groups-rings-fields

[next obsidian]

So alpha(x) is a root of x’s minimal polynomial, so it’s in there since L is normal

[gusty thistle]

i had an idea where if complex conjugation was an element of the galois group then it generates a subgroup of order two. Is that correct too?

[next obsidian]

Not necessitarly

It could be trivial

Take like Q(sqrt(2))

Complex conjugation is the identity

[gusty thistle]

oh yeah I assume I have a non real root

[next obsidian]

But if it’s non-trivial, then it’s order 2

Because like, lol

So it generates a subgroup of order 2

[gusty thistle]

So then if you have a non real root the galois group cannot be cyclic of order three right

[next obsidian]

Yeah that’s one way to see it

[gusty thistle]

cool

thanks

[next obsidian]

I think you could see it in different ways too

But yeah this is a good proof

[gusty thistle]

nice

[next obsidian]

It might be the easiest

[gusty thistle]

yes lol hehe

[next obsidian]

Okay well wait

This only works if it isn’t embedded in R

[gusty thistle]

I am going to think about what u said about Galois => Complex conjugation in Galois group

[gusty thistle]

yes iknow

assuming non-real root and polynomial over Q

Suppose I have an Galois extension K over Q, is there a nice way to show that the complex conjugation map is in its Galois group?

[coral shale]

in this convo, is conjugate a+-bi or conjugate roots

[gusty thistle]

the first

I will be more explicit and write complex conjugation

[tribal moss]

The extension being Galois implies that it is normal, so every non-real element of K is root of a minimal polynomial over Q whose roots are all in K.

[gusty thistle]

yes

[tribal moss]

This polynomial has real coefficients, so its non-real roots come in conjugate pairs.

[gusty thistle]

the polynomial associated with the Galois extension K/E ?

[tribal moss]

The minimal polynomial of whatever non-real element of K we're looking at.

[gusty thistle]

K might have only real elements

oh yes but then complex conjugation is the identity

[tribal moss]

In that case complex conjugation is the idententy on K, which is certainly in the Galois group.

[gusty thistle]

ok I understand

how does then follow that the complex comjugation map is in the Galois group ?

[tribal moss]

Do you agree that under these assumptions K is closed under complex conjugation?

[gusty thistle]

yes

[tribal moss]

So complex conjugation restricts to a bijection K -> K.

[gusty thistle]

yes

[tribal moss]

And this bijection is still a field homomorphism.

[gusty thistle]

wow

yes

therefore it must be in the Galois group

thanks

And the reason this only works fro Galois extensions is that if an extension is not Galois, then the extension may not be normal. And thus the complex conjugates of every element in the extension may not be in K :]

Insightfull thanks

U guys are the best < 3

[coral shale]

did we show this converse

[gusty thistle]

sorry i was trippin

idk wether the converse even is tryue lol

[coral shale]

if u adjoin Q with one of the roots of x^4 - 2 is what im thinking

[gusty thistle]

It is not I thinkl

[gusty thistle]

yeah

u adjoin a root of a minimal polynomial with multiple roots (one of them not i) and then u adjoin i for instance

then complex conjugation is an automorfism

but the extension is not normal

and thus not Galois

[tribal moss]

Hmm, I'm not sure that sounds convincing. Adjoining i doesn't mean we get something closed under complex conjugation.
For example, let omega be a complex third root of 2. Then Q[omega,i] is isomorphic to Q[cbrt2,i], but the latter doesn't contain two different cube roots of 2 -- if it did, it would also contain sqrt(3), but Q[sqrt3][i][cbrt2] is a tower of proper extensions of degree 2, 2, 3, so it has degree 12 over Q, which Q[cbrt2,i] cannot have.

[coral shale]

is this true?
For a given galois extension, the minimal number of elements you need to adjoin to the base field to create it is equal to the size of a minimal generating set for the galois group?

[next obsidian]

I don’t see how adjoining i doesn’t make it closed under conjugation.

Let a + bi be in L, and assume i in L. Then a + bi -2bi = a - bi in L

[tribal moss]

How do you know 2bi is in L?

(Or bi or b?)

[next obsidian]

ChmonkaS

So true, tfw real numbers but only access to rational coefficients

[pastel cliff]

this is a lil silly but im a bit confused about what it means for an elt to be idempoteeent

actually whats an example of a non-identity elt that is idempotent

[next obsidian]

Consider any ring R x S

For R and S nontrivial

Then (1,0) and (0,1)

Actually any non-trivial idempotent will give you a decomposition like this

So this is literally every way they occur

[pastel cliff]

so if R is just a ring and not the product of rings its only idempotent element is 1 right?

[next obsidian]

And 0

This is what tells you a ring is connected

But making sense of that would be impossible rn, so just consider it a shitpost

[pastel cliff]

no no no i just finished my topology hw dont say that word

just doing these

no spoilers tho

[next obsidian]

Lmfao

The thing I just said

Is (b)

[pastel cliff]

[next obsidian]

About how this is the only way this gets realized

Okay wel like

The connected thing is also this

But I did literally say (b) is what I meant

[coral shale]

bump

[agile burrow]

If I'm understanding this correctly, it can't be true. If you have a finite separable extension, then it's generated by a single element by the primitive element theorem. But there are certainly non-cyclic Galois groups

[coral shale]

oh ofc, nvm me

[agile burrow]

brain is exploding

central extensions, cohomology, projective representations

all connected.....

[chilly ocean]

[pastel cliff]

ok i lied any nudge on b)

[chilly ocean]

Do you know the general form of the Chinese remainder theorem? You can use it here, but it might be a bit overkill.

[elder wave]

try the most obvious map from R to the cart. product you can think off

[chilly ocean]

"Cross product" .

[elder wave]

german literal translation moment

nvm even in german it's the wrong word i can't even speak my own language

wait actually i'm not sure the map i suggested works here

[pastel cliff]

r -> (re, r(1-e))?

[elder wave]

it works if we're in a boolean ring / every element is idempotent

which is already commutative and i think the map might rely on that somewhere

[pastel cliff]

what were you suggesting

[elder wave]

the map you said

still worth a shot i guess

[chilly ocean]

Idempotents are defined as commuting with everything, so maybe it still works.

Alternatively, all rings are commutative.

[tribal moss]

What was "the most obvious map" supposed to be an answer to?

[elder wave]

Nitezbas question

[tribal moss]

Ah, there was the cartesian product in the question. I just can't read, ignore me.

[elder wave]

Yeah I think this might still work out

[chilly ocean]

The map does work.

[elder wave]

Sniped

[chilly ocean]

The only tricky part is surjectivity.

[elder wave]

I bruteforced the inverse

[chilly ocean]

Given (ae, b(1 - e)), you want to find an r with re = ae and r(1 - e) = b(1 - e). It's practically staring you in the face at this point.

(It looks at me while I rest.)

[agile burrow]

I'm still internalizing it, but I'll be able to talk about this eventually

[south patrol]

lol

[toxic zephyr]

the answer to this is D apparently but i can't see why. what is the multiplicative identity here?

to me all 5 seem obviously true

so i'm clearly missing something

[delicate bloom]

it does have an identity for multiplication, you can think of it coming from the chinese remainder theorem, it'll have to be 0 mod 2 and 1 mod 5, so the identity is 6

just test it out for yourself, 2*6=12=2 mod 10, etc

[toxic zephyr]

oh whoa that's crazy. i guess i didn't test all the elements lol
but it makes sense mod 5. is S isomorphic to Z_5?

[delicate bloom]

yeah but be careful cause there isn't a ring homomorphism from Z_5 to S in Z_10 since the identity isn't preserved

[toxic zephyr]

could we not define f(1)=6 in S?

like 0 to 0, 1 to 6, 2 to 2, 3 to 8, 4 to 4
havent checked if it would work but that would be my guess

[delicate bloom]

I think I phrased it kind of poorly, but I mean the map f(x)=6x to map elements of Z_5 to Z_10 which has its image as S is not a ring homomorphism because you necessarily must map 1 to 1 to be a ring homomorphism

but that doesn't stop S and Z_5 from being isomorphic, it was more just a side comment

[delicate bloom]

maybe I should elaborate more on my original point, the chinese remainder theorem gives a ring isomorphism from Z/10Z ~= Z/2Z x Z/5Z, so you can look at the elements of S and see that in Z/2Z they all map to 0 while in Z/5Z the map is all of Z/5Z. So S is the direct product of the zero ring with Z/5Z, which is isomorphic to just Z/5Z alone, since the trivial ring doesn't contribute any structure.

so you don't have to really check the numbers individually

[toxic zephyr]

ah okay that makes sense

thank you

[delicate bloom]

cool, yeah you're welcome

[gusty thistle]

Ok so the problem is that a might not be in L then?

Because if it were then bi would be in L right?

So a way to construct an extension over Q which has complex conjugation as an automorphism but is not Galois would then be the to adjoin a complex algebraic root of degree 4 and it’s complex conjugate right

Because then it is closed under complex conjugation

But the other two roots of the minimal polynomial of the adjoined element are not in the extension and thus the extension is not normal

[hollow mica]

why is an object that satisfies a universal property unique?

I'm trying to see this for the tensor product specifically but a general argument would be nice

I found this proof online

Let A, B be two universal objects. The universal property gives a unique map A → B compatible with the universal property, and also a unique map B → A compatible with the universal property. The compositions of these maps in both orders are necessarily id_A, id_B, again by the universal property.

I don't understand why the composition has to be the identity morphism

[south patrol]

You can apply the universal property with the same universal object and arrows twice, essentially.

[tribal moss]

Hmm, that might work, but it's not immediatey clear to me that adjoining two of the roots won't give you the rest of them too. A simpler way would be to adjoin e.g. the real cube root of two. Then complex conjugation is (trivially) an automorphism, or if you want it less trivial also adjoin i.

[hollow mica]

could you elaborate

[tribal moss]

Because the universal property demands that there be a unique map satisfying such-and-such, and both the identity and the composition A -> B -> A satisfies it.

[hollow mica]

If f: A -> B is the unique map compatible with the universal property, and likewise for g: B -> A, then we know that g ◦ f is compatible with the universal property (for the object A)?

[tribal moss]

Yes. I'm not sure if you have a generic definition of "universal property", but this is easy to see for concrete universal properties such as "product" or "equalizer".

[hollow mica]

f_1 is the unique morphism generated by the universal property, same with f_2 except it assumes B is the free object

so we have f_1 ◦ g_1 = h_1 and f_2 ◦ g_2 = h_2

[hollow mica]

what exactly are you claiming about f_2 ◦ f_1 here?

are you claiming that f_3 must be the identity here?

[tribal moss]

That is satisfies what ever the "unique morphism" guaranteed by the universal property is promised to satisfy.

[hollow mica]

because I don't see that

[tribal moss]

I don't understand your diagrams. Which of these arrows are given, which are produced by the universal property?

What is T?

[hollow mica]

first diagram assumes A is the the universal object, and f_1 is the morphism generated by the universal property

T is just the object you are constructing the free object from

[tribal moss]

What are g1 and h1?

Which universal property are you talking about?

[white grotto]

why is det( of a matrix rank 1 M + the identity) = trace(M) +1

how does it make sense

[hollow mica]

A universal object (on an object T) is a pair (O, m), an object O and a morphism m: T -> O such that for all maps p: T -> U, there exists a unique morphism r: O -> U such that r ◦ m = p

This is the definition I came up with when trying to extend the universal property for tensor product, sorry if it is wrong

[hollow mica]

The first diagram has O = A and m = g_1, the second has O = B and m = g_2

and then T is the same for every diagram; I am trying to prove that O is unique up to isomorphism for a given T

[tribal moss]

Okay thanks. It looks like this ends up defining the universal property of "being isomorphic to T", but it should work all the same. Let me take a look...

[tribal moss]

So in this diagram we have g1=h2 and h1=g2 because the choice of map from T to A /B is part of the data of an "universal object".

[hollow mica]

wait why?

[tribal moss]

It's perhaps better if I restate it with different letters, because the subscripts are making me confused.

[hollow mica]

sure

[hollow mica]

first diagram corresponds to universal object with data (A, g_1), second diagram is for (B, g_2)

[tribal moss]

We have T, and we have a: T -> A which is universal, and also b: T -> B which is universal.

Now we apply the universal property of A to b, giving us p: A -> B such that pa = b.

Also apply the universal property of B to a, giving q: B -> A such that qb = a.

Then pqb = pa = b.

[hollow mica]

ohhhhhhhhhh, you can apply the universal property to a particular morphism, not just a particular object

I guess applying it to a particular object doesn't even make sense, because you have to be talking about some morphism

[tribal moss]

Yes, the universal property you defined says "for all maps".

[hollow mica]

and similarly qpa = a, so A and B are isomorphic

[tribal moss]

The final step is to apply the universal property of B to b, which says there is exactly one map r: B->B such that rb=b.
But both r=id_B and r=pq satisfy that, so they must be the same.

[hollow mica]

oh right, need to show they are the identity

ok this makes sense

thanks

[tribal moss]

Note that (T, id_T) also satisfies your universal property, which is why I said the property you managed to define is that of "being isomorphic to T".

This scheme works for many different universal properties, but it is a bit tricky to formalize exactly what it is they have in common that makes it work. The trouble in the case of tensor products is that the bilinear map we start with doesn't even live in the same category as "vector spaces with linear transformations".

[hollow mica]

I just read the general definition of universal property on wikipedia; it involves a second category and a functor between them

oh

[tribal moss]

(I have seen a general definition once, managed to agree that it mostly works, and then promptly forgotten it again).

[hollow mica]

fair

[hollow mica]

oh so does the functor in question (the one relevant for the universal property) go from the category of "vector spaces with linear transformations" to the category of "vector spaces with bilinear maps"

wait

that second category doesn't make sense

hm

[tribal moss]

To be honest I'm not sure myself how to phrase the tensor-product universal property as an instance of a more general concept.

[hollow mica]

ok I will read https://math.stackexchange.com/questions/3646913/universal-property-for-tensor-product-in-an-arbitrary-category

ok the solution is to talk about these things called "multicategories", where morphisms map from a list of objects to another object (similar to how n-linear forms would)

so is there really no categorical analog then?

have I reached the peak of generality for tensor products

ok I found this, which defines the tensor product over modules, so it should be the same

I am wondering how Bilin_R(M, N; -) can be interpreted as a functor from the category of R-modules to the category Set

[wooden ember]

im not sure how to use Zorn's lemma in the end. Namely im taking my poset to be all the modules containing f(X) for which i can lift f, and i get that if i show that there is a maximal element in this poset that the maximal element must be Y by part 2), but im struggling to show my chains have an upperbound

namely because we only have existence and not uniqueness of lifts, i have no coherency of my lifts on a union over a chain

[gusty thistle]

Yes good one

Tyvm

[wooden ember]

oh god im an idiot i see what i have to do

am i applying Baer's criterion correctly by saying that FracR is an injective R-module when R is a PID?

Namely if i have a map g:(p)-> Frac(R) for (p) an ideal of R, i can lift this to h:R-> Frac(R) by h(r)=g(rp)/p right?

[oak gazelle]

how should i prove statements that assume a field characteristic to be a certain number

i haven't really worked with fields much but i have to learn about lie algebras for a research project i joined at my uni

and if i need more knowledge on fields, what's a good reference

[wooden ember]

The characteristic assumptions are there because division by n is involved

Namely, Tr(A-lambda*I_n) is just tr(A) -lambda * n

And you want this to be 0

So solving for lambda you have to divide by n

And this only makes sense if the characteristic of the field doesn’t divide n

[tribal moss]

I went back over my notes from the last time I understood this, and I think I've got it now.

Bilin_R(M,N;-) is indeed a functor from R-modules to Set.
Its action on objects is: For each R-module P, it gives you the set of all bilinear maps from M×N to P. As a set, this doesn't have any particular structure; it's just a random bag of things.
We get structure from its action on morphisms: If we have a morphism of R-modules f: P -> Q, then we can compose it onto a bilinear map M×N -> P and get a bilinear map M×N -> Q. This composition action is a map in Set from Bilin_R(M,N; P) to Bilin_R(M,N; Q), and we declare it to be the image of f under the functor Bilin_R(M,N,-). We can check that the conditions for being a functor are indeed satisfied.

Now the idea of the universal property of tensor product it something like this: Bilinear maps are cumbersome to work with because they don't live in a category. But if we can represent each bilinear map by an honest map in our category or R-modules, then things get much nicer.

It turns out that this is what the tensor product does: A bilinear map from M×N (to something) is exactly the same as an ordinary R-linear map from M⊗N (to the same something). Not only can we go uniquely from a bilinear map to a map-from-M⊗N (which is what the usual diagram form promises us), but we can also go from a map-from-M⊗N to a bilinear map simply by composing (which is so obvious that it is rarely stated explicitly).

What really makes this into a universal property is that these two equivalent concepts react in the same way to being composed with R-module homomorphisms. In the concrete setting of bilinear maps and tensor products this is almost too obvious to state, but if we want to generalize to other universal properties we'd better state it explicitly.
The technical way to do this is to require there is a natural isomorphism between the functor Bilin_R(M,N; -) and the functor Hom(M⊗N, -). Both of these are functors from R-modules to sets, and the naturality condition says exactly that if we have a bilinear map b: M×N -> P and some f: P->Q, then we can either compose m with f and then look for the map M⊗N->Q that corresponds to it or find the corresponding M⊗N -> P and compose that with f -- it makes no difference.

The natural isomorphism gives us, among other things, a bijection between Biliin_R(M,N; M⊗N) and Hom(M⊗N, M⊗N). One of the things we find in Hom(M⊗N, M⊗N) is the identity, and its counterpart in Bilim_R(M,N; M⊗N) is finally what becomes the ⊗-on-vectors bilinear map that goes with the tensor product.

In this entire discussion, M and N have been two fixed modules. The fact that the whole construction is also functorial in M and N is not captured at this level of abstraction.

[wooden ember]

am i being dumb or is this super simple and does not require whatever the hint they gave is supposed to mean

cant i just express F_p[Z/pZ] as F_p[x]/(x^p) and use correspondence theorem

in which case the ideals are just those generated by [n]\in Z/pZ

wait i see it

im a fool

that quotient isnt right

alright nvm me

[tribal moss]

In F_p[Z/pZ] we're only considering Z/pZ as an additive group, right?

[wooden ember]

yeah

this is such an ugly ring

[tribal moss]

So it should be F_p[x]/(x^p-1).

[wooden ember]

wouldnt it be F_p[x]/(x^p-1,x^2p-1,...,x^(p-1)p-1) which i am struggling to express as a principal ideal?

[tribal moss]

x^{2p}-1 is (x^p-1)(x^p+1), so it's already in the ideal.

[wooden ember]

ah right of course that's the factorisation

lol

alright it's not all that bad then

[ruby sundial]

isnt p-1+p+1=2p

not 2p-1?

[tribal moss]

$x^{2p}-1 = (x^p-1)(x^p+1)$.

[cloud walrus]

Troposphere

[ruby sundial]

oh lol

you wrote that correctly first time

my bad

[hollow mica]

this is a lot for me to understand so I will think about it throughout the day 👍

[oak gazelle]

ty vm

[wooden ember]

Am I correct in saying that Ext^i_R(M,M) = 0 for i>0 and M for i=0 where R is the module F_2[x]/(x^2+1) and M is the submodule (x+1)/(x^2+1)?

[oak gazelle]

is my understanding correct: EndV is the set of linear transformations from V to V which is a vector space, and with the 'usual product operation' (which I'm assuming is just the composition of linear transformations?), EndV is a ring. EndV with the bracket operation and the product operation, EndV is a lie algebra

wrong image

[wooden ember]

i took a long exact sequence of cohomology by starting with 0->M->R->F_2->0,. Then I noted that since R is free all the Ext groups Ext(R,M) will be 0 so that Ext^i(M,M)=Ext^(i+1)(F_2,M) for i>0. Next i computed Ext^i(F_2,M) suing a projective resolution and deduced it was 0 everywhere. And finally I computed Ext^0(M,M) as Hom(M,M)=M

oh sorry

[wooden ember]

yeah EndV is a ring with product given by composition, and it has an induced lie algebra structure

(like any ring)

[oak gazelle]

the product operation in a given ring will always satisfy the three axioms for a lie algebra?

how does a ring always induce a lie algebra

i dont know much abstract algebra so i'm trying to get caught up here, so i have a lot of gaps in knowledge

i thought that you needed to have a vector space structure before you can even consider the structure being a lie algebra

also can anyone help me figure out how to work with a bilinear map

how do i prove that a map is bilinear

you prove that it is linear in 'each of its arguments'?

[chilly ocean]

You prove that a map is bilinear by checking the definition of bilinear.

[chilly ocean]

Commutators. You need a vector space with a multiplication (an algebra), though, to get a Lie algebra.

The product operation itself in a ring will be associative, so it's not going to give you a Lie algebra.

You need to take commutators of ring elements to get anything resembling a Lie bracket.

[oak gazelle]

but then the ring is still not a vector space, right?

so how can it induce a lie algebra

little narwhal said "and it has an induced lie algebra structure (like any ring)"

[chilly ocean]

A general ring will not have an induced Lie algebra structure.

Lie algebras are in particular vector spaces.

[oak gazelle]

yes that's why i was confused

i think little narwhal was saying that any ring has a product operation given by composition? rather than any ring inducing a lie algebra

[chilly ocean]

Any ring just has a product operation. It may not be composition.

They are talking about specific rings.

Endomorphism rings of vector spaces. Here the operation is composition.

Endomorphism rings of vector spaces are vector spaces with a nice multiplication operation, so they can be given a Lie algebra structure using commutators as per my earlier message.

[oak gazelle]

thank you, that helps me understand it better

i may be in over my head on this research project with my professor and a grad student, but i want to catch up, so do you think its doable to just learn algebra concepts as i go along?

i already know quite a bit of linear algebra

so that isn't an issue

[chilly ocean]

I don't know you very well, so I couldn't say for certain. Why not just go for it?

[oak gazelle]

he wants me to read the first two chapters of this lie algebra and representation theory book by humphreys

thats true you dont really know me xd, but you're right i might as well just try

ill prob be in this chat for help a lot tho

[chilly ocean]

It could be very fun and you might learn a lot.

[oak gazelle]

yuup and im gonna get to see how real research is done, even if im not actively participating yet

[elder wave]

i've heard great stuff about humphreys

sounds like a nice opportunity if you're willing to put in some work

[oak gazelle]

i for sure am willing!

ty guys

[wooden ember]

Suppose i have two projective modules P and Q that map surjectively onto M. Is the fibered product of P and Q over M projective?

[oak gazelle]

not to get in the way of your question narwhal, but whenever anyone can, what is the goal of this matrices section? i think it's trying to describe the transformations in EndV by matrices, but the notation i don't really understand

like a matrix $e_{ij}$ where the matrix has 1 in the (i,j) position and 0 elsewhere

[wooden ember]

any endomorphism can be identified with a matrix after a choice of basis

[oak gazelle]

?

yea i knew that bit from a linear algebra book i read

[wooden ember]

once this identification is chosen, you have a basis for gl(V) in terms of elementary matrices

namely the e_ij

so to understand how the bracket works in gl(V) you just evaluate it on these elementary matrices

[oak gazelle]

and delta_jk in the image is just the identity matrix right?

and also e_ij is just standing for an arbitrary elementary matrix right?

[wooden ember]

delta_jk is the kronecker delta

[oak gazelle]

thats what i thought

[wooden ember]

it is worth 1 when j=k and 0 otherwise

[oak gazelle]

so the identity matrix

[tribal moss]

The Kronecker delta is also the elements of the identity matrix, but that is not particularly useful property here.

[oak gazelle]

ohhh

[wooden ember]

we know that e_ije_rs = delta_jr e_is

(try to prove this)

thats how they get the formula

[oak gazelle]

i actually gtg for a bit but i will prove that when i get back, should get me moving on from this section

ty ty

[wooden ember]

just bringing this back here

[pastel cliff]

is this proof good? unsure bc i never use surjectivity of the map

[wooden ember]

no this doesnt work

you want to show that phi(1_R) is an identity for all elements of S

you have only shown it for elements of the form phi(r)

so if phi wasnt surjective it wouldnt work

[pastel cliff]

oh but then that's all im missing

[wooden ember]

yup

[pastel cliff]

since phi is surjective every s \in S can be expressed as phi(r) for some r in R

also wanted to check this - the "it follows" part feels like bs bc how do we guarantee we dont get additive inverses in there

[wooden ember]

Yeah unless I’m being stupid that argument doesn’t work. What you’d instead want to do is to grab k_1 and k_2 to be the minimal naturals such that a_k_1 and b_k_2 are non zero. Then the term of degree k_1 + k_2 in the product fg has coefficient a_k_1 * b_k_2 which isn’t 0

[pastel cliff]

wdym minimal naturals

oh like indices

[wooden ember]

yeah

[pastel cliff]

this is still after multiplying f and g though right

[wooden ember]

?

[pastel cliff]

oh you literally said it

[wooden ember]

i mean yeah im showing fg isnt 0

[pastel cliff]

sorry im having a goober moment

[wooden ember]

np

[pastel cliff]

and that works bc it shows that there will always be at least one non-zero coefficient right

[wooden ember]

yup

namely the lowest degree coefficient

[versed orbit]

okay a quick question, what is an ideal that is guaranteed to be a subring? in swedish we call them "real ideals" but that seems a bit wonky in english. can't seem to find anything about the enmglish translation currently so if anyone knows what i'm talking (that would atleast make one of us XD ) then please tell me

[next obsidian]

It depends on what a “subring” is

For a lot of people a ring has to have a 1

And a subring has to contain that same 1

But then if an ideal has 1, it contains a•1 = a for all a in the ring, so it has to be equal to the ring

If you don’t require the existence of a 1, then all ideals are subring

So I’m not really sure what a “real ideal” is supposed to be!

[versed orbit]

so like if you have an ideal that is not allowed to be the ring itself,

[next obsidian]

I’m not sure what you mean like

Ohhhh

If you want an ideal to not be allowed to be everything

We call it a proper ideal

Because it’s properly contained in R

[versed orbit]

ahhh right yeah thats the word i was seeking thank you ^^'

[next obsidian]

I don’t think we have a terminology for an ideal which could also be everything

I’ve only ever heard that called “ideal”

[versed orbit]

haha same here

i just couldn't find the translation since the wiki pages differed to much in structure but thank you for the help 🙂

[wooden ember]

could someone give me an example where a projective module doesnt cancel from a direct sum? I was persuaded this was always possible earlier and thought i proved it but i see now my proof doesnt work, and im starting to think that this is false

by cancel i mean P oplus M = P oplus N => M = N

[next obsidian]

Ah fuck

[wooden ember]

i sure hope it's true

[next obsidian]

I don’t think so

[wooden ember]

because if it's false it fucks everything up

ahhhh

[next obsidian]

But there’s something very similar to this which is

https://en.m.wikipedia.org/wiki/Schanuel's_lemma

So like this is true

[wooden ember]

lmao it's kind of in the context of this

that i need my thing

[next obsidian]

Yeah I can’t produce an example I’ll think about it

But it feels not true

It’s weird because you can’t have the iso of P + N and P + M only send P to P

[wooden ember]

cause basically im trying to prove the analogue of schanuel's for an exact sequence L -> P_n -> ... -> P_0 -> M -> 0 and K -> Q_n -> ... -> Q_0 -> M -> 0 and for that i really need that cancellation property

to do induction on schanuel

[wooden ember]

yeah i thought this was true earlier because of projectivity

but then realised i done goofed

[next obsidian]

From Sham

[next obsidian]

This is true

[wooden ember]

yeah ik

[next obsidian]

Look at Matsumura

[wooden ember]

dont give a proof though

it's a graded exercise so i dont want actual help on this specifically

just to see if i can get my cancellation property working

thanks

[wooden ember]

the frick

[next obsidian]

Oh

[wooden ember]

that's a weird example

[next obsidian]

So Sham gave a more down to earth thing, well, sorta

Let M be projective and not free

Let N be such that M (+) N is frre

Then form N (+) M (+) N (+) M (+) …

Call this P

Then M (+) P is free

This is the eilenberg swindle

Have you seen it before?

[wooden ember]

nope

[next obsidian]

Okay well the point is P is free

Cuz it’s a direct sum of N (+) M which are all free

Yeah?

[wooden ember]

sure yeah

[next obsidian]

But if you put an M in the front

[wooden ember]

oh but M isnt free

[next obsidian]

Just change where you draw your parentheses

[wooden ember]

yeah that's fucked

[next obsidian]

Now it’s a direct sum of M (+) N

[wooden ember]

yeah i see

[next obsidian]

Okay so the point now

From M (+) P

Transfer one N to the left

(M (+) N) (+) P is what you get

[wooden ember]

yeah i see

[next obsidian]

So now

You have M (+) P ≈ M (+) N (+) P

M is not free

But M (+) N is

[wooden ember]

thanks that's a more understandable example

[next obsidian]

The other one should be with fg modules

[wooden ember]

i hate that this exists though

[next obsidian]

The example Sham gave

So it shows even finiteness isn’t good enough

[wooden ember]

so sad

aight back to square one i guess

[pastel cliff]

what's the point of having this proposition

like what insight do we gain from that

[wooden ember]

a whole lot of insight

you now know how to express the elements of the ideal generated by S in terms of the elements of S

which you couldnt do a priori

[next obsidian]

Yeah it’s huge

You have to use this description in proofs all the time

[wooden ember]

hurb a durb glurb ive been overthinking this

i just figured it out

[next obsidian]

Real

It do be like that

[pastel cliff]

am i losing my mind

is M_n(R) a field?

[wooden ember]

is it an integral domain?

[pastel cliff]

no...?

[wooden ember]

then it's not a field either

[south patrol]

Well

[pastel cliff]

but

[south patrol]

so i think the key thing here is that

not all rings are commutative

[pastel cliff]

"show that the only two sided ideals in R are 0 and itself"

isnt this true iff R is a field

[south patrol]

.

[wooden ember]

this is only for commutative rings

[south patrol]

that is probably the point of this problem tbh

[wooden ember]

yeah

[pastel cliff]

oh huh

so im not crazy this isnt a field right

[wooden ember]

no you're not crazy lol

it's very much not a field

unless n=1

[south patrol]

yeah as hinted, for n > 1 it is non-commutative and has zero divisors

but perhaps it's easiest to see that it's not a field because you can have matrices with vanishing determinant

[wooden ember]

fair enough im not sure why my mind jumped to zero divisors

[south patrol]

I mean lol there are multiple obstructions anyway so dw

[pastel cliff]

in that case how can i show that those ideals are unique

actually i'll just try it brb

ok i need to show that if I in Mat(R) contains any nonzero element it must contain every matrix

which is easy enough to think about but idk how to formalize it

oh

let A be in I

then (A^-1) I must be in I as well bc I is an ideal

[wooden ember]

this isnt true

and A isnt necessarily invertible

[pastel cliff]

ah

[wooden ember]

try instead to show ||every elementary matrix must belong to I||

[pastel cliff]

not reading that yet but i def need to do something something identity in I

[wooden ember]

alright ill let you think

[pastel cliff]

my only idea is that it def has additive inverses in it but i dont think you gain much from taht

[tribal moss]

Is this supposed to be a two-sided ideal?

[pastel cliff]

oh it specifies left ideals i hadn't noticed that

wait no ignore that

it is two sided ideals

[tribal moss]

So the ideal is closed under arbitrary row and column operations.

[pastel cliff]

so if i let I be an ideal with arbitrary nonzero element, can i go on to say something like "since I is an ideal, $id_{1,1} \cdot I = I$ ? where that's just the matrix with a 1 in the 1,1 coord

[cloud walrus]

stμ₂dying

[wooden ember]

you can and you can do even better

you can do e_11 I e_11 = I (e_11 is the standard notation for id_1,1)

[pastel cliff]

what does doing it on both sides do

#linear-algebra

[wooden ember]

try it on a matrix

see what happens

[pastel cliff]

you just get the single element of the top left corner

or whatever e_ii is

since \bR is a field, that lone element does have a multiplicative inverse

[pastel cliff]

do this with the matrix that just has the mult inverse of that element in the correct index

i see

[pastel cliff]

ok diff question

trying to show that $\Sigma^{\infty}_{i=0}r_ix^i$ is a unit if and only if $r_0$ is a unit in $R$.

[cloud walrus]

stμ₂dying

[pastel cliff]

i have the forward direction but idk how to do the other implication

[delicate bloom]

I'd write out the formula for the inverse in terms of the convolution directly and see that it involves division by r_0

well, preferably discover this formula on your own by multiplying by another series and work it out to see

[next obsidian]

Build up your inverse step by step

You have basically infinitely many equations to solve but they’re described inductively, it’s useful to work out how it works

Also fun to see what the difference between this and polynomials are

[oak gazelle]

how should i start proving this

[chilly ocean]

Matrix multiplication.

You should have the expression for the entries of a product of matrices somewhere.

[oak gazelle]

im just confused about the requirement for the matrices to have 1 in the i-j position

oh omg

im so dumb i just remembered something

[next obsidian]

Common experience doing math

[peak hornet]

what does it mean to restrict a group action?

\

this is the problem

not really looking for hints on the problem at least atm just trying to figure out what that phrasing means

[quiet pelican]

It mean that instead of considering the effect of G on all of A, we only consider it’s effect on the orbit of a

[next obsidian]

Namely it says that this is closed in the sense that like it will send things in that set into that set

[peak hornet]

ok, thank you!

also another quick question

Show that the collection of all orbits $O_{G}(a)$ for $a \in A$ forms a partition of A.

[cloud walrus]

failingphysics

[peak hornet]

i'm a little confused about whether we're fixing $a$ here and considering the action of all of the elements of $G$ on $a$, or fixing some $g \in G$ and considering how it acts on all of the elements of $A$

[cloud walrus]

failingphysics

[quiet pelican]

So show that every a is in some orbit, and that if two orbits have nonempty intersection, then they are the same

[quiet pelican]

We’re not fixing either in this second question

[peak hornet]

so then

we are considering the action of all elements on G

on all elements of A?

[quiet pelican]

Yes

[peak hornet]

that is really weird

not sure how to wrap my head around that atm

so then what is the difference between this and transitivity?

[quiet pelican]

Transitivity means that there is only one orbit in the partition

But even if there’s more than one, they partition the set

[peak hornet]

would you be willing to check my idea for a proof of this problem then

[quiet pelican]

Ok

[peak hornet]

ok, so first we have to show that every element of $A$ is in a partition. choose $a \in A$: we need to find $b \in A, g \in G$ such that $g \circ b = a$. choose any $g$ and $b = g^{-1} \circ a$; then the required condition is satisfied.

[quiet pelican]

The slightly quicker way to note this (that’s a particular case of your proof, which does work) is that a = ea

[peak hornet]

oh that's smart

ok that makes sense

[quiet pelican]

Also, this partition implies that there is only one, I would replace it with a partition

[cloud walrus]

failingphysics

[peak hornet]

thanks

the problem i'm proving rn is work we did last week in class

the first one is an hw question but it was linked to this class problem

now we have to show that if two orbits have a non-overlapping intersection, then the two orbits are identical. assume $\exists a, b \in G: g_1 \circ a = g_2 \circ b$. Then we left multiply by either $g_1^{-1}$ or $g_2^{-1}$ to show that $a \in O_G(b)$ or $b \in O_G(a)$, and since the inclusion holds in both directions the orbits must be identical?

[cloud walrus]

failingphysics

[peak hornet]

i'm not 100% sure about this one because i feel like we're just showing that if the two orbits have a non zero intersection then the element in that intersection exists in both orbits

never mind i see why it works

acutally no

[quiet pelican]

My proof would probably be something like Suppose $c \in O(b)$ and $c \in O(a)$ then $hb = c = ga$ so that if $d\in O(a)$ then $d = ka = k g^{-1} h b$ so that $d$ is in O(b) and similarly we see the opposite inclusion

[cloud walrus]

Micos

[quiet pelican]

I think this needs a bit more, it seems to almost assume the result you’re proving in order to show that a in O(b) and b in O(a) implies that O(a) = O(b)

[peak hornet]

i think what i'm thinking of right now is something lik

once you get $b \in O_G(a)$, you also have $g \circ b, g^2 \circ b, \dots$ and so the orbit of $b$ is clearly included in the orbit of $a$

[cloud walrus]

failingphysics

[quiet pelican]

The exact proof of that statement is basically half the proof

[peak hornet]

yeah i was trying to state it relatively intuitively

i like your proof better though

it feels a lot more explicit to me

thank you so much i really appreciate your help

[peak hornet]

i also had an idea for this if you wouldn't mind looking it over? if not that's cool too you've already been really helpful

[quiet pelican]

Ok

[jovial kelp]

hey this might be a stupid question, but are units and multiplicative inverses of a ring considered the same thing?

[quiet pelican]

Units are things with multiplicative inverses

[jovial kelp]

awesome, than you

[peak hornet]

so intuitively, identity holds trivially and the second condition holds because if you consider the orbit of any element in $O_G(a)$ and the orbit of $a$ itself, the two have to be identical

[cloud walrus]

failingphysics

[peak hornet]

actually idk about the second condition

may have to think about that more

actually

if the orbit of any element in $O_G(a)$ is identical to the orbit of $a$ itself

[cloud walrus]

failingphysics

[peak hornet]

then the orbits of elements in $O_G(a)$ are identical and so there is only 1 unique orbit, so that the group action must be transitive when restricted to the orbit

[cloud walrus]

failingphysics

[peak hornet]

and this holds from the discussion of the class problem earlier - if two orbits have a nonempty intersection then they are identical, and clearly the orbit of any element $a$ in $A$ has a nonempty intersection with the orbit of an element $a_n \in O_G(a)$

[cloud walrus]

failingphysics

[peak hornet]

if that's wrong please don't tell me the answer i want to try to figure it out myself

[quiet pelican]

Yes, that works

[peak hornet]

thank you!!!

is there a better proof than this? i feel like it was a little bit of a hack

[quiet pelican]

If your definition of transitive is has one orbit, I don’t think so really (it’s been a bit since I did group actions, so I can’t quite remember what’s the definition and what’s something we prove is equivalent)

[peak hornet]

my definition of transitive is, $\forall a, b \in O_G(a) \ \exists g \in G: b = ga$

[cloud walrus]

failingphysics

[peak hornet]

actually

i don't think our definitions are equivalent

[quiet pelican]

Is that the definition applied on A or on O(a)?

If it’s on O(a), then they are the same

If it’s applied to A, it looks to me like a trivial definition, ie every action satisfies it

[peak hornet]

yeah that’s what I’m thinking that that is the definition we were given applying on A

which is why I’m a little confused

[mint seal]

Probably a lame question but is it true that the algebraic closure of Q is isomorphic to the subset of C {a + bi | a, b are real algebraic} ?

seems like it's true but I'm a little puzzled no source I've seen ever characterizes it that way

I guess if a + bi = z is algebraic, you can solve for a and b to see they're algebraic too

[oak gazelle]

how do we deduce that the dim of sl(l+1, F) = l + (l+1)^2 - (l+1)

[delicate bloom]

there are l of the h_i and to count the e_{ij} you have (l+1)^2 places in the matrix but you remove the l+1 places along the diagonal

[tribal moss]

Yes, If z is algebraic over Q, then so is its complex conjugate (since they're roots in the same polynomial), and half their sum and difference gives you the real and imaginary part, so those are algebraic.

[oak gazelle]

i see, why is the h_i necessary for the basis of the special linear algebra

[delicate bloom]

no clue I just looked at the e_ij and h_i parts and counted them, sorry

[oak gazelle]

oh i see

i figured it out

ty tho btw

[delicate bloom]

cool

[abstract spear]

Also, I have seem this notation a few times now but was wondering if someone could tell me if my suspicion is correct

Do Q ^x and R* both mean just mean exclude 0 or does it mean to exclude all units that make that make the set invalid as a group?

[next obsidian]

you remove all non-units

for fields like Q this means taking out only 0

I imagine gl() as a function in r means you've fixed a field like say, the real numbers

and you consider the function sending a number r to gl_R(r)

[oak gazelle]

i keep running into roadblocks agh

i have no idea how to start on showing that even dimensionality is a necessary condition for existence of a non-degenerate bilinear form satisfying f(v,w) = -f(w,v)

"necessary condition" means iff basically right

so even dim --> existence of non of a ....

and vice versa

[chilly ocean]

A necessary condition is something that is implied.

Not an if and only if.

If A implies B, then we say B is a necessary condition for A.

[oak gazelle]

ohhhh yes that makes sense, you need it to imply the other thing

oh oops

if you have B then you will have A definitely

because A implies B?

[next obsidian]

no

[chilly ocean]

That is not what A implies B means. It means that if A holds, then B holds.

B can still hold if A is false.

[next obsidian]

saying B is a necessary condition for A to hold is saying that if A holds, then we necessarily need that B holds

it's like saying a necessary condition to being wet is that there is water

it's impossible to be wet without water

this is a kind of shit example but it's all I could come up with in like 1 second lol

[delicate bloom]

I won the deep sea diving competition, proof: I'm wet (whoops this is a necessary but not sufficient example :P)

[chilly ocean]

It's a general fact that skew-symmetric matrices have even rank. It isn't too difficult to prove, you can google it. If you have non-degeneracy, then the rank must be the dimension of the space; therefore, the space is of even dimension.

Alternatively, det(A) = det(A^T) = (-1)^dim det(A).

Non-degeneracy means you can divide by det(A). This gives (-1)^dim = 1, implying that the dimension is even.

I think you can use this determinant argument to prove the statement I wrote above.

[abstract spear]

Okay thanks

[small bramble]

this might be a really dumb question but - is it possible to prove in a ring that isn't unital that if y is nilpotent then so is -y? If not, is there a counterexample?

[lethal dune]

Show by indiction that (-y)ⁿ=±yⁿ depending on the parity of n.
n=1 obvious
assume for n=k, k even then
(-y)^(k+1)=(-y)^k(-y)=y^k (-y).
But now y^k(-y)+y^(k+1)=y^k(-y+y)=0 so
(-y)^(k+1)=-y^(k+1)

for for odd k as well

@small bramble

[small bramble]

alright, thanks

[little root]

Do there exist finitely generated Z-algebras that are torsion free but not free (as Z-modules)?

[mint seal]

is a Z-algebra another way of saying a ring?

[little root]

No

Wait

It’s late maybe it is •__•

Yes I suppose lol

[mint seal]

haha all good just glad I'm understanding

[little root]

I think Z[x, y] / (2x + 3y) maaaaybe works

As a counter example

[mint seal]

that looks good to me. Finitely generated, not free, and no torsion

[little root]

Is there an obvious reason why it’s torsion free?

[mint seal]

It should follow from deg(fg) = deg(f) + deg(g)

and negative degrees aren't in this ring

[little root]

Wait in this case torsion just means n*f is never 0 for a nonzero f and an integer n

[mint seal]

yup

[little root]

Oh hmm wait I’m not seeing how that formula applies sorry

[mint seal]

I think it's actually unnecessary to mention degrees

since as you said we're just multiplying by integers, the degree won't change

[little root]

I can try to prove that 2x + 3y is irreducible maybe? Then if you 2x + 3y doesn’t divide f then it won’t divide nf

[mint seal]

yeah that's good

you can use the above fact about degrees to see it's irreducible

since deg(2x + 3y) = 1

so if it's a product of two factors, those have degree 0 and 1

[little root]

Ah that’s a simple proof

Thank you :D

[mint seal]

Actually I guess more needs to be said?

the degree 0 guy needs to be a unit

is 2x + 2 considered irreducible in Q[x]?

it's 2(x + 1) and 2 is a unit in Q

so I guess it is irreducible, which feels a little weird

[woeful sage]

so yk how tensors are like $$T : V \cross V \cross ... \cross V \cross V^* \cross ... \cross V^* : \to F$$ where F is V's field

[cloud walrus]

Neamesis

[woeful sage]

does it have to be that way or can they be like $$T : X_1 \cross X_2 \cross ... \cross X_n : \to F$$ Where F is Xi's field (all Xi's have the same field) but they can have different dimensions

[cloud walrus]

Neamesis

[woeful sage]

X_i

Oh damn I meant just -> not :->

[hidden haven]

No, a tensor is defined to be the former

[woeful sage]

yes but, what sort of object would the latter be? (also how would one do a masters at cmi? (is it good for a masters in physics?))

I guess I'm asking because i dont really know the mathematical motivation behind tensors, only the physical motivation (kinda)

[delicate orchid]

“I love linear maps!!!”

[wooden ember]

just putting this back here in case someone has an answer

[tender bough]

I'm thinking of some thing fun. For a ring R, R\{0} with multiplication is a monoid. But R with multiplication is also a monoid -- the added zero respects the multiplicative identity 1 and does no harm. Sooo... I think a monoid can actually be extend with an arbitrary number of zeros as long as the multiplication is closed on the set of zeros (otherwise associativily might not satisifed, say there are 0' and 0'', think about 0' x 0'' x non-zero). When the multiplication on the set of zeroes is commutative, the "zeroness" of those zeroes actually form a partial order. For example, I can have a sequentially stronger zeros, 0, 0', 0'', 0''' ... 0^(n), with 0^(i) x 0^(j) := 0^(i) for i > j.

just some thoughts

[south patrol]

Well R \ 0 is a monoid only if R has no zero divisrs really

Also how would your zeros work?

It seems there would be nothing about them to make them "0" instead of any other symbol

Or they would all coincide

Depending on what you mean

[tender bough]

not really? from nlab

[delicate orchid]

is this not just nilpotent elements lol

[south patrol]

Yes but that is R

We both mentioned R \ 0

[wooden ember]

multiplication isnt closed if R isnt an integral domain

[south patrol]

Yes

[delicate orchid]

changed my mind, it's closer to a set of absorbing elements

[wooden ember]

i think they mean define 0_1 so that 0_1 * a = 0_1 for all a in your monoid, and then inductively define 0_i to satisfy 0_i * a = 0_i for all a in your previous extension

[south patrol]

Pain

[south patrol]

Hm

I mean sure but then idk what is interesting about it

[delicate orchid]

0_1a = 0_1 and 0_2a = 0_2 => 0_10_2 = 0_1 = 0_20_1 = 0_2 so all these zeros are just equal anyway

unless I'm missing something?

[south patrol]

So based

[wooden ember]

nah 0_10_2 isnt 0_1 i think

it's meant to be 0_2

[south patrol]

Oh ye

[wooden ember]

you only absorb elements from your previous extension

[tender bough]

WTF, I thought by definition a ring was closed under multiplication. Like this answer
https://math.stackexchange.com/a/1361467

[delicate orchid]

I see I see

[south patrol]

Yes

[wooden ember]

no but like

[south patrol]

Well that message was ambiguous

[wooden ember]

if you remove 0

[south patrol]

Once again we were talking about R \ 0

[wooden ember]

and consider the multiplication in that set

if you have zero divisors this multiplication isnt closed

so you dont get a monoid

[wooden ember]

im not sure either

[tender bough]

wow that's cool, I learned something about integral domain today

[south patrol]

That is more or less the definition

[tender bough]

yeah, I learned about integral domain today.

😳

[south patrol]

Yee fair

Also like

You can have no zero divisors without being commutative ig so like

Kinda more general actually

[tender bough]

was actually studying analytic number theory when I tried to characterize a Dirichlet character, and that's when I started thinking about monoids, because a Dirichlet character by definition is a monoid homomorphism from ((\mathbb{Z}, \times)) to ((\mathbb{C},\times)). Notably, as a consequence, the character also respects 0.

[cloud walrus]

Mattuwu

[tender bough]

I wish we had a name for monoids with both zero and one

reminds me of how {0, 1, 2, 3...} with gcd or lcm is a semilattice (thus a monoid) with 0 and 1 being the top and the bottom.

but it's different here because 0 is a not top in (Z, x) or (C, x)

[gusty thistle]

I want to determine all the generators for the multiplicative group (Z/pZ)* for p a prime where p is less than or equal to 19. Is there a clever way to do this apart from checking every element?

[south patrol]

Well once you've found one generator for each p there is a way to get all the others

[gusty thistle]

oh rlly

I just wrote a python script xD

[wooden ember]

is there any text where it is standard to consider a weight space of a lie algebra representation to not only include simultaneous eigenvectors but also simultaneous generalised eigenvectors?

because looking through books ive realised my prof has been using non standard definitions

[delicate orchid]

The text on lie algebra reps I'm reading now by Brian Hall does this (if you're referring to weight vectors)

[wooden ember]

im referring to straight up weights

for us they're 1 dimensional reps such that there is a N_x for each x with (rho(x)-lambda(x)id)^N_x * v = 0 for a fixed v

[tough raven]

Isn't that a completely multiplicative arithmetic function?
Or is that what a Dirichlet character is?

[formal ermine]

what does $\bZ^r$ in the context of groups mean

[cloud walrus]

illuminator3 👻

[chilly ocean]

I know $\bZ_r$ would mean {0,1,2,3,...,r} under addition

idk if it is the same thing

[formal ermine]

it's not

[cloud walrus]

Avina

[formal ermine]

also that's wrong

[chilly ocean]

bruh how do i use the bot?

[formal ermine]

it'd be 0,1,...,r-1

[chilly ocean]

just is notation diff then

oh yeah is r-1

yeah you're right

it doesn't matter if it is prime though

[formal ermine]

yeah but they use both Z^r and Z_k in the equation

[formal ermine]

confused addition with multiplication, mb

[chilly ocean]

yeah look at a textbook idk

i have never seen that before

[formal ermine]

we never defined it in the lecture

r, d_i \in \bZ_+

[chilly ocean]

bruh

yeh no idea

[delicate bloom]

direct product of r copies of Z, called the rank

[chilly ocean]

I thought up of this statement, could someone tell me if it is true or not?
Suppose we have groups G, H, and function $\phi(n)$ which maps G to H. If we prove $\phi(n)$ is one to one, and $|G| > |H|$, it must be onto, since every $x \ in G$ maps to 1 element in H

[cloud walrus]

Avina

[formal ermine]

if |G| > |H| then phi cannot be injective

[south patrol]

Indeed for sets S,T, saying |S| > |T| is equivalent to saying there is an injection T -> S but no injection S -> T

(Also why was it written as phi(n))

[formal ermine]

@chilly ocean $$\bZ^r = \underbrace{\bZ \times \ldots \times \bZ}_{\text{r times}}$$

[cloud walrus]

illuminator3 👻

[formal ermine]

it was that simple lol

is there an intuitive sense to the semidirect product?

[south patrol]

nice latex illuminator

[south patrol]

Well I think it's good to think in terms of the internal semidirect product to get intuition for it

[formal ermine]

what's the internal semidirect product?

[cloud walrus]

potato

[formal ermine]

what does N intersect H = 1 mean?

[south patrol]

Well slight abuse for {1}, or {e}, or whatever your preferred notation for the trivial group is

[formal ermine]

ah ok

[chilly ocean]

Ik I thought of that but then I saw a statement that $Aut(Z_n)$ is an isomorphism with $U(n)$ and $|Z_n| > |U(n)|$

[cloud walrus]

Avina

[formal ermine]

that's not possible

for two groups to be isomorphic to another there needs to be a bijective homomorphism between them

[chilly ocean]

I mean my textbook says it is

[formal ermine]

and a bijection requires |X| = |Y| for f : X -> Y

[formal ermine]

Aut(Z_n) has immensely many elements lol

more than Z_n

[pastel cliff]

what's wrong with this proof

[chilly ocean]

I did not make this up

[formal ermine]

where does it say |Z_n| > |U(n)|?

[quiet pelican]

|Aut(Z_n)| =/= |Z_n|

[thorn delta]

its a little sus

sus reasoning

[tough raven]

Is there a vector space $V$ and a sequence of endomorphisms $T_1, T_2, \dots$ such that $T_1^2 = 0, T_1 \neq 0, T_{i+1}^2 = T_i$?

[cloud walrus]

Raghuram

[runic hemlock]

sure. take V to have a basis in bijection with (0,1). Define T_i on the basis element [x] to be [x-2^{-i}] if this is in (0,1) and 0 otherwise.

[rotund aurora]

The content of 2x^2+1/2 x is 1/2 ?

[south patrol]

Nothing

this is quite a nice problem actually

because there is a nice way to prove this

[chilly ocean]

[south patrol]

oh that's what i was gonna say lol

yes nice

[ripe basalt]

How can I show that if $G$ is a finite group such that for every $n > 0$ there are at most $n$ elements of order dividing $n$ then $G$ is cyclic?

[cloud walrus]

*-algebra

[ripe basalt]

Keith Conrad has a similar result in theorem 3.3 https://kconrad.math.uconn.edu/blurbs/grouptheory/allgrouporderncyclic.pdf but he assumes $G$ is abelian

[next obsidian]

by induction you can show proper subgroups are all cyclic, and from there that there is at most one subgroup of every order

you can then assume G isn't cyclic and show via some smart counting that there aren't enough elements in the group

IIRC

[pastel cliff]

what's the "point" of PID's - ik they're obv important but the definition seems kinda random out of context

[pastel cliff]

ah ig it ties back to this right

[agile burrow]

I'm sure other people can say more, but to me the idea is that PIDs have a lot of structure that you lose in general settings, and they behave a lot like the integers. The two biggest things that come to mind immediately are that PIDs automatically have unique factorization and that there's a very nice structure theorem for finitely generated modules over PIDs

[delicate bloom]

@smoky cypress you using ML to play? I'd join but I'm in a call rn and working on stuff

[smoky cypress]

??

Oh wrong chat lmao

[hidden haven]

And the fact that any submodule of a free module over a PID is free (this is equivalent to being a PID)

This and the structure theorem are the facts that I've seen used the most in homological algebra

[quiet pelican]

No, just choose a random element, and map it to 1 mod 3 and its inverse to 2 mod 3, and everything else to 0 mod 3, this has that property but isn’t a Homomorphism for most groups

[next obsidian]

Wtf does mod 3 mean lol

[chilly ocean]

it's true when f is a homomorphism

it doesn't have to be

[formal ermine]

how many subgroups must a normal series contain? otherwise couldn't I just say $1 = G_1 \normal G_0 = G$ and conclude that $G$ is always solvable?

[cloud walrus]

illuminator3 👻

[next obsidian]

It’s not about the number

There’s more to a normal series than that you have a series of inclusions which express each thing normal in the other

You also require some condition on the quotient groups

[formal ermine]

huh

[next obsidian]

There’s a lot which are equivalent, but they probably fall in simple, abelian, or solvable

[formal ermine]

we defined this as the normal series of a group G where G_i are subgroups of G_0

[next obsidian]

The last one doesn’t make sense for a trivial series like you made

[formal ermine]

then said G is solvable iff this series contains abelian groups only

[next obsidian]

Well okay for a normal series you don’t need anything

But for a solvable group you need a normal serires whose composition factors are eg abelian

[formal ermine]

ahhh

yes

I read the text wrong

mb

[next obsidian]

[formal ermine]

but wait still

nvm

yes

thanks

[next obsidian]

I feel that

[formal ermine]

so G is solvable if D^n(G) = 1 for some n?

[next obsidian]

Idk what that means

[chilly ocean]

Deez nuts.

[next obsidian]

Is that like some upper / lower central series?

[formal ermine]

we called it the ith iterated commutator group

[next obsidian]

Okay yeah that’s the like

Lower central series or something

I think that’s equivalent IIRC but it’s not like obvious they are or anything

I think it’s clear this implies solvable

Because the quotients given by this filtration are a bunch of abelianizations

[formal ermine]

idk what that means

[next obsidian]

So all the composition factors are abelian

Abelianizariin of G is just defined to be G/[G,G]

It has the property that it’s abelian

[formal ermine]

ah

[next obsidian]

But showing a solvable group has this eventually trivialize isn’t like immediate

[formal ermine]

I was thinking about the other way

[next obsidian]

First thing I would do is induction on length of the series

That’s always helpful

And you can just do it for free since it’s trivial to prove the base case

After that idk, figure out how [,] works with quotients or some shit

[formal ermine]

waitttttttttttttttttt

we have D^(i+1)(G) is normal in D^i(G) and

D^i(G)/D^(i+1)(G) is abelian

[next obsidian]

That’s what I said

[wooden ember]

It’s called a derived series

[next obsidian]

[wooden ember]

If I’m not mistaken

[next obsidian]

.

[wooden ember]

And lower central will be the ascending chain of central lifts

[formal ermine]

what's a filtration

[next obsidian]

What’s the upper central series?

[wooden ember]

For nilpotent groups

[next obsidian]

Just like, any series

[wooden ember]

I’m pretty sure

[next obsidian]

Wait so is it like

Terminating upper central iff lower central?

Are those both for nilpotent groups?

[wooden ember]

Ah no nvm

The central lifts is upper central

[next obsidian]

I thought one of them is like

[wooden ember]

And the commutators for nilpotent groups is lower

[next obsidian]

G^n = [G^n-1,G]

G_0 = G

[wooden ember]

So [G_n,G]

Yeah

[next obsidian]

Yee

These terminating are equivalent to nilpotent rifht?

[wooden ember]

This doesn’t make any sense because lower central series starts at the top and upper at the bottom lol

[wooden ember]

No just the second one

Wait

No you’re right both of them

I always get confused lol

[next obsidian]

I’m so swag

[wooden ember]

Yeah cause nilpotent groups are just a bunch of central extensions or smth

[next obsidian]

¯_(ツ)_/¯

I just remember proving like 6 things are equivalent

With Shamrock

[wooden ember]

Lmao

It’s easier to remember with nilpotent Lie algebras imo

[next obsidian]

Like product of p-groups, central series, product of Sylows, normalizers grow

Etc etc

[wooden ember]

Yeah fuck nilpotent groups

[brittle pebble]

can anybody tell me what does "x" means in Z/(p) x Z/(p)?

it's probably multiplication.... but i don't know what the outcome of it is. For example, what is Z/3 x Z/3 ?

[next obsidian]

It’s the product

Cartesian product with the induced group structure of a product of groups

Let G and H be groups, then G x H are pairs (g,h) with g in G, h in H, make this a group by setting
(g1,h1)•(g2,h2) = (g1•g2,h1•h2)

[brittle pebble]

ohhhh i c

[next obsidian]

If you’re considering Z/(p) as a ring then it’s he same thing

And you define addition similarly

[brittle pebble]

so you can't really do G dot H if their order differs?

[elder wave]

if you have arbitrary groups their operation might not fit

[next obsidian]

What does that mean

You can take the product of any two groups

Their order is entirely irrelevant

[brittle pebble]

can you give an example of product of two groups with different orders?

[elder wave]

i think i misunderstood the question

[paper flint]

Z/2Z×Z/3Z?

[elder wave]

i thought they meant G dot H as in taking elements of different groups

[next obsidian]

The thing I described works for literally any two groups

[next obsidian]

Okay but this also is false

G and H always embed into G x H

If you want to take the product of arbitrary groups without embedding into something that doesn’t make sense

And if you want to make sense of it wrt any embedding then there you go

[elder wave]

no i meant as in somehow taking g*h in one of the coordinates

[next obsidian]

But that’s not like even well-defined so it’s a bad thing to consider

[elder wave]

yes i jsut misunderstood

[next obsidian]

Do (G x H) x Z/42069

Now you can multiply g and h in one coordinate

[elder wave]

[celest cairn]

How do I simplify this polynomial? I know how to do it in Z[x,y] but not just Z[x].
Is this possible?

[next obsidian]

This shit doesn’t even make sense

That isn’t a valid expression

[celest cairn]

K didn’t think so lmao

[next obsidian]

Where are you even pulling all these things you keep asking to compute splitting fields of and stuff

Do you just like sit down and write a random expression down

[celest cairn]

Yeah lmao

[next obsidian]

And try to find its splitting field

[celest cairn]

I’m in 11th grade

[paper flint]

Based ngl

[strong valve]

@hidden haven how did you get verified?

[formal ermine]

lmao

[brittle pebble]

I got it finally................. I misunderstood "cartesian product" I thought {0,1,2} x {0,1,2} = {(0,0), (1,1), (2,2)} but instead its {(0,0), (0,1),...,(2,2)} fuc

[next obsidian]

Yup!

What you were thinking of is called the diagonal

It’s a subset of S x S for any set S

Consisting of stuff of the form (s,s)

Where s is something in S

Reading that out loud be like “ess ess in ess ess ess ess”

[zenith trellis]

i'm sorry i did not read what you wrote just beneath that... but there is a way to prove directly psi(xy) = psi(x)psi(y) for any (x,y) in G² once you identified a = phi(e)^(-1)

you may choose wisely a1 a2 a3 b1 b2 and b3 to use the property of phi

[hidden haven]

Bought the mods for $2

[paper flint]

Your payment is due Moldi...

[rustic crown]

[next obsidian]

[coral spindle]

Your reasoning that phi(x^-1) = phi(x)^-1 implies phi(xy) = phi(x)phi(y) assumes the conclusion near the start and is therefore invalid. In fact, this is not true at all, as for instance the map f : G -> G given by f(x) = x^-1 satisfies f(xy) = f(x)f(y) only when G is Abelian.

Oh wait hold on, you're using the property above!

Jeez I can't believe I missed that. Sorry for that. I'll take a second look.

[chilly ocean]

I have a problem: "Let $\phi$ be an automorphism of $D_8$. What are the possibilities for $\phi(R_{45})$?

What actually is $R_{45}$?

[coral spindle]

As Silfer suggests you can indeed choose your elements wisely to find that this property holds, if you'd like to continue down this route. An extra hint: Since you know that phi(e) = e you can engineer things nicely.

[cloud walrus]

Avina

[coral spindle]

I don't know this notation either. You should look at the lecture notes or book that you are studying to find the answer to that.

Perhaps though, this is just the Rotation by 45 degrees.

[chilly ocean]

oh yeah, that could be it

[iron bolt]

does anyone know how to put in a more rigorous way that if u twist the coners of a rubiks cube clockwise or anticlockwise must be congruent to 0 mod 3 and if not 0 is not legally solvable

[eager willow]

There is a group of 'legal' moves which is a subgroup of the group of all the moves including the illegal twists. Making an illegal move changes the coset of the legal moves to no longer include the identity (the solved state)

Therefore all you have to do is prove that if you twist a corner counter clockwise and make any number of legal moves followed by a clockwise twist, you get a 'solvable' state, or an element of the subgroup

[iron bolt]

how do i do that

[eager willow]

Reduce down to a few cases and prove one by one how to solve those few states

[iron bolt]

like kociemba algorithm?

[eager willow]

All I know mathematically about the Rubik's cube is what I've taken the time to prove and solve myself. I know of Keith Conrad's paper as well but I haven't read it.

[iron bolt]

oh okok

thank uu

for ur time

[eager willow]

If you think about it, any algorithm you know of to solve the top two layers is sufficient to prove that you need only to consider illegal moves that happen on the last layer

Because following any illegal move you can just apply the algorithm to solve the first two layers back

[iron bolt]

then its easy to prove witch are the not solvable states

[eager willow]

Basically, an algebraic approach is only going to offer so much insight into what's going on in what amounts to a single group with a well understood set of generators. End of the day you have to work with the algorithms you know of to solve the cube.

[iron bolt]

can i just use oll to prove them ?

[eager willow]

Yes, take every combination of twisting one clockwise and another counterclockwise and use OLL-PLL to just solve it with your own hands. That's as rigorous of an approach as any.

[iron bolt]

but like in case of the cubes bigger than the 3x3 the parity of the cubes rely on if the cube size if its odd or even how can u prove that those parity exists when ur at 3x3 stage?

[eager willow]

Well the way you do the last layer may lead to more confusion. I know of algorithms which fix the corners and permute only the edges, and I've always solved the corners first and then the edges for last layer, proving parity is a non-issue

[iron bolt]

which method do u use

like if u reduce the big cube into a 3x3 stage

u can get parity