groups-rings-fields

[latent anvil]

But since the degree of the minimal polynomial is 3, the extension has degree 3

[cloud walrus]

wstayman:

[latent anvil]

Yes, that's correct

You should look at your textbook or notes

Anyway, use the same logic for Q(5^(1/4))

[cloud walrus]

wstayman:

[latent anvil]

Yes

And so why can't i 5^(1/4) be in there?

[cloud walrus]

wstayman:

wstayman:

wstayman:

[latent anvil]

No

Why would you have that?

F in the case is Q

And K is Q(5^(1/4))

[chilly ocean]

That's what's written as a definition in my notes

[latent anvil]

What have you tried furbo?

[cloud walrus]

wstayman:

[red imp]

how do you pronounce "R/I"

in the context of quotient rings

[mild laurel]

R mod I

Or R quotient I

[red imp]

thanks

[shrewd halo]

Must a finitely generated group where the orders of the generators are finite, be finite? I could see it going both ways but I can’t even think of an example where the order of the group exceeds the product of the order of the generators so maybe it can’t even go above that

[upper pivot]

sure, take free product of like, Z/2Z and Z/2Z

(this is a counterexample, group is infinite here)

[shrewd halo]

Oh so just tautologically you consider words in 2 or more generators

That’s simple enough

[upper pivot]

yeah basically

[shrewd halo]

Then I guess the question remains for finite groups, does there necessarily exist a set of generators the product of whose orders are the order of the group or lower***

[upper pivot]

hmm not necessarily for non-abelian groups

[shrewd halo]

Yeah that was my thought too

[upper pivot]

the quaterion group should be a counterexample i think

[shrewd halo]

Just looked that up, you’re right. Good call

[chilly ocean]

I don't understand what R is and how to interpret R, I just know it is a commutative ring

Like for first bullet point, I already don't know why 1-1 imply isomorphism

[upper pivot]

do you know what a ring is?

[chilly ocean]

yea

is abelian group on + and • satisfy identity distributivity and associative

[cloud walrus]

JohnDoeSmith:

[chilly ocean]

how does injective imply surjective though

[upper pivot]

the important part is with its image

f is always surjective onto its image, by defination

[chilly ocean]

heres how he defined f

I don't understand it

like how its surjective

what if R is like the real numbers or something

[upper pivot]

we are not saying its surjective to R, but to its image, which is {n*1_R| n\in Z}

[chilly ocean]

ohh

For the second bullet point, the subring {n*1_R | n in Z} would actually be a finite set right?

because presumably 1_R = m*1_R + 1_R if I am correct

[upper pivot]

its not finite

oh sorry for the second

yeah

[chilly ocean]

cool

thanks

[upper pivot]

np

[chilly ocean]

for cyclic group like in this q, is it safe to assume d>0

like could I say wlog d>0 and why?

I sort of think you can but I just want a concise way to explain

[smoky cypress]

0 doesn't divide any number other than 0

[chilly ocean]

nvm

i guess "order" is defined to be positive

[smoky cypress]

n can't be 0 because a group can't have 0 elements: it needs to have an identity

Yeah order is always positive

[chilly ocean]

is there always a generator in any group?

and if so,

what if there is more than 1 generator

[smoky cypress]

If a group is generated with 1 element then it's called a cyclic group

Klein four group is the smallest example of a group that is not cyclic

So not all groups have a single generator

[chilly ocean]

is klein 4 Z/4Z?

[smoky cypress]

No

Z/nZ is the cyclic group of order n: any cyclic group of order n is isomorphic to Z/nZ

[chilly ocean]

alright, Z/nZ is not cyclic though I think, since for instance Z/6Z has generator [1]6 and [5]6

[smoky cypress]

Um

Z/nZ is always generated by 1

[chilly ocean]

yea though 5 also generates

if its +

[smoky cypress]

Well, the definition says that it's generated by a single element

It doesn't mean that it's generated by only that element

[chilly ocean]

oh, it doesn't say if the element is unique

i see

[smoky cypress]

Yeah

[chilly ocean]

i misinterpreted the def

[smoky cypress]

In fact, the cyclic group of order n always have other elements that generates it

[chilly ocean]

ok i understand now

thanks

Z/pZ doesn't though

[smoky cypress]

Um

[chilly ocean]

wait

nvm

theres 2

[smoky cypress]

Every non 0 element in Z/pZ generates it

[chilly ocean]

oh right

..

[smoky cypress]

By lagrange's theorem

For the cyclic group C_n generated by x, x^i will generate C_n if and only if gcd(i,n)=1

[chilly ocean]

ic

[shrewd halo]

How does your book define normal?

[spice bay]

@graceful peak which part are you stuck on?

For the forward direction, you need to prove that the group axioms hold. For the reverse direction you should think through what it means for a product of right cosets to be a right coset, and then think about what it means for a right coset to have an inverse.

[shy bluff]

Ok so this is alittle bit long but I have these 2 pages

In the 2nd image, they say "Start with 1; reading the cycles from right to left, find the first cycles that moves 1 (1 -> 7)"

What do they mean by "Find the first cycle that moves 1 (1 -> 7)?

[spice bay]

Permutation notation is always a pain and is not standard across the board, which leads to this type of confusion

The idea is to take the permutation (1 4 3 6)(2 8 7)(2 4 8 5)(1 7) and "apply" it to 1

What this means is that you start at the right, and find the first cycle containing 1, in this case it is (1 7)

So you know that 1 is sent to 7, but we're not done with the permutation. Now we need to take (1 4 3 6)(2 8 7)(2 4 8 5) and see where it sends 7

(2 4 8 5) doesn't contain a 7, so we can skip it. Then we get to (2 8 7), which sends 7 to 2

Finally we need to see where (1 4 3 6) sends 2, but it doesn't contain 2 so it sends 2 to itself

Therefore 1 gets send to 2.

One way to think of this type of permutation notation is like function composition. You think of each cycle as a function, and the permutation written out as their composite. Then in order to apply it to an element, you read the functions from right to left. Hope this helps! 🙂

[shy bluff]

Um

What do you mean by "apply" it to 1?

[spice bay]

See where it sends 1

[shy bluff]

where (sigma)(tau) sends 1?

[spice bay]

Overall yeah that's what you're trying to find, but you can decompose (sigma)(tau) as a product of cycles (which I'm saying you can think of as a composite of functions) and then see where 1 is sent by starting with the rightmost cycle

[shy bluff]

So our rightmost cycle in the expression (sigma)(tau) is (1 7)

[spice bay]

Yeah

[shy bluff]

Err

I get how to go and do the first part where you treat it as a compositiont of sigma and tau

Like sigma(tau(x)) and get a new table and then make a new cycle from that

But I don't understand how they're getting it directly from the cycle notations

[spice bay]

In step (1), they write the cycle notation for sigma, followed by the cycle notation for tau

But this needs to be simplified, and the way you can do this is how I described - start with 1 and see where it is sent, by reading the cycles from right to left

[shy bluff]

So we see that 1 is sent to 7

And then we walk to the left and look for the next cycle that affects 7?

[spice bay]

Yeah exactly!

[shy bluff]

Which we see to be 2?

[spice bay]

Yeah therefore (sigma)(tau) sends 1 to 2

[shy bluff]

And then we see that if we go further left there are no more cycles that affect 2 so we conclude that (sigma)(tau)(1) = 2

Oh ok I think I see it now

And you continue that to get the one line notation

And from the one line notation you get the cycle notation for (sigma)(tau)

So they're not directly getting the cycle notation for (sigma)(tau) from the 2 cycle notations written side by side

[chilly ocean]

https://youtu.be/MpKG6FmcIHk this video explains the notation very well imo

Very clearly explains how to read it

[shy bluff]

Wait but is what I said correct? that they're not getting the cycle notation directly just by writing it side by side, but that they're writing it side by side to generate the one-line notation to then go and get th ecycle?

[chilly ocean]

I’m not sure what you mean with that

[shy bluff]

Like they're not directly going from (1 4 3 6)(2 8 7)(2 4 8 5)(1 7) to (1 2 3 6)(4 7)(5 8)

[spice bay]

What do you mean by directly though?

[shy bluff]

like you look at the former and get the latter

without any intermediary steps

[chilly ocean]

I don’t think it’s as simple as looking at it and knowing, no

[shy bluff]

ok yea because thats' what I htought that they were doing lol

[spice bay]

You can think about these cycles as certain elements of the symmetric group, and in this case (1 4 3 6)(2 8 7)(2 4 8 5)(1 7) = (1 2 3 6)(4 7)(5 8) is a valid expression in the group

It's not easy to simplify these things without working through where elements go though

[shy bluff]

Ah

Ok

Yea that's what I thought

[chilly ocean]

The video I sent has a good algorithm for simplifying those I think

[shy bluff]

I thought that they had some sort of magic that let you easily go through it without working it out

Yea I'm watching it right now

[spice bay]

There are some easy cases, for example (1 2)(3 4) = (3 4)(2 1) because the cycles are disjoint

And you can do stuff like (1 2)(2 1) = id is the identity element, fixing every element

[shy bluff]

oh tht's pretty cool

this is some uh interseting stuff lol

[spice bay]

Yeah! Glad you like it 🙂

[chilly ocean]

I’ve always wanted to give this tip

[shy bluff]

this is uh my introduction to groups and rings class lol

[chilly ocean]

Read cycle notation like a manga

[shy bluff]

I

ok

[spice bay]

lmao @chilly ocean I love that

[shy bluff]

[wise grove]

How do you like algebra so far @shy bluff ?

[chilly ocean]

I've been struggling with the following problem for a while now

Let $f \in \mathbb{F}_p[x]$ be a polynomial of degree $ d \geq 2$ and consider the Froebenius map on the quotient ring $R := \mathbb{F}_p[x]/(f)$; $\Phi(g) = g^p$

[cloud walrus]

Azu:

[chilly ocean]

Assume $f$ is separable. Prove that $f$ is irreducible if and only if $rank(\Phi - id_R) = d -1$

[cloud walrus]

Azu:

[chilly ocean]

I have already managed to do the forward direction, assuming f is irreducible. I have no idea how to do the backwards direction though

[shy bluff]

@wise grove fun so far

[kindred mist]

I think I have a proof that every unital PID has a lattice of ideals just like Z’s (I is contained in J iff J’s generator divides I’s generator). I didn’t want to spam the chat with the proof so could someone just tell me if that statement is correct?

[steady axle]

@chilly ocean your problem is interesting. Can you tell me how did u get the forward implication?

[chilly ocean]

@steady axle assuming $f$ is irreducible, let $\alpha$ be a root of $f$, so that $R$ is isomorphic to $\mathbb{F}_p(\alpha)$; we do all work in the simple extension for brevity

[cloud walrus]

Azu:

[steady axle]

Actually I did get till here but could not proceed hereafter

[bright agate]

Hi guys who wanna help me?

[chilly ocean]

[bright agate]

@chilly ocean in questions y

[chilly ocean]

that is definitely not abstract algebra lol

[bright agate]

Well i don't know what the hell abstract algebra is sorry...@chilly ocean

[chilly ocean]

you should read the channel description then

@steady axle continuing, we know the Froebenius endomorphism fixes the base field, so the rank(blah) is less than or equal to d -1

Consider an arbitrary polynomial in $\alpha$ of degree $ 1 \leq n < d$

[cloud walrus]

Azu:

[chilly ocean]

Call this polynomial $g(\alpha)$

[cloud walrus]

Azu:

[steady axle]

@chilly ocean yes I am with u

[chilly ocean]

I think there may have been an error in my earlier proof, I'm re-doing it right now

[steady axle]

K

[chilly ocean]

sorry lol

@steady axle I think I got it

We write $g$ explicitly; $g(\alpha) = c_n\alpha^n + c_{n-1}\alpha^{n-1} + ... + c_1\alpha + c_0$

[cloud walrus]

Azu:

[chilly ocean]

Suppose $\Phi$ fixes $g$; then $(\Phi - id)(g) = c_n\alpha^n(\alpha^{(p-1)n}-1) + c_{n-1}\alpha^{n-1}(\alpha^{(p-1)(n-1)-1} + ... + c_1\alpha(\alpha^{p-1} -1) = 0$

[cloud walrus]

Azu:

[chilly ocean]

The monicized version of $f$ must divide this polynomial

[cloud walrus]

Azu:

[chilly ocean]

That polynomial has degree $d$, so we have that $d|(pn)$ since the degree of the above polynomial is $pn$

[cloud walrus]

Azu:

[chilly ocean]

um

I don't know how to proceed from here, actually

I want to prove $d|(p-1)n$ or $d|(p+1)n$

[cloud walrus]

Azu:

[red imp]

for which rings $R$ is it true that $R={0}\sqcup U \sqcup Z$ where $U$ are the units and $Z$ are the zero divisors

[cloud walrus]

Bobbicals:

[latent anvil]

Sounds kind of hard

This doesn't imply localness, take Z/2Z × Z/2Z

Minimal prime ideals consist of zero divisors, but I don't think that's helpful here

The set of zero divisors (plus zero) will be closed under multiplication, but maybe not addition

If R is an integral domain then it's a field, but I don't think this property is preserved when quotienting so I'm not sure if prime implies maximal

as I said above, minimal prime ideals are only zero divisors, so if the dimension of R is 0 then maximal => minimal and so any element of R which isn't a unit is contained in a minimal prime ideal, and thus is a unit

Any artinian ring satisfies this I think?

[chilly ocean]

@steady axle following up: $\alpha-1$ is a common factor of all the terms in that polynomial, so we can divide it out and the leading term will have degree pn -1, and $\alpha$ will still be a root, so $d|pn-1$ but this implies that $d|1 \implies d = 1$ since $d|pn$

[cloud walrus]

Azu:

[chilly ocean]

however, $d \geq 2$

[cloud walrus]

Azu:

[latent anvil]

is this for the irreducible => rank is d-1 part?

[chilly ocean]

yes

[latent anvil]

Do you know the classification of finite fields?

[chilly ocean]

yeah, they're all isomorphic to F_p^n for some n

[latent anvil]

actually you don't need the classification for what I had in mind

since f is irreducible, the quotient ring is an integral domain

thus x^p - x has at most p roots

so the kernel is dimension 1, and the image is dimension d - 1 by rank nullity

[chilly ocean]

oh

is my proof above wrong?

I think it is

[latent anvil]

I didn't look at it, it just seemed long 😅

I can take a look

[chilly ocean]

lmao it should have been d <= pn and d <= pn -1

not divides

[latent anvil]

Yeah I was about to say

[chilly ocean]

what the f*** made me make that trivial mistake?!

[latent anvil]

lol

It happens

[chilly ocean]

@latent anvil so, how do you jump from x^p -x having at most p roots to the kernel having dimension 1?

[latent anvil]

The set of roots is exactly the kernel

And so the kernel has size <= p

We know it's contains Fp also, so the kernel is Fp

Fp is 1 dimensional

[chilly ocean]

ahhh I see

Thank you!

ah, you omitted writing phi and id

[latent anvil]

Yeah sorry

[chilly ocean]

It's ok

[chilly ocean]

is sending Z to Z/mZ an example of ii)

[latent anvil]

Yes

[chilly ocean]

ty

[red imp]

thanks for the response @latent anvil, it wasn't a problem that I was given but while working through an example my teacher assumed that if a nonzero element is not a unit then it is a zero divisors

and I was like "what the heck" so I thought there might be some nice rule to tell you when you can use it

the ring in question was the set of upper triangular 2x2 matrices with entries from Z_2

[latent anvil]

Oh well that is actually easy to see

Your condition is true for any finite ring

If r is an element of R, r is not a zero divisor (or zero) iff the multiplication by r function R -> R is injective

It's a unit iff that function is bijective

For a finite set R, a function R -> R is injective iff it is bijective

[red imp]

:O thanks

[steady axle]

@chilly ocean we still have to prove the other way right?

[chilly ocean]

@steady axle not anymore

[steady axle]

@chilly ocean how

[chilly ocean]

Somebody else gave me the answer, I can post it if you want

[steady axle]

yeah I saw that answer but it answers only the forward direction right?

[chilly ocean]

No, the other direction was on a different server lol

but it uses the Chinese Remainder Theorem

[steady axle]

oh ok. you can post it here

btw which server is that if i may ask

[chilly ocean]

Do you want an invite?

[steady axle]

that will do

[chilly ocean]

it's another math server

i need to add you first

[latent anvil]

Suppose I have single variable irreducible integer polynomials f1(t),...,fn(t) (which aren't unit multiple of one another). Is the ideal (f1(x1),...,fn(xn)) of Z[x1,...,xn] prime?

[latent anvil]

You have a map from Z[x1,...,xn] into C sending xi to a root of fi, but it's not clear to me that this has the right kernel

[chilly ocean]

for part 3

what does it mean to be associative and closed

f(a) in H, f(b) in H, then f(ab) in H or f(a)f(b) in H?

and for associative, isn't it just inherited from the fact that the entire H is associative

[sour plume]

Yeah, associativity is inherited; that's why some people don't include associativity as something you need to check separately in the definition of a subgroup

[solemn rain]

if x and y is in H ---> xy is in H

this means closed undder the opepration

[chilly ocean]

ok

so do i need to show closure?

to show f(K) is a subgroup

[sour plume]

Yeah

[chilly ocean]

so i assume I use homomorphism property there

[sour plume]

sounds güd

in particular because you have nothing else that makes f special except this property

[chilly ocean]

ok

ty

[chilly ocean]

what part of this problem explains why bijective is needed

i tried to prove this and i never used bijective

[mild laurel]

an inverse function doesn't exist unless its bijective?

[scarlet estuary]

f needs to be bijective for f^-1 to exist

[chilly ocean]

ok

but

the inverse function is defined, so what part of not being bijective contradicts f^-1(f(g)) = g?

[scarlet estuary]

the inverse doesnt exist

if f isnt bijective

[mild laurel]

But its not well-defined unless f is bijective

[scarlet estuary]

a function is invertible if and only if it is bijective

nothing fancy about groups here, this applies to all functions.

[mild laurel]

i.e., if you have g and g' \in G such that f(g) = f(g') = h, then what is f^{-1}(h)

[chilly ocean]

@scarlet estuary I understand this concept, but im just hypothetically saying suppose f^-1 is a function is defined this way, without calling it an inverse, and I am asking why it cannot exist

[mild laurel]

is it f^{-1}(f(g)) = g or f^{-1}(f(g')) = g'?

assuming that g \neq g'

[scarlet estuary]

zoph gave a counterexample in the case that f isnt injective

if f isnt surjective then

there exists a y in the codomain such that y is not equal to f(x) for any x

so what is f^{-1}(y)?

[chilly ocean]

oh ok

i see

yea thanks

[scarlet estuary]

anyway there's a billion proofs of "bijective iff invertible" online

you can just google them

[chilly ocean]

yea i know that, its just when I proved this I didn't use anything about bijective

so I thought it was weird

[scarlet estuary]

you're implicitly invoking bijective-ness when you start talking about the inverse

if you have "the inverse exists", then yeah, thats all you need

as far as "applications of bijectivity to this question"

[chilly ocean]

the problem already uses the bijective assumption right

[scarlet estuary]

yep, in its statement

[chilly ocean]

k, i understand then

ty

[chilly ocean]

what does the triangle symbol mean

and G/N

I'm just curious, I don't think I can solve it

means N is a normal subgroup of G and G/N is a quotient group

ah ok

[undone orbit]

@graceful peak I think the answer should be that yes, G is finitely generated

Pick a finite set of generators for N, say n_1, ..., n_t and h_1N, ..., h_sN generators of G/N. Now each element g of G lies in some coset of N which can be written as v(h_1, ..., h_s)N for some word v

in other words g = v(h_1, ..., h_s)n for some n in N, using the generators n = w(n_1, ..., n_t) for some word w, thus g = v(h_1, ..., h_s)w(n_1, ..., n_t)

so, unless I'm missing something, this would mean that G is generated by n_1, ..., n_t, h_1, ..., h_s

[hot bridge]

I need help for questions

[shrewd halo]

@hot bridge do you understand the question?

[hot bridge]

Am I created this permutation

will it be random

i dont understand

[shrewd halo]

Do you know about cycle notation?

[hot bridge]

yes i know cyle rotation

for examle (132) means 1-->3 3-->2 and 2-->1

but i dont understand this question What should we do ?

[shrewd halo]

Can you think of two other permutations that, when composed, produce this permutation such that one of them has order 3 and the other has order 4?

[hot bridge]

When I write two cylcle with 4 and 3 order will it give all of f ?

[shy bluff]

So the set Fun(X, X) is all functions that take in 2 elements from X?

[errant sierra]

I need to find how many homeomorphisms there are of Z into Z10, can someone tell me how to start this problem?

[solemn rain]

@errant sierra

how would any homeomoprhism look like

notice that both are cyclic

[errant sierra]

we would need to take a look at the generators of Z and divisors of 10 right @solemn rain

[solemn rain]

@shy bluff why 2 elements?

Fun(X,X) is the set of functions from X--->X

Z = <1>

[errant sierra]

-1 and 1 generate Z and the divisors or 10 are 1,2,5,10

[shy bluff]

Oh

[solemn rain]

not the divisors

[shy bluff]

Ok I see

So um

[solemn rain]

the generatos

Z_10 = <7>

^ verify this later

[shy bluff]

Is Fun(X, X) basically like... "Name of set (Inputs, outputs)"?

[solemn rain]

( i am using Z_10 = Z/10Z )

[shy bluff]

So like if it were X x X -> X then it'd be Fun(X x X, X)?

[solemn rain]

@shy bluff Fun(X,X) = { f | f is a function from X to X }

[shy bluff]

Oh ok

Thank you!

[solemn rain]

np

so for a homomorphism

f: Z---> Z_10

f(1^k) = 7^j

thats for any function since any element in Z is 1^k for some integer k and any element in Z_10 is 7^j for some integer j

now notice that since f is as homomorphism u need |x| = |f(x)|

so u need to map generators to generators

now try to find ur homomoprhisms using all this

[cloud walrus]

slimvesus:

[solemn rain]

idk if thats well defined

or if its a homomorphism

[cloud walrus]

slimvesus:

[smoky cypress]

Well for a homomorphism $\varphi:\bZ\to\bZ/10\bZ$, the map is uniquely defined by the value of $\varphi(1)$.

[cloud walrus]

Whoever:

[smoky cypress]

lol

and there are 10 possible values for phi(1), so there are 1 different maps

[shy bluff]

@solemn rain Does this proof look about right?

[solemn rain]

yea

[shy bluff]

Okie

Thank you!

[solemn rain]

np

[errant sierra]

thank you guys

[solemn rain]

np

[smoky cypress]

btw

In your original question

You asked for a "homeomorphism", which is different from a "homomorphism"

homeomorphism refers to something in topology, and not algebra

[shy bluff]

Uh @solemn rain how would I prove that the identity function Id_X is an identity for o?

I assume that the idnetity functino Id_x is a function such that

Id_x(x) = x for any element x in X

but what do they mean by "Id_x is an identity for o"

[solemn rain]

is an identity under composition

f(id(x)) = id(f(x))) = f(x)

for any f

for all x

[shy bluff]

Ooh

Ok

Thank you

[solemn rain]

np

[shy bluff]

I was going to go and d osomething like id((x o y)(a)) = (x o y)(id(a)) = (x o y)(a)

For any x, y in Fun(X, X) and a in X

[solemn rain]

just show id(f) = f(id) = f

for any f in Fun(X,X)

[shy bluff]

Dose this make any sense?

[solemn rain]

yea u also need surjection but okay

[shy bluff]

I'm trying to proev that a function f is left-invertible if and only if it's injective

So uh

Surjection is for right-invertibliity

[solemn rain]

yea mb

[shy bluff]

But uh this looks right it's proving that it's left-invertible if injective?

[solemn rain]

yea

[shy bluff]

but this doesnt' work for empty set case?

[solemn rain]

what empty set case

[shy bluff]

if X is empty

Then there's no x_1, x_2, or y

Err

The entire proof doesnt' seem to really work?

[solemn rain]

if X is empty then whats in Func(X,X) ?

[shy bluff]

Yea that's what I mean

Oh wait

Actually yea

That makes sence

You can't have Func(X, X) be empty

[solemn rain]

if X is empty

[shy bluff]

Or rather, if X is empty

[solemn rain]

the only possible function is x-->x

[shy bluff]

Then Fun(X, X) is empty

Right?

[solemn rain]

were x is in the empty set

no

x-->x

[shy bluff]

Identity function?

[solemn rain]

yea

its well defined

[shy bluff]

Ah ok

Oooh ok I see

Thank you!

[solemn rain]

its bijective

if H and K are subgroups of a group G

H intersects K is a subgroup of G

if H and K are of order m and n respectively

what do u think would be the order of H intersects K

( think of lagranges theorem )

@graceful peak

H intersects K is a subgroup of G

also a subgroup of H

and also a subgroup of K

the order of H intersects K must divide H and K

if H has order n and K has order m

then order of H intersects K must be lcm(m,n)

[hot bridge]

so Ha intersect Hb will be have 6 element

because H=Ha=Hb

?

@solemn rain

[solemn rain]

no

Ha=H iff a in H

a is not necessarly in H

[hot bridge]

but Ha is subgroup

it must be ncluded inverses element

then Ha =H

but i dont understand too

what did you find the answer @solemn rain

this question dont want to A intersection B

we have to find Ha intersection Hb

[solemn rain]

Ha and Hb are subgroups of G of order 6

[hot bridge]

yes

then

?

[chilly ocean]

does there exist elements in certain groups such that a*b = a, where b is not the identity?

[solemn rain]

ab=a ---> a^-1ab=a^-1a --> b=a^-1a = 1

@chilly ocean

[chilly ocean]

oops

i mean in rings

hm i think in rings you can do this right?

ab - a = 0

then use distributivity

[mild laurel]

2 * 3 = 2 in Z/4Z

[chilly ocean]

oh right

a cant be a 0-divisor

thanks

[mild laurel]

Uh what

2 is a 0 divisor here

[chilly ocean]

no what i mean is

a can't be a 0 divisor otherwise things like 2*3 = 2 can happen

[mild laurel]

Yeah, for this to be possible, a must be a zero divisor, since ab - a = 0 implies that a(b -1) = 0 like you described

[shy bluff]

Ok so part c) here

I did part a and part b

And I'm somewhat? stuck on part c)

So far what I have is that, for the left-invertible case

• Let f be an element of Fun(X, X) which is left-invertible.
• By b) we know that if it is left-invertible it is injective

What I think that I have to do is show that it is not necessarily surjective

And then show that there must be an element of X that f does not map to?

Does that sound about right?

[solemn rain]

yea

@shy bluff

[shy bluff]

oh but it has to be an arbitrary f

hrm

I'll think on this some more and try some stuff in the morning

thank you for confirming that I'm heading in the right direction though!

[solemn rain]

np!

[shy bluff]

Hrm what about the function x^2

Since X is infinite, x^2 is in X

It's uh injective but not surjective I think

[mild laurel]

That makes no sense

What if X is the set {cats, dogs, 1,2,3,4,5,6,....}

what's cats^2

[shy bluff]

Whatever your group operation is

Maybe it is too late to think about this

[mild laurel]

X isn't a group

[shy bluff]

Oh this is true

[chilly ocean]

anyone enlighten me on how to approach this set of problem

I know what all of them means, I just cant connect together

[mild laurel]

just play around with it

take some ring with characteristic 0 like Z, take some ring with characteristic p like Z/pZ

see if you can do the things they describe

[chilly ocean]

ok, ill try

[mild laurel]

formulate hypotheses

try to prove your hypotheses

[latent anvil]

improvise, adapt, overcome

[chilly ocean]

if char(R) = p is R finite

[mild laurel]

No

Uh, maybe the easiest example is taking (Z/pZ) and taking an infinite direct product of it with itself

[chilly ocean]

what do you mean though? I thought in Z/pZ there was always p elements

[mild laurel]

Yeah, but like, you know what (Z/pZ) x (Z/pZ) means right

[chilly ocean]

Oh

i see

[mild laurel]

Oh

easier example is just

F_p[x]

[latent anvil]

lol

[mild laurel]

oh well

[chilly ocean]

whats the notation mean

?

[mild laurel]

Polynomials with coefficients in Z/pZ

[chilly ocean]

oh ok

no, for clarification in F_p[x] how do you generate all elements?

[latent anvil]

what do you mean?

[chilly ocean]

Like the way they define char(R) is related to f:Z->R and whether its 1-1 or smallest element that is in kernel

if R was F_p[x], how do you construct something like x^2+x+1

[latent anvil]

I'm having trouble understanding what you mean by construct

If R is F_p[x], then x^2 + x + 1 is an element of R

[chilly ocean]

oh, it doesn't have to be surjective

ok my bad

[latent anvil]

ah I see yeah

it will only be surjective for a small number of rings

Z and Z/nZ for any positive integer n

[chilly ocean]

actually i heard of power set, but not power ring

[latent anvil]

yeah you can put a ring structure on the powerset of any set

it'll have characteristic 2

don't worry about it though

[chilly ocean]

I mean so far i did iii) and iv)

nvm

i only did iv) which is f:Z/pZ->Z is a homomorphism

or it may not be..

like I don't even know what are the typical ring homomorphism

ok, I know f:Z->Z/pZ is ring homomorphism

so iii) is true ii) is false, leaving i) and iv)

Ok, I cannot think of how to find out if i) and iv) is true

1 looks true but I don't know how to prove, and 4 looks false but then same problem

can't prove

[mild laurel]

Well, if you think its false then you need to come up with a counterexample

[chilly ocean]

No

for iv) it says there can exist

but then I'll have to prove there can't exist

thats why i could do ii) and iii)

[latent anvil]

bodied zoph

LHC, what have you tried for iv?

like, suppose you have a map f : S -> R?

why does this feel wrong?

[chilly ocean]

i don't know what to try

I want to relate Z-> R to R->S

but how do you do that

Z -> R is 1-1 is what i know

R must be infinite, bu then as shown from F_p[x] this can also be infinite

[latent anvil]

so there's two things I think would be useful here

the first is to think about the more concrete definition of characteristic

the characteristic of R is the order of 1 in the additive group of R

the second is that there's only one map Z -> R, so Z -> R must be the composition of the maps Z -> S and S -> R (if you have a map S -> R)

[chilly ocean]

ohh

now it makes more sense

thanks for reminding me of the unique map thing I forgot about it

but how do you show the map is unique, our prof didn't show it

he just stated it as proposition

[latent anvil]

prove it

[chilly ocean]

ok

[vale coral]

Say that i want to show that the fifth cyclotomic polynomial is irreducible in F_2(X). I am given that x^2+x+1 is the only irreducible polynomial in F_2[x]. I have prooved this by proving that [p] generates the group (Z/nZ)*, where n is 5 and p is 2 in this case.

However, can it be proven by checking if x^2+x+1 | fith cyclotomic polynomial? If so why / why not?

[shy bluff]

@mild laurel what if I have that f is some function that takes the nth element of x, and maps it to the 2n-th element of X

Or does that also not work because it assumes ordering of your set

But uh assuming that a set is ordered I believe that that is injective but not surjective

[shy bluff]

Like let f:X -> X be the function that maps f(x_n) = x_{2n}, whereby n denotes its position in the set X

Since X is infinite 2n is defined and an element of X

And this is injective because f(x_i) = x_{2i} = f(x_j) would require i = j

[chilly ocean]

how to prove i)?

[brisk granite]

map everything to 0?

[chilly ocean]

the issue with that I think is f(1_R) = 1_S and 1_s ≠ 0_s

please I need help, this is due very soon

I just realized i) is false,
can anyone confirm, because I used the example f:Q->Z/pZ
and f(1/3+1/3+1/3) ≠ f(1/3)+f(1/3)+f(1/3)

[chilly ocean]

solved

[steady axle]

does this follow by definition of injective object and considering cokernel of d_i-1

[steady axle]

@vale coral you mean that X2+x+1 is the only irreducible polynomial of degree 2

Consider the fifth cyclotomic poly. If it is reducible what should be the degrees of it's factors?

[vale coral]

Let us name the fifth cyclotomic poly q. Deg(q) = 4, so then say f,g are factors of q then deg(q) = deg(f)+deg(q) right? @steady axle

[steady axle]

right. but you can say more. do you know what the polynomial is?

what is your definition for nth cyclotomic polynomial

[steep hull]

Also degree of product is sum of degrees

[smoky cypress]

wrong channel

put it in #prealg-and-algebra or #precalculus

[vale coral]

@steady axleMy definition is: \prod_{1\leq k \leq n, gcd(k,n) = 1} (X-e^(2\pi i k/n))
So the fifth poly becomes x^4 + x^3 + x^2 + x + 1

[latent anvil]

me too thanks

do you have, like, a question?

[tiny yoke]

can i make it like z5xz5?

[latent anvil]

no

for example, the coset of (1,0) has infinite order

[woven delta]

yeah the rank of ZxZ/H has to be exactly 1 by tensoring with R and applying rank nullity lmao

and then you just have to compute the torsion

[latent anvil]

You can just write down a basis

but I didn't think op would understand that

[vale coral]

If I have the congruence x^3-x+1 ≡ 2 (mod 5) and I wanna rewrite it to a congruence of the form x ≡ y (mod 5) for some y \in Z. How do I get rid of the exponents. I understand the case where there is a coefficient e.g. the congruence was 2x ≡ 3 (mod 7, here I would just have to multiply by the inverse of 2 mod 7. If someone could explain for the exponentation part it would be amazing.

[woven delta]

hmm I can only think of very ugly ways to do this

but also you can't necessarily do this obviously

[scarlet estuary]

yeah theres no general method here

and i cant see of an immediate path in this case

[latent anvil]

The lhs factors as (x + 2) (x^2 + 3 x + 3)

Seems worth mentioning

[woven delta]

I think you want to look at x^3-x-1 \cong 0 mod 5 though

[latent anvil]

Oh right that still factors

[woven delta]

x-2 is a factor of that

[latent anvil]

(x + 3) (x^2 + 2 x + 3)

Yeah

[woven delta]

lol

[scarlet estuary]

yeah thats the only thing i could see but fuck modular factoring

[latent anvil]

lol

[vale coral]

It is a congrunce in a system of congrunces. Not sure if that helps?

[latent anvil]

Von Hage, compare with the equation x^3 - 3x + 1 = 2 over the real numbers

you can't get it into the form x = something

[vale coral]

x^3−x+ 1≡ 2 (mod 5)
2x^2 ≡ 1 (mod 7)
x ≡ 7 (mod 11)
is the entire thing and I have to use CRT on it in the exercise.

[woven delta]

lmao

[latent anvil]

Honestly I would probably do the first one explicitly

And find that x = 2

Like, just plug in all 5 possible values of x

[woven delta]

there are some little tricks you should think of I guess

like fermat's little theorem and shit

[scarlet estuary]

yeah honestly for small modulos

[woven delta]

but they don't apply here so much

[vale coral]

It is from an old exam set so I think I have to use some tricks to make the system applicable of CRT.

[scarlet estuary]

ah

[latent anvil]

I don't see a way to do that for the first one except doing it explicitly tbh

[scarlet estuary]

i was gonna say "for small modulos just pretend youre in Z/nZ and test all possibilities"

i think this still holds actually

[woven delta]

yeah

[vale coral]

Aight. For this one 2x^2 ≡ 1 (mod 7) would it work to take inverse of 2 mod 7 and multiply the equation with that and then do sqrt on both sides?

[latent anvil]

No

Kind of

What does sqrt mean, is the problem?

Yes you should first reduce to x^2 = 4 mod 7

[vale coral]

ye

Can one then take the square root of both sides

[latent anvil]

And in this case you know what the solutions are because 4 is a perfect square over Z

[vale coral]

and get x = 2 mod 7

[latent anvil]

No

You could have x = 5 mod 7

Also, how would you solve something like x^2 = 6 mod 13?

You can't take the squareroot of 6

[vale coral]

That is true

[latent anvil]

(i didn't actually check if that has solutions, but sometimes it will)

So you can say x^2 = 4 mod 7 has at most 2 solutions

And then you can say "oh 2 and -2 are solutions"

Which tells you x^2 = 4 mod 7 implies x = 2 mod 7 or x = -2 mod 7

Does that make sense?

[vale coral]

Yep

ah yes it make sense

Thanks for the help

[latent anvil]

np

So to recap

For the first one, you really just need to try all the possibilities

Which sucks but there's only 5 of them

For the 2nd, multiply by 2^(-1) = 4 and observe that x^2 = 4 mod 7 implies x = 2 mod 7 or x = -2 = 5 mod 7

For the last one, it's already x = 7 mod 11

So you get two possible systems

[vale coral]

One system would be with x = 2 mod 7 and the second system would be with x = -2 = 5 mod 7 right ?

[latent anvil]

Yes, exactly

And then you can apply CRT to get two possibilities for x mod 5711=385

[woven delta]

actually the fact that x^2=4 has only two solutions is a bit nontrivial if you don't know Z/7Z is a field

[latent anvil]

Yeah I originally said "by, uh, reasons"

[woven delta]

lol

[latent anvil]

if you know inverses exist you're good though

and that's just bezout's lemma

[woven delta]

yeah

[vale coral]

Doesn't it follow that Z/7Z is a field

because 7 is a prime number ?

[latent anvil]

yes

[woven delta]

yeah that's an iff

[latent anvil]

What's an example of a quadratic with too many solutions mod n?

[scarlet estuary]

what about Z/1Z liquid

[latent anvil]

Oh is that the field with 1 element?

[woven delta]

yes

[latent anvil]

[woven delta]

lmao

I only work in Z/0Z tbh

the best field

[scarlet estuary]

Z isnt a field you fucking idiot

[woven delta]

[woven delta]

lmao

[latent anvil]

From my department lounge chat

[woven delta]

just shoves in morally

[latent anvil]

I love the word morally

It means so little

[woven delta]

yeah

[scarlet estuary]

Z's morally a topological space

this is the motivation behind the p-adics

the one they dont want you to know

[shy bluff]

So for this question, I'm trying to go and prove it in the direction of assume that pi is order 2 and show that it cannot have any cycles larger than 2

Now if pi has a cycle that's longer than 2 then it has a cycle of length k that takes on thef orm (a_1... a_k)

And when we do pi^2 we see that all cycles other than this cycle map to their origin? Map to their respective identities? that is, all 1-cycles nothing changes and all transpositions are returned back to the original form

But I don't know how to deal with this (a_1... a_k) term

Like I know that this term would only go and map any input element twicee

So it'd map whatever you throw at it to a_3

[latent anvil]

I think that makes sense

If there's a cycle of length > 2, there's some element which you apply the permutation to twice and not get that element back

A good exercise (more general than this) is to determine the order of an arbitrary element in terms of its disjoint cycle decomposition

[shy bluff]

What do you mean

[latent anvil]

Suppose you have an element π in S_n

And you know the cycle decomposition of π

So e.g. π is a 2 cycle times two 3 cycles (all disjoint)

What's the order of π?

[shy bluff]

So order of pi is the smallest n such that pi^n = 1

So uh

I'm gonna guess this

6?

[latent anvil]

that's true in this case, but why?

[shy bluff]

2 * 3 = 6, we have 1 2 cycle and we have 2 3 cycles

[latent anvil]

And in general, if you have n_i number of k_i cycles, what's the order?

[shy bluff]

All cycles are disjoint

So that means that you can cycle the 3 cycles back to their initial positions with 3 appliacitons of pi

[latent anvil]

What about if you multiply π by a disjoint 6 cycle? What's the order of this new element?

[shy bluff]

hrm

Does this have to do with uh LCM?

that's what I'm guessing

[latent anvil]

Yup!

[shy bluff]

Oooh

[latent anvil]

It's exactly the lcm

Of the sizes of all the cycles which appear

[shy bluff]

so the order of the permutation is equal to the LCM

Ooh

That's pretty cool

[latent anvil]

And your problem is immediate from this

Right?

[shy bluff]

Yes

I dont' think that I can actually use that result in her ebecause we have'nt learnt it yet

But tht's interesting!

[latent anvil]

Oh yeah I just think it's worth knowing about

[shy bluff]

Yea

It is

Learning about it beforehand will make learning it in class in a month much easier haha

[opal osprey]

Is there a generalization of the prime numbers to more general algebraic structures with the notion of a product and the notion of a multiplicative identity?

More specifically, this question came up naturally to me while studying some ODEs. And I was thinking about a way to generalize the natural numbers to the real numbers.

Or maybe even the complex numbers or more general Cayley-Dickson constructions if I could.

But for now, let's just stack to the question of trying to generalize the prime numbers to the real numbers.

[stone fulcrum]

Interesting that you should ask, as this question is very central to abstract algebra.

[scarlet estuary]

https://en.wikipedia.org/wiki/Prime_element

though perhaps https://en.wikipedia.org/wiki/Irreducible_element is a better analogue

(the naming is for historical reasons)

[opal osprey]

Ohhhh

But in particular, could I generalize the prime numbers to the field of the real numbers using either of these two definitions?

[stone fulcrum]

We have a common structure called a ring. You get addition, subtraction, and multiplication, but you don't get division for every element.

Elements in rings factor. When they don't factor anymore, they're called irreducible.

[scarlet estuary]

anyway, if you use either of these definitions, R doesnt really have anything of the sort

[stone fulcrum]

Sadly not this way, since the real numbers are a field, and doesn't have elements that don't divide.

[scarlet estuary]

yeah indeed, fields in general dont really have any structure comparable to primes

but rings do

this is fundamental to galois theory

[opal osprey]

That's sad

[stone fulcrum]

Real polynomials on the other hand, do

Since they're rings

[opal osprey]

Because I was here trying to think about properties of the prime numbers that don't rely entirely on the idea of division.

[scarlet estuary]

thats what irreducible is meant to capture

[opal osprey]

Because what is really important to me, is the fundamental theorem of arithmetic.

[scarlet estuary]

not a product of two nonunits

[opal osprey]

So

[scarlet estuary]

FTA is very... R biased

[opal osprey]

Can I find a infinite subset of the real numbers, such that every real number can be written as a finite product of those?

[scarlet estuary]

its not much of a "fundamental theorem of algebra" so much as "fundamental theorem of the algebra of the real/complex numbers"

[opal osprey]

That was my first attempt to generalize prime numbers.

Srry

Fundamental theorem of arithmetic

[scarlet estuary]

ah okay

[opal osprey]

Lmao

[scarlet estuary]

i was kinda confused

[stone fulcrum]

It's not a well formed question, since R itself can be that infinite subset

[opal osprey]

Silly mistake

[scarlet estuary]

anyway, yeah, theres a lot of deep theory in ring theory about exactly those questions

[opal osprey]

What about a countable subset of R?

[stone fulcrum]

Ooh now you're talking

[scarlet estuary]

i'm assuming you mean

countably infinite

anyway, no, since that would be a bijection between that set and R

[stone fulcrum]

Set theory strikes again

[scarlet estuary]

well, not directly a bijection but

it would lead to one

[opal osprey]

oh

true

[stone fulcrum]

Wait how? Haha I think I got lost

[scarlet estuary]

theres a bijection between the naturals and finite strings of naturals

hence if you have a countable set, theres a bijection between that set and finite products of elements of that set, which then surjects into R

[if we assume the condition]

[opal osprey]

what about a proper uncountable subset of R?

maybe that could be a thing

[scarlet estuary]

sure, take R \ {1} for example

[opal osprey]

nooooooo

[scarlet estuary]

as long as it contains 0

[opal osprey]

I need more restrictions in order to make this problem interesting

[stone fulcrum]

I like where your head is at, and there's definitely something interesting here

[scarlet estuary]

and of course there are uncountable sets where the converse is true

such as [0, 1)

you cant write 2 as a product of elements of this interval

no matter what you do

[opal osprey]

what other properties do the prime numbers have that could be used to generalize them to the field of the real numbers?

[scarlet estuary]

if we could do something like that, we'd have already done it

[opal osprey]

besides FTA

[stone fulcrum]

Needs to be better than countable, but worse than a proper subset

[opal osprey]

Continuum Hypothesis lmao

[scarlet estuary]

the definition of a prime element in a ring is

p | ab implies p | a or p | b

then p is a prime element

this isnt true for any nonunit/nonzero element of R so

yeahhhhhh

[stone fulcrum]

You're definitely more interested in an irreducible element. Has no non-unit, non-associate factors

[scarlet estuary]

well actually

let me state that more precisely

R is a field so every element is, formally, a unit

except 0

so uh

this means that something like the FTA holds, yes

but... its a trivial statement

:/

as in it doesnt say anything

https://en.wikipedia.org/wiki/Unique_factorization_domain

we call R a UFD, but trivially so

more interesting is UFDs that arent fields

[opal osprey]

Oh :/

[stone fulcrum]

I feel we're straying pretty far. You wanted to talk about R

[scarlet estuary]

i mean the point is that there isnt really a good notion of primality

in R

[stone fulcrum]

Ultimately the problem with R is that every element "looks the same" when it comes to division

[scarlet estuary]

except 0

but that doesnt help

[stone fulcrum]

Screw 0, gotta be a snowflake

[opal osprey]

Maybe I could better formalize this problem if I gave myself more time

and knew more about modern algebra

I literally just thought about this 10min ago and thought it would be a fun concept lmao

[scarlet estuary]

there are certainly other properties of the primes you could try and apply to R but

besides properties like "is equal to 2, 3, 5, 7, 11, ..." or whatever

which are obviously "cheating"

i dont think theres a good way to make them apply to R

without applying to all nonzero elements of R, or none of them

¯_(ツ)_/¯

[opal osprey]

Maybe with some more restrictions this could be made a more fun non trivial problem

but idk

surely would need more time in modern algebra to deeply think about this

anyway

thanks

thanks y'all @stone fulcrum , @scarlet estuary

[stone fulcrum]

Np thanks for asking! Come back with more

[mossy dagger]

Hi, I was having trouble proving a certain result. If the polynomials (x² - a), (x² - b), and (x^2 - ab), are irreducible in F then √b ∉ F(√a). To get a better understanding I've tried thinking of examples and came up with, the case F = Q (rationals), a = 3, b = 5. This amounts to proving √5 ∉ Q(√3). I tried proceeding by contradiction and assuming √5 = c + d√3 where c, d ∈ Q but can't seem to proceed. Could someone please give me a hint?

[brisk granite]

assume sqrt(b) = c + dsqrt(a). Then, 0 = c^2 + d^2a - b + 2cd sqrt(a).

so, now, you now either c or d is 0?

so, see what happens in each case

[mossy dagger]

Oh alright, this makes sense. I was initially trying to arrive at a contradiction by showing one the polynomials to be reducible. Once I couldn't proceed even in the example case I realized I was missing something more fundamental. Thanks a lot, this clears things up!

[brisk granite]

np

Could I have a bit of help with this Q?

I've been told this result is false, but I've made a proof for it: Let K be the splitting field of x^4 + ax^3 + bx^2 + cx + d over Q. If Gal(K/Q) is Z/4Z, then a = c = 0.

proof: Suppose Gal(K/Q) is Z/4Z. Then, K has a quadratic subfield — Q(\sqrt(D)). Since [L:Q(sqrt(D))] = 2, there exists alpha such that L = Q(D, alpha) and alpha^2 belongs in Q(\sqrt(D)). [Q(alpha):Q] = 4 since Z/4Z only has one group of order 2. Then, L = Q(alpha).

Let alpha^2 = a + b(sqrt(D)). Then, (alpha^2 - a)^2 - b^2D = 0 and alpha^4 -2a(alpha^2) + (a^2 -b^2D) = 0. Hence, L is the splitting field f(x) = x^4 -2a(x^2) + (a^2 -b^2D).

so, is there something wrong with the proof or is the result true?

[somber bramble]

please invoke latex, this is really hard to read

[brisk granite]

ok

gimme a min

[somber bramble]

(just put some $ around the formulas)

[brisk granite]

yeah, but gotta use mathrm and stuff for things like Gal(K/Q)

I've been told this result is false, but I've made a proof for it: Let $K$ be the splitting field of $x^4 + ax^3 + bx^2 + cx + d$ over $\mathbb{Q}$. If $\mathrm{Gal}(K/\mathbb{Q})$ is $\mathbb{Z}/4\mathbb{Z}$, then $a = c = 0$.\

Proof: Suppose $\mathrm{Gal}(K/\mathbb{Q})$ is $\mathbb{Z}/4\mathbb{Z}$. Then, $K$ has a quadratic subfield — $\mathbb{Q}(\sqrt{D})$. Since $[L:\mathbb{Q}(\sqrt{D}] = 2$, there exists $\alpha$ such that $L = \mathbb{Q}(\sqrt{D}, \alpha)$ and $\alpha^2$ belongs in $\mathbb{Q}(\sqrt{D})$. $[\mathbb{Q}(\alpha):\mathbb{Q}] = 4$ since $\mathbb{Z}/4\mathbb{Z}$ only has one group of order $2$. Then, $L = \mathbb{Q}(\alpha)$.

Let $\alpha^2 = a + b(\sqrt{D})$. Then, $(\alpha^2 - a)^2 - b^2D = 0$ and $\alpha^4 -2a(\alpha^2) + (a^2 -b^2D)$ = 0. Hence, $L$ is the splitting field $f(x) = x^4 -2a(x^2) + (a^2 -b^2D)$.

so, uh, is the proof wrong?

@somber bramble btw, is this better?

[cloud walrus]

PolyBeanDip:

[somber bramble]

does being the splitting field of a polynomial of that form exclude being the splitting field of a polynomial that is not of this form?

[brisk granite]

oh

hmm

[somber bramble]

provided the arithmetic at the end is correct I don’t see any other mistakes but it’s been a while since I did galois theory

so I might be missing sth obvious (e.g. that the subfield is ℚ(√D) is the only possibility, right?)

the existence of K follows from the fundamental theorem of galois theory, right?

uh I mean of the subfield of K

[brisk granite]

it follows from "the galois correspondence" according to my textbook

[somber bramble]

yea that’s the fundamental theorem

or at least we called it so

[brisk granite]

ah ok

[somber bramble]

I have a question again. I asked it very similarly earlier but I think I said something unclearly and only now noticed, when I wanted to look at the answers more closely, so I’ll ask it again but be more specific :P

[brisk granite]

does being the splitting field of a polynomial of that form exclude being the splitting field of a polynomial that is not of this form?
actually, I think this isn't true. So, the proof doesn't work. Eg: x^3 + x^2 + x + 1 and (x - 1/3)^3 + (x - 1/3)^2 + (x - 1/3) + 1

[somber bramble]

I have a sequence of free groups $G_n$ whose generators I will write as $x_\alpha$ where $\alpha$ is a binary sequence of $n$ elements. So e.g. $G_2 = \langle x_{00}, x_{01}, x_{10}, x_{11} \rangle$. I also have injections $i_n \colon G_n \to G_{n+1}$ defined via $i_n(x_\alpha) = [x_{\alpha0}, x_{\alpha1}]$. Finally, $G = \lim_{\rightarrow} G_n$. I need to argue that $G$ is free.

Now in the previous discussion the assumed situation was that the bases of the free groups I have could be chosen coherently (i.e. in such a way that the injections restrict to maps between the bases). But as far as I can tell this is not the situation here, as there is no element $x$ such that $\langle x, [a,b] \rangle$ is the whole of $\langle a,b \rangle$. In that case the proof was pretty straightforward because you could choose as the basis just the direct limit of the bases. I’m not really sure how to procede here. As before I don’t actually have to prove this, providing a reference would be enough.

[cloud walrus]

Sascha Baer:

[somber bramble]

so like some general theorem would suffice

previous discussion https://discordapp.com/channels/268882317391429632/496784958430380033/707254617549045861

[hot lake]

isn't that false

ah I need to think more on it

[somber bramble]

it should be correct

it’s the fundamental group of the exterior of alexander’s horned sphere

which is supposedly free

[hot lake]

if it's true it would be the free group on countably many generators, and it would be equal to its derived subgroup, which seems obviously false to me

[somber bramble]

ah hm maybe I managed to confuse my advisor into lying to me

I also misremembered a claim from another paper here’s what it says there

so not free but “locally free”

which if my googling is correct means that every finitely-generated subgroup of it is free, but it need not be free itself?

[hot lake]

yes that's obvious

[somber bramble]

hm, yes, it is

I agree

[hot lake]

Q is my favourite non-free locally free group

I'm not sure what it means to not have finite rank for a locally free group though

[somber bramble]

(with some “abelian” thrown in there you mean?)

I suppose it means that there is no finite generating set?

[hot lake]

ah maybe

[somber bramble]

that much is true, anyway

[hot lake]

yes

[brisk granite]

May I have some help with problem 13?

So far, I've been able to show all Z/4Z extensions of Q are splitting fields of some polynomial of the form x^4 + ax^2 + b

So, now I think we can do something with Kaplansky's Theorem

But, uh, I haven't been able to do anything beyond this

may I have a hint?

[hot lake]

you mean Kronecker's theorem ?

[brisk granite]

Uh, no

[hot lake]

I don't know of any Kaplansky's theorem

[brisk granite]

this one

anyway, maybe this isn't the way to go

cuz I have no concrete idea about what to do with this theorem

[hot lake]

well you have to use it

if you know that every potential extension can be generated with a polynomial of that kind

[brisk granite]

Btw, the converses of points 1 and 2 are true as well

Do I also have to use the sum of squares theorem?

well you have to use it
I'm not sure how tho

[hot lake]

is it true that any Z/4Z-extension can be generated with such a polynomial ?

[brisk granite]

I think this proves it

[brisk granite]

Um, Zef, you still there?

[simple agate]

can anyone suggest some reading material for understanding/solving this please?

I know I have to use smith normal form and tried reading up on it in dummit & foote and had trouble understanding it/relating to my question

[stone fulcrum]

Row reduce it, lol. That will give you the kernel of your group

[soft elm]

Can i say any Aut(G) for G of order n is isomorphic to S_n

[vale coral]

Hello I have a the following question:
Say S = F_p[X] / <X^3 - X + 1>.
I need to show that S is not a domain if p = 5.
To do this I was thinking about finding a zerodivisor in S, because if one exists S is not a domain.
The elements in S can be expressed as b0 + b1X + b2X^2.
However I am unable to find a zero divisor from this?
I know that x^3 - X + 1 is reducible if p = 5. I have a feeling that this fact is useable however I don't see how.

[upper pivot]

take Z/pZ, Aut has order p-1

well what does it mean to be a zero divisor in the reduced ring

in terms of the ideal and original ring

@vale coral

[soft elm]

sorry guys big wrong

thanks john

[upper pivot]

(oops i gtg, but basically if you have x^3-x+1 is reduable in F_5, then x^3-x+1=f(x) g(x), what happens when u reduce this mod)

np

[soft elm]

aut(G0 is a subgroup of sym(G) tho

[upper pivot]

yep

AutG has an action on G, so use cayleys

[kindred mist]

Does anyone know where I can find some resources on how commutative rings can be embedded into fields or skew fields (this is mentioned in K-theory, no?)

(e.g. even a recommendation for a K-theory pdf/book would be appreciated)

(yes I have tried googling for articles and pdfs, not much luck)

I'm looking for how specifically commutative unital rings can always be embedded into fields (If I have that right)

[mild laurel]

Uh, at least for that last statement, you need to assume that your ring has no zero divisors, since fields don't have zero divisors

[kindred mist]

Ok yeah I forgot about that

[mild laurel]

But then, in this case, all you need is to just embed your domain in its field of fractions

[kindred mist]

So this

thanks!

Just going to use this to try impose some conditions on a lower bound for the smallest PID containing a given ID then

hoping that will encode some information about the original ring that wasn't obvious before, I get that sounds vague but yeah

[red imp]

where does one go after a first course in algebra

[mild laurel]

there are tons of ways to go

galois theory/commutative algebra are probably the most popular choices

@red imp

[red imp]

:o

commutative algebra sounds good for me, ring theory has grown on me

[upper pivot]

you should do it then, atiyah macdonald is a very fun book

[red imp]

thanks I was wondering which book to buy

because even if I don't actively do any algebra study after my course I just want to have a book on my shelf that I can look at when I feel interested

[kindred mist]

commutative algebra is nice ❤️ has some very powerful tools like localization

very useful for AG too

There are some good algebra pdfs and other course material online (I like math doctor bob's vids, also theres the channel: Introduction to Commutative Algebra)

[brisk granite]

one of the problems I'm doing asks me to find 6th degree polynomial with no radical solutions.

So, that means I'm looking for a polynomial (with splitting field K) such that Gal(K/Q) is S_6 or A_6?

how do I do this?