groups-rings-fields

[supple rampart]

Normaliser of a ={ x: x commutes with a}

U can check that!!

[sharp sonnet]

???

[upper pivot]

(A not being in the center ofc for my arguement)

[supple rampart]

By that definition i think N(a) is normal

[upper pivot]

it is not. i just showed you a counter example

[solemn rain]

N_G(H) = { g in G |gHg=H }

[sharp sonnet]

real algebruh hours rn

[supple rampart]

So S must be a subgroup

For N(S) to be normal

[upper pivot]

no, that still doesnt do it. take the subgroup <A> with A not in center in SL again

[sharp sonnet]

did you read what i wrote at all?

[supple rampart]

I can prove this

Given S to be a subgroup

[sharp sonnet]

go ahead

[upper pivot]

i mean if u assume S is a normal subgroup it works

[supple rampart]

no just subgroup

[upper pivot]

dude i just showed you a counter example...

[stone fulcrum]

I read the above definition as "every S in the group such that"

I think I should have read it as "every S in the subset S such that"

[upper pivot]

actually nvm that might not be true lol w/e (deleted a wrong statement oops)

[supple rampart]

Ueah ueah

[upper pivot]

but anyhow my counterexample holds

[supple rampart]

Its wrong then....at some stage..i have to assume commutavity..

In order to do the proof

[upper pivot]

well all subgroups of abelian groups are normal so lol

[supple rampart]

U reapeated wht i just said

[sharp sonnet]

in abelian groups the concept of normalizer makes little sense

(the normalizer will always be the whole group, which is normal)

[supple rampart]

If H is subgroup of Z(G)

[sharp sonnet]

i mean yeah

subgroups of the center are normal

and the normalizer of normal subgroups is the whole group

which is normal

[stone fulcrum]

To be fair, take a look at that definition provided above. It's not very clear that we're taking S to be a subset of S

[sharp sonnet]

[supple rampart]

One need to assume furthur conditions on S in order to be a normal subgroup

[sharp sonnet]

i am not even sure if there is a characterization of when the normalizer of a subset is normal

[supple rampart]

Even the single elment need to have only two conjugates in G for N(a) to be a normal subgroup

Man...i forgotten some theorem i think.

But i dont agrre...that its meaningless to define normaliser in non abelian groups.

U have the class equation waiting for u dude

[sharp sonnet]

it is, because the normalizer of every subset of an abelian group is the whole group

so you don't gain any information

[supple rampart]

U said on non abelian

[sharp sonnet]

i didnt

i just noticed, that you did

[supple rampart]

Oh" little "is used in negative sense

[sharp sonnet]

so let me be clear: it only makes sense in non abelian groups

[supple rampart]

Grammar dude

Didnot notice that english is not my first language

[supple rampart]

When does one learn catagory theory
In which level?

[sharp sonnet]

level 5 or 6

on a more serious note: when you need it

[solemn rain]

i think u need it firstly at AG or/and AT

basically dont learn it

untill u have to thats what i got as an answer to urq uestion

from the ppl who know

[sharp sonnet]

algebraic topology is the best place to learn it, because that is when the ideas historically first arose

[solemn rain]

but wait im using df

and i keep seeing arrows

in the modules section

is that category theory?

[sharp sonnet]

not really

you can use category theory everywhere, but it is just a different vocabulary

[solemn rain]

the meme that basicaly say that any math problem literally can be solved by categories

how true is that

[woven delta]

Lol

[sharp sonnet]

and just because you draw diagrams, call maps arrows and use the word functor does not mean you are doing category theory

[woven delta]

^

[sharp sonnet]

the meme is that category theory has no theorems

[woven delta]

Also mo2men please don't repeat things you've heard and don't actually know about when people ask advice

That's bad form

[sharp sonnet]

a lot of "category theory" you will encounter is just re-phrasing of ideas

which can be useful, but you should always be skeptical why you are doing that

[solemn rain]

the things ive heard were frfom here 😐

[supple rampart]

So a GENERALISATION

??

[woven delta]

That doesn't matter, you still shouldn't do it

[sharp sonnet]

you can read things like "X is just Y in the category of Z", but that is just different modes of thinking

[solemn rain]

mmmm

cool af

[woven delta]

It's usually not tbh

[solemn rain]

idguess thats subjective for like the ppl who know this stuff

[woven delta]

Idk I tend to find it annoying when people say X is just Y in Z

Like maybe it's true but it's usually not helpful

[solemn rain]

isnt that super cool tho?

like seeing fields intersect

[sharp sonnet]

ye, abstraction can be an extremely useful tool

[solemn rain]

2 different fields intersect?

like im talking about

coolness now not usefulness ;D

[woven delta]

Like I think an example where this is good is like "abelian groups are just Z modules"

[sharp sonnet]

but at some point in your mathematical life you learn that just abstracting for abstractions sake is not worth it

[woven delta]

^

[solemn rain]

i really dk what abstract means even til now

im that stupid

[supple rampart]

@sharp sonnet
I said thst thing but they made fun why is that

[sharp sonnet]

it's saying "X is just Y in Z"

well, not really, but wtv

[supple rampart]

About abstraction being a very powerful tool

[sharp sonnet]

what?

[woven delta]

That's such a vague statement

[brisk granite]

this problem has rly stumped me. may I have some help with it?

I'm guessing F < K

But I'm not quite sure how to prove it

if we assume the cubic f(x) is depressed, than f(x) = x^3 + px + q and g(x) = x^2 + qx -p^3/27

[hot lake]

that's a pretty vague question

how exactly is g popping up as part of solving f = 0 ?

if you did what I think you did then you fell in a pretty evil trap there

[brisk granite]

oh, I can show you how my book does it

derives the cubic formula that is

if you did what I think you did then you fell in a pretty evil trap there
wdym?

[hot lake]

thinking that F is a subfield of K

you poor innocent soul

[brisk granite]

lmao

if it's not that, then idk what relationship it could be

[hot lake]

it's also possible that the writer of the question thinks that F is a subfield of K

because

it seems like if there is any justice in the world, it should be so

but alas

[brisk granite]

how do you know it doesn't have to be?

[hot lake]

if F = Q(sqrt(D)) then it is Q(sqrt(-3D)) that is a subfield of K, iirc

a quick way to check if I'm drunk or not

is to go work on the cubic (x-3)(x+1)(x+2) for example

[brisk granite]

uh, K is Q then so, that isn't a great example?

[hot lake]

well if you find that F is bigger than Q then you can't have F < K

[brisk granite]

oh, I see lol

oof

I see now

[hot lake]

alpha and beta aren't in K in general

they are in KF

[brisk granite]

yea

[hot lake]

and combining them properly gives you roots that are in K

so it's hard to tell what answer they expect

[brisk granite]

if F = Q(sqrt(D)) then it is Q(sqrt(-3D)) that is a subfield of K, iirc
do you think this is what was expected?

[hot lake]

"K cap F = Q" isn't much of a relationship

and that one is awkward to state

but it is the most precise statement you can make

[brisk granite]

wait, what if we say f is irreducible

do you think that changes things?

[hot lake]

no

you can also say that if the base field contains the cube roots of unity then F < K I guess ?

[brisk granite]

if F = Q(sqrt(D)) then it is Q(sqrt(-3D)) that is a subfield of K, iirc
uh, does this even work for the example you gave?

[hot lake]

I think it does because D = -3 in that case

[brisk granite]

you can also say that if the base field contains the cube roots of unity then F < K I guess ?
I don't rly see why this is true

[hot lake]

so Q(sqrt(9)) = Q

and that's a subfield of F

which is also Q

[brisk granite]

yep, D is -3

my b

you can also say that if the base field contains the cube roots of unity then F < K I guess ?
uh, zef, why is this apparent tho?

[hot lake]

I don't remember what's the nicest way to show it

it follows from "if F = Q(sqrt(D)) then it is Q(sqrt(-3D)) that is a subfield of K, iirc" anyway

except that instead of Q you put Q(sqrt(-3))

there probably is a formula that expresses -3(alpha^3 - beta^3)² as the square of some expression in the roots of the cubic ?

that's what it should boil down to

[brisk granite]

it follows from "if F = Q(sqrt(D)) then it is Q(sqrt(-3D)) that is a subfield of K, iirc" anyway
how hard do you think this should be to prove?

[hot lake]

it's as hard as expressing -3(alpha^3 - beta^3)² as the square of some expression in the roots of the cubic

[brisk granite]

there probably is a formula that expresses -3(alpha^3 - beta^3)² as the square of some expression in the roots of the cubic ?
one last Q. soo, why does this imply Q(sqrt(-3D)) < K

[hot lake]

because the D in F = Q(sqrt(D)) is (alpha^3 - beta^3)²

now I'm having doubts

[brisk granite]

because the D in F = Q(sqrt(D)) is (alpha^3 - beta^3)²
alright, I need to think about this

[hot lake]

and so if you find -3(alpha^3 - beta^3)² = h(x1,x2,x3)² where x1,x2,x3 are the roots

[brisk granite]

wait, shouldn't the splitting field F just be Q(alpha^3)?

[hot lake]

then sqrt(-3D) = h(x1,x2,x3) is in K

[brisk granite]

but the root of my quadratic is alpha^3?

[hot lake]

Q(alpha^3) = Q(sqrt(D))

yes

= F

[brisk granite]

yea, that's why I said it's F, not K

oh

[hot lake]

wait which one is F and K again

[brisk granite]

ok, maybe im confusing myself with the letters

either way, I think we agree

[hot lake]

who decided to call F the splitting field of g lol

[brisk granite]

yeah, that's kinda wack

[old storm]

In the textbook I'm reading, this is stated in the Groups chapter. Could someone tell me why this is always the case?

[oblique river]

I'm thinking of two numbers x and y

all I'm going to tell you is that their sum is 15

what numbers am I thinking of?

[solemn rain]

15 and 0

[old storm]

I mean sure but how would you prove that rigorously

or prove it

in the first place

it seems like a very big claim

[oblique river]

as long as the group is not the trivial group (in which case the only product is e*e = e) this is true

and not really that big of a claim

xy = (xyz)*z^(-1) for any element z

so if I just give you the product xy

how do you know if it was (x times y) or (xyz times z^(-1))?

just pick any z which isn't equal to y^(-1)

and those are two different "factorizations" of the element xy

there's the "proof" I guess

[old storm]

hm I guess I kinda see what you mean there

what's up with primes though

[oblique river]

primes have nothing to do with this

also, more generally, for any finite group G, if you write out the multiplication table for G

every single column will have every single element of G appearing exactly once

which is to say that if |G| = n

then there are exactly n ways to write any element of G as a product of two other elements

so as long as G isn't the trivial group, this shows that you can't recover x and y from xy

[old storm]

I was just thinking that with primes you only have one factorization, would that still not count because you can have different orderings?

like if p is a prime and p is your product, it could be 1p or p1

I guess the group could be uh the integers?

[oblique river]

the integers don't form a group under multiplication

[old storm]

o right

bc inverse right

[oblique river]

(which is why the primes aren't relevant here)

[old storm]

ahh

ok I gotcha

[oblique river]

this same thing is true in the group of nonzero rational numbers under multiplication where the "primes" still exist

but now, you can write 5 as 5*1

but also 5/2 * 2

[old storm]

neat

[oblique river]

so you can't recover x and y from xy

[old storm]

I'm still kinda new to this stuff so I'll try reading that message with the |G| = n more carefully, thx

for multiplication table, do you mean the cayley table?

[oblique river]

yeah

just like, all possible products of two elements

arranged in a grid

[old storm]

right

[oblique river]

(btw that comment is still true for infinite groups, it's just that then your cayley table is infinite and it's not really as clear how to arrange it as a "table")

[old storm]

you mean the unrecoverable comment?

[oblique river]

I meant the cayley table comment in particular

but yes, for any nontrivial group, it's impossible to recover x and y from xy

and it's exactly because of the "x*y = xyz * z^(-1)" issue

[old storm]

you said "then there are exactly n ways to write any element of G as a product of two **other **elements." I'm going to look into this a bit more, can you confirm that "other" is necessary?

[oblique river]

I didn't mean anything with that word

there are n ways to write an element of G as a product of two elements of G

where n = |G|

is what I meant

[old storm]

ok good so not other

ty

[kindred lichen]

The law of composition is not a function in the reverse direction. That is, given an element x of a group G, we cannot map it uniquely into the product space. As was said before, if we have the product (a)(b)=x, then we can assign x to the pair (a,b) in the product space. But we can also write (ac^-1)(cb)=x hence x can also be associated with the pair (ac^-1,cb) in the product space. It's probably not the best way of thinking about it but it is a way

[smoky cypress]

tldr, law of composition is not injective and therefore inverse of the law of composition doesn’t exist

[supple rampart]

G={e}
(e,e)→e
e→(e,e)

[smoky cypress]

Lol nontrivial group of course

But what is the group $\brk{x,y|x^3=y^2=1,xy=yx}$

[cloud walrus]

Whoever:

[mild laurel]

think about it

[smoky cypress]

Hmmm all I can say that it's a group with 6 elements

And I have the cayley table written out explicitly

Is it S_3

[mild laurel]

is the group abelian?

[smoky cypress]

Lmao no

[mild laurel]

why not?

[smoky cypress]

symmetric groups aren't abelian for n≥3 I remember

But that group up there is abelian

[mild laurel]

Yeah

so it can't be S_3

[smoky cypress]

Yeah

[mild laurel]

There's only one other group of order 6

[smoky cypress]

Oh bruh

I see the isomorphism now

Is it

Z/6Z

Fantastic

[mild laurel]

indeed

[smoky cypress]

Thanks

[chilly ocean]

Had to proof that if f extends to a continuous map $E^{i+1}\to X$, then f represents the zero class in $\pi_{i}(X; x0)$

[cloud walrus]

Stephen:
Compile Error! Click the reaction for details. (You may edit your message)

[chilly ocean]

My attempt: Every element in $\pi_{i}(X,x_{0})$ it's a class of a map $(S^{i},) \to (X,x_{0})$. Via the extension $\tilde f$ of $f$ we would have that the same class is produced via $\tilde f\circ i:S^{i}\to E^{i+1} \to X$. But now we recall that $E^{i+1}$ is contractible hence an element in $\pi_{i}(X,x_{0})$ that comes as image of an element in $\pi_{i}(E^{i+1},)$ as to be $0$.

[cloud walrus]

Stephen:

[chilly ocean]

does this sounds ok?

[shy bluff]

What do they mean by "Either G is cyclic or G is isomorphic to (Z/pZ) x (Z/pZ)"? Isnt' (Z/pZ) x (Z/pZ) cyclic?

[wind steeple]

no it isn't

why would it be cyclic

[mild laurel]

Every element in (Z/pZ) x (Z/pZ) has order p, so there can't be a generator

[shy bluff]

Um

I don't quite follow

(Z/pZ) is cyclic, correct?

[bleak abyss]

Yes

What's the proposed generator of (Z/pZ) \times (Z/pZ)?

[shy bluff]

Wait is a generator just one elemnt of (Z/pZ) x (Z/pZ)?

[bleak abyss]

Yes. So like, in the case of (Z/pZ), 1 is the generator

[shy bluff]

Oooh ok

So you'd nee to have 2 different elements to ggenerate (Z/pZ) x (Z/pZ) right?

[bleak abyss]

Yeah exactly

[shy bluff]

Oh ok

That was my misunderstanding

[bleak abyss]

Sorta like how R^2 needs two elements of a basis

[shy bluff]

Yea

Ok so a generator is limited to being a single element of the ggroup

Thank you!

[bleak abyss]

Exactly. In general you can talk about a "generating set"

[shy bluff]

Ah

So the generating set for (Z/pZ) x (Z/pZ) would be like {(1, 0), (0, 1)}?

[bleak abyss]

A generating set, but yes

[shy bluff]

err yea

[upbeat juniper]

I don't see why the degree is required to be odd

is it used somewhere in the proof?

[sour plume]

In the real numbers, a counterexample would be the polynomial x^2 + 1

So yeah it's probably used

[brisk granite]

I require the added constraint that a and b are non-zero, right?

[wind steeple]

no

[chilly ocean]

no

[brisk granite]

oh ok

what if my R is F_2 and b = 0 and a = 1?

b = 0 = au = u. But u is supposed to be a unit.

nvm, Im dumb

[brisk granite]

this is how my book defines integral domain

but, uh, there is no reason an integral domain R has to contain a multiplicative identity, right?

so, in that case, doesn't the above question not work?

[sour plume]

Maybe this is one of these books which always works with "Rings with identity" but just calls them rings

It's pretty common; but in any case, yeah, units of rings only make sense if your ring has a multiplicative identity

[upbeat juniper]

In the real numbers, a counterexample would be the polynomial x^2 + 1
ohh so it's to ensure that g has deg >= 1 so it has a root?

[brisk granite]

It's pretty common; but in any case, yeah, units of rings only make sense if your ring has a multiplicative identity
ok, cool

[soft elm]

how can i prove that there always exists an element of order n in a group of order n^k where n is prime

[steady axle]

pick element from grp

its order divides the order of grp

consider cyclic subgrp generated by this element

[soft elm]

how can you say its order can be p

it can be p^2

if k>2?

[steady axle]

yes you are right but then you can do something to fix this

consider cyclic subgrp generated by this element
@steady axle !

[soft elm]

the cyclic group will have order p^2 then

how can i say theres always a cyclic subgroup of order p

i have to prove the existence

ik the order divides p^k

[steady axle]

think about this for a moment

[delicate bloom]

take an element of order n^k and raise it to the n^(k-1) power. Now when you raise this new element to the n power you get the identity

just kidding

[soft elm]

lmao

wait mero that might work actually,from lagrange order of any element is p^m where m<=k now
(x^p^(m-1))^p=e

lol

you werent kidding

[smoky cypress]

[delicate bloom]

surely I'd know if I were kidding

[smoky cypress]

https://discordapp.com/channels/268882317391429632/496784958430380033/709577307345518592

[soft elm]

i found something else

if a group has order as a multiple of a prime

then it has an element of order p

https://en.wikipedia.org/wiki/Cauchy's_theorem_(group_theory)

tis stronger

[delicate bloom]

you'll enjoy the sylow theorems

[kindred rivet]

Sylow is mega cool

[olive mirage]

Definitely one of the coolestand most unexpected results

[vast gale]

is there a change of signs when you change order of a dot product

?

does -xVector * qVector = qVector * xVector?

[smoky cypress]

No

For dot product is commutative

[vast gale]

I figured

[mild laurel]

If your inner product is on a complex vector space, you usually complex conjugate it

[vast gale]

That might be why

then

is there a way to write latex in here?

[smoky cypress]

$(x_1,\dots,x_n)\cdot(y_1,\dots,y_n)=\sum_{k=1}^nx_k\overline{y_k}$

[cloud walrus]

Whoever:

[smoky cypress]

@vast gale

[brisk granite]

@hot lake, I remember we talked a few days ago about a question I was working on, and I was wondering where I could read more about the result you were talking about. I tried looking for it in dummit and foote but I couldn't find it. (sorry for the ping)

[red imp]

how does one determine whether the ideal generated by a polynomial is a prime ideal

for example I have the ideal $\langle x^7+x^5+1\rangle$ in $\mathbb{Z}_2$

[cloud walrus]

Bobbicals:

[red imp]

what I'm really trying to do is determine whether $$\mathbb{Z}_2[x]/\langle x^7+x^5+1\rangle$$ is an integral domain

[cloud walrus]

Bobbicals:

[mild laurel]

If the ring is a field (which it is here), and your ideal is principle (which it is here), its the same as asking if the polynomial is irreducible

[red imp]

:O

[steady axle]

i.e. k[x] is pid

also how do you get black or white background in texit

i always get the default grey

[delicate bloom]

,tex --color white

[cloud walrus]

Your colour scheme has been changed to white

bertwit:

[red imp]

is there a nice way of determining whether something is irreducible in Z_2

[mild laurel]

not really in your case

[red imp]

big sad

[mild laurel]

its easy to check that it has no roots

so it must be the product of a polynomial of degree 2 and degree 5, or degree 3 and degree 4

but you can just write down all the irreducible polynomials of this degree and show thats not the case maybe

[red imp]

wew

my teacher put that question on a test and idk how he would have expected us to have done it in a reasonable amount of time

thanks for your help

there are some others that fucked me over as well

$T$ is the set of square 2x2 matrices over $\mathbb{Z}_5$ of the form $\begin{pmatrix}a&b\0&0\end{pmatrix}$

[cloud walrus]

Bobbicals:

[red imp]

is there a factor ring T/I that is isomorphic to Z_3?

[oblique river]

so you're thinking of T as a ring under matrix multiplication?

[red imp]

yeah

[oblique river]

what did you try on this problem

I don't think there are too many ways you can go

[red imp]

absolutely nothing I was like "uhhh no because T/I can't commute" but I had zero idea about anything

idek if T/I commutes or not

[oblique river]

I mean

[hot lake]

T/I is commutative at least when I = T

[oblique river]

if T/I were isomorphic to Z/3Z

then it would be commutative

[hot lake]

but there is something much simpler you can say

[oblique river]

yes, indeed

what is the size of T

[red imp]

I basically don't know anything about quotient rings, including how to find their cardinality

well I know things but I have no level of comfort with them

[oblique river]

we're not goign to be finding any cardinalities of quotient rings

[steady axle]

@red imp take a moment and think about what buncho said

[red imp]

how does the size of T relate to the size of T/I

[oblique river]

this is a pretty fundamental fact about quotients of rings and groups

I think you should spend some time reviewing lagrange's theorem

[red imp]

is lagrange's theorem the one that involves groups

we haven't covered groups yet, only ring theory

[oblique river]

the order of T is always divisible by the order of T/I

oh, that's odd

have you not talked about this for rings?

if |T| is finite then |T| = |I| * |T/I|

[red imp]

wew

I CTRL+F'd through my lecture notes and it doesn't mention lagrange until groups

[oblique river]

maybe you didn't use that name

but that fact that I mentioned above

is pretty fundamental

[red imp]

okay I'll try to cram it into my little brain

[hot lake]

how can you cover rings before groups

[red imp]

idk take a look

[steady axle]

lecture 32

[red imp]

damn my lecturer's a little cunt

giving us a question that we needed future content for

is there not another way of going about it?

[oblique river]

so I tried looking in the rings section

to see if this was discussed there

before it was discussed for groups

but I can't find anything

[red imp]

massive RIP

[oblique river]

so, here is another approach

that doesn't require any cardinality concerns

so there is always a surjective map from R to R/I

namely a --> a + I

*surjective homomorphism

so we can try to show that there are no surjective homomorphisms from R to Z_3

oh sorry, T instead of R

I'm just used to calling all rings R haha

[red imp]

oh that's a good idea

although I never would have thought of it

[oblique river]

although even this is kinda subtle though

I would show that any element t in T is equal to 3*t' for some other element t'

where by 3*t' I mean t' + t' + t'

this is not too hard to do: the matrix (a, b) (i'm just omitting the 0s on the bottom row since they're not too relevant right now)

is just equal to (2a, 2b) + (2a, 2b) + (2a, 2b)

so suppose we had some homomorphism from T to Z/3Z

then the element (2a, 2b) of T must map to something

let's call that something x

(so x = 0, 1, or 2)

but then (2a, 2b) + (2a, 2b) + (2a, 2b) must map to x + x + x

but x + x + x = 3*x = 0 no matter what x is

so therefore (a,b) must map to 0

therefore the only homomorphism T --> Z/3Z is the trivial homomorphism which isn't surjective

so like, that is an argument but seems like a lot of work to come up with that approach and that particular method

also just fyi I really don't understand these lecture notes. the order of everything is just... off... to me

[red imp]

my lecturer isn't the one who wrote them, he just follows them. maybe he somehow forgot that we don't know about lagrange's theorem yet because we learn things in a weird order

[oblique river]

yeah that's possible

imo, lagrange's theorem is a pretty important fact and I would talk about it earlier than later

how accurately is he following the notes?

maybe he talked about it in-class?

I've heard of things like this, like "I'm gonna talk about stuf fin class that's not in the book and then test you on it so you better come ot every class and take notes"

[red imp]

he literally puts the notes up on the screen and reads from them, he's completely useless

[oblique river]

oh god that's so bad

i'm sorry :(

[red imp]

;(

okay one (or maybe two) more questions

[oblique river]

what are your thoughts

[red imp]

I know that every ideal in F[x] is principal

but beyond that nothing

[oblique river]

have you ever encountered the ring F[x,y] before?

[red imp]

nope

[oblique river]

so one way to think about it is the ring of 2-variable polynomials

another way to think about it is "polynomials in the variable y, where the coefficients are polynomials in the variable x"

so the fact that F[x] isn't a field means that you can't conclude immediately that all the ideals in this ring are principal

do you know any examples of nonprincipal ideals in other polynomial rings?

like, do you know an example of a ring R and a nonprincipal ideal in R[x]?

[red imp]

not that I can think of immediately

[oblique river]

can you give me an example of any ring R with a nonprincipal ideal?

(also, I did find this: http://prntscr.com/spx157)

(example 7)

[red imp]

ohh yeah

[oblique river]

and right below this

[red imp]

I remember reading that now

[oblique river]

http://prntscr.com/spx1j2

see the last sentence

[red imp]

so if we just took the set of all polynomials in F[x,y] such that the constant term is 0

that would be a non-principal ideal

[oblique river]

yep

[red imp]

awesome

[oblique river]

although honestly i think this is another terrible test question

it's basically "do you remember one line from the notes"

when was this test btw

[red imp]

today

I handed it up a few hours ago

that was my submission email

[oblique river]

reasonable complaint

[red imp]

one last question

remember the ring T of those matrices? one question asked if there are any non-trivial ideals in T

and I was like "nah" just because I couldn't think of any

[oblique river]

I can think of at least one

[red imp]

oh wait it said non-trivial maximal ideal

I didn't even read that word at the time

god knows what a maximum ideal even is

[oblique river]

oh

I think the answer is still yes

[red imp]

in the lecture notes when defining a maximal ideal it uses the term "proper ideal" which isn't even predefined

does it just mean an ideal and not a left- or right-ideal

[scarlet estuary]

proper in the set of subsets

i.e. not the whole ring

[red imp]

oh right

okay so it's just "maximal" in the sense that it's the biggest you can find

[scarlet estuary]

yep, except for the ideal composed of the entire ring

(which is pretty trivial and unimportant)

[red imp]

which means that the question of whether something has a maximal ideal basically comes down to whether or not it has a non-trivial ideal

[hot lake]

did your T ring even have a unit ?

[red imp]

nop

not that I could see at least

[hot lake]

oh your definition of ring doesn't require the multiplicative identity

[oblique river]

I think it does have a unit

(1 0)

[hot lake]

but it does say that (R,+) is an abelian group

[red imp]

(1,0) is only an identity when multiplied from the left hand side

I think

[hot lake]

yeah something like that

and (1,1) is only an identity when multiplied from the other side

.... wait that's impossible

no they are identities for the same side

or any (1,x)

that ring is weird

[oblique river]

oh you're right

I don't normally work with noncommutative rings

it's a left identity but not a right one

[hot lake]

i don't normally work with rings without a multiplicative unit

so you got a definition of ring that talks about abeliangroups without having ever talked about groups before ?

[red imp]

does it actually

I'm sure it only puts that in as a remark

oh it's because we've seen the definition of groups in our previous classes but never worked with them extensively

like in linear algebra using matrix groups

but not actually doing any group theory

[vast gale]

Question when you Have a pure Quaternion ( W Scalar = 0 ) why is the Conversion from Quat to Axis Angle Only Pi?

the value of the angle is locked at Pi

Ex Quaternion [x, y, z, w] [ 0, 0, 1, 0 ] = Axis-Angle {[x, y, z], angle (radians)} { [ 0, 0, 1 ], 3.1415927 }

[brisk granite]

what is k in this question?

[chilly ocean]

A field I suppose

[brisk granite]

ok

So, that means k[x_1,x_2,...,x_n] is UFD?

[left copper]

short question: does every equivalence relation induces a congruence?

[sour plume]

If your definition of congruence is like on wikipedia, i.e. "an equivalence relation on an algebraic structure (such as a group, ring, or vector space) that is compatible with the structure in the sense that algebraic operations done with equivalent elements will yield equivalent elements.", then no, not necessarily

For example on the integers, you could define an equivalence relation by saying "a and b are equivalent if they have the same sign", i.e. all positives are equivalent, all negatives are equivalent, and zero is equivalent to itself

But this is not compatible with addition, for example; 3 is equivalent to 1, but (3 - 2) = 1 is not equivalent to (1 - 2) = -1

[left copper]

right, thanks

[sour plume]

np!

[left copper]

that cleared some ambiguity

[woeful flint]

hey guys, I have a question that states:

Let $\phi : R \to R^\prime$ be a surjective homomorphism and and $I^\prime$ an ideal of $R^\prime$. Show that the preimage $\phi^{-1}(I^\prime)$ is an ideal of $R$ and that it contains the kernel. I may have done something wrong but i didnt use surjectivity anywhere in its proof. Thoughts?

[cloud walrus]

Kraft Macaroni:

[chilly ocean]

Show your proof, surjectivity should be needed.

[woeful flint]

If $x,y \in I = \phi^{-1}(I^\prime)$ then $\phi(x), \phi(y) \in I^\prime$ and as $I^\prime$ is an ideal $\phi(x) + \phi(y) \in I^\prime$ and so $\phi(x + y) \in I^\prime$ i.e. $x + y\in I$. If $i\in I$ and $r \in R$ then $\phi(i)\in I^\prime$ and $\phi(r)\in R^\prime$ and as $I^\prime$ is an ideal $\phi(r)\phi(i) = \phi(ri) \in I^\prime$ i.e. $ri\in I$. So $I$ is an ideal.

We have that $\ker \phi \subset I$ as if $k \in \ker \phi$, $\phi(k) = 0$ and as $0\in I^\prime$ we must have that $k \in I$.

[cloud walrus]

Kraft Macaroni:

[woeful flint]

sorry it took long didnt see the message

[sour plume]

I think that should be fine; iirc, if you want to show that the image of an ideal is an ideal, you need surjectivity of the morphism, but the preimage should always work

Quick math proof: Consider the induced morphism $R \to R'/I'$. The kernel of this morphism is exactly the preimage of $I'$, and hence automatically an ideal

[cloud walrus]

Lartomato:

[woeful flint]

Yeah i was sure that i didnt need it

thanks

[knotty mason]

I'm confused about seperability and inseperability. I understand that you need a char =/= 0 field for inseperability

I found this example p(X) = X^6+X^5+X^4+2X^3+2X^2+X+2 over Z/3Z

[olive mirage]

What are you confused about, exactly?

[knotty mason]

this polynomial is not irreducible, it factors as (X^2 + 1)(X^4 + X^3 + X + 2)

maybe even factors more over Z/3Z

[olive mirage]

yes, finite fields like Z/3Z are perfect, which means that every finite extension of them is separable

[knotty mason]

oh

[olive mirage]

so if you want an inseprable extension, you need a finite characteristic infinite field.

[knotty mason]

thank you

So to answer your question

I want to work out an example of an inseperable field extension

and i want to compute its degree and the number of homomorphisms from L -> k^bar

[olive mirage]

Would you like me to give you one?

[knotty mason]

and I believe these will be different

yes

[olive mirage]

So, take the field $F_3(t)$ of rational functions in the variable t over the field $F_3$

[cloud walrus]

unknownknight:

[olive mirage]

($F_3$ here just being Z/3Z)

[cloud walrus]

unknownknight:

[olive mirage]

Are you happy with that, or do you need some more explanation?

[knotty mason]

I understand that

[olive mirage]

So then the polynomial x^3-t is irreducible

but when you let, say, $\gamma$ be a root of it, then it suddenly factors as $(x-\gamma)^3$

[cloud walrus]

unknownknight:

[knotty mason]

I see. (x - gamma)^3 = x^3 - gamma^3 because we are in char 3

[olive mirage]

Exactly!

and recall that you can grab the repeated roots of a polynomial f by calculating $gcd(f, f')$

[knotty mason]

let k = F_3(t), P(x) = x^3-t, L = k/P. I think L has degree 3 and there is only one homomorphism from L into k^bar

[cloud walrus]

unknownknight:

[olive mirage]

and what has gone awry here is that $f' = 0 $ even though f isn't constant.

[cloud walrus]

unknownknight:

[knotty mason]

oh interesting

[olive mirage]

So, here is where there is another subtlety

L is actually isomorphic to k

in general, k(t) has both infinitely many subfields and infinitely many extensions isomorphic to itself

[knotty mason]

woah, what a weird field

[olive mirage]

but, for example, k(t^2+1) is a subfield of k(t) that is also isomorphic to it

So yes, weird things happen with function fields because of this.

[knotty mason]

this is great

[olive mirage]

but, if you buckle down, and push through it, what you should really be looking for is not just a a homomorphism from L into a kbar, but one that fixes k as a subfield of L

and once you add that restriction, then you're right, there is only one.

[knotty mason]

oh that is a detail i had not considered at all

[olive mirage]

(and that corresponds exactly to the fact that $\gamma$ is the only root of our polynomial in kbar, so it is the only place I can possibly send it)

[cloud walrus]

unknownknight:

[knotty mason]

I see

that does make sense

[olive mirage]

It is a detail that you just never think about when you're used to thinking about, e.g., Q, because Q doesn't have any subfields at all, much less one isomorphic to itself.

[knotty mason]

One thing I am a bit unsure about is [L:k] - I define this as the vector space dimension of L over k - I want to say it's 3 because the polynomial has degree 3 but because the weirdness of gamma and stuff I'm not certain

[olive mirage]

You're right, it is degree 3.

[knotty mason]

will i simply have [L:k] = 3 since P is irreducible

[olive mirage]

1, $\gamma$, and $\gamma^2$ are linearly independent over k

[cloud walrus]

unknownknight:

[knotty mason]

I get it! I think I finally understand a bit about inseperability - thanks very much!

And what you said about fixing k also just clicked for me, that's why I am reading about k-homomorphisms and not just homomorphisms

[olive mirage]

No problem. Fortunately, every field of characteristic zero, and every finite field is perfect, so tehre are no inseparable extensions at all, so that means you don't encounter these things "in the wild" too often.

[knotty mason]

yeah it seems like a very obscure part of the theory and I think that's why it's been hard for me to grasp until now

[olive mirage]

really, the reason you need to talk about it at all in an intro course/text on this topic is that if you don't, you can't do any of the proofs, because you need this hypothesis. It just so happens that this hypothesis is automatic for most cases you care about .

Of coruse there are times in research where this comes up.

If you're interested, I think a very helpful exercise is to look at the polynomial $f(x) = x^p-x-t$.

[cloud walrus]

unknownknight:

[knotty mason]

over F_p(t)?

f'(x) = p x^{p-1} - 1 = -1 so the gcd(f,f') = 1 i think this one is seperable?

[olive mirage]

So, suppose one of the roots is $\gamma$

[cloud walrus]

unknownknight:

[olive mirage]

what are the other roots?

and then, if you know what a Galois group is, what is the Galois group of this polynomial?

[knotty mason]

wait does fermats little theorem apply, x^p = x ? or would that only be true of constants

[olive mirage]

that is only true of constants

[knotty mason]

I wanted to pull in roots of unity, $(\zeta_p^i \gamma)^p = \gamma - t$ is almost another root but not

[cloud walrus]

skymoo:

[olive mirage]

there are no pth roots of unity other than 1 in characteristic p

because x^p-1 = (x-1)^p

[knotty mason]

good to know!

[olive mirage]

yeah, you'll often come across the fact that you can't meaninguflly take square roots in cahracteristic 2 and cube roots in cahracteristic 3

[knotty mason]

if a^p = a then gamma + a is another root, so we can just take a = 1,2,..,p-1

[olive mirage]

bingo

so that is p roots of a degree p polynomial

so bam

[knotty mason]

it's interesting that they are all of the form gamma + a, usually the there roots would be combinations of powers of gamma

I guess the galois group would be the cyclic group for this reason, but I'm not really sure

[olive mirage]

Yeah, so now prove the opposite: If $\gamma$ and $\gamma+1$ are conjugate (e.g. the roots of the same irreducible polynomial) then the characteristic of the field is finite

[cloud walrus]

unknownknight:

[knotty mason]

Let P have degree d. this one's a bit tricky, I'm thinking of expanding out P(gamma + 1) and taking away everything from P(gamma) leaving me with a degree d-1 polynomial = 0

which shouldn't happen in char 0

[olive mirage]

Hint: Think about the Galois group.

[knotty mason]

oh that's so neat, if the galois group is cyclic of order m then the roots are gamma, gamma + 1, gamma + 2, .. and then we must get back to gamma so we gain a relation some m = 0

[olive mirage]

exactly, otherwise we get infinitely many roots, and whoops

If you want, I can drop a collection of 100ish algebra questions like that I put together to help grad students study for comps

(I think most books in algebra have terrible problems)

[golden pasture]

oh yes that would be great!

[knotty mason]

That would be great! I probably wont do all 100 but I found these extremely helpful for my understanding

[olive mirage]

Enjoy! There are a lot of problems I really love in there

and also some problems about flat modules.

[knotty mason]

awesome, and thanks again for helping me earlier!

[olive mirage]

no problem, this is easily one of my favorite courses to teach/help with (-:

[chilly ocean]

wow thats nice, saving that

[smoky cypress]

It only has 20 problems

But still

Nice

Thank you!

[olive mirage]

it has like 105 problems doens't it?

peeks

[smoky cypress]

Oh

It does have 105 problems

I'm dumb

For each section the number starts from 1 again

I see

Sorry

[olive mirage]

anyway, I have no shortage of abstract algebra questions if you ever really want more about a particular topic, I likely have them laying around in an old test or homework set or whatever.

[solemn rain]

@olive mirage

a problem in ur pset: : If all proper subgroups of a group are cyclic then the group is abelian

consider Q_8 ( the quaternion group )

counterexample?

[olive mirage]

perfect

[solemn rain]

got it

. Let G be a group and let K be a subgroup of G. Give necessary and sufficient conditions for K to be
the kernel of a homomorphism from G to G.

[olive mirage]

it's definitely one of those results that just sounds very believable

[solemn rain]

K has to be a normal subgroup?

[olive mirage]

Z is a normal subgroup of R, but it is not the kernal of any homomorphism from R to R

so necessary, but not sufficient

[solemn rain]

ohh

it has to be from G to G

missed that

i was abotu to say R --> S^1

okay let me think :d

@olive mirage can i just say G/K must be a subgroup of G?

also?

[olive mirage]

Well, not be a subgroup, but be isomorphic to a subgroup

[solemn rain]

yea yea

i meant that

got it

is that right?

[olive mirage]

So now question: is that enough to prove that K has to be a direct summand of G?

That is, does it follow that G = H x K for some H.

[solemn rain]

no

[olive mirage]

Example?

[solemn rain]

well idk about examples but like

H is a subgroup

wait idk wait

gcd(|H|,|K) need not to be one

hence cant be direct product

right?

wait what no

XD

sr

[olive mirage]

Q_8 x D_8 is a perfectly fine direct product

[solemn rain]

yea yea

[olive mirage]

(even if no one agrees what D_8 is)

[solemn rain]

yea

if H is normal

and they intersect in identity

[olive mirage]

So have we said enough to conclude that H is normal?

[solemn rain]

wdym

[olive mirage]

so in other words if f:G -> G is a hom with kernel K, is there always a normal subgroup H in G such that G=H x K

you seem skeptical, but if its not true, give me a counter example (-:

[solemn rain]

phi(G)?

G = phi(K) x K ?

[olive mirage]

Yeah, that's what I'm asking

[solemn rain]

so yes

the answer is yes

[olive mirage]

is the image of a homomorphism automatically normal?

[solemn rain]

phi(K) is normal in G yea

[olive mirage]

It is actually not true that the image of a homomorphism is always a normal subgroup

[solemn rain]

realy

yea

phi(K) is normal in G if K is normal

right?

[olive mirage]

Wait, isn't phi(K) trivial if K is the kernel?

I think you meant G = phi(G) x K

[solemn rain]

fuck xd

yea

[olive mirage]

but the theorem you want about the image of a normal subgroup being normal requires the homomorphism to be onto

[solemn rain]

yea yea

okay idk

idk why phi(G) is not normal in G

[olive mirage]

Maybe look for a counterexample in a class of groups that has a famously small number of normal subgroups?

[solemn rain]

okayy

im really bad at constructing homomoprhisms from my mind

but id say like

A_5 has only 2 normal subgroups , so if there is a homomropsih the kernel would be either 1 or A_5 ---> any subgroup would have to intersect in 1 atleast hence it fails the conditions to be a direct product

so id say A_5

is a counterexample

with any homomoprhism

[olive mirage]

Well, sadly, all nontrivial homomorphisms from A_5 to A_5 are onto

But how about a homomorphism from S_5 to S_5 with kernel A_5

[solemn rain]

how did u get that?

yea

yea ur right

but how did u know all homs are onto? ( in A_5)

[olive mirage]

Exercise: If G is a finite simple group, prove every non-trivial homomorphism f:G \to G is onto. [Hint: consider the kernel]

[solemn rain]

fuck yeaaaaaa

yea yea

yea cool

yea yea i got it

tysm

cool pset

im going to be a nuisance for you checking for answers : sorry

😐

[olive mirage]

haha no problem!

[solemn rain]

@olive mirage

10. Prove every group of order 105 has a subgroup of order 21.

we look at sylow 5 subgroups?

[olive mirage]

Warning: this will require understanding semidirect products.

[solemn rain]

oh okay can u solve it for me

i understand them like fairly weakly

[olive mirage]

So, what do the Sylow theorems say here, about the number of each of these subgroups?

[solemn rain]

they are congrunt to 1 mod 5

the number

and must divide 21

@olive mirage

[olive mirage]

what about the other primes?

[solemn rain]

n_7 = 1 mod 7

and n_7 must divide 15

n_7 = 15

n_5 = 21

[olive mirage]

yeah, so if there is no subgroup of order 21, what does that tell us about these numbers?

[solemn rain]

n_3 = 7

idk how what im writing is true

they overcount the order of the group

[olive mirage]

so, why can't n_5 be 1?

[solemn rain]

it can

yea i forgot to write all cases lol sorry

so

n_7 = 15 or 1

n_5 = 21 or 1

and n_3 = 7 or 1

[olive mirage]

Claim: If n_7 = 1 OR n_3=1 then the group has a subgroup of order 21.

[solemn rain]

why

cuz it completes?

the elements? idguess?

[olive mirage]

so remember that n_p=1 <==> the Sylow subgroup is normal.

[solemn rain]

yes

[olive mirage]

so say n_7 is 1, then we have a normal subgroup, call it H, of order 7

[solemn rain]

ok

[olive mirage]

and we have a subgroup (not necessarily normal) K of order 3.

[solemn rain]

yea

[olive mirage]

Claim: HK = {hk| h in H k in K} is a subgroup of order 21

[solemn rain]

yea thats true

okay

what semidirect product did u use tho?

[olive mirage]

that's in general true of the product of an aribitrary subgroup and a normal subgroup.

well, that HK is exactly what a semidirect product is.

but I will admit, this is not the qeustion I thought it was 😛

[solemn rain]

isnt that an 'internal' product?

or external*?

iam fuzzy with the terms

if u can educate me herfe

[olive mirage]

it's an internal semidrect product

[solemn rain]

whats a semidirect product

[olive mirage]

I fear that is deeper tahn I can go in text chat.

[solemn rain]

(g,h)*(a,b) = (ga,phi(h)b) something like that

i dont remember correctly

where phi is some auto or homo

i really dont remember xd

okay i will review for that later

tysm

[olive mirage]

I would say Chapter 4 of Dummit and Foote on group actions is a great thing to really, really focus on if you have it available to you. It goes in depth on group actions, which make understanding all of this much easier.

[solemn rain]

thats how i learnt

group actions really

am i bad lol?

[olive mirage]

I don't think so!

[solemn rain]

i am using hoffman now for lin algebra

as i learnt algebra b4 lin algebra which is supposedly bad

anyways ty for the pset agian

[olive mirage]

Absolutely!

I'm currently drilling through notation trying to figure out how to make Euler-Maclaurin seem lucid and natural 😛

[solemn rain]

gl

[gusty plinth]

Is there a simple way to determine if an element of a permutation group is a member of a subgroup of that permutation group?

[stone fulcrum]

Haha got an example?

[gusty plinth]

im thinking about iwth the rubik's cube

and thistlewaite's algorithm

is there any easy way to verify if an element is in G1 instead of G0

[stone fulcrum]

Oof haha. So first, the rubik's cube doesn't form a permutation group. It's pretty difficult to talk about the individual elements in general

[gusty plinth]

The rubik's cube is a permutation group

hmm I guess it's not exactly a permutation group because 1/12 random permutations are invalid

[stone fulcrum]

To be more specific, it's not a group that contains all of the permutations of a set

It's a subgroup of a permutation group, but so is every other group so ehh

[gusty plinth]

haha so probably not the best avenue to go down

I've been mulling over how to implement this algorithm for ages

[golden pasture]

oh lol this is terribly hard

but basically you make sure that each piece can be reached to the correct spot first

then check parity of appropriate pieces

[upbeat juniper]

am I right to say that for a polynomial f(x) with roots a1, ..., an, its splitting field is F(a1, ..., an)?

it seems like the only way to get a different splitting field is if the roots lie in some other field

[mild laurel]

That's right yes

[upbeat juniper]

I see thanks

[delicate bloom]

the name "splitting" field comes from the fact that the polynomial splits into the linear factors because you have those roots

[upbeat juniper]

yep

[chilly ocean]

does anyone have a recommendation for an abstract algebra textbook (around first course level) if my focus will be completing exercises?

[upbeat juniper]

I'm confused regarding when ppl refer to a splitting field without reference to a particular polynomial

some take it to mean a 'splitting field over some polynomial' while others see it as 'splitting field over some set of polynomials', but aren't these 2 definitions different?

for e.g. Q(2^(1/2), 2^(1/3), ...) is a valid splitting field under the second defn but not the first?

[stone fulcrum]

@chilly ocean
Dummit and foote is the common suggestion

[steady axle]

i would suggest artin

[chilly ocean]

which one moves faster?

[steady axle]

dummit and foote has too many exercises

[chilly ocean]

I know the basics and would like to do harder exercises

[steady axle]

then go fir herstein but i have only read group theory from that book

[chilly ocean]

hmm ok

I'll skim each of each of the three and see what I like

thanks

[upper pivot]

herstein is good for problems yeah

i recommend agains df

its too slow and unfocused imo

artin has nice problems too, and it has nice topics

[golden pasture]

jacobson

😛

herstein is good for problem but crippling notationwise

[rich patrol]

Theres like 5000 intro alg books so just see what fits

[golden pasture]

yea

[knotty mason]

the example of non-normality i have is Q(cbrt(2))

and of inseperability I have x^3 - t over F_3(t)

they have a strong similarity

in the insep. case one root is a triple root

in the non-normal case something weird happened too, but we only got one root

i don't know if there a connection between the two that im missing or not

[upbeat juniper]

doesnt the nonnormal case have 3 distinct complex roots?

if it's just one root the polynomial wld be (x-cbrt(2))^3 right?

[knotty mason]

the problem is that when we form the quotient Q[X]/(X^3 - 2) we cannot express the complex roots in terms of X

so we need to add cube roots of unity as well to fix this

[upbeat juniper]

oh yeah my bad, I forgot while normality was for a moment

[knotty mason]

they are such awkward concepts

I get why we need them though

[upbeat juniper]

haha are you just learning this stuff

[knotty mason]

kind of - I tried to learn galois theory before but ignored these parts and it didn't work out

[upbeat juniper]

funny enough I just started learning this like yesterday

[knotty mason]

oh cool

[left copper]

is there something like a notion of a "dual" of a vector space for magmas, semigroups, monoids and other group-like structures?
I know that set+operation like structures don't have scalars, but there could be defined some "fuctional" that maps monoids to some semigroup of functions (they are isomorphic) that take numbers as inputs/outputs. All in all, that "functional" could map whatever elements of a monoid to some field.

[solemn rain]

if G is an abelian group

consider Hom(G,Z) where Z is group of integers

i think thats how far u can go with the concept of dual for groups

[left copper]

oh interesting!

[solemn rain]

i think u should read on

pontryagin dual

i have no knowledge on this tho so sorry

[left copper]

wow thats really advanced stuff

[solemn rain]

idk 🤷‍♂️

i read about duals in hoffman kunze lin algebra

Prove or disprove: The set of squares in a group is a subgroup.

counterexample : any non abelian group ?

[uncut girder]

@left copper

[left copper]

wow thanks @uncut girder

[left copper]

@uncut girder my question arose from somewhat similar example. I had magma G = (G, & ) , a nonempty set S and operation * : G^S x G^S --> G^S, (f * g)(s) = f(s) & g(s). Question was to show that (G^S, *) is magma and that magma (G^S, *) is commutative (associative, with neutral element) iff (G, &) is commutative (associative, with neutral element)

I tried to check if the two magmas are isomorphic, or at least find epimorphism from G to G^S (that would ensure the answer to the whole question, since homomorphic image of a magma (that is com., ass. or with neutral) is also a magma (com. ass. or with neutral), but couldn't find an appropriate isomorphism)

[shy bluff]

Question about part c here, what is ker s? I know that it's the set of all elements that map to identity, but can det (phi) even go to 0?

[oblique river]

det is a homomorphism GL_n(Q) --> Q^\times

not to Q

[shy bluff]

hrm

Can you give me an example of an element that would be in A_n?

[oblique river]

do you understand what I meant by saying that it was a homomorphism to Q^\times?

[shy bluff]

That for x, y in GL_n(Q), det(xy) = det(x)det(y)?

[oblique river]

yes that's true

what is the operation on Q^\times that makes it into a group?

[shy bluff]

With regular matrix multiplication on the left and multiplication under the rationals on the right

Multiplication?

[oblique river]

great!

multiplication, not addition

[shy bluff]

Yes

[oblique river]

so, what is the identity element?

[shy bluff]

Oh the identity is 1

Not 0

[oblique river]

yep :)

[shy bluff]

So A_n is the set of all x in S_n that would send s(x) to 1?

[oblique river]

yes

the things with determinant 1

[shy bluff]

But is'nt 1 also part of the image?

[oblique river]

of course, the identity is in the image of every homomorphism

[shy bluff]

Ah wait that's not a problem

Yea sorry

Image is just everything that something can map to

And the kernel is specifically the identity

[oblique river]

it's the things that map to the identity, yes

[shy bluff]

Err is specifically everything that maps to identity

[oblique river]

you should try to compute some examples on your own here

try S_3

and look at the permutations (12) and (123)

write out the corresponding matrices

and find their determinants

S_3 has 6 elements, see if you can figure out how many of them (and which ones) are in A_3

[shy bluff]

oke

Thank you!

So if we call an element of S_n t, then Phi(t) creates a permutation matrix with number of swaps = |t|, i.e, the number of columns to swap

And then because s = det(phi), it just becomes (-1)^|t|

So then ker(s) is just all elements of S_n of even order?

[oblique river]

nope

you should actually write out the matrix corresponding to (12)

as an element of S_3

what you said at first about "number of swaps" is relevant

ker(s) is the elements of S_n which require an even number of swaps

but the number of swaps doesn't really have anything to do with the order

for example, (12) and (12)(34) both have order 2 in S_4 but one of them is 1 swap and the other is two swaps

[shy bluff]

I get uh $\begin{bmatrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1\end{bmatrix}$

[cloud walrus]

Liria ^(;,;)^:

[oblique river]

yes

what's the determinant of that

[shy bluff]

It's -1

[oblique river]

great

so (12) is not in A_3

(even though it has even order!)

[hot lake]

I'm curious how do you define the determinant of a matrix in the first place

[shy bluff]

Oh

But (123) is in A_3

[oblique river]

yes

[shy bluff]

And (13) is not in A_3

hrm

[oblique river]

correct

[shy bluff]

Does this have somethingg to do with the number of cycles too?

[oblique river]

yes

well, kind of

not just any cycles

you can write any element of S_n as a product of transpositions (2-cycles)

and it's the number of those transpositions

that matters

[shy bluff]

Oh

We didn't learn this 😔

[oblique river]

the bit about trasnpositions?

[shy bluff]

Yea

[oblique river]

I mean, that's not a problem

because you're defining A_n this way

like, just because there's another way to think about A_n doesn't mean you're not equipped to think about A_n this way

but you definitely understand the other point of view when you said earlier it's about how many flips you need

each flip is just a transposition

[shy bluff]

Yea

[hot lake]

how have you defined determinants without using signs of permutations ?

[shy bluff]

Uh

[oblique river]

you can define determinants inductively using the row expansion thing

[shy bluff]

Yea that

[oblique river]

or you can define determinants as "the unique alternating multilinear map on (R^n)^n which is normalized so that det(I) = 1"

[shy bluff]

No we did'nt learn the latter one

[hot lake]

ah

[shy bluff]

Most people in this course are expected to have taken the advanced first and second year linear algebra courses where I'm guessing gthat hte ydefine it differently 😔

How can you count the number of such transpositions though?

[oblique river]

it's hard

but it's not really relevant to this problem if that's what you're asking for

I guess it's actually not too hard. given any permutation, you can write is as a product of transpositions by just like