groups-rings-fields

[hot lake]

ah well that changes things

I missed that assumption

[wooden ember]

itll be easier if i send you a photo of what i did gimme a minute

off of the motivation that this was shown earlier:

[hot lake]

I get cognitive dissonance from the difference in how much you care about both parts of the proof

[wooden ember]

it's not really a quesiton of how much i care it's more about how i write my exercises

i dont write all that i calculate in my head but sometimes i need to write the computations

so the proofs can seem disproportionate

[hot lake]

it's good but it's weird how you put so much detail on how to get (23) form (12) and not on how to get (12) from sigma and tau

[wooden ember]

and i could clean them up

[hot lake]

which is the harder part in my opinion

[wooden ember]

i maen again the exercise isnt complete

[hot lake]

how so

[wooden ember]

i still need to go from the ai's to i's

which was why i asked my question in the first place

[rustic crown]

have you seen how conjugation affect cycles?

[hot lake]

^^^^^^^^^^^^^

[wooden ember]

not formally only through experience

[rustic crown]

g (a1 a2 ... an) g^-1 = (g(a1) g(a2) .... g(an))

[wooden ember]

oh i see

[hot lake]

well you know that you can get any permutation of {a1 ; ... ; ap}

but {a1 ; .... ; ap} = {1 ; ... ; p}

[wooden ember]

right of course

i have all permutations of a that's enough

im a bit shtupid

since S_p is just permutations on p elements not specifically 1,2,...,p

lmao

[hot lake]

and your question about if you have a group morphism from <S> to <T> that's bijective on S, then is that group morphism bijective

well there are counterexamples

[wooden ember]

yeah as det gave

it did feel ishy

fishy*

[novel parrot]

looking at the sylow theorems

but i dont understand why we can take Q = P_1

would Q be a sylow p group itself?

[rustic crown]

i'm guessing that Q is a sylow p-subgroup and you're acting on the set of sylow p-subgroups by conjugation

number of Q-conjugates of a sylow p-subgroup P is [Q : N(P) intersect Q]

you can choose Q to be the representative of the orbit containing Q

[novel parrot]

it says that Q is just any P subgroup

so just wondering how they can let it be p1 ?

[rustic crown]

Pi are representatives of the orbits, right?

There would be one orbit which contains Q

infact that orbit has size 1 since number of Q-conjugates of Q is [Q:N(Q) intersect Q] = 1

that's what we're trying to show, number of Q-conjugates of Q is 1 but number of Q-conjugates of P is divisible by p where P is any sylow p-subgroup different from Q.

O1, ..., Os are the different orbits. We can number them arbitrarily. One of this orbit must contain Q, wlog it was O1, but then the only representative of O1 is Q which we call P1

[novel parrot]

oh! that makes sense!

but one thing

why is the only representative for the orbit containing Q, only Q ?

@rustic crown

[rustic crown]

what is even representative?

say a group G acts on a set X. And consider some orbit O.

what is a representative of O?

[novel parrot]

some g(x)

[rustic crown]

what's g, what's x?

[novel parrot]

g is something for the group, x from the set

[rustic crown]

yea, so O_x = {g*x | g in G}

but notice that any element of O_x equally represents O_x

O_{gx} = {h * (gx) | h in G} = {(hg) * x | h in G} = {g' * x | g' in G} = O_x

so a representative is just an element of the orbit O

if you know one element, you can recover every other just by multiplying by elements from G

now in the actual problem, we say that the orbit of Q is set of all Q-conjugates of Q, which you proved (probably in that lemma 19) that has size 1. so O_{Q} = {Q}, so Q is its only representative

(if you're confused about that lemma, just use orbit-stablizer theorem. |O_P| = [Q : Stab(P)]. Now Stab(P) = {q in Q | qPq^-1 = P} = {q in Q | q in N_G(P)} = [N_G(P) intersect Q])

[novel parrot]

Q is acting on Conjugates of Sylow P subgroups

$q (gPg^{-1}) q^{-1}$ for some g in G and all q in Q

[cloud walrus]

ActiveChapter

[novel parrot]

that would be an orbit right?

[rustic crown]

no

that would just give you all sylow p-subgroups

this is pretty much the content of sylow's second theorem

i'm saying the orbit containing P would be the set {qPq^-1 | q in Q}

(here the Q is the group which acts on the set X = {sylow p-subgruops})

[novel parrot]

ok ive understood this

but not why there is an orbit containing Q and why the whole orbit would be just Q

[rustic crown]

Q is a sylow p-subgroup

X = set of sylow p-subgroups

Q is in X

[novel parrot]

Q is just a p group

so not in X ?

[rustic crown]

oh i see, i had to guess what Q was before.

[novel parrot]

sorry, i shouldve shown this

[rustic crown]

they're saying we did that calculation for an arbitrary p-subgroup. so it should also work for some particular p-subgroup say P1

does that make sense?

[novel parrot]

so we need to take Q as a sylow subgroup to get that result?

[rustic crown]

yup

but that size of orbit in equation (4.1) would work for any p-subgroup Q

[novel parrot]

yep

[rustic crown]

ype, if none of Pi contain Q, then [Q:N(Pi) intersect Q] is divisible by p for every i.

which contradicts the sum is 1 mod p

[novel parrot]

yep

[rustic crown]

[novel parrot]

You are a life saver!

[rustic crown]

[wooden ember]

so my question is regarding ex 8 (ex 7 is in the image for reference). Ive completed the exercise because i already knew the standard way to show A_4 has no subgroup of order 6, but I couldn't figure out another way to show this using this knowledge that it is isomorphic to the group of rigid motions of the tetrahedron. Any ideas?

[rustic crown]

i feel like this is a rephrasing of some argument, but still think its nice enough to be said.

let that subgroup act on the tetrahedron with vertices labeled 1, 2, 3, 4. Now what is the stablizer of 4? if we fix 4 at the back, we can only do {(1), (123), (132)} to move around the front triangle by rotating it. So the stablizer must be subgroup of that and has size either 1 or 3, 
-it can't be 1, because in that case orbit must have size 6 and there are only 4 vertices of the tetrahedron!
-so those 3 are precisely the things in the stablizer. which means orbit of 1 2 and 3 are the same thing. but then orbit of 4 must have size 2, and it can't include any of 1, 2, 3 as once you include one, you need to include others which is a contradiction!

since i didn't use the word "even permutation" i'll call it a win.

[wooden ember]

that's a nice one but i dont think it's what the book was suggesting as group actions were briefly addressed in section 1.7 and they are delved into more in depth in chapter 4 (right after these exercises). This here is section 3.5

it's a shame authors dont publish solutions to their exercises so you know what was going on in their head

[prisma ibex]

yea this is nice, especially with the previous proposition. I'd make sure you know how to do this without the action of A_4 on the cube, but you probably get this lol

nice solution

[ivory dust]

How do I apply this line of thinking

to the polynomial in Q[x]:

$f(x) = (x^4-2x^2-1)^2 - 5$

to prove its solvable?

Should I first expand?

or I was thinking maybe let x^2 = y

then do f(y) solvable

but idk if that implies f(x) solvable

$f(x) = x^8-4x^6+2x^4+4x^2-4$

if I expand

$f(y) = y^4-4y^3+2y^2+4y-4$

[cloud walrus]

Ramtin

Ramtin

Ramtin

[rustic crown]

its solvable if you can solve it

[ivory dust]

😩

[rustic crown]

$((x^2 - 1)^2 - 2)^2 - 5$

[cloud walrus]

det

[rustic crown]

so consider the extensions

[ivory dust]

F = Q (right?)

then B = F(something)

B = F(sqrt5, sqrt2)?

err

B1 = F(sqrt5)

B2 = B1(sqrt2)

would be better?

[rustic crown]

$\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{5}) \subseteq \mathbb{Q}(\sqrt{5}, i) \subseteq \mathbb{Q}(\sqrt{5}, i, \sqrt{2 + \sqrt{5}}) \subseteq ...$

[cloud walrus]

det

[rustic crown]

that 2-sqrt{5} would be negative. so you would also require i

[ivory dust]

will it include 5^(1/4)

[rustic crown]

that's usually not very easy to say

[ivory dust]

okkk

tysm

Hm wait whered the i come from

shouldnt it be i*sqrt(2+sqrt5))

not just a lone i

[rustic crown]

usually just get extension for each root, and then use the fact that composites of radical extensions are radical extensions

here i was trying to get a small looking extension

so Q(sqrt(5)) contains both +- sqrt(5)

now in the next step we want to include +- sqrt(2 +- sqrt(5))

[ivory dust]

Ohh

wait i have an idea would this work?

[rustic crown]

you can just include both separately, but i wanted to include 1 and use i to get other. but got too lazy to finish it

[ivory dust]

Q < Q(sqrt5) < Q(sqrt5, a) < Q(sqrt5, a, b)
a^2 = 2+-sqrt5
b^2 = 1 +- a

[cloud walrus]

det

[ivory dust]

Ohh

yeaa

would my idea cover those cases?

so itd just be Q(sqrt5, a, b)

[rustic crown]

a^2 = 2+-sqrt5
what does this mean? should i take + or -?

[ivory dust]

both i guess haha

if thats not allowed i guess i could write it as a1, a2, b1, b2

[rustic crown]

a shouldn't mean 2 different things... you need to include both

[ivory dust]

Oh i see

okk tyvm

[rustic crown]

[ivory dust]

so in my version which kinda poorly written thered be 2 answers for a, and 4 for b

which would match urs

okok i get it

[rustic crown]

yep

[short pulsar]

Hey! I asked a question in the dynamical systems channel yesterday, unfortunately I didn't get an answer. I suspect the question has quite a lot to do with number theory and or abstract algebra too. Please head over and look at it if you feel like it. At the time of writing, it is at the bottom :/

[wooden ember]

i completed exercise 3 but i probably did it wrong as I didnt use the fact that G is abelian. What could I be missing?

exercise 1 also showed that $\bigcap_{g\in G}gG_ag^{-1}$ is the kernel of the action for transitive G acting on A. So i used it along with 2 to show G is faithful which gives us $\sigma(a)\neq a$ for $\sigma\in G-{1}$, $a\in A$

[cloud walrus]

Little Narwhal

[wooden ember]

Then the bijection just followed from the map $\sigma \to \sigma(a)$ being a bijection: surjective because G is transitive, injective because G is faithful

[cloud walrus]

Little Narwhal

[wooden ember]

but nowhere does G being abelian factor in here

[dusty river]

I think there's a mistake here, I don't see how you deduce sigma(a) neq a for all sigma and a

[wooden ember]

$\sigma(a)=a$ iff $\sigma(a)=id(a)$ iff $\sigma = id$

because G is faithful

[cloud walrus]

Little Narwhal

[dusty river]

Yes

And both of the left statements have a (for all a)

But when you negate all 3 statements

it becomes (there exists a such that not ....)

not (for all a, not ....)

[wooden ember]

suppose there exists a such that $\sigma(a)=a$ then $\sigma(a)=id(a)$ so $\sigma = id$, a contradiction

[cloud walrus]

Little Narwhal

[wooden ember]

is this incorrect?

[cloud walrus]

Coldilocks

[dusty river]

look at the effect of (12) on the set {1,2,3}

It fixes 3, but isn't identity

[wooden ember]

Ahhhh i see

it only has the effect of the identity locally right

thanks

[dusty river]

Yeah

[wooden ember]

so i suppose it is at this step that we must use the fact G is abelian

[dusty river]

Should be

[wooden ember]

ah i got it

Suppose $\exists a\in A$ with $\sigma(a)=a$ for some $\sigma\in G-{1}$. Then $\sigma\in G_a$. So by the previous exercise $\bigcap_{\tau\in G}\tau{\sigma}\tau^{-1} \subset \bigcap_{\tau\in G}\tau G_a\tau^{-1}=1$. Since G is abelian, $\tau\sigma\tau^{-1}=\sigma$ for all $\tau\in G$. So $\sigma\in \bigcap_{\tau\in G}\tau{\sigma}\tau^{-1}$ which is a contradiction.

[cloud walrus]

Little Narwhal

[dusty river]

That's neat

[wooden ember]

so id like to say that $G_a$ is isomorphic to $S_{n-1}$ which is transitive on ${1,2,...,n-1}$ so $G_a$ must be transitive on ${1,2,...,a-1,a+1,...,n}$ but i dont know how to say it properly since the sets being acted on in both cases are different so it's not as simple as just saying the groups are isomorphic

[cloud walrus]

Little Narwhal

[gritty sparrow]

What you really need is a pair (f,b) of a group isomorphism f:G_a->S_(n-1) and a bijection of sets b:A\{a}->{1,2,…n-1} that are compatible, ie: b(g.a)= f(g).b(a)

[unique juniper]

dont really see how they both should be normal

[mild laurel]

What does it mean for a subgroup to be normal?

[unique juniper]

if nothing happens to the subgroup under conjugation

ive just tried considering the amount of elements

i can see that we have either sylow 7 or sylow 5 to be normal

if we have a normal sylow 7 group, we could have 21 sylow 5 groups

[mild laurel]

If you let P be the sylow 7 subgroup and Q be a sylow 5 subgroup, if you assume that P is normal, then PQ is also a subgroup of G. Can you show that this subgroup has to be normal in G?

[unique juniper]

ok yep, i got it thanks!

[ivory dust]

if a is a root of a polynomial w coefficients in F(r) then is it a root of a polynomial with coefficients in F?

trying to explain/handwavy proof that constructible numbers are algebraic (in? over? Q)

[maiden ocean]

Well F(r)(x) = F(r, x) right?

[ivory dust]

but i dont x has to be in F(r)

err

idk

[maiden ocean]

x is an arbitrary second variable right

[ivory dust]

yea

i agree w ur statement

[maiden ocean]

Yeah so you're basically asking if a being a solution to a 2 variable polynomial in F implies it is a solution to a 1 variable polynomial in F, right?

(you have to fudge this a bit because its technically F(r)[x] but ill leave the details to you)

[gritty sparrow]

Is r algebraic over F? That changes the answer and given the context seems likely

[maiden ocean]

Oh yeah in the algebraic case its different

(you can show this through transcendence degree arguments)

[gritty sparrow]

I mean, algebraic over algebraic is algebraic can be done through pretty elementary means like saying that F(r,a) is finite over F

[maiden ocean]

Yeah I mean thinking about the general case

[gritty sparrow]

If r is transcendental, then x-r has a solution r which is not algebraic over F

[ivory dust]

i saw this argument online

if xo is a root of polynomial in Fk then its a root of polynomial w coeff in Fk-1 and so on until Q

where Q < F1 < F2 < ... < Fk is a tower of fields

Fi = Fi-1(sqrt(ri))

for some ri where sqrt(ri) not in Fi-1

and the argument was if x0 is a root of polynomial in Fk then its a root of polynomial in Q

hence x0 (constructible #) is algebraic over Q

but i dont understand this line

[ivory dust]

this

[maiden ocean]

Is r_i in F_i-1?

cause then sqrt(r_i) is the root of r_i^2 and these extensions are all algebraic

[ivory dust]

like if y is a root of $f(x) = sqrt(2)x^2+5$ then is y a root of another polynomial in just Q? ratrher than Q(sqrt2)?

[cloud walrus]

Ramtin

[gritty sparrow]

This is because an algebraic extension of an algebraic extension is algebraic. Fk(x0) is algebraic over Fk. Fk is algebraic over F_(k-1) so F_k(x0) is algebraic over F_(k-1)

[ivory dust]

Ohhh

i know that theorem

truee

and any finite extension is algebraic right?

or is it the otherway around

[maiden ocean]

Yeah so if x_0 is algebraic over F_k it is algebraic over Q

Finite extensions are algebraic yes

[ivory dust]

but algebraic ext arent necessarily

finite?

[maiden ocean]

Yes

[ivory dust]

ooh ok ok tys

😄

[unreal portal]

I asked this question a few days ago and haven't made any progress, asking again and hoping someone else will know what to do lol.
I'm trying to compare ideals in $\mathbb{Z}[t, t^{-1}]$, so my instinct is to construct a Grobner basis for them. I do this by mapping the ideals to the ring $R=\mathbb{Q}[x,y]/(xy-1)$. The problem is that I'm doing this in sage, and I don't know the 'right' way to do the computation. Should I do the computation all in $\mathbb{Q}[x,y]$, then take the quotient of each generator, or find the basis directly in the quotient ring? I've tried both in sage and they seem to give different answers, which is why I'm a little confused

[cloud walrus]

cgodfrey

[unreal portal]

if I do

R.<x,y> = PolynomialRing(QQ)
Rq = R.quotient([x*y-1])
ideal1 = Rq.ideal([2*x^2 - 4*y + 1, 3*x^3-2*x^2+y^2-1])
ideal2 = R.ideal([2*x^2 - 4*y + 1, 3*x^3-2*x^2+y^2-1])
print(ideal1.groebner_basis())
print(ideal2.groebner_basis())
print(x*y in ideal2.groebner_basis())

I get the output

[1]
[y^3 - 319/4*y^2 + 45/8*x + 51*y - 9/2, x^2 - 2*y + 1/2, x*y + 1/6*y^2 - 1/4*x - 2/3*y]
False

So the ideal in the quotient ring is the entire space, while the ideal in the parent ring is generated by those three polynomials. But since xy isn't in that ideal, I feel like that suggests if I take the quotient of each generator, I would end up with a different ideal

lemme know if anyone has ideas on how I might be going wrong

[sullen island]

heya, im trying to understand this remark by explicity defining $\beta$, but its turning out to be surprisingly hard, would appreciate any help on seeing this fact (which seems to be 'trivial' according to the text)

[cloud walrus]

xy

[fossil shuttle]

choose bases tbh

[sullen island]

this is what i've got so far

[fossil shuttle]

ok lol i think you're overthinking this. if it helps, the only space you need to choose a basis for is $P$

[cloud walrus]

diligentClerk

[sullen island]

??!?!

[fossil shuttle]

Haha. Ok. Let's say you have a basis for $P$, some $\left{p_i\right}$. Start by figuring out where these should go. where can you send $p_i$ such we have $\varepsilon(\beta(p_i)) = \gamma(p_i)$ for each $i$?

[cloud walrus]

diligentClerk

[fossil shuttle]

think Axiom of Choice

if it's not obvious you do need to use the assumption that $\varepsilon$ is surjective.

[cloud walrus]

diligentClerk

[sullen island]

if i use axiom of choice, dont i need to specify a basis for $B$ anyway?

[cloud walrus]

xy

[fossil shuttle]

no. i promise you do not need to choose a basis for $B$ or $C$ here.

[cloud walrus]

diligentClerk

[fossil shuttle]

ok

forget vector sapces

just replace these with sets

lol

[fossil shuttle]

$P, B, C$ are all sets with no extra structure.

[cloud walrus]

diligentClerk

[novel parrot]

why does this kernal contain H ?

[fossil shuttle]

$\varepsilon$ is surjective, $\gamma$ is some map $P\to C$

[sullen island]

P to C lol

[cloud walrus]

diligentClerk

[sullen island]

yup

[fossil shuttle]

can you find $\beta$ in this case?

[cloud walrus]

diligentClerk

[novel parrot]

anyone know?

is it just that the stabilizer of H is just H so the kernal must be something less than H?

[sullen island]

by axiom of choice, there exists $\beta'$ mapping $P$ to $B$ ? and then we linearly extend $\beta'$ to obtain a linear transformation $\beta$ ?

[cloud walrus]

xy

[fossil shuttle]

yep.

[sullen island]

but thats not explicitly "finding" $\beta$

[cloud walrus]

xy

[fossil shuttle]

nope.

that's fair

maybe "find" was the wrong word lol

[sullen island]

im not so convinced that it proves the statement - linear transformations are defined by their action on bases

so in particular we have to specify bases in order to prove existence of $\beta$

[cloud walrus]

xy

[fossil shuttle]

yes

well

there exists a basis for $P$

[cloud walrus]

diligentClerk

[fossil shuttle]

by the axiom of choice

[sullen island]

axiom of choice only gives a choice function but it may not be `linearly extend-able' to give beta

uhmm

[fossil shuttle]

you can prove from the basic properties of a basis that every function from a basis for $P$ into $B$ extends in a unique way to give a linear transformation

like

[cloud walrus]

diligentClerk

[sullen island]

oh right.... right

[fossil shuttle]

let $v$ be a vector in $P$. then $v$ can be written in one and only one way as $\sum a_i p_i$. so choosing where to send the $p_i$ tells you where $v$ has to go

[cloud walrus]

diligentClerk

[sullen island]

so doing it interms of what i did here was overthinking it

[fossil shuttle]

yeah, that's not necessary

this is a great property of bases btw

good one to keep in your back pocket

[sullen island]

alright, thank you 🙂

[chilly ocean]

Guys, Isn't the answer correct?

[dusty river]

nope. Why are you working with 2 different fields?

[delicate bloom]

you can't prove by example

the field F is totally generic, you can't appeal to trying to reduce in F_2 or something like that

[dusty river]

They seem to be claiming that it is false

or not

[chilly ocean]

Isn't it same as proof by induction ?

[dusty river]

nope, proof by induction handles all cases

not a specific one

[delicate bloom]

well, maybe it helps to see a degree 4 that is reducible that doesn't have a root

having a root really just means it has a linear factor

a degree 4 can factor into two irreducible quadratics, so it doesn't necessarily have a root in this case

but degree 3 is too small of a degree to factor into stuff where none of the terms are linear

[vestal snow]

Let pi: X --> Y be a qcqs morphism of schemes and let F be a quasicoherent sheaf over X. Show that pi_* F is a quasicoharent sheaf over Y

What is the module action of Y on pi_* F?

[vast quiver]

so what is the data for a scheme morphism?

going forwards you have a continuous map X—>Y

[vestal snow]

Oh lmao I just figured it out I think

The map of ringed spaces is in the other direction

kind of

Thanks

[vast quiver]

yeah, backwards you have a sheaf morphism O_Y->pi_* O_X

[oblique matrix]

Let G be a group containing at least one subgroup of a fixed finite order s. Show that the intersection of all subgroups of G of order s is a normal subgroup of G. [Hint: Use the fact that if H has order s then so does x^{−1}Hx for all x∈G.]

I'm stuck on this problem. Conjugation permutes the subgroups of order s, but I'm not sure how to formally show it.

[vestal snow]

Show that x_x^-1 is a bijection on such subgroups

for all x

In order to do so, come up with an inverse

Do you see what the inverse must be?

[oblique matrix]

that part was fine since conjugation between subgroups is always a bijection

(group multiplication always injective and onto their image)

[vestal snow]

No, I meant show that this function is a bijection on the set of all subgroups of order s

[oblique matrix]

ah i see

Ok

Let f:A->A where A is the set of all subgroups of order s be defined by f(B)=xBx^{-1}

this is well defined because conjugating subgroups preserve order

[vestal snow]

Yes

[oblique matrix]

injective because if B\neq C, then f(B)\neq f(C)

[vestal snow]

Don't try and prove injectivity + surjectivity

Try coming up with an inverse

[oblique matrix]

(if there is an element in one and not the other that element must be in one of the images but not in the other)

f^{-1}(B)=x^{-1}Bx

[vestal snow]

Yep

[oblique matrix]

and this is also well defined

[vestal snow]

Yes

And verifying that these are inverses is straightforward

[oblique matrix]

f(f^{-1}(B))=B for every B\in A

[vestal snow]

And the other way around

[oblique matrix]

yes

ok so it permutes the subgroups in A

so to show normality of N

Let N be the intersection of all N_i with s elements

We want to show gNg^{-1}\subseteq N

but N=\cap N_i =\cap gN_ig^{-1}=g(\capN_i)g^{-1}=gNg^{-1}

so they're actually equal

[cloud walrus]

AAAAAAAAAAAAAAAAAAAA

[oblique matrix]

Ok Thank you for your help!

[vestal snow]

No problem

[ivory dust]

if $\alpha \in F$ but $\sqrt(\alpha) \notin F$ then are elements of F in the form $a+b\sqrt(\alpha)$, a,b $\in F$

[cloud walrus]

Ramtin

[ivory dust]

or are elements of $F(\sqrt(\alpha))$ in the form $f(\sqrt(\alpha))/g(\sqrt(\alpha))$ where f, g $\in$ $F[x]$ g non zero$

[cloud walrus]

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

[ivory dust]

im assuming yes

cuz like sqrt2 isnt in Q

but 2 is

so elements in Q(sqrt2) are a+bsqrt2

[ivory dust]

also im assuming this is always true actually, but the form above is more simple

[gritty sparrow]

you mean elements of the field $F(sqrt(\alpha))$ right?

[cloud walrus]

saketh

[ivory dust]

omg yea sorry

for both

cases

[gritty sparrow]

Well in this case both of the ways you described the elements are actually equivalent, so they are both correct

[ivory dust]

Is cuz alpha^n neN is always in F

so polynomial of f(sqrt(alpha)) all the even degree terms

become absorbed as a term in F

?

but wb the denominator g(sqrt(alpha))

[gritty sparrow]

Yes, and the odd degree terms can be absorbed as part of bsqrt(alpha) for some b in F

Yoy can do the same with g, and by rationalising the denominator you can get rid of g entirely

[ivory dust]

Ohh

f/g is same as a+bsqrt(alpha) / c + dsqrt(alpha) which is same as x+ysqrt(alpha)

[gritty sparrow]

For example 1/(1+sqrt(a)= 1-sqrt(a)/1-a

[gritty sparrow]

Yes

[ivory dust]

okk tyys

sm

[gritty sparrow]

Np. Btw a similar thing can be done for F(t) whenever t satisfies a polynomial over F. That is F(t) which consists of quotients f(t)/g(t) can be transformed into just polynomials r(t) where the degree of r is less than the degree of the polynomial t satisfies in F

[ivory dust]

ohh

like Q(pi)?

but Q(pi) cant be reduced i think

[gritty sparrow]

pi doesn’t satisfy a polynomial over Q so it can’t be reduced

[ivory dust]

less than or equal?

oh i think i get it

like f(x) = x^3 - 2 over Q[x]

so f/g = r where deg(r) < 3? or <= 3?

😄

[gritty sparrow]

Fuck i typed “<“ and 3

And it made that into a heart

[ivory dust]

LMFAO

[gritty sparrow]

Anyway, I’m deleting that, but you get what I’m saying right?

[ivory dust]

yess

i woulda guessed <= tho

but i think i get why its <

cuz if its irreducible

F / ideal of polynomial with that root not in F

has basis w deg(that polynomial) elements?

basis w that degree*

[gritty sparrow]

Yes exactly, this is pretty much the reason.

And the fact that the ideal generated by an irreducible polynomial is maximal in F[x] proves that this is a field

[ivory dust]

so if the degree of the irreducible polynomial is 3

then F(x) can be represented as a+bx+cx^2 where x is the root of that polynomial?

and a,b,c e F

[gritty sparrow]

Yes

[ivory dust]

so a basis would be {1,x,x^2} dimension 3

okk

i get it

tys

[gritty sparrow]

The basis is {1,x,x^2}

[ivory dust]

woops

yea

that made 0 sense

since a,b,c all in F

💀

[gritty sparrow]

Yeah

[ivory dust]

if w = e^(i2pi/7)

then is E = Q(w) = a+bw?

and if phi(w) : E -> E is automorphic

with phi(w) = w^3

what does it mean when it asks me

prove order of phi is 6?

does that mean phi^6(w) = w? or phi^6(a+bw) = a+bw?

i.e., phi^6 is id map?

[small bison]

E should be a degree 7 extension of Q cause (x^7 - 1)/(x - 1) is the min poly of w

so the elements are linear combinations of 1, w, w^2, ..., w^6

but yeah the question asks to show that phi as an element of the group of automorphisms of Q has order 6

ie phi^6 = id

[ivory dust]

Ohh

so E := { f(w)/g(w), f,g e Q[x]} = { a0+a1w+...+a6w6, ai e Q} since w is root of irreducible polynomial f(x) = (x^7-1)/(x-1) over Q (cyclotomic polynomial degree p =7? shows its irreducibile by eisensteins?)

[small bison]

yeah eisenstein works

and the f(x+1) trick

[ivory dust]

tyyy

is it true

if phi automorphic

phi^n(a+b) = phi^n(a) + phi^n(b)

[velvet vigil]

is there any difference between a groupoid and a magma or are they the same thing

[kind temple]

they’re different

[hidden haven]

They are very different

Magma is a set with a binary operator

Groupoid is a category with all morphisms being isomorphisms

You can define groupoids without categories, I think there's a definition on wikipedia

[mild laurel]

maybe some people call them both names

[kind temple]

groupoid is also a set with a unary operation of inverse and a partial function G x G —> G. some extra rules are needed to define the behavior of the two operations, but that’s it

[velvet vigil]

yeah i read on a few web pages and it looks like the terms are used interchangeably? idk im confused

[hidden haven]

You need a bunch of conditions on that partial function though

[velvet vigil]

what partial function?

[hidden haven]

No that's the standard definition of groupoid but it seems what you're talking about is really just magma

[kind temple]

the groupoid “operation”

groupoids have inverses. magmas may not have inverses.

[velvet vigil]

but the wiki says

This article is about the algebraic structure. For groupoids in category theory, see Groupoid. For other uses, see Magma (disambiguation).

Algebraic structures between magmas and groups.
In abstract algebra, a magma, binar[1] or, rarely, groupoid is a basic kind of algebraic structure.

so concerning group theory, is it the same thing?

[hidden haven]

The French are back at it again 😎

[kind temple]

noice

[velvet vigil]

the pic gets blurred when i zoom

can u send a link

[hidden haven]

There's no share link on talk pages for some reason 😵‍💫

But it's saying that groupoid has 2 meanings

One meaning is the same as magma

But that's basically never used

So whatever text you are referring to, might indeed be using them interchangeably

But magma is the standard term for a set that has a binary operation

[velvet vigil]

whats the second meaning?

[hidden haven]

This

[hidden haven]

Or this

Both are equivalent

[velvet vigil]

not too familiar with category theory but does that mean there are functions between two structures that dont change them?

[hidden haven]

No, there are abstract categories in which the objects are not structures

So just refer to what coycoy said

[velvet vigil]

okay

[hidden haven]

Like it's a completely different kind of structure. You have a set G, and set of types T, where each element of G has a type (a,b) where a and b are in T

[velvet vigil]

hm

[hidden haven]

Then the multiplication is allowed only when you have something of type (a,b) and something of type (b,c) to get something of type (a,c)

I'm just restating the category definition lol

But yeah completely different structure

[velvet vigil]

okay

is there any recommended text for this?

i mean abstract algebra/group theory?

[hidden haven]

You can look at the pins in #book-recommendations

They probably won't cover groupoids though

Those would usually be studied in some algebraic topology book would be my guess

[velvet vigil]

hah the pins are kinda hard to follow

[hidden haven]

Lol yeah look for sloth king daminark's message that starts with dummit and foote

That includes brief reviews of some commonly recommended abs alg books

[ivory dust]

if A and B are points in F^2, F is a field

L(A,B) (line thru A and B) is described similarly to R^2 right?

[fossil shuttle]

ronnie brown has a book called "Topology Via Groupoids" or something and he shills it on mathoverflow in essentially every post he makes

[gray comet]

Clerk would know, since he's Ronnie Brown

[hybrid dawn]

How detailed should my understanding of rings be to understand basic module theory?

[unique juniper]

All of the basic results about rings

[vestal snow]

Let f: X -> Spec A be a projective morphism and let A be a noetherian ring. Then is X locally noetherian?

I don't want the proof, just want to know if this is true

seems like it should because proper, hence finite.

[gritty sparrow]

I think it should be because if X has a closed embedding into P^n(A) then we can cover X with open affines that look like quotients of open affines of P^n(A) and all of those rings should be noetherian

[vestal snow]

Thanks

[vestal snow]

Why does $H^{n-1}(P^n,O(m)^{\oplus j})$ and $H^n(P^n,G)$ being finitely generated imply that $H^{n-1}(P^n,F)$ is finitely generated?

[cloud walrus]

Have a Banana

[gritty sparrow]

I think he is using the fact that A is noetherian to say that $H^{n-1}(P^n,F)/kernel of the map$ is a submodule of $H^n(P^n,G)$ which is noetherian as an A module hence it is finitely generated. And we also know that the kernel of the map is finitely generated as an A module. From these we can get that $H^{n-1}(P^n,F)$ is finitely generated

[cloud walrus]

saketh

[opal osprey]

Is it true in general that for $\mathbb{Q}_{p}$ we have that
\
\
$0 = \inf\left{\dfrac{1}{n+1} ; \vert ; n \in \mathbb{N}\right}$ ?

[cloud walrus]

MisterSystem

[mild laurel]

What does inf mean in Q_p

[opal osprey]

Oh

That's right

It is not ordered

To be more explicit

[warm saffron]

$0=\inf{1/(n+1):n\in\mathbb N}$ is correct

[opal osprey]

I am trying to find an example of an ordered non-archimedian field such that this doesn't hold.

[cloud walrus]

logician_pdx

[opal osprey]

But really, I can't find any good examples

[mild laurel]

You have to be in char 0 right? Otherwise some of the terms won't make sense

If you're in char 0, then Q is a subfield and contains all the numbers you care about and the non archimedean must restrict to a non archimedean on Q which must be the Q_p metric for some p

So I don't know if there are any examples

[opal osprey]

Oh

Found the example I was looking for

I was not thinking about Q_p lmao

But Q(t), which is the field of all rational functions with the order introduced in this book I am reading

I am sorry

It's been a while since I've read this book and I got messed up

[mild laurel]

Hm what's the ordering and metric?

[opal osprey]

Let $r(t) \in \mathbb{Q}(t)$ be a rational function, i.e $r(t) = \dfrac{p(t)}{q(t)}$ where $p,q$ are rational polynomials.
\
\
Then, we say that $r(t) > 0$ iff the coefficient of greatest degree of $p(t) q(t)$ is positive.

[cloud walrus]

MisterSystem

[opal osprey]

This defines an ordered field as he proves in the book

But is not Archimedean

Because if you take any natural number $n$. We have that $r(t) = t \in \mathbb{Q}(t)$, but $t-n > 0$

[cloud walrus]

MisterSystem

[mild laurel]

Ah okay I see

Yeah I was thinking about non archimedean as in non archimedean valuations on fields

[opal osprey]

So $\mathbb{N}$ is unbounded

[cloud walrus]

MisterSystem

[opal osprey]

Yeah

I got a little bit carried away

I was trying to find this example of ordered non-archimedian field again

Because a friend of mine asked me for an example

But I am Brazilian and I don't know some math terms in English

And I messed up some definitions lmao

[somber vale]

huh I discovered something neat: if you treat integration as a binary operation on $\mathbb{C}(x)$, you actually get a quasigroup, with the operations defined as ```latex
\begin{align*}
a * b &= \int a,db \
a \mathbin{/} b &= \frac{da}{db} \
a \mathbin{\backslash} b &= \int \frac {1}{a},d\left(\frac{da}{db}\right)
\end{align*}

[cloud walrus]

wooloowoo

[somber vale]

the right division is pretty much to be expected, although I found the left division property much more interesting

[chilly ocean]

Interesting

[somber vale]

well I'm not exactly sure if it's a quasigroup because of the Latin square property but it is at least a magma with left and right division which is neat?

I think it probably is

nvm this is wrong lol

I think it should actually be latex \begin{align*} a * b &= \int a\,db \\ a \mathbin{/} b &= \frac{da}{db} \\ a \mathbin{\backslash} b &= \int \frac {1}{a}\,db \end{align*}

[cloud walrus]

wooloowoo

[thorn delta]

I'm stuck on the last part of this proof. How do we even know if x^-1 is in B?

[mild laurel]

Isn't that one definition of a valuation ring

[thorn delta]

all i know is that x is in B or x^-1 is in B. To say x^-1 is in B requires assuming the conclusion

[maiden ocean]

Theres an error in this i think

there is no restriction map on A

Omega is the algebraic closure of A[x^-1]/m' so we compose to get a map A[x^-1] -> A[x^-1]/m' = k' -> Omega right

@thorn delta

this is the projection composed with the inclusion

so it extends to a homomorphism g: B -> Omega

since x^-1 is in A[x^-1], g(x^-1) is given by composing the inclusion with the projection, and thus clearly sends x^-1 to 0

right

[thorn delta]

okay i see so this works if we just omit the restriction step

[maiden ocean]

Yeah

so x^-1 is in the kernel of g = the maximal ideal of B and thus x^-1 is a non-unit in B

thus x is not in B

[thorn delta]

ye, makes sense

[maiden ocean]

This is one of the bigger errata in the book i think

[thorn delta]

had tunnel vision thinking there could not possibly be an error in AM

ive only found one other (which happens twice)

[maiden ocean]

the really bad one is when they fuck up the A-algebra structure on the tensor product lol

[thorn delta]

yes!

thats the one i was thinking of

[maiden ocean]

the other ones are pretty teeny tiny

like typo stuff

@dusty river lurk

[mild laurel]

o yea whoops I misread it said A subseteq B and didn't think more about it

[wooden ember]

just to be sure when they say conjugacy classes of equal size in H, they are speaking of conjugacy classes of H (So orbits of H on itself under the action of conjugation)?

that's the only way i can get it to work

[unique juniper]

i think its just

orbits of G under conjugation that are contained in H

[wooden ember]

because this is exercise 9 that they refer to so what I say is that G acts transitively on K, and then the distinct orbits of H acting on K (which are just the distinct orbits of H acting on itself contained in K) have equal cardinality and there are $\left|G:HC_G(x)\right|$ such orbits.

[cloud walrus]

Lilki Narki ✓

[wooden ember]

works easily that way

yeah i think im correct considering the exercise that follows

[novel parrot]

i dont understand that last paragraph

and how it contains A

and how F(A) is just linear combinations of A

[dusty river]

Functions from A to R are like tuples of length |A| with entries in R

For example if A has 2 elements, then f is a function which assigns to each element of A an element of R

So you can view it as a pair (f(a_1), f(a_2)) where A = {a_1, a_2}

a \in A is identified with the tuple that is 1 at the coordinate corresponding to a, and 0 elsewhere, ie the function which maps a to 1 of R and everything else to 0

All functions become R-linear combinations of these f_a's

Because any function f has finitely many non zero entries, and 0s elsewhere

so you can write it as an R-linear combination of the f_a's

[hidden haven]

Think of how you do all this for vector spaces, this is exactly that

[novel parrot]

ive understood that elements of A correspond to these functions

when it says all linear combinations, is it sums of these functions

[hidden haven]

Yeah, pointwise sums (scaled by elements of R, pointwise again)

[novel parrot]

so like, $f_x = f_{a_{i}} + \cdots + f_{a_{i}}$

[cloud walrus]

ActiveChapter

[hidden haven]

It's just doing this: (2,3,4) = 2(1,0,0) + 3(0,1,0) + 4(0,0,1)

[cloud walrus]

Milkychocci ✓

[hidden haven]

f(a) is an element of R

So this is an R linear combination

[novel parrot]

ooh

do we have to use all elements of A for every x in F?

or let some r's be 0?

and F(A) doesnt contain A but rather something isomorphic to A?

[hidden haven]

No, only some subset, it says there are some a_1,...,a_n in A such that...

[hidden haven]

Yeah same thing. You just identify a with f_a and abuse notation

[novel parrot]

oh yes

[novel parrot]

okay

[hidden haven]

in this case A should really be a subset of F

(ie the identification is important)

There can be many ways of identifying an arbitrary set B as a subset of F, but you fix an identification then talk about this because freeness depends on the identification

[knotty mason]

How do I work out the size of the splitting field of a polynomial? Galois theory

[wraith obsidian]

Are you having problems with a specific polynomial?
I'm not too knowledged about Galois theory so I don't know whether a systematic process exists but I'm sure a little context would help.

[bleak perch]

If im given this cycle, how would I compute sigma^2? would I just apply the cycle decomposition to itself?

[chilly ocean]

ye

[bleak perch]

Okay, when I did that I got ( 1 5 4) (7 8) (9 10), does that sound correct? (fixing 2,3 6)

[urban ice]

yeah

i mean, apply it to each number twice and you'll see it works

[bleak perch]

Ok awesome

I have a final question about 3e, I understand that we are trying to find the order of sigma^18 conjugated by tau. What exactly is tau? Is it just equal to sigma?

[urban ice]

tau is any permutation

[bleak perch]

I'm not sure what you mean to be honest. I get that tau is any permutation, but If its not defined to something how can I conjugate? I know I'm missing something

[urban ice]

You don't actually care about what tau is

[chilly radish]

It's some permutation of 10 elements, you don't care what it does specifically

[urban ice]

Two questions :

• what happens when you multiply the conjugate of a and b?
• what happens when you conjugate identity?

[bleak perch]

In the case of cycles, I know that two elements are conjugate iff they have the same k-cycle. If I were to apply tau, I know I would get something of the sort (tau(n), tau(n+1), etc), but thats to the extent of my knowledge here

[urban ice]

Just write down what i told you to try and see what happens

[bleak perch]

I'm really not sure

[urban ice]

about what?

@bleak perch what did you write

[bleak perch]

I dont know how to apply the two questions you asked to the problem. Ive been trying to think, but to no use

[urban ice]

that will come later

what are your answers for those two

[bleak perch]

if you multiple the conjugate of a and b isnt it (a+b)(a-b) = a^2 - b^2, or am I tripping

[urban ice]

not in this context

in this context, the conjugate of an element x is τxτ⁻¹

[bleak perch]

ok then if u conjugate a,b isnt it τaτ⁻¹ τbτ⁻¹ and the inner taus cancel?

[urban ice]

yup

so the product of the conjugates is the conjugate of the products

in other words, what's (τστ⁻¹)ⁿ ?

[bleak perch]

τ σ σ_2 σ_3 ... τ⁻¹

[urban ice]

no, to the power of n

[bleak perch]

so just (τσ^nτ⁻¹) ?

[urban ice]

yup

and now, what's the conjugate of id?

[bleak perch]

isnt it still the identity?

[urban ice]

yup

So in short

we want to find when $(\tau\sigma^{18}\tau^{-1})^n = \text{id}$

[cloud walrus]

Syst3ms

[urban ice]

What we just showed is that this is equivalent to finding when $(\sigma^{18})^n = \text{id}$

[cloud walrus]

Syst3ms

[urban ice]

Oh look, no taus!

[bleak perch]

So, now we just have to find some n, that makes that equation true?

[urban ice]

basically

since you (hopefully) know that the order of σ is 12

you're really just trying to find n such that 18n is a multiple of 12

shouldn't be too hard

[bleak perch]

least common multiple right?

[urban ice]

yes, but please tell me you don't need a calculator to do this

[bleak perch]

i know that the lcm of 18 and 12 is 36

[urban ice]

the question is about n tho

[bleak perch]

yeah thats what Im trying to piece together

[urban ice]

dude

try values

stop trying to approach this in a calculatory way you're just losing track of what you're doing

this is like trying to solve 12/x=6 by doing three algebraic manipulations, when the question is asking "12 divided by what equals 6?"

[bleak perch]

2? lol my bad man ive been up for 25 hours my brain is fried

[urban ice]

yes, as an answer to both questions

and i advise you get some sleep

18*2=36 which is a multiple of 12

so it reduces to the identity, so 2 is the order

[bleak perch]

Jesus man I can't believe that flew over my head

[velvet vigil]

in a D3 Dihedral group, why do we not consider the rotation about the sides? like a vertical flip

[chilly radish]

Cuz that's not a symmetry of the triangle

[wooden ember]

Surely the last exercise is an error?? I can prove the forward direction but there are plenty of counterexamples to the backwards direction. Take the center of D8 for example, the set {1,r^2}, then 1s is not equal to sr^2.

or even a counterexample you can generate from the previous exercise: just take g and g of order >2 in the center of G

what should the exercise actually read???

adding the condition g=h^-1 makes it work

yep looked up an errata i thought so

[novel parrot]

what does expanding x+y in each variable mean?

i dont understand how we are adding these maps as they are not homomorphisms

[mild laurel]

Multilinear basically means it's an R module homomorphism in each entry

[urban ice]

(another way to see multilinear is that if you fix all inputs but one, then the resulting function is linear, ie φ(a,λx+y)=λφ(a,x)+φ(a,y))

[novel parrot]

the map from vi to phi(...vi...) is the homomorphism right

[mild laurel]

Where you fix all other inputs? Yeah

[vestal snow]

Can someone help me understand the difference between line bundles and invertible sheafs?

It seems to me that a line bundle is something that is locally a free sheaf of rank 1 and an invertible sheaf is something which is a tensor factor of O_X

[novel parrot]

so why does $\varphi(v_1, \cdots, x,y ,\cdots ,v_n) + \varphi(v_1, \cdots, y,x ,\cdots ,v_n) = \varphi(v_1, \cdots, x+y,x+y ,\cdots ,v_n)$

[cloud walrus]

ActiveChapter

[vestal snow]

But the book I'm using seems to use them interchangeably

[mild laurel]

The terms with double x and double y are zero due to the alternating property

[gritty sparrow]

They are equivalent (I don’t remember if there is some condition on the equivalency though, but it is probably mild)

[novel parrot]

i know the double x and y will go to 0

but not why x,y and y,x sum to x+y,x+y

[vestal snow]

Is there an easy proof of this?

I can kind of see it being true

[gritty sparrow]

I managed to find a proof of it on the stack project after a bit of googling, the proof seems more complicated than I remember it being, but here: https://stacks.math.columbia.edu/tag/01CR

[mild laurel]

Just write it out. You have that f(x+y,x+y) = f(x,x+y) + f(y,x+y) since the function is linear in the first coordinate and then do the same thing for the second coordinate

@novel parrot

[gritty sparrow]

Though that may be because they have a weird 3rd definition of invertible moduule

[novel parrot]

but f isnt a homomorphism?

[mild laurel]

Again, f is a homomorphism in the first coordinate if you fix the second coordinate

So it's true that f(x+y,z) = f(x,z)+f(y,z) for any z

[novel parrot]

okk

i see it now

[vestal snow]

What does it mean by "if j is the k^th element of J"?

[sturdy marsh]

J is an ordered subset of {1,2,..., n}

so it has a k^th element

For example if J = {2,4,5,7}

then 5 is the third element of J

[vestal snow]

I don't see why this would result in a complex

The composition should be (-1)^something res_{{U_I},{U_K}}

I don't think this should be 0 in general

[sturdy marsh]

that is not the composition

you need to sum over a bunch of things

[wooden ember]

so i can understand why Aut(H)=1, but I don't understand what it has to do with corollary 14, which states that for any subgroup K of G, K is isomorphic to each of its G-conjugates.

[rustic crown]

i'm sure they just mean any automorphism will take that order 1 element to another order 1 element and the order 2 element to another order 2 element. but this forces that this map is the identity. maybe send the screenshot of corollary 14 for more clarity

[wooden ember]

i mean yeah that's how i thought about it too, i just didnt really get what it had to do with corollary 14.

reference

oh i know

"conjugate elements [...] have the same order"

i suppose that's what they were referring to, i didnt see that was a part of corollary 14

[rustic crown]

i guess, but not all automorphisms are achieved via conjugation

Z2 is too smol to be thinking like this lol

[wooden ember]

yeah lol

[untold cloud]

hi, guys, for a product of two ideals $IJ = {\sum^n_{i=1} a_i b_i : a_i \in I, b_i\in J}$, are $a_i$, $b_i$ fixed? I mean can I change $a_i$, $b_i$? So for example, can I write $IJ = {\sum^n_{i=1} a'_i b'_i : a'_i \in I, b'_i\in J}$ ?

[cloud walrus]

cuiyuze0728

[untold cloud]

thanks

[frank fiber]

let G a group, why Hom(Z^n,G) is a subgroup of G^n?

[maiden ocean]

Have you seen the fact that Hom(X oplus Y, Z) = Hom(X, Z) oplus Hom(Y, Z)

[frank fiber]

no

[maiden ocean]

Well it kind of makes sense right

Like if i have a map f: X -> Z and a map g: Y -> Z

it defines a map X oplus Y -> Z by sending (x, y) to f(x) + g(y) right

lmao why did i say (f(x), g(y))

[frank fiber]

yes

[maiden ocean]

Right so like

then Hom(Z^n, G) = Hom(Z, G)^n right

(up to isomorphism)

[frank fiber]

it makes sense

[maiden ocean]

So it suffices to show that Hom(Z, G) is isomorphic to G, no?

Because then Hom(Z^n, G) cong Hom(Z, G)^n cong G^n

[frank fiber]

yes

[maiden ocean]

Uh sorry, ok so we dont have abelianness actually so we dont quite get the nice isomorphism through direct products but we do have that Hom(A oplus B, G) is a subgroup of Hom(A, G) oplus Hom(B, G). if we have a map f: A oplus B -> G we can get maps f_A: A -> G by sending a to f(a, 1) and likewise f_B: B -> G by sending b to f(1, b).

so we can define a map Hom(A oplus B, G) -> Hom(A, G) oplus Hom(B, G) by sending f to (f_A, f_B). and to check injectivity just use the fact that if (f_A, f_B) = (g_A, g_B), f(a, b) = f(a, 1)f(1, b) = g(a, 1)g(1, b) = g(a, b)

@frank fiber sorry about the delay i am trying to explain this because this is a very important and useful property of Hom

But does this make sense? like we dont necessarily have an isomorphism of Hom(Z^n, G) with Hom(Z, G)^n but we know that its a subgroup of Hom(Z, G)^n

[gritty sparrow]

Sorry if this is a bit irrelevant, but if G is not abelian I’m pretty sure Hom(Z,G) is not even necessarily a group

[upper pivot]

it should be right, under multiplication of maps: (fg)(x) = f(x) g(x)

[frank fiber]

yes

[gritty sparrow]

fg(xy)=f(xy)g(xy)=f(x)f(y)g(x)g(y) but we can’t commute these terms to make it f(x)g(x)f(y)g(y) so it isn’t q homomorphism

[upper pivot]

well ig its like, maps from Z^n so these should commute in this case

or no nvm

yeah ok i dont see why this is a group lol

[gritty sparrow]

Yeah so I think the question should just be rephrased to give a bijection from this set to G^n which I think should still work out

(Ie f maps to ( f(e1),f(e2),..f(en) ) should still be a bijection atleast. (ei being the element (0,0,..1,0,0,..0)

[maiden ocean]

You dont get surjecitivty when G is non-abelian because since the domain is abelian the image has to commute as well I think

the image is pairs of commuting elements in G

[maiden ocean]

anyway so we know that Hom(Z^n, G) can be embedded as a subgroup of Hom(Z, G)^n (assuming that Hom(Z, G) has the structure of a group at all lol)

so it suffices to show that Hom(Z, G) cong G

you should try to construct the isomorphism on your own. remember that Z is generated by 1 so every homomorphism from Z into G is determined by the image of 1

the direct thing saketh talked about works too and strictly speaking this line of argument is not valid because G is non abelian so you cant really invoke the isomorphisms if u want to guarantee the group structure is preserved (it is tho lol) but like in any case where Hom(Z, G) has a group structure (which should be most of the time you need to do stuff like this) its better to think about the properties of Hom sets e.g factoring through direct sums, Hom(Z, G) cong G, etc

These are just good things to know

[gritty sparrow]

Ah yes you’re right, I’ll probably need the free product of Z or some other such nonsense to get things to work

[maiden ocean]

Yes

The reason this works out nicely in Ab is that direct sums are coproducts in that category

but the coproduct in Grp is the free product

so sadly things do not factor quite so nicely

[kindred mist]

I have an extension $F = K(u,v)$ of a field $K$ of prime characteristic, where $u$ is separable over $K$. I am trying to prove that $F$ is a simple extension (of $K$). We can assume that $E:= K(u)$ is the maximal separable extension $S$ (intermediate to the extension $K\subset F$, otherwise, $S$ is a finite dimensional separable extension of $K$ and thus equals $K(w)$ for some $w\in F$, then $u\in K(w)$, so that $F = K(w,v)$). If I can show that every intermediate field to the extension $K\subset F$ either contains $E$ or is contained in $E$, then I am done (then there will be only finitely many intermediate fields to the extension $K\subset F$). Does anyone have any hints for how to proceed from here (I have found some other things but idk that any of them are helpful, e.g. that $v$ is purely inseparable over $E$, I think how to proceed from here might have something to do with factoring minimal polynomials but I haven't made a lot more progress than this)

[cloud walrus]

𝓒𝓸𝓾𝓷𝓽𝓪𝓫𝓵𝓮

[kindred mist]

(I want to say that embedding into a normal closure of F won't really help either, if F were separable over K we would be done already)

(I know what to do if v is separable or purely inseparable over K, but not in the more general case)

[gritty sparrow]

I believe in the proof of the primitive element theorem we only need all but one of the elements to be separable, so we can directly say that this extension is simple

[kindred mist]

Do you know where I can see this proof? Maybe the one in Hungerford?

[gritty sparrow]

It is the same as the usual proof that shows that the primitive element of F(u,v) is some linear combination tu+rv. Is this the proof of the primitive element theorem you know? In any case this proof should be in milne’s field theory notes online. (I don’t know about hungerford, I haven’t read it)

[kindred mist]

I want to say that Hungerford did something different (I am following Hungerford), but I'll check out Milne, thanks

[gritty sparrow]

Np

[kindred mist]

Yeah Milne 5.1, awesome

Nearly done with section 6 of 9 in Hungerford Field/Galois theory now

[fickle pewter]

Hi, can I ask a question? Is {0} a proper ideal?

I need to find an example that fit to a definition which says it should be a proper ideal. But I can't find any example beside {0}

[gritty sparrow]

Yes it is

[fickle pewter]

oh, okaii. Thank you so muchh. I've read some books but there are different idea for proper ideal lol.

[gritty sparrow]

Well, a proper ideal is just an ideal that isn’t the whole ring usually. (So my answer actually uses the definition of a ring where 1!=0)

[fickle pewter]

oh got it. it helps a lot.

Can you explain about 1!=0?? I often see it in a journal but don't understand it

[gritty sparrow]

I’m kinda curious, what is the original question?

[fickle pewter]

the definition?

[gritty sparrow]

So if we don’t have that rule, we can have a ring given by the set {0} where 0x0=0 and 0+0=0, you can check that this will satisfy all the other properties of a ring, so that 1!=0 is to exclude this

[fickle pewter]

I see. So, that's the function of that notation. I'll remember it.

[maiden ocean]

I've never seen that defn of a ring

In a field yes, we usually require that 0 and 1 are distinct

But the trivial ring is certainly a thing

[fickle pewter]

the one I sent?

it's from a journal and I have a task to apply the definition on integral domain

[maiden ocean]

No the notion that 0 and 1 have to be distinct

it excludes the trivial ring which is a bit odd

Unlike with fields there arent really many reasons to define {0} to not be a ring

[dusty river]

Some people do that. There's no real disadvantage to not having it is there?

It is not the zero object of Ring, and without it you can define all ideals to be proper

[maiden ocean]

Its the terminal object

[dusty river]

yeah

[maiden ocean]

It doesnt matter too much

[dusty river]

but is that as useful?

It lets you define maximal ideals as actual maximal ideals lol

[maiden ocean]

what do you mean maximal ideals as actual maximal ideals

[dusty river]

maximal ideals are actually maximal proper ideals

[maiden ocean]

It lets you say that every ring has a maximal ideal i guess

Yes but you still have the ideal consisting of the whole ring

[dusty river]

Yeah I am saying it excludes that

so you don't have to say proper

[maiden ocean]

how does saying 0 neq 1 in a ring do that

[dusty river]

Though ideal addition is no longer a thing

[dusty river]

The full ring is no longer a kernel

So you might want to redefine ideal so that it coincides with kernels

[maiden ocean]

Oh I guess if you define ideals as kernels of ring homomorphisms then sure

[dusty river]

Yeah

[maiden ocean]

It doesnt really matter too much but it seems like it would get annoying

Like exact sequences in Ring for example

[dusty river]

Yeah that's true

Just make them exact sequences in Rng

[maiden ocean]

[gritty sparrow]

Ring is not an abelian category

Anyway, I didn’t really think about the convention that much, I just said some shit

[dusty river]

oh yeah you can't have exact sequences in Ring anyway

You can in Rng though because ideals are subrngs

subrng lmao

[maiden ocean]

Oh yeah

Oh well no one can agree what rings are anyway

[gritty sparrow]

I’m pretty sure Rng won’t be abelian either

[dusty river]

Yeah but at least you would have kernels and quotients

That should be enough to define exactness

Like do it concretely

[dusty river]

I don't see what the issue is. A ring is exactly an abelian group with an extra operation. Just that we often work with distributive unital associative rings.

[gritty sparrow]

I think he was just joking about the fact that we have just been arguing about the def of a ring for a while

[maiden ocean]

No we work with distributive unital associative commutative noetherian rings

[fickle pewter]

ohh

[maiden ocean]

No others exist

[dusty river]

😌

[ivory dust]

So im asked to find degree of extension field Q($\sqrt$(1+2*2^{(1/3)})$) over Q

[cloud walrus]

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

[ivory dust]

ugh im so bad at latex

anyway

if a is a root of that polynomail --> a^2 = 1+ 2cuberoot2
so:
f(a) = a^6 -3a^4+3a^2-17 = 0

is f irreducible over Q?

if so then its degree is degree of [Q[x]/(f(x)) : Q[x]] = df = 6?

but not sure if f is irreducible or how to show that (over Q)

Eisensteins doesnt work

(i think)

[small bison]

Try building a tower

Like Q -> Q(2^1/3) -> your field extension

[ivory dust]

Q(2^1/3):Q = 3

[small bison]

Then you just have to figure out if a is in the intermediate one or not

If it is then it’s a degree 3 extension cause they’ll be the same

Otherwise it’s a degree 6

[ivory dust]

then Q(2^1/3, sqrt(1+a))?

a e Q(2^1/3)

[small bison]

Should be like Q(2^1/3, a) assuming a is the thing in the parentheses that you typed earlier

[ivory dust]

wait f(x) in mod 3 is

x^6 - 2

[ivory dust]

ohh

wouldnt a easy way be to show

f(x) mod p = 3 irred ---> f(x) irred in Q

then Q/ideal generated by f has degree of f over Q

[small bison]

Hmm not sure if that’s true

Like x^2 - 2 is irreducible over Q

But not in Z/2Z

[ivory dust]

it is

Let f∈Z[x] be a non-constant polynomial and let p be a prime number which is not a divisor of the leading coefficient of f. I need to prove that if f is irreducible over Zp, then f is irreducible over Q.

[hidden haven]

Yeah it's not an iff

What ramtin said is right

[ivory dust]

x^6 - 2 mod 3 for 0,1,2

is not 0

so no linear factors

[hidden haven]

You might be able to work with this, it's (x²-2)³ in F_3

So you get no factorizations of the type 3+3

[ivory dust]

but it can have quadractic, cubic, type factors?

[hidden haven]

Only possible ones are of type 4+2

And that actually finishes it off

[ivory dust]

Wait why is this not possible

[hidden haven]

Didn't get this

[ivory dust]

this

[hidden haven]

Because no such factorizations are possible over F_3

Of type 3+3

x²-2 is irreducible

And F_3[x] has unique factorization

[ivory dust]

so x^6 - 2 cant factor as (x^2-2) is what ur saying :o?

[hidden haven]

No

By 3+3 factorization I mean a factorization into 2 cubics

[ivory dust]

2 cubics that have roots in F3 or Z3

cuz no roots

ohh

[hidden haven]

(x²-2)³ is q factorization of type 2+2+2

F3 and Z3 are the same thing

[ivory dust]

yea sorry

[hidden haven]

Unless you're talking about 3 adics

[ivory dust]

so 3+3 is not possible, why is the other cases not possible?

2+2+2, 2+4?

1+5?

jk i get 1+5

[hidden haven]

Suppose 2+4 were possible

Then what could you say about the degree of the extension

[ivory dust]

its either 2 or 4?

ERrr

[hidden haven]

Not necessarily

But 2 or 4 divides it

[ivory dust]

the degree of extension [Q/f : Q] is what u mean right

[hidden haven]

No, Q(a)/Q

The extension in the problem

[ivory dust]

o

[hidden haven]

You get that 2 or 4 divides that degree, because adjoining a means adjoining a root of f, ie a root of one of its irreducible factors

[hidden haven]

Sorry this was correct lol

Is that fine?

[ivory dust]

Ok wait so from first glance the most the degree could be is 6, and least is 1 (obviously not 1 since they are different fields)

[hidden haven]

oof there's one detail I missed, hopefully it works out lol

[ivory dust]

and if f(x) in F3

[hidden haven]

Also the degree divides 6

[ivory dust]

factors as 2+4 degree polynomials

i.e. quadratic * quadratic^2

that are both irreducible?

then Q(a)/Q has degreee 2, or 4

[hidden haven]

No, in F_3 the factorization is 2+2+2

But when you bring it back to Q (if possible)

It could turn out that a 2+2 part of it is irreducible

So it won't necessarily be quadratic² in !

In Q

[ivory dust]

o ur saying f(x) = x^6 - 2 = (x^2-2)^3

?

[hidden haven]

Yes

If there were a 3+3 or 5+1 factorization, it would reduce mod 3 to a factorization of that kind again

But it can't

[ivory dust]

wait so f is reducible.. so my idea wouldnt work regardless

[hidden haven]

It isn't reducible

It is reducible mod 3

[ivory dust]

oh

[hidden haven]

But you get information from how it reduces mod 3

[ivory dust]

ohh

[hidden haven]

Mod 3 reduction was used to eliminate the cases of 3+3 and 5+1

And now we have to eliminate 2+4

[ivory dust]

Ohhh

[hidden haven]

And you can do that by degree arguments, check what factors the degree of the extension should have

And that eliminates 2+4

[ivory dust]

damn this question lowkey hard

[small bison]

Would it be easier to check that a isn’t in Q(2^1/3)

[ivory dust]

maybe ext fields is easier

LMFAO

i was hoping showing f(x) irred over Q would be easy

[small bison]

I’d just do it some dumb way like write a as a linear combo and square both sides

[hidden haven]

I don't think there's a uniform way to check that but maybe you can just grind it out in this case

My solution is complete right?

[ivory dust]

Yea

[hidden haven]

[ivory dust]

it shows f irred over Q

so

Q/(f(x)) over Q has degree (deg f)

[hidden haven]

Yes

But that's not relevant

Oh after f irreducible

Lol nvm

[ivory dust]

maybe its easier

to just do tower of field

like

Q:Q(cuberoot3) = 3