groups-rings-fields

[upper pivot]

but yeah its probably a good idea to focus on these

rather than DRP

[errant wedge]

Huh I feel like itd be easier than like 446 since no real final

im taking real anal in winter 🙏

[upper pivot]

oh yeah i mean i havent had a course with a final in years, so i forgot how much time it takes tbh

[errant wedge]

topology kinda counts tho right

its not algebra at least so im not doing entirely algebra

[upper pivot]

true

[errant wedge]

Also what iso theorem are you applying 🥲 im not seeing it

[barren sierra]

(R/I)/J(R/I) = R/(I + J)

for the middle isomorphism (R a ring, I, J ideals)

[dull ginkgo]

Hello chat are we doing Gaussian integer shit

[barren sierra]

How is this easier for this case of finite Galois extensions (vs finite extensions in general)

is the only way to do this to just still prove there exist finitely many intermediate extensions => such alpha exists?

or is there something easier?

[errant wedge]

What am I doing wrong here?
[\Z[i]/(3)=\Z[i]/3\Z[i]]
[\cong[\Z[x]/(x^2+1)]/3[\Z[x]/(x^2+1)]]
[\cong[\Z[x]/(x^2+1)]/[3\Z[x]/(x^2+1)]]
[\cong\Z[x]/3\Z[x]]

[cloud walrus]

Sara

[rocky cloak]

I don't know if it's necessarily easier, but it would be sufficient to construct an element not fixed by any automorphism.

And for an infinite field then I guess you could just refer to a vector space not being a finite union of proper subspaces.

[rocky cloak]

As per the formula, you're not adding J and I. You've just written R/I / JR/I = R/J

[barren sierra]

So in that second line, R = Z[x], I = (x^2 + 1), J = (3)

[errant wedge]

Right sorry shouldve specified, where is the error in the above? I used that (R/I)/(J/I)\cong R/J, but Z[i]/(3) isnt isomorphic to Z[x]/(3) right

[barren sierra]

the error is in the second line to third line

[errant wedge]

Ok, so in general it's not true that n(R/I) cong (nR)/I

?

[barren sierra]

what does n(R/I) even mean

[errant wedge]

Right sorry not for any ring since that doesnt make sense, but here 3(Z[x]/I) is the ideal of multiples of 3 in Z[x]/I

[errant wedge]

so quotienting by
3(Z[x]/(x^2+1))
the ideal of multiples of 3 of (elements of Z[x] mod x^2+1))
is different from quotienting by
(3Z[x])/(x^2+1)
the ideal of (multiples of 3 of elements of Z[x]) mod x^2+1
I'm having trouble understanding why this is, sorry

I know it's wrong im just trying to logic through why

[crystal vale]

If H_1 and H_2 have a finite index in G then so has H_1 intersection H_2.

We have the result that any right coset relative to H_1 intersection H_2 is the intersection of a right coset of H_1 with a right coset of H_2.

And I proved this result by showing (H_1 and H_2)a = H_1a and H_2a, is it correct?

Then using that result we know H_1 and H_2 have finite index therefore by cardinality argument we have finite right coset of H_1 and H_2.

[tribal moss]

That looks right, but you may want to expand slightly on "by cardinality argument".

[tribal moss]

(R/I)/(J/I) cong R/J is only true when J contains I.
Here you have J = (3) and I = (x²+1), which are not contained in each other.

[knotty badger]

Yes, if J does not contain I, it’s unclear what the notation J/I would even mean

[tribal moss]

It can still reasonably mean the image of J under the projection R -> R/I.

[knotty badger]

Ooh, that’s a good point!

[tribal moss]

So in contrast to @barren sierra I think the step from line 2 to line 3 actually does work.

[crystal vale]

I meant if we can write any coset of H_1 and H_2 as intersection of right coset of H_1 and right coset of H_2, then let number of right coset of H_1 is a and number of right coset of H_2 is b then atmost right coset of H_1 and H_2 is ab

[tribal moss]

Yes. (By the way, it's a bit confusing to write intersection as "and". Writing "intersect" or "cap" would be less confusing).

[knotty badger]

Mhm, when I see “and” in the context I’m more likely to read it as “considering both simultaneously”

[crystal vale]

Okay thank you

[cloud lynx]

hello, why is Q[X,Y]/(X) ismorphic to Q[Y]?

[knotty badger]

At one level - this is just because you can define an explicit ring isomorphism between them

[cloud lynx]

and how does it look like?

the ring isomorphism

[knotty badger]

In one direction, you send a polynomial p(Y) in Q[Y] to the equivalence class p(Y) + (X) in Q[X, Y]/(X)

One can verify this is a ring homomorphism

And moreover that it is bijective - thus, its inverse is a ring homomorphism too, so it establishes a ring isomorphism

Of course, the way I prefer to do this is with universal properties, but that’s because I like category theory

[cloud lynx]

mmh is there more a intuitive approach?

[knotty badger]

Well, it’s intuitive for me

But I know category theory, so I cannot guarantee it’ll be intuitive for you

I’m willing to go over it, though, if you want

[cloud lynx]

I dont even know how Q[X,Y]/(X) looks like- like ok the elements of this are p(y)+(X) but its too abstract for me

would be very helpful:((

[knotty badger]

Ok, let me try this!

So you’re right - looking at what Q[X, Y]/(X) “is” feels pretty abstract, right?

[cloud lynx]

yes

[knotty badger]

The categorical approach starts by changing our perspective

Instead of looking at what the ring “is”, we look at what the ring “does”, how it relates to other rings, how to use it

In particular, we will seek a nice alternative description of ring homomorphisms Q[X, Y]/(X) -> R, for R an arbitrary ring

Can I just ask - are rings assumed to be commutative, here?

[cloud lynx]

in this case yes

[knotty badger]

Wonderful, that simplifies things

[cloud lynx]

but what do you mean by what the ring does?

[knotty badger]

It turns out we can find one! They naturally correspond to ring homomorphisms Q[X, Y] -> R which send X to 0

Do you see why?

[knotty badger]

So perhaps first - if I have a ring homomorphism Q[X, Y]/(X) -> R, how can one obtain a ring homomorphism of this form?

[cloud lynx]

no that was always a Mystery for me why it is the kernel

[knotty badger]

I see I see

[cloud lynx]

so why we send X to 0

[knotty badger]

So - are you familiar with the quotient map Q[X, Y] -> Q[X, Y]/(X)?

[cloud lynx]

isnt it the canonical epimorphism?

[knotty badger]

Yes!

[cloud lynx]

then yes

[knotty badger]

It may help to draw a diagram here

We start with a ring homomorphism Q[X, Y]/(X) -> R

And we want to obtain a ring homomorphism Q[X, Y] -> R

How might we do this?

[cloud lynx]

with homorphism theorem ?

[knotty badger]

.

[cloud lynx]

or in the generel the universal property says that if (X) is subset of the kernel there exists always a ring homomorphism from Q[X,Y]->R

[knotty badger]

It’s a lot simpler than you’re imagining

[knotty badger]

We have this

[knotty badger]

And we have this

Try drawing a diagram, I really recommend it!

[cloud lynx]

I did wait i am thinking

i dont know:( so the diagram has to comute so if we set X=0 (idk why) then all terms with X in it are gone so what is left are only the Y terms

so we can chose Q[Y]=R

i dont knwo

[knotty badger]

We can just compose them, right?

We start from Q[X, Y]

Apply the quotient map, to get to Q[X, Y]/(X)

And then apply the ring homomorphism Q[X, Y]/(X) -> R

This gets us a ring homomorphism Q[X, Y] -> R

Now, can you check this ring homomorphism sends X to 0?

[cloud lynx]

oh that is what you meant yes

[cloud lynx]

no i cant because i dont know how R looks like

[knotty badger]

You can, actually!

Try following X through the composition

[cloud lynx]

sending only X?

then its gonna be zero because it is the equivalent class 0

because it is a ring homomorphim

[knotty badger]

Yes!

[cloud lynx]

it sends 0 to 0

[knotty badger]

Yep yep yep

[knotty badger]

So, it is true you have no idea what R looks like

And yet, we still deduced where X gets sent

Anyway, do you agree we have a way to turn a ring hom Q[X, Y]/(X) -> R, into a ring hom Q[X, Y] -> R sending X to 0?

[cloud lynx]

yes

by setting all X terms to zero then there are only Y terms left so Q[Y]

[knotty badger]

Right, I haven’t mentioned Q[Y] yet but that’s a good instinct

[knotty badger]

Now, see if you can do the reverse!

[shell pilot]

Show that a nonconstant polynomial in Z[x] that is irreducible over Z is primitive.

Just looking for a hint. I know that the only units in Z are 1 and -1 and for a polynomial to be primitive means that the gcd of the coefficients of the polynomial is 1.

For the polynomial (call it f(x)) in Z[x] to be irreducible means that it can be written as the product of g(x) and h(x) where g(x) or h(x) are units of Z[x] and g(x) and h(x) are also elements of Z[x].

[cloud lynx]

what do you mean?

[knotty badger]

So, take a ring homomorphism Q[X, Y] -> R that sends X to 0, and make a ring homomorphism Q[X, Y]/(X) -> R

[cloud lynx]

ok i try

just applying the homomorphism theorem ?

[knotty badger]

What is that…?

[cloud lynx]

if we have a ring homomorphism R->R'

call it p

then R/kernel p isomorphic to R'

[south patrol]

Only if it is surjective

[cloud lynx]

no

yes

right

i mean R'=image p

[dull ginkgo]

Does Jacobson density theorem extend to semisimple modules?

[knotty badger]

Yeah, we don’t want an isomorphism, just a homomorphism

[knotty badger]

And indeed, the kernel of the homomorphism Q[X, Y] -> R may be larger than (X)

[cloud lynx]

oh i didnt consider it right

[cloud lynx]

be looking at the ring homo Q[X,Y]/(X)->Q[X,Y]->R?

[knotty badger]

Well, I don’t see how that works

Which ring homomorphism Q[X, Y]/(X) -> Q[X, Y] are you talking about exactly

[cloud lynx]

how it looks like?

p(X,Y)+(X)->p(X,Y)

[dull ginkgo]

What’s up?

[cloud lynx]

i dont know know:(

i thought by just composing it

[knotty badger]

The issue is this is not well-defined

[cloud lynx]

oh

[knotty badger]

(By the way, if this is getting too long, you can back out if you want to)

[cloud lynx]

oh yes you re right because we are looking at equivalent classes

can u tell me how? i really dont know:(

[knotty badger]

So, your idea is kind of correct

We’re given $\phi : Q[X, Y] \to R$

[cloud walrus]

Pseudonium

[knotty badger]

And what we do is define $\tilde \phi : Q[X, Y]/(X) \to R$ by

[cloud walrus]

Pseudonium

[knotty badger]

$\tilde \phi(p(X, Y) + (X)) = \phi(p(X, Y))$

[cloud walrus]

Pseudonium

[knotty badger]

Indeed, we’re essentially forced into this choice

One has to check this is “well-defined”

And that this is a ring homomorphism

It’d be worth checking these things yourself

You do something very similar in the proof of the homomorphism theorem, in fact

[cloud lynx]

ok I will try

thank you for your patience and commitment to help me^^

[knotty badger]

Ofc ofc

But don’t feel like you have to continue if you don’t want to, equally

[cloud lynx]

i have to do it lol

for the exam

[knotty badger]

The exam…?

[cloud lynx]

one exam exercise is to tell wether a Ideal is prim max or nothing in a Polynomial Ring

[cloud lynx]

on group ring field theory

but it is in September so I have still some time left

[knotty badger]

Oh cool

[knotty badger]

I do wish the universal property of quotient was stated more explicitly

It’s not hard to understand once you’ve seen it

Just that ring homomorphisms R/J -> S naturally correspond to ring homomorphisms R -> S that send J to 0

But it’s a little hard to come up with this on-the-fly in the middle of solving a bigger problem

[knotty badger]

Same with most other universal properties

[tribal moss]

I'll take the opportunity to state that I actually agree with bringing universal properties in here. :-)

A quotient is the universal solution to the problem of finding a homomorphism that kills such-and-such elements.

[knotty badger]

Indeed, and one can even generalise further if they wish

If you have a relation E on a ring R, you can look at ring homs f : R -> S which respect E, meaning that whenever xEy, f(x) = f(y)

You can then ask whether there’s some ring K such that ring homs K -> S naturally correspond to ring homs R -> S respecting E

And it turns out there is though the construction is pretty complicated

You take the closure of E under:

• Reflexivity
• Transitivity
• Symmetry
• Ring Addition
• Ring multiplication

This gets you an equivalence relation E’. You then define an ideal J by looking at the equivalence class of “0” under E’. Then the ring K = R/J works

So in this sense, quotients are about “imposing conditions” on maps out of an object - you find some quotient such that maps out of the quotient not satisfying any extra conditions correspond to maps out of the original object satisfying the imposed conditions

The conditions get baked into the structure of the quotient

[tribal moss]

It might be less footwork to say take the ideal generated by { x-y | xEy }.

[knotty badger]

Ooh yes that is neater

A similar thing happens for all algebraic structures - given a relation E on the structure X, one can consider maps X -> Y which respect E

This turn out to be in natural correspondence with maps Q -> Y not satisfying any further conditions for some quotient Q

And the way you get it is by closing E under reflexivity, symmetry, transitivity, and the algebraic operations in your structure, to get an equivalence relation. Then quotienting X by that equivalence relation, and transporting the structure onto the quotient

This is how normal subgroups, ring ideals, Lie algebra ideals etc always arise - they’re the equivalence class of “0”

Finally, there are examples outside pure algebra where one has to worry about “closing up” a relation to form an equivalence relation - for example, attaching an n-cell to a topological space

In this case I find the universal property easier to work with, because it only mentions and uses the original relation, and not the “closed up” equivalence relation

[kind temple]

how do we see that the kernel of the induced map is the zeroth reduced homology group?

[south patrol]

this is the map C_0(X)/im(d_1) -e-> Z and the kernel is then ker(e)/im(d_1) = reduced homology in degree 0

[kind temple]

does -e-> mean that it is induced by e?

[south patrol]

ye

[kind temple]

do kernels and quotients just work like that? i don't recall that fact

[south patrol]

Yes, it's a simple computation if u wanna check

[kind temple]

so if you have a homomorphism f : G --> H and a normal subgroup N of G with quotient map p : G --> G/N, then the induced map has kernel ker(f) / N?

provided that ker(f) contains N

[south patrol]

Sure

This is fortunately somewhat trivial as the map sends xN to f(x)

[kind temple]

oh right

that makes it more clear

thanks

[south patrol]

Another way to see this is that you can form a section of epsilon by sending like 1 -> sigma (for some fixed sigma) and so epsilon then can be identified with a projection onto a direct summand of C_0

[kind temple]

how is the identification used?

if i am interpretting you correctly, this is what you are saying, where gamma is a section of epsilon

how does that help show that ker phi is the zeroth reduced homology

[dull ginkgo]

I’m using left module because I’m actually a sane human being.

Anyway I am trying to prove this via induction

But I am a bit stuck on the base step

Assume we have x in U, Ann(x) = J, and another element y such that Jy = {0}

Because U is simple, U = Rx. Thus y = kx for some k in R.

Thus Jkx = {0}, so Jk = J (absorbed on the right for left-ideal J)

But I’m unsure how to show y = f(x) for some R-linear endomorphism f especially if k isn’t central

[dull ginkgo]

Actually

The left ideal Ann(x) is maximal, thus if Ann(y) contains Ann(x), then they must be equal.

[south patrol]

Tbh I now realise it probably isn't that helpful though it makes the calculation simpler ig and gives you a way to view reduced as a direct summand

[lone niche]

I can think of two ways to do this. One way is to define the induced map H_0(X)->Z abstractly via the universal property of the quotient, and then check that the reduced homology H_0(X) satisfies the universal property of the kernel of that abstract map.
The other way is to define explicitly the induced map H_0(X)->Z by m(x+im(d_1))= epsilon(x), where x is in ker(d_0)=C_0(X). Then you can check directly what the kernel of that map is.

[kind temple]

this kind of stamps it out i think

but yea, thats how i was getting the induced map, i saw that the map C_0(X) --> Z descended to the quotient

[fallen ruin]

Hi all, i'm using Aluffi to review some field theory and i'm completely stuck on this... I can't see how the sum $\alpha_i$ is immediately helpful. Intuitively what its saying makes sense to me... i tried looking at the case where d=3 and multiplying out all the linear factors to get f(x), so all the coefficients would be in terms of the $\alpha_i$'s (products/sums) but not sure where to go from there. A hint would be really appreciated thank you (this is not for any homework, this is for self studying just for fun)

[cloud walrus]

zhulrak

[nimble folio]

Let $R$ be a ring. An (additive) abelian group $M$ is an \textit{\underline{almost left $R$-module}} if there is a function $R \times M \rightarrow M$ which satisfies all of the left $R$-module axioms except for Axiom (4): we do not assume that $1m = m$ for all $m \in M$. Prove that $M = M_1 \oplus M_0$, where $M_1 = {m \in M:1m = 1}$ and $M_0 = {m \in M: rm = 0 \hspace{4pt} \text{for all $r \in R$}}$ are subgroups of $M$ that are almost left $R$-modules (in fact, $M_1$ is a left $R$-module).

[cloud walrus]

clubsoda14

[nimble folio]

I think I can prove this on my own but shouldnt it be $M \cong M_1 \oplus M_0$ instead? If that is not the case, how would you show $M = M_1 \oplus M_0$?

[cloud walrus]

clubsoda14

[fallen ruin]

It would be equality I believe because both M_1 and M_0 are subsets of M

[nimble folio]

Okay got it

[barren sierra]

this would be an "internal" direct sum

as yes both are subsets

[nimble folio]

Okay got it

My first guess was to show theres a bijective map between the two but I can try showing inclusion both ways

[nimble folio]

I'm a bit iffy on the difference between internal and external sums, can you expain?

one of them is just the direct product right?

[barren sierra]

they're the same

but we say "internal" in this case when both components are subsets of the original

it's really a matter of perspective

the external is like taking two modules and combining them to make a new module

the internal is when you want to say "hey this module is the direct sum of these two submodules of the bigger module"

[nimble folio]

Thats a really good way to put it

Thanks

[barren sierra]

almost no one really makes the distinction

[barren sierra]

hence here in my opinion = and \cong are interchangable in a sense

[nimble folio]

So you could prove this by showing theres a bijective map between the two sets rather than showing inclusion both ways

[lone niche]

Hint: Look at the constant coefficients of the polynomials that would result from factoring the polynomial into a product of polynomials.

[fallen ruin]

Thank you... I was able to figure out if you suppose it is reducible then the 2nd coefficient of the factors would not be in k.. for example if one of the factors was degree 3 you would get $x^3-(\alpha_1+\alpha_2+\alpha+3)x^2+(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3)x-\alpha_1\alpha_2\alpha_3$ but $\alpha_1+\alpha_2+\alpha_3$ isn't in k so we have a contradiction, i.e. no such factors can exist and f is irreducible

[cloud walrus]

zhulrak

[lone niche]

Whoops yes sorry. I mixed up the trace and the norm's positions.

[fallen ruin]

no worries i appreciate the hint! this one was bothering me for a while lol

[toxic zephyr]

so the correspondence theorem gives us a one-to-one correspondence (bijection?) between the ideals of R/I and the ideals of R that contain I? for example, if R=Z and I=nZ, then the ideals of Z/nZ are mZ/nZ where m divides n. and the ideals that contain nZ are the ideals mZ where m divides n. this is pretty basic and simple to reason out for R=Z, but the correspondence tells us this idea generalizes very nicely?

[barren sierra]

yup!

[toxic zephyr]

and so the question of "what are the ideals of R that contain I" can be answered by simply finding the ideals of R/I? and vice versa, i suppose (i'm not sure which is easier in general)

[barren sierra]

depends on the problem lol

[toxic zephyr]

damn thats so cool though

oh wait so then by the FHT R/K ~ S (ker f = K and f: R-> S onto)*, then the ideals of S all correspond to ideals of R that contain K?

does this also give another way to sort of "interpret" maximal ideals? if M is maximal, then R/M is a field which has no proper ideals. then of course R can have no proper ideals that contain M? and similarly, if R has no proper ideals containing M, then R/M can't have any proper ideals so it's a field?

[barren sierra]

yea M is maximal if and only if R/M is a field

do you know what integral domains are and what prime ideals are?

[toxic zephyr]

yeah that's a fact that i've always sort of found hard to understand intuitively. R/P being an integral domain iff P is prime is pretty intuitive in contrast. i'm wondering if this idea of fields having no proper ideals and maximal ideals having no proper containing ideals is the intuitive connective glue between them.

[barren sierra]

it is

[toxic zephyr]

damn that's awesome. correspondence theorem the mvp

ty @barren sierra (:

[errant wedge]

J is an ideal of R/I? What does J(R/I) mean

[cloud walrus]

Sara

[errant wedge]

wait this is true in general its just check defns whoops

[cloud walrus]

Sara

[crystal vale]

Let G be a finite group, f be an automorphism of G and set I = { g in G | f(g) = g^-1 }. Suppose that | I | >(3/4)|G|. Show that G is abelian.

Since | I | > (3/4)| G |, we can say that G = I^2.

And from it we have f^2 = 1.

Any hint?

If we show that G = I then it will be Abelian but I don't think it is a necessary condition because I is not a subgroup

[toxic zephyr]

what are the elements of Z[x]/(x^2,2x-1) like? obviously bc of the x^2, we can represent any element with just a+bx. but 2x=1. which means 2x^2=0=x. a+bx=(2a+b)x=(4a+2b)x^2=0
so how the heck does this work? is this actually just the trivial ring? is there a way to show that? like an algorithm to write a linear combination of x^2 and 2x-1 to get 1?

[knotty badger]

Well, ring homomorphisms from it to R correspond to

Picking an element r in R such that r^2 = 0 and 2r = 1

So then 1^2 = 4 r^2 = 0

So 1 = 0

So yeah it’s the trivial ring

[knotty badger]

Here R is an arbitrary ring

Universal property of quotient ftw

[toxic zephyr]

then i assume we should be able to find a linear combination to show 1 is in the ideal. is there a smart way to go about it?

[knotty badger]

Well you can show x is in the ideal

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

Then you just do 2(x) - (2x - 1) = 1

[toxic zephyr]

I see that's smart

[knotty badger]

So I guess 4 (x^2) - (2x + 1)(2x - 1) = 1

Is the linear combination you want

[toxic zephyr]

checks out

my quest to find a quotient of Z[x] that can't be represented as a matrix ring fails again.
if it's a principal ideal, then it seems we just need a matrix with minimal polynomial same as the generator. but I can't find a non-principle ideal that doesn't just quotient to 0 or a Z/nZ ring.
I wonder if one can prove it's impossible or if there's something relatively simple I'm not seeing

[knotty badger]

I guess you’d have to study all possible quotients of Z[x]

I.e. all possible ideals of Z[x]

I think it’s noetherian right

So ya just need to worry about finite subsets of Z[x] really

Generated by p_1(x), …, p_n(x)

And see whether all of these can be represented as matrix rings

We know that for a single polynomial you can

Cause companion matrix

[knotty badger]

I imagine ya probably want to look at the ideal generated by the leading coefficients of these polynomials

[toxic zephyr]

yeah. since F[x] is a PID, I believe any quotient should be matrixable. but Z[x] isn't, so I'm trying to just look at all ideals generated by exactly two elements (to start). what can we say about them? is there a way to find pairs of polynomials who generate a proper ideal? what's the "smallest" ideal we can generate (so the quotient is larger and more interesting)

[toxic zephyr]

yeah the coefficient on the leading terms is probably the most important aspect

my gut conjecture is that all quotients are either iso to a matrix ring or just Z/nZ but imma play around with it first

[lean sail]

sorry to interrupt the discussion - quick question for anyone here that has read Gallian's contemporary abstract algebra. any idea of roughly how many problems to do per chapter? he has more than 50 exercises per chapter. what would be a good amount of exercises to do for an undergraduate first course in abstract algebra?

[coral spindle]

What precisely are you looking for when you say you don't want it to be a matrix ring?

Because every ring is a matrix ring – of 1x1 matrices over that ring!

The ring Z[x]/(2x+1) might interest you.

[knotty badger]

Yeah ig this does depend on what kind of matrix ring you want

[coral spindle]

That's why I'm asking

[toxic zephyr]

basically we can't just map x to some matrix A. or something. I suppose really just a quotient like a+bx but we can't just write a specific formula for how x behaves. honestly idk I'm just sorta messing around reviewing ring theory ||instead of studying for quals||

yeah I was messing around with Z[x]/(2x-1) and that was really interesting. apparently iso to Z[1/2] which is all rationals of the form a/2^n right?

[coral spindle]

Yes that's right, this is known as the ring of Dyadic rationals

[toxic zephyr]

my next attempt is an ideal generated by quadratics p1,p2 with coprime leading coefficients. this should mean we can always eliminate all terms x^2 and higher

[coral spindle]

Indeed we would

Then you'd have representatives of degree 1

[toxic zephyr]

or at least one generator of degree one right? and the other of the form x^2+stuff
right?

[coral spindle]

Well the ring is generated by the image of x in any case

But the point is that when choosing representatives of elements in the quotient, you can always choose one of degree less than 2.

This is why I thought you'd like Z[x]/(2x-1) – you cannot do the same thing

[toxic zephyr]

oh I think I misunderstood when you say representative you don't mean generator right

[coral spindle]

No! I mean representatives of the cosets of the ideal in the quotient ring

So that e.g. the numbers 0, 1, ..., n - 1 are representatives of the elements of Z/(n)

[toxic zephyr]

okay yeah then we're on the same page. that was the goal to be able to WLOG assume elements of the form a+bx or something

[coral spindle]

Yeah and ofc in general you can't do this

[knotty badger]

Yeah then I think the ring of dyadic rationals does this

Cause Cayley Hamilton e.g.

Like you can never express 1/2^n as an integer linear combination of 1/2^k for k < n

So it won’t be a matrix ring in your sense

I think

[coral spindle]

The sense that hasn't been defined

[knotty badger]

The sense I can guess

[toxic zephyr]

i think the kind of sense I mean is that if we can WLOG assume coset representatives are degree 1, then we might be able to represent each coset with a 2x2 matrix.

[coral spindle]

Here is my best shot at formalising what you mean.

You are looking for a subring R of the quotient such that the quotient is a free R-module, and you want to therefore embed the quotient as a subring of End_R(the quotient)

And this will truly be a matrix ring iff the rank of this module is finite.

Indeed Z[x]/(2x-1) is free as a Z-module but it is not of finite rank.

[toxic zephyr]

whereas something like $\bZ[x]/(x^2-n)$ is?
and we can identify each coset with a matrix

$a+bx\mapsto\m{a&nb\b&a}$

[cloud walrus]

inconspicuous old man & mime

[toxic zephyr]

this stuff I'm not as familiar with. it's been a long time if I've ever even learned it

[coral spindle]

Yes, it has a free basis 1 + (x^2 - n), x + (x^2 - n)

[toxic zephyr]

that makes sense

[coral spindle]

Free modules are just modules with bases.

Bases exactly as you're familiar with in linear algebra.

The point is precisely the same, that you can identify the module with R^n

Not every module has a basis, of course.

[knotty badger]

Oh wait this seems to use classification in some way

[toxic zephyr]

yeah that's stuff I'm good with. more the embedding and End_R

[knotty badger]

Since Z is a Euclidean domain

[coral spindle]

Hm I guess there is a wrinkle here that I didn't consider, that there might not be an embedding of the quotient into the endomorphism ring

No in fact this must hold

Let me just make this clear @toxic zephyr

Any R-endomorphism of a free R-module of rank n can be written as an nxn matrix

If our quotient is an R-algebra... ok I didn't define this, but the point is that multiplication in the quotient will be an R-linear map

We have a map from the quotient into M_nxn(R) which just tells us how every element of the quotient acts as a map on the basis

This is an injective map since we can always recover the element by looking at how it acts on the identity.

Does this give you a better idea of the picture?

[knotty badger]

(Yoneda!)

[coral spindle]

Like I can give a simpler example

C is an R-algebra, and ofc since we're working with vector spaces, it is free.

I choose the basis 1, i

I can then see how any element of C acts on C, and represent it as a matrix

This is the same process, just specified.

[toxic zephyr]

yeah that does help a lot actually

[knotty badger]

Oh wow this is extremely neat

[toxic zephyr]

this is sort of where the idea came from. I noticed that in Q[x]/(x^3-2) we can write multiplication as basically matrix multiplication (each element as an operator of multiplication defines a linear map)

[coral spindle]

Yes indeed any nontrivial quotient of k[x] will be finite rank

as a k-module

(Because k[x] is a PID, you fill in the gaps I'm sure)

[toxic zephyr]

yeah exactly

alright it's almost 4am for me lol
thanks for all the help with my silly little algebra adventure @coral spindle @knotty badger

[dull ginkgo]

Integral extension moment

[rain grove]

Im gaving trouble with some definitions in my book, so when we defined groups generated by elements; <a> we said they include their inverses. But what if a has infinite order? Then <a> is not a group no?

[delicate orchid]

<a> is the smallest group containing a by definition, so yes it's a group

[coral spindle]

<a> = {..., a^-2, a^-1, e, a, a^2, ...}.

[rain grove]

ohh ok so <a> is a^m for all m in Z, not just a a^2 a^3 ... a^n = a

[coral spindle]

Yes

[coral spindle]

Exercise: prove this from the definition Wew gave you.

(It's not a tricky one, it's one of those things everyone should do at least once)

[rain grove]

it includes {a,1}, because its a group it contains a^{-1}, so now {a^-1, 1, a} and then it also has to be closed so it contains all a^k and a^{-k}

[tardy hedge]

Oh Hell Nah… he got it right!

[delicate orchid]

u gotta stop doing this quip dawg... it's bad for ur heart...

[tardy hedge]

😢

[delicate orchid]

and MY soul

[tardy hedge]

It makes me happy …

[rain grove]

I can't put into words why if an element a has infinite order then its inverse also has infinite order

[tardy hedge]

If a^-1 had finite order then a has finite order

Multiply it

a^-1 times a^-1 finite times = e

[rain grove]

Oh wait I already proved for finite order with that forumla ord(a^{-1}) = ord(a)/gcd(ord(a),-1)

So now I can just say that if order is infinite the order of inverse its not finite so then its infinite

[coral spindle]

True, but it's much simpler than that

If a^x = e for some x > 0 then (a^x)^-1 = a^-x = e^-1 = e

I.e. a is of finite order iff a^-1 is of finite order

[rain grove]

Well for finite I had to show its the same

Does that show its the same?

[coral spindle]

It does in fact.

It proves more generally that a^x = e iff (a^-1)^x = e.

[rain grove]

Why is x also the smallest positive number that (a^-1)^x = e?

[coral spindle]

I said nothing about the least, this is a statement about all numbers x.

The numbers x such that a^x = e are exactly the same numbers such that (a^-1)^x = e.

So the least such number is obviously the same!

[rain grove]

Hmm ok

[nimble folio]

Can't seem to figure this out, I was able to prove that $M_1 \oplus M_0 \subseteq M$ by letting $m_1 + m_0 \in M_1 \oplus M_0$ \textemdash multiplying by $r$ yields $r(m_1 + m_0) = rm_1 + rm_0 = rm_1$. Since $rm_1 \in M_1$ and $M_1$ is a submodule of $M$, it must be the case that $M_1 \oplus M_0 \subseteq M$.

My trouble is showing the reverse inclusion: If $m \in M$, my guess was to assume that $m = x + y$ for some $x,y \in M$ (because $M$ is additively closed). This didn't really lead anywhere however, or maybe it does.

[cloud walrus]

clubsoda14

[nimble folio]

I'm sure the answer is obvious but I've hit a wall

[hazy lion]

if m is not 1•m you can consider their difference. Note M_1+M_0 subseteq M is immediately the case! The check you made is not necessary. They’re just subsets of M. The thing you actually have to check to state the direct sum inclusion is that (1) they are sub-almost modules (2) their intersection is {0}

[nimble folio]

Sorry, why is M_1 + M_0 subseteq M immediately the case? Is it because m - 1m = x for some (possible nonzero) x in M?

[hazy lion]

If A and B are Abelian subgroups of M, A+B is contained in M. Or even just if A,B are subsets. It’s because if a in A and b in B, a+b is in M.. because +:M x M -> M takes elements of M to element of M

[nimble folio]

Lol I should have known

Thanks

I'm still confused on the reverse inclusion

Showing that M_1,M_0 are sub-almost modules and M_1 ∩ M_0 = {0} implies that M subseteq M_1 + M_0 ?

[errant wedge]

Whats the intuition behind the second and third isomorphism theorems? Looking for a nice explanation if it exists, in the vein of first iso, where you have some ring hom and make it injective on the image by quotienting by the kernel, forcing the things it sends to 0 to be the 0 of the quotient ring

[urban geyser]

(Can you guys lemme know when you’re done so I can send in my question? Thank you!)

[knotty badger]

They follow from the first one

[errant wedge]

Hmmm ok

[knotty badger]

So for the second one

Let H be a subgroup of G, and N be a normal subgroup

There is an inclusion map H -> G, which is also a group hom

There is a quotient map G -> G/N, which is also a group hom

Composing gets you a group hom H -> G/N

Then just apply first iso to this group hom

[errant wedge]

Ah ok, that makes sense

[knotty badger]

For the third one, you want to use the universal property of the quotient

Let N be a normal subgroup of G, and M be a normal subgroup of G, contained in N

[tribal moss]

(But the way, beware that the numbering of the isomorphism theorems varies wildly between textbooks, so just saying "the second and third isomorphism theorems" may not clearly tell a reader what you're asking about).

[errant wedge]

Good to know, ty

[knotty badger]

Then group homs out of G/M correspond to group homs out of G that send M to 0

But the quotient map G -> G/N is a group hom out of G that sends N to 0, so in particular sends M to 0

So by the universal property of the quotient, you get a group hom G/M -> G/N

Then apply first iso to this group hom

[errant wedge]

tyty

[knotty badger]

Np, cat theory definitely clarified a lot of things about the iso theorems for me

[knotty badger]

Oh ig we’re done

[urban geyser]

Thanks!

finally be able to step into these channels to ask a hopefully-meaningful question :3 first time trying to do an actual proof for myself so humour me if it’s something stupid

Regarding how to prove Lagrange’s theorem, that $\mathrm{Order}(H) ; | ; \mathrm{Order}(G)$. If I understand correctly, we’re proving that, given some group $G$ and a subgroup $H$ which contains the unique identity $e$, the cosets cover the set such that the intersections of any such partitions is empty, which I decomposed into two statements: $\$

(1) that multiplication by $g$ does not merely return another distinct element in $H$ or, formally, that $h_{1} \neq gh_{2}$ for any $h_{1}, h_{2} \in H$, and $\$
(2) that multiplication by $g$ does not merely return the same element as if $g = e$, or formally: that $h_{1} \neq gh_{1}$.
Also, that these partitions are bijective to one-another.
$\$

The proof I was given for the injection part of it is straightforward. Injection would mean there’s a duplicate. in other words, that $gh_{1}=gh_{2}$. By the definition of injection, $gh_{1}=gh_{2} \Rightarrow h_{1}=h_{2}$. Assume $h_{1} \neq h_{2}$, and this just gives the contrapositive of the definition. Taking the left inverse of both sides $g^{-1}(gh_{1}) = g^{-1}(gh_{2}) \Rightarrow h_{1}=h_{2},$ which contradicts our assumption.

But doesn’t this proof already assume that an inverse exists? I.e. that (left cosets) of a group $G$ have an inverse element? This would be easy if the partitions $gH$ of $G$ were groups, but they aren’t since only $H$ has the identity element as mentioned above. Doesn’t this make the proof basically tautological on the definition of an injection? I can also provide the proof given for why the intersection of the subgroup $H$ and the proceeding coset $g_{1}H$ is empty, which also makes use of inverses (which here actually makes sense since $H$ is actually a group and by definition includes inverses).

[cloud walrus]

thimg

[coral spindle]

Your statement (1) is false in general.

I'm not really clear on what you're trying to do here. Are you trying to prove that the cosets partition the space?

The way you've written this makes it unclear to me what you're claiming.

@urban geyser can you please clairfy

[urban geyser]

Yes, give me a second to get back home...

[coral spindle]

The proof you've written here makes no sense to me I'm afraid.

It just doesn't prove what you seem to say it does.

[urban geyser]

It’s not yet a proof of Lagrange’s theorem, it’s just a question on a part of the proof. For statement 1, which is written more generally below, this is the proof I was given:

[coral spindle]

So what's the problem you're having with this here

[urban geyser]

That was the proof for why the cosets don’t overlap, but I was actually wondering how one would prove that there are no duplicate elements in the cosets (i.e. that theyre the same size) which i interpreted as saying there’s a bijection between them. The proof I was given seems tautological though on the injection part

[coral spindle]

OK so no, the proof you wrote way up there does not show that

[urban geyser]

yiss, thats what i figured

[coral spindle]

And I don't see how you conclude it's tautological? I'm not sure you are referring to the right concept there, I think you mean circular.

[urban geyser]

But i was confused since i was given that

[urban geyser]

oops yes

[coral spindle]

But no I mean you say:

I.e. that (left cosets) of a group $G$ have an inverse element?
What does it mean for a left coset to have an inverse? This isn't meaningful.

Unless you are talking about a particular situation, there is no meaning to the inverse of a coset.

[urban geyser]

doesnt the “proof” say $g^{-1}(gh_{1}) = g^{-1}(gh_{2}) \Rightarrow h_{1}=h_{2}$?

[cloud walrus]

thimg

[coral spindle]

Yes, and?

g is an element of G

it is not a coset of H in G.

[urban geyser]

oh oh oh

[coral spindle]

This is simply a correct fact about the action of a group on itself.

[urban geyser]

I think i see the reasoning then

okay ill keep working through it, thanks

sorry

[coral spindle]

Feel free to ask here once you've worked things out

[urban geyser]

Regarding part of proving Lagrange’s theorem, that $\mathrm{Order}(H) ; | ; \mathrm{Order}(G)$. We’re proving that, given some group $G$ and a subgroup $H$ which contains the unique identity $e$, the cosets cover the set such that the intersections of any such partitions is empty and that these partitions are bijective to one-another, the former of which I decomposed into two (still-unproven) statements: $\$

(1) That multiplication by $g$ should not merely return another distinct element in $H$ or that $h_{j} \neq gh_{i}$ for any $h_{i}, h_{j} \in H$
and $\$

[\begin{tikzcd}
{} && {} \
& {h_{i} ; ; ; ; ; ; h_{j}} \
{} && {}
\arrow[no head, from=1-1, to=3-1]
\arrow["H"{description, pos=1}, no head, from=1-3, to=1-1]
\arrow["g"{description}, from=2-2, to=2-2, loop, in=55, out=125, distance=10mm]
\arrow[no head, from=3-1, to=3-3]
\arrow[no head, from=3-3, to=1-3]
\end{tikzcd}]

(2) That multiplication by $g$ does not merely return the same element as if $g = e$, or that $h_{i} \neq gh_{i}$:

[\begin{tikzcd}
{} && {} \
& {h_{i}} \
{} && {}
\arrow[no head, from=1-1, to=3-1]
\arrow["H"{description, pos=1}, no head, from=1-3, to=1-1]
\arrow["{g=e}"{description}, from=2-2, to=2-2, loop, in=55, out=125, distance=10mm]
\arrow[no head, from=3-1, to=3-3]
\arrow[no head, from=3-3, to=1-3]
\end{tikzcd}]

$\$

Non-injection from $H \to gH$ implies a duplicate, that $gh_{1}=gh_{2}$. By the definition of injection, $gh_{1}=gh_{2} \Rightarrow h_{1}=h_{2}$. Assume $h_{1} \neq h_{2}$, and this just gives the contrapositive. Taking the left inverse of both sides $g^{-1}(gh_{1}) = g^{-1}(gh_{2}) \Rightarrow h_{1}=h_{2},$ which contradicts our assumption.

Does this assume $\exists g^{-1} \forall g \in G$? That cosets of a subgroup $H$ have an inverse element? This could be if the partitions $g_{1 \leq n \leq k}H$ of $G$ were themselves groups, but only $H$ has identity. Does this make the proof circular?

[cloud walrus]

thimg

[urban geyser]

sorry in advance for the notation, i ran out of space

So i used the kinda pretentious (for this simple question) \forall and \exists and whatnot

But I think I see now that since g is in G, not in H, it by definition of a group it has an inverse, the inverse exists—even if g is in a coset of H and may not form a group by itself.

[sour spear]

Does anyone know a paper where I can look for solving polynomials over p-adic fields? Someone told me it is possible to get every root of a polynomial by a formula over a p-adic field

[urban geyser]

I realize now what I meant to say is that the choice of g outside of H which generates the coset by multiplication from the left has an inverse, and i see now that it trivially does because thats what a group implies… sorry for the longwinded way of putting it, but i was also trying to run through the proof myself by typing it out

i hope my updated rephrasing of the question makes sense even if it’s redundant by now

[coral spindle]

(1) as written is still false.

You haven't restricted the element g in any way.

[urban geyser]

multiplication by g_{1} i should say, no?

[coral spindle]

Listen, I don't know why you're using this approach. Why not prove that the cosets partition G directly? That is, prove that if gH and g'H are two cosets, either gH = g'H or gH n g'H is empty.

[coral spindle]

g_1 is not defined anywhere in the statement

[urban geyser]

its just what i was given

[coral spindle]

I'm not sure I buy that, considering as it's written, this is wrong!

I'm also unclear on what the proof is supposed to be showing. It shows that g. defines an injection H → gH, but this has no relation to what's being claimed above.

I just don't know what's going on here

Can you try and explain the problem you are trying to solve, or the misunderstanding you're having, from the top?

[urban geyser]

i was told the proof works in 4 steps: 1.) choose a subgroup H in G, 2.) Cover G with cosets of H, 3.) Show they dont overlap, 4.) Show that theyre the same size

[coral spindle]

Yes, this is how the proof of Lagrange works.

[urban geyser]

I interpreted 4 as saying theres a bijection from coset to coset so i just tried seeing how to prove its injective first (and i still havent talked about surjection)

[coral spindle]

There is indeed a bijection from gH to g'H for any g, g', OK.

Great

Is that the part you are struggling with?

[urban geyser]

and I generate the (left) cosets of H by picking some element g_{1} outside of H by definition, and multiplying all elements of h of H by it

and then showing that intersection of H and g_1 H is empty

[coral spindle]

No this is incorrect.

H is a coset of H in G

You are missing the coset H

I'm not entirely sure what you mean by this in the first place, since you are just restating the definition of cosets

A coset of H in G is a subset of G of the form gH for some g in G.

Was this not made clear before? Have you seen a different definition?

[urban geyser]

I’m not sure i understand this then

[coral spindle]

What part of it is confusing to you? Please elaborate.

[urban geyser]

Well, what i have written down doesnt confuse me at all

[coral spindle]

It is a proof that when g is not in H, we have that H and gH are disjoint.

[urban geyser]

but i don’t understand what you mean

[coral spindle]

OK but you say you don't understand it, so can you explain what you're not understanding.

g_1 H is indeed a coset of H in G

so is H. H = eH is a coset of H in G.

Is that the confusion?

[urban geyser]

yeah… g=e is one of the cases, is that not (2) in my latex image?

[coral spindle]

(2) is false for some elements g of G, so I don't know what it means.

Are you saying that for any g in G, either (1) or (2) holds?

Because this makes sense to me.

[urban geyser]

yeah

[coral spindle]

There we go! So you need to say at some point "for all g in G, one of the following holds:" and this would be clear.

[urban geyser]

sorry, i just wrote “decomposed the former”

ill try to be better about that

[coral spindle]

OK so I now understand what you are trying to say in your message.

[urban geyser]

yiss

so lemme run from the top of how im like, sketching it out

[coral spindle]

What you write is still incorrect

The actual dichotomy should be that, for all g in G, one of the following holds:

1. For all h in H, gh is not in H.
2. For all h in H, gh is in H.

You have assumed that if gH = H, then gh = h for all h in H. But this is false.

[urban geyser]

I’m not sure i follow, and im not sure i have the tools to either right now

could you elaborate?

[coral spindle]

OK, the point is that your dichotomy is false.

You claim that for all g in G, either (1) or (2) holds.

This is incorrect.

[urban geyser]

i think i see—technically there’s a third case, where multiplication of h by g doesn’t just return some gh in H—in which case that’d be when there’s no intersection between the two cosets

please keep in mind i learned this material today be gentle

[coral spindle]

Why not simply use this correct dichotomy

[coral spindle]

If you want clarification on any point just ask and I will be happy to elaborate

[urban geyser]

idk i just started doodling and this is what came out, but im sure as i learn more ill find smarter ways to phrase and prove things

[coral spindle]

OK well I hope I have pointed out the errors here and you can try a different approach

[urban geyser]

if i were to rewrite this in terms of the three cases (1. returns a distinct element in H, 2. returns the same element in H. 3. Returns any element not in H) would that at least be technically correct?

even if we could “bundle” up 1 and 2 into one statement?

[coral spindle]

What precisely is the claim, this is a bit too imprecise for me

Are you saying that, for all g in G, exactly one of the following hold:

1. For all h in H, gh is in H and gh =/= h.
2. For all h in H, gh = h.
3. For all h in H, gh is not in H.
   Because yes, this is true.

[urban geyser]

yes

precisely

[coral spindle]

Why not simply combine 1 and 2 to say instead:

1. For all h in H, gh is in H
2. For all h in H, gh is not in H.

[coral spindle]

This is precisely the dichotomy I presented above.

[urban geyser]

i just found it easier to talk about the “unpackaged” version since i already have a proof of gh is in H and gh =/= h, and i know that 2. for all h in H, gh=g, it wouldnt work once you get past the first coset which we picked as having the identity in order to be a subgroup, so that was also easy to imagine. it would only leave the 3rd case where gh is not in H, meaning the intersection is empty

[urban geyser]

i understand. but i have no idea yet how to express or prove the “packaged” statement in one swift movement without dissecting it into the two cases

[coral spindle]

Did you have any other details you wanted to discuss?

Maybe I could show you the usual way that people prove that the cosets of H partition G for an example.

[urban geyser]

yes please

[coral spindle]

hit the enter button too early

Proposition: for all $g_1, g_2 \in G$, we have either $g_1H = g_2H$ or $g_1H \cap g_2H = \emptyset$.

\bigskip
Proof.
If $g_1H \cap g_2H = \emptyset$ then we are done, so assume that there is some $x \in g_1H \cap g_2H$.
Then by definition there exist $h_1, h_2 \in H$ such that $g_1h_1 = x = g_2h_2$.

Now suppose $g_1h \in g_1H$ is any element. Then $g_1h = g_1h_1h_1^{-1}h = g_2h_2h_1^{-1}h$, which is by definition in $g_2H$. So this proves that $g_1H \subseteq g_2H$. The same argument in reverse shows that $g_2H \subseteq g_1H$ so the two sets are equal. QED.

[cloud walrus]

Boytjie

[coral spindle]

Combined with the fact that g in gH, we have proven that the cosets of H partition G.

[urban geyser]

a lot more direct, yeah…

[coral spindle]

Indeed.

[urban geyser]

this is the first time i did anything remotely close to abstract reasoning humour me

im fresh out of linear algebra

[coral spindle]

Have I not been humoring you?

[coral spindle]

This is the benefit of working directly from the definitions of things. The statement of this proposition is just the 2nd requirement in the definition of a partition

It's always worth, when you are asked to prove something, to write down the definitions of everything involved, at least until you are fancy enough at proofs to keep them in your head

[urban geyser]

ill be honest i didnt know the definition of a partition

[coral spindle]

Well.

[urban geyser]

i knew that cosets were one but i havent seen this yet

so yes makes sense

[coral spindle]

It is always a good idea to know the definitions of the things you are asked to show!

[urban geyser]

i technically wasnt asked to show this, at least not yet, but i wanted to get a head start meow

but yes true

now my only hope is that im not so stupid that i wont get through it

[coral spindle]

I don't think there's even a hint that that might be true

It's just a matter of practice

[urban geyser]

ur proof is a lot closer to this for sure

[coral spindle]

No my proof isn't mentioned here. It's previously proven in the book. They even say:

Since the different left cosets of H constitute a partition,
Which is what I proved.

So in fact the proof you're looking for is before this photograph you've taken

[urban geyser]

yiss

bc they also mention “cancellation laws” which is what i think i was doing with the inverses and all that

[coral spindle]

Yes, they are omitting the proof that x |-> ax is a bijection H → aH

[urban geyser]

yiss which is what i got stuck in the weeds over

wow, i read some more on partition and yeah, this was almost obviously true 😦

equivalence relations define partitions. left cosets are equivalence classes. left cosets partition G. Then literally every other requirement is just a definition of a partition (covers G, doesnt overlap, bijection)

i think i understand it more of why this works by doing this nitty-gritty, but for an actual proof it’s so simple just using equivalence and partitions, although admittedly i dont think id understand it if i only read this version of the proof

[knotty badger]

Yeah I much prefer the equivalence class route

[urban geyser]

in my feeble defense i havent learned that yet 🥺

[knotty badger]

Oh

The equivalence relation you want is just

[urban geyser]

but i did vaguely hear that equivalence relations and partitions define each other

yeee h = gh

[knotty badger]

$x \sim y \iff x^{-1} y \in H$

[cloud walrus]

Pseudonium

[urban geyser]

wait

i misspoke

u know what i mean

[knotty badger]

Yes they’re equivalent notions

[urban geyser]

no wonder i looked silly

[knotty badger]

They’re not exactly the same

But there’s a way to translate between them

[urban geyser]

literally took me like an hour or two to reason out my sketch of a proof

yeesh

shouldve taken 2 mins

its ok i hope, most people went through this phase right? sorry, irrelevant convo we can talk elsewhere

[tardy hedge]

I too constantly doubt myself

So, you’re not alone in that at least

[hidden wind]

i often have this taking hours for a "shouldve taken 2 mins" thing too

i think as long as you're getting something down to paper and are in fact thinking then it's no problem and this time is probably well spent and the experience you gain will be beneficial later even if it don't immediately lead to the solution you're looking for

the real downer is when you're just completely stuck

[amber burrow]

I wanted to ask: the fundamental theorem of finitely generated abelian groups can be written as some product of cyclic groups, but also Z^m at the end?

Like what is the Z^m doing?

Also isn’t Z infinite, but the theorem applies to finite groups?

[hidden wind]

the form with Z^m applies to finitely generated groups

Z itself for example, while not a finite group, is generated by the finite set consisting simply of 1

Z^2 by {(0,1), (1,0)} etc

[barren sierra]

finite groups are finitely generated of course

[next obsidian]

Proof?

[dull ginkgo]

wikipedia is being WRONG (i think)

[next obsidian]

Chmowned

[dull ginkgo]

multiplication by a ring element is not a linear endomorphism unless that element is central when projected into R/Ann(U)

[next obsidian]

Why?

Linearity only requires distributivity no?

[dull ginkgo]

Endomorphism of left R-modules

[next obsidian]

Wut

U is a right R-module and you’re looking at the map
U -> U which maps u to u•r

[dull ginkgo]

I am instead using left modules because I am sane

same difference

[next obsidian]

All you need to do is verify that u + v maps to u•r + v•r

Which is true cuz (u+v)•r = u•r + v•r

[dull ginkgo]

It is an endomorphism of R-modules so it would have to be R-linear

i.e f(xr) = f(x)r

[next obsidian]

fuck you

Actually this is why it’s a right module I think?

f(x) = x•r

Uhhh

[dull ginkgo]

It would make no difference

Here:

[next obsidian]

f(x•s) = x•s•r = x•r•s

1984

Maybe it means a Z-linear endomorphism

[dull ginkgo]

it definitely doesn't

[next obsidian]

¯_(ツ)_/¯

[next obsidian]

Yeah

[dull ginkgo]

In particular

[next obsidian]

Does U being simple somehow make it work?

[dull ginkgo]

End_R(U) is the centralizer of R's embedding into End_Z(U) by left multiplication

(It would be R^op embedding by right multiplication)

[next obsidian]

Yeah I don’t thinj being simple makes it work by any magic

[dull ginkgo]

if U is centralizer, then Ann(x) is left (right) maximal for each element x

[next obsidian]

Rest in pepeloni

[dull ginkgo]

Ann(x) contains maximal two-sided ideal Ann(U)

we need centrality modulo Ann(U)

Maybe assume Ann(x) = Ann(y)

y = rx for some r (LEFT MODULES BECAUSE I AM NORMAL)

Ann(x)r = Ann(y) = Ann(x)

so at least it absorbs r on the right

[dull ginkgo]

This wikipedia page is wrong then QED

Then isn't THIS wrong

The base case is what i described

[dull ginkgo]

Also schur's lemma would say r is a unit

[tough raven]

"endomorphism of U" must mean additive endomorphism (or possibly D-linear endomorphism where D = End_R(U)).

[chilly radish]

Yea this is poorly phrased but I don't think they mean R-linear (given the rest of the article)

U can also check isaacs

[chilly radish]

No it wouldn't. Just that if the inverse is given by multiplication by s, then rs-1 is in the annihilator.

[errant wedge]

Is there a nice way to do this

[tough raven]

Hint: ||Z_4[i] = Z[i]/(4) and Z[i] is a PID||.

Wait, subrings, not ideals.

[rocky cloak]

The ring is extremely finite, so just look at subrings generated by one element

[chilly ocean]

Wouldn't that be a proof tho? I'm really confused by what the author meant to say by "Without proof".

[rocky cloak]

They mean just think about it and then list them out without a formal proof of that being everything

[errant wedge]

Just means no justifcation needed, not that im not allowed to use proofs

[errant wedge]

so all of these are generated by a single element?

Is there no like general trick/something to look for, say, if I were asked this on an exam

Is it just generate and hope I got them all

[rocky cloak]

My definition of ring includes 1. If you don't include 1, then there's a few more subrings

Still, you just sort of think about what's the subring generated by one element, then from these what happens if you add in another element, etc

[dull ginkgo]

I didn’t check Isaac’s but I think it meant D-linear operations (i.e, in the double centralizer of R^op in End_Z(U)

Because the proof works out if you consider that and would make sense

[surreal dagger]

If we have a surjective ring homomorphism between R and R´, then we can identify ideals from R´ with ideals from R containing the Kernel.
I think im not grasping exactly why it is the way it is with the condition to contain the kernel, here are some loose thoughts:

• We can identify R´ with R/ker(f) since its a surjective map.
• The homomorphic image of any Ideal under a surjective map is an Ideal.
• The projection map into R/ker(f) and taking the preimage of an Ideal in R/ker(f) gives us the bijection. The preimage of such an Ideal automatically contains ker(f). That way it makes some sense that we consider Ideals from R containing ker(f).

If I dont demand that the kernel is contained in the Ideal, it can break: consider Z->Z/2Z and the Ideal 3Z in Z which does not contain 2Z.
Now tying everything together, whats a more natural perspective to see this?

[knotty badger]

So, preimages of ideals are always ideals, and in the case of surjections, images of ideals are always ideals

Moreover since the ring hom is surjective, applying preimage then image is the identity

One need only check that applying image then preimage is also the identity - this is not true on arbitrary subsets of R, but it is true on ideals

Indeed this is why kernels are ideals - they are the preimage of the ideal {0}

You can apply what I’m saying to the canonical quotient map R -> R/ker f if you prefer

Which is also surjective

[surreal dagger]

Right

[surreal dagger]

So would you just say: It just breaks if we dont demand that the kernel is contained in the Ideal?
Im fine with that

[knotty badger]

Ah yes, it’s only true on ideals that contain the kernel

I forgot that

[surreal dagger]

Yep thats what I want to know about

[knotty badger]

I guess the universal property of the quotient gives another perspective on this

Identifying ideals with kernels of ring homs

Ring homs out of R which are zero on I and on ker f naturally correspond to ring homs out of R/ker f which are zero on I/ker f

But if I contains ker f, you can shorten this to

Ring homs out of R which are zero on I naturally correspond to ring homs out of R/ker f which are zero on I/ker f

[chilly ocean]

If you are asking for intuition. Think of ideals as sets of relations you can impose on your ring. When you quotient by an ideal, you are basically forcing those relations to become true in the new ring. After you impose some relations, you can only impose some more relations, the ones you already imposed will always stay true. The kernel is the relations you imposed. The ideals that contain the kernel are the sets of relations bigger than the ones you already imposed, so they turn out to be the same thing as the sets of relations you can impose on your new ring.

[knotty badger]

Ooh yes “ideals are relations” is an excellent perspective

It’s even how one can derive the notions of ideal, normal subgroup etc

They’re always the equivalence class of 0

[chilly ocean]

For groups and rings and some other nice things yes, in general ideals are not the same thing as relations, but that's just an aside.

(and by relations I really mean congruences)

[knotty badger]

Oh, they’re not?

Yes I mean congruences too

I was under the impression ideals were always these

[chilly ocean]

No, for example for monoids, there's also a notion of ideal, which is weaker than congruences.

[knotty badger]

Ooh I see!

What’s the notion? And - why is it useful?

[coral spindle]

The notion is exactly as you'd expect: the 'super' closure property, and it is useful because the structure of ideals tells you about some properties of the semigroup. See: Green's relations.

[chilly ocean]

Something about model theory iirc

[knotty badger]

Telling me it’s “exactly as I’d expect” is not exactly helpful

Because what I would expect is congruences

[coral spindle]

I then immediately said what it is

[knotty badger]

Uh idk what the super closure property is

[coral spindle]

The two properties that ring ideals are that they're a subgroup, and they are 'super' closed under multiplication.

[knotty badger]

Wait I thought this was about monoids

[coral spindle]

The semigroup ideal property is just the latter.

Potato potato.

[knotty badger]

I still don’t really understand…

Could you just say what it means for a subset S of a monoid M to be an ideal?

In the sense that is different from congruences

[coral spindle]

Ideals of monoids are not related to congruences.

[knotty badger]

Why is this so hard lol

[coral spindle]

I'm typing it out right now

A subset S of a monoid M is an ideal if for all m in M, mS is a subset of S, and S is nonempty.

[knotty badger]

Oh I see

Right, interesting

[coral spindle]

So it is just the super closure property of ring ideals, see

[chilly ocean]

Aren't you able to obtain a congruence from an ideal by setting everything in it to 0?

[coral spindle]

You can

[knotty badger]

Eh… I kinda see what you mean, but I would’ve understood a lot quicker if you’d just said the definition

[coral spindle]

But there is no longer the perfect correspondence

[chilly ocean]

I've thought of them as "special congruences"

[knotty badger]

And wdym by the semigroup?

You mean forgetting the identity?

[coral spindle]

The theory of monoids and the theory of semigroups are so close that usually people just talk about semigroups.

[knotty badger]

Then - this kind of sounds like “semigroup ideals”, no?

Or “semigroup congruences”, even

[chilly ocean]

I thought you needed a constant in your theory to talk about ideals?

[coral spindle]

Like I said they are not congruences

[coral spindle]

Why do you say that?

[knotty badger]

They’re not congruences under the monoid structure

But maybe they are under the semigroup structure…?

[coral spindle]

They're not

[knotty badger]

I see

Hmm, why aren’t these subgroups of the monoid

[coral spindle]

Why would they be subgroups?

You have no guarantee of inverses, and the only ideal that contains the identity is the whole monoid.

[knotty badger]

Ah, the issue is they don’t have the identity

I see

[coral spindle]

No that's not the only issue

[chilly ocean]

The place where I originally read about them required that iirc, but I guess maybe they can be generalized.

[knotty badger]

Right but you would need the identity to be a subgroup

That feels like the main issue

[coral spindle]

No these would be 'unital' subgroups.

[knotty badger]

Huh

[coral spindle]

You can have subgroups where the identity is not the identity of the monoid.

As it happens we do understand the "nonunital" subgroups of semigroups in some sense, they are special H-classes, where H is Green's H-relation.

[knotty badger]

Oh, I didn’t realise

I thought it would be a requirement to have the identities coincide

Interesting

[chilly ocean]

Depends on context.

[knotty badger]

Clearly, yeah

[coral spindle]

That's a very restrictive class of subgroups – the subgroups of the group of units of the monoid.

[chilly ocean]

Depends on whether you define identities as a property or a structure.

[knotty badger]

Uh will these be groups necessarily?

Can you not have a sub monoid that’s not a group?

[coral spindle]

I mean not really? Semigroup theorists talk about group H-classes all the time.

For example, group H-classes determine the simple modules of a monoid in characteristic 0.

[knotty badger]

E.g. N is a submonoid of Z surely

[coral spindle]

Will what things be groups?

[knotty badger]

It sounded like you were saying that any sub monoid containing the identity would be a subgroup

[coral spindle]

No not at all.

[knotty badger]

Ok well that wasn’t very clear

[coral spindle]

Sure

[knotty badger]

.

[coral spindle]

The subgroups of the monoid containing 1 are the subgroups of the group of units of the monoid.

Is this unclear?

[knotty badger]

Uh I think so

[coral spindle]

Sure

[knotty badger]

Or wait

It is clear

Uh

Lol

Sorry sorry

[dull ginkgo]

We can still have congruences of a submonoid iff gM = Mh for some h. We don't necessarily need inverses

[knotty badger]

I ain’t an algebraist or anything

[coral spindle]

I'm not sure what you mean here lol

[dull ginkgo]

Are we talking about congruence/quotient relations of a monoid

[coral spindle]

No we're mostly just talking about ideals

But I'm still not clear on what you mean

[knotty badger]

Hmm, so I think my takeaway is that “congruences = ideals” is mostly but not always true

For the purposes of group and ring theory I think it’s ok

[coral spindle]

No this is a special thing about groups, really.

[knotty badger]

Is it not true for rings too?

[coral spindle]

Like the fact that ideals exist really stems from normal subgroups

The additional condition is just an extra restriction.

[knotty badger]

And it’s true for vector spaces and modules

[coral spindle]

Because they're groups

[knotty badger]

Sure forgetful functor and all

But they’re not just groups

[coral spindle]

As a semigroup theorist at my institution put it, to paraphrase, groups are very "rigid" and semigroups are "floppy."

I think this is a good way to see it

[knotty badger]

Yeah I don’t really come across or study semigroups myself

Interesting to know though

I appreciate groups more lol

[coral spindle]

Yes, semigroups suck.

I tease my semigroup theorist friends that they don't actually study semigroups.

They study semigroups with the extra restrictions that blah blah blah

[knotty badger]

Eh I didn’t say they suck

I’m sure they’re useful, I just haven’t come across their uses

[coral spindle]

Most semigroups are simply completely boring

No, semigroups are roundly not useful I'm afraid.

[knotty badger]

Well idk I won’t necessarily trust blanket statements from you

[coral spindle]

Thanks

If you look at certain very restricted classes of semigroups, they get useful and interesting, but the average semigroups is pretty uninteresting (see above link).

Fortunately I have evidence to back up my facts.

[knotty badger]

It’s not that deep

[chilly ocean]

Why does being nilpotent of degree 3 make them boring? Most finite groups are 2-groups

[knotty badger]

Ooh what’s a 2-group?

Wait

[chilly ocean]

p-group for p=2

[knotty badger]

Ohhh is this p-groups

Nice

Yeah lol I remember seeing group numbers of each size

Jumps up enormously at powers of 2

[chilly ocean]

wait actually its open whether most finite groups are 2-groups sorry

[knotty badger]

I see

[coral spindle]

2-groups are actually very interesting! But this isn't the point.

What is the only restriction on a semigroup? That it is associative. 3-Nilpotence makes the associativity condition trivial.

So in some sense "most" semigroups are trivial semigroups.

We cannot use Green's relations to analyse them, and that's the tool that semigroup theorists use.

They just look flat.

[chilly ocean]

you mean 3-nilpotent semigroups are the same as 3-nilpotent magmas?

[coral spindle]

I don't know what a 3-nilpotent magma is, but I can imagine it's the same thing

But it's all explained in the link I put above.

Like to compare and contrast here, the only group that satisfies this condition is the trivial group.

[chilly ocean]

Still not convinced why they are boring, but I guess you know more about semigroup theory than I do

[coral spindle]

Ben Steinberg calls them "weeds"

to get a perspective on what big name semigroup theorists think of them

[chilly ocean]

Do you have a source for that?

[coral spindle]

Lmao sure I'll find the paper

It's in his joint paper with Rhodes, titled "The q-theory of finite semigroups," in the introduction.

This quote that groups are "gems" is one that sticks with me, my supervisor likes it too

[chilly ocean]

Are there non-nilpotent semigroups that semigroup theorists consider uninteresting?

[coral spindle]

Uh, probably? I don't know

You're kinda missing the point here though

Like I said, all the standard tools of semigroup theory fail for 3-nilpotent semigroups, because the one single axiom does not restrict them in any way!

There is no structure that you analyse there

They are uninteresting because they satisfy the requirements trivially.

It's like how the discrete topology is an uninteresting topology

[chilly ocean]

I see

[delicate orchid]

such a fake structure

[hidden wind]

i like gems

[surreal dagger]

Thank you, so, to some extent, what youre saying is that there is a isomorphism between:
R/A and R´/A´ given that we have a surjective map between R and R´ and the Ideals A and A´ correspond to each other with A containing kernel of the map.
Where we R´=R/ker(f) if we unravel it a little bit.

Regarding relation: Do you mean that given a map between two Sets we can always identify elements if they have the same image. We can form the quotient S/~ that has e.g. Ring structure iff it was a ring homomorphism and the 0 in the quotient corresponds to some ideal?

[coral spindle]

Yeah this is the trick: things are equal iff their difference is 0, so to identify things in the image it suffices to say what gets sent to 0 in the image.

You’re correctly noticing that we can do the same thing for sets and get an ‘isomorphism theorem’ for sets!

[surreal dagger]

Alright, so given any Ideal I that contains ker(f), we get exactly one map f* between R´ and R/I and this map gives us exactly one isomorphism R`/ker(f*) = R/I
And this way we identified Ideals containing ker(f) with Ideals of R´. Thank you too

[knotty badger]

Yay

[surreal dagger]

Good stuff

This is fun thinking about it like that

[knotty badger]

Yeah the first isomorphism theorem holds for basically all algebraic structures i know of

You apply the universal property of subsets and quotients

And use that a bijective homomorphism is an isomorphism

[coral spindle]

Because it’s a theorem of universal algebra

[knotty badger]

What’s neat is that this also holds for compact hausdorff spaces

And that’s a hint that you can view them algebraically

[hidden wind]

god i'm so bad with even the simplest cardinality arguments 😭

[arctic trail]

Indices are unironically really based

shoutout to [G:H] and the isomorphism theorems for relating indices of different groups

[knotty badger]

Lol you could read this as talking about physics indices

[arctic trail]

ah no I don't care about physics

[knotty badger]

I see

[arctic trail]

Physics lowered my GPA in undergrad

[knotty badger]

Hm, right

[arctic trail]

It was just so boring

[knotty badger]

I’m sorry it got presented that way..

[arctic trail]

the professor just did variable manipulations

[knotty badger]

Huh, that sounds like what algebra felt like to me

Which was the part of math I didn’t like in my degree

[arctic trail]

And the course for that class was really big

like Classical mechanics (Newtonian Lagrangian Hamiltonian), Special relativity, Minkowski geometry, using tensors for no good reason because everything we did we could've just done with matrices, + quantum mechanics. This isn't 1 semester worth of material

[knotty badger]

Ok that is a lot for one semester

I guess I went into uni doing math but ended up switching to physics

[arctic trail]

Does $\varinjlim S_n$ have a name?

[cloud walrus]

Trivial Lemma

[south patrol]

S_oo

I believe

[knotty badger]

Makes sense

Generated by the pairwise transpositions in N

[chilly ocean]

How to show that the ideal (1+sqrt(-5), 2) of Z[sqrt(-5)] is not the whole ring?

I did it by taking the quotient, but I wonder about alternative ways to do it.

[coral spindle]

Well an ideal is the whole ring iff it contains the unit. So we can try arguing that this ideal doesn't contain the unit by looking at integral linear combinations of the generators.

But put frankly, using first iso is a cleaner solution.

[chilly radish]

In this case the direct method is also really quick but generally can be messy

[coral spindle]

The way you would do the first thing will work something like arguing that it has no irrational component, so it must be a Z-multiple of 2, but then 1 is odd.

[terse crystal]

The quotient ring is Z[x]/(x+1,2) and (x+1,2) is a prime ideal thus not the whole ring. (We know prime ideals of Z[x] are (f(x)): f irreducible and (f(x), p): f(x) irreducible over Z/pZ)

[crystal vale]

End(Q, +,0) is isomorphic to Q ?

[terse crystal]

Yes, bf(a/b)=f(a)=af(1)

[crystal vale]

Okay thank you

[rain grove]

I manage to prove that for a and b != 1 in groupG and bab = a, if ord(a) = 2k+1 then b^2 = 1.

What is an example when ord(a) = 2k and order b is not 2?

[terse crystal]

How did you prove the first part

[rain grove]

something like this

[coral spindle]

In the quaternion group Q_8, consider a = i and b = j

[crystal vale]

But what if k = 1?

[terse crystal]

Q_8 is brilliant, I also find out that a=(12), b=(123) in S_3 works

[rain grove]

This example only exists in non-commutative groups?

[coral spindle]

I wanted to do C_4 \rtimes C_4, but semidirect products are too lofty. Q_8 is just a quotient of this.

[coral spindle]

If the group is commutative, then bab = bba = a, so bb = 1.

[rain grove]

Okk yeah ty

[hidden wind]

whew how come i never payed much attention to cayley graphs before now, these things are lovely

[coral spindle]

Haha what if we saw them as spaces what if we studied the metric properties of cayley graphs

[south patrol]

Hyperbolic

[coral spindle]

pure hyperbole, surely

[south patrol]

One of these days I should learn some GGT

It seems to be what half the topology department does lol

[coral spindle]

It feels like a good time to be in GGT

Lots of students in it too

[hidden wind]

i know the free group over two elements is isomorphic to a hyperbolic reflection group (that of an ideal triangle)... but i suppose it generalizes to more elements as well?

[coral spindle]

Yeah there are other groups which are reflection groups on hyperbolic spaces, and somehow this is related to a property of the cayley graph.

[hidden wind]

The Cayley graph of a free group of finite rank, with respect to a free generating set, is a tree on which the group acts freely, preserving the orientation. As a topological space (a one-dimensional simplicial complex), this Cayley graph Γ(F) is contractible. For a finitely presented group G, the natural homomorphism defined above, φ : F → G, defines a covering map of Cayley graphs φ* : Γ(F) → Γ(G), in fact a universal covering. Hence, the fundamental group of the Cayley graph Γ(G) is isomorphic to the kernel of φ, the normal subgroup of relations among the generators of G. The extreme case is when G = {e}, the trivial group, considered with as many generators as F, all of them trivial; the Cayley graph Γ(G) is a bouquet of circles, and its fundamental group is F itself.

this thing? from wikipedia

[coral spindle]

The Cayley graph?

Uh yes?

I was thinking of hyperbolicity of a space, defined by (amongst other things) the delta-slim triangle condition.

[hidden wind]

sounds lovely

[chilly ocean]

here space means metric space?

[coral spindle]

Yeah indeed

You need some additional conditions but yes, metric spaces

[delicate orchid]

we need to recognise that triangles need not be delta-slim to be beautiful...

[surreal dagger]

Im looking for abstract algebra books or other exercise resources with lots of problems that have a computational vibe to it ranging from easy to hard.
It should cover groups, rings, fields and maybe even modules.
Im thinking of exercises where you have to reason about concrete objects using the general theory.
Of course one option is to just go through different books and pick the exercises myself, but if some sources already exist I would be happy if someone can share.

[dull ginkgo]

I would recommend Jacobson but unfortunately there is little to no exercise resources for solutions

[velvet steeple]

Wait why is this

It's an abelian group under addition, I agree

But associative and distributive under mult. is not "monoid which distributes over addition"

A monoid under mult. is associative and has a neutral element

[coral spindle]

Yes they are listing additional assumptions of a ring

We also require the ring to have a unit, typically.

(I.e., a multiplicative neutral element)

[velvet steeple]

I usually only require that for a field

[coral spindle]

Well this is not standard. Usually we call such things "rngs" or "rungs."

That is, a rIng without Identity is a rng

In most cases, if people are talking about rings, they mean it contains a unit.

[velvet steeple]

Ok

[coral spindle]

Here, they are talking about this.

[chilly radish]

(This heavily depends on the field)

And personal preference but yea most people assume rings have 1

[chilly ocean]

If you don't require identity, then you can see it as a semigroup which distributes over addition.

[coral spindle]

It also depends on the language. For example, in French a field is not necessarily commutative, but in English it is.

[chilly radish]

True

[velvet steeple]

So you say R is a "commutative field" there.. Sounds weird

[chilly radish]

Intro algebra texts will also sometimes not assume unitality to do things in the proper generality.

Personally I resent the name "rng" for being annoying to pronounce and not informative

[coral spindle]

Yes, in french they do indeed say corps commutatif!

I also find this funny

[coral spindle]

True, I should mention that people also just say "nonunital ring" which is the clearest in my opinion.

[velvet steeple]

"areeee ennn geee"

[chilly radish]

Yea that's my preference

[coral spindle]

People do indeed say it like that sometimes, but I also hear people say "rng" and "rung" exactly the same way

One of my bugbears with algebra terminology is that people will say "R-algebra" and not specify if they mean unital, associative, or even commutative. I was in a talk recently where they had implicitly assumed they were all commutative the whole time, and I had no idea.

It was very confusing...

[chilly ocean]

- Who are you?
- You really want to know?
- Yes!
- You are an R-algebra.
- Do you have the slightest idea how little that narrows it down?

[coral spindle]

Too bad, it's all you get

[rain grove]

It ended up being useful doing that... I used (b) part where I show they are equal to lcm if b != a^k when proving that a permutation order is lcm of all orders of its disjoint cycles

But there might be an easier way still

[knotty badger]

I see!

Yeah this is why one looks at multiple proofs

It’s not that there’s One True Proof

[rain grove]

:)) yepp

[surreal dagger]

Basic Algebra 1? Thank you, I will check it out.
I think my goal is to become quick with routine type of problems and computations, so that im comfortable with working with the objects directly and my mind does not have to delegate much RAM to such problems when I encounter them in the scheme of a different problem.
Thats why I think it would be nice if there where a big list of problems of one type in ascending difficulty.

[chilly radish]

I would go insane

[rain grove]

If X is a set what is <X>?

[chilly radish]

In what context

[rain grove]

X subset of a group

[chilly radish]

The subgroup generated by X

i.e. the smallest subgroup containing X

[crystal vale]

Which book do you use?

[rain grove]

A book my professor wrote

Idk if its obvious that <X> = <X^{-1}> or am missing something

[delicate orchid]

Yeah it’s obvious

[crystal vale]

Is it available on the internet?

[rain grove]

You want to see it?

[crystal vale]

Yes

[hidden wind]

oh i just entered here with a related question

i’ve always seen groups generated by a set S to be defined as the elements that are products over S and over the inverses of the elements of S

why do i never see anything about words over S but not their inverses?

in the abelian finitely generated groups i suppose it’s clear that they’re the same, but in general?

[chilly radish]

It's because this isn't a group

Generally

You can look at the monoid generated by a subset S of a group, but idk how much people care about that

[hidden wind]

yeah the case of, like, N is clear, but i don’t see why there can’t be interesting cases as well

[chilly radish]

There probably are, it's just not something that comes up naturally often

[hidden wind]

hm

[chilly radish]

Someone somewhere surely cares about this

[hidden wind]

wondering btw as i’m trying to formulate the conditions for a edge-coloured graph to define a group, as that seemed like fun and is indeed so far

[hidden wind]

oh and wait and yeah i’m wrong they’re not the same for infinite abelian finitely generated groups… i forgot about Z woops

[nimble folio]

This might be a silly question

Is the first isomorphism theorem a universal property? if not, why isnt it?

[coral spindle]

People talk about strings over an alphabet all the time, just not really in group theory, because they're not groups. They are, however, monoids, and in fact are the free monoid in the same way that the free group is free.

[coral spindle]

No, it's just not in the correct form. In an intro category theory book you can see examples of universal properties.

A universal property of quotient groups is that it is the initial object in the category of groups H with a map out of G that annihilates the chosen normal subgroup.

You can prove that quotient groups satisfy this via the first iso, though, but it needs a couple extra steps.

[knotty badger]

Not by itself but you can use universal properties to prove it

That’s the perspective I have at least

And I do like the proof when phrased in that way

[nimble folio]

Yea I'm trying to get better at understanding universal properties

Since they keep showing up

[knotty badger]

Mhm mhm

They’re really just alternative descriptions for maps into or out of an object

[delicate orchid]

They keep showing up cause u got pseudo in the chat 😭

[knotty badger]

Which ones do you know?

[delicate orchid]

Here I do think it’s a good perspective for first iso tho

[coral spindle]

They're the first thing you learn when studying category theory. The definition is very simple: it is a property of an object in a category that describes that object up to isomorphism. It's not a mystical special thing. If you'd like to learn more category theory, I recommend Leinster's "Basic Category Theory."

[knotty badger]

I mean, you can use universal properties to define things

But it doesn’t mean you have to

[nimble folio]

i've seen them for quotient groups, localization, and free modules