groups-rings-fields

[bitter rover]

This isn't true unless F is also continuous

[night tartan]

no - famously, there exists a counterexample to f(x+y) = f(x) + f(y)

(think hamel basis!)

so u'll have to use more facts than just that

[dull ginkgo]

That’s by definition of field automorphism

To be specific, the ring of R-linear endomorphisms of R is just R, where the endomorphisms are right multiplication

R is a 1 dimensional vector space over itself who’s basis element is 1

So yes lol

[tardy hedge]

God showing $a_n^{r+1}b_{m-r}= 0$ for 2 i) was so annoying for me

[cloud walrus]

kiand123

[tardy hedge]

All the indices and everything

At least it gave me good review on the nuts and bolts of how polynomials multiply

But UGH

[ebon gyro]

ahaha i just graded my commutative algebra students on something similar

yeah these are good to do in full though

[tardy hedge]

Hahahaha

Yeah. I knew it was good to actually work on the details even tho i didnt want to

[dull ginkgo]

Jacobson radical? In MY commutative algebra class? That is noncommutative filth and must be purged! For shame.

[cloud walrus]

yeshua

[serene dune]

is there anything else other than the form P \oplus Q

[kind temple]

no, that is all of them, if you mean P and Q are subgroups of Z

[serene dune]

yeah i meant that

can u give me any example which is not of that form

[kind temple]

no, i just said that is all of them lol

[serene dune]

ok

im blind

[inner steppe]

How am I even supposed to solve this?

[lone niche]

Check if they have the same roots.

[delicate bloom]

if they happen to be equal, you know from the left that (x+3) divides it

[inner steppe]

Then what?

[serene dune]

number of subgroups of $\Bbb Z_p \times\Bbb Z_p$

[cloud walrus]

yeshua

[serene dune]

number of subgroups of $\Bbb Z_p \times\Bbb Z_p \times\Bbb Z_p$

[cloud walrus]

yeshua

[serene dune]

for this the subgroups can be of order 1,p,p^2
out of which two are trivial

how to determine the rest ?

any hint, pls just give me the minimal amount of hint

[rustic crown]

a subgroup of (F_p)^n is automatically an F_p-vector space

so you want to find subspaces inside an n-dimensional space over k = F_p

[delicate bloom]

then use that to answer the second question

[winged void]

I have a question regarding the following theorem

\begin{proposition}
Let $R$ be a commutative ring, $a \in R$, and $f \in R[X]$. Then there exists a $q \in R[X]$ such that
[f = q \cdot (X-a) + f(a).]
\end{proposition}

why this holds only if its a commutative ring

what would be the problem if it was not

[cloud walrus]

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

[regal zodiac]

Hello, could someone tell me why any element of homK(F) extends to an element of HomK(L, algebraic closure of L) ?

[rustic crown]

if you have a map F --> Lbar, then you can extend it to any algebraic extension L/F

[serene dune]

m dimensional subspace inside n dim space

$\frac{\prod_{k=0}^{m-1} (r^n - r^k)}{\prod_{k=0}^{m-1} (r^m - r^k)}$

amirite

[regal zodiac]

But here i just have a map F -> F don't i?

[rustic crown]

yee r being the number elements in the field k (in our case r = p)

[serene dune]

yep i missed that

[rustic crown]

yee, you have a map tau: F --> F which you can compose with the inclusion F --> Fbar = Lbar

[cloud walrus]

yeshua

[rustic crown]

what if you use this for f = (X-a)(X-b), that should force q = (X-b) right?

[regal zodiac]

Oh okay i see now, thank u so much

[rustic crown]

but then we have (X-a)(X-b) = (X-b)(X-a) which would mean ab = ba

[serene dune]

made a mistake

$\frac{\prod_{k=0}^{m-1} (r^n - r^k)}{\prod_{k=0}^{m-1} (r^m - r^k)}$

[cloud walrus]

yeshua

[rustic crown]

ah right

its {m linearly indep vectors}/GL_m(F_p)

[serene dune]

so to find the total number of subgroups i have to count all possible dimensional subspace right?

starting from 0 to n

[rustic crown]

yee

[serene dune]

$\sum_{m=0}^{n} \binom{n}{m}r = \sum{m=0}^{n} \frac{\prod_{k=0}^{m-1} (r^n - r^k)}{\prod_{k=0}^{m-1} (r^m - r^k)}$

[cloud walrus]

yeshua

[serene dune]

is this a sin to do ?

[rustic crown]

not really, but that notation isn't conveying anything lol

ig better to write |Gr(m, n)(F_p)|

where Gr(m, n) is the grassmannian of m-dimensional subspaces of an n-dimensional space

[serene dune]

woah

what is grassman doooooin

:orangecuh:

[rustic crown]

[thorn jay]

Because the factor theorem only holds for commutative rings

f(X) - f(a) has a as a root, so if the factor theorem holds it must be divisible by X - a

If it doesnt then the factor theorem need not hold

In fact this statement is equivalent to it

[rustic crown]

Gr(d, n)(k) is the standard-ish notation for d-dimensional subspaces of k^n. It's called the grassmannian of d-dimensional subspaces.

[serene dune]

there's a section called grassman ring in Hoffman kunze, but i skipped it

:orangecuh:

[regal zodiac]

Could someone explain what is the intuition behind the definition of a polynomial being solvable by radicals ?
Like i knew since a long time that being solvable by radicals means that the roots can be expressed in terms of the coefficients and classic operations but why is this equivalent that the extension K / splitting field of f is contained in a radical extension ?

[thorn jay]

Well, if you can express all roots of the function using only nested radicals, then you can have a finite tower of simple field extensions which just take an element of the base field and takes a radical of it (F(nroot(a)) for some a in F), and eventually all the roots of the polynomial must be contained in it, meaning that the splitting field is container in it

[regal zodiac]

Hum i see but i just struggle with a small detail, If we denote by a1.....an the roots of f, why can't we merely consider the following chain of extensions :
F , F(a1) = F1 , F1(a2) = F2 .....
Then Fn will be the splitting field of f over F and thus F / splitting field of f is radical
Why do u need the contained in a radical extension?

[thorn jay]

Because that tower isnt a tower of radical extensions

Q -> Q(sqrt(1 + sqrt(2)))
Is nor a radical extension, while
Q -> Q(sqrt(2)) -> Q(sqrt(1 + sqrt(2)), sqrt(2))
is

And Q(sqrt(1 + sqrt(2))) ≠ Q(sqrt(1 + sqrt(2)), sqrt(2))

You cannot obtain Q(sqrt(1 + sqrt(2))) by only performing radical extensions

It'll always yield a bigger field containing Q(sqrt(1 + sqrt(2))

[regal zodiac]

Yes okay i understand now ! My mistake was that i was assuming that if a1 is a root of f there exists n such that a1^n is in F but it's not True as u showed with sqrt(1+sqrt(2))

Thank u so much @thorn jay

[thorn jay]

Ofc! :3

[regal zodiac]

Last question, If now i got that the F -> splitting field of f is contained in a radical extension, does it means that each root of f can be expressed with classical operation on the coefficients ?

[thorn jay]

Yes

At least, they can be expressed as a combination of rational numbers and radicals

[regal zodiac]

Yes i can see it now, thanks !

[thorn jay]

[inner steppe]

Is there any advantage to think this way α(m∘n)=α(m)+α(n) and not α(m+n)=α(m)∘α(n)
Is α(m+n)=α(m)∘α(n) the wrong way of thinking?

[thorn jay]

Yes because
a(m + n) = m + n + 3
a(m) • a(n) = a(m) + a(n) + 3 = m + n + 9

Also please define •

[inner steppe]

Sorry this was what I was supposed to post, the first screenshot was part of the solution lol

[thorn jay]

Thus a(m + n) ≠ a(m) • a(n)

[rocky cloak]

alpha is supposed to map from (Z, °) to (Z, +). So then the first one is the correct definition of homomorphism. The second would be if it was going the other way

[inner steppe]

but is a bijection so technically shouldn't this work too (Z, +) to (Z, °)?

[inner steppe]

Assume we don't know alpha

[thorn jay]

No

[inner steppe]

I meant is it possible to find a bjection alpha such that

[rocky cloak]

No, just because something is a bijection doesn't make it a homomorphism.

[inner steppe]

α(m+n)=α(m)∘α(n) and would it be considered an isomorphism

[thorn jay]

I dont get your question

You're asking multiple things at the same time

And i think you've changed your question

So please state clearly what you want to know

[inner steppe]

No I didn't change my question, at least not on purpose. Lets assume that I never posted the first screenshot and are given the following question. Can I find a bijection b(m∘n)=b(m)+b(n) or b(m+n)=b(m)∘b(n) to show that they are isomorphic or is one wrong even if I do end up finding such function and why? As in who is to say that b(m∘n)=b(m)+b(n) correct and not b(m+n)=b(m)∘b(n) assuming it is even possible to find two different functions b for each cases

Btw I am not claiming that b should be the same function in both cases.

[rocky cloak]

You can find such a b that's fine. It would simply be the inverse of alpha

[inner steppe]

That would also prove isomorphism right?

[rocky cloak]

Yes, you just need one isomorphism in either direction

[inner steppe]

In that case why should someone pick the other route?

b(m∘n)=b(m)+b(n) and not b(m+n)=b(m)∘b(n)

[rocky cloak]

I mean you just have to pick one

[thorn jay]

You pick the easiest

For example for an isomorphism between (R, +) and (R+, •) i'd pick
e^x : (R, +) -> (R+, •)
Instead of
ln(x) : (R+, •) ->(R, +)
because i generally find e^x less finicky than ln(x)

(R+ denotes the positive reals)

[inner steppe]

Let H, K be subgroups to a finite group G. If |H| and |K| are relative prime and a, b ∈ G, which values are possible for |aH ∩ bK |?

This is how I was thinking if H=K and a=b then |H| + |K| and if H!=K and a!=b, all of a sudden it is not as easy.

it is my understanding that aH and bK build cosest but this is all that tells me aH ∩ bH = ∅.

I am stuck

[thorn jay]

It tells you nothing right?

Well at least
|aH ^ bK| = |b^-1aH ^ K|

So all the possible values are
|gH ^ K|

(^ denotes set intersection)

[inner steppe]

not sure if I understand this one

[rocky cloak]

Guess also worth thinking about how HnK is a subgroup of K

[inner steppe]

At least always one

because of identity element but other than that it depends on if they have similar elements or not

but still really not getting anywhere with this assuming I on the right direction

[rocky cloak]

I mean (aH)nK doesn't contain identity unless a is in H

[inner steppe]

yes but then in that case only the second statement is true

considering aH doesn't need to be a group.

[rocky cloak]

What's the second statement?

[crystal vale]

A cyclic group which is infinite then it has infinitely cyclic subgroups.

Let G = < x>, then all <x^i> are cyclic subgroups and are all distinct, right?

[inner steppe]

it depends on if they have similar elements or not

[rocky cloak]

Well, I mean it always depends on that. Thats what the intersection is after all

[inner steppe]

yes but what were you implying when you said

[rocky cloak]

Well maybe you want to think about a connection between (aH)nK and cosets inside the group K

[south patrol]

Hello

Finite direct sums

Otherwise you can do silly things

Like if M_1, M_2,... are all nonzero modules then their direct sum isn't Noetherian (even if all the M_i are) because you have the ascending chain like

M_1 (+) ... (+) M_n

[dull ginkgo]

Ah true, not artinian either cuz M_n (+) M_n+1 (+) M_(n+2)…?

No wait it’s direct sums

[south patrol]

This still works

[inner steppe]

Sorry I have now thought about for a long time and not getting anywhere important. The only time (aH)nK and cosets inside the group K are the same when aaK

[rocky cloak]

So either the intersection is empty, or aH contains an element k of K. Then aH = kH.

Now, is there a relationship between (kH)nK and k(HnK)?

[inner steppe]

Sorry I am just not seeing where we are going with this. What do you mean by kH?

I have no clue about the (kH) n k but second one can only have one element k since given that |H| and |K| are relative prime it means that the subgroup |H ∩ K| must divide both |H| and |K| since it a subgroup of both. Thus |H∩K|=1.

I also btw have a hint in the question c ∈ aH ∩ bK then aH = cH

but it is really not telling me anything

Because

[rocky cloak]

kH means the coset. You know things of the form {kh | h in H}

Think about what it would mean if kh is in K

[inner steppe]

I don't know if this is what you were saying earlier but the only time aH = bK is if h=k and a=b. If my understanding it correct searching for aH ∩ bK is the same as identifying H ∩ K and |H ∩ K| is true only for identity element as said before so the only element in common is c1 thus the order is 1. The other possibility is 0 because a!=b.

Is this correct?

[rocky cloak]

That it's either 1 or 0 is correct.

I'm not sure I'm totally following your reasoning

[inner steppe]

I am not sure either but here is an example. Lets say G = {1,2,3,4,5,6,7,8,9} and Let H={1,2,3,4} and K={1,4,5,6}

When is aH = bK, well the only time that will be true is if we pick 1 and 4 , as in when h=k. So now it is just about identifying H ∩ K because if we ever pick a != b then aH=bK will never be true. |H ∩ K|=1 because |H| and |K| are relatively prime. Thus the only possible element in aH ∩ bK is ce where c = a=b and e is the idenitity element 1

Is this a correct or clear reasoning?

[tardy hedge]

Not sure how to use ex1

For 2 i)

Im not sure how to create new nilpotent elements from the statement in 1

[rocky cloak]

Well, I can't say I'm following this.

Like your example doesn't appear to be a group, but ignoring that aH will never equal bK, but maybe you're talking about how they can intersect...

[inner steppe]

but they only intesect if aH = bK

Yeah I know it is not a group at all

But I just sort of wanted to work with actual examples but probably not the best one.

[rocky cloak]

Well, just picking a and b the identity, H and K still has {e} as intersection even though H and K are not equal.

Like I think maybe you mean something else when you say aH = bK

[inner steppe]

When we say aH ∩ bK

Don't we want the elements where ah = bk, no?

As in the elements both of the sets have in common.

If this going to be true then a=b and h=k. Is what I am saying totally wrong so far or unclear? If wrong then I am really back to square one (confused).

[tardy hedge]

I have gotten that $a_{n-1}b_{m}+a_nb_{m-1} = 0$ so using ex 1 we can say $a_{n-1}b_{m}$ cannot be a unit

[cloud walrus]

kiand123

[tardy hedge]

Can this help me in showing $a_{n-1}$ is nilpotent?

[cloud walrus]

kiand123

[tardy hedge]

Tbh i didnt want help on this question but its taking me too long

I need to get to the other stuff

This wasnt even an assigned question but i wanted to do it cause polynomials confuse me

[grave sedge]

Multiply this by a_n

[finite turtle]

why does Z(G) = 1 lead to C_G (H)=H here? more context be helpful?

[grave sedge]

If C(H) were larger it would be all of G

Which means x would commute with everything

[tardy hedge]

Why? That just kills the first term

[grave sedge]

I'm going off the hint given with the exercise

[tardy hedge]

Oh

[grave sedge]

It should give a(n)^2b(m-1)=0, right?

[tardy hedge]

Yea

We showed that already

[grave sedge]

Have you done the induction thing? If yes, what's a(n)^(m+1)b(0)?

[tardy hedge]

ive done the induction thing

I know an is nilpotent

[grave sedge]

Ah ok

[tardy hedge]

Im just needing help on the very last part

[grave sedge]

Then a(n)x^n is also nilpotent

[tardy hedge]

I have an nilpotent now use ex 1 to show the others are

[grave sedge]

And f-a(n)x^n is unit+nilpotent

Now you can induct on n

[rocky cloak]

Well if for example b=k is the identity and a = h^-1 you would have ah = bk

[tardy hedge]

Bruhhh

Yea i didnt think to go back to the whole polynomial

[finite turtle]

why is this exactly?

[tardy hedge]

Thanks matt

@rocky cloak how long do u think is a good amount until asking for help on problems? Id like to try to figure stuff out myself as much as possible but at the same time i dont have all the time in the world and i need to get going on stuff

[inner steppe]

but still it would only be identity element they will share

but I get your point, how else do I explain it then

[grave sedge]

H is contained in C(H) because it is abelian, and H has prime index

I.e. the only subgroups containing H are H and G

[rocky cloak]

I think it probably depends on the problem, and wether you're making any progress or not.

I guess you just have to weigh how much time you have / how long it takes before you get frustrated.

I guess if you're very stuck. Then I guess it's a good idea to ask for a hint, and then go back to spending time for yourself.

[inner steppe]

Given f(x) = x^2 + x + 1 ∈ Z_2[x]

I was asked to draw the addition table for Z_2[x]/ f(x)

I included x^2 but they didn't in the table, why? How else are we gonna represent polynomials like x^2 + 1?

[inner steppe]

I get that x^2 = -(x+1), but how can we ever get x^2 by using addition?

[rocky cloak]

x^2 + 1 = (x^2 + x + 1) + x

so can be represented by x

[hidden wind]

i’m trying to show this: if A is a ring with exactly one prime ideal, every element of A is either nilpotent or a unit

and i’m simply not sure how to go about it…

i know the ring must be local, but that’s about all i have

and i’m not sure if that’s even relevant

[wraith cargo]

uhhh is this true?
Wouldn't p-adic integers be a counterexample

or am I tweaking 💀

[rocky cloak]

It is relevant if you know what units in a local ring is

[dim widget]

That has 2 prime ideals

[wraith cargo]

ah right the zero ideal

[dim widget]

If you have one prime ideal by definition it is the nilradical

[hidden wind]

ok thank

i’ve learnt all my (very little) basic ring theory from Clark’s Elements of Abstract Algebra but i’ve just recently realized some of the terminology there is nonstandard

for example Clark considers the whole ring itself as a maximal ideal

[rocky cloak]

.... What do they call maximal ideals then?

[hidden wind]

proper maximal ideals

[rocky cloak]

That's... So pointless and stupid

[hidden wind]

haha

[rocky cloak]

At least "maximal proper ideals" would make sense as term

Even though it's unnecessarily long

[hidden wind]

ok nvm i misremembered, he does exclude the entire ring when defining maximal ideals, but he doesn’t for prime ideals

i suppose that’s slightly less of a crime

[rocky cloak]

Yeah, the naive definition of prime does include the entire ring after all. It's just convenient to exclude it

[hidden wind]

i suppose it’s analogous to considering 1 as a prime in Z

[rocky cloak]

Indeed

[hidden wind]

this other problem seems suspiciously easy

given ideals I = (8, 12) and J = (x^3 + 4x^2 + 4) both in Z[x], i’m asked to find a generating sets for I+J, IJ and sqrt I

it seems so straightforward i’m overcome with the dread that i’m misunderstanding something big time

[unkempt stream]

is sqrt(I) the nilradical

[hidden wind]

yes

clearly I = (8,12) = (4) as no polynomials of positive degree divides 8 or 12, and gcd(8,12) = 4

[unkempt stream]

yeah GCD domain moment

IJ should be simple lol

[hidden wind]

so i can just use the identities (f)+(g) = (f,g) and (f)(g) = (fg) right?? what am i missing

[unkempt stream]

I forget specifically if Z[x] is also a GCD domain

[hidden wind]

for the first two at least

[unkempt stream]

Oh yeah i forgot, if R is a GCD domain then so is R[X]

[hidden wind]

for the sum I+J i’ve removed subtracted some linear combinations in the two generators to arrive at (x^3, 4)

and the product i guess should simply be (4x^3 + 16x^2 + 16)

[unkempt stream]

that's an equivalent ideal

(4, x^3 + 4x^2 + 4) = (4, x^3) so whatever looks nicer

[hidden wind]

ok thanks i think i just really need a break

[unkempt stream]

or you can do some conceptual exercises that don't require you to explicitly write shit down

[gusty thistle]

For each prime ideal p of a ring R you get a field by localising at p and the taking the quotient at the maximal ideal (or equivalently by taking the fraction field of the quotient). I was wondering if you get isomorphic fields for each prime ideal or do you sometimes have two primes with non isomorphic fields ?

[rotund aurora]

Just take R=Z

[hidden wind]

yeah def that too

[ebon gyro]

you know that the nilradical is the intersection of all primes - what can you say about the nilradical as an ideal and what can you say about nilpotents in a local ring

[hidden wind]

omg hi ari

[ebon gyro]

hiii

omg sorry i read your dms and they were very cool i will respond to them

[hidden wind]

i’m so bad at abstract algebra but i’m having a lot of fun with it

also right i completely forgot about that characterization of the nilradical, thanks will have to have a proper look at this

i’m still rather confused by the basic objects to have a look at when studying rings and their ideals

[golden turtle]

I must have a fundamental misunderstanding of this question as I thought this was how you always solve these types of problems

What is the mistake and is this close to the proper approach?

[rocky cloak]

The mistake is just like described in the yellow box.

But it is very close to correct. Just account for that small error and you're good

[golden turtle]

But I thought that they were disjoint excluding identity

[rocky cloak]

Why?

[golden turtle]

Because two sylow subgroups are either identical or disjoint excluding identity I thought

[rocky cloak]

Maybe have a look at the symmetry group of a hexagon

[golden turtle]

I don’t see how it isnt true since if P and Q are distinct p sylow subgroups their intersection is a subgroup and must have order dividing p by Lagrange so

[vapid vale]

you may be recalling the disjointness from when they actually have prime order, and thus their intersection can only be trivial

8 is not prime

[rocky cloak]

So... What

[golden turtle]

So they only intersect at the identity

Oh I see

8

[rocky cloak]

Well like hk said, they can just intersect in some smaller subgroup

[golden turtle]

🤦🏽‍♂️

[vapid vale]

lol

[golden turtle]

Fuck me

[dull ginkgo]

Sylow groups are silly

[dull ginkgo]

Prime cyclic groups intersect only at the identity, which with some Sylow arguments actually let you rule out possibilities by realizing how much of the group the Sylow groups take up

[gusty thistle]

Lol

Yea

😭

Thanks

Ive got to start thinking more yarn

[marsh fiber]

Skill issuing hard on this. First I tried finding a pair of coprime ideals to factor I so I could use CRT to decompose R/I, but I couldn't find a nice pair, since e.g. (1-X,X-Y).(X,Y) is too big (includes X-X^2) and the same thing happens for other "obvious" factors. I also tried gaming out the relations and found that all elements of R/I were equiv a+bx+cy and tried to find an iso that way but all choices I could think of didn't respect the operation properly. Not really sure what to do, though my best guess is I'm not seeing a good factorisation of I

(already posted this in help but what the hell)

[wraith cargo]

maybe find a surjection R -> S with kernel I

[bitter rover]

What can you say if y - xy = 0?

What can you say if x^2 - xy = 0?

[marsh fiber]

does that logic work on R/I? I thought about it but i was wary of zero divisors

[bitter rover]

Is that directed at me? "Divisible by blah" becomes "equals 0" in the quotient.

[marsh fiber]

i mean so because y-xy=0 in the quotient it doesn;t follow y or 1-x are 0 in the quotient right

i did think about doing that and using f(0,0) and (1,1) somehow

but i couldnt really justify it

[bitter rover]

A polynomial ring is an integral domain if and only if the original ring is.

(And every field is an integral domain)

[grave sedge]

||Y and 1-Y|| should be the idempotents which give the factorization

[wraith cargo]

I think they were talking about the quotient ring?
Tho idk exactly what their issue is

[rocky cloak]

I think you can modify your choice of ideals slightly to
||(1-X, X-Y).(Y, X^2)||

[bitter rover]

Same.

[marsh fiber]

yeah I don't see why x-xy=0 (in the quotient) should tell me anything useful about x or y or whatever

I can't use domain facts because that;s no longer a domain right

[marsh fiber]

fml

[rocky cloak]

Really how I found these were just considering a homomorphism to QxQ[x]/x^2 though. And seeing what the kernel would be when composing with projections into Q and Q[x]/x^2

So that's probably the easiest way

[marsh fiber]

didn't occur to me to reason backwards i just sat there trying to think of something from the ether

ty so much

[inner steppe]

Let G = U(Z18) be the group of invertible elements in Z18.
G acts on the set X = Z18 through multiplication. Determine all orbits under this action.
Answer;
The orbit for x is defined as Gx := {gx : g ∈ G}. All calculations are done modulo 18 and yield:
G1 = {1, 5, 7, 11, 13, 17} G2 = {2, 10, 14, 4, 8, 16} G3 = {3, 15} G6 = {6, 12} G9 = {9} G0 = {0}

I get that {1, 5, 7, 11, 13, 17} is the inveritble set in Z18. But according to the definition of an orbit but shouldn't gx be part of the set U(Z18)? Why instead teh module.

it doesn't explicitly say in the definition of orbits that gx needs to be in G either

[marsh fiber]

orbits of subsets of groups don't have to be part of the subset

the oribt of 1 is everything you get multiplying U(Z18) by 1 which does jsut get you back U(Z18) yeah, but viewed this time as part of the set being acted on rather than the group

[inner steppe]

Why does need to be in module 18? Does an orbit gx where g in G need to be in G as well?

I guess the set of elements we can reach by applying g doesn't need to be a subset of X. But I really don't get what you mean here btw "orbits of subsets of groups don't have to be part of the subset"

[marsh fiber]

No, the orbit is a part of the set the group acts on. So we think of the "group" Z18 and the "set" Z18 as different things

we could have e.g. the set of points on a circle or whatever, 18th roots of unity

[inner steppe]

but in that case {2, 10, 14, 4, 8, 16} isn't really a part of U(Z18)

[marsh fiber]

U(Z18) is the group, Z18 is the set. The invertible elements act on the set, and that oribt is also aprt of the set

[inner steppe]

Omg

It totally makes sense now. I thought the other way around for some reasons lol

that X = U(Z18) and G = Z18

[marsh fiber]

oh yeah that wouldn;t work

that can;t be a group since 5 or whatever isn;t invertible

wait 5 is totally invertible haha i gave an awful example. but you know what i mean

[inner steppe]

Thanks for the answer, I get what you mean. Gotta rethink my solution now haha

[marsh fiber]

sorry is uz18 the noninvertible elements under multiplication

otherwise that just isn't true

[inner steppe]

no worries, in the answers as you can see they only did G1, G2, G3, G6, G9, G0

but aren't supposed to do it for all g's in G basically [0, 18]?

[marsh fiber]

I suspect you get some duplicates

probably G4=G2, etc

[inner steppe]

but is there like a proof or as it obvious to see or do I need to try all and see? But in any case don't we have to specify them even if they are duplicates

[marsh fiber]

there's some theorem about like order of element times size of orbit=size of group or something but i dont really remember my group theory like at all

oh, orbit-stabiliser is what i was thinking of. probably

[icy totem]

We know that a finite group G (whose order is multiple of a prime p) has a number of elements of order p that is multiple of p-1 and congruent to -1 mod p.

Do you know other theorems about the number of elements of order p?

[inner steppe]

We know that g in the group G has order 680 and we are looking for the order of g^72.

(g^72)^n = g^72n = 1, we know that g^680=1 which means that k=680 is the lowest number where g^k=1. So I couldn't really continue because I still cannot understand why the solution is not 72n=680 <=> n = 680/ 72 but obviously this is wrong because the fraction is not an integer.

[pine shuttle]

I’m having a hard time with Hungerford 3.1.7 - prove that any ring R with more than one element such that for each nonzero a in R there is a unique b in R such that aba=a prove:
(a) R has no zero divisors
(b) bab=b
(c) R has an identity
(d) R is a Division Ring
I’m stuck on part (c), showing that there is an identity element

[bitter rover]

You have left and right cancellation if there are no zero divisors

Like, if R has an identity then (2) tells you what it has to be

Confirm that it is

[pine shuttle]

Don’t you need to first establish the existence of multiplicative inverses?
Sorry I’m a little rusty on these things

[bitter rover]

No, that's sufficient but not necessary. The integers have cancellation, too, for example, and no non-trivial inverses

If a ring has no zero divisors and ab=ac then a(b-c) = 0 implies either a is 0 or b=c

[pine shuttle]

I appreciate it, that was the thing I was missing

[bitter rover]

It's why it's called an integral domain, after all!

[sharp sonnet]

how did you define nilradical?

also what does the \approx mean

ok, i have no idea then

[kind temple]

i would guess this means the localization of R at a

well, like, {1,a,a^2,...}

but idk, just a guess

[restive birch]

okay i came across this question:
Prove that if H is a normal subgroup of G of prime index p then for any subgroup K ≤ G, either K≤H or G = HK and |K : K cap H| = p.

it feels like its wanting me to use the second isomorphism thm which was just proved in the book, but that yields |H : K cap H| = p, which isnt the result I want. it might be a typo but i cant produce a counterexample

[rocky cloak]

The second isomorphism theorem tells you that
HK/H is isomorphic to K/(K cap H)

[restive birch]

ohh i read that wrong and got K and H in the second part swapped

thanks!

[inner steppe]

I sort of did in a weird way and assumed that since 72 doesn't divide 680, then there should be a number 680k=72n. In order for g^72n=1 then we need it to be divisible by 680k because (g^680)^k. But I don't really get which theorem they are refering to by "well known theorem", I can't really find it. But another question I have is, isn't it possible that there can be another number k who is not divisible by 680 but g^k=1?

[rocky cloak]

Imagine k was not divisible by 680.

Then we can find q and r such that k = 680q + r with 0<r<680.

Then 1 = g^k = (g^680)^q * g^r = g^r

But since r<680 that contradicts 680 being the order of g.

[inner steppe]

Thanks, it makes sense, but which theorem were they talking about?

[rocky cloak]

That's the theorem, the one I just described

[inner steppe]

Does it have like a name you now X's theorem where X is the name of the founder?

[rocky cloak]

Not that I'm aware of

[regal zodiac]

Isn't this wrong without the assumption of f being irreducible over Q?

For instance , If we take f(x) = (x+1)^3(x+1)^2 in Q[x], then the splitting field is just Q(i) and Gal(Q(i)/Q) is cleaely not S5

[rocky cloak]

Yeah, you need f irreducible

[regal zodiac]

Thanks

[inner steppe]

In the solution they suspected that x^k where k=1,2,3,13,26 because we have (3^3)-1=26 elements and by langrange order of subgroup need to divide a group.

My question is how does x^k=1 guarantee that it is the primitive element since the definition of primitive element is an element who can generate the entire multiplicative group except 0.

I don't really see the correlation.

[tardy hedge]

Q4 is just applying Q1, and i) and ii) in q2 i think right

Basically it follows pretty straightforwardly from those other questions

Do i have any intuition for it tho? No. Lol

[dull ginkgo]

It’s kinda just induction and “knocking down” the higher coefficients

Because multiplication of polynomials is nice for the leading coefficient, then you can work down

[tardy hedge]

Thanks. Yeah

[dull ginkgo]

This idea works for both the zero divisor (extending the torsion case) and the unit one

Because in both cases

We are knocking down the degrees. We are stating that the product of two polynomials is a constant

[dull ginkgo]

Maybe try considering what happens when we say f(x)g(x) has degree less than both f and g? What happens to the first few leading coefficients?

You can also write f(x) = f_1(x) + f_2(x)x^k where k is the degree of the product :3

The ultimate conclusion btw is that all the coefficients for x^n greater than k are zero divisors

Which is why the inequality deg(fg) <= deg(f) + deg(g) is an equality for integral rings

[tardy hedge]

Not even really sure if i understand the statement for Q8

Yea i dont know

[chilly ocean]

Take the set of all prime ideals of A with inclusion order, then show that it has minimal element

It is kind of a poset

[tardy hedge]

Ok right because not every two prime ideals can be ordered like that

Not every pair is comparable with inclusion

[chilly ocean]

Yes therefore we are taking the inclusion

[tardy hedge]

Therefore?

[chilly ocean]

Ignore

[tardy hedge]

Ok thanks. We use zorns lemma here i suppose …

Maybe not idk

Ill just try it

[chilly ocean]

It is Atiyah, right?

[tardy hedge]

Well isnt just intersection gives u minimal elements

Yeah

[chilly ocean]

Is the intersection of prime ideals prime ideal ?

No, right

[tardy hedge]

Oh, i didnt think of that

I need to think of why no

[chilly ocean]

Okay think

You can easily construct a counterexample

[tardy hedge]

Yeah in Z intersect (3) and (7) gives u (0) but (0) not prime ideal

[dull ginkgo]

We are considering things element-wise, but you can probably pass through / invert a relation on their prime ideals containers

since ab in P if a in P or b in P is made two way by primality

[tardy hedge]

Im not sure what u mean tbh

I still gotta read ur polynomial comements they seemed insightful

Btw happy canadian thanksgiving. Lol

[dull ginkgo]

Trying to think of a way to describe it intuitively vs symbolically

[chilly ocean]

No (3) intersect (7) is not (0), it is (21)

And (0) is prime ideal in Z

[tardy hedge]

Oops

[unkempt stream]

don't take anything I say as gospel, most of these exercises I haven't done before and I'm just frankensteining ideas together intuitively til they make sense

[tardy hedge]

Lol i truly didnt think that thru

[unkempt stream]

minimal elements in respect to inclusion

that's a really fucky way of saying it

really it's that there are minimal prime ideals in respect to inclusion

you can use zorn's lemma on the "opposite order"

[tardy hedge]

Duh 🙄

[unkempt stream]

where A <= B iff B is a subset of A

[chilly ocean]

Yes I think that works

[chilly ocean]

Any problem ?

[unkempt stream]

and then it's chain complete, because unions are suprema for inclusion, intersections are infima (which become suprema in the opposite order), and intersection of prime ideals is prime :3

[tardy hedge]

Wdym

[unkempt stream]

Pretty good intro to the Zariski topology too btw

[chilly ocean]

I mean you react Duh, so I don't get it

[tardy hedge]

That was to myself lol

Cuz i made a silly mistake

[tardy hedge]

Ya im gonna work on those ones soon

[unkempt stream]

me avoiding symbolics like the plague

[chilly ocean]

7, it is interesting I think we assumed A is a commutative ring with identity, right?

[unkempt stream]

Closed sets are like the "spectrum" of ideals, sets of prime ideals containing an ideal. the whole ab in P <=> a in P or b in P thing gives finite unions of these "spectra" also are spectra (take the product of the ideals), while the arbitrary intersection thing is immediate

[tardy hedge]

Yea

[chilly ocean]

Now can you show in the Boolean ring every prime ideal is the maximal ideal

[tardy hedge]

True

[tardy hedge]

Sounds cool

[chilly ocean]

That exercise is generalization

[tardy hedge]

Yeah !

[unkempt stream]

I'm not too aware of the actual application of it being a topology is (or any given set theoretic structure, such as monotone classes, sigma algebras, etc.) but I guess you have an excuse to use topology jargon lmao

[unkempt stream]

tbf the prime part is redundant, you can also just consider ideals in general. Mainly it's due to intersections preserving primality.

And that's pretty easy since the auxillary inclusion is a "forall P_i for i indexing"

@tardy hedge btw is r(a) the radical of the principal ideal (a)

[chilly ocean]

No they denote the ideal by a, it is not an principal ideal generated by (a)

[unkempt stream]

???

(a) is the principal ideal

[chilly ocean]

No

a is an ideal

[dull ginkgo]

OH

throwing off my chi using lower case

[chilly ocean]

I = r(I) if and only if I is the intersection of prime ideals

[dull ginkgo]

I can think of a god awful localization proof assuming Boolean prime ideal theorem

[marsh fiber]

is there in general a good way to compute radicals of ideals?

[dull ginkgo]

Wdym by compute

Like given generators?

[marsh fiber]

write as generators, given generators of original ideal

yeah

the specific thing I'm looking for is r(y-xy, x^2-xy), which I already found, but I was wondering how to it for other ideals

[dull ginkgo]

Hmmm I’m not too sure, we’d be finding (f(x))^n = formal sum of generators

So you’d need the formal sum of generators to be a power in the first place

I haven’t done this before, I’ll think about it

[mighty kiln]

Even if the given generators have no radicals, the radical can have new elements

Say (x²+y², x²-y²) ⊂ k[x,y]

So I doubt there's a "nice" algorithm for that

[dull ginkgo]

oh yeah, 2x^2 and 2y^2 would be in the ideal, so if k is a non-char 2 field, then so is x^2 and y^2. Thus x and y are in the ideal, so the radical is the whole ring

Wait no

1 isn’t there

It’s just (x,y)

[dull ginkgo]

Honestly this is probably some grobner basis evilness

So play it by ear/intuition for easy problems

[crystal vale]

Any hint for part 33 a, ?

[dull ginkgo]

I will ask what you think so far

You can show every ideal is contained in some M_c :)

[crystal vale]

Tbh I have no idea

[dull ginkgo]

Would you like a hint?

A pretty strong one?

[unkempt stream]

The map sending f(x) to f(c) for a given c is a ring homomorphism into the field R

[crystal vale]

I think I need to work on this

[crystal vale]

Yes evaluation mapping, right ?

[unkempt stream]

If M is maximal, then C([0,1])/M is a field...

[crystal vale]

Yes

[unkempt stream]

I have class in a few mins, i'll come back with both ways

As one way...

Units of C([0,1]) cannot have a 0, and proper ideals cannot contain units

[tardy hedge]

When we are talking about a chain (in reference to zorns lemma stuff), we can compare each element pairwise, but does that mean we can order the elements completely like x < y < z …

Yes

[stone vault]

Yes, a chain in a poset forms a totally ordered set by transitivity

[crystal vale]

Yes

[crystal vale]

So every element in proper ideal I, has 0 at some point

[unkempt stream]

Assume J is an ideal of C([0,1]) such that for each c in [0,1], there is an f in J such that f(c) =/= 0

[night tartan]

^ is probably the best way to approach the problem

we know that if our ideal contains a unit, then it's the entirety of R

what is a unit in R here? well it's just a function that is never equal to 0

intuitively, i take some function (which might have 0s at say a,b,c)

find a function that isn't equal to 0 at those points

and add up the squares of those functions

now that doesn't quite work but u can adapt that into a solution

[crystal vale]

Yes someone suggested that use the compactness of [0,1]

[dull ginkgo]

PARTITION OF UNITY JUMPSCARE

[crystal vale]

Now I got it thanks @dull ginkgo @night tartan

[dull ginkgo]

Nice!

[night tartan]

nw

[elfin wraith]

Can I check that im not being dumb, because either im missing some special case or my lecturer is doing something weird,

For my homework im asked how many elements are in $\mathbb{Z}_4[x]/\langle 2x^2 + x\rangle$ and obviously the usual method would just be find a monic representative of the ideal and then division algo to justify it just being polynomials of degree 1 if this was over say like $\mathbb{F}_3$, but this quotient ring isnt even an ID never mind a ED so how on earth can we use the division argument? She has done this calculation for $\mathbb{Z}/\langle x^3+1\rangle$ in the lecture notes, but since neither of these ideals are prime we dont have an ID so im not super happy about using the division argument

[cloud walrus]

Nope

[elfin wraith]

is there some like special case that im missing where monic polynomials over $\mathbb{Z}_4$ dont have zero divisors in such a way that prevents us using the algorithm? I just dont see how we can always have $f(x) = (x^3+1)q(x)+r(x)$ when we have zero divisors to contend with

[cloud walrus]

Nope

[rocky cloak]

So it is still possible to divide by monic polynomials, but I'm not sure that will help you so much here.

A hint could be that if y if nilpotent, then 1+y is a unit.

[elfin wraith]

Im sure there is some other way to go about it, and ill definitely take that hint as something to think about and see what I can do, but this is the calculation shes provided in the notes and im just not sure that its properly justified, (lemmas 1 and 2 are just the division algorithm)

Like I dont think im wrong in saying that x^3+1 isnt prime so this isnt an ID, so like why can we apply the algorithm like that? Can we even do so?

[rocky cloak]

The division algorithm has nothing to do with being prime.

Basically you can think of it like if f has degree n, so
f(x) = fn x^n + ...
then
f(x) - fn x^n-3 (x^3 + 1)
has lower degree. Then just iterate that until you get a polynomial of degree less than 3

This uses that x^3 + 1 is monic though (or at least that the leading coefficient is a unit)

Which is not true for 2x^2 + x

[elfin wraith]

I was just mentioning primality for R/I being an ID, i thought we could only apply the algorithm in euclidean domains which are necessarily integral domains no? Im sure im just missing something here, this lecturer is a very accomplished algebraist so I cant see her making a basic mistake like that twice, but im not seeing what it is

[rocky cloak]

The division algorithm works for monic polynomials over any commutative ring

In a euclidean domain you can divide arbitrary elements, but you can always divide by monic polynomials over any ring

So basically if F is a field F[x] is euclidean because all polynomials are monic (up to unit)

[elfin wraith]

Im still not quite seeing why we dont run into zero divisor issues with f = qg + r though, is this related to the fact that units cant be zero divisors or something?

so 1 always being a unit is why monic polynomials are fine?

[rocky cloak]

Yeah, that's right. A monic polynomial can never be a zero divisor.

I mean if
f = fn x^n + ...
for nonzero fn, and
g = x^m + ...
Then
fg = fn x^n+m + ...
which is nonzero

[elfin wraith]

Yeah ok i was just being kinda dumb there, its quite obvious when you write it out like that

[rocky cloak]

Everything is obvious when you know it 😉

[elfin wraith]

Youre not wrong haha, just got a bit lost there and couldnt see the wood for the trees I think

[hidden wind]

is there a better way to do (2) here than explicitly write out the elements because that's what i'm trying currently and it's so ugly

also i completely forgot i was cooking so now i have pasta with garlic and chilli burnt black mmmm

[coral steeple]

Can I get a small hint on proving that the commutator subgroup is normal? I've been trying to show that the conjugate of any commutator x^-1y^-1xy is a product of commutators but I'm completely stuck and have no other ideas.

I also tried showing that congruence mod (the commutator subgroup) is really a congruence on the underlying group. No dice.

[rocky cloak]

I'm not sure I see how you're getting something superugly. Like isn't it just

p/s * m/t = pm / st
pm / s = p/s * m/1

[hidden wind]

ah yeah i'm def overcomplicating, considering for ideals I in A a general element of IM as a finite sum of products in I by M

ok so i won't need such a double sum excellent excellent actually that makes perfect sense i'm just way too tired 😭 have been rushing to get through this assignment so that i can ditch studies for a week and run off to rome

thanks jagr

[rocky cloak]

A hint could be that conjugation is a group homomorphism, and a commutator is just a product of stuff

[coral steeple]

Of course I should have thought of looking at conjugation as a homomorphism. I'll think about that, thanks!

[golden turtle]

I am trying to see how the proof of example (1) goes, and it's basically laid out in the explanation already

but all I see is that Psi is a left-inverse of ell

I don't see why ell is surjective necessarily

i can send an image incase my question doesn't make enough sense

[rocky cloak]

If you know that elements in the tensor product looks like you can just directly show it's surjective

[golden turtle]

they're a sum of constants*basic tensors

so then I can just apply the tensor product operation properties to show that it is surjective probably

actually do we even need constants

any element in the tensor product is just a finite sum of decomposable tensors, I think

[rustic crown]

you can say more

if M is a cyclic module generated by m, then M⊗_R N has elements that look like m⊗n, where n in N varies.

i.e. don't need to look at finite sums, just elementary tensors are enough

[golden turtle]

why do we think that M is cyclic generated by m? I might be missing something

[rustic crown]

i gave a tiny bit general result

in our case, M = R and m = 1

so all elements in R⊗_R N actually look like 1⊗n

[kind temple]

this is because for a scalar s,
s(u (x) v) = (su) (x) v = u (x) (sv)

[golden turtle]

yeah I think so its surjective because say we want to map onto r otimes n then we map ell(rn)=1 otimes rn = r (1 otimes n)= r otimes n

but this is if every element is just an elementary tensor instead of considering finite sums

[kind temple]

linearity

[golden turtle]

is ell linear?

ell(r1n1+r2n2)=1 otimes r1n1+r2n2 = 1 otimes r1n1 + 1 otimes r2n2 = r1 otimes n1 + r2 otimes n2

[kind temple]

the tensor is

[golden turtle]

does this work to generalizing to finite sums

[kind temple]

in one coordinate the tensor is linear, and you can write each of the summands as 1 (x) m

so the sum is also of the form 1 (x) m

[golden turtle]

yeah I think
1 otimes (r1n1+....+rn n_m) = 1 otimes (r1n1) + ... + 1 otimes (r_mn_m) = r1 otimes n1 + ... + r_m otimes n_m

[restive birch]

okay for some reason this is stumping me, can i get a small hint?

show that if G₁, G₂ are subgroups of G with G₁ a normal subgroup of G₂ and G₂/G₁ abelian, and N is any subgroup of G, then (G₂ cap N)/(G₁ cap N) is abelian

@ me if you responds please

[dull ginkgo]

TENSOR PRODUCT JUMPSCARE

[dull ginkgo]

Consider the center as a subgroup lol

[restive birch]

hmm i dont see how that helps, let me look at it for a minute

oh wait hang on

[rustic crown]

doesn't (G2 n N)/(G1 n N) embed into G2/G1?

[restive birch]

does it? i looked at that for a minute

[rustic crown]

the kernel of G2 n N --> G2 --> G2/G1 is G1 n N

[restive birch]

hang on

okay yea its elements in G₁ and N so yea, i see it

thanks

[rustic crown]

[dull ginkgo]

Oh that also works too lol

[restive birch]

would you mind outlining where your hint was trying to point me?

[unkempt stream]

it ultimately was a roundabout way of what det said so dw about it

[restive birch]

man how tf did we get from giving a set multiplication to whatevers going on here

math moment

[thorn jay]

Mfw symmetry is structurally a mess

[olive parrot]

in GAP is there an O(1) way to write Difference(List(c1, x -> x^-1), AsList(c2)) = []; where c1 and c2 are conjugacy classes of a subgroup of a symmetric group and have the same cycle structure? Although i dont know how to prove this expression's correctness Representative(c1)^-1 in c2 and Representative(c2)^-1 in c1; is a bit faster but is still O(n).

To simplify, given two conjugacy classes of a subgroup of a symmetric group that have the same cycle structure and consequently the same number of elements, is there a quick way to determine if the elements in the first conjugacy class are the inverses of the elements in the second class? I highly doubt there isn't a fast group theoretic approach for this problem but I'm not sure what to do

[icy totem]

Sorry I think I didnt understand, how can you have two different conjugacy classes both having elements with same cycle decomposition?

[inner steppe]

Can someone please explain this to me. Are they claiming that every finite group G whose order is greater than 2 is cyclic?

[restive birch]

no

theyre saying if G is a finite group of order >= 2

then any one of the statements being true

means that the other 2 must also be true

so theyre either all true or all false

[icy totem]

You probably read "the following statements are equivalent" as "the following statements are true". Note that they dont mean the same thing

[olive parrot]

that's what GAP is telling me

gap> Filtered(second_cycle_group_classes, k -> CycleStructurePerm(Representative(k)) = [ 1, 1,,, 1,, 1 ]);
[ (3,27,33)(7,18)(12,39,20,15,37,47,13,44)(14,22,46,41,40,16)^G, (3,33,27)(7,18)(12,39,20,15,37,47,13,44)(14,22,40,16,46,41)^G ]```

to be specific the use case is the rubik's cube

gap> U := ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19);;
gap> L := ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35);;
gap> F := (17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13)( 8,30,41,11);;
gap> R := (25,27,32,30)(26,29,31,28)( 3,38,43,19)( 5,36,45,21)( 8,33,48,24);;
gap> B := (33,35,40,38)(34,37,39,36)( 3, 9,46,32)( 2,12,47,29)( 1,14,48,27);;
gap> D := (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39)(16,24,32,40);;
gap> cube := Group(U, L, F, R, B, D);;
gap> r := Random(cube);;
gap> IsSubgroup(SymmetricGroup(48), cube);
true
gap> IsSymmetricGroup(cube);
false
gap> ConjugacyClass(cube, r) = ConjugacyClass(cube, r^-1);
false
gap> CycleStructurePerm(r) = CycleStructurePerm(r^-1);
true```

[icy totem]

I dont understand this type of code sorry, I can just give you group theory answers

[olive parrot]

hm i think this is only true with symmetric groups

[inner steppe]

Aha so they just wanted to say that if one is true then the rest are true too

[olive parrot]

Correct me if im wrong

[icy totem]

Yea ofc, because elements of an arbitrary group dont have a cycle decomposition (beside maybe the cycle decomposition of them seen as permutation of G).

[icy totem]

Exactly!

[olive parrot]

It has a cycle decomposition but isn’t symmetric idk the word for it

Two different conjugacy classes in this group have the same cycle structure

[icy totem]

Wait are you conjugating only by elements of your subgroup?

[olive parrot]

yeah

[icy totem]

Ahh ok

[olive parrot]

I mentioned it was a subgroup of a symmetric group in case it was relevant

Can you quickly determine if all elements in one conjugacy class are inverses of all the elements in another

[icy totem]

Im thinking about it but I dont know any statement that deals with this

Even tho it doesnt seem diffficult

[olive parrot]

Difference(List(c1, x -> x^-1), AsList(c2)) = []; works

But is super slow and bad

It should be pretty clear what GAP code is doing

[icy totem]

By any chance is your subgroup exactly the alternating group? Because then there is the answer

[olive parrot]

My subgroup specifically is the Rubik’s cube group so no I don’t think so

You’re given that they have the same cycle structure in case that helps

[icy totem]

The rubik cube group is a subgroup of the alternating group tho

[olive parrot]

It is?

Ok well even then it’s still not alternating itself

[icy totem]

Because every element of the rubik group is the product of 2 4-cycles no?

[olive parrot]

^ here is the exact definition and six generators

[icy totem]

But still I think MAYBE I have the asnwer, I just have to check one moment a pdf ok?

[olive parrot]

Oki

The use case is in searching elements of conjugacy classes of a specific order and im noticing that these inverse conjugacy classes duplicate my search’s result with an inverse permutation

I want some way to determine if a conjugacy class only contains inversed elements to avoid redundant computation

[icy totem]

Ok last question, such conjugacy classes have the cycle decomposition of a rubik's cube move e.g. your conjugacy classes are basically sets of moves of the cube?

[olive parrot]

Ok well my actual use case is for a subgroup of the Rubik’s cube group not the entire group I omitted that to simplify the problem statement

But idt it should change anything

Hopefully

I should have said that earlier sorry xD

[icy totem]

If yes, then Im 99% sure that the elements that conjugate a certain element g with its inverse are Just product of 2 disjoint transpositions, that is they are basically moves that swap 2 pairs of boxes of the cube. I hope this helps even tho its not a complete solution.

So you just have t check that the product of these "special" permutations with an element ofyour conjugacy class does/doesnt give an element of the other class, I dont know if this is computationally better or not

[olive parrot]

I haven’t formally studied math so I’m not sure exactly what that means xD I’ll do research and try to understand it though

Another idea I had was to simply just check if the inverse of one element of a class is in the other

It seems to work with a few tests something tells me if just one element is jnversed then all of them are

But it’s an ok solution because it’s still O(n) and idk if it’s correct

[icy totem]

Yea that sounds good

Also sorry if i confused u (its 1 am for me lol) im tired as hell

[olive parrot]

Nw

[icy totem]

Yea thahlt should be right

[olive parrot]

How do you know

[icy totem]

That exactly was an exercise I did a while ago in a paper, I dont know of i remember corrrectly tho.

Let me check

No ok I didnt remember well

Forget what I said lol

[olive parrot]

😿

[icy totem]

I definitely have to go to bed

Lol

[tardy hedge]

poop

[dull ginkgo]

[vagrant zinc]

Why do you guys think Gallian's book is a bad book?

Quick question, is that I was looking at it and I saw that the Gallian book seemed relatively bad, as for the exercise section, I do not know if my thinking is correct, but the examples are very good, I do not know, whether to consider this book to study their examples.

[tardy hedge]

ikr

[sonic coral]

i dislike the notation for certain things but that doesn’t say anything on the content in the book

[vagrant zinc]

I also don't like the note that is also something negative I guess, but it fulfills its function, although the list of exercises is not that I like it either.

[coral steeple]

@rocky cloak I proved the commutator subgroup H of G is normal: if m is in H, and g in G, then g^-1mg=m(m^-1g^-1mg) is in H. What was the proof you were thinking of when you gave me your hint? I couldn't work it out.

[rapid junco]

i am trying to prove Lemma 26 but I am stuck, the one direction i have down pat, but if K/H is normal in G/H then what trick do I use to show that K is normal in G?

this is what some old notes say, but Arguing similar to above I don't quite see the solution.

[rapid junco]

got it

probably not the standard way but oh well

[restive birch]

okay this problem states:
Let $1 = N_0 \unlhd N_1 \unlhd N_2 ... \unlhd N_r = G$ and $1 = M_0 \unlhd M_1 \unlhd M_2 = G$ be composition series. Show that $r = 2$.

[cloud walrus]

jan Mo Lakeli Tu Katuwe Pa

[restive birch]

it says to use the second isomorphism thm

i have no idea where to start

just a nudge would be nice

[serene dune]

,w jordan holder theorem

[cloud walrus]

Results provided by WolframAlpha

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

Requested by bcegkmqsw

[restive birch]

i know

this exercise wants me to prove that

its the first step

[crystal vale]

I don't want a hint, I am just not sure what exactly the author asked, give an example of a non-trivial commutative ring in which every element is of square 0, so it wants a non-trivial ring means there exists elements x,y such that xy≠0 and r^2 = 0 for all r in R.

Right? Don't give a hint

Yes the ring has no unity because then it will be zero ring

[kind temple]

right

[rocky cloak]

If we write [x, y] for the commutator, then
g[x, y]g^- = [gxg^-, gyg^-]

[crystal vale]

I don't see how 1 helps me to prove 2

[warm ember]

boolean rings are commutative

use induction

boolean rings have char 2 also

for the inductive step you use 1

[crystal vale]

Yes

[crystal vale]

So first I have to show its cardinality 2^k

Oh it is something every non-zero element has additive order 2 so it must be isomorphic to (Z/2Z)^k

So it has cardinality 2^k

Right?

[night tartan]

yeah roughly speaking

from the previous part, we take any e in R not equal to 0 or 1

then we induct based on the fact that R is isomorphic to Re x R(1-e)

[night tartan]

(and then stare hard at what Re and R(1-e) are, and realise that they're both boolean rings too)

[trail cave]

In 17b, how do I show that for all h in H, h^-1 is in H?

this is the last correction I need to make for my homework but I'm struggling

[mighty kiln]

Any element in H has finite order

[trail cave]

I know, but I don't know where to go from there. Here's my work

I was looking back at it and I don't really understand what I was trying to argue

[mighty kiln]

Any element h of finite order has a power that is h^-1

[crystal vale]

Because then Re ≠ R and R(1-e) ≠ R so we can induct on their cardinality

[crystal vale]

Yes they are Boolean ring

Thank you ❤️ Kevin and LY

[night tartan]

nw!

[serene dune]

i would try to write like this

[trail cave]

Why does there exist an inverse of h in H though?

H isn't a group

Or, not assumed to be

[serene dune]

so H to H is a sort of bijection(not so sure here but i am hoping) just by the closure of the set

[trail cave]

honestly I understand why this makes sense but I think this technique is too elegant to put in my hw lol

[serene dune]

no, lol, this is primitive

u surely can use this type of argument

feel free to prove the injection and subjection part

[trail cave]

I just mean rather that our lecture/textbook hasn't used this kind of technique with functions at all so it's like pulling it out of a hat; I see why it makes sense, though, so I suppose there's no harm in writing it out. Thank you

[tardy hedge]

Why is if ab subset p a prime ideal, then a subset p or b subset p? a and b are ideals

This seems obvious what am i not seeing

[rocky cloak]

Assume b is not a subset of p. Then there is x in b, but not in p.

Now for every y in a think about the element yx

[bitter rover]

Write down all the definitions involved. Put them next to each other on a piece of paper, right in front of you, like a mathematical mise en place.

[tardy hedge]

Thanks yea so the fact that p is prime isnt relevant i guess

[rocky cloak]

....?

It is kinda the only thing that's relevant

[tardy hedge]

I thought at first just the ideal structure was relevant

K one sec

It forces y to be in p

[rocky cloak]

a and b can be arbitrary sets though

[tardy hedge]

Ok, also there is a lemma regarding intersections of ideals and primes i should review that

[inner steppe]

Is this really an only one way implication?

Then how come they can take the modules of the terms separately? In other words how come they use Fermat's theorem term by term?

[south patrol]

I mean if a = b mod n and c = d mod n then a + c = b + d mod n

i.e. addition is well-defined mod n

[bitter rover]

The implication says that you're free to replace a number with any other number it's congruent to (when adding or multiplying), and congruence will still hold.

The opposite direction isn't even true with equality.

Like, if a + c = b + d it's not the case that a = b and c = d.

[inner steppe]

Sounds logical lol, I was just a little confused why they took the module for the terms separately

I feel like I understand what they did but not completely

[bitter rover]

Replace congruence with equality.

If a = b and c = d then a + c = b + d

Hopefully that's not suspect.

But if a = b then it gives you, e.g.,

a + c = b + c

If a = b and c = c then a + c = b + c

(Just to illustrate how that's an application of the statement involving both variables at once)

By doing "both at once" you're proving simultaneously that a + c = b + c and c + a = c + b

Which...for a commutative operation like + that's not so interesting.

[inner steppe]

I was talking about this post

[inner steppe]

I mean in the polynomial equation. I am not seeing the relation sorry.

[bitter rover]

The statement "If a = b and c = d then a + c = b + d" licenses the fact that you can "add a number to both sides", which you hopefully find uncontroversial.

[rocky cloak]

They're using the above picture in exactly the way the direction of the arrow indicates. If that's your confusion...

[bitter rover]

Here we want "If a ~ b and c ~ d then a + c ~ b + d, which licenses "adding a number to both sides" with respect to ~.

So that we can "pretend" that ~ behaves like = with respect to addition, at least.

[inner steppe]

if C = 14x^39 mod 19 then we can replace with C in the equation. Isn't that what they are doing?

[bitter rover]

Take x^57

Well:

x^57 = (x^19)*(x^19)*(x^19)

But x^19 ≡ x for all x (mod 19).

So if x^19 ≡ x then (x^19)*(x^19)*(x^19) ≡ x*(x^19)*(x^19). Repeat that twice, you get

(x^19)*(x^19)*(x^19) ≡ x*x*x
                     ≡ x^3

[bitter rover]

Yes. Maybe I misunderstood what you were confused about. It sounded like you didn't get why proving that + and * are well-defined WRT modular congruence licenses doing things term-by-term.

[inner steppe]

Thanks for the help anyways, it means a lot but so when given an equation or whatever if I can see that a term mod m can be simplified then I can do that and put it in the equation right? I just need to double check so I don't make mistakes like this.

[bitter rover]

Godspeed!

[tardy hedge]

For 16 …

Draw pictures? It looks like Spec(Z) has its closed set be finite

[rocky cloak]

I guess, find the poset of primes, and get a little creative.

[tardy hedge]

Open sets of Spec(Z) consists of all prime ideals of Z with finitely many removed

[inner steppe]

How do I find the roots here?

[tardy hedge]

Im not sure what u mean by find the poset of primes

[rocky cloak]

Like which prime ideals are contained in each other

[inner steppe]

Isn't it like irreducible in F, but in the answer they claim that g(2a) = 0 but I get 2a

But still I wanna know how the to get to that conclusion

[rocky cloak]

You could take a general element, plug it in and see what it needs to be a root.

[inner steppe]

are you telling me to try out all elements in F and see if they give 0?

[rocky cloak]

Like calculate g(b alpha + c) = 0 and solve for b and c

[inner steppe]

I am sorry, how is that gonna make the problem any easier?

[rocky cloak]

Because you'll be able to solve it...
I'm not sure what you mean by easier, easier than what?

[inner steppe]

I mean I don't understand what you are telling to do at all. I might be very slow but what is supposed to be b and c ?

[rocky cloak]

Like
g(b alpha + c) =
(b alpha + c)^2 + b alpha + c + 2 =
b^2 alpha^2 + 2bc alpha + c^2 + b alpha + c + 2 =
b^2 alpha + b^2 + 2bc alpha + c^2 + b alpha + c + 2

so b^2 + c^2 + c + 2 = 0
and
b^2 + 2bc + b = 0

So from the second equation either b=0 or b+2c+1=0.

Putting b=0 in the first has no solution. So then b = 2+c, which you can plug in to get c=0

You could also just brute force the 9 possible options I guess

[tardy hedge]

I thought prime ideals in Z arent contained in each other. Besides (0)

[rocky cloak]

Yup, so for Z it's pretty easy

[tardy hedge]

Oh i see

Yea i guess somehow (0) is near everything since its closure is everything

Or smth like that

[rocky cloak]

Like the prime ideals correspond to irreducible closed sets. So drawing them as some geometric object containing other closed sets would make sense

[inner steppe]

so b^2 + c^2 + c + 2 = 0
and
b^2 + 2bc + b = 0

Where dod you get the second equation, perhaps from f(x)?
How did you simplify the first equation and removed any term that was multiplied with alpha?

[rocky cloak]

They're the coefficients of alpha and 1 in the equation right above.

[inner steppe]

Continuation of the question,
How come you are allowed to do that
so b^2 + c^2 + c + 2 = 0
and
b^2 + 2bc + b = 0?

How come you could create two polynomials = 0 from one polynomial coefficient

[rocky cloak]

We know that
d alpha + e = 0 means d=0 and e=0

Otherwise alpha = -e/d, but alpha is not in Z3

[inner steppe]

Sorry but how do we know that d alpha + e = 0 . Which also got me thinking why did you even pick "b alpha + c". Was it random?

[rocky cloak]

We were looking for roots of g inside F.

So I took an element in F, i.e. b alpha + c and put it into the equation g(x) = 0

[inner steppe]

yes

so you could also have just picked b tho in F?

right but might be stupid question but why b alpha + c.

[rocky cloak]

Because that's what the elements of F look like.

They are exactly
0, 1, 2, alpha, alpha+1, alpha+2, 2alpha, 2alpha+1, 2alpha+2

I.e. b alpha+c

[tardy hedge]

So Spec(R) just consists of {(0)} ?

Thanks this is new to me so weird to think about

[inner steppe]

Yes I understand that but I was asking why did you pick " 2alpha+1" instead of 2. Probably a stupid question but I don't understand.
I also don't understand how we know that d alpha + e = 0

[tardy hedge]

Ok with the polynomial rings damn its been so long since ive reviewed ideals of polynomials

[rocky cloak]

I don't understand the question. When did I pick 2 alpha + 1 instead of 2?

[inner steppe]

b alpha + c = 2 alpha + 1 where b=2 and c=1?

Just an example

Ik we could have picked anything 0-2

oh wait

nevermind lol

but still confused about this one tho d alpha + e = 0

[rocky cloak]

So I take the equation
g(b alpha + c) = 0
We simplify the left hand side, which gives us some equations which we solve

The solution happens to be b=2, c=0

[inner steppe]

now I am following but how come you are allowed to separate
b^2 alpha + b^2 + 2bc alpha + c^2 + b alpha + c + 2
to
so b^2 + c^2 + c + 2 = 0
and
b^2 + 2bc + b = 0?

I get that the terms that contain alpha is spearated with those that don't

but it doesn't really add up for me

[rocky cloak]

Because if
d alpha + e = 0
then d=e=0

[inner steppe]

Aha okay so you did

(b² + 2bc + b)α + (b²+c+2) = 0

This has solutions only if e=d=0 in d alpha + e = 0

God that is such a smart move compared to the solutions given. Thank you so much for the help!!

[graceful dome]

So in order to prove this I just have to show the subsets of G seperated by the order of their elements partitions G and form a subgroup right

Sorry I forgot the actual question

Cause then lagrange's tells me that the order of the subgroups divides the order of G and since they partition G, the sum of the elements must equal the order of G

[rocky cloak]

I don't think there's anything that needs to form a subgroup, you just need to show that phi(d) is 0 when d doesn't divide n

And that G[d] partition G I guess. If that's not immediately clear

[graceful dome]

but if I dont show each G[d] is a subgroup, I cant apply lagranges

[rocky cloak]

But G[d] isn't a subgroup

For example the identity don't have order d (unless d=1)

[graceful dome]

right

Okay good point, so I shouldnt even be paying attention to lagranges here then

[rocky cloak]

You should use Lagrange to argue what the order of an element can be though

[graceful dome]

oh

so I assume there exists an element whose order doesnt divide o(G)

then define a cyclic subgroup of G with that element as it's generator

then lagrange's would tell me that would be impossible so phi(d) must be 0 when d doesnt divide n

then it would also show G[d] partitions G which basically ends the proof

[tardy hedge]

What would be the best way of explaining why if R has exactly one prime ideal, every element of R is either a unit or nilpotent?

I am thinking specifically of showing that if an element is not in the ideal, then it is a unit

Whats the best way of explaining that part?

[coral spindle]

Isn't k[x] a counterexample to this? (x) is its only prime ideal, but x is not nilpotent. Am I missing something here?

Oh wait are you including, like (0)?

Do we care about nonzero prime ideals?

[tardy hedge]

This is from a/m. I believe they include (0) to be a prime ideal

[coral spindle]

OK good, I see my mistake then

[dull ginkgo]

nilradical :3

And units are elements lying outside of any proper ideal

[tardy hedge]

Yea

But its one prime ideal not one proper ideal

[dull ginkgo]

Maximal ideals are prime

Ergo there is one maximal ideal (local)

So any element is either in the nilradical, the one maximal ideal here, or is a unit (outside of ALL proper ideals, all a subset of aforementioned maximal nilradical)

[tardy hedge]

are we implicitly using the fact that every ideal is contained in a maximal ideal

[dull ginkgo]

Y e s

[coral spindle]

I mean you could see it that way, but equally the ring must have a maximal ideal, which is prime, so that must be the unique maximal ideal

[dull ginkgo]

Problem is also the nilradical as defined via nilpotency needs Boolean principal ideal theorem or an equivalence to show every ideal is contained in a prime ideal in the first place

So that’s to be expected :3

[tardy hedge]

ok thank u

To show A/R field -> A has one prime ideal

I just did: R must be maximal, and we know R is intersection of all prime ideals, so there must be only one prime ideal or else R wont be a maximal ideal

intersect p < p right

[dull ginkgo]

Ye

[tardy hedge]

thx

[dull ginkgo]

Well it is p, since it’s an intersection of ONE prime ideal

[tardy hedge]

Yeah and if it was intersection of more than one than it wouldnt be a maximal ideal cause it would be contained in one of the p's

[dull ginkgo]

The most difficult part here is the intersection of prime ideals being the nilradical

[tardy hedge]

Yeah

Ive been really wanting to review that proof

[dull ginkgo]

My preferred method is localization which behaves “opposite” to quotienting in terms of what happens to ideals

That’s a good thing to look into

Since units are elements OUTSIDE of ideals, so if we “turn” some elements into units (pray to yog sathoth there’s no zero divisors) the ideals that contain any of those elements we are localizing become the whole ring

[cloud walrus]

Austin

Austin

[golden turtle]

so I don't understand what I am doing wrong

[terse yarrow]

Hi, can someone explain Galois theory to me in Fortnite terms please?

[rocky cloak]

Note, that the list of composition factors is the same, but they don't necessarily appear in the same order.

If M1 and N_r-1 don't intersect, then M1N_r-1 must be a normal subgroup bigger than M1.

[golden turtle]

so it must be M2

[rocky cloak]

Indeed, which is all of G

[golden turtle]

oh okay so going back to the first application of 2nd iso I think,
MN_(r-1)/N_(r-1) iso to M1 is the assumption

but MN_(r-1) is all of G, if M1 cap N_(r-1)=<e>

so

G/N_(r-1) iso to M1

I want to say that this implies N_(r-1) iso to M_2

[rocky cloak]

Well, you would want to show that N_r-1 is iso to M2/M1 in this case

And you'd want to use 2nd iso again

[golden turtle]

Oh because we already have M2 is M1N_r-1 so we want N_(r-1) is M2/M1

[golden turtle]

say that we show this then that makes N_(r-1) iso M1 right?

[rocky cloak]

No, if you show that N_(r-1) is isomorphic to M2/M1, then you will have shown it's isomorphic to M2/M1

[golden turtle]

do we not know anything about M2/M1 from the composition series?

[rocky cloak]

We know it's a simple group

It's one of the composition factors

[golden turtle]

well and for the proof to go through we need it to be M1 but I guess we have not shown that it is

[golden turtle]

well and it can't be M2 since if it was then M1 is the same as M0

if it is M0 then M1 is M2 which also cant happen

so since its one of the composition factors it must be M1

[rocky cloak]

Yeah, it's none of those things because it's M2/M1

[golden turtle]

[rocky cloak]

The composition factors being M1 and M2/M1

So it's the second composition factor

[golden turtle]

I might be getting confused I guess but I thought the goal was to show N_(r-1) isomorphic to M_1 to show that r=2

And so if we have N_(r-1) isomorphic to M2/M1, then we want to show that M2/M1 is isomorphic to M_1?

[rocky cloak]

If you can show that N_(r-1) is simple, then r=2

Consider G = Z/6
and the two series
0 < 2Z/6 < G
0 < 3Z/6 < G

[golden turtle]

well if we've shown it isomorphic to M2/M1 then it is simple by the properties of the compositions series

[rocky cloak]

Indeed

[golden turtle]

but why does it being simple imply r=2?

[rocky cloak]

Well N_r-2 is a normal subgroup, so N_r-2 = 1, so r-2 = 0

[elfin wraith]

Im slightly concerned that this isnt even an algebra issue and im just forgetting/not seeing something super basic from like calculus but ill ask anyway,
H(K[x_1,..,x_n];x) is the Hilbert series of a polynomial ring in n indeterminates over an algebraically closed field, I agree that this is the sum of monomials in the ring but the textbook im reading says this is also prod{i=1}^{n} 1/(1-x_i) and im not sure at all where thats coming from, it actually states that before the sum of monomials part, which at least to me, is far more clear from the definition

[rocky cloak]

1/(1-x) = 1 + x + x^2 + ...

That's indeed from calculus

[elfin wraith]

I do now feel embarrassed for even asking that, I dont even have the excuse of not doing calc for 4 years because thats the classic lauraunt series trick and I did CA last sem...

well thanks none the less!

[golden turtle]

and for this question when it says "show that r=2 and that the list of composition factors is the same", it doesn't suffice to just show r=2 I have to also show further something about the N_i isomorphic to the M_i?

[rocky cloak]

You have to show that the composition factors are the same.

This is not showing Ni isomorphic to Mi though. You seem to be confused about that

[golden turtle]

I must be, is it showing that they are actually equal and not just isomorphic?

I've shown that M2/M1 iso to N_(r-1) now

using 2nd iso

like you suggested

[rocky cloak]

The composition factors of the series
1 < M1 < M2
is M1 and M2/M1
The composition factors of
1 < N1 < N2
is N1 and N2/N1.

So N1 needs to be isomorphic to either M1 or M2/M1 and N2/N1 needs to be the other one.

[rocky cloak]

This example is somewhat useful

[restive birch]

i was just trying to do this, curious to see how you do it

i just ended up skipping it

[golden turtle]

Oh I thought the composition factors were just M1, M2, themselves

[restive birch]

no

this is for D&F 3.4.9 right

[golden turtle]

well I still did that part fine

ya a d&f exercise

[golden turtle]

so N_(r-1)/(N_(r-1) cap M_1) is iso to N_(r-1)M1/M1

[golden turtle]

and N_(r-1)M1 is iso to M2

[rocky cloak]

No the composition factors are the simple groups that appear as subquotients in the filtration

[golden turtle]

so since N_(r-1)cap M_1=<e> then you're done

[restive birch]

wait how did we get these

[golden turtle]

from the earlier discussion, if M1 cap N_(r-1) =<e> M_1N_(r-1) is normal and larger than M1

so it must be M2

[restive birch]

ah

wait sorry where did we get M_1N_(r-1) normal from

is this one of the iso thms

[golden turtle]

its the product of two normal subgroups

[restive birch]

oh, 2nd iso?

[golden turtle]

I don't think it is an isomorphism theorem

could be wrong

but just the product of normal subgroups is normal

i think easy to prove

[thorn jay]

It isn't

But it is a property called congruence-permutability

It is equivalent to the existence of the term
p(x, y, z) = xy^-1z

[golden turtle]

so for the first case we get real equality of N_(r-1)=M_1 and then it remains to show N2/N1 isomorphic to M2/M1?

[rocky cloak]

Yes, though I'm not sure I'd say it remains to show.

Like if M1=N1 and M2=N2, then M2/M1 = N2/N1, that's just how equality works

[golden turtle]

I know its dumb but I'm just getting a little confused on how specifically I showed M2=N2

[rocky cloak]

They are both defined to be equal to G

[golden turtle]

oh

right

🤦‍♂️

so it really only is the 2nd part of the 2nd case, we have M2/M1 iso N1 and need to show N2/N1 iso to M1

[inner steppe]

I don't even have a clue on how to even begin

[warm ember]

finite fields are unique

and of order prime power

[inner steppe]

You mean |F| =p^r

But I still don't understand how to apply this on the question.

[warm ember]

is 6 a prime power?

[inner steppe]

no

[warm ember]

is there a field with 6 elements?

[inner steppe]

not sure

[warm ember]

finite fields always have prime power number of elements

[inner steppe]

all finite field, like all?

[warm ember]

yes

[inner steppe]

But still how do we derive how many distinct fields exist with |F| = n

[warm ember]

finite fields are unique

[inner steppe]

meaning?

[warm ember]

there is a unique field with p^k elements

[inner steppe]

so there is no other field with the same size as another field?

if finite?

[warm ember]

up to isomorphism

yes

[inner steppe]

so there can be two fields of the same order if isomorphic?

[warm ember]

yes isomorphic fields always have same order

[elfin wraith]

To make it clear why we can only go up to isomorphism, think about F_3, the field on 3 elements and Z_3, the field of intergers mod 3

Specifically, what the elements of each actually are

[inner steppe]

I am not sure I understand what you are saying.

[elfin wraith]

Any isomorphism is a bijection remember, this is already true of bijections

[elfin wraith]

Im trying to get at why we can only say theyre the same up to isomorphism rather than saying theyre equal, its a subtle difference in fairness, but I think helps illustrate why we say unique up to isomorphism

Because you can come up with a few different contrusctions for these finite fields, and theyll look subtley different, but you can really feel that they are the same

[bitter rover]

https://people.math.osu.edu/cogdell.1/6112-Mazur-www.pdf

[rustic rapids]

For this i got x1=x2=x3 = 0 for the homogenous. x1 = 1/3, x2 = 2/3, x3= -4/3 for the latter. For (a) I believe the solns are the same. But for (b) - (d) is the only way to do them is to find the modular inverse pertaining to x1,x2,x3 in for each field? (the homogenous is just all the multiples of p right?)

[rustic rapids]

i got (b) (1,0,0), (c) no soln, (d) (5,3,1). For x1,x2,x3 respectively mod p. Would this be right?

[chilly ocean]

Sounds correct

[restive birch]

the following holds, right?:
if $H$ finite, $H \leq G$ is the unique subgroup of $G$ of order $H$, then $H \unlhd G$ (so that in particular $<(1 2)(3 4), (1 3)(2 4)> \unlhd A_4$)

[cloud walrus]

jan Mo Lakeli Tu Katuwe Pa

[restive birch]

because i think $gHg^{-1} \leq G$ right

[cloud walrus]

jan Mo Lakeli Tu Katuwe Pa

[restive birch]

and $H \to gHg^{-1}$ should be a bijection

[cloud walrus]

jan Mo Lakeli Tu Katuwe Pa

[restive birch]

unless im missing smth

[vapid vale]

yes

[restive birch]

thanks

[serene dune]

$\varphi(h) = ghg^{-1}$ is a homomorphism

\ \

$H \cong gHg^{-1}$