groups-rings-fields

[sturdy marsh]

then F (x)_(O_X) G is the sheafification of the presheaf U ---> F(U) (x)_(O_X(U)) G(U)

right?

[vestal snow]

Yes

[sturdy marsh]

so they have the same stalks

[vestal snow]

Yes

I think some change of scalars argument would need to be used

[sturdy marsh]

what's the definition of the stalk

[vestal snow]

Which one?

Universal property or the construction?

[sturdy marsh]

the construction

[vestal snow]

(f,U)_p

Or in this case

It would be (f (x)_O_X(U) g, U)_p

elements of that form

[sturdy marsh]

no I meant the colimit

it's the colimit of all the F(U)

for U containing p

where do the structure maps live

[vestal snow]

What are the structure maps again?

[sturdy marsh]

F(U) ---> F(V) for V \subset U

the stalk is the colimit of that diagram

[vestal snow]

I have no idea

I don't know what the category is called

But its the category where scalar multiplication agrees with restriction

[sturdy marsh]

actually you know what, it's probably better to do it directly

idk if what I was going for works well

Write down a map from F(U) \otimes O(U) G(U) ---> stalk

for all U

[vestal snow]

Okay

Done

f (x) g goes to (f (x) g, U)

[sturdy marsh]

induce map from colimit ( of abelian groups)

[vestal snow]

Right

Wait

[vestal snow]

won't the colimit just be the stalk?

Can you make this more precise?

[sturdy marsh]

the domain will be the stalk of the tensor product

the codomain is the tensor product of the stalks

[vestal snow]

Oh then the original map from F(U) (x)_O(U) G(U) to F_p (x)O_X,p G(p) will be f (x) g maps to (f,U)_p (x) (g,U)_p

[sturdy marsh]

right

[vestal snow]

And the map from (F (x)_O G)_p would send (f (x) g,U)_p to (f,U)_p (x) (g,U)_p

And this would be an abelian group morphism

[sturdy marsh]

right, and verify that this is a morphism of O_X,p modules

[vestal snow]

By hand?

[sturdy marsh]

all of the morphisms from F(U) (x) G(U) ---> stalk are morphisms of O_X(U)-modules

[sturdy marsh]

any appendage you like

[vestal snow]

So that makes (F (x) G)_p --> stalk an O_X,p module?

[sturdy marsh]

you need to prove it

but yes

which is easy

pick germs

[vestal snow]

Oh wait

Yeah its easy

[sturdy marsh]

and do it at the U-level

[vestal snow]

Using the construction

I need to stop trying to do everything using some universal property

[sturdy marsh]

once you have tensor hom for sheaves im pretty sure you can try fiddling around with it and prove this

but yeah the direct way seems easy enough

[vestal snow]

@sturdy marsh I think there is a problem with the argument

For each U, we get an O(U) module homomorphism F(U) (x)_O(U) G(U) --> F_p (x)_Op G_p

And we though of these as abelian group morphisms to get an abelian group morphism from their colimit

However, we don't know if the colimit where the morphisms are just group morphisms is the same as the colimits where the morphisms are abelian groups which agree on multiplication action by O_X

The latter is what we define to be (F (x)_O G)_p

[sturdy marsh]

"the colimits where the morphisms are abelian groups " ?

@vestal snow

[vestal snow]

I'm having trouble trying to write down what I mean

[sturdy marsh]

to be clear, f_U : F(U) \otimes G(U) ---> (tensor product of stalks) is an O(U)-module homomorphism

the map f: Colim (F(U) \otimes G(U)) ---> (tensor product of stalks) is a priori a map of abelian groups

[vestal snow]

So (F (x)_O G)_p is the universal object such that if there is some map from each F(U) (x)_O(U) G(U) into some other module M which behaves nicely with the restriction maps AND the multiplication, then there is a unique map from (F (x)_O G)_p into M

However, in the argument, we considered the colimit with the universal property that if there is some map from each F(U) (x)_O(U) G(U) into some other module M which behaves nicely with the restriction maps, then there is a unique map from the colimit into M

[sturdy marsh]

try constructing this object, the colimit is taken in the category of abelian groups

and then you give the colimit an O_X,p -module structure

[vestal snow]

Is my definition of (F (x)_O G)_p wrong?

Or in general, given an O_X module M, is M_p defined by thinking of it as only an abelian group and then adding on the O_X,p structure?

[sturdy marsh]

I think the definition is okay, taking stalks probably commutes with forgetting the O_X -module structure

[vestal snow]

How do most books define it?

[sturdy marsh]

I would know if I read a book

[vestal snow]

Maybe I should ask @next obsidian

Just want to make sure my definitions are correct

[next obsidian]

Colimit as an abelian group

[sturdy marsh]

or you could just try proving that the construction satisfies the property

[next obsidian]

Gets a structure as module over the colimit of the rings

That’s all I know

[vestal snow]

Got it

So Brofib was right

[next obsidian]

Afaik

[vestal snow]

Thanks

[next obsidian]

I said I think you can show a colimit of modules has a module structure over the colimit of the ring via tensor product

I think this is actually the good way to do this

Like a left A-module structure is a map A (x) M -> M where (x) is the tensor product of abelian groups subject to a rule which means its associative

If that map is phi, the way it works is that you define am = phi(a (x) m)

You can really easily show this is equivalent to a module structure

The rule for associativity is that like,
A (x) A (x) M -> M is equal, when the two paths are like

id (x) phi followed by phi

Or multiplication on A (x) id

Followed by phi

Anyway, if M_i is a system over A_i

Then from the maps A_i (x) M_i -> M_i you get a canonical map

colim (A_i (x) M_i) -> colim M_i

Since tensor product commutes with direct limits you get a canonical map

(colim A_i) (x) (colim M_i) -> colim M_i

Blah blah verify it’s “associative”, it follows by just taking direct limits on the square we want to commute

Anyway, the point is this gives you a canonical way to give a module structure, I guess you could verify it’s the same as the “hands-on” definition you’ve made

but I think it justifies why like... this is the right way to do direct limits of modules over different rings

Like if you do the direct limit in Ab, you get a canonical way to give it an A module structure from the various A_i-module structures of the M_i

[vestal snow]

Brofib how do I verify that the map I got from (F (x) G)_p --> F_p (x) G_p is injective?

[sturdy marsh]

write down an inverse

[vestal snow]

Yup that's what I thought

Will it be the map induced by F_p x G_p --> (F (x) G)_p?

[chilly ocean]

Everytime I see your name I read Have a Banana at first.

[steady axle]

If a group $G$ is generated by set $X$ such that $gxg^{-1} \in X$ for all $x \in X$ and $g\in G$. I want to prove that $[G,G]$ is generated by $xyx^{-1}y^{-1}$. any hints?

[cloud walrus]

bert

[delicate orchid]

I’m pretty sure that means that G is the normaliser of X, that’s probably an ok starting point

[vestal snow]

Wdym?

[steady axle]

how does that help ?

[viscid pewter]

what's y

@steady axle

what's y

[chilly ocean]

Quick question : when proving that every group of prime order is cyclic, does it follow from the corollary of Lagrange's theorem which says that the order of every element of a (finite) group divides the order of the group ? So say if |G| = p where p is prime we know that the order of every element is 1 or p. But since p >=2 and the only element of order 1 is the identity, when there must be an element of order p thus G is cyclic, is this decent as a rough proof?

ye

My point is :in a general case where |G| = n just saying that the order of every element is a divisor of n doesn't suffice to say that there is an element of order n right?

Lagrange's theorem doesn't imply existence of such elements, yes.

Although there is a Cauchy theorem, that says for a finite group G with prime dividing order G there exists element x in G of order p.

[viscid pewter]

not every group is cyclic, so it's clear

[sturdy mirage]

what does this expression mean? $\mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$ is not a splitting field over $\mathbb{Q}$. the way i understand a splitting field, is that you need to relatae it to an actual polynomial, or a family of polynomials.

[cloud walrus]

reking

[sturdy mirage]

so if there is no polynomial mentioned, what then?

[chilly ocean]

Yes

you need a polynomial

So you want to find if there is a polynomial such that this extension is a splitting field for it

[chilly ocean]

Didn't know it was named after Cauchy but thanks

[viscid pewter]

something something sylow also

[sturdy mirage]

So maybe what they meant with this sentence, is that there exists no such polynomial

[final pasture]

Cauchy follows easily from the Sylows anyway

[chilly ocean]

Yes

You can see that minimal polynomial over Q for cbrt2 is x^3 - 2, which has complex roots

So it doesn't split into linear factors over Q(sqrt2, cbrt2)

[daring ibex]

yes very quickly from sylow

consider a prime dividing order of group, this has a p-sylow group. Take any nonunit element

[hidden haven]

(assuming the theorem that splitting fields are normal extensions)

[daring ibex]

The cyclic group that it generates must have order dividing p^n for some n \in N

U can take it from there

[viscid pewter]

two qs, first one is a quick sanity check

[daring ibex]

qq

[viscid pewter]

if a group is a permutation group, the natural action is gonna be faithful, right?

[hidden haven]

Yes

[viscid pewter]

ok

actual question

[daring ibex]

wdym by natural action

[viscid pewter]

ykwim

stuck on part d

i've proved the backwards direction but not the forwards; ie. i need help proving that if a transitive group G is primitive on A, all the stabiliser subgroups of the elements are maximal

so my thinking is we should suppose some subgroup H strictly containing some Ga, strictly contained within G, and seek a contradiction

and my blind guess is that the image of a after the action of H is a block B, and H is like GB, the stabiliser

a non-trivial block, hence contradiction

but i can't prove it

let the image be B, then there's h in H such that h(a) = b in B where b != a...

i get the feeling this should be really simple

if B is a block, then h(B) = B, as h(a) = b and then the definition of a block forces this

combining two different elements of H, h and i with i(a) = c

then as H is closed, hi is in H, so (hi)(a) is in B

hmmmmm

[steady axle]

y is an element of X

[viscid pewter]

brilliant

sorry, i'm kinda busy rn, i may or may not get to your thing later

[steady axle]

ok np

[hidden haven]

Suppose g is in G but not in G_a. Let H be the subgroup generated by G_a along with g. You need to show that H=G. Then H doesn't fix a as g doesn't, so the only block H has is the whole group. You can prove that H has to act transitively on A (if x couldn't map to y under H, then orbit of x under H would be a block). Then you can decompose the action of any element of G on A into the action of an element of H and the action of an element of G_a, and by faithfulness of the action, it would mean that any element of g is a product of elements of H so is in H. The way you decompose is you see how what g maps to a, ie say g(b) = a. By transitivity of the action of H you get existence of some h so that h(b) = a, and then g(h^-1) fixes a.

[carmine fossil]

Here's an alternate proof:Let H contain G_a but not be G for some a. The claim is that The set H.a={h.a|h in H} is a block.
Note that g(H.a)=(gH).a
Let (gH).a inter H.a be nonzero.Then
g(h_1).a=h_2.a for some h_1,h_2 in H. This implies (h_2)^-1 g h_1 is in G_a which is in H,implying g is in H and gH=H

[hidden haven]

Yeah orbits will always be blocks

But how do you know it's a non trivial block?

[carmine fossil]

You mean how do I know H.a is nontrivial?

[hidden haven]

(in fact it won't be, because my proof says that it will be the whole group)

[carmine fossil]

Because H has more Elements than G_a,which means there is a element x in H not in G_a,implying x.a and a will be different elements

[hidden haven]

By non trivial I mean not equal to A

[carmine fossil]

I see

[hidden haven]

If it's equal to A you don't get a contradiction

[carmine fossil]

Ok,My "proof" is part of your proof

[mystic jungle]

R is a commutative ring with unity. Prove that R is a field if and only if R - {0} is a group under multiplication

i feel like ive done this before under the guise of different wording? idk

I showed R is a field iff ideals in R and R and 0

[rustic crown]

Well, only thing you need to notice is the identity of R\{0}

[stone lake]

strange, I'm pretty sure sources use exactly that as a definition too

[rustic crown]

yea, the only difference is the packaging, the actual content is same.

[mystic jungle]

what do you mean by packaging, could you guide me through it

[rustic crown]

Okay, so if R\{0} is a group, then what can you say about the identity of this group?

[mystic jungle]

identity is nonzero?

[rustic crown]

yep, can you tell me exactly which element of R that is?

[mystic jungle]

1

[rustic crown]

yep!

this means that for any element a in R\{0} there is an element b in R\{0} such that ab = 1. This is precisely the condition for it to be a field!

The other direction is also pretty easy. When I said its the same content with different package, I mean the contents are the axioms. Try to write down the field-axioms and compare that with Commutative Ring with Unity and Group axioms of R and R\{0}.

[viscid pewter]

ok, stared at this a bit, got bored, came back, and now i actually get it. ty

i guess i was just on the wrong track entirely

[mystic jungle]

what are the group axioms of R and R{0}

thats R for ring not real numbers right

[rustic crown]

Commutative Ring with Unity and Group axioms of R and R\{0} (respectively).

[rustic crown]

yep

If R is a commutative Ring with unity, then
saying "Every non-zero element is invertible" is same as saying "R\{0} forms a group under multiplication".

So let's say we're given that R\{0} form a group under multiplication. Notice that 1 and 0 can't equal in R, otherwise R\{0} is the empty set and can't be a group!

But since 1 * r = r for every element of R, its also true for elements of R\{0}. This shows that 1 is the multiplicative identity for the group R\{0}.

Now since every element of a group is invertible, this gives you that R forms a field!

[chilly ocean]

Quick question if anyone can confirm: when it is said there are 2 groups of order 6 up to isomorphism namely the cyclic group Z6 and the group S3, what is exactly meant by this group? In my lecture notes, we explore S3 as the symmetric group on 3 elements : namely the symmetries on an equilateral triangle but I suppose we can also regard S3 as the group of permutations on {1,2,3} right?

[delicate bloom]

yeah

rotating a triangle is the same as cycling through the elements of the set

reflection of the triangle is the same as exchanging two elements of the set

[chilly ocean]

Yeah I figured there was a correspondance between the two

[delicate bloom]

I guess explicitly labeling the vertices of the triangle as 1,2,3 makes it more obvious

[chilly ocean]

So S3 is the same as the dihedral group D3

[rustic crown]

yep, that "same-ness" is called isomorphism of groups.

[chilly ocean]

Yeah gotcha

So back to the original question, this is different than Z6 because the latter is non cyclic

[rustic crown]

yep!

if two groups are isomorphic and one is cyclic. Then so must be the other.

[chilly ocean]

Perfect, thanks

[mystic jungle]

so here you are proving the backward direction right?

i just gotta shwo the forward direction

[rustic crown]

right

[mystic jungle]

Suppose R is a field, then by definition every non zero element in R must be invertible and since R has unity we know that for all a in R a * 1 = a

what can i say about a * 1 = a to imply that R{0} is a group under multiuplication

[rustic crown]

just verify all the axioms.
you need to show the operation is closed, so given two non-zero elements, the product is non-zero. 1 is the identity and associativity are directly inherited. lastly, inverse of a non-zero element is non-zero so this will prove the forward direction.

[chilly ocean]

If I have a permutation which has 4 cycles (including those of length 1) in its cycle structure like (1 8 6)(2 10 9 4)(3 7)(5) then its sign is given by (-1)^(10 - 4) = (-1)^6 = 1

Am I correct?

[rustic crown]

yep

sign of each cycle would be (-1)^{length - 1}

and sum of lengths is going to be 10... those -1s will accumulate to give -4

[chilly ocean]

ah right, because its a product of length-1 transpositions each of sign -1

Makes sense, thanks for the clarification ^^

[chilly ocean]

Also, how can I determine the number of permutations with cycle shape [2^2, 1] ?

It's a product of 2 transpositions

I thought it was just 5C4 but the answer is 15..

Nvm figured it out

[mystic jungle]

i ge tthat but like aren't i just showing that R/{0} is a group under multiplication ? instead of using the fact that R is a field to show that its a group under multiplication

[rustic crown]

Um how will you show that inverses exist and product of non zero elements is non zero without using that R is a field.

[mystic jungle]

OH

right right

Suppose R is a field, then by definition every non zero element in R must be invertible and since R has unity we know that for all a in R a * 1 = a. Since multiplicative inverses exist and the product of non zero elements is non zero we can show that R-{0} is a group under multiplication. The operation is closed as already stated, we know that the ring has unity so identity is 1, associativity is inherited from ring R, and we know inverses exist since R is a field. Therefore R-{0} is a group under multiplication.

Now suppose R-{0} is a group with operation multiplication. Since R is a ring we have that 1 * r = r for all r in R, and since R-{0} contains all non zero elements we also have that 1 * r = r for all r in R-{0}. Therefore we have multiplicative identity 1 for R-{0}.
Therefore if we have that 1 * r = r for all r in R-{0} with identity 1 it implies that inverses exist and all non zero elements are invertible which further implies that R is a field.

i basically took ur explanations and compacted it

is that good enough or do i need to use less words

[small bison]

imo a little too wordy

for field => R-{0} is a group, you just need to justify why multiplication is closed and why inverses exist

cause the multiplication is already associative with an identity for a ring

[next obsidian]

R\{0} is always a monoid, aka its associative and has an identity

To be a group, it’s then equivalent everything has a multiplicative inverse, which is the definition of a field

You can finish it off in like two lines

Without needing to argue both directions separately

[daring ibex]

I'm having a stupidity moment rn . Direct sums, external and internal are only really talked about when talking about abelian groups, right? Otherwise it's just internal and external direct products.

[thorn delta]

yea, direct sums are coproducts in the category of abelian groups, whereas the weak direct product of groups doesn't have any categorical significance that I know of

[daring ibex]

yeah that's what i thought

It's just a book

used \oplus for general groups and it sent me down this rabbit hole thinking i missed some major thing . Though it's just notation for a weak direct product of groups

[thorn delta]

tbf they would coincide for products of finitely many groups

but \oplus does seem like weird notation for nonabelian groups to me lol

[daring ibex]

ye it was a general family

[daring ibex]

yeah i was really doubting myself for a moment there

[thorn delta]

What's an example of $R$-modules $A$ and $B$ such that $A$ is not iso to $B$ but $A \oplus R \cong B \oplus R$?

[cloud walrus]

kxrider

[small bison]

maybe like a countable dimension free module

actually maybe that doesn't work

[golden pasture]

Isnt A\otimes_RR=A

i think you meant like

$$\frac{\mathbb Z}{2\mathbb Z}\otimes_{\mathbb Z}\frac{\mathbb Z}{5\mathbb Z}=0$$
$$\frac{\mathbb Z}{3\mathbb Z}\otimes_{\mathbb Z}\frac{\mathbb Z}{5\mathbb Z}=0$$

[cloud walrus]

ari 十年生死两茫茫，不思量，自难忘。

[small bison]

we want direct sums tho

this is the "easiest" example off the top of my head: https://math.stackexchange.com/questions/1761068/is-it-possible-for-r-oplus-m-and-r-oplus-n-to-be-isomorphic-to-each-other

[golden pasture]

oh frick i cant read

[chilly ocean]

[thorn delta]

Thank y’all

[fair obsidian]

Out of curiosity, I was wondering if the following example also works: Take A to be an infinite dimensional real vector space, say with countably many basis elements $e_n$, and R to be the ring of linear maps from A to A. The action of R on A is to apply the linear map. Then I think B = 0 works, as R should be the isomorphic to a product of countably many copies of A giving
$$B \oplus R \cong R \cong \prod_{n = 1}^\infty A\cong A \oplus \prod_{n = 1}^\infty A \cong A \oplus R$$
The part I'm not sure about is whether my isomorphism between R and a countable product of A's makes sense. I think the map which sends a linear operator $T: A \to A$ to the sequence of vectors $(Te_1, Te_2, \dots)$ gives the isomorphism.

[cloud walrus]

The_Vman

[marsh fractal]

can anyone explain to a noob why finite integral domains are fields

[final pasture]

I don't understand what the question is asking.
What is a "pattern of disjoint cycle structure" ? 🤔

[final pasture]

let A be a finite integral domain and let a € A be nonzero.
Show that x |-> ax is injective, using the fact that A is an integral domain.
Since A is finite, it is then bijective, meaning that 1_A has an antecedent through f, which will be an inverse for a

[marsh fractal]

ok i havent seen antecedents but thanks ill give them a google

[final pasture]

I may have translated it badly

I just mean that there exists some x € A such that f(x) = 1_A

[hidden haven]

1_A has an inverse image under f

[final pasture]

^

[marsh fractal]

ah ok cool i thought it was a math term that makes sense

[final pasture]

while you're here Moldi, can you help ? 👀

[hidden haven]

I have no clue lol

[fair obsidian]

My interpretation would be that things like a 2-cycle, 3-cycle, and a disjoint pair of 2 cycles, etc. are different types of patterns

[final pasture]

oh 🤔

[hidden haven]

yeah I think they are just asking for possible cycle structures

[final pasture]

ok yeah makes sense ig, thanks

[fair obsidian]

No problem

[final pasture]

Yeah I think it should be an iso.

It will be injective because if a linear operator sends a basis all to 0, then it must be the zero map

oh wait 🤔

will it be surjective though

[marsh fractal]

in an integral domain why would the multiplication by a nonzero element be injective? i mean it makes sense but how would i deduce it from axioms

[hidden haven]

yes you can send the bases to anything under linear transformations

[final pasture]

yeah right, that's what I thought first and then I brainlagged

[final pasture]

assume ax = ay for some x, y € A, that means that ax - ay = 0, so a(x-y) = 0, and since A is an integral domain and a is nonzero, x-y must be zero, so x = y

[fair obsidian]

Awesome, thanks!

[marsh fractal]

ty!

[hidden haven]

You can also do the finite integral domain is a field thing from pigeonhole principle which is nice

[final pasture]

(injectivity => bijective is kinda using pigeonhole principle, no ?)

[hidden haven]

For any non zero a, take the set of all its powers, which will have to have repititions, deduce that some power has to be identity

[hidden haven]

oh right it is too

[final pasture]

oh yeah, it's the same proof than the one to show that every element has finite order in a finite group, nice

[hidden haven]

Yeah

[final pasture]

For this, the answer would be: 2-cycle, 3-cycle, 4-cycle, identity, prod of 2 disjoint cycles, and their respective signatures would be -1, 1, -1, 1 and 1 ?

[hidden haven]

Yes

[final pasture]

thanks

[vestal snow]

Is the main idea here that the terms we take the colimit over on the right are exactly the same as the terms we take colimit over on the left (with possible repetitions on the right, but that does not change the colimit)?

Also I am using that the stalks of the presheaf are the same as the stalks of the sheafification

[vestal snow]

The actual verification seems like a straightforward, but complicated, diagram chase so if anyone can confirm there is nothing else to this problem I'll skip it

[latent anvil]

I think there's an extra step is replacing the stalks of π^-1 G with the stalks of the presheaf you sheafify to get the inverse image sheaf

Because those are sectionwise colimits

and so you're then taking nested colimits and it will be the same result by the thing you're sketching out

[vestal snow]

Isn't this what I said here?

[vestal snow]

Here

[latent anvil]

Ah sorry I missed that, my bad

Then yes I agree with your proof sketch

[vestal snow]

Thanks

[chilly ocean]

@plain sequoia sorry pinging and dunnno if you are still here but was the nagpaul book any good ?

[plain sequoia]

I am surprised that you digged up such an old message but yeah I liked that book

[chilly ocean]

Oh thanks, a friend recommended it to me and i was skeptical since no one apparently has heard of it

thank you!

When talking about permutation groups on n elements and their cycle notations, is it true that there are as many cycle shapes as there are partitions of n?

A5 is the alternating group of degree 5?

Afaik each 3 cycle is a commutator

[gritty sparrow]

Yes this is what they want you to do

Yeah seems a bit trivial, but i can’t think of what else it could be

[gritty sparrow]

Since we don’t write the 1 cycles usually, I suppose this isn’t the case

[chilly ocean]

Right because I have this quite confusing result from notes which says that every cycle notation defines a unique partition of n called the cycle shape or cycle structure of n

[gritty sparrow]

Well, if we do count the one cycles, it will be true

[chilly ocean]

Actually, nvm a cycle shape (or a partition) can define several permutations

So for instance in S5 [2, 1*3] is given by 5C2= 10

If I’m not mistaken

Those are different permutations but with the same cycle shape in S5

[gritty sparrow]

Yes ofc. But the note was about cycle shapes and partitions right?

[chilly ocean]

Yep exactly

So as many cycle shapes as partitions if we count the 1 cycles

[hidden haven]

By one cycles do you mean the shape of the identity permutation?

Because we do count that as a cycle shape

[gritty sparrow]

Is it a shape if you can’t see it?

[hidden haven]

[gritty sparrow]

[latent anvil]

Yes

[final pasture]

I love this emote

[hidden haven]

same

complaining to your mom about what you just did vibe

[final pasture]

y e a h

"Moom, sis' ate my part of the chocolate cake "

[gritty sparrow]

[latent anvil]

Let $A$ be a commutative noetherian local ring

[cloud walrus]

shamrock

[latent anvil]

What can I say if all the associated primes of $A$ are height 0

[cloud walrus]

shamrock

[latent anvil]

Thinking

so like, I know that these are always associated primes of A

Since they're the minimal elements of its support

wait isn't this always true lol

Oh no sorry it's true in a reduced ring I think

Because zero divisors = union of minimal primes

So "not much"

Oh well hmm in my setup I also know A_p is a field for height 0 p

That's also neat

[hidden haven]

Yeah you just have to calculate xyx'y' where ' is inverse

[final pasture]

how lazy of you

[hidden haven]

So yeah just write that product in cycle decomposition notation @flint crater

[final pasture]

you could have typed ^{-1}

[hidden haven]

Heck no

[final pasture]

[hidden haven]

Or maybe it was x'y'xy?

I don't remember the definition of the commutator but yeah whatever your book says

[final pasture]

the usual def is the first you gave

[hidden haven]

[final pasture]

(but it's the same thing anyway)

(just take inverses of x and y and you get the other def)

[hidden haven]

I'm guessing you'll get some 3 cycle, and by varying the 123 and 145 you'll be able to get any 3 cycle, so you will generate A_n itself

[hidden haven]

Ye ik but I hope there's a standard definition to avoid ambiguity like in all other math

[final pasture]

nah, there's always some authors to take different definitions lol

[hidden haven]

Yeah, so the rest should follow

[final pasture]

see wiki lol

[hidden haven]

Lmao

(but yeah I was making a joke on how there's no standard and a bunch of ambiguity in all other math too)

Limit points

[final pasture]

there's also the 2 def of the charac polynomial

[hidden haven]

[final pasture]

sorry I can't write

[hidden haven]

what

[final pasture]

some authors define it as det(XIn - M) and some define it as det(M-Xin)

[hidden haven]

Lmao

Right

[final pasture]

one def make the polynomial unitary

[hidden haven]

Mfw det is undefined

[final pasture]

and the other def gives that the evaluation at 0 is always equal to det(M)

(while in the other def, it is the case only in even/odd dimensions (don't remember which one lol))

[hidden haven]

In even

Ig

[final pasture]

yeah right that's even iirc

[final pasture]

Has @rustic crown ever been defined ?

[hidden haven]

I would hope so

[final pasture]

[rustic crown]

?

[hidden haven]

Just wanted to ask how discrete math tuts are going

[rustic crown]

they're just going 😛

[hidden haven]

oof

maybe people here might know some cool things to do in tuts

[rustic crown]

[chilly ocean]

what tuts

King Tut

[final pasture]

tuts like in tutoring ?

[chilly ocean]

I do

[hidden haven]

det is teaching assistant for the discrete math course

[final pasture]

oh nice

[chilly ocean]

shika

[rustic crown]

i have taken about 3 tuts by now... in the start when kv was teaching induction, told them a few ways to solve recurrences. when kv was doing pigeon hole, told them about Minkowski's Theorem, then did some simple stuff about generating functions. When kv was doing bijective and other combinatorial proofs, told them about Euler's Pentagonal number theorem.

the only problem is that they don't speak much during the tuts, so idk what they're feeling

[hidden haven]

online tuts are just sad

[rustic crown]

Hanny offered me to TA group theory in Aug-Nov

[final pasture]

yes ? That's like the third time you type that without context in the last few days I think lol

[hidden haven]

are you gonna do it

[rustic crown]

yep

[hidden haven]

yay

[rustic crown]

its some nptel course

[hidden haven]

wait grp thy in Aug-Nov?

oohh

[final pasture]

nptel = ?

[hidden haven]

i heard those pay many money

[rustic crown]

(yes >.<)

[hidden haven]

indian govt online undergrad level courses

[final pasture]

oh

nice

[hidden haven]

I wanna TA one just for the money tbh

[final pasture]

wait are they the courses you were looking at the other time @paper flint ?

[rustic crown]

lol

[hidden haven]

our college dont pay shit for TAing

[rustic crown]

but its fun to TA lol

[paper flint]

Yes

Ah yes

Moldi is from CMI

[carmine fossil]

You mean they don't pay you at all?

[paper flint]

Would you be TAing Hanumanthu's course?

[carmine fossil]

Or the pay is pathetic?

[final pasture]

CMI = ?

[paper flint]

Chennai Mathematical Institute

[rustic crown]

oh you know Hanny?

[paper flint]

I saw his lectures before, I think he's great!

[hidden haven]

yes very fun and also very good for learning

[rustic crown]

(i didn't like his lectures 😶 )

[paper flint]

Wait, det at CMI too?

[hidden haven]

they pay 6k/sem which is

[rustic crown]

6k is less than 100 usd

[hidden haven]

also I liked hanny's lectures

[paper flint]

My teachers set a bar so low it is easy to exceed it

[hidden haven]

they depend on the course hes teaching

but i liked his topology course

not so much his linear algebra course

[rustic crown]

it was practically Munkre 😶

[hidden haven]

I dont read the recommended textbooks so i dont care

[final pasture]

what's a decent salary for one person in India ?

[final pasture]

[carmine fossil]

50000 a month?

[hidden haven]

yeah that seems legit, middle class kinda salary for 1 person lol

[rustic crown]

same lol

i didn't read books before college >.<

[carmine fossil]

tfw D&F is not in the recommendations

[final pasture]

imagine reading

[hidden haven]

or ig you could say 20k/month is like a living wage?

[final pasture]

nerds

[hidden haven]

and we get 6k/4 months

[carmine fossil]

I don't think that is a livable wage in a city

[rustic crown]

better than 0k/4 months 😛

[hidden haven]

ig

tho my brother lives in a PG (he works in a different city) for less than 10k/month

[rustic crown]

i have to grade 2 assignments and a quiz

for 7 students >.<

[final pasture]

det being like "hey, every positive real number is bounded below by 0 !!"

[tough raven]

Wait @hidden haven you've TAed a course at CMI?

[paper flint]

Was it ever offered online?

[hidden haven]

lol 7 students is such a nice number i wanna go back to alg3 when i used to grade 8

[hidden haven]

yes

[hidden haven]

no it wasnt nptel, was a cmi course

[rustic crown]

i mean the work is pretty fun... so i would be doing something similar with master|Y| but without money 😛

[carmine fossil]

grading looks fun

[final pasture]

too many indian people I feel excluded 😔

[rustic crown]

aww

🥺

[hidden haven]

grading is annoying tbh (tho for small enough groups its fun)

[paper flint]

NPTEL never offers a fucking course on general topology

[tough raven]

Hmm
This should move to #chill or #math-general IG

[hidden haven]

tutorials are the best part, but having to take them online is very sad because as det said, you cant guage response at all

[hidden haven]

right lol

[carmine fossil]

Imagine having tutorials

[final pasture]

(for general top I think Lee's topological manifold is nice, from the small part I've read of it, it seems decent and more readable than Munkres)

[paper flint]

I'd probably learn it from Knapp when the time comes

[final pasture]

works too

[hidden haven]

actually you can probably look at cmi course playlists (they are unlisted but wink wink)

[paper flint]

Oh I generally don't play them on Swayam anyway

I'll look it up, thanks!

[hidden haven]

how will you look them up when they are unlisted

[paper flint]

CMI website might have links somewhere I suppose

Like course webpages or sth

[rustic crown]

when i saw cmi website, i wondered if it was a real institute

[hidden haven]

nah its not so organised lol profs upload them on their accounts and mail the link to the relevant batch

this is india bro

[paper flint]

CMI is still good

God

The ISI website

Jamia's website

They look like they haven't been updated since 1995

[rustic crown]

lol

[hidden haven]

all good developers were stolen by google and went to the US

[paper flint]

[rustic crown]

(btw why did we shift from general channel again ?)

[paper flint]

[hidden haven]

idk

[final pasture]

#groups-rings-fields is an attractive fixed point

[carmine fossil]

Is this the part where we switch servers

[hidden haven]

we can move to #point-set-topology if this is too much spam

[stone lake]

I followed Hanumanthu's NPTEL courses on abstract alg

Agreed, he's good

[rustic crown]

I attended his module and galois theory after attending a same course from Upendra and reading Aluffi... so felt kinda worse.

[hidden haven]

> upendra
> kinda worse
:pepega:

[stone lake]

You two are CMI students?

[hidden haven]

Yes

[hidden haven]

Why is pepega not ?

[stone lake]

Which program? BSc?

[rustic crown]

okay... very bad 😄

[hidden haven]

Yeah

[hidden haven]

Lol I meant there'll be very few profs who won't get you the exact same comparison

Upendra best

[rustic crown]

[final pasture]

imagine reading a cat theory meme seriously

[hidden haven]

?

[final pasture]

this

[hidden haven]

meme?

[rustic crown]

Aluffi is a meme lol?

[final pasture]

I was trying to make a joke, but ig it wasn't funny lol

[rustic crown]

this was funny lol

[zinc cloud]

i feel like this the best place for it, only a quick one but are we allowed to say that the identity of a ring is a unit of the ring?

ive got it written in my notes that " 1 is a unique hile there may be other units of a in U(R) units should not be confused with the indentity 1"

so idk

might be a dumb question ik but i just wanna clarify my definition

[delicate bloom]

the identity is a unit, but not all units are the identity

like -1 in Z is a unit

it's clearly not the identity, they just want you to not be confused and think the word 'unit' means 'identity' I think

[zinc cloud]

ok gucci i get it, i think the wording of his definition was slightly confusing

thank you!

[steady axle]

For $(a_1 \dots a_k) \in S_n$, the order of centralizer of $(a_1\dots a_k)$ is $k(n-k)!$ right?

[cloud walrus]

bert

[carmine fossil]

Should be

[steady axle]

how do find the size of its normalizer?

[hidden haven]

Use the fact that $\sigma (a_1,\cdots,a_k)\sigma^{-1} = (\sigma(a_1),\cdots,\sigma(a_k))$. I think you should get that the normalizer is exactly the centralizer in this case.

[cloud walrus]

Moldilocks

[steady axle]

That's what I thought because $(\sigma(a_1),\cdots,\sigma(a_k))$ should again be a power of $(a_1,\cdots,a_k)$ so $\sigma $ is some power of $(a_1,\cdots,a_k)$ and the claim follows.

[cloud walrus]

bert

[hidden haven]

Yeah

[steady axle]

do you know how to do this @hidden haven ?

[hidden haven]

I saw it yesterday, couldn't figure out anything

[chilly ocean]

I have a question about conjugates in general: in a given group G if two elements are conjugates say h = kgk^-1 , we can construct an isomorphism from G to G given by f(x) = kxk^-1 that particularly takes the element g to its conjugate h. How can we say, in precise terms, that conjugates are "basically the same elements" in a group? I get how isomorphisms characterise the "same-ness" of algebraic structures but can we also say the same for specific elements within a group?

[hidden haven]

That's exactly what sameness is, there's an isomorphism of G to itself mapping g to h

[viscid pewter]

isomorphisms take elements of the same order to elements of the same order, for a start

if they generate before then they'll generate after

the map just preserves everything

[hidden haven]

If you think of an isomorphism as a renaming of elements then g and h can really be viewed as the same element

Where you just relabel everything in G without changing the operation

And g is relabeled to h

[viscid pewter]

it just means that the two elements act exactly the same way, there's a sorta arbitrariness to how you're labelling them

if you want actual terminology for it, saying they're conjugate might be the terminoology

if ykwim

[chilly ocean]

Yeah in my head I just picture that if we get the same group table for both groups whenever they're isomorphic, just different labels for elements

[tough raven]

Outer automorphisms exist tho

[viscid pewter]

oh f

i'm a fool

[chilly ocean]

shouldn't it be inner automorphism for the ones given by conjugation?

[tough raven]

What I meant was that a "terminology" for this should also apply to elements related by an outer automorphism

[chilly ocean]

What does it mean for two elements to act the same way? When you apply the isomorphism, the whole group can get changed, so they act the same but towards i think possibly different elements

@chilly ocean Can I dm you?

[chilly ocean]

Only if you either tell me a subgroup of the free group on 2 generators isomorphic to the free group on 3 generators or prove no such subgroup exists

[cloud walrus]

Ledog

[next obsidian]

Subgroup exists for all free groups on n generators

Oops

That’s what ledog proved using topology

Elegiggle

[tough raven]

Is there any integral domain S such that R (real numbers) is the field of fractions of S?

[hot lake]

S=R ?

[tough raven]

Oh

Is there a ... non-trivial example?

I don't know what that should mean

[hot lake]

with the axiom of choice, probably

stuff like a maximal subring not containing 1/2 ?

[chilly ocean]

How can we directly know the cosets of H in S4? I get that there are 6 of them as the index is |S4 : H | = |S4|/|H| = 4!/4 = 6

[oblique river]

that slide seems pretty self-explanatory

you know what H is

and you can multiply H by different elements

like (12)H and (13)H

can you explain what you are confused about?

[chilly ocean]

So S4 has 4! = 24 elements, why wouldn't we multiply H by say (1324)?

That is what confuses me really

[oblique river]

you could

I don't see why you think you have to multiply by (1324)

or what that has to do with |S4| = 24

[chilly ocean]

So pick any 6 elements of S4 and form the 6 cosets of H right?

[oblique river]

no

because two elements can give the same coset

[chilly ocean]

sorry 6 elements not in H*

[oblique river]

for example, if h is an element of H, then hH = H

still no

if a is an element of bH

then aH = bH

you could look at (1324)H, but that would be the same as (12)H

which I know because the slide tells me that (1324) is an element of the coset (12)H

[chilly ocean]

right, that clarifies my confusion

thanks

[oblique river]

in general, there aren't going to be unique coset representatives

in this example, what it looks like they did was the following:

[chilly ocean]

being in same coset is an equivalence relation

[oblique river]

(12) isn't in H, so (12)H is a coset of H

[chilly ocean]

in general we'd start with simple elements like 2 cycles etc. until forming the 6 distinct cosets

[oblique river]

calculate (12)H and notice that (13) isn't in there

so then write down (13)H

and finally (14)H

[chilly ocean]

Yh totally makes sense

[oblique river]

but now we can't take (23)H because that's already covered

👍

[chilly ocean]

a = bh so h = b^-1a so b^-1a is in H so aH = bH. Is this proof correct?

[oblique river]

essentially, yes

I might include one more step at the end, saying b^(-1)a is in H, so b^(-1)aH = H, and then multiply on the left by b

[glossy pulsar]

Anyone got some experience with Galois Theory that could help me with something?

I'm trying to proce that if we have the following extensions K < L < E, that if K < E is normal then L < E is normal

[oblique river]

what is your definition of "normal"?

[glossy pulsar]

The definiton we're using for normal is the following: for any irreducible f in K[X] which has a root in E, it holds that f splits into linear factors in E[X]

So basically, any irreducible either has no roots in E or all of them are in E

[oblique river]

were you going to say what you had tried so far?

[glossy pulsar]

I've done a simlar exercise for seperable extensions

Using the fact that K[X] < L[X]

But I don't see how information about irreducibles in K give information about irreducibles in L

[oblique river]

if f is an irreducible in K[x], f need not be irreducible in L[x]. But what can you say about f in L[x]?

or maybe I should ask, "what can you do to f in L[x]"

[glossy pulsar]

You could factor it in L, and try to work further with any non-linear terms

[oblique river]

not sure what you mean by the second part

but the first part is correct

if f is irreducible in K[x], it need not be irreducible in L[x], but it can be factored into irreducibles in L[x]

Is every irreducible in L[x] of this form?

that is, if g is an irreducible polynomial in L[x], is it the factor of some polynomial coming from K[x]?

[glossy pulsar]

I would presume yes

Though I don't know how to prove such a thing

[oblique river]

proving that is basically the key to answering this question

[glossy pulsar]

Understood

[oblique river]

basically, to prove that K < L is normal, your proof structure is going to have to look something like this:

1. start with an irreducible g in L[x] which has a root r in E
2. find some irreducible polynomial f in K[x] which has r as a root (because we need to use the fact that K < E is normal, which means we need an irreducible poly over K and some root in E)
3. somehow relate f and g so that we can relate the roots of f to the roots of g

[glossy pulsar]

Thanks for the help

I think I got it, if I'm not mistaken g will be the minimal polynomial of r in L

[oblique river]

that's always a feature of this kind of setup. if g is irreducible over some field F, and r is a root of g, then g is the min poly of r over F

[glossy pulsar]

then you can use the fact that the minimal polynomial in L g_K is also a part of L[X] to show that g | g_K

as g_K linearises, g must as well

[oblique river]

that's right. (this g_K is what I called f.)

[glossy pulsar]

Thanks

I've been going quite mad not being able to find it

It's this small 'do it yourself exercise" that I've been stuck on for way too long

[oblique river]

I know the feeling

[chilly ocean]

$G = { \begin{bmatrix}
a & b \
0& 1\
\end{bmatrix} : a,b \in \mathbb{Z}_{7}, a \neq 0 }$

Given this group, I have determined its conjugacy classes

[oblique river]

pls fix your latex 😭

[daring ibex]

[chilly ocean]

Woops

[cloud walrus]

Laïka

[chilly ocean]

Now fixed

[oblique river]

smh not even using \left\lbrace and \right\rbrace

[chilly ocean]

So I've determined its conjugacy classes

$C_{1} = \left\lbrace \begin{bmatrix}
1 & 0 \
0& 1\
\end{bmatrix} \right\rbrace$

[oblique river]

tiny braces 😭

[chilly ocean]

Idk how to get them bigger 😐

[oblique river]

i told you

[chilly ocean]

ah

[oblique river]

\left\lbrace and \right\rbrace

[chilly ocean]

missed that

[cloud walrus]

Laïka

[oblique river]

😌

[chilly ocean]

$C_{2} = \left\lbrace \begin{bmatrix}
1 & b \
0& 1\
\end{bmatrix}, b \in \mathbb{Z}_{7} \right\rbrace$

[cloud walrus]

Laïka

[chilly ocean]

$C{3} = \left\lbrace \begin{bmatrix}
a& b \
0& 1\
\end{bmatrix}, a \neq 0, b \in \mathbb{Z}{7} \right\rbrace$

[cloud walrus]

Laïka

[chilly ocean]

Now I'm asked to determine a normal subgroup H

I've looked at the solution - its says that a normal subgroup is a union of conjugacy classes... OK. Now the reasoning is as follows, we should take elements from C2 in H, otherwise, the order of H would not divide that of G (which is 42)

[oblique river]

C1 and C2 have overlap

presumably in C2 you mean b \neq 0

[chilly ocean]

Yup

[oblique river]

and also presumably in C3 you mean a \neq 1

[chilly ocean]

Yeah

So that the order of C1 is 1, the order of C2 is 6 and the order of C3 is 5x7=35

That corresponds to the class equation of G so yeah

But now I'm wondering, when finding the normal subgroup H, if we only take the union of C1 and C3 we get that the order of H is 1+35 = 36 and 36 doesn't divide 42

So that's how we justify the argument about taking the union with C2 as well?

[oblique river]

I'm not sure what you are trying to justify

two normal subgroups are C1 and C1 U C2 U C3

namely {e} and the whole group

you are asked to find another one

the only remaining options are C1 U C2, C1 U C3, and C2 U C3

[chilly ocean]

Yeah basically I was trying to understand why C1UC3 wasn't a normal subgroup (not a subgroup at all actually)

[oblique river]

two of those can't be subgroups because their size doesn't divide 42. This means that C1 U C2 could be a subgroup, but you actually have to prove it

I don't think you have to say anything other than "36 does not divide 42"

[chilly ocean]

Cool, thanks ^^

[daring ibex]

Is this really "the" dual of A? There are many exponents the abelian group can have, no?

And i don't think like Hom(A,Z_m) and Hom(A,Z_{km}) will be isomorphic, so what gives

[oblique river]

if A has exponent m then those will be isomorphic

that's kind of the point haha

[daring ibex]

yeah, i mean if they weren't then this wouldn't be well defined. But how are they isomorphic, it doesn't feel right hm

[oblique river]

for an abelian group A and positive integer m, let A_m denote the subgroup of A of elements with exponent m, i.e., A_m = {a in A | ma = 0}.

Lemma: Suppose that A and B are abelian and A has exponent m. Then the image of any homomorphism A --> B is contained in B_m.

[oblique river]

i think it feels very right :P

also it would still be well-defined even if they weren't isomorphic

[daring ibex]

im probably just thinking about this wrong

[daring ibex]

i mean it wouldn't be well defined to say "the" dual group, no?

[oblique river]

the definition doesn't say A dual = Hom(A, Z_{km}) for any k, it says A dual = Hom(A, Z_{m})

what do you mean? this is defining what "dual" means

the definition clearly doesnt depend on any choices

[daring ibex]

I think we're talking about diff things when we both say well defined

[oblique river]

I think "well-defined" is a well-defined notion

[daring ibex]

What I mean is that it's not right to say "the" dual group when there are many exponents for a group, and each can come up with a dual group, if they are not isomorphic

[oblique river]

oooooooooooooooohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

I see where you're coming from now

yes, I was thinking of "dual as an exponent-m group" as sort of one thing but you're right that that's not really a good notion

you are correct that this is talking about "dual" in a way which shouldn't depend on m, but the definition does appear to depend on m

you are right, I was misunderstanding you

[daring ibex]

ah i see i see

[oblique river]

prove my lemma and then think about why that makes this well-defined

[daring ibex]

but in this case they are isomorphic, right

so it does make sense to say the dual group

[daring ibex]

yes ill do that, thank you

[oblique river]

what are "they"? Hom(A, Z/mZ) and Hom(A, Z/(km)Z)?

[daring ibex]

yes

[oblique river]

yes they are isomorphic and that's what my lemma proves

at least, that's what my lemma proves after you think about it

[daring ibex]

loll ill give it some thought, thanks buncho

[chilly ocean]

Does it have to be an exhaustive union of conjugacy classes? ie. do we have to take all elements of say C1 and C2, or can it be some elements of C1 with some elements of C2?

[oblique river]

what other kind of union is there?

the union of A and B means everything in A and everything in B, you don't get to pick and choose

maybe you should review the proof of why a normal subgroup must be a union of conjugacy classes

[chilly ocean]

Makes sense, I just came across another answer where they took a normal subgroup with b anything and a either 1 or 6 in the above definition of G

[oblique river]

so then your conjugacy classes were wrong

[chilly ocean]

I get that it the normal subgroup has to be closed under conjugacy, every element in this subgroup must contain all its conjugates

[oblique river]

that's right

[chilly ocean]

I'll have to review but thanks anyway ^^

[hot lake]

the correct definition of dual is Hom(G,Q/Z). But if G has exponent m then any morphism into Q/Z has image into the elements of order dividing m, so that's the same as Z/mZ

[daring ibex]

i don't see how that is true oof

[oblique river]

Not sure what you mean by “the correct definiton”. Both yours and the definition mirza’s source gave are correct, and they are equivalent.

[oblique river]

You can prove these definitions are equivalent again by using my lemma. In this case, you just need to understand the structure of Q/Z

[daring ibex]

Hm i did actually prove ur lemma but i'm not sure how it shoes the homsets are isomorphic

[oblique river]

Maybe you should also think about that

Maybe you should think about what (Z/kmZ)_m is

(In my notation)

[daring ibex]

oh at a glance i think that'd be iso to Z/kZ

mebe not actually

[oblique river]

How could it be? It’s the elements of Z/kmZ which have exponent m

But something in Z/kZ has exponent k

[daring ibex]

yeah i was being funky for a moment there

[latent anvil]

Algebroni

[daring ibex]

Oh it'd be Z/mZ then i believe

[oblique river]

Indeed

[sinful mirage]

how is this being implied?

what does restricting to V mean?

I thought it means by letting lambda=0,but that's not allwoed

[next obsidian]

In general restriction a function means you have a function like
f:X -> Y

If you had a subset S < X

You could look at the function
g: S -> Y given by
g(x) = f(x)

All you’re doing is changing the domain of the function f to something smaller

[sinful mirage]

yes but i can't see how to apply this idea here

why would the domain be a priori not V in the above proof?

[next obsidian]

Isnt it because

The function is being defined on this semi direct product?

And we have a copy of V inside there as like just the first coordinates?

[sinful mirage]

right,but if we want to restrict it to v,we need to set lambda=0

cause V=semidirect prod,when lambda=0

but lambda can't be zero

or I can't see how the restriction implies what is being written

[next obsidian]

Well I’m not 100% sure what’s going on but the domain of sigma are tuples like

(x,y) yeah?

[sinful mirage]

yes

[next obsidian]

Isnt V just the set of (x,0)?

[sinful mirage]

yes,this is what I think so too

[next obsidian]

I don’t see how the second coordinate being 0 means lambda is 0

[sinful mirage]

but then how does the next step make sense?

because the second coordinate is lambda

[next obsidian]

No, that’s the image of (0,1)

[sinful mirage]

what is (0,1) here?

[next obsidian]

Isnt that just an element in the domain?

Okay I’m gonna try to read through the entire thinf

[sinful mirage]

ok like you said it makes sense that I could restrict sigma to only V by setting the second coordinate to zero

so sigma_V should eat things like (v,0)?

where v is any v in V

[next obsidian]

Right

[sinful mirage]

but i can't see how they get the conclusion on D

[next obsidian]

Also, sorry but I had class!

[next obsidian]

I’m like 10 mins late oops

[sinful mirage]

okay,no prob

[old zenith]

so i was reading the infinite napkin, and the following question popped up: why do we call the number of elements in a group the order of the group? why not call it the cardinality of the group, analogous to how we would call sets?

[scarlet estuary]

you can call it cardinality as well, but its less common

order of a group is intimately connected to the order of its elements

(lagranges theorem)

[old zenith]

i mean

if you're going to have the order of a group and the order of an element i think the terminology is going to get quite overloaded

[scarlet estuary]

theyre basically the same concept

the order of an element is the order of the subgroup generated by that element

[old zenith]

ok that explanation makes more sense then

[scarlet estuary]

that does raise the question of why we dont call both cardinality

i guess thats just historical

[hidden haven]

Also ig it's the distinction between the group and it's underlying set

[final pasture]

for the group itself, it's not that important ig, but imagine calling order of the elements "cardinality". Depends on the setting you're working in, but usually elements themselves are going to be set, and that wouldn't always be clear about which cardinality you speak (the cardinality of the element as an element of the group or as a set)

[hidden haven]

Why do you need to change elements to cardinality? You can just call it order of elements and cardinality of group, and order of an element = cardinality of subgroup generated by it would be a theorem it's more just historical probably as nami said

[final pasture]

Why do you need to change elements to cardinality?
because this:
theyre basically the same concept

I like the fact that they're called the same

[hidden haven]

Hmm I don't think you need to have the same name for corresponding things but ye subjective ig

(take cardinality of underlying set of group and order of group for example, they're basically the same concept too)

[final pasture]

fair

[next obsidian]

I’m a

[final pasture]

[uncut girder]

Conjugation be like uwu^-1

[chilly ocean]

Conjugation by a self inverse is exactly uwu

[rustic crown]

notice by rational root theorem that the polynomial has no rational roots. (you just have to try t = 1 and t = -1). This means the given cubic is irreducible!

Now K contains r, r^2 -2 and by vieta it also contains the third root. Which shows K/Q is splitting field of that polynomial! hence normal.

Gal(K/Q) has size 3. Which means its isomorphic to Z/3Z. You would know that you can send any root to any other root to get such an automorphism... so indeed the map sending r to r1 is an automorphism!

btw there is a very clever way to do the first part!

plug t = x + 1/x, then you get
x^3 + 1/x^3 + 3(x + 1/2) + x^2 + 1/x^2 + 2 - 2(x+1/x) - 1
= x^3 + x^2 + x + 1 + 1/x + 1/x^2 + 1/x^3

if r = s + 1/s then s is 7th root of unity and so is s^2 which means r^2 - 2 = s^2 + 1/s^2 is also a root of that polynomial!

the good part about the above solution is that it lets you construct more similar polynomials!

sum of roots = -1

sum of roots of x^n - ax^n-1 +- .... is a

this is easily seen by multiplying (x-r_1)(x-r_2)...(x-r_n). vieta is just these relations between weird sum/product of roots and coefficients of the polynomial

[final pasture]

Is there a proof of this that doesn't require checking by hand that some map is bijective ? Like using the "real" first iso theorem (the fact that V/U is canonically iso to W through L )

I'd say that the canonical projection of V in V/U is a bijection between subspaces of V containing U and subspaces of V/U, and then compose to get L: V/U -> W should work

but that still requires to check the first claim

ig we can't avoid to atleast check the claim I'm making about the canonical projection 🤔

[gritty sparrow]

Isn’t this a fact that would come after you prove a theorem like the first isomorphism theorem?

[final pasture]

which 1st iso theorem lol ? Because Knapp names the theorem on my screenshot 1st iso theorem, but the 1st iso theorem i'm familiar with is more like the factorization result

this

[gritty sparrow]

Ah wellI assumed that in the book you put in the screenshot, quotients hadn’t been discussed yet. That is an interesting question though, I can’t think of any way that avoids some sort of actual bijection checking

[final pasture]

I don't care about bazooking the question with stuff not covered yet, if that avoids details checks, I'm just asking because I'm curious and I feel like that is the kind of result that could be proved by purely relying on already proved results and avoiding details checks

[gritty sparrow]

Putting a lot of effort to be lazy, huh

[final pasture]

exactly

and anyway I already did the checks lol

[rustic crown]

say f : (a) --> (b) which sends S containing U to L(S)
and g : (b) --> (a) which sends T to L^-1(T). (indeed, the inverse image contains the kernel U).

since L is onto, we get for free that f(g(T)) = T. Now given S containing U, g(f(S)) = L^-1(L(S)) = S + U = S.

so both fg and gf are identity.

This is the usual proof. so you want to prove its a bijection but not want to do the work?

[upper pivot]

i mean think about the subspaces of V/U

and those of W

[gritty sparrow]

Ideally you would have an isomorphism in Set, rather than a peasant bijection

[rustic crown]

they same

[gritty sparrow]

You fool, how can they be the same when they feel so different

[rustic crown]

feels the same to me

[final pasture]

definitely not the same, one is way more abstract nonsense than the other

[rustic crown]

what exactly do you want to avoid doing in this proof?

[gritty sparrow]

Yeah, i could never understand the motivation behind ZFC

[final pasture]

s a m e

[gritty sparrow]

[rustic crown]

who are you calling 😶 ?

[final pasture]

since L is onto, we get for free that f(g(T)) = T. Now given S containing U, g(f(S)) = L^-1(L(S)) = S + U = S.
this line lol

Like it's definitely not hard

(even though Knapp writes half of page to prove that lol)

[rustic crown]

F

[final pasture]

But I wonder if there's a slicker proof than just doing the checks

[final pasture]

in particular, if we can use the fact that we already did the checks for this ^ map to avoid redoing them for the new map

Now given S containing U, g(f(S)) = L^-1(L(S)) = S + U = S.
Also this deserves more details, doesn't it ?

like it's clear that S C L^-1(L(S))

but it's not clear that there's an equality, just by writing what you did, except if I'm missing something

[rustic crown]

if x in L^-1(L(S)) then L(x) = L(s) for some s in S which means L(x-s) = 0 and x - s in U so x in S + U

[rustic crown]

S + U contained in L^-1(L(S)) is equally clear

[final pasture]

yeah

[gritty sparrow]

Well, if you are gonna write this line you may as well just say x is in S and be done with it (not that it matters ofc)

[final pasture]

but I think that deserved to be said, it's not that obvious

I mean it is but eh ig you get what I mean

[rustic crown]

i have seen this too many times to think about it

[final pasture]

Anyway det any idea on how to avoid these details of the usual proof ?

idk why I'm spending so much time on that lol

[rustic crown]

g(f(S)) = L^-1(L(S)) = S + U
this line is kinda useful on its own so i would want it to be there in the proof 😶

[final pasture]

[rustic crown]

mostly you would also want to know what the bijection is instead of just knoing that a bijection exists.

[gritty sparrow]

This does seem really optimistic, this is quite a fundamental/basic theorem after all you need to get your hands dirty somewhere

[final pasture]

I see the 1st iso theorem (the one on the factorization) as the actual fundamental part here and I was expecting to be able to deduce this result from it, but yeah ig I can't without atleast showing that the canonical projection is a bijection between subspaces of V containing U and subspaces of V/U

[rustic crown]

you have to somewhere use the fact that the map is onto and that S contains U and all that... so idt if you can make it any smaller

i wouldn't expect to use first iso theorem because here we only get a set map and not a linear map.

(ig you can say poset maps (inclusion preserving)... but yea no other nice structure)

[final pasture]

Yeah ok, fair ig, thanks for the answers

[rustic crown]

[gritty sparrow]

[rustic crown]

[gritty sparrow]

[rustic crown]

[final pasture]

[rustic crown]

[carmine fossil]

[gritty sparrow]

[final pasture]

[rustic crown]

[hidden haven]

[gritty sparrow]

[final pasture]

[rustic crown]

[gritty sparrow]

[chilly ocean]

Anyone has already seen this notation : $[x]_{n}$?

[cloud walrus]

Laïka

[chilly ocean]

I'm wondering if that means "reduced x modulo n"

[gritty sparrow]

I actually saw this notation on this server somewhere, but I haven’t seen it anywhere else. I have also seen it be used for the nth component of a vector x , and for the nth term of a formal series as well.

[rustic crown]

yea feels like a local notation when the group/ring Z/nZ isn't defined yet.

[chilly ocean]

That's where I was coming to

Z/nZ is isomorphic to Z_n but not equal. I see a lot of textbooks, maths forums basically saying that they're equal

Z/nZ is the set of cosets of nZ in Z so they're of the form k +nZ. And Z_n is just the n reduced integers modulo n, so 0....n-1. It amounts to the same thing, but the structure of sets are different I suppose?

[gritty sparrow]

Well wether they are equal pr not really depends on a very trivial choice for your definitions. In some books they quite literally are equal, not that it matters ofc

[rustic crown]

[x]_n is the notation of the equivalence class containing x under the equivalence relation a ~ b iff n | (a-b). notice as a set this is precisely x + nZ

[chilly ocean]

n|(a-b) so a and b have the same remainder so they're in the same equivalence class

I see

Thanks

[rustic crown]

[chilly ocean]

Anyway in an abstract class like this, we no longer case about how elements are presented (whether equivalence classes like [x]_n or as integer remainders modulo n ie 0,1...,n-1)), it is the same-ness of their properties that matter

[rustic crown]

yep

[gritty sparrow]

Exactly

[vestal snow]

Can someone explain the germ analogy here?

[chilly ocean]

Everytime I see your name I read Hahn Banach at first.

[vestal snow]

Wdym?

[sinful mirage]

can anyone help me understand the sigma_v D sigma_v=lambda D part please?

[tough raven]

IDK about using the "real" first iso theorem
But you can very slightly avoid some work IG
By considering image and preimage mappings
For any f : X -> Y
img(preimg(T)) = T intersect range(f)
And preimg(img(S)) = "closure" of S under the equivalence relation x ~ y ⇔ f(x) = f(y). (In the case of a linear transformation, this is just S + ker(f).)
From this, it follows that both are idempotent i.e.
for T ⊂ range(f), img(preimg(T)) = T and
if S already contains y whenever x in S and x ~ y, then preimg(img(S)) = S.
(In fact, they form a Galois connection between P(X) and P(Y).)

If S is a subspace of X, then the closure condition just becomes ker(f) ⊂ S.

So the fixed-point properties become proofs that the two maps are mutually inverse. (You also need to note that img, preimg preserve subspaces though.)

Of course, depending on how much of this you write out, this could be more work than "checking it by hand" 😅.

[tough raven]

Oh

[final pasture]

Thanks for the answer anyway

[tough raven]

Not sure if the existence of factorisations can help prove that preimg(img(S)) = S for certain S
At least, not without checking something by hand at some point.

[sinful mirage]

actually i'm a bit confused even earlier

[hot lake]

where are you confused ?

[sinful mirage]

so in the first place,why is [sigma(g),sigma(g)]=[g,g]>

then what do we denote by (0,1)?

and why does restricting that yield the equation given

these are my confusions

[hot lake]

so

sigma is an automorphism of g, so [sigma(x),sigma(y)] = sigma([x,y])

[sinful mirage]

right

but why sigma[x,y]=[x,y]

[hot lake]

and it's involutive so sigma² = id

[sinful mirage]

but sigma(x) sigma(y) isn't sigma^2(x,y)

or?

[hot lake]

to show that sigma([g,g]) = [g,g] I guess you proceed by the usual double inclusion

pick x,y in g, sigma[x,y] = [sigma(x),sigma(y)], which is in [g,g] because sigma(x) and sigma(y) are in g

that shows sigma([g,g]) is contained in [g,g]

then pick x,y in g, [x,y] = sigma²([x,y]) = sigma([sigma(x),sigma(y)]), which is in sigma([g,g])

that shows [g,g] is contained in sigma([g,g])

oh wait they actually had a proof

sigma([g,g]) = [sigma(g),sigma(g)] is obvious because sigma is an automorphism I think ?

and sigma(g) = g

because sigma is a bijection from g to g

[sinful mirage]

but why is sigma(g)=g?

it is only for real forms

[hot lake]

I mean g as in the whole lie algebra

not a particular element of it

[sinful mirage]

ah

[hot lake]

g is a seemi direct product

its elements are written as pairs (v,z) for v in V and z in C

the lie bracket on it is defined by [(0,1),(v,0)] = D(v) or something

you would need to check the definition I'm not sure if it's that or - D(v)

and [(v,0),(w,0)] = ([v,w] (bracket in V) , 0) = (0,0) = 0

[sinful mirage]

yes i'm not sure what D(v) is either

[hot lake]

they literally tell you what D is

D(e1) = 2e1 and D(e2) = i e2

it's a good old complex linear map from V to V

[sinful mirage]

right,but how is this related to the bracket at all?

a bracket is [x,y]

[hot lake]

because in the definition of that semidirect product they use D

[sinful mirage]

so [(0,1),(.,0)]=D(.)?

D defines the bracket?

[hot lake]

the definition of the bracket on the semi-direct product uses the action of the right piece on the left piece

you absolutely need to go look at how the semi direct product of two lie algebras is defined

here C acts on V by D

and the lie bracket on the V part is 0

so here we should have [(v1,z1),(v2,z2)] = [z1 D(v2) - z2 D(v1), 0]

lie bracket on g

[sinful mirage]

this is the def you are referring to?

[hot lake]

I think so yeah

here the lie bracket on V is trivial so the set of derivations of V is just the set of complex linear endomorphisms of V

[sinful mirage]

so in this formula,which is the map rho in our case?

[hot lake]

it's the map that is C-linear and maps 1 in C to D in End(V)

[sinful mirage]

ok,right so D is a map in End(V),cause it eats(e_i) and spits out e_j

right

[hot lake]

well if it eats e1 it spits out 2e1 and if it eats e2 it spits out i e2

[sinful mirage]

but i don't see how this follows from the picture with rho

[hot lake]

do you have a picture with the

[sinful mirage]

i'd say ([v_1,v_2],...)

[hot lake]

I suspect your definition has it mirrored

[v1,v2] = 0

[sinful mirage]

why?

[hot lake]

because they say "on the abelian lie algebra V =C²..."

abelian means [v1,v2] = 0 forall v1,v2 in V

[sinful mirage]

oh

right

[sinful mirage]

ok so [(v_1,z_1),(v_2,z)2)]=[0,[z_1,z_2]+ \rho(v_1)(z_2)-\rho(v_2)(z_1)]

[hot lake]

no

[sinful mirage]

and i should figure out how rho acts to get to your def

idk what i'm missing

[hot lake]

your text and the definition you found on the internet disagree on who should be on the left and who should be on the right

I think

[sinful mirage]

this is my text def