this stuff with a bigger picture mindset, I'd be missing out on a lot of subtle, yet important, details

[next obsidian]

You just have to be really really careful about it and hope honestly

That’s the type of stuff it helps to have someone grade you on who knows more, but without that you have to just self grade

Take a proverbial microscope and at each step ask “wait, HOW do I know this”, play devil’s advocate and purposely try to find every reason to conclude your proof is bullshit

If you mean not being able to figure out the small details when in the process of writing the proof even if you have a good idea of the overall big picture (or at least what you think is the big picture on how to solve it), I can’t help you with that haha, if you figure out a solution pls let me know lmao

[vestal snow]

Alright, I guess I'll just keep at it

one more thing

can you help me out with a small step in one of my proofs?

[next obsidian]

Maybe, I can at least try

[vestal snow]

basically, I want to prove that any ring A is flat (when tensoring by itself)

[next obsidian]

Do you know what a natural isomorphism is?

If not, that’s totally fine you don’t need it

But conceptually it gives a like 1 line proof (tho you have to show the claim)

[vestal snow]

one of the equivalent conditions of flatness is that if $f:M' \rightarrow M$ is injective, then $f\otimes 1:M'\otimes A \rightarrow M \otimes A$ is injective

[cloud walrus]

Have a Banana Bitch:

[next obsidian]

Yup

[vestal snow]

Yeah I think I know what you mean

how would I show the injectivity of $f \otimes 1$?

[cloud walrus]

Have a Banana Bitch:

[vestal snow]

I have the following

[next obsidian]

So I think this is where you want to show A (x)_A - is naturally iso to the identity functor, but hit me with what you got right now

[vestal snow]

I'm trying to prove that the kernel is trivial

[next obsidian]

Yeah, so I would not go about this on elements

[vestal snow]

and this is what I have

[next obsidian]

Do you know that M (x) A iso to M

For all M?

When you tensor over A

[vestal snow]

yes I do

[next obsidian]

Do you know the specific map?

[vestal snow]

$a \otimes m$ goes to $am$ and then you extend linearly

[cloud walrus]

Have a Banana Bitch:

[next obsidian]

Yup

So

Morally f (x) 1 should be just exactly the same as f right?

[vestal snow]

ah

[next obsidian]

You can show a square commutes

Involving f, f (x) 1

And then the ISO’s

Of M (x) A

This is just showing the details of “M (x)_A - is naturally isomorphic to the identity functor”

But from there this let’s you transfer all statements about injectivity and surjectivity of f (x) 1

[vestal snow]

See I thought about doing that and now that you explain it that makes perfect sense

[next obsidian]

Into those with f

Yeah so to do so

You need to verify the square actually commutes

Which is why I asked if you know the specific map

Since to show it commutes you want to show it on the simple tensors

But you’ll find it ends up being pretty immediate IIRC

[vestal snow]

the reason that I wanted to do it using a more element oriented way was because I was having trouble proving $f \otimes 1$ is injective in other kinds of problems as well

[cloud walrus]

Have a Banana Bitch:

[next obsidian]

So this is general advice

[vestal snow]

I'm not that comfortable with free modules and the notion of equality seems very strange

[next obsidian]

Equality is strange

That’s good to question it, I still am thinking about it