groups-rings-fields

[round niche]

ye i want to know if that is right combinations

for that finite field

[blazing hazel]

how many elements has F_2^3?

[round niche]

8

right

[blazing hazel]

I think it is better to count subspaces of F_p^n directly, take any non-zero element (p^n -1) and then you have to remove all (non-zero) F_p mutiples of that

[round niche]

can you show example

what you mean

[blazing hazel]

in F_3^2, you have the space given by (1,0). But it also contains (2,0)

so every vector subspace contains 2 elements (ignore for a second the 0)

Since all subspaces contain the same ammount of elements

just divide the number of non-zero elements in the field by that number

i.e. (9-1) / 2

[round niche]

so in my case its (0,0,0) , (1,0,0) and so on

[blazing hazel]

ignore (0,0,0) for the moment

as it is in all subspaces

[round niche]

ah yes so 7

[blazing hazel]

yup

[round niche]

ok so

is it the sum of combinations of 7

7 choose K

[blazing hazel]

0-dimentional = 1

1-dimentional = 7

3-dimentional = 1

[round niche]

so there is 7 1-dimensional subspace

is that what you are saying

[blazing hazel]

yup

[round niche]

ok i get that

what does the 3-dimensional

how did yo uget 1

[blazing hazel]

the entire space is the only space of dimention 3

[round niche]

oh im retard

so its total of 9 subspaces

[blazing hazel]

you are missing 2-dimentional subspaces yet

[round niche]

so the ones in xy xz and yz

[blazing hazel]

you can convince yourself that there are 7

[round niche]

is it bc

the 7 are not scalar multiples

of each other

[blazing hazel]

it is because to determine a plane you need an equation ax + by + cz = 0, and you count possible a,b,c

[round niche]

oh ye its any plane that could be made

not just yz xz and xy

so this means there are 16 possible subspaces

[round niche]

thank you my friend for the help

[blazing hazel]

np

🙂

[cobalt pilot]

Follow up question from my inquiry yesterday. Say that I have have two ideals of a commutative ring A with identity. Let's call them I and J. Assume that J does not contain I. I know that if J DID contain I, then f(J) would be an ideal in A/I, where f: A to A/I is a ring homomorphism. Now, since J does not contain I, can we say that f(J) is not an ideal in A/I or do we not know enough? If it's not an ideal in A/I, how would one prove that (preferably from the definitions if possible as this is towards the beginning of the course for me)?

[mild laurel]

Be careful of saying that f is just a ring homomorphism

There are usually many ring homomorphisms from A to A/I

The one you care about when you say these types of things is the natural surjection from A to A/I

[cobalt pilot]

Yes, that's what I ment. f(a)=a+I where a in A

[wind steeple]

If J is contained in I, f(J)={0} so this is an ideal

If J is arbitrary ideal, f(J) is an ideal bc f is surjective

[cobalt pilot]

Hmmm. If A = Z and I = 3Z and J=5Z. Then I is not in J but 5Z is not an ideal in Z/3Z, right?

Well, I guess most ideals just becomes the whole of A/I

which is technically an ideal

[mild laurel]

It is an ideal

5 gets mapped to 2, 10 gets mapped to 1, 15 gets mapped to 0

So it's all of Z/3Z

[cobalt pilot]

Thanks. I've been thinking about this a bit wrong.

[solemn hollow]

to prove that $\varphi$ is a homomorphism do I just test all possible pairs? or is there a better way

[cloud walrus]

EpicGuy4227:

[mild laurel]

Do it in terms of the generators

[edgy fog]

All right, so max plus algebra is a real party

Just did some reading on it because a friend is taking a class in it. Anyone have some additional resources on further study?

[mild laurel]

Are you talking about abstract algebra?

[edgy fog]

I would argue that of the threads available, then max plus algebra, or as its sometimes called max tropical semiring, would fit into this subject best.

[cloud walrus]

Honeabear:

[static robin]

i'm trying to learn the Pythagoras's table of a set E and i'm confused on how i can deduce the properties of my Binary operation
i can only tell if it's commutative or not

https://cdn.discordapp.com/attachments/319630866915655692/626087398929072138/unknown.png

for example this is a commutative binary operation.

[mild laurel]

Associativity is pretty hard to check for

There's https://en.wikipedia.org/wiki/Light's_associativity_test

[static robin]

OOHH I didn't know we can check that from the table

what about invertible elements ?

[mild laurel]

I mean we can't really

Or at least

We can by just check a * (b * c) = (a * b) * c for all triples a,b,c

[static robin]

i guess that can be easily done if the elements give as a result the identity

[mild laurel]

But that's super hard

[static robin]

ok

so i can only check for commutativity ?

[mild laurel]

You can check for associativity

Both in the way I just described

And in the link I posted

[static robin]

and other than that

?

[mild laurel]

Commutivity is just much more straight forward to check for

Inverses can be done in the way you describe

But you have to actually confirm that you have an identity element first

[static robin]

yeah ofc

no other sneaky way ?

i just liked how commutative is so easy to check for

gonna check for associativity if i can get my head around it

Thanks !!!

[mild laurel]

Yeah, it's mostly because the table is binary

So checking a binary property like commutivity, or identity element is pretty straightforward

If you had a 3D table that gave you a * (b * c) and (a * b) * c for all elements a,b,c , then associativity would be easy to check for too

[static robin]

yeah i don't think i can do that

talking about the 3D table here

[woven delta]

Associativity is hard

See elliptic curves

You usually prove associativity because of function composition tbh

[static robin]

Thanks !!!! will check

yeeeh that method is hard :((((

i haven't even fully grasped it yet

[static robin]

i have f being this :

and i have a group G

i need to prove that f is a morphism iff G is an abelian group (commutative)

now i've done the morphism part, and i think that f is also an endomorphism

but i'm stuck in the commutative part

can i say that x²=e

?

scratch that, new idea can i say this : $xy=f(f(x)f(y))=((x^{-1}y^{-1})^{-1}=(y^{-1})^{-1}(x^{-1})^{-1}=yx$ ?

[cloud walrus]

life > Random:

[sharp sonnet]

what are you actually trying to show? an endomorphism is by definition a homomorphism f: G -> G, and you already know that your function is a homomorphism?

[static robin]

no i need to show that f is a morphism if and only if G is commutative

[sharp sonnet]

but you said you have done that?

[static robin]

no i said i've done the morphism part

aka i still need to show that if f is a morphism then G is commutative

i'm pretty sure my proof is right

i just need confirmation

[sharp sonnet]

ok

your proof is correct

[static robin]

Eyyy thanks !!

[cobalt pilot]

Who knew Bernie Sanders was that good at Abstract Algebra!?

[static robin]

What you mean ? @cobalt pilot

who's bernie

Might have sent him to gulag

never know

shh

i have this problem : let G be a finite set with * an associative binary operation. we suppose that every element is regular in G show that G is a group
can i use two different function to solve this ?
or do they have to be the same

regular means that $xa=xb \implies a=b$ and $ax=bx \implies a=b$

[cloud walrus]

life > Random:

[analog oracle]

is there anyone here who is proficient with GAP?

[analog oracle]

https://math.stackexchange.com/questions/3368494/generating-and-extending-groups-with-gap

[idle horizon]

@static robin To show it's a group the best bet is to show it satisfies the group axioms. That's it contains an identity, inverse for all elements and the operation is associative (which is given).
Not really sure what you mean by functions tho here.

And ofcourse, closure.

[bitter mauve]

is there a galois theory study group going on rn?

this channel always has some of those stuffs

[static robin]

@idle horizon i can use the following function to prove it :

[fading wagon]

@bitter mauve yeah there is

[static robin]

$f:\bbN \to \bbR$ with $f(n)=x^{n}$

[cloud walrus]

Iosif Stalin:

[mild laurel]

@fading wagon there is?

[fading wagon]

dm

probably the algebraic nt discord server we are in

[mild laurel]

What?

[fading wagon]

"Algebraic Number Theory Reading Group" I guess

[mild laurel]

That's not really galois theory

[fading wagon]

I guess

[thorn delta]

I need to prove that the product of every nonzero element in the field $\bbZ / p \bbZ$ is -1. Its clear that $p-1$ and $1$ are their own inverses, but how do you know that there aren't other elements with order 2 as well?

[cloud walrus]

kxrider:

[mild laurel]

Why is it a problem if there are other elements with order 2?

[fading wagon]

Inverses go in pairs, pair them up

Suppose x^2=1, then (x+1)(x-1)=0

then?

[woven delta]

Oh that's clever actually

[thorn delta]

zopherus, if a and b are two more elements with order two on top of p-1 and 1, then the product of every element would be -ab. The only way that could be -1 is if a and b are inverses of each other.

Either x+1 or x-1 is zero?

[woven delta]

And this argument only works for Z_n where n is prime

I like that

[thorn delta]

hm how so?

[mild laurel]

You can show a cyclic group has at most one element of order 2

[woven delta]

Because Z_n is an integral domain iff it's a field

(well, finite commutative rings are integral domains iff fields)

[mild laurel]

Yeah idk, using the fact that x^2 = 1 implies that (x + 1)(x-1) = 0 is a lot more than you need to use here I think

[woven delta]

That's true, it's just nice

[mild laurel]

You don't need the entire ring structure of Z/pZ, you only really need the multiplicative group structure

[woven delta]

I guess the standard stuff is you pair up 1 and p-1 and 2 and p-2 and so on

p-1 is even so that's good

[mild laurel]

Hm? You pair up every element with its inverse

[woven delta]

Oh, oof

Yeah, you do that

[thorn delta]

its says to solve the problem by pairing up elements with their inverses, but I still have the confusion that i mentioned earlier

[woven delta]

And 1 and p-1 are self inverse

And everything else goes to 1

[mild laurel]

@thorn delta There's really no problem

[unkempt socket]

kx, the hint that Element gave was really good

[woven delta]

And 1 is 1 and p-1 is -1

So you're done

[mild laurel]

Who cares if you have more than one element of order two

Okay this is a lie maybe you do care

[thorn delta]

i know i understand elements proof. im just curious what y'all are thinking about.

[woven delta]

Lmao I shouldn't be thinking about math right after I wake up

[mild laurel]

Yeah, I think you actually have to state that Z/pZ^* is cyclic

[woven delta]

I make dumb mistakes

[mild laurel]

So it can have at most one element of order 2

[woven delta]

I like elements argument

For that

[thorn delta]

it has two elements of order two tho

[woven delta]

You don't

[thorn delta]

err would you say 1 has order 2?

[woven delta]

I think I had to prove a more general version of this fact

No

The order is minimal

[mild laurel]

The usage of order is pretty bad tbh

People use it either way pretty often

But yeah, I mean strict order 2

[thorn delta]

im trying to think why cyclic groups have at most one element of order 2. I feel like I've proven that before but uhhmm can't quite remember.

if a is a generator, and b^m = 1 for some b = a^n, then we have that the order of a divides mn...

nah i don't think that would help

[queen vine]

if a is a generator and b^2=1 with b!=1, we have b=a^k for some 0<k<n, so a^(2k)=1. But then 2k has to be a multiple of n and since 0<k<n we have that 2k=n, so there is exactly one element of order 2 (b=a^(n/2)) if n is even and none if n is odd. In general there are phi(d) elements of order d if d|n.

[mild laurel]

Giving out answers isn't how you help people learn

[tacit pebble]

How could I find the number of roots of a permutation of a given cycle signature. For example, if I have 3 5cycles in S15?

[mild laurel]

What does root of a permutation mean?

Square root?

[tacit pebble]

So for example if a^2=(12345)(678910)(1112131415) what would be examples of a? Or how would I go about finding the number of a’s?

a^2 is a composed with a

I know that if you have 1 15cycle you can find a root by jumping over elements so for example the root of (12345) in S5 is just (15234)

[mild laurel]

Well for your first example

What happens if you find the root of each cycle separately and then combine them?

[tacit pebble]

Hmm, well maybe I should try the simple example of just S5 and then try disjoin in S15, so how many roots in S5 for an arbitrary permutation

Just 1 right?

[mild laurel]

Depends on how long the cycle is

[tacit pebble]

Ok I think I figured it out

Thanks

[latent anvil]

Tfw your first day of graduate algebra doesn't cover category theory 😔

[bleak abyss]

Disgusting

Lol my AG prof last class was like yeah you guys seem to like categories and universal properties a little bit too much

[latent anvil]

Apparently my first AG lecture had the prof talk about how great universal properties are

I had a one time TA obligation and am sad I missed it :(

Algebra was just: groups exist, as do homomorphisms, automorphisms, the cyclic groups, etc...

Nothing interesting yet

Manifolds was pretty cool just because Jack Lee seems fun. Mostly high level stuff about why we would care about manifolds and not embedding them

[latent anvil]

Wanted to share something I thought was neat

In my ring theory class, in the first week of polynomial rings, we had the problem to show K[x, y]/(xy) and K[x] × K[y] are not isomorphic

My solution at the time was something nasty involving looking at the zero divisors and computing some products or powers of polynomials

But now after a summer of AG, I can see a really geometrical proof

Spec is a functor, so it sends isomorphic rings to homeomorphic spaces. Thus it suffices to show that Spec (K[x, y]/(xy)) and Spec (K[x] × K[y]) are distinct

By a problem in Hartshorne I did, the spectrum of a product is disconnected (this part is easy on its own), so it suffices to show that Spec (K[x, y]/(xy)) isn't disconnected

This should be intuitive, since it's the coordinate axes

Suppose we had a disconnection of Spec(K[x, y]/(xy)), say by disjoint nonempty closed sets C = V(I), D = V(J). Then wlog (x) is in C, and this implies (y) is in D, since if (x) and (y) were in C, then for any P in the spectrum, since x*y = 0 is in P, either x or y is, so I <= (x) <= P or I <= (y) <= P. This would imply C was the whole space, so it's not disjoint from D.
Thus (x) is in C and (y) in D. But then I <= (x) <= (x, y) and J <= (y) <= (x, y). Then (x, y) is prime as its maximal, and so C and D share a point, a contradiction

[static robin]

Let G be a set with an associative binary operation with :

Show that G is a group

so i need to show that there is an identity

and prove it's uniqueness

and then show that every element is invertible.

i got this so fat

far*

$a=ax=ya$ and $b=bx'=by'$ so existence is automatic ?

[cloud walrus]

Iosif Stalin:

[static robin]

$ab=abxx'=yy'ab$

[cloud walrus]

Iosif Stalin:

[static robin]

idk how to show the uniqness and the invertible elements :((

[wind steeple]

K[x,y]/(xy) isn't an integral domain while K[x]xK[y] is, isn't it ? @latent anvil

[bleak abyss]

@wind steeple nope, (0,1)*(1,0) = (0,0)

[wind steeple]

Ups

[woven delta]

Wait, you're taking manifolds with Jack Lee? @latent anvil

I forgot he was at UW

I thought you just meant using his book

[latent anvil]

Yup! It's pretty awesome

He's retiring this year too

[bleak abyss]

Oh wow he's a lot older than I thought

[fringe nexus]

To show that for $n \ge 5$, $A_n$ is the only subgroup of N such that $|S_n/G| \le n$, why can we conclude that from the kernel being $A_n$?

[cloud walrus]

Victoria:

[tender mist]

@fringe nexus I suppose because A_n is a maximal subgroup?

[fringe nexus]

oh right

thats it

[tender mist]

@fringe nexus didn't they consider the case k=n anyway?

[fringe nexus]

huh?

I dont think they did

[tender mist]

So it's still missing right?

(I left out a "why" above, sorry lol)

[whole bridge]

not sure if this is more algebra or number theory, but am reading diamond and shurman's first course on modular forms and was wondering if someone else here has been through it

thinking i might just skip chapter 4 or at least very quickly skim it since it seems mostly independent of the later sections and i don't care so much for analytic number theory in general

[bleak abyss]

It's something I'd like to go through in the future

[whole bridge]

it is super duper cool, the first two chapters are easy but chapter three (which i just finished) is much more dense

i would recommend u be really comfortable with complex analysis and have had a course or two in algebra (not necessarily number theory, but it would help too)

[bleak abyss]

Right now I'm doing a reading group on Silverman's Arithmetic of Elliptic Curves

But yeah hopefully the next thing will be Diamond and Shurman

[latent anvil]

I am not smart

[whole bridge]

silverman's books are so hard 🤘😔🤘

[latent anvil]

This was about a flat modules thingy

I got confused because I thought I verified a certain sequence of tensor products was exact

When it isn't in general

I verified it's exact in the middle...which it always is because the tensor product is right exact

[mild laurel]

@whole bridge hey

What do you mean you don't care about analytic number theory

[whole bridge]

exactly that 🤔

i just cant do it lol

[mild laurel]

I believe in u

I'm studying analytic number theory so you have to too

[cobalt pilot]

I sense a new theorem: If Zopherus studies analytic number theory then anyone else can as well.

Not quite sure how to prove it though

[mild laurel]

It's a pretty trivial proof honestly

I'm an idiot so anyone else can do it too

Qed

[whole bridge]

it's not about smarts

definitely u are an enormous idiot

[mild laurel]

yeah you're the biggest idiot i know

[whole bridge]

it's a matter of class

[mild laurel]

yeah i don't really know anything about classes in set theory either

[whole bridge]

[mild laurel]

i should register for jmm

[whole bridge]

wtf dude come on LOL

wtf u doing

[mild laurel]

None of my group has

its not due til october 5th right were okay

[whole bridge]

[mild laurel]

what

[cobalt pilot]

No place in this discord for math memes?

[mild laurel]

#chill

[cobalt pilot]

Thanks

(Asking the important questions)

[cloud walrus]

QuickMaffs:

[tender mist]

Think about how φ_bar is defined, i.e. about what is φ_bar(xN) for each coset.
Then suppose xN=yN, and see why φ_bar is independent from the representant

[latent anvil]

That's the thing you need to prove

Suppose xN = yN and prove phi(x) = phi(y)

[tribal pasture]

Oh okay, I will try that then

Oh I get the point, that is how it is defined. That is φ_bar should sent each fibre to its image that if xN = yN then phi(x) = phi(y)

[latent anvil]

I'm not sure what you mean

Are you able to prove xN = yN implies phi(x) = phi(y)?

[tender mist]

You'll see that the equivalence relation induced by the kernel of an homomorphism corresponds to the equivalence relation induced by the homomorphism by xRy iff f(x)=f(y)

What shamrock is saying basically, which is an if and only if

[tribal pasture]

@latent anvil I think yeah since $xy^{-1} \in N \implies \phi(xy^{-1}) = \phi(1) \implies \phi(x) = \phi(y)$ since $N$ is the kernal

[latent anvil]

Yeah, that's exactly it

[cloud walrus]

QuickMaffs:

[latent anvil]

And that tells you the map is well defined

[tribal pasture]

Oh okay perfect

Thanks!

[latent anvil]

The point Mat is making is good though

Do you understand how normal subgroups and quotients relate to equivalence relations and quotients of sets?

[tribal pasture]

Yes since the quotient is basically the cosets of a normal subgroup

[latent anvil]

Yup. The point is that when that normal subgroup is the kernel of φ, the equivalence relation it induces on G is exactly the same as φ(x) = φ(y)

[tribal pasture]

And cosets are basically the partition of a group which is defined by an equivalence relation

OHHHHH

Thanksss!!!!!!!

And thats why the quotient group is also isomorphic to the image of a homomorphism whose kernel is N

[latent anvil]

You should've seen at some point that every equivalence relation on a set X is of the form x ~ y iff f(x) = f(y) for some f : X -> Y

Right?

[tribal pasture]

Indeed

[latent anvil]

Well when we try and generalize equivalence relations to groups, we run into problems (the equivalence relation might not respect the group structure)

So the right kind of equivalence relations to consider are the ones of that form for f : G -> H a group homomorphism

And what we were just discussing is that this perspective is the same as the one with normal subgroups

[tribal pasture]

So basically relations that are equivalent relations and also respect the composition law?

[latent anvil]

Yeah, pretty much

And if you have such a relation ~ on G, the set { g in G : g ~ 1 } is a subgroup of G

(and it's normal!)

Algebra: Chapter 0 has a good discussion of this imo

[tribal pasture]

So are all normal groups of this form?

[latent anvil]

Yup

Given N, take ~ to be the relation x ~ y iff xy^(-1) in N

Or equivalently (by normality) x^(-1)y in N

[tribal pasture]

So what can I say about K and N, if they are both normal, according to this? K=N? K isomorphic to N?

[tender mist]

The interesting thing is that this sort of explains the definition of a normal subgroup. If you take an equivalence relation and impose that it is compatible with the group structure, you get for the equivalence class [e] the defining properties of a normal subgroup

[tribal pasture]

TRUEE TRUE!

[tender mist]

The same thing works for equiv relations compatible with ring structure and ideals

[tribal pasture]

So in this sense, normal is just a group theory word for equivalence relations on groups

[latent anvil]

Exactly

[tribal pasture]

So what can I say about K and N, if they are both normal, according to this? K=N? K isomorphic to N? What about this though?

[steep hull]

Well try to see why left/right cosets being equal is necessary/sufficient for defining a normal subgroup

[latent anvil]

You can't say anything if you don't assume more about K and N

A group can have multiple normal subgroups

[tribal pasture]

But they both are of the form {g in G: g ~ 1}?

[latent anvil]

For different relations ~

[tribal pasture]

Oh okay perfect!

[latent anvil]

Also, you should work this out on paper

[steep hull]

Is there some sort of bound for the number of normal subgroups

[latent anvil]

The stuff I'm saying

@steep hull you can get info for specific groups by sylow theory but I'm not sure how much you can say without more information

There's at most 2^(|G|-1)

[steep hull]

Wait p-sylow groups are conjugate and hence only normal if there is one

[latent anvil]

You can get better estimates for sure but they'll all be super high

Yes, but by looking at the sylows you can deduce info about the group

Like for example if there is only one sylow, and thus a normal subgroup

Which can be checked by just looking at the prime factors and divisibility (sometimes)

[steep hull]

Yeah. Can anything else be deduced

[tribal pasture]

@latent anvil But if as you say, that if N,K are normal only if we have different relations, then when I say that NK is isomorphic to G only if N and K are normal, what I actually mean is that there are equivalence relations R_1, R_2 which are compatible with group structure, such that under R_1 N is [1] and under R_2 K is [1]?

[steep hull]

Via sylow

@tribal pasture If N is normal, then there exists a group H=G/N s.t. N*G/N=G, meaning that each element in g=hn uniquely

[tribal pasture]

If u dont mind me asking, was my interpretation in terms of equivalence relations correct?

[steep hull]

Yeah essentially the group G/N is what you get by killing all of N (setting it to e). You only get group structure if you’re able to flip cosets around.

However note that the above only works if N and G/N are both normal

To prove the normality implies well-defined quotient group result, show that if hN=nhN then hN=Nh

The other direction is easy

[tribal pasture]

h being the element in a normal group H?

[steep hull]

h is a coset representative for now

Bad notation

Then show that hN=Nh implies that the coset representatives form a group

[whole bridge]

do u understand the idea of a coset?

[tribal pasture]

Yeah!

[whole bridge]

ok so when u construct the set of cosets, say G/H, this does not necessarily have to form a group

in fact, it usually doesnt

[steep hull]

I keep screwing up notation

@tribal pasture it’s fixed

[whole bridge]

however, the quotient is a group if and only if H is a normal subgroup. this is the whole point

[tribal pasture]

Yeah I was thoroughly explained above. I was merely confused about the notation used.

[steep hull]

I’m outlining the proof though

[whole bridge]

ah, i see. sorry i didnt read any of it

[tribal pasture]

Thanks for the explanation!

[steep hull]

@tribal pasture do you understand how to prove it now

[whole bridge]

from dummit and foote

[tribal pasture]

@steep hull Yes thank you!

[steep hull]

@whole bridge whoops I did this on my own

Do you recommend the book given that I understand the main ideas pretty well

[whole bridge]

i think dummit and foote is good if u like a lot of exposition

it can be dry and the linear algebra part sucks, but it talks a lot and goes through lots of calculation

i recommend cross referencing with lang's book if u want to go through it since lang has slicker proofs sometimes

[steep hull]

Thanks for the advice!

I’m skimming through it and I’m pretty comfortable until the cohomology/AG/rep theory parts

[whole bridge]

well, that is what the book is for

[bleak abyss]

Ugh Dummit and Foote

[uncut girder]

Dammit my foot

[tribal pasture]

Can somebody tell me how they concluded the thus part? I understand the implications before that (as in how a=d, c=f) but not sure how were they able to infer that b=0 in h

The form of K is as given above thats matrices with 1 on diagonals, and an arbitrary number on the right upper corner and all others 0

[latent anvil]

Take an arbitrary matrix

It's equivalent to one of that form

Because hK = h'K iff their 12 and 23 slots are equal

[tribal pasture]

I am just concerned about how they took b = 0 in h. All the previous implications are clear to me.

[latent anvil]

Start with b nonzero

Look at the matrix which has that entry deleted

Call these h and h'

I claim that hK = h'K

Do you see why?

[tribal pasture]

Yes, but I can also take b=e?

And with that I still have the desired result

[latent anvil]

What's e?

Just an arbitrary real number?

Because then yes

[tribal pasture]

indeed.

[latent anvil]

The point is that none of the matrices of this form are equivalent

For different a, b

We've found a canonical representative of each equivalence class

And this gives you a really concrete way to multiply the cosets together

[tribal pasture]

Yeah I understand from the part that the equivalence between differents b's are accounted by the definition of K. I am just not sure in the method they provided, how did the "thus" allowed them to assume b=0

[latent anvil]

Try and prove that those two sets are contained in one another

Clearly the one on the right is contained in the set of all cosets

So suppose you have an arbitrary element of H/K

What does it look like?

{{1,a,b},{0,1,c},{0,0,1}} K

Right?

Sorry

I meant to include the possibly nonzero b there

Let h = {{1,a,b},{0,1,c},{0,0,1}}. We want to show hK is in the set on the right

Do you follow so far?

@tribal pasture

I haven't proven the equality yet

[tribal pasture]

Yeah but why are you assuming b=0?

[latent anvil]

I'm not

Oh sorry I made a typo

I edited it but it didn't go through

Sorry!

Does that look getter?

[tribal pasture]

Yep

[latent anvil]

Let h' = {{1,a,0},{0,1,c},{0,0,1}}

[tribal pasture]

Oh yes I am aware in this case

[latent anvil]

The computation in the screenshot says hK = h'K

And clearly h'K is in the set on the right

[tribal pasture]

So b can be anything right?

[latent anvil]

Yes

But whatever it is, we'll have hK = h'K

[tribal pasture]

So basically different a,c defines the different coset, whereas the variation in b defines the coset itself

I can also consider hK = Kh correct?

Yeah I can

Thanks for help!

[latent anvil]

I'm not sure if that's true

I don't think it's false

I just don't want to check it

The thing about variation in b versus a, c is right though

[tribal pasture]

I mean because hK = Kh is equivalent to K being normal although this appraoch will become messy

[latent anvil]

Yeah, I meant I wasn't sure if it was normal

[steep hull]

Yeah for matrices the left/right coset interpretation is definitely the most reasonable for calculations

[brisk granite]

If $G$ is solvable and $H \leq G$, then does that mean $H$ is solvable?

[cloud walrus]

Cabbage:

[latent anvil]

I think yes. Do you know what the derived series for a group is?

@brisk granite

[brisk granite]

no

what is a derived series

?

[latent anvil]

Given a group G, we define the commutator subgroup [G, G] of G to be the subgroup generated by { ghg^(-1)h^(-1) : g, h in G }

We also use the notation G' for [G, G]

Then G' is contained in G, and G'' is contained in G', and so on

So we get a chain G >= G' >= G'' >= ... >= G^(n) >= ...

Where each prime means "take the commutator subgroup"

[fading wagon]

wait...

I think the commutator subgroup is wrong there [EDIT: FIXED]

[latent anvil]

G is solvable if and only if this series is eventually the identity

Oh you're right

Sorry, I just gave the generators

it is 3:30 am right now and I am liable to slip up after doing AG for hours

@brisk granite does that definition make sense?

[brisk granite]

yrp

*yep

[latent anvil]

Lemme try and remember why this is equivalent to solvability

How do you define it? Cyclic composition factors?

[fading wagon]

not yet

The quotient groups are abelian?

[latent anvil]

Oh so this implies solvability

[fading wagon]

yeah

[latent anvil]

Because G' is normal in G and G/G' is abelian

[brisk granite]

why are the quotients abelian tho?

[fading wagon]

that's part of the definition

[latent anvil]

We've gotten rid of the commutators

@brisk granite do you mean why are the quotients in this series abelian?

[fading wagon]

wait, you can prove that? Hold my keyboard

[brisk granite]

@latent anvil yes

[latent anvil]

Let [g, h] = g h g^(-1) h^(-1). Then gh = [g, h] hg

Yeah?

And each [g, h] goes to the identity in the quotient

Does that make sense @brisk granite ?

[brisk granite]

yep

[latent anvil]

So that shows "finite derived series" implies "solvable"

[brisk granite]

I see

[latent anvil]

Also, just to check, what's your definition of solvable?

[brisk granite]

A group is solvable when when it has a solvable series — a normal series where G_i+1/G_i is abelian

[latent anvil]

Now suppose G is solvable, so you have G = G_0 >= G_1 >= G_2 >= ... >= G_(n-1) >= G_n = 1 a normal series where the factors are abelian

[fading wagon]

yeah, checks out that the quotient groups are abelian

[latent anvil]

I think our indices are flipped but w/e

Ah okay

So check this out

suppose H <= G is normal with G/H abelian. Then G' <= H

The idea/proof is that if G/H is abelian, you have to kill off all the commutators [g, h]

Then you inductively prove that the ith derived subgroup G^(i) is contained in the ith term of the abelian series G_i

If we assume G^(i) <= G_i, then G^(i+1) = (G^(i))' <= G_i', and since G_i/G_(i+1) is abelian, G_i' <= G_(i+1)

Thus the nth derived subgroup is contained in G_n, which is trivial

@brisk granite how's that?

If you know this definition of solvable, the original claim is immediate, since at each step the ith derived subgroup of H is contained in the ith derived subgroup of G

So if G is solvable, it's derived subgroup terminates, and so H's must as well

[languid moss]

Does e have no solution

I ended up getting x=30 if assuming 3 does divide 5

I got it. It’s not a solution

I want to make sure I have the right concept. If the GCD does divide the number I am given then I will have to find x? If the GCD is 1 it will always have an x

I will have to find x using Bèzout’s identity

[wind steeple]

#elementary-number-theory

[languid moss]

Okay

[wild sapphire]

How do you guys think of Abstract Vector Spaces? I'm at the point of transitioning from thinking about vectors as arrows or number lists to this more abstract (and proper) definition with the axioms.

I just feel all over the place with it

[latent anvil]

what about them?

You might want #linear-algebra btw

I'm happy to help here though

General vector spaces are useful when studying geometry, in more ways than your might expect. First off, suppose you have a linear transformation from R^n -> R^m. The kernel of that map is a vector space, but it doesn't have a natural choice of basis, i.e. there's no canonical way to look at it as l-dimensional Euclidean space

Similarly, if you have a plane in three dimensions at an angle, we can think of that as R^2, but once again we have to choose a basis, and there's no obviously correct way to do that

But they're also useful when doing fancier geometry. If you have a surface, like the sphere, at any point of your surface there's a "tangent space", which is vector space that we think of as all the tangent vectors at that point. For a sphere, this can be thought of as the plane tangent to that sphere at that point

This was supposed to be examples of vector spaces in more general contexts but I guess they're still just arrows here

They're just harder to represent as tuples of numbers

[delicate bloom]

I think it's a mistake to try to force it as turning vectors into just an array of numbers, because there's the entire associated structure to go with it, it's just once you've gotten comfy with the pointy arrows in space you begin to understand how it can be decoupled for solving different kinds of problems than it might originally seem

like for instance, you can look at adding polynomials as a vector space and get a handful of nice results to play with this way, the easiest entry point would be vandermonde matrices imo, although there's a lot to talk about there

you can consider balancing chemical equations as molecules representing vectors with the components different atoms

there's this kind of game with lights that you turn off and on and you can only turn on a + shape of lights uhh

https://www-assets.yoyogames.com/system/blogs/featured_images/000/000/386/original/ChipFlipTechBlogFeature.png

you can consider these as vector spaces over F_2

trying to give some non geometric examples, there are many more once you start to think that way you can turn many things into a vector space

[fading wagon]

polynomials

integrate f(x)g(x) dx from -1 to 1 as inner product

[stone fulcrum]

@wild sapphire
Vectors were taught to you as arrows that you can add, and scalar multiply.

You can generalize this nicely, by forgetting about the "arrow" bit. ANYTHING that you can add together and scalar multiply can be considered "vector like" and has the same properties.

Great example is polynomials. You can add any two polynomials together, and you can multiply them by a constant. Polynomials form a vector space. They have a "basis" (1,x,x²...) and a "dimension" (degree of highest polynomial you're including)

However they don't have a length. Note you don't need lengths for the general vector space

[potent lynx]

why is everything so easy and cool in math

xd

[mild laurel]

"so easy"

oh sweet child

[potent lynx]

xdd

whats hard about math? literally nothing

jk jk

[stone fulcrum]

Oh that's a great example I didn't think about earlier. Vector spaces, and linear transformations between them @potent lynx

[potent lynx]

ye

thats what i was refering to

i have a bad question

why is the study of vector spaces called 'linear algebra'

why not just regular algebra lol

like 'vector space theory' xd

whats so special

[stone fulcrum]

The linear transformation is very linear

So we call it linear algebra lol.

[potent lynx]

lol

and whats up with mattrices

why is there no like 'matrices' to help up with groups or other structures

[stone fulcrum]

Matrices are the linear transformations!

[potent lynx]

oh fuck ye

you spoiled it btw

but ok xd

[stone fulcrum]

As well, there is a way to encode groups into matrices. We call it representation theory

[potent lynx]

cool

af

[stone fulcrum]

Flammablemaths just did a video where he creates a matrix alternative to the complex numbers

[potent lynx]

uh oh

i will go see it

[stone fulcrum]

It's pretty good but very lengthy if you know linear algebra already

[potent lynx]

i will try lol

[stone fulcrum]

Skip to the end imo

[potent lynx]

ye

tysm

for replying to my stupid questions

[stone fulcrum]

They're not stupid! Grats for trying to understand the things that most people just assume

[potent lynx]

lol ty

[woven delta]

It's not a matrix alternative lol

The material form of the complex numbers is literally re^i\theta

[stone fulcrum]

Ooh, wonder if he's going to find e^it for his matrix i?

[latent anvil]

The reason that we have matrices for vector spaces is that we have bases for vector spaces

You can choose to send the basis vectors anywhere you want, and that uniquely defines a linear map

So we can use a matrix to just write down where those basis vectors go

For more general algebraic structures, there's not going to be an analogue. All vector spaces are "free", which is very weird

[tribal pasture]

AoC reaccs onli

[fickle brook]

y'all i'm actually kinda fucked

i'm stuck on a problem from my own pset

i guess i'm gonna reclassify its difficulty later 😂

the problem is as follows:

let G be a nontrivial group with Aut(G) = {id_G}. Show Aut(G^n) iso to S_n.

i know that Aut(G^n) is at least as big as S_n as the previous part of the same problem involved showing the former contains an isomorphic copy of the latter (S_n can be embedded into Aut(G^n) as coordinate-shuffling autos)

but i'm drawing a blank on how to prove there are no other autos

or even if there actually aren't any

[bleak abyss]

So my instinct here would be to take some generic automorphism and compose with one of the coordinate permuting ones to sorta "undo it modulo an automorphism of G"

[fickle brook]

yeah but

hmm

what to compose with

[bleak abyss]

Yeah I'm thinking about how to execute this correctly

Okay so, if Aut(G) is trivial, then G is abelian

Since otherwise conjugation would be non-trivial

And since G is abelian, x->x^{-1} is an automorphism, so every element in the group has order 2

Not sure if this is relevant or not

[latent anvil]

That's clever

It lets us do linear algebra

G is a vector space over F2, so pick a basis

Now we're working with an infinite direct sum of Z/2Z

[bleak abyss]

And then fuck with the basis if it's high dim yeah

Wait does that mean that groups with trivial automorphism group are either trivial or Z/2?

Sounds kinda strong

[latent anvil]

I think so???

Because otherwise you can permute the factors

Wild

[bleak abyss]

Lmao so in this case G = Z/2 since we assume non-trivial

[latent anvil]

I love arguments where you accidentally classify all such cases

[bleak abyss]

Okay this is funny

[latent anvil]

Anyways, the automorphism group of G^n is GL(F2, n), right?

These will all be permutation matrices

[bleak abyss]

Yeah that takes care of it I guess

Jeez

[latent anvil]

I wonder if there's a nicer way...

[bleak abyss]

Yeah I think my original idea is probably what was intended

[latent anvil]

Sorry for interrupting lol

[bleak abyss]

You basically just say okay if \sigma isn't a shuffle automorphism

[cloud walrus]

Dami:

[bleak abyss]

Oh wait a second here's a question

[cloud walrus]

Dami:

[bleak abyss]

Yeah it definitely is

[latent anvil]

Yes

[bleak abyss]

Okay this is convenient, since the idea is that if an automorphism isn't a shuffle automorphism, then this behavior will manifest on the generators

[latent anvil]

It could shuffle some stuff and not others though

[bleak abyss]

Yeah but that gives me at most two coordinates to worry about

[fickle brook]

if anyone comes up with a solution please DM me with it so i can include it in my pset's solution key

[bleak abyss]

@fickle brook we did kinda nuke this problem earlier, turns out the only non-trivial group with trivial automorphism group is Z/2

[fickle brook]

oh

i mean

ok i'm gonna reclassify this as a 5-star problem lol

[upbeat plank]

I'm assuming there's another way to do it since it's consistent with ZF that there are other groups with trivial automorphism groups

[bleak abyss]

Okay so

[fickle brook]

wait what

[bleak abyss]

My thing required taking a basis

So in principle it invokes choice

And achristensen feels this fact shouldn't require choice

[upbeat plank]

Yeah I could be wrong of course, but it just doesn't feel terribly choicey

[cloud walrus]

Dami:

[bleak abyss]

These should both be homomorphisms

[fickle brook]

oh and uh

a quickie

i need an example of three groups G, H, K such that G normal in H and H normal in K but G not normal in K

i'm having trouble coming up with a c/e

[latent anvil]

C2 in K4 in S4

[bleak abyss]

K4?

[latent anvil]

Or C2 in K4 in D4 iirc

Klein four

[bleak abyss]

Oh Klein

[upbeat plank]

I think V4 is standard

[latent anvil]

I had to come up with this for my problem sets, it's a problem on week 3

[upbeat plank]

or just V?

idk

[latent anvil]

@fickle brook here you go

[cloud walrus]

Dami:

[latent anvil]

Feel free to steal any content from https://math.berkeley.edu/~tb65536/algebra_materials/index.html, I worked on them and think they're pretty good

[upbeat plank]

You still need to check that sigma takes one factor and sends it entirely inside another, or I don't see how this works

[fickle brook]

oh and another quickie

[bleak abyss]

I mean yeah no claim of being nearly done yet, still trying to formalize this in my mind

I feel this should be super easy but it's evading me

[fickle brook]

i remember doing it but forgot how to do it

G is a group, H and K are finite subgroups, show |HK| |H cap K| = |H| |K|

[latent anvil]

See week 5

You let H×K act on HK by (h, k) * x = h x k^(-1)

Iirc

The stabilizer of the identity are all pairs (h, k) where h = k

Which is the same size as H \cap K

[bleak abyss]

That works too, I was gonna do something fairly direct, like yeah enumerate with possible duplicates and find out how badly you overcounted

[latent anvil]

That's better if they don't know group actions yet

totally unrelated but the trivial k-algebra has no maximal ideals but does admit a homomorphism into k, and that's fucking me up

[bleak abyss]

The trivial k-algebra is what?

[latent anvil]

Zero ring

Wait no

It doesn't!

[bleak abyss]

Uh, depending on who you ask, homomorphisms have to preserve multiplicative identities

[latent anvil]

Yeah, sorry

Just realized. Okay, this is good. I'm glad I posted something stupid

thanks lol

[bleak abyss]

np

Anyway yeah I should prob get back to my own Silverman stuff but good problem Ann!

[latent anvil]

Lmk if you have any other stuff about designing group theory homeworks Ann, I spent a ton of time on mine this summer

[fickle brook]

oh all i have is a homebrewed 30-problem pset

about a year old

i'm cleaning it up rn

[latent anvil]

Is using orbit stabilizer okay for the HK problem?

[fickle brook]

idk probably is

[potent lynx]

any hints for this problem:

if G is a group and H and K are subgroups

prove that |H.K ... something like that

= |H| * |K|

?

[latent anvil]

we just did this lol

H, K finite subgroups

do you know the orbit stabilizer theorem @potent lynx?

[potent lynx]

oh fuck

sorry

no

i dont

[latent anvil]

no you're good, don't worry about it

My argument wouldn't work for you then anyways

[potent lynx]

ye

ty

[latent anvil]

Let me try and translate it to something without group actions

@fickle brook that problem is false

Aut(C2) is trivial but Aut(C2×C_2) = S3 ≠ S2

[fickle brook]

oh

dfsklg

hecc

[latent anvil]

@bleak abyss

fyi

Classifying groups such that Aut(G) is trivial could still be a good problem

Or you can ask them to determine the order of Aut(G^n)

[bleak abyss]

Oh yeah

[latent anvil]

Which is like (2^n-1)(2^n-2)...(2^n-2^{n-1}) if you think about it in terms of choosing n linearly independent bitvectors

[bleak abyss]

Swap two 1s is trivials

Lol we got meme'd on

[latent anvil]

I posed it to my friend who's good at group theory, and he immediately said "If Aut(G)=1 then G=1 or G=C_2" with no justification

[fickle brook]

i'm just gonna remove that problem from the pset lmfao

[potent lynx]

whats Aut(G) XD

[fickle brook]

wait fuck

fuck

now there are no 5-stars

😂

[latent anvil]

I can help with that

[fickle brook]

ha

hhaha

[latent anvil]

What do they know?

[fickle brook]

hahahahahahahahah

hahahahaa

hahahahahaaahaa

hhahaha

[bleak abyss]

"Classify all finite simple groups up to isomorphism"

[fickle brook]

jhgjsdkfgsdgsdf

wow

[latent anvil]

Look at the tricky problems on https://math.berkeley.edu/~tb65536/algebra_materials/index.html

They're pretty hard

[potent lynx]

Let G be a group of order 57. Assume that G is not a cyclic group.

Then determine the number of elements in G of order 3.

[bleak abyss]

But yeah I think knowing which groups have trivial automorphism group is probably something that makes sense for many group theorists to just know off the top of their head

[potent lynx]

any help with this?>

[bleak abyss]

Instinct is Sylow

[latent anvil]

^

[potent lynx]

ok thats something i dont know yet

[bleak abyss]

If there's a normal subgroup of order 3, then I think that just requires it to be cyclic

Right?

So then

[latent anvil]

What do you mean? All groups of order 3 are cyclic

[bleak abyss]

it = group of order 57

[fickle brook]

aight so

hm

yeah

[bleak abyss]

Because then the semidirect product is trivial

[fickle brook]

i think i'm gonna replace this Aut(G^n) problem with "find a c/e to this statement"

and keep it as the lone five-star in the pset

[latent anvil]

Yeah I think so sloth

Oh! For five stars

Fixed point free automorphisms are great

[bleak abyss]

Yeah if you have a normal subgroup of order 3, obv there's a normal subgroup of order 19

Then yeah HK is the direct product

So that's done

[latent anvil]

Call an automorphism φ : G -> G a fixed point free automorphism if φ(x) = x implies x = 1. Classify all pairs (G, φ) where |G| = 12 and |φ| = 6

[bleak abyss]

Now otherwise

[latent anvil]

@ann if you need a harder problem

[fickle brook]

φ(x) = 1 implies x = 1.
isn't that just triviality of kernel aka injectivity

[bleak abyss]

The number of subgroups of order 3 has to be 1 mod 3 and has to divide 19... So it has to be 19

[latent anvil]

Oh yeah lol, should be φ(x) = x

[bleak abyss]

But each subgroup of order 3 gives two elements of order 3

[fickle brook]

and what's |phi|

[bleak abyss]

So there should be 38 elements of order 3 @potent lynx

[latent anvil]

Order in the group of automorphisms

[bleak abyss]

There might be something easier

Okay so there's an element of order 19

Oh oh

Yeah that's way easier

There can't be two subgroups of order 19

You'd blow up the group

So there's a single subgroup of order 19

57-19 = 38

The remaining elements have to all have order 3

gg

[latent anvil]

You're saying there's only 1 nineteen sylow bc otherwise there's >19 of them?

And that's too big

Makes sense to me

[bleak abyss]

Not even Sylow

Or I mean secretly it's Sylow but really it's just that you can't have a group of order 57 with two order 19 subgroups because of the whole |HK| business

Intersection is trivial because shit's cyclic

[potent lynx]

sylow is grad stuff right?

[stone fulcrum]

Not necessarily. Just is included in some group theory courses if they go deep enough

[mild laurel]

My first algebra course covered sylow, and still did ring and field stuff

[latent anvil]

I feel like sylow and group actions should be in a first course in group theory

well of course I feel that way, that's why I include them in my first course in group theory

[mild laurel]

I mean I'd say that covering rings and fields are more important

[latent anvil]

Oh yeah definitely

We do those w/galois second quarter

[mild laurel]

Oh right UW is on quarters, yeah so it'd be hard to cover rings and fields in the first quarter then I guess

[latent anvil]

Yee

We do five weeks of basic group theory (subgroups, quotients, lagrange), which is the content of the normal undergrad group theory course, then five weeks of group actions/sylow/Jordan Holder/classification of finite abelian groups. Then in the next quarter 6 weeks of rings and 5 weeks of fields/galois

[bleak abyss]

Chicago's undergrad was groups first quarter, rings and basic modules second quarter, fields/Galois third quarter

[latent anvil]

I'm trying to include modules by putting them on the problem sets

Because the ta did it last year and I learned stuff from those problems

But I don't think anybody going to actually do them

The normal undergrad algebra course at the uw does rings first, then groups, then fields

[bleak abyss]

What do you do third quarter of algebra?

Lmao

[mild laurel]

Wait really?

[latent anvil]

Yeah, as of last year

It's weird

@sloth this special course only runs winter/spring

[bleak abyss]

Is the special one honors undergrad? Grad?

[latent anvil]

So people coming in have already done a quarter of honors analysis and know how to do hard math

It's a special undergrad seminar type thingy

Pretty informal, I organized it and manages to have them offer it for credit last year

students give lectures and instead of graded homework you present problem solutions to peers

[bleak abyss]

I see

[latent anvil]

Let A, B be finitely generated k-algebras

And ψ : A -> B a homomorphism of k-algebras

Is ψ^(-1)(m) maximal in A for all m maximal in B?

This feels like it should be false

It's true, wtf. Ψ' : A/ψ^(-1)(m) -> B/m ≈ k is surjective

And I think only B needs to be finitely generated, since that's how I know B/m ≈ k

[bleak abyss]

Uh

[latent anvil]

Nvm, got flipped around

Sorry

[bleak abyss]

So k here I'm guessing is algebraically closed

[latent anvil]

I think this still works though

Yeah, my bad

[bleak abyss]

So that's why you're able to cite Zariski

[latent anvil]

Sorry, I unswapped A and B

[bleak abyss]

I think you want both to be finitely generated

[latent anvil]

What do you mean cite zariski? That finitely generated k-algebras which are fields are k?

[bleak abyss]

Yeah

And oh wait no now that I think about it only B needs to be fg yeah

[latent anvil]

Yee

[bleak abyss]

Okay so

I think you actually need psi to be surjective here

And then at this point it has nothing to do with being k-algebras

Since otherwise how do you know that the last map is surjective?

[latent anvil]

I don't think you need ψ to be surjective. That past map is surjective because any map from a k-algebra into k is surjective

r = r * 1 = r * φ(1) = φ(r * 1)

Does that make sense?

It's a general fact that the preimage of a maximal ideal under a surjective map is maximal, but this is better than that

[bleak abyss]

Oh tru it's not 0 map

Okay okay I'm happy

[latent anvil]

This is why working with finitely generated algebras over an algebraically closed field is so nice

Maximal spectrum is a functor from fg-k-alg -> locally ringed spaces

you don't need to deal with spec or generic point weirdness

[bleak abyss]

Yeah for sure

I guess I'll do some Silverman ranting here

Gotta actually stay focused lmao

[latent anvil]

This is elliptic curve stuff?

[bleak abyss]

Yeah, though chapters 1-2 are basically a crash course in the needed AG

[cloud walrus]

Dami:

[bleak abyss]

So in particular, rational maps a priori aren't defined everywhere

[latent anvil]

k bar (V1) being the field of fractions of the homogenous coordinate ring of V1?

[bleak abyss]

Yeah

The specific description here is somewhat odd. One is just, define it as the function field of an affine part (and on affine varieties it's the field of fractions of the coordinate ring)

The other is to take the quotient of homogeneous functions of the same degree where the denominator isn't identically 0 on V, and identify f/g with h/k if fk = gh

[hot lake]

why do you say the rational map isn't defined everywhere ?

oh hmm

[bleak abyss]

Denominators can vanish

It's like thinking of rational functions on C, they're defined where the denominator is non-zero

We don't allow the denominator to vanish everywhere but somewhere is fine

[hot lake]

well for every point on V1 there should be an tuple of functions defined on a neighbourhood on P

but it may be the case that no tuple works globally

and you go from one tuple to the other by multiplying everything by the same rational fraction

[bleak abyss]

Yeah I think Silverman might get to that later

For now he's just allowing functions that are defined not everywhere 🤷

[hot lake]

maybe he expected it to be read as [f1...fn] defining the map and not f1...fn themselves, and then he didn't include the "forall f1...fn such that the map is [f1...fn] and for any point P where everything is defined, f(P) = ...

afterall [f1...fn] is an equivalence class

oh wait lol

[bleak abyss]

So, he said in a remark right afterwards that we may be able to fix points where shit's not defined

[hot lake]

remark 3.1

yeah

well okay then

so yes rational maps may not be well defined everywhere

only morphisms 👀

[latent anvil]

no rational maps, only morphisms

this doesn't mean anything it just popped into my head

[bleak abyss]

Okay now I'm back, my aunt called

[bleak abyss]

Hmm something I'm not seeing

Okay so

[cloud walrus]

Dami:

Dami:

Dami:

Dami:

[bleak abyss]

So no I'm stupid

When you multiply everything though by the lcm, "part of the lcm" goes toward killing denominators, and the degree just checks out

[cloud walrus]

Dami:

[bleak abyss]

No need to play some k_i/k_i game

I guess one more thing to be sure of

Oh okay actually it's easy

But just to write it out

[cloud walrus]

Dami:

[chilly ocean]

Nvm, I answered my own question

Okay, another quick question

If I show that this particular group is closed under inverses, do I also have to show that the inverses are in the set itself? The proof I did shows that implicitly, but should I explicitly state that?

[bleak abyss]

"Closed under inverses" usually means the inverses are in the same set

How you'd specifically go about phrasing that is a bit context dependent

[chilly ocean]

Ah

Alright, ty

A subgroup can be trivial, as in only one element, correct?

(Assuming the element is the identity element of course.)

Like let's say Z/3Z, modular addition. Would 0 in that group be a subgroup? It's not empty, has the identity, is binary and associative, and has its inverse

[latent anvil]

yes

[chilly ocean]

Wait... I answered my own question again

This is why I shouldn't be trusted with a math degree

[latent anvil]

Any other questions?

Quick, ask about whether there are any nontrivial zeroes of the Reimann zeta function off the critical line

[chilly ocean]

(If there are, wouldn't the hypothesis be disproven?)

[latent anvil]

(yes, that is the joke)

(if you ask it you'll answer it)

[chilly ocean]

XD

Well, time to find out. If you see a black person win the Abel Medal soon, that's me.

[bleak abyss]

Future doxx

[latent anvil]

Does this version of the nullstelensatz hold? The maximal ideals of k[x1,..,xn] whose quotient is k are of the form (x1 - a1, ..., xn - an) for (a1,...,an) in k^n, even if k is not algebraically closed

[latent anvil]

Yes and it's easy to prove. Look at the map f : k[x1,...,xn] -> k which mods out and then goes through an isomorphism to k. Then for each xi, let ai = f(xi), so that f(xi - ai) = 0. Then (x1 - a1, ..., xn - an) is contained in the kernel of f, but it is also maximal, so it equals the kernel of f.

[chilly ocean]

I'm an idiot

They wanted me to find all the elements that commute with everything in a particular set

But I was given that the group was abelian...

And I did it by hand

[delicate bloom]

I hope it was a finite group

[chilly ocean]

@delicate bloom I would still be doing it if it wasn't. I can't supertask.

[latent anvil]

you'll never make it in math then

[chilly ocean]

D:

[delicate bloom]

can't supertask

why even try?

[chilly ocean]

At least I'll make it in statistics

[latent anvil]

I heard that in the last quals exam at my school, students had to sum the harmonic series by hand

[delicate bloom]

tough

I guess you could rearrange it so that it gives a finite value, but you'd still be stuck summing infinitely many terms

[latent anvil]

yeah the trick is to use sacred geometry

[bleak abyss]

It's infinite ez

Harmonic series diverges

Boom done

[latent anvil]

wtf

but the numbers keep getting smaller??

[chilly ocean]

An easy proof is to compare it to a smaller series that diverges

[latent anvil]

fam I am kidding

[bleak abyss]

I think shamrock is being ironic here

[chilly ocean]

Oh XD

[bleak abyss]

Imitating the typical calc student

[latent anvil]

I appreciate the intent though

[bleak abyss]

p-adically though..

[delicate bloom]

still diverges

[chilly ocean]

Is shamrock Jacobian?

Jkjk

[latent anvil]

no I am not

[bleak abyss]

Psychopotence is actually jacobian and is trying to cover it up by being hypervigilent about jacobian alts

[chilly ocean]

(Everyone without a pfp that mainly hangs out in the advanced mathematics I think is jacobian, no matter how improbable)

[latent anvil]

(sloth does not suspect I am a psychopotence alt, good)

[chilly ocean]

[bleak abyss]

@delicate bloom presumably you're gunning for how multiples of p aren't units?

Or oh

Also the norm blows up

[delicate bloom]

1/n is large p-adically

[bleak abyss]

I was making a super naive statement lmao

[delicate bloom]

for large powers of p

kinda oscillates

1/p^n I should say is what's large

1/(1+p^n) is a unit very nearly 1

[bleak abyss]

Basically just being like "tfw a_n -> 0 => \sum a_n converges"

[chilly ocean]

I do the dumbest overthinking in this class

[delicate bloom]

ahh cauchy sequences |a_{n+1}-a_n| -> 0

[latent anvil]

I now have a profile picture

[bleak abyss]

Shamrock you should also be a sloth

[latent anvil]

hmm

[bleak abyss]

Some of the greatest mathematicians were sloths

Mikhail Slowmov

Lasloth Lovasz and Babai

[latent anvil]

Grothensloth

[bleak abyss]

Alexander Slothendieck

[latent anvil]

Lol

[chilly ocean]

So, closed under products, has the identity...how to prove it's closed under inverses...

[bleak abyss]

The list goes on

[chilly ocean]

Wait

Dumbass again, nvm

Got it