groups-rings-fields

[prisma ibex]

If you're talking about Langlands over function fields of curves over finite fields then almost everything is known

[formal ermine]

ooo I know all of those words

[prisma ibex]

Over number fields the situation is pretty bad

[formal ermine]

just not in that context

[prisma ibex]

some portion of the Langlands correspondence for GL_2 over Q is basically what Wiles and Taylor proved around modularity for elliptic curves (and Fermat's last theorem)

there hasn't been a whole lot of progress beyond that in the number field case

err I should say there hasn't been much progress in getting both sides of the correspondence (both automorphic to Galois, and also Galois back to automorphic) beyond GL_2

there are a lot of results on the automorphic to Galois direction for pretty general groups through the cohomology of Shimura varieties

it's the Galois back to automorphic direction that is the hardest direction (in my opinion at least)

in the function field setting this is also true, we know the automorphic to Galois direction for literally any G but only know the Galois back to automorphic direction for like, G=GL_n and a few other isolated cases, but definitely not general G

the case G=GL_1 is class field theory and this is completely solved, see Langlands doesn't have to be intimidating!

[rotund aurora]

Is this Deligne and the Weil conjectures?

[prisma ibex]

no this is a different result

it's due to work of Lafforgue

[rotund aurora]

Ah k

[formal ermine]

which one

[prisma ibex]

Drinfeld did this for GL_2, Lafforgue did this for GL_n, the other Lafforgue did the automorphic to Galois direction for general G (there are two Lafforgue brothers, it's funny that they both did Langlands)

[formal ermine]

don't they have another brother

what happened to that guy

[prisma ibex]

oh yeah I forgot about the third Lafforgue

Thomas is now a teacher in a classe préparatoire aux grandes écoles at Lycée Louis le Grand in Paris

neat

[formal ermine]

ng do you have a brother

[prisma ibex]

no

[rotund aurora]

u sure

[prisma ibex]

[formal ermine]

"check again in 9 months"

[south patrol]

Gotem

[foggy inlet]

I need help understanding quasitriangular hopf algebras
specifically, I don't really understand what the quasitriangularity property would look like explicitly
for example
$$(\Delta\otimes 1)R = R_{13}R_{23}$$
If we assume just for the example that $R = \theta^{\mu\nu}\partial_\mu\otimes\partial_\nu$, then with
but if $\Delta(\partial) = \partial\otimes 1 + 1\otimes \partial$, you get something like

$\theta^{\mu\nu}(\partial_\nu\otimes 1\otimes\partial_\nu + 1\otimes \partial_\mu\otimes\partial_\nu) = (\theta^{\mu\nu}\partial_\mu\otimes 1\otimes \partial_\nu)(1\otimes\theta^{\mu\nu}\partial_\mu\otimes\partial_\nu$
,
Idk how to relate the two sides

[cloud walrus]

Quantumnessie

[fast stratus]

My attempt

[red marlin]

Try expressing both sides of the equation in a more compact form using the properties of the given elements and their operations, also see if there is any patterns or common terms to establish a connection between the sides.

[celest furnace]

Is G Abelian?

[fast stratus]

No

That is where i’m stuck

[celest furnace]

Probably use that H is normal somehow

[foggy inlet]

I realized my error there; The equation seems to hold nicely if you write R as the exponential of what I wrote there

However, I don't understand how to generalize a bit to when the coproduct is no longer just that simple form

If the coproduct is perturbatively expanded as $\Delta(\partial) = \partial\otimes 1 + 1\otimes \partial + \kappa \partial \otimes \square$ (let's assume the new term is the D'Alembertian) the rhs would then be something like $$R_{13}R_{23}R_\kappa$$ with $$R_\kappa = \exp [\theta^{\mu\nu}\partial_\mu \otimes \kappa\square\otimes \partial_\nu]$$

[cloud walrus]

Quantumnessie

[foggy inlet]

so we'd need to "absorb" the new term into some other old term somehow, but idk how and it doesn't seem to appear that nicely in the literature

[pastel cliff]

what do we usually mean by "extending" something

im thinking in particular of extending homomorphisms to tensor products

[red marlin]

Try simplifying the RHS of the equation by using properties of the given elements and their operations then look for patterns or common terms to establish a connection between the RHS and the given expression for $R$ also you may need to absorb the new term involving $\kappa \square$ into another term to simplify the RHS and make it easier to relate to $R$.

[cloud walrus]

mrfeynman

[red marlin]

It means broadening or generalizing a math object or operation.

In algebraic extension, youd find ways to extend homomorphisms to tensor products of algebraic objects like fields or rings, it might involve defining new rules to handle the added structure.

[dreamy jewel]

Let $\mathbb{Q}_p$ be the p-adic numbers and $\mathbb{Z}_p$ be the p-adic integers, is the quotient $\mathbb{Q}_p/\mathbb{Z}_p$ compact? I would think not, right?

[cloud walrus]

CelesteCrow

[pastel cliff]

is there any intuition for the minimal polynomial of an algebraic element of a field extension

[south patrol]

Wdym

[formal ermine]

the smallest polynomial which has that element as a root lol

[frigid lark]

Try raising the element to a couple of powers, might give you some hints

[south patrol]

Pls clarify the q lol

Like we seem to have interpreted it in 3 different ways xd

Intuition for why it exists vs what it is vs how to calculate it

[wraith cargo]

Ask vague question
Do not reiterate
Leave chat
massive chad

[rotund aurora]

polynomials are central part of algebra, it's like asking for the motivation of algebra lol

[lethal dune]

what question lol

[frigid lark]

^

[obsidian loom]

Hi

I'm having trouble understanding what conjugation is

I know the definition but is there any helpful interpretation of it?

[rotund aurora]

Z_p is compact

idk if that says anything, I'm topo noob

like Q_p is def not compact

[pastel cliff]

i had to relocate lmao

one sec

[rotund aurora]

an automorphism? Look up conjugation in the symmetric groups and see what is actually going on

[pastel cliff]

i guess i meant more like relative to minimal polynomials of a matrix

[rotund aurora]

mmh if I had to quess, I'd say no because something like the sequence $a_n=\sum_{i\geq -n}p^i$ still makes sense in the quotient, has no limit points in $\Q_p$ and is very "far" from $\Z_p$

I wonder if quotients of a non-compact topological space by a compact subspace can ever be compact. I think no?

[pastel cliff]

and if the minimal polynomial of an algebraic element encodes any information in the same way the minimal polynomial of a transformation does

[cloud walrus]

Croqueta

[grand cliff]

Learning about systems of imprimitivity.
If a group action admits a nontrivial system of imprimitivity then that action is imprimitive.
A corollary of this lemma would be that any simple groups can only act primitively.

Is there a way to characterise when the induced action of a normal is not transitive?

Wait, I don't think my conclusion is correct. Since it's just saying that "if we have a normal subgroup" then the orbits are a system of imprimitivity

So in fact, this test does not tell us anything about simple groups

[lethal dune]

yes they are same infact

ℚ (a) be extension of ℚ the 1,a,a²,...,aⁿ forms a basis then you treat "multiplication by a" as a Q linear endomorphism and look at it's minimal/characteristic poly, which in face is the minimal poly of a

[oblique river]

The quotient of a noncompact group by a compact subgroup is automatically noncompact

Nothing special about the p-adics needed

[oblique river]

That’s right

[proper moth]

I'm thinking write f as (1 2 3 4)(5 6)? Then I think the cyclic subgroup generated by f has 7 elements? (e, 4 arising from the first cycle, and 2 arising from the second?)

But I don't think that's right

[formal ermine]

it can't have 7 elements

[proper moth]

oh, because that's prime?

[formal ermine]

7 doesn't divide 6!

[proper moth]

ah yeah

what's the correct way to think about this problem?

[proper moth]

ah, got it I think. I think really I'm wanting to find the value of n for which f^n = e

which I guess is the LCM of the orders of the disjoint cycles. I suppose that's the number of elements, and the elements are then just e, f, f^2, ..., f^(n-1) (?)

[woven obsidian]

yep

[karmic moat]

then use binomial theorem on (x+1)^(n-1) + (x+1)^(n-2) + … + (x + 1) + 1, which leaves us with x^(n+1) + nx^(n-2) + n(n-1)/2 x^(n-3) + … + n, and this polynomial is eisenstein, taking n as the prime

is that right?

and then because phi(f(x)) = f(x+1), since the above polynomial is eisenstein, it’s irreducible, thus x^(n-1) + … + 1 is irreducible

thus we conclude that [Q(ε) : Q] = n-1

i think i need to also show that x^(n-1) + … + 1 is minimal

[formal ermine]

yes

[formal ermine]

it's monic and irreducible

thus minimal

[karmic moat]

yeah right

ty!

[obsidian loom]

Hi

I'm having trouble with 1)

I don't know any definitions that relate orders before and after homomorphisms

[formal ermine]

phi(g^n) = phi(g)^n

[obsidian loom]

ok i did know that haha

but that implies that ord(g) = ord(phi(g))

[formal ermine]

no it doesn't

what if there exist a smaller n for the order of phi(g)

[gilded osprey]

I am trying to find a ring, N, and an element of that ring, n, such that n is left invertible (with multiplication) but not right invertible.
Obviously N cannot be commutative, and all the other non-commutative rings I have tried (quaternions, square matrices) have no elements that are only invertible on one side

Any suggestions to point me in the right direction? I'm pretty new to rings unfortunately, and so I don't know a whole lot of non-commutative ones off the top of my head yet

The most I have been able to find online is dealing with matrices acting on infinite vector spaces, but don't really know how to conceptualize that (this is what I found online: https://math.stackexchange.com/questions/2396264/element-may-not-have-2-sided-inverse-in-infinite-ring?noredirect=1&lq=1)

[formal ermine]

can't you take like

R^N

form its end ring

right shift has a left inverse but not a right inverse

[tribal moss]

"end ring" = "ring of linear transformations under addition and function composition"

[gilded osprey]

TBH I am still trying to learn about endomorphism rings

[gilded osprey]

Ah, that makes sense

[gilded osprey]

So, are you thinking having a vector, and the shifts move the elements of the vector around?

[tribal moss]

The right shift is the transformation defined by

T(a,b,c,d,...) = (0,a,b,c,...)

[gilded osprey]

Would the vector need to be infinite then for this to work?

[formal ermine]

I'm thinking of it as a sequence

[tribal moss]

Yes, Illum meant $\bR^{\bN}$, not $\bR^n$ for some number $n$.

[cloud walrus]

Troposphere

[gilded osprey]

Oooh whack, I have not seen that notation yet. That makes more sense though. I'll think through this for a bit and come back if I get stuck or have other questions. Thanks guys!

[obsidian loom]

$g^n = e_G \implies ord(g) = e_G$ \ $\phi(g^n) = \phi(g)^n = e_H \implies ord(\phi(g)) = n$

[cloud walrus]

trayan_b

[obsidian loom]

i get that e_G and e_H are different but I don't see how the different identity elements changes the number n

[formal ermine]

your last implication is wrong

how do you know that n is minimal?

[obsidian loom]

by definition of order

being the minimal n that makes g the identity

[formal ermine]

yes, but how do you know that n is minimal

[obsidian loom]

so I just have to say: let n be the least natural number such that g^n = e?

(at the beginning of my answer)

[formal ermine]

that's not what I'm trying to say

for your last implication

you need to know that n is the smallest number such that phi(g)^n = e

but how do you conclude that?

[wicked zephyr]

Hi

[obsidian loom]

i get you know

i have no idea

[wicked zephyr]

Every subgroup of a free group is free. Is the converse true? If the group is non-free, it has a non-free subgroup?

[tribal moss]

Itself?

[wicked zephyr]

i mean, a proper subgroup, of course

[rotund aurora]

you can take a product of a free group and a non free group, and things like that

ah wait "it has"

thought you were asking if all would be non-free

[tribal moss]

(Q,+) is not free, but I think all its proper subgroups are.

[lethal dune]

Don't think that quite works

[tribal moss]

I though wrongly. The additive group of dyadic rationals is not free.

[lethal dune]

ye any localization might do

(maybe)

[tribal moss]

But does the dyadic rationals have a non-free subgroup?

[lethal dune]

itself?

[rotund aurora]

cant you take 4-adic rationals?

[tribal moss]

Yeah haha. Proper subgroup sorry.

[tribal moss]

That wouldn't be proper -- e.g 1/2³ = 1/4² + 1/4².

[rotund aurora]

lmao right

this is interesting tbh

[lethal dune]

ok so does this argument work? Any subgroup of Z[½] is a subgroup of Q so either cyclic ( free) or dense in R. So for any subgroup of Z[½] to be dense, it has to contain larger and larger power of ½ⁿ?

[tribal moss]

Hmm, a priori the smaller and smaller elements our group contains might not have unit numerator.

[lethal dune]

but that's fine?

[tribal moss]

3Z[½] would seem to be a proper non-free subgroup.

[rotund aurora]

true

[lethal dune]

right

[rotund aurora]

the numerators always divisible by 3

[lethal dune]

Ok here's a dumb one, Z/p is simple and trivially satisfies all proper subgroups are free

[tribal moss]

🎉

[obsidian loom]

Just to confirm, a consequence of Lagrange's Theorem is that every element in a group G has order which divides |G|. Is that correct or am I mistaken?

[formal ermine]

yes

you look at the subgroup it generates

[obsidian loom]

oh

now i understand why

ty

[cloud walrus]

MyMathYourMath

[lethal dune]

pi1 of RP2

[solar glacier]

it should be pi1 of RP2#RP2

[south patrol]

Pi1 of klein bottle right

[lethal dune]

Mb yes

[solar glacier]

but what explicit group is this? i.e., iso to what well known group

[lethal dune]

It’s just that

[solar glacier]

is it semi direct pf Z with itself

[lethal dune]

You can say its Z*Z/2

[cloud walrus]

MyMathYourMath

[lethal dune]

No

Haven't I already answered this before? Feels like Déjà vu

[solar glacier]

yeah but im still a bit confused :/

[lethal dune]

I remember reducing it to <x,y | x² >

which is the free product of ℤ* ℤ/2

[solar glacier]

Ah so its free product of Z with Z/2?

[lethal dune]

hopefully what I'm saying is right

[solar glacier]

lol

[lethal dune]

But I'll wait for someone else to confirm

[uneven scaffold]

Show that R is a local ring if and only if the following property is true: for any x in R, either x or 1 - x is a unit. For the first implication if we assume that R is a local ring, then R has only one maximal ideal call it m. Now the way I see this is that Since R/m is a field, m must contain all the possible non-units of R as it's the only one maximal ideal. So each x in R \ m would be a unit right? I'm not sure how I can show this either x or 1 - x business though. Any hints?

[lethal dune]

(R,m) local then notice that m is the Jacobson radical

[uneven scaffold]

Haven't covered Jacobson radicals at this point. Is this needed here?

[lethal dune]

You have already showed R\m is the set of all units.

[lethal dune]

you don't need it but it's helpful

say x and 1-x both are non unit then where should they belong?

[uneven scaffold]

In m, but then 1 - x + x = 1 would also be in m and so m = R contradicts?

[lethal dune]

yup

[uneven scaffold]

Thanks:)

[kind jacinth]

M2(R), the 2 × 2 matrices with entries in R. I need to find all ideals.
why would a matrix of form something like
0 0
0 1
not be an ideal?

because
0 0
0 -1 would generate same thing?

[solar glacier]

i wanna say M2(R) has no nontrivial ideals.. i could be wrong thohgh..

[kind jacinth]

it doesnt

but why is that?

[karmic moat]

prove by contradiction

suppose you have
a b
c d
contained in a non trivial ideal

see if you can find contradiction

hint: ||recall that elementary row and column operations can be expressed via matrix multiplication ||

[kind jacinth]

first need to make sure i know what an ideal is. So if i understood correctly, if i take an element in the ring (say a) and an element in the subtring (say b) ab and ba must appear in the subring

[karmic moat]

i assume here by “ideal” we mean “two-sided ideal”

[kind jacinth]

for it to be called an ideal. right?

[karmic moat]

yeah

[kind jacinth]

ok and in this case if id take an element in 0 0
0 1
and an element in m2(R) clearly i will get something not in my subring generator 0 0
0 1
i guess this part makes sense now

but proving by contradiction is a good exercise

[kind jacinth]

question. If instead of R i'd have Z would this still hold?

[formal ermine]

is there any nice classification of topological fields?

[lethal dune]

what about M_2(2Z)?

[lethal dune]

why would fields have a canonical topology?

[kind jacinth]

i think it should have more than the non-trivial ideals because its not a field but i am not sure why

[formal ermine]

huh?

[lethal dune]

wym by a topological field? Topological group which is a field?

[lethal dune]

I was saying M_2(2Z), is this an ideal of M_2(Z)?

[kind jacinth]

ohh. Yes it shoul dbe

all are i think right? M_2(2Z/3Z/4Z/...)

and the maximal ideals would be M_2(nZ) where n is a prime number?

[lethal dune]

no

what's M_2(Z)/M_2(pZ)?

[kind jacinth]

if M_2(pZ) is the maximum ideal M_2(Z)/M_2(pZ) would be a field.

[lethal dune]

non necessary in non cummutative setting

[kind jacinth]

ahhh

[lethal dune]

potato

[formal ermine]

the usual definition. a field equipped with a topological space whose multiplication and inversion is continuous

[lethal dune]

then you can just give all of them the discrete topology so

[formal ermine]

right I see

[lethal dune]

there's no canonical way to topologise

[proper moth]

Reading Pinter's algebra book. In it, for the proof of Cayley's theorem, he defines a function pi_a(x) = ax, for all a in G (pi_b, pi_c, etc), the set of which is called G*

Then he's proving G* is a subgroup of S_G. For proving closure he writes:

this proves pi_a * pi_b = pi_(ab), but don't you need to show that pi_ab is in G*, which requires that ab is in G?

oh I see I think. G is a group, so since a,b are in G, so is ab

[kind jacinth]

they take {0, 1, a, b} as a field of 4 elements and do the cayley table for it. Why cant {0, 1, 2, 3} be a field of 4 elements

?

[kind jacinth]

i guess its because it screws up for multiplication

not all rows have identity

how does the multiplication table represent a field?

field means identity must appear everywhere. Identity for multiplication is 1 but first row doesnt have it

[formal ermine]

what why?

0 has no inverse

[kind jacinth]

then identity must appear in all rows except 0?

or this isnt even a requirement?

[formal ermine]

yes

by definition

[kind jacinth]

also how would i fill the multiplication table myself?
I can fill the first row and first column.
then I know that i should fill 1 where (1,1) is since identity * identity = identity

how would i fill the rest though such that it satisfies a field?

like how would i know a*a = b and not 1?

[delicate orchid]

that would force a*b = b which hopefully you can see the problem with

[kind jacinth]

because only b*identity should give back b? and a isnt identity?

[delicate orchid]

yes exactly

[kind jacinth]

ok i get it now

thanks!

[kind jacinth]

help with this? ker(theta) is all elements of Z110 that map to the 0s (identity) in Z130 right?

but how do i find these elements (the generator)

[south patrol]

wdym

these are just symbols

[kind jacinth]

figured that one out. the question was "Let {0, 1, a, b} be the ground set of a field with four elements"

[kind jacinth]

i am not sure about this one though. i know that since 1->13 thus
110 ->130 (which is zero) so 110 is one of the things in the kernel that map to 0

but how do i figure out all elements?

actually does anything besides 110 map to 0?

[delicate orchid]

110 = 0 -> 0 not 130, but yes it's in the kernel

and yes there are other elements that map to 0, 13 is not coprime to 130

[kind jacinth]

yeah i think i figured it out. its 10Z_110

what do they mean find "all" homomorphism. this function either is a homomorphism (if its onto) or it isnt

<6> would map to <15>

1+<6> to 1+<15>
...
5+<6> to 5+<15>
then id have to map <6> to 6+<15> as well

[delicate orchid]

by find all homomorphisms they literally mean find all homomorphisms

every map phi such that phi(x)phi(y) = phi(xy)

[kind jacinth]

where x, y would be elements of Z/6Z right?

[fleet pelican]

yes

[kind jacinth]

so like i need to find such elements such that the property above holds?

[fleet pelican]

functions

[kind jacinth]

like a function that maps Z/6Z to Z/15?

[fleet pelican]

in this case, it is phi(x+y)=phi(x)+phi(y)

[kind jacinth]

and once i define this function the property phi(x)phi(y) = phi(xy should hold for any x y right?

[delicate orchid]

if phi is a homomorphism, sure

do you know about homomorphisms and the image of generators?

[kind jacinth]

homomorphism and generators yes. i am not sure what u mean by the image of generators

[delicate orchid]

I mean you can do this problem without it, I was just asking

[kind jacinth]

but yes i did cover it

[cloud walrus]

MyMathYourMath

[formal ermine]

what are I and J

[solar glacier]

ideals

[formal ermine]

uh not just some k

wait are they proper ideals

[solar glacier]

im not given wether or not theyre proper

[formal ermine]

then ignore what I said

[cloud walrus]

MyMathYourMath

[delicate orchid]

yes

[solar glacier]

ok cool thansk

[delicate orchid]

I^k is the additive subgroup generated by products of k elements of I, so just pick a k times lol

[kind jacinth]

i was thinking of defining my function like
theta(a+6Z) = a+15Z
for a=0 theta(6Z)=15Z etc..

[delicate orchid]

well yes you need 6Z = 15Z because a group homomorphism maps the identity to the identity

[delicate orchid]

here's my hint: A homomorphism is uniquely determined by the image of the generators

[white oxide]

i'm struggling to see how these two definitions are equivalent, can somebody help me out?

because if g(x) is a unit isn't it still a polynomial

oh

it's because it's in F and not in F[x]?

[delicate orchid]

it's the non-zero elements in F

[white oxide]

ah got it

thx!

[white oxide]

what does a trivial factorization mean? f(x) x 1 in F?

[coral shale]

yes

[south patrol]

Not sure what you mean by "1 in F" etc

But no, a trivial factorisation is a factorisation equivalent to that factorisation i.e. a factorisation of the form like

f(x) = g(x) a1 ... an where a1,...,an are units

[white oxide]

my fault i was being sloppy with notation

ye that's what i meant

[cloud walrus]

MyMathYourMath

[south patrol]

Yes

Probably the simplest is just to take radicals of both sides and note sqrt(I^k) = sqrt(I) lol

[solar glacier]

ir just a brute force way using set inclusion

[south patrol]

ye

ok i mean the most like generalisable way for proving these things lol

[solar glacier]

is this the brut force way

[cloud walrus]

MyMathYourMath

MyMathYourMath

[south patrol]

well mentioning (as you did twice) that a is in R is redundant but yes

[solar glacier]

forces $a \in \sqrt{J}$ not $a^{nk}$

[cloud walrus]

MyMathYourMath

[solid olive]

Is there a faster way to multiply cycles then by converting them into tableaus and then doing the composition from right to left?

[south patrol]

uh so like I'll show by means of an example

for calculating (23)(124) for example (in S4), the way I would compute is to go one by one: 1 is sent to 2 by the first, then ultimately to 3. 3, in turn, is sent to 2. 2 is sent to 4. 4 is sent to 1. So this means it's the cycle (1, 3, 2, 4)

You pick an element and follow it around until you reach the element again

Here I was lucky in that I didn't have to pick anything else but yeah

[solid olive]

And if you don’t specify what a your in what do you do

[south patrol]

wdym "what a"

[solid olive]

What S•

[south patrol]

oh

Well I assume you do know considering you're the one doing the multiplication right

But anyway, you only ever have to check the numbers which are among those we wrote down

Because all the others are fixed

[solid olive]

I assume is the s of the highest number

[south patrol]

Like, do you calculate (23)(12) we only have to see where 1 2 and 3 going

Regardless of whether this is as an element of S3 or Sn for n = 1207105971

[solid olive]

But the damn book were using never specifies

Okay so I’m right about that at least

Next when have multiple disjoint cycles and you want to write them as a product of transpositions you basically convert each cycle

For example if I were to write (156)(234) as a product of transpositions I would find the product of transpositions of (156) and multiply that by the product of transpositions of (234) right?

[delicate orchid]

Sure that would work

But only because they’re disjoint

[solid olive]

Okay what if they aren’t disjoint…

Would you just delete the repeated transpositions?

[delicate orchid]

You’d calculate their product first, and then decompose whatever that is

you do right to left don’t you ok

[solid olive]

Okay and finding the product in the disjoint case didn’t work because there disjoint…got it

[delicate orchid]

For example, (12)(23) = (123) = (13)(12)

Sorry if I’ve done that wrong I’m having to multiply these permutations in reverse order to what I’m used to

[kind jacinth]

A bit confused by the function. Since its Z2xZ2 shouldnt theta(1) = a tuple? Say (1,1)?

[delicate orchid]

The identity of Z_2xZ_2 is (1,1) but we just write the identity as 1

[kind jacinth]

Oh and what about the theta(x)=2? Is it like (2,2)?

[delicate orchid]

If you want to be explicit:
[1_{\bZ_2\times\bZ_2} = (1_{\bZ_2}, 1_{\bZ_2})]

[cloud walrus]

Wew

[delicate orchid]

God that was hard to do on a phone keyboard

[kind jacinth]

Yeah makes more sense now

[kind jacinth]

?

[delicate orchid]

Now this one I’m not 100% on

(2,2) is just (0,0) in Z2xZ2

Going off the “unique ring homomorphism” I’m guessing this maps x -> (1,0) and 1 -> (0,1)?

Wait no that can’t be right hmmm

Oh does it literally mean 1+1? Lol

What an oddly phrased question

Regardless you don’t actually need to know what the map is to do the question

[kind jacinth]

Wat do you mean by this?

[delicate orchid]

2 = 1+1

[kind jacinth]

Like (1,1) and they do 1+1?

[delicate orchid]

That’s literally the only thing I can come up with as to what they mean by that

Which is weird considering the ring is char 2 and all that

[kind jacinth]

I know that Z2 would be a maximal ideal of Z but yeah Z2xZ2 is different.

[delicate orchid]

Z2 isn’t an ideal of Z, 2Z is

I’m assuming you’re using Z2 as short hand for Z/2Z

[kind jacinth]

Oh right

[delicate orchid]

Am I mistaken there?

[kind jacinth]

No

So should i first come up with a function that does the given mapping?

Also z2[x]? Z2 only has the elements 0,1 can i have sth else? Or is it like if x =2 then its 2mod2 and so on

So 0,2,4,... (2Z) map to 1 and the rest to 2?

[delicate orchid]

It’s polynomials with coefficients in Z2

Like 1+x^2+x^3

[kind jacinth]

Ohh it has to do with polynomial rings. Havent done that yet so ill do that first before attempting this. Thanks!

[delicate orchid]

Maybe they’ll explain what 2 means there

[sonic coral]

x^4 + 1 is irreducible over Q right

actually wait

yeah it is

[south patrol]

Ye

It is a cyclotomic bad boy

[delicate orchid]

Wtf!

[south patrol]

But also we know how it factors over R

[sonic coral]

it’s not enough to say that it’s irreducible bc it has complex roots right

[south patrol]

No

[sonic coral]

since x^4+2x^2+1 is reducible over Q

[south patrol]

Yh you need another kind of argument

Since it clearly has no roots, you can just check it's impossible to have it as a product of two quadratics

[delicate orchid]

The root argument only holds for degree 2 and 3 due to partition related reasons

[sonic coral]

that makes sense ^

[south patrol]

Or another way is to see how it factors over R and like

Any factorisation over Q gives a factorisation over R

I imagine there are other, more ad hoc methods tho

[sonic coral]

well i clearly can’t use eisenstein since the +1 or mod p irreducibility

so i tried shifting it over and then using eisenstein but maybe i didn’t try enough shifts

[south patrol]

Yeah it is a bit trickier

[sonic coral]

so i guess i’d have to look at the product of two quadratics

[south patrol]

But honestly factoring as a product of two monic quadratics will work nicely

Applying Gauss's lemma, we can assume those quadratics are in Z[x] and monic too

[sonic coral]

i supposed it factored as (x^2 +ax + 1)(x^2 + bx +1) and found a=-b and b^2 = 2, so the case of 11=1 doesn’t work in Q, so i guess i’d need to check the -1-1 case

[south patrol]

Yup nice

[sonic coral]

got it. thanks

[south patrol]

No problemo

[sinful nebula]

Any hints on C?

[south patrol]

Well what I would try: there's a simpler condition than (CD)^n = C^n D^n you can try to satisfy

[sinful nebula]

as in

(CD)^2 = C^2 D^2

?

[south patrol]

Well okay it is hard to communicate my idea without spoiling it

But try to find some simplee condition on C and D which implies that (CD^n = C^n D^n for all nx

[sinful nebula]

hmm

is it C^2 D^2 = D^2 C^2

[sinful nebula]

more precisely C^n D^n = D^n C^n

[south patrol]

Even simpler

And bear in mind that you need smth that breaks the previous bit lol

[sinful nebula]

ugh

i have no idea how you can have a matrix that does not commute but still follows for the case where n=2

[frigid lark]

Do matrices form a group under multiplication?

[sinful nebula]

non-singular ones

either a set of entirely singular ones or a set of entirely non singular ones?

[frigid lark]

Can you have two singular matrixes A, B where AB = 0, but BA is not 0?

[sonic coral]

i’ve managed to get it to work for n>=2 but not n=1

[sinful nebula]

no?

[south patrol]

That is what I meant Parrot aha

And yes, it is possible

[frigid lark]

Ahh, did I say too much?

[south patrol]

Eh seems foin

[sonic coral]

i don’t see how that works for the n=1 case though, bc that’s what I also came up with

oh wait

it’s C^n D^n not the other way around

[sinful nebula]

Surely if A and B were singular matrices with AB = 0 then A = 0B^{-1} = 0 so A = 0, then A is not singular

I have the matrices A = [0 1//0 0] and B = [1 0// 0 0] but just from brute force

[sonic coral]

that works

[sinful nebula]

Yeah but I don't understand why

[chilly ocean]

If f is ring homomorphism, then f(1)=1 isn't it?

[prime sundial]

i have that E/F is a field extension with degree n, and want to show that if d divides n, there is a unique field extension K, E/K/F. how do i start this?

does the tower rule guarantee the existence of such a field extension?

[royal cosmos]

Let V and W be quadratic spaces of dim 2 and discriminant -1. Show that V and W are isomorphic.

What do we think?

[thorn delta]

is the degree of K/F supposed to be d?

[prime sundial]

yes

though it really doesn't matter since if n = d * m, then the degree of K/F can be either d or m

ok i think this may not actually be that difficult if i just take an ordered basis of E and restrict it to the first d elements

and this should be unique up to reordering of the ordered basis of E

ok this may not be that trivial, unless i can somehow get 1 into the basis

[thorn delta]

well, you can always complete {1} to a basis of E/F, but the span of d elements of your basis need not be a field

[prime sundial]

yeah that's the issue i'm running into

1 is clearly a nonzero vector in the space so it can be a basis element, i just don't know how i'd go about making a basis for K from the basis for E

[prime sundial]

maybe there's something to be said for the injection map needing to be a homomorphism

i can maybe say E = F(\alpha) for some alpha of degree n, and let K = F(\alpha^(n/d))

ahhh that's not necessarily the case

[long nebula]

cursed cursed cursed I hate algebra

[prime sundial]

if i have an extension E/F of degree n, then does that mean there exists an irreducible polynomial with degree n in F[x]?

we're working specifically with finite fields

[frigid lark]

Yes

I think you can apply some counting argument

[prime sundial]

then if we let that polynomial be p(x), E must be the splitting field of p(x) right?

[frigid lark]

No

[prime sundial]

i guess i meant to say it's a splitting field for some polynomial of degree n

[long nebula]

No, not all extensions are splitting fields

[prime sundial]

or is that still not the case

argh ok thanks

[frigid lark]

It is

[long nebula]

(Normal extensions are splitting fields)

[frigid lark]

X^p^n - X

For finite fields

[long nebula]

Oh we're working only with finite fields

Okay yea every finite field looks like F_{p^n}, which is the splitting field of x^(p^n) - x, you're right

[frigid lark]

I can only say this statement is true for finite fields

[prime sundial]

i'm trying to show that if d divides n, then there is an intermediate field extension K of degree d over F. can i play around with the roots of that splitting polynomial somehow? or am i on the wrong track

[long nebula]

hmm do you know any galois theory

[prime sundial]

not yet

[long nebula]

I think you can do it by playing with the roots of the splitting polynomial and grouping them by the degree of their minimal polynomial

[prime sundial]

we also haven't done separability yet

[long nebula]

Have you learned yet that every finite field is isomorphic to F_{p^n}

Which is the splitting field over F_p of x^(p^n) - x

[prime sundial]

i think we have yes

yes this all sounds familiar

[long nebula]

Okay yea then you should (I think) be able to show that if d divides n, there's a factor of x^(p^n) - x which has degree d, and then think about the splitting field of that

I'm not 100% confident though

[prime sundial]

ugh this is rough

[oblique river]

what is this a corollary of

[long nebula]

just that finite fields look like F_{p^n}, are galois over F_p, etc

I can't remember the proof without galois theory because the proof with galois theory is simple 😭

[prime sundial]

so i can say the extension E is a splitting field for some p(x), so E = F(a_1, ..., a_r), p(a_i) = 0. this feels too general and like im not actually going to be able to construct K

[oblique river]

i mean the proof with real galois theory is even simpler

x^4 + 1 has galois group Z/2 x Z/2 over Q; its galois group modulo any prime p is a subquotient of that group

[long nebula]

Oh I meant the proof of what maximo is trying to do

[oblique river]

however galois groups over finite fields are always cyclic, therefore the galois group of x^4 + 1 has size < 4

oh

sorry lol thought you were still talking about that corollary

[long nebula]

hahahaha

[prime sundial]

kind of a side thought

but if E is the splitting field for p(x), can there be repeated roots in p(x)? as in, can the factorization over E have (x-a_1)(x-a_1)...

i don't see why not right away

[oblique river]

if p(x) is irreducible, then no (assuming that you're over finite fields still)

[prime sundial]

i am

and alright

so it splits into exactly n linear factors with n distinct roots?

[oblique river]

sorry i'm coming in late here but have you proven that the only finite fields are F_{p^n}?

[prime sundial]

i hate algebra

yes

[oblique river]

okay so then instead of working with some "abstract" finite fields, let's just limit ourselves to these

[prime sundial]

ok that's what eric suggested earlier

[oblique river]

The base field F is F_q for some q = p^k

the extension is E and it must therefore be F_(q^n)

[prime sundial]

is this n the same as the degree of the extension

[oblique river]

yes

[prime sundial]

ok

[oblique river]

now if d | n, what can you say about F_(q^d)?

[prime sundial]

i don't know. what should i be looking for? i'm sure we've talked about it in class but i'm quite weak with extensions and fields in general

[oblique river]

(if you want you can write this all in terms of p: F = F_(p^k), E = F_(p^(nk)), and K = F_(q^d) = F_(p^(dk)); maybe that looks better to you)

I mean, that is the field that you are trying to show exists

like, F only has a single field extension of degree d

so instead of trying to show that some random field extension of F of degree d lives inside of E

start with the only field extension of F of degree d that exists

the only facts that you really need here are that 1) the only finite fields are F_(p^k) and 2) F_(p^a) is contained in F_(p^b) iff a | b

it sounds like you've proven the first one already, and i guess the second one is sort of what you are trying to prove now but in more generality, but maybe you've already proven this special case

[prime sundial]

ok this is going to be a silly question

when we're saying it's contained, we're saying there's an isomorphism to a subfield?

they're modulo different things right

[oblique river]

not sure what you mean by "modulo different things"

up to isomorphism, there is exactly one field of size p^k; I'm saying that you should just identify all of those fields

[prime sundial]

so i associate F_p with Z/pZ, so saying F_p is contained in F_q makes me a lil uneasy

maybe i'm making it too specific

[oblique river]

no

F_q is not Z/qZ

[prime sundial]

ok that's the notation we use in my class

sorry for the confusion

[oblique river]

Z/qZ is not a field unless q is prime

above, I used q to represent a prime power, p^k

[prime sundial]

so by F_{p^n} we're literally just saying "the field with p^n elements"

and that's it

[oblique river]

yes

what else would it be?

Z/p^nZ isn't a field unless n = 1

[prime sundial]

ok i think i understand why F_{q^d} would be contained in F_^{q^n}

so $F \simeq F_{q}, E \simeq F_{q^n}$ and since $d | n, K = F_{q^d}\le F_{q^n}$
uniqueness is given for free too

[cloud walrus]

maximo

[oblique river]

that's right

[prime sundial]

ok that sounds good. thanks a ton

[oblique river]

np

i think the thing going forward here to keep in mind about finite fields, which feels a little strange is like

[prime sundial]

thank you eric, kxrider, and parrot tea as well

[oblique river]

because the theory of finite fields is so rigid

[prime sundial]

just stick to F_{p^n}?

[oblique river]

you can just work with concrete examples at all times

yeah

[prime sundial]

makes sense

[oblique river]

like you dont have to say "let E be a finite field" because then it's hard to get leverage

[prime sundial]

i kept jumping into vector spaces cause extensions, but this is simpler yeah

[oblique river]

you can say "let E = F_{p^n}"

and now you know lots of things, like exactly what its subfields and extensions are, and a minimal polynomial over F_p

[thorn delta]

I think ive missed the memo or something. were we given that E is a finite field?

[prime sundial]

not when i first posted the question

i missed that part, that's on me

sorry about that

another silly question, E = F_{q^n} because we have n basis elements and q options for the coefficients on each right

is that justification enough

[oblique river]

that's one way to think of it yeah

or just like, the degree of F_(p^b) over F_(p^a), assuming a | b, is b/a

which you can see using the tower law since both of those are extensions over F_p

[prime sundial]

im sure it could be done with (F_p^n)s again but that way seems succint

[oblique river]

yeah

[prime sundial]

oh tower law ok

yeah

[modest urchin]

hey guys i have a problem

Let $K$ be an extension over $F$ and let $G=\operatorname{Aut}(K / F)$ be its Galois group. Prove that $K$ is always Galois over $G^{\prime}$.

[cloud walrus]

Ray 永遠是炸銀絲卷有多好

[modest urchin]

G' is the correspond field of G in galois theory

to give this a prove, I want to claim that $G' = Aut(K/G')$ , then i stucked

[cloud walrus]

Ray 永遠是炸銀絲卷有多好

[formal ermine]

what is this supposed to mean

do you mean the fixed field of G?

[indigo ridge]

how would I be able to find the center of a group with the only given being its cayley or multiplications table?

[delicate orchid]

see which elements x satisfy xy = yx for all y in G

I believe this equivalent to the row of the cayley table corresponding to x being equal the transpose of the column corresponding to x

[rustic crown]

[long nebula]

I just wanna make sure I did this right: the example shows the formula doesn't hold because 6 ≠ 3(3)/1?

[indigo ridge]

im getting a minor thing wrong the problem is I cant pinpoint it, for some reasom I suspect it is (c) can anyone help me out which part did I get wrong?

[formal ermine]

yes

[long nebula]

ty illum

[long nebula]

part d

[indigo ridge]

what's wrong with it?

am I missing a number

[long nebula]

one of your answers doesn't work

[indigo ridge]

hm why is that?

[long nebula]

How did you get them?

[digital raptor]

Does the polynomial representation of a finite field form a polynomial ring?

[hot lake]

you wrote in part a) that x7³ = x5 then wrote in part d) that x7³ = x7

[long nebula]

A finite field is a quotient of a polynomial ring

[formal ermine]

wdym

[long nebula]

[\mathbb F_{p^n} \cong \mathbb F_p[x]/(x^{p^n}-x)]

[cloud walrus]

mollifiERIC

[indigo ridge]

I did it two ways, the first way is I broke it into cases case 1: where x^2 = e and case 2: where x^2 = x2. the second way i did it was brute force and chekced every single number x1, x2 e.t.c

[long nebula]

how did you get the two cases?

also yeah ^ what zef klop said

[indigo ridge]

I used the assumption that x^2 is either x2 or e for all elements of G

[digital raptor]

Is the quotient of a polynomial ring a polynomial ring?

[indigo ridge]

from the table

[long nebula]

Oh that's what you mean I see

And then how did you get that x7^3 = x7?

[formal ermine]

no

it's a splitting field in this case

[long nebula]

polynomial rings always have an infinite # of elements (unless they're trivial I guess)

[indigo ridge]

(x^2)(x) = x7 ===> (e)(x)=x7 ====> x=x7

[long nebula]

but x7^2 ≠ e

[indigo ridge]

yeah ik

[long nebula]

Oh I see what you're doing, that's a good strategy, but that only narrows down the possibilities; it doesn't tell you if those things actually are solutions

[digital raptor]

If we have a finite field of order p^n then the elements can be represented as polynomials of degree less than n with coefficients in GF(p) so it seemed like this polynomial representation formed a polynomial ring (considering it represents a field)

[long nebula]

So once you get x5 and x7, you have to check both to see if they actually fulfill the equation or not

[indigo ridge]

yeah I know it comes up with a silly result for part A

I did i chekced all by brute force

[formal ermine]

what is a polynomial ring for you?

for me it is R[x1, ..., xn]

[long nebula]

Yeah x7^3 ≠ x7, so you shouldn't include it in part d

[formal ermine]

also it means the splitting field of x^p^n - x over Fp, like eric already said

[digital raptor]

Honestly I thought it was just a ring where the elements are polynomials

[indigo ridge]

Ok gotcha

you were right @long nebula

thanks!

[formal ermine]

i.e. it's not "represented by polynomials", but the roots of a polynomial with coefficients in Fp

[long nebula]

no problem!

[long nebula]

I mean it can be represented by polynomials, it's just a quotient of the polynomial ring

[long nebula]

Usually polynomial ring means like R[x] or R[x1,x2,..,xn] or something

[digital raptor]

Which is necessarily infinite?

[long nebula]

Yeah, unless your ring is just zero

Since you have x, x^2, x^3, x^4, ...

So an infinite number of elements

[digital raptor]

Sorry to pester but why is it necessarily infinite? The polynomial representation of a field seems to form a ring and the elements are... polynomials

[hot lake]

then the operations are not the usual operations on polynomials

[frigid lark]

Equivalence classes of polynomials

It's a quotient ring

[long nebula]

^

You're identifying polynomials together

[hot lake]

um

[long nebula]

Wait did I say something dumb

[hot lake]

yeah to make a field you want to quotient by an irreducible polynomial

and you quotiented by the product of all of them

[long nebula]

Oh wait

Oops 🤦

[hot lake]

so the chinese remainder theorem says you get a big product of copies of Fp²

[south patrol]

oop

[oblique river]

And F_p!

[long nebula]

Okay technically my original statement was still true

But oops

[hot lake]

hmm ah yeah there are some Fp too

[digital raptor]

Okay thank you all

[warm shoal]

How is x^2H = H?

[hot lake]

but Fp² still is a splitting field of the thing

[warm shoal]

I get that H is the identity of G/H

[hot lake]

x² is in H yes ?

if h is in H, maybe hH = H

[warm shoal]

wow

is that true for only normal subgroups>

like when H normal G

[hot lake]

you need normality on the previous equality

(xH)² = x²H

[warm shoal]

why?

[hot lake]

because if H is not normal then G/H isn't a group so you shouldn't be talking about multiplying classes

but you can still kinda talk about (xH)(xH) = xHxH and you need normality again if you want to have the x and H commute to get x²H

[warm shoal]

oooooo gotcha

so when H is normal

it's true that

(Hx)^m = Hx^m = H

[hot lake]

(Hx)^m = Hx^m yeah

[warm shoal]

thank ya

[hot lake]

and that might also be free from the definition of the group operation in G/H

since usually you use normality there to argue that (xH) * (yH) = xyH is well-defined

[kind jacinth]

what?? how is the product a zero polynomial

what do they mean by that? p(x)*q(x) is 0??

also by "polynomials in Z12[x]" im assuming they mean the range can only be elements from 0 to 11 (so its 3+3x^3 mod 12)

[formal ermine]

yes

[kind jacinth]

oh ok i think this answers my question.. product is 12

which is 0

[formal ermine]

polynomial with coefficients in Z/12Z

[kind jacinth]

right?

[formal ermine]

yes all of the coefficients will be 12

hi powotato

[kind jacinth]

arent these cosets? Z/12Z is sth like {<12>, 1+<12>, ..., 11+<12>}?

[formal ermine]

ye

cyclic group of order 12

[south patrol]

Z/12Z = Z_12 to most people lol

[kind jacinth]

so its just the elements {0, ... 11} right?

[south patrol]

Viewing Z_12 as just {0,1, ..., 11} with certain operations is a bit hackier and i only see it used to introduce it when you've not done quotients

[formal ermine]

12 adic integers

[kind jacinth]

ah but {<12>, 1+<12>, ..., 11+<12>} = {0,...,11} right?

[south patrol]

Well

{[0],...,[11]}

if [a] is defined to mean a + 12Z ye

though when you are working with them you often just say liek 1 or smth

[kind jacinth]

alright thanks

[chilly radish]

Imagine completing wrt a non prime ideal

[gilded osprey]

Given a finite (potential) ring and the tables for its two operations, is there an easy way to verify distributivity to in fact show it is indeed a ring?

I ask since I am tasked with creating a field with 4 elements, and am starting by constructing a ring. From the tables it is clear to see commutativity, identities, and additive inverse, but am stuck on showing distributivity without just enumerating all combinations and explicitly showing it

[lethal dune]

I mean unless you know your operation is already distributive, I don't see any other way to show other than showing explicitly

though if you can show your operations are actually taken from a ring then you can just claim they are distributive as they are in the original ring

for example Z as a subring of Q satisfies the distributivity

[gilded osprey]

Yeah, basically what I've ended up doing is shortening how much I need to explicitly show. E.g. I don't need to show any expansions involving 0 such as aa=a(a+0)=aa+a0=aa is automatically going to be true
Also since addition is commutative, I only need to show one "direction" of an expansion such as aa=a(b+1) also shows for aa=a(1+b)
This actually shortened it down considerably

[long nebula]

where are we depending on the characteristic of F being not two?

[white oxide]

are there no units in R[x] where R is a ring except for 1 (assuming R is a ring with unity), since for any x^n we must have x^-n as an inverse

[long nebula]

If R is an integral domain, then we have deg (fg) = deg f + deg g, meaning that f can only have a multiplicative inverse if it's of degree zero, i.e. it's in R and invertible

[white oxide]

how do we know that deg(fg) = deg f + deg g

[long nebula]

I'm not sure about the case where R is not an integral domain

[white oxide]

oh wait

[long nebula]

If f = sum a_i x^i and g = sum b_i x^i, then a_(deg f) x^(deg f) * b_(deg g) x^(deg g) = a_(deg f)b_(deg g) x^(deg f + deg g) which is nonzero as long as we're working in an integral domain

[white oxide]

ig that's obvious

gotcha

thx!

[long nebula]

np, I'm not sure about the general case though when R can have zero divisors

[elder wave]

it's not true in that case

take polynomials s.t. the constants are inverse to each other in R and choose the other coefficients s.t. the multiplication yields zero

[long nebula]

Can you give an example

I was trying to think of one but couldn't 😭

[elder wave]

take some Z/nZ[X]

[long nebula]

yea

[elder wave]

non prime obviously

[long nebula]

yea

[lethal dune]

you are using to claim that (xi-xj) and (xj-xi) are indeed different

otherwise it's fixed by all transposition and hence by S_n itself

[long nebula]

ohhh

ty!

[lethal dune]

fun fact, D is a square or not in the field determines whether the Gal group of a cubic is S_3 or A_3

[long nebula]

Just learned that! :)

[south patrol]

Poggies

[formal ermine]

separable irreducible cubic*

🤓

[lethal dune]

sure

[kind jacinth]

why does this imply d is a zero polynomial?

jus because its smaller than d(x)?

[fleet pelican]

i think you're leaving important information out of your question

[barren sierra]

^^^

[kind jacinth]

sorry should send the whole proof

[barren sierra]

I think I know the proof

[kind jacinth]

its about gcd

[barren sierra]

but I want to make sure I have the right context in my head

so just send

[karmic moat]

same lmao

[barren sierra]

whole proof plz

[delicate orchid]

d(x) is chosen to be the smallest non-zero polynomial in S, you’ve found that b(x) is both in S and that it’s less than d(x) - to avoid a contradiction b(x) must be zero

[kind jacinth]

ahh right. r(x) > 1-a(x)r(x) and there is a negative sign

so clearly b(x) is smaller than d(x). alright thanks!

[delicate orchid]

The ordering is given by degree, so no that’s not the reason

You know the degree of b(x) must be less than d(x) by properties of the division algorithm

[void cosmos]

in ZxZ , the subring (a,a) is a subring that is not an ideal that serves as a counterexample for the statement: (a+I)(b+I) =ab+I is well defined over R/I where I is a subring

is that true

[south patrol]

Yeah

i think

[delicate orchid]

It’s not that it’s not well defined it just that it’s not equals lol. I think a = (1,0) b = (0,1) is a counter example

[void cosmos]

okay

i did another example

i think it was (5,3) (2,3) something like that

with a' and b' something similiar

a = a' and b=b' (as in relation)

but ab not a'b'

is the proof of two finite fields having same elements isomorphic

[delicate orchid]

I was thinking (2,0) and (1,2) if (1,0) (0,1) didn’t work

[delicate orchid]

What?

[void cosmos]

does it go like: just show that a finite field by lagrange theorem must be a splitting field for a poly (x^p^n-1)) hence by uniqueness of splitting fields this follows?

[formal ermine]

idk what you mean by lagrange theorem but yeah this is how it usually goes

[void cosmos]

lagrange: x^|G| = 1

x in G

[delicate orchid]

All elements of a finite field must satisfy that polynomial by Lagrange

[void cosmos]

so we have x^(p^n-1) = 1 which gives tit

okayy

<3> is maximal in Z[i] right?

or wait

fuck i cant remember how i did that lmfao

[formal ermine]

huh

it's x^p^n - x

over Fp

[formal ermine]

oh

[delicate orchid]

He missed the -1 out first time round dw about it

[void cosmos]

Z[i] is a pid so if i show 3 is prime then im done right?

[formal ermine]

the proof I was thinking of uses the frobenius endomorphism

[formal ermine]

oh lmao didn't see that

[void cosmos]

is this true boys and girls

sorry , im revising for an exam

[delicate orchid]

Perhaps it is ladies and gents

[void cosmos]

my answers

[formal ermine]

What

are you jerking off to math

[void cosmos]

no clue what ur saying

[agile burrow]

You could also check if x^2 + 1 is irreducible mod 3

[void cosmos]

it isnt so ig i got it correct

yea i did it using a norm argument

does this work?

[delicate orchid]

U could also simply know that primes in Z are primes in Z[i] if and only if they’re 3 mod 4

[void cosmos]

showing 3 is prime (irred) in Z[i]?

yo p important question

R={m/2^n | m in Z , n in N}

isnt this the localization of Z wrt 2??

or am i missing smth

the prof said no?

or maaybe she didnt read it correctly

[delicate orchid]

No

Where’s 1/6

[void cosmos]

wait why does it need to have 1/6?

[delicate orchid]

I’m assuming you mean the ideal generated by 2 here btw

[void cosmos]

no

or wait

[glossy crag]

Depends on what is meant by "wrt 2"

[void cosmos]

like im INVERTING 2

[agile burrow]

What actually is the terminology for when you invert {1, x, x^2, ...}

[delicate orchid]

Z(x) lol

So that’s Z(1/2)

[agile burrow]

Because when people say they localize at a prime, that means they invert every element in the complement of the prime ideal

[void cosmos]

yea it was a true or false problem and i said its true cuz of bijection between prime ideals(between ring and localization) but if R is a pid then S^-1R is a pid so prime -->maximal

and 3 is prime in Z soo

is this wrong

[agile burrow]

I'm confused

[delicate orchid]

The problem is that simply isn’t the localisation away from 2

[void cosmos]

well wouldnt this be the localization wrt to S with S = {1,2,4,..}?

[agile burrow]

Yes

[void cosmos]

so then this works?

[delicate orchid]

Yeah but is that what you mean by “respect to 2”?

[void cosmos]

maybe it was this shitty textbook i read

but that was the notation

[agile burrow]

I think I don't understand what your justification is justifying

[delicate orchid]

Fairs

[void cosmos]

localizaing wrt an element x is localizing the set S with {x,x^2,..}

[void cosmos]

there is a correspondonce between prime ideals here and there

by the natural map

so i just used that

• the fact that this is a pid cuz Z is a pid

so i get maximal

but ig this shouldnt been 3

it should have been [3]

the equiv cclass

[agile burrow]

You're asking if the ideal generated by 3 is maximal in Z localized at {1, 2, 4, ...}

[void cosmos]

ye

[agile burrow]

Ok, so I just didn't understand your question. Then yes, it suffices to show that (3) is prime in the localization, which it is because it is prime in Z and doesn't intersect the multiplicative subset

[void cosmos]

cool af

[solar glacier]

i should know this.. theres a rule with maximal ideals in PIDs i think lol im brain farting hard

[void cosmos]

i am doing good so far..

[void cosmos]

yea

maximal is prime and prime is maximal in PIDS

[solar glacier]

yes

[void cosmos]

but ofc god had to nerf me so i flunked the proof of first iso by swapping the order of the copmosition

instead of f o pi i did pi o f

yo please tell me q6 is false

with R being any ring with identity and S being any ring with no identity

(its loading)

[delicate orchid]

PauseChamp

Bro on that dial up modem

[void cosmos]

lol chill kiddo i play league on 60 ping

uk im fast

[delicate orchid]

I play league
That’s a block

[void cosmos]

6 true or false

sorry for the food markings

i was hungry

this shit is false af

right?

[delicate orchid]

I’m going to complain about the typesetting instead of being helpful

[void cosmos]

😠

[delicate orchid]

All rings have an identity trivial

[void cosmos]

Rank A+ in Garfield Kart - Furious Racing

[delicate orchid]

Yeah u jealous???

[void cosmos]

tf is this

is this a kids game

[delicate orchid]

……….. this is a MANS game

[void cosmos]

yea garfield

lmao

[delicate orchid]

Irregardless of your heresy, one direction is obvious, what did you do for the other?

[void cosmos]

the other is false?

my counterexample sa

was

A being having no identity and B having an identity

then AxB has no identity

but AxB/Ax{0}

has

[delicate orchid]

Very nice

[void cosmos]

cuz this is iso to B by first iso theorem

IT WORKS?

[delicate orchid]

I can’t see anything wrong with it

[void cosmos]

thats what she said

the prof.

hopefully it works

so i fucked up two questions , the subgroups of finite fields are cyclic ( saw it on andre weils basic number theory but just skipped it it was too boring ) and i mismatched the first iso theorem proof

yet i still proved the map was well-defined lmfao

hopefully i get any partial credit

[delicate orchid]

Wait I thought this was a practice test

[void cosmos]

no

it was a test i took

in algebra

it was to test me if i can get tested on the grad algebra

cuz i am not a maht major

i taught all this by myself so they dont know if im stupid or not

[formal ermine]

you have to take a test to be able to take a test?

[delicate orchid]

Ok as long as you’d submitted it before asking stuff here

[void cosmos]

yea employee of the month here

[formal ermine]

moamen do you major in engineering

[void cosmos]

yes

cs

cyber security

[formal ermine]

disgusting

[delicate orchid]

U know I got those accolades… STACKED

[void cosmos]

yea... nothign to say man it was

a middle esatern affair

do u know the steortype of asian parents

forcing their sons to be doctors?

same thing here

was this a long test?

it was for 2 hours

[delicate orchid]

Mid length

[void cosmos]

damn it took me exactly like 1 hour and a half

but ig alot of the proofs are standard like

u wont be too clever

[slender estuary]

Okay, let's uh. @delicate orchid

[delicate orchid]

What we doing here boss

[slender estuary]

Well, I was having trouble defining rings

right

Using very simple definitions

we need to start from a simple list, then explain what each means

[delicate orchid]

Ok I’ll go get a list of the ring axioms

[slender estuary]

Yea

[delicate orchid]

are you confused about any of them?

[slender estuary]

So, does the set

have to form a group under these operations?

• , *

because i've heard that

[delicate orchid]

these imply the set is a group under addition (hopefully you can see that these are just the properties of a group's operation)

the third point implies it's an abelian group under addition

[slender estuary]

Okay yeah that makes sense.

So a ring is a set that forms a group under addition.

simple terms

[delicate orchid]

yes

[slender estuary]

and, that's it?

Do we have more?

[fleet pelican]

it's usually abelian too

[delicate orchid]

well there are 2 other bullet points so yes

[slender estuary]

Multiplication is associative

[delicate orchid]

we have a second operation called "multiplication"
which satisfies the other bullet points

[slender estuary]

And distributes over addition

[delicate orchid]

yes

[slender estuary]

But, it does not form a group under multiplication, because multiplication is not commutative in this set?

[delicate orchid]

it doesn't form a group under multiplication because there aren't any inverses

[slender estuary]

So, what about a set that forms a group under addition and multiplication? That, is not a ring

yes?

[delicate orchid]

the rings you're looking at will also have a "multiplicative identity" called 1 that satisfies 1*x = x*1 = x
but there doesn't need to be an x^-1 such that x*x^-1 = 1 !!!

[delicate orchid]

if it's an abelian group under both addition and multiplication it's called a field, which is a type of ring

[slender estuary]

because it wouldn't exist under this set?

the 1?

PR

OR

[delicate orchid]

ok we're doing an example

[slender estuary]

Sorry, i think i know

wait

[delicate orchid]

take the integers Z = {...,-2,-1,0,1,2,...} these form a ring under the usual + and *

[slender estuary]

ok

sure

[delicate orchid]

but, can you find an element in Z, lets call it x, such that 2x = 1?

[slender estuary]

You can't

that's impossible

[delicate orchid]

exactly, so 2 doesn't have a multiplicative inverse! 2^-1 doesn't exist in Z

but this is ok, because we don't need 2^-1 for Z to be a ring

[slender estuary]

Oh right i know what you mean now.

You mean,
x^-1 such that x*x^-1 = 1

is implying that, if you dont have an inverse for every element then it does not satisify the group axiom

[delicate orchid]

yes, exactly

but just because it doesn't satisfy that axiom doesn't mean it's not a ring

(sometimes you even have rings without a "1" in them, so it doesn't even make sense to ask this question! But that's a different conversation)

[fleet pelican]

Z is a very good example of a "basic/generic" ring

[delicate orchid]

an initial example if you will ;)

[slender estuary]

So, it does form a group under addition for it to be a ring. -- this is a must
But, it does not have to form a group under multiplication, it's optional.

[delicate orchid]

yes, precisely

[slender estuary]

but, this isn't included in here.

[delicate orchid]

it just has to satisfy the ring axioms - which just so happen to imply it's an abelian group under addition

[slender estuary]

which is a little confusing to me

[delicate orchid]

the ones in red are exactly the axioms for a group

closure, associativity, identity, inverses

[slender estuary]

ye

[delicate orchid]

so it is included in there! Just a little bit hidden