groups-rings-fields

[wooden ember]

because in that case H is the elementary abelian group of order p^3 in which it's not necessarily true that there is a characteristic subgroup isomorphic to Z_p X Z_p and I can't really think of another approach

to clarify: I can find a subgroup isomorphic to Z_p X Z_p, just not a normal one

[gritty sparrow]

Am I misunderstanding something? H_0 is abelian so all of its subgroups are normal, and there is a subgroup isomorphic to Z_pxZ_p

[wooden ember]

sure but im looking for a subgroup normal to P

the idea in the |H_0| = p^2 case was that since H_0 is itself of the form required and characteristic in H which is normal in P, H_0 was normal in P giving the result. But when |H_0|=p^3 im failing to find a subgroup of it of order p^2 necessarily normal in P

[gritty sparrow]

I see

[wooden ember]

any ideas

[gritty sparrow]

Nope, I’m still thinking about it

[wooden ember]

welp ima go to bed maybe a fresh mind will help me better tmrw

[gritty sparrow]

Ok, I think I have an idea: Induct on the statement that P has a normal subgroup Z_pxZ_p and that Z_pxZ_p has a cyclic p subgroup that is normal in P. The case where P/Z is cyclic, ie P is abelian will work out in pretty much the same way. In the other case, by induction Hbar has a subgroup of order p that is normal in P/Z, call it T. If |H_0| is p^2 then we are done without anything related to T as H_0 works and contains Z and Z is normal in P bc it is in the centre. Otherwise the preimage of T is a subgroup of order p^2 and since all the elements will satisfy x^p=1, the preimage of T looks like Z_pxZ_p. Since T was normal in P/Z, its preimage will be normal in P. Also it contains Z which is normal in P for the same reason as before

[terse crystal]

I think the irreducible components of V(y^4-x^2, y^4-x^2y^2+xy^2-x^3) are {V(x+y^2) ,V(x,y)}, not {V(x+y^2) , V(x-1,y-1), V(x-1,y+1)}

E is a splitting field of a polynomial f over F then for any middle field E/K/F of course E is a splitting field of f over K by definition

[gritty sparrow]

(1,-1) and (1,1) are solutions right? Because the first polynomial vanishes obviously and the second becomes 1-1+1-1 as well. Also V(x,y) is a subset of V(x+y^2)

[terse crystal]

Yes I made a mistake he is right

[wooden ember]

Thank you that does seem to work!

[urban acorn]

Hmm. In general given a - say integral domain - R, let's define an equivalence relation $\sim \subset R \cross R$ such that $a \sim b \iff a = bu$ for some $u \in R^{\cross}$. We have a monoid structure on $R/\sim$ given by multiplication of representatives.

[next obsidian]

You need to do \sim

~

Doesn’t work

It might work if you put \ before it but I’m not sure

[cloud walrus]

All groups are abelian

[urban acorn]

Okay, thanks.

So my question was, how interesting is this monoid?

like, we can tell whether R is a UFD using it, right?

I think that's equivalent to saying it's a free monoid.

[novel parrot]

will not having a in P also be a contradiction

[chilly ocean]

why?

law of excludded middle homie

[novel parrot]

what does this mean

[viscid pewter]

it means the middle is an exclusion zone

[chilly ocean]

either a^(n-1) in P or a in P but first one cannot be

so a in P

not sure what would that contradict but anyways you havent shown the entire pic

[novel parrot]

what if i said, a(a^n-1)

a cant be cuz less than n

then a^n-1

[chilly ocean]

a non zero tho?

[novel parrot]

yeah

?

[novel parrot]

"there is a smallest power n, a^n in P"

[chilly ocean]

so if you take a it doesnt cotradcit minimality

wait

can you show the whole thing maybe

[novel parrot]

ok

[chilly ocean]

I guess the thing is, the minimality of n is such that a^n = 0

because 0 always in an ideal

[novel parrot]

no

[chilly ocean]

why

[novel parrot]

a^k = 0 which will be in P

so we just look at the least k which is in P

not 0 necessarily?

[upper pivot]

so the idea is

we know there is some a^n in P since a is nilpotent

so the set of a^k in P is non empty

so it has a minimal element(with respect to the power).

[novel parrot]

yes

[upper pivot]

doesnt need to be 0 as you said.

[novel parrot]

if a^{n-1} a = a^n in P and n is minimal

wont these both be contradictions

[gritty sparrow]

No

[chilly ocean]

hmm yeah like what if you have a^2

oh wait bad exmaple

[upper pivot]

well you are saying n is minimal, and basically the arguement shows n=1

[gritty sparrow]

They never said 1 was less than n

[novel parrot]

if a^2 was in P

we cant show that P is prime from this yea?

[gritty sparrow]

Isn’t P assumed to be prime? I don’t get what you’re saying exactly

[novel parrot]

yes mmm

i also have another question

in the next section

I dont understand the contradiction there

we wanna show that P is prime?

[upper pivot]

We wanna show a is in P

[gritty sparrow]

They wan’t to show that the set of nilpotent elements contains the intersection of prime ideals. The way they are doing that is by taking a non-nilpotent element and showing that there is a prime ideal not containing it. So they have made a construction for something not containing a, called P, and they wan’t to show that it is prime

[novel parrot]

i mean

so showing that P is not a prime, is a contradiction to it being in S yeah?

[gritty sparrow]

No, there are many elements of S that aren’t prime

[novel parrot]

it says that x,y not in P but xy is in P yes?

so P contains a power of a, a contradiction yes?

[chilly ocean]

I guess this arguemnt is supposed to show that n=1?

ye john says so too then ye

[upper pivot]

I’m confused are we on the first part of the proof?

[novel parrot]

no

second part

[upper pivot]

Where we show that the nilradical is contained in all prime ideal

Oh oops

Yeah disregard what. I said then

[novel parrot]

sorry im confused abotu the contradiction in part 2

second paragrapgh

[gritty sparrow]

Basically, the proof shows via contradiction that any maximal element of S is a prime ideal

And all elements of S contain no powers of a, so a maximal element of S is a prime ideal not containing a

[novel parrot]

so x,y is not in P then xy is not in P IF P is in S yeah?

then we take reverse?

[gritty sparrow]

x,y not in P then xy not in P if P is a maximal element of S

[novel parrot]

oh okey

[gritty sparrow]

The fact that P is maximal is important, for example if A is not an integral domain, (0) is in S, but is not a prime ideal

[novel parrot]

Okey

i understand

thanks

[upper pivot]

(ill add one thing, there is kind of easier to show if you are familiar with fraction rings. take S = {f^n} and so S^{-1}A is a non-zero ring iff f is not nilpotent, which contains a non-trivial prime ideal P (so this cannot intersect S). This is kind of just rephrasing that proof but i have always found this neater)

[wraith blaze]

How do I start?

[scarlet estuary]

wtf that was a putnam problem

anyway, think of it lik ethis

say G, H are both proper subgroups of the same group and neither are subsets of the other

it is a known fact (can you prove this?) that their union is not a group

do you see why the conclusion follows from this?

hint for proving the fact: ||take g in G not in H, and h in H not in G|| and then ||consider gh||

[dusty river]

I think Lagrange also works

Oh wait only for finite groups

[scarlet estuary]

i think the proof i suggested is more "fundamental"

in that it might help build intuition for how subgroups work

and ||how restrictive of a condition invertibility is||

[wraith blaze]

Let me think..

[delicate orchid]

weird seeing group theory in a putnam problem

conicidentally that's the first putnam problem I've been able to do

[dusty river]

Ye also that's kinda how you prove Lagrange in the first place

[south patrol]

that proof also reminds me of showing that if U,W are subspaces of a vector space V, then U cup W is a subspace if and only if one of the subspaces is contained in the other lol

I suppose it's the same thing just because a vector space is an additive group

[delicate orchid]

yup, in fact that's kinda a stronger result because of the extra structure vector spaces have

[ivory dust]

in tower: Q<Q(sqrt2)<Q(4th root 2)

basis diagram would be like

{1, sqrt2} for Q<Q(sqrt2)

and {1,4th root2, sqrt2, 3/4th root 2} for Q<Q(4th root 2)

but how would u write basis for

Q(sqrt2)<Q(4th root 2)

is it

{1, 4th root2}?

or {1, 3/4th root 2}

?

[wraith blaze]

I don't know what to do with gh

[scarlet estuary]

suppose G U H is a group

then gh is in G U H, right?

howd it get there

from G? from H?

can you see where to get a contradiction

[wraith blaze]

Oh

I can feel a contradiction, something like gh violating closure property or something but I can't put my finger on it wait a second

What do you think of this?

[scarlet estuary]

right thats the idea

if gh is in G, then since g is in G, g^-1 is as well

but then g^-1 (gh) = h is in G

(by associativity)

contradiction

and we get the same thing if gh is in H

hence gh cannot be in G U H, hence G U H cannot be closed, hence G U H is not a group

can you extend that to prove the original statement?

solution: ||if G U H is a group, one must be a subset of the other, but since theyre both proper, their union (which is just the larger group) cant be the full group||

[wraith blaze]

As union of two groups is not a group (if they're not subsets of each other), that means the union of two subgroups isn't a group either as subgroups are groups

Idk if that's correct

[scarlet estuary]

union of two subgroups is not a group UNLESS one is a subset of the other

thats an important unless

but can you see why it doesnt matter here?

why cant one of the groups be a subset of the other?

[wraith blaze]

Yes, I forgot to type that assumption

Because the union of those two groups become one of the groups involved if one is a proper subset of the other one

[scarlet estuary]

right

which means the union cant be the full group, because ||they were assumed to be PROPER subgroups||

[wraith blaze]

Yep, thanks :)

[dull prawn]

so what do we think about the case of 3 subgroups? I think that's supposed to be the harder part

unless theres an easy example im missing

[wraith blaze]

Yeah that's what I was about to ask

I'll try to do it first

[scarlet estuary]

there is an easy example

whats the usual small group you think of when the "easy" small groups dont work?

answer: ||klein 4-group||

[dull prawn]

oh of course

[scarlet estuary]

a note: this generalizes, ||a group is the union of 3 proper subgroups iff it has the klein 4-group as a quotient||

see https://www.jstor.org/stable/2316854 (obvious spoilers for the answer)

[wraith blaze]

What about this?

[scarlet estuary]

have you done the computations

[wraith blaze]

From the second one I thought that |a|*|b|=|a+b| but that was wrong apparently

I tried this a long time ago, I'll try to do it again

One minute

[chilly ocean]

ok minute passed i'll ask my thing cause ive been struggling with this too long. Let $F/K$ be field extension and $E_1, E_2$ intermediate fields. Show that $$|E_1 E_2/K| \leq |E_1/K| + |E_2/K|$$ where by $|F/K|$ I mean trans degree

[cloud walrus]

Ledog

[wraith blaze]

For a) |a||a+b|=|b|
b) |a||b|=|a+b|
c) |b|*|a+b|=|a|

Not sure what to make of this

[dusty river]

"minute passed"
Take transcendence bases of E_i, B_i, then E_1E_2/K(union of the B_i) should be algebraic, so |union of B_i| ≥ transcendence degree (not sure if this works)

[ivory dust]

The galois grp of E/Q where E is splitting field of f(x) = x^4+5 over Q has order 8

and E=Q(i, 5^{1/4}, w) where w^8 = 1 8th root of unity of 1

I showed all this did everything, but now im stuck

at what the group structure is

cuz theres alot of order 8 groups

One big thing I know tho based on doing the multiplication table of Gal(E/Q) = {phi1, ... phi8}

is its non abelian group

So its not Z2xZ4, Z8, E8 = {1, g1, g2..., g8 s.t. gi^8 = 1, gi gj = gj gi},

I know theres D8 which is the grp of symmetry

and i see theres this group Quarterion? group never heard of it

so i guess its either one of those

[south patrol]

the quaternion group is basically {1,-1,i,-i,j,-j,k,-k} under multiplication

i believe

[ivory dust]

Oh ok

do you think this galois grp is that? or D8 any thoughts?

im kinda confused what root maps to what root

[south patrol]

(i only know about groups, no galois theory, can't help further i'm afraid)

[ivory dust]

okk nps ty

[south patrol]

what's the multiplication table like?

it's either D8 or quaternions if it's non abelian, probably D8

[wooden ember]

The only non abelian groups of order 8 are D8 and Q8: try maybe finding a presentation for your group? (I dont know anything about galois groups this is all i can really say. Maybe see if it has more than 1 cyclic subgroup of order 4?)

Could I get a hint on this one? Im thinking that I probably need to work by induction and write the group explicitly as a semidirect product and show the only possible homomorphism defining that semidirect product is trivial but im failing to find anything in that direction. For reference exercise 56 states that every finite group for which eveyr proper subgroup is abelian is solvable (and the few exercises before classified numbers n such that all groups of a certain order are cyclic iff the order is one of these n).

[ivory dust]

okk ty

ill post the multiplication table maybe in a bit if i dont figure it out

and someone can provide some isnght

[dusty river]

Do you need to write D8 or Q8 when I learned gal thy we'd usually leave stuff in semi direct product form whenever we could, usually the small galois groups (at least the ones that usually appear in psets, like the one you posted) are semi direct products of cyclic groups

Anyway you shouldn't need to figure out the full multiplication table to identify them

Some partial information should be enough

[dusty river]

Lilki narki already mentioned this

[ivory dust]

Ohh

its non abelian so its not direct product of cyclic group

but im assuming semi direct product implies it may not be abelian

[dusty river]

Yeah

[ivory dust]

the question requires me to write out the multiplication table

[dusty river]

Enjoy

[ivory dust]

64

🙃

i think its Q8

cuz its not D8

and non abelian

[cursive temple]

can i get a hint on this

[small bison]

Induction maybe

[cursive temple]

on deg f?

[small bison]

Yeah

[cursive temple]

seems a bit not nice

but probably works

[small bison]

Shouldn’t be too terrible

[gritty sparrow]

Just do repeated division with remainder by g

[small bison]

Euclidean division + strong induction

[cursive temple]

thonk

thanks

ill be back if i get stuck again

yeah this is clear enough thanks for the help

[ivory dust]

if w^8 = 1

is w^2= i

and

how would you write splitting field of f(x) = x^4 + 5 over Q in terms of the factors in terms of w?

like i have (x-i^(3/4)5^(1/4)) ...

[kind temple]

if $G$ is a group and $s\in G$, when is $S={g\in G:gsg^{-1}=s}$ the trivial group?

[cloud walrus]

c squared

[kind temple]

if s is the identity, then its the whole group

but i cant really see when it would be trivial

[viscid pewter]

when the centraliser of s is trivial ¯_(ツ)_/¯

[delicate orchid]

first thought I had is that s^ns(s^n)^-1 = s^(n-n+1) = s so it would have to be when s is order 2

not sure if that logic is sound or not

[kind temple]

bruh lol

ig that works

[viscid pewter]

oh, wait

consider the inverse of s

there we go

[gritty sparrow]

There probably is no good general answer to your question, there are only inferences you can make about the group. Many groups will never even have such an element that makes this true

[viscid pewter]

no, i think that's the point

if g = s^-1, gsg^-1 always = s

so only the trivial group?

[gritty sparrow]

Huh?

[viscid pewter]

i am very tired and may be making an obvious mistake

[kind temple]

s is already in S

and S is a subgroup, so s^-1 is in S

[gritty sparrow]

I don’t get what your logic is here, what are you showing exactly?

[viscid pewter]

well if s is in S then S is non-trivial unless s is 1

but if s is 1 then the whole group is in S

so for S to be trivial, the whole group must be trivial??

[kind temple]

no, sss^{-1} = se = s regardless of what s is

[gritty sparrow]

Ah I see what you are saying

[viscid pewter]

i mean yeah it works for s as well

s is in S

[kind temple]

no, s is in G

[viscid pewter]

s must be in S also

by definition

[kind temple]

i just showed you that s is also in S

[viscid pewter]

yeah, exactly

so if s is in S, then S is only trivial if s = 1

[gritty sparrow]

So your q is, when is S trivial, notice s is in S, that is what kaisheng is saying

[kind temple]

yes

[gritty sparrow]

So S is always nontrivial

Unless G is trivial

[kind temple]

but when is S trivial?

that is what i was asking

[viscid pewter]

bruuuh

when G is trivial

that's it

[delicate orchid]

yeah, that seems right

[kind temple]

oh fruck

[delicate orchid]

otherwise you have that s^-1 in S, thus it's not trivial

[kind temple]

wait my brain literally is not moving rn

[viscid pewter]

it be like that sometimes

[kind temple]

so s is in S regardless of what s is. and S is trivial if and only if G is trivial

[viscid pewter]

yes

[delicate orchid]

yup

[kind temple]

ok. cool. thanks guys

[delicate orchid]

wait what if s^-1 = s

[viscid pewter]

it still works

[south patrol]

The centraliser of s is trivial when the conjugacy class of s is G ig is important

[gritty sparrow]

Well that is overkill

[south patrol]

And e is in its own conjugacy class

[delicate orchid]

oh yes of course it does

[south patrol]

So G = {e}

[kind temple]

S should be the stabilizer of s tho right?

[delicate orchid]

{e} is trivial not {e, s}

I'm just being dumb

[kind temple]

unless those two words are synonymous

[viscid pewter]

centraliser, i think?

[delicate orchid]

stabiliser under what group action 🤨

it's the centraliser yeah

[kind temple]

conjugation

[gritty sparrow]

He is using the orbit stabiliser relationship

[viscid pewter]

well that's not standard

that i know of

[delicate orchid]

then yeah that's the stabiliser under conjugation

[south patrol]

But ye I think orbit-stab is easiest here

[viscid pewter]

it is??

[south patrol]

idk how that's overkill lol

[viscid pewter]

in what world

[gritty sparrow]

Kaisheng’s argument was easier

[delicate orchid]

orb stab 🔮 🔪

I prefered Kaisheng's argument, very little machinery

[south patrol]

Oh sure sorry ye fair

I like it more now

:)

[delicate orchid]

:)

[viscid pewter]

fun times

[next obsidian]

:)

[kind temple]

so just to clean it up, if G is trivial, then S is trivial, trivially.
if G is not trivial, then s = e or s != e. if s = e, then S = G is non-trivial. if s != e, then s in S is again non-trivial, which proves the contrapositive

[south patrol]

Seems good

[bleak perch]

Hey, I have a question, how would I compute the order of the group of units in Z_{3136] ?

[kind temple]

i wanna say you need to find all elements coprime to 3136, no?

[bleak perch]

yeah i figured it out, used the euler totient function

On another note, I was going through some example proofs, and Got stuck on this particular problem, Let R be a commutative ring with 1. Assume that R has at least 3 elements and that x^2 = x for all x ∈ R. Prove or disprove: There exists a field F so that R is a subring of F. How would I prove this?

[mild laurel]

It says prove or disprove, how do you know its true?

[bleak perch]

I suppose thats my problem.. I'm unsure if its true or false

[mild laurel]

That's how it is sometimes, try doing both

Maybe try to come up with some examples of R that fit the conditions and play around with those so you can get some intuition for whether it should be true or not

[next obsidian]

Have you made progress?

[bleak perch]

Just reading up on definitions, I should be able to handle it, If I have any questions I'll ask

[terse crystal]

false

[kind temple]

counter example?

[dusty river]

||what are the idempotents in a field||

[gritty sparrow]

You still need a counter example to show that such an R exists, otherwise we’ll have vacuous truth I suppose

[kind temple]

oh. i see.

[thorny flame]

how do I find the order of 7 in mod 160?

[dusty river]

Do you know how to write a while loop

[gritty sparrow]

160=32x5 and the order of 7 mod 5 is 4, the order of 7 mod 32 will divide 16, finding it out should be doable by hand.

Then take the lcm I suppose

I think the order is 4

If my calculation is correct that is

[thorny flame]

thanks

[dusty river]

Units digit of 7⁴ will be 1

Oh wait

That's what is needed nvm lol

[gritty sparrow]

Yeah I just verified my calculations, I think 4 is the order

7^3=343=23 mod160, and 23x7= 161=1 mod160

[novel parrot]

why is (d) a proper ideal of M ?

[mild laurel]

zero divisors are not units

[novel parrot]

why does that help

it could not be the whole ring

but not completely contained in M

[mild laurel]

I'm not sure what you mean

the ideal (d) is contained in a maximal ideal, and there is only one maximal ideal in this ring which is M

[novel parrot]

why is there only 1 maximal ideal

[mild laurel]

Read the previous sentences

[novel parrot]

ok

so sqrtQ is contained in all prime ideals which are then contained in a bunch of maximal ideals, but by assumption, these so called maximal ideals are just sqrtQ yeah?

[mild laurel]

The radical of Q is exactly the intersection of all prime ideals, but since the radical of Q is maximal, there can't be any other prime ideals

[chilly ocean]

What's a good introductory book on Group Theory, or perhaps a PDF from a course online?

[terse crystal]

f(x)=x^2-x can only have at most two roots in any field

[kind temple]

right. in this situation u would have at least 3. saketh makes a good point tho. does such a ring even exist?

[dusty river]

Yeah there are non trivial rings where everything is idempotent, they're called Boolean rings

F_2[x]/(x²) works I think

[kind temple]

no idea what those symbols mean but sweet

[terse crystal]

Shouldn’t it be F_2[x]/(x^2-x)?

[dusty river]

oh yeah lol mb

[delicate orchid]

Yeah cause in F_2[x]/(x^2) you have x*x = 0 which isn’t x

[terse crystal]

Btw is your name written in white😂

[dusty river]

It's {0,1, x, x+1} where addition is defined such that a+a=0 for all a and multiplication is done distributively keeping the idempotence in mind

[dusty river]

Mine?

It's the not very ppl role

[terse crystal]

😂

I was thinking defining a multiplication on the abelian group Z/3Z making it have this property. I think defining 2*2=2 works right?

Never mind

[dusty river]

2=1+1

[delicate orchid]

Could violate distrubtivity I’d have to check

[terse crystal]

It doesn’t work

[dusty river]

Yeah

[delicate orchid]

Yeah it does

[terse crystal]

0=2(1+2)=2+2=1 false indeed

[delicate orchid]

2 TIMES 2= (1+1)*(1+1) = 1+1+1+1 = 1 =/= 2

[dusty river]

All Boolean rings have characteristic 2

[terse crystal]

Yeah thanks

[terse crystal]

Since 2x = (a+x)^2-a^2-x^2=a+x-a-x=0 right

Sorry a=1

[dusty river]

Yep

Then it works using x+1 = (x+1)² = x² + 2x + 1 = 3x + 1

[kind temple]

sry to interrupt. is there like a fancy name for the notation you used here?
like, does F[x]/(p(x)) have a name or something?

[delicate orchid]

It’s a quotient ring

[dusty river]

It's a quotient ring

Bruh

[delicate orchid]

Lol

[dusty river]

vibing

[kind temple]

lol thanks

[delicate orchid]

Surprised you haven’t come across them yet

I learnt them right after ideals

[kind temple]

haven’t done much AA yet. ik a lot of the standard undergrad group theory, but nothing really past that.

and ik LA, but ig that’s separate?

[hidden haven]

LA is part of AA

[delicate orchid]

^

Study of vector spaces

[terse crystal]

I don’t know whether students should learn linear algebra especially learn linear algebra before learning algebra

When talking about one single linear transformation,or Jordan normal form or something, it seems like just special cases of finitely generated module over PID to me

[chilly ocean]

I mean, they are, but modules aren't covered in first aa class anyways usually, yet lin alg has bunch of applications.

like how else would the syllabus look like if you didnt have linear algebra first? you can't study pretty much any analysis, first aa course wouldnt have as many examples as well...

[terse crystal]

Yeah makes sense.

[novel parrot]

isolated prime is when the associated prime does not contain anyother associated prime yeah?

[terse crystal]

According to Wikipedia yes

Btw I don’t think this concept matters... it’s the rest of associated primes called embedded primes that matter. Sometimes not having embedded points can induce good properties...

[thorny flame]

can a group have more than one identity element?

[chilly ocean]

no

[scarlet estuary]

this is an easy proof

give it a shot

suppose $e_1, e_2$ are both identity elements and show that $e_1 = e_2$

[cloud walrus]

Namington

[scarlet estuary]

big hint: ||consider the product e_1 * e_2 and apply the identity property||

[delicate orchid]

classic proof

[urban acorn]

flashback to learning linear algebra from Axler before knowing any real math

[thorny flame]

I see thanks

[novel parrot]

right ok

last part, why is detB = 0?

we assuming that R is a domain?

or is there another way without R being a domain

<@&286206848099549185>

[chilly ocean]

what page is this?

i'd guess it's because det B = (det B)1 = (det B)(some linear combination of v_i's) = 0

[novel parrot]

ah

but

dont we need R to be domain

page 691

[chilly ocean]

i don't see where we need to assume R is a domain

[novel parrot]

ok me too

[chilly ocean]

[novel parrot]

lol

[novel parrot]

does this mean (phi(I)S) generated over S

or phi(I) generated over S

[rustic crown]

same thing?

phi(I) is contained in phi(I)*S

and phi(I)*S is contained in (phi(i))

[novel parrot]

dont we need S to have 1

[rustic crown]

rings have 1

[gritty sparrow]

I think we should say that phi(I)S is an ideal already, and is the ideal generated in S by phi(I)

[rustic crown]

rngs are weird, so you're saying that ideal generated by phi(I) need not even contain phi(I)?

[novel parrot]

i mean im fine with having a 1

just wondering lol

[rustic crown]

yea i think saketh is right

we usually define I * J to be the ideal generated by all products i*j

[novel parrot]

yep ok that works

thanks

[rain oxide]

If anyone ever runs in to this again, the "fractional distance" in the context of reed Solomon codes apparently means the number of points that the two functions disagree on in a subset of the domain

It's also sometimes called the relative distance

I think abstract algebra was probably the wrong place to put this, it was just in the middle of field/group stuff in a tutorial so I assumed wrongly

[frank fiber]

Let p a prime, what is the minimal polynomial of $e^{2 \pi i/p}$ over $Q$?

[cloud walrus]

Or x1

[mild laurel]

x^(p-1) + x^(p-2) + ... + x + 1

[chilly ocean]

I thought there was a shit in polynomials theory where that polynomial is not irreducible

,w cyclotomic polynomial reducible

[cloud walrus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by 88ddda#7872

[dusty river]

That polynomial won't be irreducible if p is not prime, but then it's not called cyclotomic. nth cyclotomic polynomial is the minimal polynomial for exp(2iπ/n)

[chilly ocean]

Ah

[unreal portal]

okay I don't know if this is a dumb question and it's just me staying up too late or what, but this seems simple yet has me stumped. Let's say that two groups are isomorphic, and I have the isomorphism $f:G\to H$ between them. Further, both of $G$ and $H$ are finitely presented groups. If I take one of the generators of $G$, say $x_1$, and calculate $f(x_1)$... what does that actually represent? Like, yeah it's the image of $x_1$ in $H$, and it respects all the group relations in the domain, but I feel like I'm still missing something

[cloud walrus]

cgodfrey

[next obsidian]

It doesn’t really represent anytbing

If you take a finite set of generators of G and then push all them through f you’ll get generators for H

And all the relations are the same, but f(x1) by itself is just kinda a guy chilling

[unreal portal]

To give a specific example, I have this isomorphism from the group on all those generators with all those relations, to the free group on two generators. And it shows the image of each generator. Is there anything significant in those non-trivial ones?

It's like, do those images convey 'the same' information as the original relations, since this is an isomorphism?

[split flicker]

holy shit

did you write this out or is this a computer @unreal portal

[crimson falcon]

Anyone can point me what steps to follow to do this problem, i'm a bit lost on galois:

Given an irreducible quartic polynomial over the integers, given that it's galois group is S4 over Q and Y a root of the polynomial

is the extension Q(Y) a galois extension? All the definitions are swimming on my head

[hidden haven]

For a Galois extension E/F, |Aut(E/F)| = [E:F]

Try using this

Actually I am not sure if that will be useful lol

Galois means normal and separable

[gritty sparrow]

Well, the minpoly of Y is the original poly, so it can’t split

I assume that’s what they want

Can’t split in Q(Y), sorry

[terse crystal]

No. Since otherwise let E be the splitting field of the polynomial f over Q, then Q[Y]=Q(Y)=E. Therefore f(x)=Π{(x-P_j(Y)):0<=j<=3} where p_j are 4 polynomials with coefficients in Q, P_0(Y)=Y . By assumption there exists g_1, g_2 from Gal(E/Q) such that g_1(Y,P_1(Y), P_2(Y),P_3(Y))= (P_1(Y), Y, P_2(Y),P_3(Y)), and g_2(Y,P_1(Y), P_2(Y),P_3(Y))= (P_1(Y), Y, P_3(Y),P_2(Y)) which is a contradiction since g_1(Y)=P_1(Y)=g_2(Y) implies that g_1=g_2

[cursive temple]

this is what is done in 44

i think f_0 is just the product of all of the r_k(x_i)- terms

but i cant see how we get the bound that deg f_0 <= deg f without having a better bound that deg r_k < q

[novel parrot]

how did they start this proof?

becuz if r's are independent

f should be 0

?

[mild laurel]

"Otherwise"

[novel parrot]

otherwise meaning we are not taking r to be independent?

[mild laurel]

Yeah

If r_1, ... r_m are algebraically independent. Otherwise

[novel parrot]

got it

[novel parrot]

i dont actually undersrtand that proof

so whats actually going on here?

isnt g just the same as f

so g is inside a ring where powers of xm are already in ?

[hot lake]

g is clearly not the same as f ?

f may not be a polynomial in the form described in the last line

the whole point there is to have a change of variable to get something into that form

for example if f(x,y) = xy

[novel parrot]

what i mean is

f(x_1,...,x_m) = f(X_1 + xm^am, .... , xm) yeah?

now they take x_i as a coefficinet for g yes?

[hot lake]

well g is f after a change of variables

[novel parrot]

?

[hot lake]

g and f aren't polynomials in the same indeterminates

[novel parrot]

yeah

but

the ring that g is from

contains f

?

[hot lake]

???

[novel parrot]

??

i dont get

[hot lake]

g is an element of k[X1...X(m-1),xm]

f is an element of k[x1...x(m-1),xm]

[novel parrot]

but that first ring contains the other yea?

[hot lake]

you can inject the first one into the second one by sending Xi to xi-xm^ai and vice versa

they are isomorphic by a change of variables

[novel parrot]

ok

[chilly ocean]

Even forgetting change of variables, those two rings are the same in some natural way, right?

[novel parrot]

i think so

[hot lake]

well there are lots of isomorphisms between them

dunno if any one is more natural than any other

[novel parrot]

reverse inclusion maybe]

but

if f(x,y) = xy

[hot lake]

if you treat Xi as completely different indeterminates that have nothing to do with k[x1...xm], one isn't a subset of the other

[novel parrot]

what were you saying about this example?

[hot lake]

well it has degree d=2

so the change of variables says to pick X = x-y²

[novel parrot]

ok

[hot lake]

and so x=X+y²

and then g(X,y) = f(X+y²,y) = Xy+y³

a polynomial of degree 3 in y

whose leading term is a pure power of y

unlike f

and so when you go back to your finitely generated ring A = k[r1,r2]

[novel parrot]

this just is xy no?

[hot lake]

you get that if r1r2 = 0 then r2 is integral over k[r1-r2²]

xy is not a polynomial in X and y

[novel parrot]

replacing X

for x-y^2

[hot lake]

then you say g is the image of f by the isomorphism between k[X,y] and k[x,y] given by the change of variable x=X+y²

it is dangerous to say "g is just f" if the answer to the question "what is the degree of the polynomial" changes between f and g

[novel parrot]

its only changing because we are absorbing the first n-1 indeterminates as coefficients

[hot lake]

the change of variable isn't induced by an isomorphism between k[x1...x(m-1)] and k[X1...X(m-1)]

[novel parrot]

also the next part

is says each term of m contributes a since term of y of g

i dont understand that

is it that each term of g comes from just 1 term of f ?

[hot lake]

no it says each monomial term of f contributes a single term of the form a constant times xm^e to g

[novel parrot]

when we say constant

we mean an element of the field

?

[hot lake]

of the form c * xm^e for some c in k

yes

[novel parrot]

okay

[hot lake]

if you look at a term c * x1^a1 * x2^a2 * ... * xm^am in f

after the change of variable it gives a big polynomial

c * (X1 + xm^alpha1)^a1 * ... * (X(m-1) + xm^alpha(m-1))^a(m-1) * xm^am

[novel parrot]

yeah xm witha large power with c

[hot lake]

and it has only one monomial that is a constant times a pure power of xm

which is c * xm^(a1 * alpha1) * xm^(a2 * alpha2) * ... * xm^(a(m-1) * alpha(m-1)) * xm^am

[novel parrot]

yup

[hot lake]

= c * xm^(am + a1 * (1+d) + a2 * (1+d)² + ... + a(m-1) * (1-d)^(m-1))

and now, each exponent can only come from one monomial term from f

because am,a1,a2 and so on are just the digits of the exponent written in base (d+1)

so given an exponent you can recover at most one possible candidate monomial in f

[novel parrot]

ok

[hot lake]

so they can't cancel each other

and that ensures that g is of the form described

c * a pure power of xm ^ a big exponent + some lower degree terms

[novel parrot]

ok

and we only care about the highest power

to make it monic

?

[hot lake]

yeah I guess next page takes care of removing the c and make it monic

[novel parrot]

ok

thanks alot

also another question

is it common to get any transcendental elements

in a k algebra

[hot lake]

yes

[novel parrot]

but if we take Q as a the field

no transcendentals?

or not many

[hot lake]

well there are lots of Q-algebras

Q[x] is a Q-algebra where x is transcendental over Q

[novel parrot]

right

[unreal portal]

this was done by a computer in sage lol

[novel parrot]

i dont really understand this question

Z is a UFD itself

and Z[i] contains Z

so any integral element in Z[i] is in Z anyway

???

[next obsidian]

No

Z is integrally closed in Q

Not in Q(i)

[novel parrot]

but it says that a UFD's are integrallly closed?

[next obsidian]

In their field of fractions

[novel parrot]

oh

okay

Is the fact that UFD's are integrally closed hard to prove?

because i dont think it was shown

oh nvm i see it

[next obsidian]

No

you basically just factor the top and bottom into irreducibles and the result pops out

[novel parrot]

so like take an a/b and we see that factors of b must divide a so its actually an element of the ring

[next obsidian]

Yeah

[novel parrot]

$(x^2 - y^3)$

[cloud walrus]

ActiveChapter

[novel parrot]

does this ideal contain any powers of y?

alone

or

i can determine if an element if an element is in that ideal if theres no remainder after division yea?

[next obsidian]

You won’t be able to division with remainder in general

You can just prove it for any ring

[novel parrot]

how?

[next obsidian]

Play around with powers of x and y and stuff

[novel parrot]

alright

[next obsidian]

Just assume that f(x,y)(x^2 -y^3) is only in terms of y

You can just chop away at f and show f = 0

[novel parrot]

alright

[delicate orchid]

ah so my proof was right, I didn't want to post it and spoil 😵‍💫

[fiery berry]

i want to prove that a transcendental element $t \in L$ in an extension $K \subset L$ is irreducible. It feels like this should be super easy to prove but i am a bit lost in formalism. I know that the polynomial ring K[x] is isomorphic to K(t) in this case and somehow want to say that since x is irreducible, t has to be irreducible aswell. What's bothering me is that to be honest i am a bit confused about polynomial rings in general. In Ring Theory one defines a polynomial Ring just as a set of mappings from $\mathbb{N} \rightarrow R$ and from the field theoretic perspective it looks more like you just append a completely new element to the Ring that is irreducible and has absolutely nothing to do with the Ring whatsoever. So a more concrete question would be why, formally x is an irreducible element of K[x]. I dont even know if this makes sense anymore.

[cloud walrus]

chrisply

[hot lake]

um what do you mean by an element of L being irreducible ??

[split flicker]

cant be factored more right?

Cant be written as product of two nonzero polynomials in K

[dusty river]

non unit

[hot lake]

who said elements of K or L were polynomials

[split flicker]

i think irreducible is only for polynomials

[hot lake]

but aren't K and L fields ?

[dusty river]

It's for any ring 5 is irreducible in Z

[split flicker]

oh

so its just analogous to prime

i thought it was polynomial analog of prime

[hot lake]

in rings, it's not the same

[dusty river]

sort of

[fiery berry]

yea just irreducible element in an integral domain

but i see that in a field this doesn't really make sense right?

[hot lake]

but in fields both notion are useless cuz everything is a unit

[dusty river]

It does, it is just trivially true

btw what you said about new elements in fields is analagous to polynomial rings too

You take polynomial ring and then field of fractions

[cloud walrus]

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

[fiery berry]

bruh im too dumb for latex

[dusty river]

oh x^n is a prime element of K[x^n] whenever K is a domain

hint ||quotient by (x^n)||

[hot lake]

ah so you have fields K(x^n) and K(x) and you want the minimal polynomial of x over K(x^n) ?

[fiery berry]

yup

[fiery berry]

ook i think i see K[x^n]/(x^n) is isomorphic to K and since K is a domain we can just say that (x^n) had to be a prime ideal and thus x^n is a prime element right?

[dusty river]

[fiery berry]

ok nice, ty then my proof shall be completed now

[lilac trench]

everyone was bad once

[frank fiber]

Let $M:K$ a field extension and $a,b \in M$, is it posible that $[K(a,b):K(a)] < [K(b):K]$?

[cloud walrus]

Or x1

[chilly ocean]

ye? take K=Q, b=2^(1/4), a=2^(1/2)?

@frank fiber

[frank fiber]

here $[Q(2^{1/4}:Q)]$ is 4 right?

[cloud walrus]

Or x1

[chilly ocean]

generally like, if you have more elements the minimal polynomial for b might be lower degree

yeah

x^4-2 is minimal

[frank fiber]

thanks

[novel parrot]

so shall i try and show that k[t] and R have the same field of fractions

and since anything integral over k[t] lies in k[t] so contains integral elements of R?

is this an ok way of doing this

or am i wrong

<@&286206848099549185>

[terse crystal]

Isn’t that k[x]—>k[x,y]/P integral and injective?

I think it is

the canonical homomorphism mapping x to x+P

[novel parrot]

hmmm

where do we go from there

[novel parrot]

maybe its not injective

x^2 goes to x^3 - y^2

x^3 to x^2 - y^2

idk

is my way totally wrong

im trying to copy these

[terse crystal]

I think it is. any f(x) from K[x] doesn’t belong to P

[novel parrot]

ok

where from there?

[terse crystal]

I can’t see how the field of fractions helps here

[novel parrot]

we need to find the integral closure in the field of fractions

if we show that they have the same field of fractions

[terse crystal]

But you don’t have to

You can simply follow the definition

[chilly ocean]

what page is this?

[novel parrot]

definition of normalization?

[novel parrot]

703

[terse crystal]

That it is integral over a subring having the form of a polynomial ring

[novel parrot]

i thought normalization was to find the integral closure in the field of fractions?

[terse crystal]

If an integral B is integral over an integral subring A then Frac(B) is algebraic over Frac(A)

[next obsidian]

It is

I’m a little confused as to what they’re getting at, it sounds like something related to Noether normalization

[novel parrot]

isnt it the same as question 2

with different ideal

[terse crystal]

He wants to find the integral closure of R in frac(R). And I think noether normalization theorem can be applied here using the fact I just mentioned.

[novel parrot]

but how does noether normalization work here

[next obsidian]

But they’re asking about the definition of normalization m

[terse crystal]

This: If an integral B is integral over an integral subring A then Frac(B) is algebraic over Frac(A)

[novel parrot]

is this the same thing

so R would be integral over a subring with <= variables

and isnt Question 4 the same as question 2

@terse crystal

[terse crystal]

Rethinking...

[novel parrot]

ok

if Q4 is the same as Q2 i can do it if i can show that k(t) = Frac(R)

[novel parrot]

any idea.. ?

[terse crystal]

Yes . I think R is normal itself

R=k[x][Y]/(Y^2-x^2-x^3)=k[x,y] where y = Y+ (Y^2-x^2-x^3) Frac(R)=k(x)[y]=k(x)(y)

Any f+gy from Frac(R) if f+gy is integral over R, using euclidean algorithm we can assume that degree of f and g are both negative, then show that f+gy=0

Let a= f+gy since a is integral over R therefore a^n=Σ{s_k a^k :0<=k<n}

for some integer n and some s_k from R

Compare the degrees on both sides

[novel parrot]

isnt it easier to use the UFD property

that they are integrally closed

because i dont follow what you wrote

[terse crystal]

I don’t know whether R is a UFD

[novel parrot]

if we could show that k[t] contains R

and their frac field is same

then done right?

[next obsidian]

Why is k[t] a UFD?

t isn’t necessarily transcendental

[novel parrot]

t = y/x

[next obsidian]

t is x/y or y/x inside the field of fractions right?

[novel parrot]

yea

[terse crystal]

Then we have different results... your integral closure would be k[t] while mine is R...

[novel parrot]

yeah ...

maybe k[t] is also contained in R lol#

[terse crystal]

Anyway my point is in this formula a^n=Σ{s_k a^k :0<=k<n} the degrees of the coefficients of the constant term (of y) are smaller than that of the right side

Since the degrees of f and g are negative and the degrees of both coefficients of any S_k are non- negative

I used degree(f+g)>=degree(f), degree(g) ( more accurately I think degree(f+g)=max{degree(f),degree(g)} ) for any f , g from k(x)

[novel parrot]

hmm

ok

i also have another question

[terse crystal]

Where is it

[novel parrot]

it says using 12 and 26.2

12 says radI = intersection of all primes contained I = radI

but how does it work?

we can do it directly instead

?

[next obsidian]

You just distribute the \cap R inside

You turn that into intersecting a bunch of primes containing IS\cap R

And by lying over stuff it turns into the intersection of primes containing I

Lying over being (2) in Theorem 26

[novel parrot]

wdym

[next obsidian]

You have (\big cap P )\ cap R

For P primes containing IS

Just turn that into

[novel parrot]

oh right

[next obsidian]

\bigcap(P\cap R)

[novel parrot]

uhm

[novel parrot]

i understand we get this

what next?

[next obsidian]

That turns into the intersection of all primes containing I

By (2)

Each P\cap R is a prime containing I

And you know each prime containing I shows up because of (2)

[novel parrot]

so $P \cap R = Q \cap R$ where Q is a prime in S

[cloud walrus]

ActiveChapter

[novel parrot]

yeah?

[next obsidian]

Wat

P is a prime in S

To begin with

[novel parrot]

i dont really know how every prime shows up

[next obsidian]

Because of (2)

Take Q a prime containing I

[terse crystal]

Rad_R(I) is {x from R : x^n belongs to I for some n}?

[next obsidian]

Then there’s a prime P in S such that Q = P\cap R

But P then contains IS

[novel parrot]

OH

ok

[next obsidian]

So P shows up here

[novel parrot]

right

yep

yes

thanks

[novel parrot]

yes

[terse crystal]

Oh so Q \cap R=P where P (Q) is a prime ideal of R (S) respectively then I is contained in P iff IS is contained in Q

Oh basically what he just said

[novel parrot]

yep thank you

[terse crystal]

After reconsideration, I think you are right. Just noticed that x=(y/x)^2-1 and y=x(y/x)...

[fickle tapir]

is the cardinality of all finite subsets of R the same as |R|?

i tried proving it but didnt get far

[terse crystal]

What is R?

[fickle tapir]

The reals

[terse crystal]

Real numbers?

[fickle tapir]

yeah

the cardinality of the set containing all finite subsets of R

[next obsidian]

I actually am seriously questioning if this set‘a cardinality is independent of the continuum hypothesis

It probably is?

Also this isn’t algebra haha

[fickle tapir]

where can i ask this?

[next obsidian]

Maybe foundations?

[gritty sparrow]

It has the same cardinality as R

[fickle tapir]

and then somehow their union also has cardinality |R|

[gritty sparrow]

Yeah that could work

[fickle tapir]

but the only straightforward bijection i can think of is the base case of size 1, with a subset with one element

with n > 1, its not clear

[next obsidian]

Oh my god

[kind temple]

let F be the set of finite subsets of N. you know that |F|<= |P(N)| = c

[next obsidian]

Hurb

[kind temple]

this is an injection from R to F…

[next obsidian]

I’m dumb lol

[fickle tapir]

wait why are we doing subsets of N?

[kind temple]

oh, i thought that was assumed. i don’t think the collection of all finite sets is a set?

[next obsidian]

Take R x … x R and map an element (a1,…,an) to the set {a1,…,an}

This surjects onto a set containing all subsets of size n right?

[fickle tapir]

i thought its a set containig sets

[next obsidian]

Like it will surject exactly onto the set of all subsets of size <= n

[kind temple]

oh i missed it, finite subsets of R

my bad

[fickle tapir]

yeah

[next obsidian]

Anyway the argument I did should give an upper bound by |R|

So they’re all |R|

[fickle tapir]

to reiterate, if we let S be the set of all finite subsets of R,
for each s in S we can enumerate the elements and map it to RxRxR
this tells us |S| <= |RxRx...R| = |R|?

Now we need to show |R| <= |S|

[next obsidian]

I mean we know the set of all finite subsets of size <= n has size |R| right?

It’s at least that big because it contains singletons

[fickle tapir]

The thing with this is that im not sure how many R's there are in the cross product since you can always find a larger finite subsets

[next obsidian]

It’s at most that big because it’s surjected onto by R^n

Then S is the countable union of these so it also has size |R|

Yeah

[fickle tapir]

yeah

[fickle tapir]

im not sure about this one

i get the singleton part though

[next obsidian]

I proved it?

[next obsidian]

Right here

[fickle tapir]

ohh

[next obsidian]

It’s surjected on by R^n

[fickle tapir]

oh wait i was thinking about the set of all finite subsets as a whole

[next obsidian]

Nah

[fickle tapir]

instead of set of all finites of size n

[next obsidian]

You wanna just chop it up into countable many sets

You could either pick size <= n or just size n

But I did <= n because the function I did naturally worked with that set

[fickle tapir]

ok, so we have established that after chopping it up, they each have cardinality of |R|

[next obsidian]

A countable union of sets of a fixed infinite cardinal kappa has size kappa

[fickle tapir]

and there are a countable number of subsets of size n

[next obsidian]

Yeah

[fickle tapir]

i only have a lemma that says

a countable union is countable

[terse crystal]

How to prove that the cardinality of direct product of countable copies of R equals the cardinality of R?

[next obsidian]

Yes

Haha

You can replace the first c with <= c

[fickle tapir]

Ok so lets say our set is called S

and S = S_1 u S_2 u S_3

[next obsidian]

Ughhhhhh

It’s an absorptive property of cardinal multiplication

Yes

[fickle tapir]

isn't that different from c though

[next obsidian]

Yes but you can replace it <= c

Just like… prove it

It has cardinality at least c

Like

Clearly

[fickle tapir]

i am given that N is the smallest infinity

[next obsidian]

Now just add in dummy sets

Until you have c many

Then you’re a subset of that

Which also has cardinality c

So you’re bounded from@below and above by c

[fickle tapir]

luckily there is a theorem that says if X is an infinte set, |N| <= |X| 🙂

i thought you had to show |N| = |R| to use the lemma or something

because it said union of c sets

but R is uncountable

so that canot hapen

are you adding empty sets in the union?

until you have c many unions

not fully

yeah

wait

if |A| <= |B|
Then is |A| u |B \ A| = |B|?

i was wondering if i could use some argument like that to add sufficiently many empty sets in the union to attain c many unions like momy said

but i want to not hand wave it

which is tricky

hmm

thanks ill try thinking about it a bit more

[next obsidian]

Just take

C copies of C itself

Or like

[fickle tapir]

how do you take copies of a cardinality

[next obsidian]

I mean

A cardinal is a set

[fickle tapir]

ohh

[next obsidian]

Yeah

I mean…

Ugh this gets annoying

Foundations is all just like

Homotopy equicalwnt

But the exact specifics can vary a bit

On how you get taught it or construct everything

[fickle tapir]

yeah i will

[terse crystal]

Let A be the set of finite subsets of R. B be the set of finite multisubsets (allowing duplicated elements) of R. Clearly R can be viewed as a subset of A and A is a subset of B. I can construct a bijection from B to positive real numbers: we know there exists a bijection g from R to (0,1). any {x_1,...,x_n} from B, g(x_i)=0.x_i1 x_i2 .... we can map {x_1,...,x_n} to (n-1).x_11 x_21 ... x_n1 x_12 x_22 ... x_n2 ...

Therefore |R|<=|A|<=|B|=|R|

[cursive temple]

should the ideal generated by some element $a$ be written as $(a)$ or $\langle a \rangle$

[cloud walrus]

Pappa

[cursive temple]

i feel like ive seen the second notation somewhere but jacobson uses the first one

[golden pasture]

first usually

[cursive temple]

so the second one is generally reserved for subgroups generated by those elements?

[next obsidian]

The latter is used for modules

So technically if you considered A as an A-module you could do that

But like, no one ever ever ever ever does it

You’re more likely to see aA

[golden pasture]

hm i jus call (a) an A-module as well

[next obsidian]

Yeah, but for a module M you use <> to talk about a subsmoidke of M generated by elements of it

So you could use that notation on a inside of A

But even when you’re considering modules and A as an A-module I always still see (a) or aA

[golden pasture]

ahhh

[cursive temple]

i know that the answer will be ap_2p_3-bp_1p_4-cp_5, where p_i is the ith symmetric polynomial

how do i find a,b and c then

[novel parrot]

If the ideal of a algebraic set is radical

Will that set be a variety

[oblique river]

the ideal of any subset will be radical, so no

[thorny flame]

does a group of prime number of order is cyclic?

[chilly ocean]

do you know lagrange's theorem?

[novel parrot]

wait

[oblique river]

👀

[novel parrot]

how do you know ideal of any set is radical

only algebraic i thought

[oblique river]

what do you mean only algebraic?

what is an algebraic ideal?

[novel parrot]

vanishing points of some polynomials

or common zeros

or roots

[oblique river]

oh, yes, an algebraic set of points

i thought you were describing the ideal as algebraic

not the points

[novel parrot]

oh

[thorny flame]

yes

[novel parrot]

the nullstellensatz only says ideals of algebraic sets are radical?

[oblique river]

yes, given any set X in k^n, the ideal of polys vanishing on X will always be radical

no

the nullstellensatz says that if X is algebraic then V(I(X)) = X, and I(X) is radical

it also says that I(V(J)) for some ideal J is the radical of J

the nullstellensatz makes no claims about I(X) for non-algebraic sets X

[novel parrot]

does it work for non algebraic sets though

is what i mean

[oblique river]

I(X) will still be radical, but V(I(X)) won't be equal to X anymore

it can't be equal

becuase V(anything) is always algebraic

so if X isn't algebraic, they can't be equal

[novel parrot]

ohoho ok

yep

oke

but -

do you know how to show V and W are varieties then

[weary terrace]

For those of you who are familiar with Burnside's theorem:
Let $\rho$ be a representation of a group $G$. If $\rho$ is a faithful representation of $G$, i.e., $\rho:G\to GL(V)$ is injective, then any irreducible representation of $G$ is contained in some tensor power $\rho^{\otimes n}$ of $\rho$.

I think the converse is also true, but I'm not sure how to prove it. Any ideas?

[oblique river]

(oh sorry, I was using V earlier instead of Z)

[cloud walrus]

RiesZ

[oblique river]

sorry but could you not interrupt an in-progress conversation?

[weary terrace]

Sorry, didn't notice..

[oblique river]

what is the definition of a variety here?

[novel parrot]

irreducible algebraic set

so ideal of it is prime ideal

[oblique river]

oh, so then what youre asked to prove there is that theyre irreducible

yes

the fact that they're algebraic is automatic

[novel parrot]

yes

[oblique river]

maybe i'm misunderstanding your question then

[novel parrot]

how can we show that they are irreducible?

[oblique river]

by showing that I and J are prime

[novel parrot]

ok

but i dont really know how to show J is prime

shall we .. reduce it first maybe

[oblique river]

that can be hard, sometimes you can take the quotient and work with that

i am a little confused thought cuz I dont think (uv + v) is a prime ideal?

because uv + v = v(u+1)

[novel parrot]

i dont really know either but v and u+1 are irreducible

[oblique river]

yeah so their product shouldnt generate a prime ideal

[novel parrot]

thats what i thought ...

[oblique river]

like if you think about Z(uv + v) = Z(v(u+1))

that's going to be where v = 0 and where u = -1

so it's like, two crossing lines