groups-rings-fields

[hidden haven]

Define f to be the constant 17 polynomial

That satisfies the given conditions

Ok no it doesn't

Define it to be x+13

It satisfies the conditions but has 1 rational root

[prime cloak]

Does anyone speak french ?

Basically G is finite H sub group of G p smallest divisor of |G| show that H is distinguished

[hot lake]

that looks like a difficult exercise

[prime cloak]

It is

[hot lake]

what chapter is that in

[prime cloak]

Group theory

It is one of the last exercice of my book

[hot lake]

iirc there is a proof using group representations but I don't remember enough to come up with it on the spot

[prime cloak]

Ok

[hot lake]

I think you can look at the set of conjugates of H

say that G acts as permutations of this

but then that set is too small ?

unless it has size exactly p, but then H is the kernel of the action ?

[prime cloak]

I looked at the " class at the left" of H

[hot lake]

let X = {gHg-1 ; g in G}

say that if g and g' are in the same right coset, then gHg-1 = g'Hg'-1 so there are at most p elements in X

if |X| < p then G can only act trivially on it or else any g with a nontrivial action would have an order that has a prime factor smaller than p. Thus H is normal and so X={H} and had size 1

if |X| = p it means that {g | gHg-1 = H} = H and uuuh

H should be the kernel of the action, which shows that H is normal, which feels megaweird because in this case you also get a contradiction

G acts on X by conjugation also I forgot to say

okay I think I'm wrong in the |X|=p case

{g | gHg-1 = H} = H doesn't mean that H acts trivially

but still the action of G on X can only basically be a p-cycle

and even that feels nontrivial to prove

[wooden ember]

the proof i know is by the action of left multiplication

[hot lake]

ah maybe

so action on the set of right cosets, which has size p

[wooden ember]

Let $\pi_H : G\to S_p$ be the permutation representation of $G$ acting on the set of left cosets of $H$ by left multiplication. Then $K=\operatorname{ker}\pi_H \leq H$ since only elements of $H$ fix $H$. Let $\left |H:K\right|=k$. Then $\left|G:K\right| = pk$ and we have $G/K \cong \operatorname{im}\pi_H \leq S_p$. Hence $pk | p! \implies k | (p-1)!$. If $k\neq 1$, every prime divisor of $k$ must be greater than or equal to $p$ but this is a contradiction to our divisibility condition, so we must have $k=1$

[cloud walrus]

𝓛’ai ℕarwhal ✓

[wooden ember]

(hence H = K is normal)

[chilly ocean]

can kummer extensions help solve the inverse galois problem for general solvable groups

or maybe nilpotent groups

if Q had roots of unity it would be so much easier to think about

idk how to get past that

oh that reminds me I know you can get some interesting weird numbers when you add roots of unity to Q. like I think Q with 18th roots of ubityy contains the 18th root of 2. so I was wondering if there was some like way of determining which numbers are contained in an extension of Q to contain nth roots of 1

oh wait maybe you don't use kummer extensions maybe you do the trick where you want Z/nZ so you take primes that are 1 mod n

[golden pasture]

i think you're thinking of the p-th cyclotomic field containing sqrt(p) or sqrt(-p) depending if p=1/3 mod 4

[chilly ocean]

but I do know that you can get the 18th root of 2

[golden pasture]

hm i doubt so since you dont know how the group "product" is like

(you may be interested in Shafarevich theorem tho)

[golden pasture]

you cant unfortunately

[chilly ocean]

oh 😿

[golden pasture]

you cant find sqrt(2) inside cyclotomic field of order 18

[chilly ocean]

yes I confused the resuky

result

the actual result I was thinking was that Q(18th roots of 1) and Q(real positive 18th root of 2) have a nontrivial intersection

it has degree 2 over Q

[golden pasture]

hmm is that true 🤔

[chilly ocean]

yes

now I realize that it was stupid for me to think they both contain 18th root of 2

cause then it wouldn't have degree 2

but yes

I assumed too much

and I thought that the galois group of nth roots of 2 was the ax+b group mod n

but that relies on Q(roots of 2) inteesect Q(roots of 1) being Q

[chilly ocean]

which would have been sexy but it fails for n=18 and 36 im pretty sure

it works for all other n I'm pretty sure

[chilly ocean]

omg you're right. my brain jumped way too far

[golden pasture]

wait uh why is this true

if you consider Q(2^{1/18}), the roots of unity is only {1,-1}

[chilly ocean]

ok well what about Q(sqrt3)

the roots of unity in that one are only +-1

[golden pasture]

and x^6-x^3+1 does not split

[chilly ocean]

but that doesn't stop it from having an intersection with 3rd roots of unity

or 6th roots I forgot

or maybe 7th roots

but like that doesn't break √3's heart

so why is 18th root of 2s heart broke and

broken

I forgot why they intersect

[golden pasture]

x^6-x^3+1 doesnt split as well, so Q(sqrt(3),\zeta_{18}) has degree 12

[golden pasture]

^

in this case

p=3

which is 3 mod 4

so sqrt(-3) appears

and in fact it does work

because

[chilly ocean]

oh yeah

I see

[golden pasture]

[x^6-x^3+1=\left(x^3-\frac12\sqrt{-3}-\frac12\right)\left(x^3+\frac12\sqrt{-3}-\frac12\right)]

which tells you adding the 18th root of unity is a degree 3 extension

[cloud walrus]

ari 亲

[chilly ocean]

ok yup

I don't know the proof for 18th root of 2

but I saw it on the internet 😿😿

[golden pasture]

sounds like a scam kek

i think what you are thinking of

is the galois closure of Q(2^{1/18}) is Q(2^{1/18},\zeta_{18})

in fact the splitting field of x^n-a over K is K(a^{1/n},\zeta_n)

[chilly ocean]

I don't understand the proof

but

I got my numbers wrong 🙁

it's actually for multiples of 8

the 8nth root of 2 intersects 8nth roots of unity at a degree 2 extension

they intersect at Q(√2)

so it's less cool actually

cause that makes sense

omg wait that's obvious

omg I'm getting a lot wrong today

[chilly ocean]

that still leaves the question of what numbers we can get in general from extensions that contain nth roots of unity

[waxen hedge]

That's not your question, but the Kronecker-Weber theorem tells you that every abelian extension of Q is contained in a cyclotomic extension of Q

[oak grove]

just to check, normality would be important here because it allows $(gN)\cdot(hN) = g(Nh)N=g(hN)N=(gh)N$?

[cloud walrus]

jan Niku

[sharp sonnet]

yes, in fact $gNhN \subseteq ghN$ is equivalent to normality and we get equality in that

[cloud walrus]

Lochverstärker

[oak grove]

oh interesting

[sharp sonnet]

to see this|| choose g=1 and you get hN \subseteq Nh and the other inclusion by inverting every element in the sets first||

[oak grove]

man cosets and quotient groups are hard to understand

still trying to get why this is what comes out

my teacher did some examples in class and got similar things but i havent been able to follow

shouldnt each gH just be everything in G if H is a subset of G?

[viscid pewter]

no

it's a specific g

so for example

if g is 1

then gH = 1H = {1 * h} = {h} = H

[oak grove]

OH

[viscid pewter]

each coset should have the same number of elements as H

[ionic ingot]

thank you!

[hidden haven]

F_(3²) and (F_3)² are very different things tho

[next obsidian]

What if we take 2 = 1 tho?

[hidden haven]

mb I didn't consider it in full generality

[wooden ember]

struggling with i=> ii. I know all I need to show is that the prime ideal is a subset of the nilradical (to get equality) but i cant get it to work. Note that I cant use the result that the nilradical is the intersection of prime ideals, only that it is a subset of them

[rustic crown]

Do you know about localisation? There is a pretty simple way to think from there, of course you can remove the stuff about localisation and give a straight proof. Another thing, what I'm saying is pretty much proving that intersection of all primes is nilradical.

@wooden ember

Okie I'm sleepy. I'll just type it up. Let R a ring and an element s say isn't nilpotent. Want to find a prime ideal that doesn't contain s. What you can do is physically make this s an invertible element, by localizing at the set S = {1, s, s²,...}. So you get a new ring which well denote R_s and here s is invertible. We also have the natural map R --> R_s. You can show that there is a correspondence between prime ideals of R that doesn't intersect S and prime ideals of R_s. But the latter is a non-zero ring, because otherwise 1=0 implying s^k(1-0) = 0 in R which is not true as s wasn't nilpotent. So the right ring must have a maximal ideal, hence a prime ideal. It's inverse image is a prime that doesn't contain s. So this shows that if an element is not nilpotent, then it doesn't lie in some prime ideal.

To remove localisation from the argument, just look at the set of ideals that do not contain any power of s. This is nonempty as (0) is such an ideal. Show that maximal element of this set is a prime ideal.

Now I sleep. Okie good night!

[inner acorn]

[unreal portal]

Anyone have suggestions on how to prove that an arbitrary open subset of an algebraic variety is quasi-compact? At least in this context, an algebraic variety is quasi-compact and locally isomorphic to a ringed space $(V(I), \mathcal{O}_{V(I)})$ for some radical ideal $I\subset k[x_1, \dots, x_n]$

[cloud walrus]

cgodfrey

[unreal portal]

I think I have to use the fact that $V(I)$ is Noetherian, but I'm not sure how

[cloud walrus]

cgodfrey

[thorn delta]

@unreal portal do you know how to prove this for Spec R?

[unreal portal]

we haven't done anything with the spec of a ring, but (I think) I was able to show it for an affine algebraic variety

oop, that should be $\varphi:X\to V(I)$

[cloud walrus]

cgodfrey

[viscid pewter]

ok so considering C\{0}
one automorphism is to map every complex number to its conjugate
i'm like 95% sure another is to map re^it to (1/r)e^it
are there any other continuous ones that aren't just a combination of those two

[thorn delta]

this doesn't look right. if you're "picking an ordering of I" and you're assuming that you can index it with natural numbers, then you've already picked a subcover. also i think there are other mistakes here like your ascending chain is descending and you descending chain is ascending, and when you take complements you don't switch the direction of inclusion

[unreal portal]

oh, hmm

I was trying to use the axiom of dependent choice for that, to construct the infinite ascending chain of open sets that don't cover V(I)

[thorn delta]

ok lets walk through this:. Suppose we have A_k, how do you get A_{k+1}

[unreal portal]

pick some U_i not already selected, and set A_{k+1} to be the union of A_k with U_i

[thorn delta]

ok ok yea ofc, i follow now. so the chain A1 \subset A_2 \subset .... eventually stabilizes at A_N, but you can't say A_N = V(I)

not unless the union of the chain of increasing open sets is all of V(I)

but DC only gave you a countable subset of your open cover, so you would have to do more work to show that that chain is a countable subcover

[unreal portal]

How could it stabilize to anything other than V(I) if the union of all the U_i is V(I)?

Or is the idea because the ascending chain is only countable, while there could be uncountable U_i?

[thorn delta]

the union of all the U_i for i in your index set is V(I), but not in this countable ascending chain

[unreal portal]

Okay, but if A_N isn't V(I), then that means there's some x in V(I) - A_N. And x is in some U_j, so we can construct a bigger open set. Doesn't this contradict the fact that the chain stabilizes to A_N?

https://math.stackexchange.com/questions/1442665/a-noetherian-topological-space-is-compact

at least that's what I'm getting from this

[thorn delta]

huh okay. ig that just wasn't clear to me without explicitly invoking zorns

[unreal portal]

yeah that's fair

you certainly had me convinced it wasn't right

[thorn delta]

i don't think it follows just from this reasoning though. zorns lemma works but may there is something weaker you can get away with idk.

Anyway, for ur proof, i noticed that you have $\overline{A_k} \subseteq \overline{A_{k+1}}$, but the order of the indices should actually be switched. Also for readability u might want to make the letter for your index set different from your ideal. other than that looks good to me

[cloud walrus]

kxrider

[unreal portal]

yeah that's definitely a typo, thanks for noticing that

[dawn quest]

isn't this wrong?

the product of a_1 and e_1 isn't defined

if I'm correct, any idea what it should say instead?

[next obsidian]

@dawn quest I believe that this means like (a1,1)

And (1,a2)

Or maybe a 0 instead of a 1

[dawn quest]

maybe

any idea where to start if it were the former

[terse crystal]

I don’t think that’s true. Maybe we can find a k-algebra having a maximal ideal that isn’t finitely generated. I am thinking let A=k[T_1, T_2,…] infinite many indeterminate and let U=spec(A)-{(T_1,T_2,…)} . This should be a counterexample right?

Closed subschemes of an affine scheme are affine but it’s not true for open subscheme

[unreal portal]

we're not up to schemes yet, so idk if that makes the difference

but that's what the assignment is

¯_(ツ)_/¯

and this is specifically the definition of algebraic variety we're using

[terse crystal]

Oh my bad, I forgot that affine algebraic varieties are associated to finitely generated k-algebra…

[unreal portal]

well if you have any nudges into the right direction, I'd appreciate it!

[terse crystal]

Oh I am a dumb… X is covered by affine open subsets U_j, any open subset U of X, U is the union of intersection of U and U_j so first we can reduce to the case where X is affine. Let X=spec(A) where A is a finitely generated k algebra (therefore noetherian) Then any open subset U is the complement of a closed subset V(I) where I is an ideal of A, therefore finitely generated. I=ΣA f_i then U is the union of principal open subsets D(f_i)

[iron barn]

do permutations in symmetric group have to be symmetrical?

[kind temple]

symmetric in what way

[iron barn]

numerically

srry bruh , just getting this

[kind temple]

[worldly dome]

(╯°□°）╯︵ ┻━┻

┬─┬ ノ( ゜-゜ノ)

[wooden ember]

Thanks! I’ve been trying to show that every ideal maximal with respect to not containing (s) is prime (I’ve already shown it exists): it’s the last part giving me trouble

[rustic crown]

not containing (s)? or not intersecting {1, s, s^2, ...}?

[wooden ember]

I was working on not containing (s) yeah cause I had an earlier result about ideals maximale relative to not containing finitely generated ideals

[rustic crown]

if P is the maximal element of that set (zorn), then say it was not prime... if a and b are outside P, but ab in P, then look at the two ideals I = P + (a) and J = P + (b)
Notice that IJ is contained in P but I and J are strictly larger than P. this is an equivalent criterion for being a prime ideal. what can you say if I and J are strictly larger?

[wooden ember]

Yeah I went around those lines but I didn’t see why IJ was contained in P though I knew it must be true

[rustic crown]

elements of IJ are generated by P*P, P*(a), P*(b) and (a)(b) all these are in P!

[wooden ember]

🤦

Of course

I must have been tired

[rustic crown]

lol

[wooden ember]

Thank you so much

[rustic crown]

[wooden ember]

So the only subtlety left is showing (s) is in IJ

I’ll think about it when I get back to work

Probably not today cause I have a lot of exercises to finish for class

[rustic crown]

wait no

i think you need to look at the stuff not intersecting {1, s, s^2, ...}

so both I and J must intersect this as they bigger than maximal element

s^n in I and s^m in J, but then s^(m+n) in P a contradiction

[wooden ember]

Or just s in I and 0 in J? So s in IJ and (s) in IJ?

Wait no lamp

Lmao*

[rustic crown]

[wooden ember]

Being stupid

[rustic crown]

we need s is not nilpotent to say that there atleast one such ideal, namely (0)

[wooden ember]

Alright ill see if I can get it to work with (s) and otherwise ill do it as you said

[wooden ember]

You can always get an ideal maximal with respect to not containing a finitely generated ideal?

Ah but I guess this could be (0) and we don’t know that we’re in an integral domain so it could not be prime

So yeah my case shouldn’t work generally I’ll go with yours thanks

[rustic crown]

if you know about localizations, this makes a lot of sense... so maybe look it up
it's like defining fields of fractions but you don't wanna make every non-zero element invertible

[wooden ember]

I’ll give it a read thanks !

[rustic crown]

[terse crystal]

It has only one prime ideal p so it’s a local ring, any element not from p is invertible. p is naturally the intersection of all prime ideals which is the nilradical, so elements of it are nilpotent…

[rustic crown]

yea... but i think the book didn't prove intersections of primes is nilradical... so we basically did that proof

[terse crystal]

Oh I see

[sour plume]

i've been asked to join a reading course on "Mackey's imprimitivity theorem", does anyone have some perspective or opinion on why this is interesting or cool

the wikipedia article seems okay but somehow i'm not really getting a feeling for what one can do with this

[brittle goblet]

Hey all, I have a question with regards to the orthocenter of simplexes. Namely, I want to show that the intersection of all the altitudinal hyperplanes is the orthocenter

But I'm a little lost as to how to do that

any help would be appreciated!

[cloud walrus]

dumbo

[void cosmos]

@ember field prove it

or disprove it

by (gH)^-1

you mean a coset

in context of a quotient group

since ur talking about inverses

right?

[cloud walrus]

dumbo

[void cosmos]

yea

it is

gj

[wooden ember]

Only if H is normal

[stoic rose]

Am I missing something or is problem 46 nuked by Sylow's theorems?

[void cosmos]

ur not

missing something

i remember solving with them

[willow mason]

yea sylow

makes it easy

[stoic rose]

Ok I was expecting it to be harder than that 😁

[fluid pelican]

yo

probs a stupid question but, what does it mean for an element to be centralised by a group

e.g. Z(P) are the only p-elements of G centralised by P

I think it means that the element commutes with all the elements of P?

[willow mason]

ye center = all elements that commute with group

[long obsidian]

Can anyone recommend a YouTube series on algebra?

Like it could just be a prof recording their graduate algebra.

[chilly ocean]

borcherds probably has something

i know he has some videos on commutative algebra

[chilly ocean]

so like the algebraic closure of Q is algebraically closed, but you can still throw in transcendental elements and get bigger fields

you can do this for C too i guess, by just taking the field of fractions in the polynomial ring

but is there any meaningful sense of what the "biggest" extension of a field is

do the extensions of a field even form a set?

[hidden haven]

No you can have arbitrarily large extensions

[prisma ibex]

they don't form a set, no

[chilly ocean]

i was thinking you could have an extension for any set

but I couldn't see how to formally define it

[hidden haven]

Take a polynomial ring over any set of variables with coefficients in your field and you get a domain, then take field of fractions. Any 2 fields you get this way are isomorphic iff the set of variables have the same cardinality

[chilly ocean]

all finire products/sums in the symbols?

[prisma ibex]

if you have any set I and any field F you can form the purely transcendental extension F(x_i)_i with transcendence basis {x_i : i in I}

where each x_i is just a formal transcendental element you adjoin

this works for any cardinality

[hidden haven]

[chilly ocean]

thought so

that's cool

wait so R is isomorphic to an extension of Q to contain R many transcendental elements?

that sounds wild

that's some nasty tangled version of R

but its so cool

[hidden haven]

No idea what you mean

[chilly ocean]

like

index a set x_i with reals

and take the transcendental extension of Q to contain all of them

like does that make sense

thats gonna be isomorphic to R right

[hidden haven]

No because the process I described doesn't create all extensions of Q

Only the purely transcendental ones

[chilly ocean]

ohhhh

yeah I forgot 😰

[hidden haven]

For example Q(√2) doesn't arise as this

[chilly ocean]

yessssss woopsy

what about the algebraic closure of Q intersect R

as the base field

[hidden haven]

Doesn't work because the extension is still not purely transcendental

[chilly ocean]

is that cause like

[median star]

https://youtube.com/playlist?list=PLelIK3uylPMGzHBuR3hLMHrYfMqWWsmx5

[chilly ocean]

π and √π and stuff

[hidden haven]

Yep exactly

[chilly ocean]

😕

[hidden haven]

You can keep taking roots

[chilly ocean]

yup

you can take the algebraic closure of Q(pi)

[hidden haven]

Pretty much because Q is not a free Z module

😌

[chilly ocean]

what fields are?

[hidden haven]

Same thing with other transcendentals tho

None, I'm not taking about Q as a field, just a group

It's relevant because we're looking at powers of π

[chilly ocean]

k I assumed

[hidden haven]

You can take any rational power

[chilly ocean]

yes

ohhhh

smarty pants

[hidden haven]

[chilly ocean]

that's not the copy of Q that I was thinking of

I was like

base field??

[hidden haven]

Ye lol

[chilly ocean]

damn so we hit a wall so quickly

what do quotients of R² by a principle ideal look like if R is a PID

they seem to be products R×R/I

[next obsidian]

No

Take Z and do like Z x Z mod the ideal generated by (a,a)

[chilly ocean]

so (1,1) generates the module Z/(a)

but then

[next obsidian]

Wut

[chilly ocean]

like 😿

(1,1) and (1,0) gives you the whole module right

[next obsidian]

How do those exist in Z/(a)

[chilly ocean]

they don't

[next obsidian]

Then how can they generate the module if they don’t live in it?

[chilly ocean]

they generate an isomorphic module

[next obsidian]

Uh

Z x Z / ((a,a)) isn’t isomorphic to Z/(a)

[chilly ocean]

no I'm talking about the submodule generayed by (1,1)

[next obsidian]

Okay

That was very unclear

Lol

[chilly ocean]

sorry

when I say it "is" a module I just mean it's isomorphic

[next obsidian]

That… is not a good habit

You could say “generate a submodule isomorphic to”

[chilly ocean]

yeah I was lazy and I didn't know if would cause confusion

[next obsidian]

But it’s like impossible to figure out what you meant just by saying “is” haha

Anyway yeah (1,0) and (1,1) generate the module

But I don’t see what the punchline is supposed to be IG

[chilly ocean]

I mean have tou ever said like "this galois group "is" Z/nZ"

[next obsidian]

There’s a difference there

You’re talking about a specific element inside Z x Z /((a,a))

Whatever this is besides the point

[chilly ocean]

yeah sorry

so (1,1) And (1,0) together generate the whole module

[next obsidian]

Indeed

But what’s the kicker?

Is there something you want to conclude from this?

[chilly ocean]

so isn't Z²/(a,a) isomorphic to Z×Z/a

[next obsidian]

What’s the proposed isomorphism? I can see it as modules I guess but as a ring things are weird

[chilly ocean]

(1,0) maps to itself, (0,1) maps to (1,1)

as rings idk

I was thinking just as modules

[next obsidian]

Oh

Yeah this isn’t a ring map

But as modules it works

[chilly ocean]

yes the ring is weird

ok so I should have been more clear from the beginning

sorry I swear I try to save 3.4 words and end up not saying enough

I'm taking R² as a module to begin with

but I said "principle ideal" as if I was treating it like a ring

so yeah I was not clear

[next obsidian]

Btw it’s “principal”

[chilly ocean]

it is?

[next obsidian]

Also I think you might run into issues if I isn’t principal

Yes

[chilly ocean]

I thought principal was the name of the person who leads the school

[next obsidian]

It also is the name of an ideal generated by 1 thing

[chilly ocean]

and principle is like "this is a good principle to live by"

yeja but that isn't math

[next obsidian]

Anyway I think you’re gonna run into issues here still

Namely I think it’s possible that R^2/I is all torsion

So that it can’t be isomorphic to R x R/J because this has a free part

Something like maybe

Z^2/((2,0),(0,3))

I think if you have the element (a,b)

Then if you multiply by 6

This ends up in the ideal so it’s 0

As like 3a(2,0) + 2b(0,3)

So this shows it can’t be isomorphic to Z x Z/J because the element (1,0) in the latter module isn’t torsion

[chilly ocean]

I guess I need to think of ideals as my pals

[next obsidian]

Does my example at least make sense?

[chilly ocean]

yes it does

sorry

[next obsidian]

No worries

[chilly ocean]

I was getting over the spelling

so yeah but your example

you did the quotient by the module generated by 2 elemtns

[next obsidian]

Yes

But your initial question didn’t state I had to be principal

[chilly ocean]

for just one element it should work

[next obsidian]

Uhhh

Idk

Maybe

You have to show there’s a free part

And then the classification of finite modules over a PID probably handles the rest

But I don’t wanna think about why that might be true

Lol

[chilly ocean]

oh that makes it easy

I didn't want to use that

my apologies for being unclear

that's something I definitely need to work on

I like use cheater words to mean stuff that's close enough to what it is in my head

[dull root]

Let $G = C_1 \times C_2$ be the direct product of two infinite cyclic groups, then if $g \in G$, the generated subgroup $<g>$ must have infinite index. I tried to show this by contradiction by supposing there were only finite cosets of $<g>$ that parition $G$, but am not having much luck. Is there another direction that is good?

[cloud walrus]

Mr.Hahn-Banach

[willow mason]

G = <g1,g2>

g^1G is a coset

g^2G another coset

so on

[terse crystal]

G=Z ×Z, g is an element (m,n). If m doesn’t equal 0, then clearly {(0,tn)<g>: t from Z} are distinct elements, similar when n doesn’t equal 0. Trivial when m=0=n

[final oasis]

I am trying to show These are the only groups of order 15

My idea.
• Show that the groups of order 5 and 3 are unique
• then using that to show they are normal which would imply they are abilean
• ofter that using the gen of the subgroups of order 5 and 3 to get that the group must be cyclic

I want to brain storm ways to get the first point

[silent osprey]

sylow

[final oasis]

Yeah they gives that group of order 5 and 3 exist

[silent osprey]

third sylow

[final oasis]

Haven't seen that one

[silent osprey]

also those groups are isomorphic

[final oasis]

I know

That part of what I am to prove
Use the product of the gen of each subgroup

[final oasis]

Is there some basic intuition to it

[silent osprey]

I think this problem was discussed recently and it was pretty clear you need the uniqueness of subgroup to prove this group is cyclic #groups-rings-fields message

[final oasis]

I saw that, it was already assumed that the subgroups of order 3 and 5 were unique from the start

You could just count the elements offer that

Then there must be an element of order 15

[silent osprey]

the product logic also works because the order can only be 3,5 or 15 and due to uniqueness order of 15 would be forced

[final oasis]

What said helps, the only thing I am missing now is an intuition for sylow theorems

And that should wrap it up

There is another problem I was thinking about

Say p is the smallest prime dividing the order of a group G, H a subgroup of index p, then is H normal

[cursive temple]

does anyone have hints for this

[hidden haven]

E is K but you add the roots of f. Break this into a tower of extensions, adding one root at a time, and use the fact that the root of a polynomial must map to a root of the same polynomial under any field map

This way you'll be able to prove that the monomorphism extends to a monomorphism E/F → E/F. Prove that such a monomorphism must be an automorphism

[cloud walrus]

dumbo

[rustic crown]

so consider a g in (G n K) and an h in (H n K) what can you say about g h g'?

else if you're cat brained... think about the kernel of the composition K --> G --> G/H

[cloud walrus]

dumbo

[rustic crown]

But do you know if ghg' is in K? Where do all 3 of these belong?

[cloud walrus]

dumbo

[rustic crown]

Yep!

Also sorry for using g' for g^(-1)... Was too lazy to type lol

[cloud walrus]

Mr.Hahn-Banach

[deep nova]

So I'm studying about permutation groups

and the definition is fairly straightforward

but we got introduced the concept of symmetric groups on n objects, which consists of all permutations of {1,2,....,n}

And it says this symmetric group; the set has elements which are functions

but I'm not sure if I understand why this is so?

why is the set of functions who are part of the permutation groups called symmetric groups?

[final oasis]

symmetry of n objects

https://www.youtube.com/watch?v=RnqwFpyqJFw&list=PL8yHsr3EFj51pjBvvCPipgAT3SYpIiIsJ&index=1&t=854s

I think you might find the answer withing a few minutes

[deep nova]

thanks

[chilly ocean]

yesss I love this guy

[final oasis]

me too

he gives good intuition

[lavish nexus]

theres <= for subgroups

is there a similar symbol for subrings?

[barren sierra]

?

What is ≤ for groups @lavish nexus

[hidden haven]

H \leq G means H is a subgroup of G

use the same symbol for subrings I guess

just define it first

[tropic spade]

The degree of an element a over a field K which is algebraic over a subfield F of K is just the degree of the minimum polynomial of a over F right?

Pretty sure my book never defined what the degree of an element is, but I thought that fraleigh defines it that way. :/

[hidden haven]

Yes

[tropic spade]

Thank goodness.

[opal osprey]

Let $V$ be a finite dimensional vector space over a field $\mathbb{K}$ (We can suppose of characteristic zero if it makes things nicer). Suppose that we have $P(x) \in \mathbb{K}[x]$ a non-constant irreducible polynomial over $\mathbb{K}$ and $T \in \text{End}_{\mathbb{K}}(V)$. Is it true that
$$
\text{deg}(P) , , \vert , , \text{dim} \left(\text{ker} , P(T) \right)
$$
I.e, is it true that the degree of $P$ divides the dimension of the kernel of $P(T)$?

I was thinking about this problem while trying to solve another problem.

Idk how to actually approach this, maybe something in the lines of strucuture theorem for finitely generated PIDs?

[cloud walrus]

MisterSystem

[hidden haven]

Don't see why this should be the case. If you take K = C and V = C^4 then I can imagine that there would be a T which has a deg 3 minimal polynomial P and then 3 doesn't divide 4

Such a T should exist because you can take the matrix to be diag(1,1,2,3)

[rustic crown]

it says P should be irreducible... not just minimal

if you take C, then deg(P) = 1 so nothing to prove

[opal osprey]

Yup

I am interested in the case of Q

and not every polynomial in Q splits, so that's more interesting.

[rustic crown]

right

[hidden haven]

oh lol yeah

[opal osprey]

Just to give some insight into why I thought of this problem

I were doing some of Serge Lang's graduate algebra textbook problems

And I was thinking about this one for a bit

I actually solved it using the primary decomposition theorem/structurue theorem for finitely generated modules over a PID

It's in portuguese, but whatever

but then like

We have more specific decompositions that apply to vector spaces

We have lemme des noyaux

Which would give us this decomposition:

Let $\mathbb{K}$ be a field and $P,Q \in \mathbb{K}[x]$ be coprime polynomials. Then, for any vector space $V$ over $\mathbb{K}$ and $f \in \text{End}(V)$ we have:
$$
\text{ker} , P(f) \oplus \text{ker} , Q(f) = \text{ker} , PQ(f)
$$
In france they call this "Lemme des Noyaux'' or something and this is basically a particular case of the primary decomposition theorem lmao.

[cloud walrus]

MisterSystem

[opal osprey]

So applying this to $(x-1)$ and $x^4 + x^3 + x^2 + x + 1$, since they are coprime in $\mathbb{Q}[x]$ and using the fact that $ker(A-I) = 0$ would give us this decomposition:
$$
V \cong \text{ker}(A-I) \oplus \text{ker}(A^{4} + A^{3} + A^{2} + A+I) \cong \text{ker}(A^{4} + A^{3} + A^{2} + A+I)
$$

[cloud walrus]

MisterSystem

[opal osprey]

and by somehow proving that the the kernel of A^4 + A^3 + A^2+A+I is divisible by 4 would solve the exercise

this is in fact the first approach I tried

And I couldn't proceed lol

[rustic crown]

okie i got a solution....

so say W = ker P(T)

now W is T-stable right... because if w in W, then P(T)w = 0 then T * P(T)w = P(T) * Tw = 0 this shows Tw in W... this was because K[T] is commutative

so we can restrict the attention to W and forget about V

on W, T satisfies an irreducible polynomial P

so what we can do is give W a K[x] module structure where action of x is same as applying T

but this will factor through K[x]/(P)

so W has a K[x]/(P)-module structure

since P is irreducible that is actually a field

so W is just a K[x]/(P) vector space

W = (K[x]/(P))^r as vector spaces

now take dim with respect to K

done

[opal osprey]

[rustic crown]

dim W = dim (ker P(T)) = r * deg(P)

[opal osprey]

That's such a nice solution btw

[rustic crown]

[opal osprey]

I will actually have to reread it

But the whole idea makes sense

We end up with a decomposition for W

that reminds me of the decomposition we get applying the structure theorem

and is similar to the argument I applied before

Thank you so much btw

[coarse stag]

what does (H : K) even mean given H is a subgroup of K

[rustic crown]

index of H in K

number of left/right cosets of H in K

size of the set K/H

[coarse stag]

shouldnt it be the other way around tho

[rustic crown]

lol right

should be (K:H) just as in the final thing which they ask you to prove

[coarse stag]

ok thats what I assumed, but just wanted to check

also I dont see how (K:H) is nessesarily finite

since if H is the identity, then (K:H) = |K|

and if K is infinite, there seems to be an issue

[rustic crown]

that's given to you

[coarse stag]

oh nvm

[rustic crown]

the problem should be saying that both (G:K) and (K:H) are finite... and in this case show that (G:H) is finite and equals the product

[opal osprey]

Ok so det, there's a few things I should check; right? So for $W$ we have the monoid of $(\mathbb{K}[x], \cdot)$
$$
\sigma :& \mathbb{K}[x] \times W \rightarrow W \
\left(\sum\limits_{k=0}^{n} x^{k}, w \right)& \mapsto \sum\limits_{k=0}^{n} T^{K}(w)
$$
Right? And this is well defined because $T(w) \in W$ for $w \in W$.
\
\
So this makes $W$ into a $\mathbb{K}[x]$ module. Now, I have the quotient map.
$$
\pi : \mathbb{K}[x] \rightarrow \mathbb{K}[x] / (P)
$$
And I have to notice that for $w \in W$ fixed, I have the map
$$
\sigma_{w} : \mathbb{K}[x]& \rightarrow W \
\sum\limits_{k=0}^{n} x^{k}& \mapsto \sum\limits_{k=0}^{n} T^{k}(w)
$$
gives us a ring homomorphism.
\
\
Since $(P) \subset \text{ker}(\sigma_{w})$ by hypothesis, we in fact a well defined map
$$
\tilde{\sigma} : \mathbb{K}[x]/(P) \times W& \rightarrow W
$$
Which is induced by the first isomorphism theorem.
\
\
I have to check so $\sigma$ factors through $\pi$ and $\tilde{\sigma}$ and I have an induced $\mathbb{K}[x]/(P)$ module structure given by $\tilde{\sigma}$.

[cloud walrus]

MisterSystem
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[opal osprey]

This is just me trying to make sure I understood the proof lol

Filling out the details

One thing tho

That I don't understand

Is how do I guarantee W is still going to be finite dimensional

over K[x]/(P) ?

That might be a silly question

But I should check that

[rustic crown]

because it is finite dimensional over K

and K[x]/(P) is an extension of K

take any finite generating set of W over K, that same set will generate over the extension

[opal osprey]

Oh yeah, you are right

[rustic crown]

also i like to think of module as giving a ring map from R --> End(V) where V is any abelian group

so we had a ring map K --> End(W)

we can use substitution or the universal property of polynomial rings to get a map K[x] --> End(W) by sending x to T

[opal osprey]

I need to apply this

[rustic crown]

now the usual universal property of quotients

to get a map K[x]/(P) --> End(W)

[opal osprey]

I think that I was struggling by (for some reason) not realizing K[x]/(P) is an extension of K immediately lol (I need a bit of sleep)

[opal osprey]

Oh yeah

That's definitely a more direct approach

[rustic crown]

you don't really need it

because End(W) is not a field

[opal osprey]

No

With the fact that W is fin dimensional over K[x]/(P)

I should be more explicit with the stuff I am refering to sorry lol

[rustic crown]

wait but it still doesn't work like that right?

to show W is finite dimensional you need to show there is a finite spanning set

W is just an abelian group apriori

[opal osprey]

Aight

Appreciate the help

[rustic crown]

[rustic crown]

i should mention that here we need T to commute with all 'scalar' maps. else this won't work

like for instance if S and T were two non-commutative operators on W, then you can't use the universal property twice to get a K[x, y] module structure. because in K[x, y] the elements x and y commute, but their images S and T won't

will have to use 'non-commutative polynomials' in that case

[wooden ember]

this is just a consequence of the universal property associated to the ring of fractions right? Or am I missing some subtlety?

[wide hound]

if i divide by a fraction in an algebraic problem, how can i solve this? for example: x^2/4/3 = 10

thank you

[gritty sparrow]

I don't see how injectivity of the map follows directly from the universal property, I think that may be the subtlety

[hidden haven]

Assuming D doesn't contain 0

[hidden haven]

This is not appropriate for this channel, pls post in one of the channels in math help

[wooden ember]

injectivity is part of the property

[hidden haven]

bruh

[wooden ember]

?

[hidden haven]

Usually injectivity isn't required, requiring it makes it so that you can't localise at all multiplicatively closed subsets

[wooden ember]

this is when R is commutative btw

[hidden haven]

yes

[wooden ember]

and D contains no zero divisors

[hidden haven]

but yeah with this definition it is a tautology

oh

[wooden ember]

i know there's some generalization maybe that's what you're thinking about

[hidden haven]

then works out ig

[wooden ember]

they define that in chap 15

maybe ill give it a look before though cause localization seems important and i dont wanna wait that long

[hidden haven]

which book are you doing

[wooden ember]

havent stopped d&f

[hidden haven]

bruh why not do AM for commie alg

[wooden ember]

ima do that when i get to chap 15 exactly

since chap 15 is commutative alg but according to activechapter it sucks

so ill switch onto AM then

this section is just an intro to rings section

anyways gotta head out thanks

[hidden haven]

ok epic

cya

[hidden haven]

onto means surjection so be careful using the word like this 😌

[rustic crown]

is AM uwu?

if not i won't read it

Aluffi was a very book

[hidden haven]

🤮

[gritty sparrow]

It is Owo but not exactly uwu.

[rustic crown]

oh, ig it's worth a try then

[upper pivot]

pls for ur own sake dont read AM

and if u do read it alongside hartshorne

[hidden haven]

[weak oriole]

Is AM good for a intro to comm alg?
I've heard it is very thin and terse

[hidden haven]

It's good, you just have to do the exercises as well because there's some important stuff in exercises

Knowing some cat theory makes comm alg so much easier 😌

[wooden ember]

to show that it is the smallest (i first showed the latter part: that F has Z/pZ or Q as a subfield) i can either show it has no non zero subfields or that every subfield of F contains it right?

cause the former is much easier to show

[urban acorn]

yeah, showing either would suffice

but they're both really the same thing

[wooden ember]

yeah thought so

showing it's unique seems harder though

[hidden haven]

How does the former suffice? It just shows that Q or Fp is a minimal subfield

[urban acorn]

I mean suffices for showing that ℚ or ℤ/pℤ are the unique minimal subfield, given that there is a unique minimal subfield

but I realise now that might not be what they meant

[hidden haven]

oh ok yeah

[urban acorn]

@wooden ember you can take the intersection of all subfields

it's easy to see it's a subfield that is contained in all subfields

[hidden haven]

any subfield must contain one and the subfield generated by 1 is Q or Fp

[wooden ember]

ah right yeah

[wooden ember]

the double brackets for formal power series is such a pain to write

just ends up looking like a wonky cousin of the ring of formal laurent series

what's special about 1/2 here? Wouldn't this work with any element of R? (since we can always get some nZ a subring that doesnt contain that element and then easily conclude with Zorn?)

[untold cloud]

Is the statement: polynomial with degree n can have at most n distinct roots true for any field? Because I do not see the anything about fields is involved in the proof.

[plucky flicker]

Hi! How can i prove that if in a ring (possibly without identity) x^6 = x, then x^2 = x

I proved that 2x = 0 for all a in R, and i want to derive that x^2 = x

Without Jacobson theorem

[weak oriole]

Prereqs?

[hidden haven]

Just basic ring theory, though even that is done in the book, just a bit quickly

[hidden haven]

nZ is not a subring of Z because it doesn't have the same unit

[hidden haven]

Yes. If k is a field, k[x] is a UFD. Some a ∈ k is a root of f(x) ∈ k[x] iff (x-a) | f(x). A degree n polynomial can have at most n distinct linear factors

[hidden haven]

For all x I assume? It's not true for individual x

[wooden ember]

this is my definition though

no requirements for having the same unit

nZ doesnt have units but Z does, yet nZ is a subring of Z

[hidden haven]

[wooden ember]

bruh d&f why

ok but so this aside, with the definition of subring given by d&f what i said would be true

from now on i think ill cross reference every definition

i swear ima fail my linear algebra and abstract algebra exams because ill have been used to conflicting definitions

why cant mathematicians be standard

so i looked through my teacher's lin alg notes to make sure: he defines rings as having a multiplicative identity and subrings as having that same identity, and additionally defines a ring morphism as necessarily mapping the multiplicative identity of the domain ring to that of the target ring

this is making me mad

at least groups have standardized definitions why do rings have to be so inconsistent

[cursive temple]

ring morphism which doest map identites to identities yikes

sounds icky

[wooden ember]

if i fail math because of d&f im gonna sue

[cursive temple]

proof that d&f is objectively bad

[wooden ember]

bruh im sure it's not just d&f though

surely there are plenty of texts with different definitions

like those who'll use the term rng and others who'll say ring without identity: that's a standard example

[hidden haven]

d&f is taking rings to be non unital by default

Which is very 🥴

[wooden ember]

reeeee

[hidden haven]

Because we have a name for those things

They're called rngs 😌

[wooden ember]

i really dont wanna just quit on d&f though 😭

but this is making me really annoyed

you dont get it i have 600 pages of exercises written: i must finish what ive started

[hidden haven]

switch onto 😌

[wooden ember]

oh anyhow looking back at the question i posted they do specify 1 \in A

so it wouldnt work for just any number

but i guess 1/3 would have still worked

or just 1/n in general

[urban acorn]

I believe later it matters less

[wooden ember]

i mean theyve already assumed R is commutative for two whole sections

[urban acorn]

because you're gonna explicitly be working on like UFDs and polynomial rings and such

[wooden ember]

fair enough

ill remember from now on that the most standard thing is for rings to have a unit

gotta head out

[weak oriole]

Even groups don't
Herstein has some different definition of isomorphism

[urban acorn]

what utter nonsense is this clown writing

[brisk crescent]

Yeah, I would agree that it's not very clear

[hidden haven]

[urban acorn]

[hidden haven]

So there can be an isomorphism between non isomorphic groups

[urban acorn]

at least he defines "is isomorphic to" standardly

[wooden ember]

I guess it’s to emphasize that an injective homomorphism gives us an isomorphic copy of our domain group inside our target group???

Still fucked though

[urban acorn]

yeah

I guess

[opal osprey]

I am trying to apply this theorem to other problems now.

More specifically

Let $V$ be a real vector space endowed with endomorphisms $J,K \in \text{End}{\mathbb{R}}(V)$ that satisfy:
$$
J^{2} = K^{2} = -\text{Id}{V}
$$
and
$$
JK=-KJ= \text{Id}{V}
$$
And I want to show $\text{dim}(V)$ is divisible by $4$.
\
\
To apply the aforementioned result, I would have to find an operator $T \in \text{End}{\mathbb{R}}(V)$ and an irreducible degree 4 polynomial $P \in \text{R}[x]$ for which $p(T) = 0$. And I am not able to find such operator nor such a polynomial.

[cloud walrus]

MisterSystem

[opal osprey]

Any ideas?

I know how to prove this result using some other tricks

But it would be nice to apply that theorem det proved yesterday...

[rustic crown]

You probably have an error in the problem JK = -KJ = 1 doesn't make sense with earlier information

[opal osprey]

Oh

I meant JK=-KJ only

[rustic crown]

because JK + KJ = 0, and multiplying on the left by J then gives J^2 K + JKJ = 0 which means -K + J = 0

J = K

[opal osprey]

Idk why I wrote down Id v there

[rustic crown]

yea

the strategy you're taking won't work

there are no irreducibles of deg > 2 over R

but you can get around that with some non-commutativity

have you heard about quaternions?

[opal osprey]

Yeah, I know them

[rustic crown]

yee nice

[opal osprey]

This problem is motivated by them really

Btw

I proved this result

[rustic crown]

so you can show that V is a vector space over them

[opal osprey]

By showing V has a quaternion structure

Yeah, exactly

I wanted to know if that strategy would work tho

[rustic crown]

yee that works

quaternions satisfy those relations... j^2 = k^2 = -1 and jk = -kj = i

[chilly ocean]

hmmm

[opal osprey]

No, I mean. I know I could use some quaternion structure on V to show dim V is divisible by 4. What I wanted to know is that showing V is given by the vanishing of an irreducible degree 4 polynomial would work too.

[rustic crown]

oh nope

like i said... every irreducible over R has deg <= 2

[opal osprey]

Sheesh, then that lemma is quite restrictive

[rustic crown]

need some non-commutativity to get around this

but the proof isn't

you can practically use the same proof

V is an R-vector space, this is same as giving a ring homomorphism from R --> End(V)

End as in endomorphisms of it's abelian group, which has a natural ring structure under composition

now j and k don't commute (in fact they anti-commute) so we can use non-commutative polynomial (or tensor algebra stuff, if you've heard about it)

we'll get a map R<x,y> --> End(V) sending x to J and y to K

the kernel is a two-sided ideal containing x^2 + 1, y^2 + 1, and xy + yx

by universal property of quotients, this will factor through the quotient giving the map, R<x, y>/(x^2+1, y^2+1, xy+yx) --> End(V)

the first ring is isomorphic to the quaternion algebra H

which is a division ring, so you can still do linear algebra and basis and everything is a thing

so V is an H-vector space, so V is isomorphic to H^r as H-vector spaces, but you can forget structure to get isomorphim of R-vector spaces

taking dim_R gives dim V = r * dim H = 4r

[rustic crown]

so only this is left to verify

[opal osprey]

Sheesh, that's kinda nice.

[rustic crown]

But notice you have a ring map, R<x,y>/(x^2+1, y^2+1, xy+yx) --> H sending x to j and y to k

[opal osprey]

I mean, H is defined R[x,y]/(x^2+1,y^2+1, xy+yx) tho. Similar to how we define C as R[x]/(x^2+1)

So that's nice

[rustic crown]

oh if that's your definition then

btw people use [...] for commutative poly and <...> for non-commutative

(or they just say, symmetric/exterior/tensor algebra)

[strong moth]

i want to prove that for any set A, A is isomorphic to itself. does this proof work: let id_A : A->A, id_A-: A->A then we see (id_A^- o id_A) = id_A = (id_A o id_A) and that (id_A o id_A^-) = id_A = (id_A o id_A). therefore id_A- = id_A and (id_A o id_A) = id_A thus A is isomorphic to itself

[opal osprey]

That's a notation I haven't seen tbh

One thing that is nice tho

About this approach

[strong moth]

does the proof work?

[opal osprey]

Is that I didn't notice how important anti-commutativity would be here.

We can define the quaternions as a Clifford Algebra over the reals, and there we naturally define it as a quotient of the tensor algebra by a two-sided ideal

And this construction naturally appears in this proof

That's nice

[rustic crown]

dunno what Clifford Algebra is tho

[rustic crown]

in any case, the proof is actually pretty simple. This map is surjective because R<x, y> --> H is. And notice R-dimension is at most 4, because we can reduce x^2, y^2, yx in terms of 1, x, y, xy. So you have a R-linear map from that weird ring to H which is surjective, dimension of left can't be smaller, this with rank-nullity proves it's an isomorphim

[opal osprey]

If you have a field $\mathbb{K}$, which we usually take of characteristic different than 2 so that symmetric bilinear forms are identified with quadratic forms, and we also have a vector space over $\mathbb{K}$ endowed with a certain quadratic form $\varphi : V \rightarrow \mathbb{K}$, then we can define a certain "universal unital associate algebra", called the Clifford Algebra $\text{Cl}(V,Q)$ for which we have:
$$
v^{2} = Q(v) 1
$$
Where in the left hand side we have $v \in V$ and the product of $\text{Cl}(V,Q)$ and on the right hand side we have the unity of this induced Clifford Algebra.
\
\
In fact, if
$$
T(V) = \bigoplus_{n \in \mathbb{N}} V^{\otimes n}
$$
Is the tensor algebra of $V$, we define
$$
\text{Cl}(V,Q) = T(V)/(, v \tensor v - Q(v) 1 , )
$$
Where $(v \otimes v - Q(v) 1)$ denotes the two sided ideal generated by elements of the form $v \otimes v - Q(v) 1$ for $v \in V$.

[cloud walrus]

MisterSystem
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[opal osprey]

This construction satisfies a nice universal property with respect to others unital associative algebras

And like

At least for me

It seems to naturally appear in Representation Theory of Lie Algebras

That's where I mainly encounter these sorts of problems lol

[opal osprey]

Thanks

[rustic crown]

lol i had to read rep theory today... but was too lazy

hope my prof doesn't kill me

he gave us a homework to read till chapter 4 of Etingof and email us the label of all the subsections that he covered in the lecture and all the subsections that he didn't cover

[opal osprey]

Sheesh, that's a good amount of work

Good luck

[rustic crown]

this looks pretty nice...

[rustic crown]

lol better than solving some bunch of exercises xD

[lethal cipher]

Correct me if I am wrong, but the automorphism group for Z_2 is just the identity right?

[viscid pewter]

it is

[lethal cipher]

So then, wouldn't the automorphism group for the direct product Z_2xZ_2 also be the identity?

[viscid pewter]

no

you can 'swap' the two instances of Z2

with each other

for starters

there's more

[lethal cipher]

Hmm, then how do I try to find said things?

Also, is a "swap" a homomorphism?

[viscid pewter]

i'm using it to mean an automorphism on the generators, which is sorta the same

let G = <a, b|a^2 = b^2 = (ab)^2 = 1>

then the automorphism that maps a to b and b to a is what i'm talking about

[lethal cipher]

Z_2xZ_2 has no generators though

[viscid pewter]

it has no single generator

it is generated by a and b

[lethal cipher]

I'm not sure I understand

[viscid pewter]

so we can take a by itself

and then a^2 = 1, so we're done

and we can take b by itself

and then b^2 = 1, so we're done

and then we can look at all the possible ways of combining a and b

but ab = ba, so there's only ab

and for example abaaababab would just be the same as a^6b^4, it's abelian

so we have four elements

1, a, b and ab

and they're generated by <a, b>

to be clear, a group may not be able to be generated by any single element

a set whose elements generate a group is generally called a generating set, and the elements are called generators

[lethal cipher]

Hmmm, we certainly haven't discussed that yet, but I do understand

So how does this fact become useful to find Aut(Z_2xZ_2)?

[viscid pewter]

so to find all the automorphisms

we can just think of all the possible mappings of the generators

we can map:
a to a and b to b
a to a and b to ab
a to b and b to ab
a to b and b to a
a to ab and b to a
a to ab and b to b

[lethal cipher]

What is a and b in our group?

[viscid pewter]

any two elements of order 2

a, b and ab are all indistinguishable

[lethal cipher]

Okay sure. So any element that is not (0,0)

[viscid pewter]

generally, for any group (Zp)^n, all the elements except i have the same behaviour

where p is prime, naturally

[lethal cipher]

Ooo, that is nifty

[viscid pewter]

yep

but yeah that's one way to get the order of the automorphism group, and then it's fairly easy to see what the group actually is

well, i say fairly easy, i think you need to compute it

look at the composition of maps

[lethal cipher]

This is a good starting point. Ty

[viscid pewter]

np, this particular type of thing is an interest of mine

[lavish nexus]

I'm thinking I can let this object be a group ring S[G] but what should ring S be?

or just take it to be Z[G]?

[opal osprey]

Hey

Is there a place where I look for a proof of the Structure Theorem for Finitely Generated Modules over a PID that doesn't rely too much on the Smith Normal Form?

The proof I currently know uses a lot of the theory of matrices over a PID

And I prefer when things are done intrinsically

[thorn delta]

isn't smith normal form equivalent to the structure theorem? Anyway i'm pretty sure Lang doesn't rely on smith normal form

[potent briar]

how do i show that the annhilator of a subset of a vector space is a subsapce of its dual

🤯

[potent briar]

im having trouble coming up with an element of the annhilator for a small example

:x

[opal osprey]

Let $S \subset V$, where $V$ is a vector space over a field $\mathbb{K}$ and denote $\text{Ann}(S) := {\varphi \in V^{\ast} , \vert , \varphi_{|S} = 0 }$.
\
\
Well, we have first to show $0 \in \text{Ann}(S)$, which is obvious and we have to show that $\forall \varphi, \psi \in \text{Ann}(S)$ and $\lambda \in \mathbb{K}$ we have $\varphi + \lambda \psi \in \text{Ann}(S)$.
\
\
This is indeed the case, $\forall x \in S$ we have that $(\varphi + \lambda \psi)(x) = \varphi(x) + \lambda \psi(x)$. Since $\varphi, \psi \in \text{Ann}(S)$, we have $\varphi(x) = 0$ and $\psi(x) = 0$. Therefore, we have $(\varphi + \lambda \psi)(x) = 0, \forall x \in S$ and so $\text{Ann}(S) \subset V^{\ast}$ is a subspace of $V^{\ast}$.

[cloud walrus]

MisterSystem

[opal osprey]

It pretty much follows by definition

Try redoing the proof by yourself

[potent briar]

makes sense

but for example, whats a member of ${(2,1), (3,4)}^0$, just to get a feel for them

(vectors on R^2)

[cloud walrus]

vik

[potent briar]

y(x) = 0, y(x) = x - x

[opal osprey]

Oh, yeah in this example we can compute a basis for the anihilator for such a subset explicitly!

[potent briar]

even more basic

i want to see a functional that annhilates both of them

just 1

[opal osprey]

That's what we will do

I just want to show you that we can basically just solve a system of linear equations

in order to find such functionals

In this case, it is pretty boring, the 0 linear functional is the only linear functional that vanishes on this set

Notice this pretty basic property

Let $S \subset V$ be a subset of a vector space $V$ over a field, then we have $\text{Ann}(S) = \text{Ann}( \text{span}(S))$.

[cloud walrus]

MisterSystem

[opal osprey]

Try to prove this

moreover

notice that (2,1) and (3,4) are linearly independent

and considering {(2,1),(3,4)} as a subser of R^2

we have that they span R^2

so Ann({(2,1), (3,4)}) = Ann(R^2)

but the only linear functional that vanishes on the whole of R^2 is the 0 linear functional

so Ann({(2,1), (3,4)}) = {0}

Do you get that?

[potent briar]

is $y(x) = x_1 - x_1$ not a linear funcitonal?

[opal osprey]

isn't that just 0?

[potent briar]

oh sure

[opal osprey]

yeah, as we have seen, the 0 linear functional is the only one that vanishes on {(2,1), (3,4)} considering it as a subset of R^2

[opal osprey]

Try doing this exercise btw

Want some other more interesting example?

Let's think about {(1,1,3), (1,0,2)} in R^3

the linear functionals that vanish on this set are not all 0

in fact, they form a 1 dimensional space

Notice that a linear functional in $\mathbb{R}^{3}$ is a function
$$
\varphi : \mathbb{R}^{3} \rightarrow \mathbb{R}
$$
such that $\varphi(x,y,z) = a_{1} x + a_{2} y + a_{3} z$ for some coefficients $a_{1}, a_{2}, a_{3} \in \mathbb{R}$.
\
\
and we want to find linear functionals such that the coefficients satisfy
$$
a_{1} \cdot 1 + a_{2} \cdot 1 + a_{3} \cdot 3 = 0
$$
and
$$
a_{1} \cdot 1 + a_{2} \cdot 0 + a_{3} \cdot 2 = 0
$$

[cloud walrus]

MisterSystem

[opal osprey]

So we need to solve a system of linear equations

which is this one

$$
\begin{cases}
a_{1} + a_{2} + 3 a_{3} = 0 \
a_{1} + 2 a_{3} = 0
\end{cases}
$$

[cloud walrus]

MisterSystem

[opal osprey]

and if we solve it...

,w x+y+3z = 0, x+2z = 0

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by MisterSystem#1236

[opal osprey]

input any of these solutions into your coefficients

and we get an example of a linear functional in the annihilator of {(1,1,3),(1,0,2)}

for example

$\varphi(x,y,z) = x + \dfrac{1}{2} y - \dfrac{1}{2} z$

[cloud walrus]

MisterSystem

[opal osprey]

I encourage you to check this linear functional is inded in {(1,1,3), (1,0,2)}

but yeah, in general, that's the way to find examples of linear functionals in the annihilator of a set.

@potent briar

[potent briar]

If $y \in S^0$ then $y(x) = 0$

so $y(\alpha x_1 + \beta x_2) = \alpha y(x_1) + \beta y(x_2) = \alpha 0 + \beta 0 = 0$

[cloud walrus]

vik

[potent briar]

so the annhilator of the subset = the annhilator of the span?

[opal osprey]

Yeah, that's basically how the proof goes

[potent briar]

well actually $y(span(S)) = y(\sum_{i}\alpha_i s_i) = \alpha_1 s_i + ... + \alpha_n s_n$

$= \alpha_1 0 + ... + \alpha_n 0 = 0$

[cloud walrus]

vik

[potent briar]

thats better

[opal osprey]

yup, there you go!

Ann(S) = Ann(span(S)) also gives you insightul things about the dimension of the annihilator of a set

[potent briar]

yeah thats the first theorem in this chapter

but i wanted to have concrete examples to make the idea more tangible first

[opal osprey]

.

[opal osprey]

.

here you go

[potent briar]

yeah im looking at it

thanks :p

[opal osprey]

np

[potent briar]

ooh right so

the Ann of subset that spans the whole space

is always {0}

interesting

[opal osprey]

yeah, that's how I found out the annihilator of {(2,1) , (3,4)} so quickly

[potent briar]

but in your example

you had only 2 independent vectors in a space of dim 3

so what if you just try to do the system of equations on the R^2 example, it just has no solution?

[opal osprey]

It has only a trivial solution

the system of equations would be

$$
\begin{cases}
2 a_{1} + a_{2} = 0 \
3 a_{1} + 4 a_{2} = 0
\end{cases}
$$

[cloud walrus]

MisterSystem

[opal osprey]

the solutions of this system of equations

[potent briar]

only 0

[opal osprey]

yup

and in general

the solution of such a system of equations

will give you information about a basis for the annihilator

and about the dimension of the annihilator

try doing this as an exercise on your own now

find a basis for the annihilator of {(1,0,1,3),(2,4,1,2)} as a subset of R^4

[potent briar]

you missed a coordinate in the second vector

or maybe just missed a comma

[opal osprey]

Oh, yeah I missed a comma

sorry

[potent briar]

ok i solved the system but i dont get how you go from that to the coefficients of the functiona

A + C + 3D = 0
2A + 4B + C + 2D = 0

A = D - 4B
B = D/4 - A/4
C = -3D - A
D = A + B

[opal osprey]

find a basis for the solution set

it will make things more straightforward

since you have already solved the system, this should be no problem

[potent briar]

basis for the solution set?

im sorry, i don't have a background in pure maths, this is my first real maths textbook

[opal osprey]

notice that we can write the solutions as C = -4A - 12B, D = A+4B

With A,B arbitrary

so notice that any solution can be written as follows

$$
\begin{bmatrix}
A \
B \
C \
D
\end{bmatrix}

\begin{bmatrix}
A \
B \
-4A - 12B\
A+4B
\end{bmatrix}

A \cdot
\begin{bmatrix}
1 \
0 \
-4\
1
\end{bmatrix}
+
B \cdot
\begin{bmatrix}
0 \
1 \
-12 \
4
\end{bmatrix}
$$

[cloud walrus]

MisterSystem

[opal osprey]

Right?

any solution of that equation can be written in this way

to the solution set, i.e the set of all solutions of the aforementioned equation

has as a basis {(1,0,-4,1) , (0,1,-12,4)}

and you know why we wanted to solve that system of equations, right?

the solutions of this equation

will give us the coefficients

of a linear functional on R^4

that vanish on {(1,0,1,3),(2,4,1,2)}

and when we solved this system of equations

we find that any such coefficients can be given as a linear combination of (1,0,-4,1) and (0,1,-12,4)

in particular

a linear functional with coeffiecients given by (1,0,-4,1) vanishes on {(1,0,1,3),(2,4,1,2)}

a linear functional with coefficients (0,1,-12,4) too

and any linear combination of these

[potent briar]

!!

[opal osprey]

$$
\varphi_{1}(x,y,z,w) = x-4z+w
$$
and
$$
\varphi_{2}(x,y,z,w) = y-12z+4w
$$
are a basis for ${(1,0,1,3),(2,4,1,2)}^{0}$

[cloud walrus]

MisterSystem

[opal osprey]

that's what we found

[potent briar]

the fact that there are two free variables (A, B) and 2 others that depend on those, is because we had 2 vectors in a 4dim space?

[opal osprey]

Exactly !

To be more precise

We have the following theorem

Let $V$ be a finite dimensional vector space over a field $\mathbb{K}$ and $S \subset V$ any subset. Then, we have:
$$
\text{dim}(\text{Ann}(S)) = \text{dim}(V) - \text{dim}(\text{span}(S))
$$

[cloud walrus]

MisterSystem

[opal osprey]

so the dimension of the annihilator of a subset

in a finite dimensional vector space ofc

is given by the dimension of the whole space

minus the dimension of the span of that subset

in our case, V = R^4 and so dim(V) = 4, and S = {(1,0,1,3),(2,4,1,2)} where dim(span(S)) = 2 since these are linearly independent

so we (without any sorts of computations)

conclude that dim(Ann(S)) = dim(V) - dim(span(S)) = 4 - 2 = 2

we don't even need to solve a system of equations to get information about the dimension

[potent briar]

right

[opal osprey]

btw, this is a consequence of rank-nullity

and I think can be even more readily seen from the more general formulation of the first isomorphism theorem

if you have seen any of these

give it a try at trying to solve this exercise

[potent briar]

hmm halmos does it in a seemingly more roundabout way

using dual basis

(the chapter right before annhilators)

[opal osprey]

Prolly isn't that much of a different argument tbh

If you have covered quotient spaces at the moment

[potent briar]

https://download.tuxfamily.org/openmathdep/algebra_linear/Finite_Vector_Spaces-Halmos.pdf#page=33

[potent briar]

no

[opal osprey]

oh sheesh

[potent briar]

what lol

[opal osprey]

I just thought this one argument using the first isomorphism theorem would be nice to state

but that's ok if the book hasn't covered it yet