groups-rings-fields

[molten silo]

ok

[prisma ibex]

does that make a little more sense?

[molten silo]

yes

still dont how to show it though

[prisma ibex]

Here's an idea for the first problem: to write down a 7-cycle, you're writing down a sequence of numbers among 1-7 that is non-repeating

among which there are 7!=5040 choices, right?

(you have 7 choices for the first number, then 6 choices for the next number because you can't repeat the first, then 5 choices for the next number because you can't repeat the first and the second, and so on)

[molten silo]

so 7! take away the repeats

[prisma ibex]

now note that, for instance, (1 2 3 4 5 6 7) and (2 3 4 5 6 7 1) represent the same permutation. So you have to divide out by rotations of your sequence

See how to finish it now?

[molten silo]

yes

[prisma ibex]

🙂

[molten silo]

For the second question i just show it is a product of even transpositions right?

[prisma ibex]

Close! Transpositions are by definition 2-cycles, and they are odd permutations

but the composition of an even number of transpositions is an even permutation

so you have to show that a 7-cycle is a product of an even number of transpositions, that's one way to do it

[molten silo]

ok

thx

How about this one

[prisma ibex]

Two elements of S_n live in the same conjugacy class iff they have the same cycle type

[molten silo]

yeah

[prisma ibex]

so the number of elements in this conjugacy class should be the number of 7-cycles you computed earlier 🙂

[molten silo]

i though so

so 720

[prisma ibex]

mhm!

[wind cypress]

My algebra course/module started today, loved it :).

[prisma ibex]

nice!

[wind cypress]

Too bad teacher didn't yet specify which exercises we are supposed to do. Tomorrow is next lecture (remote lectures only due to corona).

[prisma ibex]

There's a nice trick for computing the sizes of conjugacy classes in symmetric groups that goes as follows:
conjugacy classes in S_n are in bijection with unordered integer partitions of n. For such a partition \lambda=n^{a_n}...1^{a_1} the size of the corresponding conjugacy class is given:

[wind cypress]

I have exercises in the material he has written, but there are so many exercises, I won't do them blindly and hope those are what we are supposed to do 🙂

[prisma ibex]

$\frac{n!}{\prod^n_{i=1}(i)^{a_i}(a_i!)}$

[cloud walrus]

nGroupoid

[chilly ocean]

I had a number theorist for my group theory class, shame we literally only mentioned exact sequences during the last lecture Sadge

[gritty latch]

still stuck on this q, any help appreciated

[prisma ibex]

For example there are 7 conjugacy classes in S_5. For say the middle one, corresponding to the partition 5=3+2 the size of the conjugacy class is 5!/(2^1 1! 3^1 1!)=20

or say for the next one, corresponding to the partition 5=2+2+1 the size of the conjugacy class is 5!/(2^2 2! 1^1 1!)=15

[molten silo]

how would you do this?

[prisma ibex]

hint: the number of elements in a conjugacy class is the index of the centralizer of any element of that conjugacy class

so the centralizer here will be an index 720 subgroup of S_7. Should be able to conclude from here. 🙂

[molten silo]

nope

not a clue mate

[prisma ibex]

Okay another hint: if a subgroup of S_7 has index 720, what can we say about the size of that subgroup?

(More generally if G is a finite group of order n, what does it mean for a subgroup H of G to have index k? What can we say about the order of H?)

[molten silo]

i dont know what is meant by the index

[stark sigil]

There's also a brute force proof as long as you're willing to relabel elements so that $g$ looks like (1234567)

[prisma ibex]

okay @molten silo we say that a subgroup H of G has index n, written [G:H]=n, if |G|/|H|=n

[molten silo]

ok

[prisma ibex]

it's like the size of H relative to the size of G

[molten silo]

this is 720 right?

[prisma ibex]

anyways if the centralizer here has index 720 that means the centralizer has order 7!/720=7

Now what do we know about groups of order 7? 🙂

[molten silo]

7 - cycle in s7?

[prisma ibex]

yes, this ends up being a subgroup generated by that 7-cycle, so it must be cyclic

said more cleanly, we know that the centralizer has to have order 7, and any group of prime order is a cyclic group

[molten silo]

oh isee

oh my god

[prisma ibex]

(maybe not the most general approach since this is sort of a coincidence owing to the fact that we're working with a symmetric group S_n where n is prime)

(the general fact nevertheless is that the centralizer of an n-cycle in any symmetric group S_n has index n and is the cyclic group generated by that n-cycle)

(more generally, the centralizer of an element of S_n is generated by that element along with the permutations which are disjoint from that element)

[stark sigil]

For n not prime, you can still prove the centralizer is cyclic just noticing that all the powers of g are included in the centralizer and there are already n of those

[carmine fossil]

Just show it has no root over that field

[prisma ibex]

Yup Icy that works!

[molten silo]

Any hints brothers

[prisma ibex]

@frail zinc If this polynomial is reducible, it must have at least one linear factor, meaning it has to have a zero in F_5. So determine whether or not this polynomial has zeros on F_5; if it doesn't the polynomial is irreducible.

@molten silo If G is a group of order pq for p and q distinct primes and H is a nontrivial normal subgroup of G, we know that the order of H has to properly divide pq

So either H has order p or order q (or order pq, but then the statement is obvious)

What can you conclude about the order of the quotient?

[molten silo]

h must have order p or q

the order of the quotient has order p or q right?

[prisma ibex]

Yup! If H has order p then G/H has order q, and if H has order q then G/H has order p

[molten silo]

ok

[prisma ibex]

but p and q are prime; what do we know about finite groups of prime order?

🙂

[molten silo]

so it must cyclic right

[prisma ibex]

yup!

[frail zinc]

@prisma ibex how do you check whether it has a zero in F_5 tho, bit rusty on this

[molten silo]

thus abelian

[prisma ibex]

Yup!

[molten silo]

thanks

[prisma ibex]

@frail zinc You sort of have to check it by hand lol

I think there are some criteria that make this easier if you're doing this over F_p for p really large but over F_5 there's not much to check

we just want to check whether t^3+2t+1=0 mod 5 for t=0,1,2,3,4

,W table t^3+2t+1 from t=0 to t=4

[cloud walrus]

Results provided by WolframAlpha

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

Requested by nGroupoid#5061

[prisma ibex]

none of these look like 0 mod 5 to me 🙂

[delicate bloom]

if I was doing it by hand like on a test I'd do -2, -1, 0, 1, 2 instead, slightly faster

[prisma ibex]

True!

[molten silo]

If you look at part (ii) you see that if h = [e] then g/h isnt necessarily abelian right

[frail zinc]

Ahh right so I can just choose any consecutive 5 numbers

[prisma ibex]

Mhm! You just need to choose representatives from each mod 5 congruence class

[frail zinc]

Ah okay, so it looks like its not reducible

[prisma ibex]

@molten silo that's right, there are examples of groups of order pq for p and q distinct primes which aren't Abelian

for example dihedral groups D_p of order 2p (where p>2)

If you instead let p=q then every group of order p^2 is Abelian: it's either Z/p^2Z or Z/pZxZ/pZ

[molten silo]

Your are so clever

[prisma ibex]

(sorry for the chat spam on algebra, I'm preparing for an algebra exam right now and it's kinda helpful talking this stuff through!)

hope this is helpful

[frail zinc]

Its appreciated lol thank you for the help

extremely helpful

[prisma ibex]

🙂

[frail zinc]

might have some more questions soon lol

[vestal snow]

I had a question about function fields and the dictionary between topological and geometric stuff. Suppose that $K$ is an algebraically closed field and $E$ is a function field in one variable over $K$. Let $F$ be a finite extension of $E$. It is well-known that there is an equivalence of categories between function fields and projective curves. Let $C_F$ and $C_E$ be the curves corresponding to $F$ and $E$ respectively. Let $p: C_F \rightarrow C_E$ be a covering map. Is the order of the group of covering transformations of this cover the same as the group of automorphisms of $E/F$?

[cloud walrus]

Have a Banana Bitch

[molten silo]

To show something is an isomorphism it must be a bijective homomorphism right?

[vestal snow]

In groups, yes

[molten silo]

thanks brother

[vestal snow]

Np

[latent anvil]

What have you tried/done so far?

[prisma ibex]

Hint: what is the determinant of the unit element in GL(n,C), and what is the unit element in C*?

Moreover, how is the determinant of a product of elements in GL(n,C) related to the determinants of the individual elements?

[sturdy marsh]

the correspondence is between curves + birational maps and function fields. The degree of the field extension corresponds to the degree of the map

the only problem is that that was how degree was defined for us

[vestal snow]

Yeah I know. I just used the correspondence to get C_F and C_E

[sturdy marsh]

oh ig you can also define degree by looking at the pullback of O(1)

[vestal snow]

I'm just wondering if the the two groups are isomorphic or not

[sturdy marsh]

dunno, usually you don't get nice covering maps

they normally have branch points

[vestal snow]

If we somehow get an equivalance of categories of covering maps and curves and function fields and morphisms, we'd be done

[prisma ibex]

you can still consider unramified covers though

if you have a smooth projective algebraic curve of genus g>0 it has nontrivial fundamental group so you're going to see some unramified covers

[sturdy marsh]

true

[vestal snow]

Assume that the genus is greater than 0

I don't really know what branched covers and unramified covers mean, but I'll try to follow along

[sturdy marsh]

you would need something like 2g_1-2 divides 2g_2 -2 or something like that

[prisma ibex]

branched just means like you're allowing some of the sheets of the cover to come together at some points upstairs

in which case we say the cover ramifies at those points (or rather above the points they project to)

otherwise we say that the cover is unramified. Unramified covers are covers in the "usual" sense

[sturdy marsh]

have a banana bitch as you do NT, it is related to ramification in the number field sense as well

[vestal snow]

Like ramified places?

[prisma ibex]

Yes, exact same idea

places of the function field correspond to points on the curve

ramification over those places in the function field sense corresponds to having branching over some ramification points in the geometric sense

[vestal snow]

Okay I think I kinda understand what's going on

[chilly ocean]

Can someone help me talk through I-adic completions. So with the (x)-adic topology the competion of R[x] should be R[[x]]

The open sets around zero are all the ideals (x)^n

the open sets around any other point is just translations, so f+(x)^n are the open sets around f

[prisma ibex]

Yup!

[chilly ocean]

Now to get the completion of R[x], we take the inverse limit of R[x]/(x)^n

If I understand correctly, the inverse limit is defined to be all coherent sequences

[prisma ibex]

Yup, the completion here is R[[x]]. The point is that given any f in R[[x]] you can consider its truncation to any R[x]/(x^n)

Conversely, if you have a coherent sequence of elements f_n in R[x]/(x^n) then this gives an element of the (x)-adic completion

but such a coherent sequence is the same as giving an element of R[[x]]

[chilly ocean]

hmm okay i am getting a little bit confused here then

a coherent sequence is a sequence a_n and a group homomorphism \phi such that \phi(a_n)=a_{n-1} ?

[prisma ibex]

Right. In this example, you want f_{n+1} in R[x]/(x^{n+1}) and f_n in R[x]/(x^n) such that the former reduces to the latter mod (x^n)

like, each time you "go up the tower" you're prescribing the next coefficient of the power series

and the coherence forces the first n coefficients f_{n+1} to match those of f_n

(I might have some indexing fuckups but you get the idea)

[chilly ocean]

so the group homo is fixed for a given completion

and we are just taking these group homos to just be inclusions

i think this might be where i was getting confused

[prisma ibex]

lol I think I'm confusing myself too haha

well let's think

We have a filtration R>I>I^2>...

passing to quotients that filtration gets reversed

which are a bunch of projections

I mean the other example to keep in mind is like

Z_p is the completion of Z at (p)

so it's the inverse limit of the rings Z/p^nZ

that is, the elements are coherent sequences (n_1,n_2,...) with n_i=n_j mod p^i for i<j

similar statement holds for R[[x]] as the inverse limit of the rings R[x]/(x^n)

[chilly ocean]

Okay

[chilly ocean]

I’ll look at this

[delicate bloom]

I think you can also do it with 0<a<1 defining |f|:=a^ord(f) for f in R(x) to get an ultrametric abs value, making cauchy sequences and getting the completion that way

basically just p-adics but without carrying, hensel should work too

[prisma ibex]

huh

[delicate bloom]

set the output to be =(a,b) and solve the system of 2 eqns

[vestal snow]

@prisma ibex just wanted to clarify: totally ramified points of a function field = branched points of the corresponding curve?

Very liberal use of =, but you should get the main idea

[prisma ibex]

yea the places over which an extension of function fields of curves ramifies corresponds to the branch points of the corresponding curves

[vestal snow]

Isn't there a difference between ramified and totally ramified though?

[oblique river]

you can be ramified but not totally ramified

if you want to think geometrically in terms of curves, if you have a map of curves of degree n, then whenever a point p has fewer than n preimages, it's ramified

but totally ramified would mean it has one preimage

[vestal snow]

Yeah I meant

that is the correspondence between ramified points and branch points or between totally ramified points and branch points?

[oblique river]

oh sorry

ramified

[prisma ibex]

pedantic point: ramification points are what happens downstairs, branch points are what happens upstairs

but yes the the points of ramification downstairs correspond to places in the function field that extensions ramify over

no need to impose total ramification

[vestal snow]

Hmm this means that I misunderstood something in one of the papers that I was reading

[oblique river]

not sure if it's related to this but there are cases where ramified implies totally ramified, for example, if the covering group is cyclic of prime degree

[vestal snow]

Ah that solves it!

[prisma ibex]

Ah good point, didn't think of this!

[vestal snow]

It's Z/p^nZ covers

[prisma ibex]

Ahh yup there you go!

[oblique river]

oh nice, tha'ts not prime degree but in fact it is also true for prime power degree

[vestal snow]

Arigatou minna

[oblique river]

yw 👍

[prisma ibex]

🥳

[oblique river]

err, not just prime power degree, but like

specifically Z/p^nZ

(but not Z/p x Z/p)

[vestal snow]

cyclic of prime power

[prisma ibex]

yea, cyclic of prime power order

[oblique river]

yep

[prisma ibex]

cool cool this is good

[vestal snow]

I'm still pretty lacking in my geometric knowledge about this stuff

[golden pasture]

for fun is there any meaning to tamely vs wildly ramified other than on the higher ramification groups

i too have no geometry

[vestal snow]

So I'm trying to intuit through it using my knowledge about function fields and the principle "if this was not true, life would be a pain so it is true"

[prisma ibex]

Kinda, yea!

[vestal snow]

I could swear you helped me on an AG problem a while back

[golden pasture]

you see
AG != geometric intuit

[prisma ibex]

I was actually going to share one of my favorite unsettling facts. So I mentioned etale \pi_1 as the correct replacement for the fundamental group. This often behaves as we would expect: if you take the affine line A^1 over an algebraically closed field of characteristic 0 then it is etale simply connected in that etale \pi_1 of A^1 is trivial: it has no nontrivial finite etale covers

this is SPECTACULARLY false if you take the affine line A^1 over an algebraically closed field of characteristic p

etale \pi_1 of A^1 in this case is an infinitely generated profinite group

so it has an utterly gargantuan supply of finite etale covers

[vestal snow]

I looked up the AG definition of branch point and it literally defined it in terms of function fields lol

[prisma ibex]

Ahahaha nice

[vestal snow]

So this was tautological

Though I guess all of math is tautological

[prisma ibex]

There are non function field definitions of ramification and branching but they aren’t really necessary here where you can just work with the associated function fields

But yea I do agree it’s a bit tautological if you define it this way

[barren sierra]

Suppose we have a ring X such that for all x in X, x^n = x for some n in N

where that n is the same for all elements in X

is it such that the ring is commutative

I've proven this for the case n = 2

but I'm struggling with the base case

[next obsidian]

wut

just start your induction at n = 2

the base case for n = 1 is false

take any noncommutative ring

also idk it sounds kinda fake imo

maybe if you're finite

[barren sierra]

wait I meant to say for some n in N >= 2

whoops

ok but for different sets that n might be different right?

[wind steeple]

for n>1 it is true iirc

[barren sierra]

so I can't do induction

[wind steeple]

but it is highly non trivial

[barren sierra]

ok that's comforting lol

[wind steeple]

for n = 3 for instance I know a proof with cumbersome algebraic manipulations

[barren sierra]

yea same

[wind steeple]

(non that high but it's technical I mean)

[barren sierra]

uses (x+y)^3 = x + y

expand that left side

do some manipulation

kinda wack

use the fact that 3x = -3x in the case when n = 3

[wind steeple]

yeah

[barren sierra]

but I can't work on it for higher values of n

much less general n

[wind steeple]

btw for arbitrary n I remember I saw this theorem has a name but I don't remember which name it has lol

[barren sierra]

I was just given this as an exercise by a prof I'm talking with 😭

[wind steeple]

I don't have any idead on how you can prove it but instead of trying to manipulate algebraic expression, maybe you can get insights of the structure of your ring with much simpler rings

for n = 3 you can factor A like A = 3B + 3C smth like that isn't it ?

I don't remember the proof

for n = 2 you can write A = eA (+) (1-e)A for all e in A, maybe you can find something similar for n>2

[molten silo]

Can someone help me with part 2

finding the kernel and image

[sullen bloom]

Do you know what kernel and image mean?

[molten silo]

i got that for the first composition that the kernel is the unit circle in the complex plane

and the image is the positive reals

i think its wrong

[sullen bloom]

That's right.

[molten silo]

you sure

[sullen bloom]

The unit circle all goes to 1, which is the identity in the reals under multiplication

And the image is the positive reals because given a real number r, pr_1(|r|, r/|r|) is just r.

[molten silo]

so i was right

[sullen bloom]

Yes, to my understanding.

[molten silo]

but then in part c we are asked to sketch the quotient

this is isomorphic to the image

how do you skecth the image

[sullen bloom]

Well the image is just the positive reals, whereas the cosets are a bunch of circles. (The unit circle scaled/rotated by some complex number)

The isomorphism comes from identifying each circle with the point at which it intersects the positive real line.

This is equivalent to identifying each circle with its radius

[molten silo]

not really understanding

oh ok

but how would i sketch that

i didnt even know the cosets are circles

do the cosets of circles represent the positive reals

[sullen bloom]

the kernel, K, is a circle of radius 1. This is the kernel because it's the set of complex numbers with magnitude 1. Cosets of the kernel have the form zK, for some complex number z. If you think of complex multiplication geometrically, you're scaling (elements of K) by the magnitude (of z) and rotating by the angle (of z). But the angle doesn't matter, because the kernel is a circle. So cosets zK are just scaled versions of K.

[molten silo]

ok

[sullen bloom]

and we assign each circle to its radius

[chilly ocean]

Can you ping me if you get an answer

Seems a very cool problem

[molten silo]

how does this relate to the image?

[sullen bloom]

the image is precisely the positive reals, which are all possible radii.

[molten silo]

yes

[sullen bloom]

so the quotient by the kernel is isomorphic to the image

[molten silo]

this makes sense

but what am i suppose to sketch

[sullen bloom]

circles

of various sizes

[sullen aurora]

It's called jacobsons theorem https://ysharifi.wordpress.com/tag/jacobson-theorem/

[chilly ocean]

Just stumbled across that myself

[molten silo]

So these circles can be anywhere on the plane

[sullen bloom]

no, they're centered at the origin

[chilly ocean]

Cool that it doesn’t even need the n to be the same for each x

[molten silo]

since only the radius had to be positive

[sullen bloom]

just scaled versions of the unit circle

[molten silo]

so no translation, just rotations and scaling

[sullen bloom]

yeah

[molten silo]

isee

[sullen bloom]

although rotations are invisible, cause, youknow, circles

[bleak abyss]

Oh damn that's a nifty theorem

[molten silo]

and for the second composition it had kernel z = 1 right?

[chilly ocean]

Pretty cool alright

Also in hindsight it shouldn’t have been to surprising

x^n=x is pretty strong condition to have on all elements

[latent anvil]

$\mathrm{Gal}(axy/Brain)$

[cloud walrus]

shamroc$\overline{k}$

[next obsidian]

$\mathrm{het}$

[cloud walrus]

Chmonkey 2.0

[prisma ibex]

R^op=Я

[chilly ocean]

Rop Rop

[golden pasture]

return oriented programming time

[molten silo]

How?

[uncut girder]

Determinant takes O_4 to +-1

But SO_4 to 1

Fill in the details

[molten silo]

what does the index mean

[latent anvil]

do you know what a coset of a subgroup is?

[molten silo]

yes

[uncut girder]

It's the number of cosets

[molten silo]

oh no

ive only proved that S04(R) is a normal subgroup

if that enough

[prisma ibex]

@molten silo I mentioned this before, the index of a normal subgroup is the size of the quotient

But yes PTY said this correctly, take the group homomorphism det:O_4(R)->{-1,+1} sending an orthogonal matrix to its determinant

the kernel is SO_4(R)

[golden pasture]

i doubt so

consider the case of permutation groyps

or wait just consider the case of any group and any element with ab=ba

it isnt then true that <a>=<b>

[frail zinc]

Not sure what you mean

[chilly ocean]

This is very not true

[golden pasture]

?

[chilly ocean]

By this I mean I cannot even think of an example where it is true

In any ring take 1 and any element x

In Z take any two distinct elements

sigh english

*then it isn't true that

[chilly ocean]

I think we all know what they meant

And they probably meant to type

Isn’t it then true that

wait so what part of this are you objecting to?

The mathematical statement is not true

which statement

ab=ba implies <a>=<b>

I am also objecting to you talking down to people for improper English

It’s not everyone’s first language

No need to be rude

what no i'm not i think they meant it's not true that

[chilly ocean]

I doubt they meant that

which is very different from 'it isn't true that?'

If something isn’t true in general, why would they give restrictions that also don’t make it true

i mean i think they knew that it wasn't true

but english is a bad language so it came across the wrong way

Suppose they knew P was not true

Why would they then say

‘Given such and such conditions P is not true’

delete the then from this sentence 'it isnt then true that <a>=<b>'

[hot lake]

did someone delete a message or something

[nova plank]

Yall are both crazy. Why are you arguing over what someone else meant?

[chilly ocean]

Swap the it and the isn’t

[hot lake]

that completely changes the meaning

[chilly ocean]

english sigh

[chilly ocean]

I thought they were talking down to someone for writing a bad sentence

[chilly ocean]

Well I mean anything complete changes the meaning of something incomprehensible

no i though you were jumping on them

misunderstanding

[chilly ocean]

Ah okay sorry

Lmao

[carmine fossil]

D_3

[rugged hazel]

What is the identity of a group Z with binary operation a • b = a^2 + b

Or is there no identity hence not a group

[carmine fossil]

There's no identity

[rugged hazel]

👍

[rugged hazel]

Can someone give an example of an infinite, non-commutative group which contains elements of every finite order

[delicate bloom]

nxn matrices with nonzero determinant

under multiplication and n>1 ofc

[rugged hazel]

That’s commutative isn’t it ?

Oh no nvm

Is this because the elements can be written as cycles

And 11 is a prime so will have to have 11 elements but S9 only has 9?

[prisma ibex]

right to get an element of order 11 you need an 11 cycle, and there's no room for this in S_9

[delicate bloom]

do you know Lagrange's theorem?

[prisma ibex]

whereas to get an element of order 10 you can compose a transposition with a disjoint 5 cycle

[molten silo]

i thought it is because 10 divides the order of s9. But 11 doesnt devide order(s9)

For this question

[prisma ibex]

yea that works

[molten silo]

Is this a valid proof

[prisma ibex]

it's still faster to just take SO_4(R) to be the kernel of det:O_4(R)->{-1,+1} but this is rather direct

[molten silo]

so it works right?

[prisma ibex]

Yup!

Not the shortest proof but it does show that you're thinking about how to apply the definitions directly

[molten silo]

Because im not sure how mapping would determine its a normal subgroup

[prisma ibex]

the kernel of any group homomorphism is a normal subgroup

[molten silo]

really

[prisma ibex]

yes

moreover any normal subgroup arises in this way

[molten silo]

so all we do is define a map to thekernel of O4(R)

using determinants

[prisma ibex]

well SO_4(R) is more or less by definition the kernel of det:O_4(R)->{-1,+1}

you just have to say a word or two about why determinant is a group homomorphism

then you're good to go

[molten silo]

thanks

[prisma ibex]

👍

[molten silo]

If you take a look at part 2 of this problems. The first composition has kernel the unit circle and image the positive reals. But for the second composition does this have kernel = {1} and image the unit circle?

[rugged hazel]

What’s an example of two groups G,H with 1-1 homomorphism ø:G—>H such that G is commutative and H is NOT commutative

[prisma ibex]

e.g. take H=S_3 and G=Z/3Z the subgroup generated by a 3-cycle

[rugged hazel]

Wait what’s Z/3Z I’m not seen that

Ive*

[carmine fossil]

What are the groups you have seen?

Do you know Z_3?

[molten silo]

is it not just the cosets of 3Z

[rugged hazel]

Z3 as in modulo

[carmine fossil]

Yes

[rugged hazel]

Yh yh I’ve seen that

[carmine fossil]

That's Z/3Z

[rugged hazel]

Ahhhh right

How’s that a homomorphism if you don’t mind explaining

[carmine fossil]

Map 1 to (123)

[molten silo]

How do i sketch C/ker{1}, where c is the complex numbers thats not zero

[golden pasture]

ker{1} as in?

[molten silo]

kernel is z = 1

this is what i got for the kernel

[golden pasture]

isn't that then C*

complex numbers under multiplication

[molten silo]

yes

This is meant to be isomorphic to the image which is the unit circle

and so how would you sketch C*/kernel = 1

[rugged hazel]

How do you do part ii)

Normal subgroup means ghg^-1 is in H

Is it referring to left, right cosets ? Or is it a v simple proof that I’m just not seeing

[solemn shoal]

For any h ∈ H, there exists unique g ∈ H

Such that aha^(-1) = g

ah = ga

[rugged hazel]

Ahhhh thank you

[prisma ibex]

(I'm pedantic about writing Z/pZ because Z_p refers to something else that number theorists like 😛 )

[rugged hazel]

Ahah no worries I’ve just not seen that notation before lol

[rugged hazel]

I started by saying

g € ker(theta)

[carmine fossil]

"By first isomorphism theorem, this is true"

[rugged hazel]

Implies theta(g) = eh

It says use fundamental theorem of homomorphisms

[carmine fossil]

Is that the first iso theorem?

G/ker Phi iso im(phi)

[rugged hazel]

Oh Yh that’s the same

But how can I prove that

As there’s no actual phi

Just two groups

[carmine fossil]

Construct a phi

[rugged hazel]

Alright let me try that

Not sure where to start :/

[solemn shoal]

What homomorphism leads to K being the kernel?

That is, φ(g) = e iff g⁴ = e

[rugged hazel]

Then do you say g ~ G, g~ker(phi) implies phi(g) = identity of H?

~ just bring contained in

[solemn shoal]

Ye

[rugged hazel]

That’s fine but then when doing that. I get phi(g) = e

Or should I be doing phi(g^4)

[solemn shoal]

What is φ(x)?

[rugged hazel]

e

[solemn shoal]

But then the kernel would be G not K

[rugged hazel]

g^4 then ?

Phi(x) = g^4

[solemn shoal]

φ(x) = x⁴ should do the trick

[rugged hazel]

Ahhh

Then to prove is it phi(g^4) for all g^4 in H then (g^4)^4 = e^4 = e

[solemn shoal]

Since φ(g) = g⁴ = e iff g ∈ K

Then K is the kernel

[rugged hazel]

Oh wait Yh I was looking at K and H the wrong way round

[solemn shoal]

Oh

[rugged hazel]

You’re right thanks

[solemn shoal]

Np

[rugged hazel]

Why are any two groups of order p prime, isomorphic ?

[chilly ocean]

they're both cyclic

same order

pretty obvious isomorphism to choose

[solemn shoal]

Any element other than the identity has order p

So if g is not e, then all elements can be written as g^n

This gives them the cyclic structure

[rugged hazel]

Any examples i can use ?

Zp under addition

[chilly ocean]

that's the only example

[solemn shoal]

Z/5Z = {0, 1, 2, 3, 4}

Choose 2 for example

[rugged hazel]

Z5 and what other group

[solemn shoal]

2 = 2
2*2 = 4
2*3 = 1
2*4 = 3
2*5 = 0

[rugged hazel]

Perfect okay thanks a lot

[chilly ocean]

backslash your *'s

prevents it from being italicized

[solemn shoal]

Ohh thx

[solemn shoal]

Here 2*n is akin to the g^n previously mentioned

[vestal snow]

What does the order of a differential mean?

In algebraic geometry

[gritty latch]

let R be a UFD and let P be a nonzero prime ideal of R such that if P′ is another prime ideal with (0)⫋P′⊆P, then P′=P. Prove that P is principal.

any help with this?

[chilly ocean]

maybe, you can choose a non-zero element x of P, factor x into stuff using the fact that R is a UFD, and then look at the principal ideals generated by that stuff x factors into

don't take my word for it

[rugged hazel]

How can you show that if G has order prime p, then G is cyclic

[chilly ocean]

lagrange's theorem

any non-identity element has order \neq 1 dividing p, and p is prime, so that order has to be p

[rugged hazel]

And how does that show it’s cyclic

[chilly ocean]

well if a group has order p

and an element has order p

what does that mean

[rugged hazel]

gotcha

what does that mean in the earlier message “\neq 1” ?

[chilly ocean]

the order of a non-identity element is gonna be greater than 1

in order to force the order to be the prime number p, it has to be a number not equal to 1 that divides p

[rugged hazel]

Okay thanks

[chilly ocean]

it's a very tiny detail but i think it's important to say

[sly crescent]

Is there an analog of the greatest common divisor for groups?

[oblique river]

not for groups, but there (kind of) is for rings

[sly crescent]

Why not?

It should be simple. Just take the highest-order common subgroup(s).

[cyan marten]

How would you define a | b for groups?

The reason it works for rings is that not every element is a multiple of another.

[sly crescent]

a is a subgroup of b

[cyan marten]

Did someone solve this?

[cyan marten]

I thought you wanted to define the GCD for elements..

Maybe what you want is just the intersection, and the "LCM" is then the subgroup generated by them. We haven't really used the group structure, though, only the fact that the set of subgroups is a lattice.

[oblique river]

in a group, every element divides every other element

so "divisibility" isn't really a useful notion

[cyan marten]

What they meant is defining "divisibility" as "inclusion of subgroups."

[oblique river]

oh I see

in which case I think it would go the other way. a divides b if <a> contains <b>, so the gcd of a and b would be the subgroup generated by a and b

[cyan marten]

Reasonable

[latent anvil]

I would also say that

The gcd is the binary supremum/least upper bound in the reversed divisibility poset of Z, and that's the same as the poset of subgroups of Z

so more generally we should take the binary supremum/least upper bound of two subgroups, ie the subgroup they generate

[latent anvil]

Yes. Do you want the answer?

[cyan marten]

Yes.

[latent anvil]

μ(G) is the number of generators of G

So it's φ(|G|) if G is cyclic and 0 otherwise

[dim escarp]

Oh kek reversed divisibility

[cyan marten]

What's the multiplicative group of the algebraic closure of Fp?

[sturdy marsh]

it's a field, so everything except zero

?

[cyan marten]

Yeah, but I meant whether it had a simple description.

[sturdy marsh]

it probably wont be any simpler than a description of the alg closure

which is just throw in all the qth roots of unity

[cyan marten]

I thought so, but then we know that, for example, all finite subgroups are cyclic..

And whenever p^n = 1 (mod d), there is a unique element of order d.

Maybe it's the Prüfer p-group

[sturdy marsh]

ig I am really unsure about what you mean by a nice description

the finite subgroups are cyclic thing is true for C* too

but I can't think of a description of C* other than C - {0}

[cyan marten]

Basically whether it's isomorphic to some group which I probably heard of in another context. I can't think of a candidate other than the Prüfer p-group (the p-Sylow subgroup of Q/Z), but I am not sure.

[sturdy marsh]

is it the prufer p-group?

you have primitive qth roots for q neq p

[latent anvil]

It should be the directed limit of Z/(p^k - 1)Z, but I think the maps in-between are weird

Maybe the choice of maps doesn't matter tho, as long as they're injective

[knotty mason]

when would you both add and divide by an ideal ? why an expression like this (R + I)/I ?

[chilly ocean]

Do you have a sufficient answer for doing each individually?

[grand cliff]

https://en.wikipedia.org/wiki/Eckmann–Hilton_argument
Does this imply that there can only exist at most 1 group per set

since all groups have unit

[wind steeple]

no

Z/4Z and Z/2Z² aren't the same

you can't have the 2d relation with every groups

[odd abyss]

I don't understand why
If we have N = { 0, 1 , 2 ...}
(N, +) is not a group of Z

[carmine fossil]

No inverses

[odd abyss]

Could you explain to me more? The inverses of what?

pls

[carmine fossil]

0 is the additive identity

Suppose (N,+) was a group,given an element a,there would be a'(in N) such that a+a'=0(a' is called inverse of a)

But there is no such a in N,such that say 1+a=0(i.e, 1 doesn't have an inverse)

[odd abyss]

a can be the same number?

like 1+1

or 2+2

[carmine fossil]

a can be anything in N

[odd abyss]

so why not 0 + 0

is = 0 it's good no ?

[carmine fossil]

Yea 0 has an inverse

But,for a group you need EVERY element to have an inverse

[odd abyss]

Do you have any examples of numbers that are inverse? (fairly simple examples)

[carmine fossil]

There is no inverse of 1 in N

Let's take Z

1+(-1)=0

-1 is inverse of 1 in Z

[odd abyss]

So in Z we have two inverse

0 + 0 = 0

and

1 + (-1) = 0

[carmine fossil]

Inverse is specific to an element

You say inverse of 0 is 0

And inverse of 1 is -1

In general, inverse of a is -a

[odd abyss]

ok thank you

For it to be a group, you must meet the 3 conditions :

-> Associativity
-> Element neutral
-> inverse

?

[carmine fossil]

Yes

[chilly ocean]

Suppose we have an exact sequence of groups M to A to B to C to N

When M and N are zero we know that B is the direct sum of A and C.

What can we say if M and N are no zero?

[odd abyss]

Why for
{ Z \ {0], . }
1 . n = n . 1 = n
1 is a neutral element ?
The neutral element it shouldn't be 0 always ?

[carmine fossil]

No,The neutral element depends on the operation

[dreamy snow]

I'm trying to prove that if a group's sylow-p-subgroups intersect only in the identity, then the number of sylow-p-subgroups is congruent to 1 mod their order. One can probably show this relatively easily with group actions. I attempted a different approach though, basically by looking at the conjugation of the p-Sylows directly.
I got the result that |Syl_p(G)| = p^k+1, so ≡ 1 mod p^k. However I'm not sure if this is a coincidence and I just accidentally fudged the proof or whether it's valid. In particular I'm not sure if the part that I marked orange is correct. Any help would be much appreciated

[weak flower]

Is it manageable to construct a polynomial over an integral domain R of positive degree that does not have any irreducible factors?

[latent anvil]

I think you'll need R to fail to be noetherian for this to work

And a lot of easy examples (eg polynomial ring in infinitely many variables do have this property)

[weak flower]

I think R actually cannot have ACC on principal ideal, because f can be factored indefinitely while the degree of the factors cannot descend indefinitely.

[latent anvil]

yes, I agree

(the thing you said is why noetherian domains have factorization into irreducibles)

[odd abyss]

Let (G, ∗) be a group. Describe all morphisms of groups Z → (G, ∗)

how to do that?

[weak flower]

And There actually might have to be a polynomial that can keep splitting out constant terms, while not leaving an irreducible polynomial behind. This sounds a bit like black magic to me

[wispy scaffold]

Z is a free group generated by 1, and you just need to study 1 \mapsto g \in G

[chilly ocean]

And follow the definition of a group morphism

[weak flower]

And There actually might have to be a polynomial that can keep splitting out constant terms, while not leaving an irreducible polynomial behind. This sounds a bit like black magic to me
@weak flower Does this mean that I need at least 2 elements with infinitely many common factors, and no maximal common factors?

[odd abyss]

1 \mapsto g \in G ?

[odd abyss]

the definition is f(a+b) = f(a) + f(b)

[chilly ocean]

and something else too

[odd abyss]

My teacher gave us this definition to find the morphism)

the "something else" 'is for isomorphism

[chilly ocean]

no

[odd abyss]

For an isomorphism, it is necessary to prove that it is a morphism + bijective (or null kernel)

[chilly ocean]

you need to have f(identity of Z) = identity of G

[odd abyss]

what do you mean by identity of Z and identity of G ?

[chilly ocean]

the element that has the property that ae=ea=a for all a

[wispy scaffold]

if f is a morphism, then f(0)=f(0+0) = f(0) + f(0) and f(0)=e

[odd abyss]

f(a+b) = f(a) + f(b)

[wispy scaffold]

doesnt matter

[odd abyss]

well I don't succed this question :/

[wispy scaffold]

wrt this, as any element of Z can be written as n = 1 + ... + 1 or - n = - (1 + ... + 1), as f is assumed a morphism, you have f(n) = f(1) + ... + f(1)

so the important thing is what is the value of f in 1

[odd abyss]

Well, you did like my teacher, but with a little less explanation and precisely, I understood nothing with him, I thought that there would be another method

f(n) = f(n-1+1) = f(n-1) + f(1)

= f(n-2+1) + f(1)

= f(1) + f(1) + .... + f(1) n times

= a* a* a* ...

[latent anvil]

@weak flower I think if an example exists one should exist with constant polynomials

[odd abyss]

why we determinate f(n) and not f(-n)

after that my teacher do

f(e) = e

f(a^-1) = (f(a)^-1)

I don't understand

so the important thing is what is the value of f in 1

How we find the value of f in 1 ?

[wispy scaffold]

your mission is not to find a single value of f in 1, but to find all morphism f:Z -> G

what you must accomplish is to find a characterisation of a morphism f:Z -> G

[odd abyss]

Ok but say that f (n) = f (1) + ... + f (1)
or
f (-n) = f (-1) + ... + f (-1)
is it sufficient?

[wispy scaffold]

describe Z

a number can be zero, positive or negative

[odd abyss]

yes

Z = { ...,- 3 , -2, -1 , 0 , 1 , 2 ,3 ...}

[wispy scaffold]

yes. We would like to avoid negative number. How can we use positive number for describing negative numbers?

[odd abyss]

Well, you pose n> 0

i dont know

[wispy scaffold]

but given z\in Z with z<0

how do we write it like a function ||(something times)|| of a positive number?

[odd abyss]

I don't know

my teacher use neutral element

😢

[wispy scaffold]

it is just changing the sign, z = -(-z) with -z > 0

and we would like to describe positive numbers in a more elemental fashion

and we write z = 1 + ... + 1 if z>0. This give us a description of Z wrt its generator, wrt 1.

[odd abyss]

isn't that what you wrote here?

[wispy scaffold]

yes

[odd abyss]

I still do not understand what is the expected answer in this type of question

[wispy scaffold]

I already wrote the expected answer

I am trying to explain in detail why you only need to check for f(1) instead of defining explicitly f

[odd abyss]

yes but must be shown

[wispy scaffold]

yes, i said that

and we did it, as any element z \in Z can be written as

1. z = 0
2. z = 1 + ... + 1
3. z = - (1 + ... + 1)

then we can notice that, given f:Z -> G morphism, we have

1. f(z) = 0 if z = 0
2. f(z) = f(1) + ... + f(1)
3. f(z) = -(f(1) + ... + f(1))

[wispy scaffold]

If we know f, it is oblivious that we know f(1). If we know that the value of a morphism f in 1, we can extend it using the above, and recover f.

[opal osprey]

So, I was studying today about maximal ideals.

And the professor pointed out that as a corollary that given a ring A and and ideal I, A/I is a field iff I is a maximal ideal, one could easily check that every maximal ideal is also a prime ideal.

But the converse is not always true.

Is there another condition I should impose on a prime ideal in order for it to be a maximal ideal?

He slightly talked about how in every principal ideal domain, a prime ideal is always a maximal ideal.

But is it a iff?

[glad juniper]

@opal osprey No it is not an iff: https://math.stackexchange.com/q/1334665/

[next obsidian]

I mean the easiest condition to impose on a prime ideal is that no other (prime) ideal properly contains it

Or that the quotient is a field

There really isn’t very nice criterion in general situations, and commutative algebra gives you a few more tools but it requires more assumptions and is a lot more advanced than what you have to work with at the moment

[light tusk]

easy example to keep in mind: in the ring k[x,y], the ideal (x) is prime and properly contained in the maximal ideal (x,y)
a way to see that (x) is prime but not maximal: $k[x,y]/(x) \cong k[y]$ is a domain (hence (x) is prime in k[x,y]), but k[y] is not a field

[cloud walrus]

Apopheniac

[narrow ibex]

Just thought of this out of the blue, if ( \varphi : R \to S ) is a ring homomorphism where ( S ) is a reduced ring, does ( \varphi ) factor through the reduced part of ( R )?

[cloud walrus]

g_ijoe

[next obsidian]

Yup!

You can just show this manually

the thing you quotient out by is the nilradical aka all nilpotent stuff

[narrow ibex]

thanks! I was just about to prove it by hand

[next obsidian]

the image of nilpotetn stuff is still nilpotent

so if S is reduced they all map to 0

in fact it factors uniquely by property of quotients

[narrow ibex]

nice, I'll go ahead and still prove it for myself, but thanks again for the heads up

[gritty latch]

how does one go about finding the number of elements in Z5[x]/(x^2+1)

is there a general way to do it for Zn[x]/(f(x))?

[gritty latch]

what does the bar mean exactly?

[carmine fossil]

$\bar{a}$ refers to the equivalence class containing a

[cloud walrus]

MoonBears-D-

[gritty latch]

hm we havent really considered vector spaces in this before

but wouldnt 2 fail in the case of Z6[x]/(2x+4)?

since that has inf many elements?

[carmine fossil]

That has finitely many elements

6 elements

[gritty latch]

but its isomorphic to Z2[x] x Z3 tho

[carmine fossil]

Prob Z_6 is not a field

[gritty latch]

oh yea it has zero divisors

so if Zn isnt a field would you always get an inf number of elements in Zn[x]/(f(x))?

or is there a counter

ig you could always break it down to an iso with an inf field somehow

but man coming up with these isos isnt easy

[hot lake]

if it were isomorphic to a field it would be a field

[gritty latch]

oh yea

but i meant like an iso with something with inf elements, like Z2[x] for eg

[odd abyss]

Are the following groups isomorphic? Why ?
I know, if I don't say stupid things, that to show that a group is isomorphic you have to show that it is a morphism [ f(a+b) = f(a)+f(b) or f(ab) = f(a)f(b)] then show that it is bijective (find the inverse, for example) [ a*a^(-1) = a^(-1) * a = neutral element or a + (-a) = (-a) + a = neutral element)
When I have something like
(R, +) and (R>0, *) for the first I experience two get out of difficulty
I have two groups ...

[carmine fossil]

Given a positive real number a,do you agree any positive real number can be seen as a^m for some real number m?

[odd abyss]

It’s for me ?

[compact needle]

If f(x) is monic, this will always be finite

[gritty latch]

for a field right?

[compact needle]

For any n

[gritty latch]

hm

ok i think i see why

so i can use the fact in the quotient x+a=0 --> x=-a and just keep reducing every polynomial?

err

[compact needle]

You have to be careful with that

[gritty latch]

and a similar thing with higher degrees ig

[compact needle]

How do you actually show that f(a)=0 -> x-a is a factor of f(x) over a field?

[gritty latch]

uh

wdym?

[compact needle]

Do you remember the argument? (it definitely requires being over a field)

[gritty latch]

but arent I just using the fact that f(x) is the 'zero' element in my quotient?

not sure how the field comes into play

[compact needle]

You're right. All you really need to use is that f(x) = 0 in Zn[x]/f(x)

So the real question is how do you make "big" polynomials "small"?

[gritty latch]

so like if i have Z5[x]/(x-3)

and x^3+3x+1

can I not just keep replacing each x with 3 and reducing mod 5

[compact needle]

You can, but that specific argument is relying heavily on x-3 being degree 1

[gritty latch]

yea

and if its not monic i need it to be a field right?

[compact needle]

Yeah, that is helpful

But how do you make x^3+3x+1 look smaller in Z5[x]/(x^2-2)?

[gritty latch]

ok i see now why this fails in the case of Z6[x]/(2x+4)

2 is not invertible in Z6

so i cant do this

[compact needle]

It's true

[gritty latch]

but then that doesnt necessarily mean it has infinite elements does it

[compact needle]

Not in and of itself. But consider the set {x^n} for all n>0.

No difference of two different x^n and x^m will ever be divisible by 2, so definitely not by 2x+4

So they are all pairwise not congruent

[gritty latch]

im not sure how divisibility comes into play here

[compact needle]

What does it mean for g(x)=h(x) in Z6[x]/(2x+4)?

[gritty latch]

ok nvm i think i see what u mean

every polynomial i have is a multiple of 2x+4

but that set has an inf no of elements not divisible by it

so inf elements

[compact needle]

I'm not quite sure what you mean

[gritty latch]

yea idk what i just said either ignore that

[gritty latch]

every poly is of the form g(x)=h(x)(2x+4)?

[compact needle]

That's what it means to be zero, right?

g(x) is zero if g(x)=h(x)(2x+4) for some poly h(x)

Two polys are then equal if there difference g(x)-h(x)=0

i.e., g(x)=h(x) in Z6[x]/(2x+4) iff g(x)-h(x)=k(x)(2x+4) in Z6[x] for some k(x)

[gritty latch]

sorry internet died

[compact needle]

No wories

[gritty latch]

yea that makes sense, i was thinking about these quotients the wrong way

so essentially in my quotient every multiple of 2x+4 acts like a zero

[compact needle]

Yup. Just like figuring out if two integers are equal in Z6 itself

[gritty latch]

yea similiar to Zn

hm ok i see

[compact needle]

Seeing that Z6[x]/(2x+4) is infinite is a lot like seeing that Z[x]/6 is infinite

[gritty latch]

ok so

[compact needle]

(or, more to the point, Z[x]/2 is infinite)

[gritty latch]

x^n-x^m is never a multiple of 2x+4 so its not in the ideal

so it cant ever be 0

[compact needle]

Yup

[gritty latch]

which gives me many unique elements in the quotient right

since i can just vary n

[compact needle]

Correct

[gritty latch]

😅

thanks alot i appreciate it

[compact needle]

Now for the other part!

[gritty latch]

also sorry i missed this, but i just take x^2=2 since it acts like a zero

then i can reduce every polynomial to a first order one

ax+b, 5 choices for 5 5 for b so 25 elements total

[compact needle]

It's true. But what if f(x)=x^2+x+1 instead of x^2-2?

[gritty latch]

i can do the same thing no?

or similar

x^2=-x-1 or 4x+4 in Z5

so i can reduce them all to first order polys

[compact needle]

Yup 🙂

And since it's monic, you can always reduce things of degree >1 down. If it weren't monic (really if it's lead coefficient is not a unit), you would have the trouble we saw above

[gritty latch]

yea so if its not a field i need to do more work

[compact needle]

All you really need is the lead coefficient to be a unit

[gritty latch]

that can happen even if i dont have a field right?

[compact needle]

For sure. 5 is a unit in Z6

5*5=25=24+1=1

[gritty latch]

right ofc

thanks again this was very helpful

[compact needle]

You're welcome 🙂

Using these arguments, it's not too hard to give an upper bound on how big Zn[x]/f(x) is when f(x) is monic and degree m

[gritty latch]

hm

is it n^m?

[compact needle]

Indeed!

[gritty latch]

[gritty latch]

if i have the product of 2 rings RxS

i can easily see the ideals of it would be in the form of IxJ

I,J ideals of R,S respect.

but apparently the max ideals are of the form IxS, and RxJ

i cant see why

[hot lake]

what would you want them to be ?

[gritty latch]

nothing in particular

i just want intuition into why that result is true

[nova plank]

Because any IxJ will always be contained in IxS and RxJ

[hot lake]

because I1 x J1 is a subset of I2 x J2 if and only if I1 is a susbet of I2 and J1 is a subset of J2, so if you want a maximal ideal there is no other choice but to have I = R and J maximal in S or vice-versa

[gritty latch]

thank you

[odd abyss]

it's isomorphism ?

[quaint ivy]

hint: see if $(\bZ/5\bZ\setminus{0}, \cdot)$ is cyclic.

[cloud walrus]

derivada.schwarziana

[chilly ocean]

just write the cayley tables

[quaint ivy]

take for instance 2 and see what 2·2, 2·2·2, etc. evaluate to

[odd abyss]

I don't understand

why 2* 22 * 22

2^1

2^2

2^3

2^4

2^5

[quaint ivy]

a group is cyclic if it is generated by a single element, so if you can write any element in Z/5Z \ {0} as 2^n for some n you'd show that it is cyclic, and generated by 2

aaand (Z/4Z, +) is also cyclic, it's generated by 1 since 2=1+1 and 3=1+1+1

so you should be able to construct the isomorphism using that

[odd abyss]

Ok thank you, but you don't need to study (Z/4Z, +), if you study Z/5Z \ {0} is enough

because Z/5Z \ {0} it's cardinality 4

and for that ,

?

Indication ?

[delicate bloom]

are you trying to say that groups are the same if they have the same cardinality? this is false

[odd abyss]

the cardinal is important here because we use the categorization of cyclic groups

We will now show that a cyclic group is determined, with a
isomorphism near, by its order. More precisely, it is isomorphic to (Z / mZ, +) if its order m is finite
(resp. to (Z, +) if its order is infinite). In addition, any subgroup of a cyclic group is cyclic.

the most important here is that the group is cyclic

if two groups are cyclic and similarly cardinal then they are isomorphic

for that

indication?

[delicate bloom]

how do you know they're both cyclic if you don't study one of them to show the isomorphism?

[odd abyss]

Z/5Z - 0 is not cyclic ?

Don't understand why you say what I said is wrong when it's true

it's not true ?

[odd abyss]

Ok, I asked the question on a math discord in French, and they told me your questions don't make sense and that I was right

This question does not make sense you have to exhibit a generator, so you have shown that it is cyclical

[latent anvil]

at no point did you actually say something was a generator lmfao

[odd abyss]

This is not what I was criticized for.. I said how to solve the question, I was told that was not it

[latent anvil]

wait why did the messages go away im confused

i swear there was more text here

[odd abyss]

In fact, I too am confused. (WhileEnglish is not my mother tongue.)
We take back, this is my last intervention i don't want to pollute the channel
So, I come, I present my idea to solve the question. I give the definition that solves it and I say
__ Me :__

Ok thank you, but you don't need to study (Z/4Z, +), if you study Z/5Z \ {0} is enough
__ First person__ : I am being answered :
how do you know they're both cyclic if you don't study one of them to show the isomorphism?
__2nd person : __What another person responds to:
I thought this was a perfectly reasonable point, but you seemed not to think so

So I don't understand English ?

[cyan marten]

If two groups are realizable as Galois groups of some extension of Q, is any semidirect product also realizable?

[oblique river]

I don't see why that should necessarily be the case -- do you have a reason to believe it's true?

[sturdy marsh]

can you even do it for direct products?

[latent anvil]

that's an interesting question

like the issue is maybe the two extensions intersect nontrivially

[sturdy marsh]

yup

otherwise the compositum would work

[latent anvil]

So if we know it's realizable how can we change the extension?

Like can I find other extensions with the same galois group?

Feels hard

[ivory dust]

Ok i asked this in #point-set-topology but maybe this section is more appropriate.
What would be the symmetry group of a parallelapiped in R3 with L=/=W=/=H
in R2 i know you can rotate by 180
and by 0 of course, so im guessing in R2 itd be order2

[sullen aurora]

I’m not sure but maybe it’s Z_2 for parrelelpipeds in all dimensions? For generic parralelpipeds

Cause you can realize any symmetry group as a subgroup of the permutation group of the vertices i think

For any given vertex, there’s only one other vertex with the same angles and whatnot, which is the oppisite one

Once you match that up, then since all the lengths are all different, it enforces the orientation of all the other vertices

So its identity and inversion only

Hopefully that was coherent

[gritty latch]

I may just be being dumb, but i cant see why the following is true:
Since the ideal has 4 elements and the ring has 20, there must be 5 cosets

Ideal is principal generated by (1,5) and ring is Z2xZ10

[carmine fossil]

Take the ideal as a subgroup and the ring as a group. then, there are exactly 5 cosets,since no of cosets=20/4

[gritty latch]

oh yea forgot about the index stuff

thanks

[carmine fossil]

Let the ring be (R,+,•)and + be the commutative operation. Then (R,+) is an abelian group

Implying (R,+) has to be Z_10 x Z_2 by the fundamental theorem of abelian groups

[gritty latch]

you mean iso to Z10xZ2?

[carmine fossil]

Yes

[gritty latch]

how does that apply here

[carmine fossil]

(R,+) is an abelian group with 20 elements

[gritty latch]

yea

[carmine fossil]

Is your ring commutative?

I don't think we can say your ring R is iso to Z_2 x Z_10 as rings in general

[gritty latch]

let R be a UFD and let P be a nonzero prime ideal of R such that if P′ is another prime ideal with (0)⫋P′⊆P, then P′=P. Prove that P is principal.

I posted this a while back but i had to leave

still no idea how to approach it

[sturdy marsh]

Pick a nonzero element in the ideal

factor it into primes

and...

[gritty latch]

im guessing it should be generated by one of the prime factors?

[nova plank]

You did get a response last time

Go back to your old message

[nova plank]

.

[gritty latch]

o

thank you

yea im unsure how to utilize the second part of the assumption

[gritty latch]

<@&286206848099549185> could use a tiny push on this q

[wind parrot]

So say x is a nonzero element of the ideal P, and it factors into primes as x= p_1^e_1...p_r^e_r. Because the ideal is prime, at least one of the p_i ||is an element of P||

[nova plank]

Let x be an element of p, then x = up1p2...pn where u is a unit and the pi's are prime. Since P is prime, one of the pi's belongs to P.... Sniped, but sending because I already typed it all

Then consider (pi) to utilize the second part.

@gritty latch

[cyan marten]

I barely know any Galois theory, but I noticed that if we take Q(a, b) and a and b are somewhat related, but not too much, we get a semidirect product as the Galois group of Q(a, b)/Q.

[chilly ocean]

If $ f: \mathbb{D_9} \to \mathbb{S_8} $ is a homomorphism, prove that $ kerf $ is nontrivial.

What happened with bot lol?

Okay so that was the problem, I saw a solution that says: If kerf was trivial, then f(\rho) (a rotation) would have order 9, which doesn't exist in S_8

And I'm not sure if that's correct, because we know that ord(f(\rho)) | ord(\rho)

And since ord(\rho) = 9, and kerf is nontrivial, we could have ord(f(\rho)) = 3 or ord(f(\rho)) = 9

And in S_8 you have cycles of length 3

[quaint ivy]

try using the first isomorphism theorem instead

||D_9≃D_9/ker(f)≃f(D_9)⊂S_8 and then S_8 would contain an element of order 9||

[chilly ocean]

But that proof mentioned above is incorrect, right @quaint ivy

?

[quaint ivy]

well it's not a proof since you can't conclude that S_8 has elements of order 9 from "ord(f(\rho)) | ord(\rho)", as you said

that fact about the order is true but it's not enough to prove this

[chilly ocean]

Yeah I meant if it was enough to disprove his proof

That's all I asked 😄

[languid meteor]

Hey all, I was just wondering what the notation (2, 1 + sqrt(-5))^2 means in relation to ideals of rings

is it { (2a + (1+sqrt(-5))b)(2c + (1+sqrt(-5))d ) | a,b,c,d are in R}

[languid meteor]

so its like {2a + (1+sqrt(-5))b + 2c + (1+sqrt(-5))d | a,b,c,d in R}

no wait thats just the same as the original ideal

[languid meteor]

ah gotcha, thank you!

[dim escarp]

Interesting notation

[sly crescent]

Have the point groups in four dimensions been enumerated?

[glossy yoke]

appears so: https://en.wikipedia.org/wiki/Point_group#Four_dimensions

[sly crescent]

That list is missing several.

In fact, it’s missing infinitely many (the step prism and swirlprism symmetries).

It also doesn’t contain any chiral cases.

[cyan marten]

Well, if ker f = 1, then we have an equality here, not just divisibility, so this is really a contradiction.

[chilly ocean]

How is that so, isn't ord(f(\rho)) = ord(\rho) only if f is an isomorphism?

And kerf = 1 only tells us its injective

[wind parrot]

No, injective is enough to say that a homomorphism preserves the order of elements

[chilly ocean]

oh i didn't know that, thanks

[chilly ocean]

Could you send me a proof of that? Since I can't find it in textbook or online

look how far you've come, slimvesus

a few weeks you didn't know what a group was

now look at you

it's obvious

well, if you know groups, then since every ring also contains a group structure, that means you know rings too

as well as modules, vector spaces, algebras, ...

🧠

[carmine fossil]

Next should be rings

Also, Slim what do you do formally? Linear algebra?

*What did you do before group theory

[opal osprey]

Besides the trivial examples of vector spaces

I don't really think I've ever seen an example of an R-module before.

I am pretty sure algebraic geometric has lots of examples of those.

Could you tell me any?

Also, are there any easier examples besides the ones that naturally arise in algebraic geometry?

[sturdy marsh]

Modules appear in algebraic anything, not just AG

anyway, an Abelian group is a Z-module

vector spaces are modules over their endomorphism rings (and this is a fairly important viewpoint)

also, if you have a vector space with an endomorphism, then you can make it a module over a polynomial ring in a natural way

Ideals of rings are modules

abelian groups with a linear group action on them have module structures over the group algebra

[opal osprey]

Yeah I mean, there are these trivial examples such as taking the direct product of a ring R n times and choosing the ring of scalars as R itself.

[sturdy marsh]

singular cohomology is a module over the steenrod algebra

[opal osprey]

But I'd like some more concrete examples

That naturally arise when studying certain things.

[sturdy marsh]

well all of the other examples come up when "studying things"

and abelian groups are wayy more complicated than free groups over Z

for example, Q

but yeah if you want a fairly rich example, look at a vector space with an endomorphism

a very careful analysis of the resulting k[x] module structure will result in recovering the rational canonical form

in general, if you have a family of operators that form a group G, then the space is a k[G] (the obvious way ) -module

[opal osprey]

Wait, so you think about a vector space V equipped with a certain linear transformation T : V --> V?

[sturdy marsh]

and this is sorta the beginning of representation theory

[sturdy marsh]

yes

[opal osprey]

I can't see how we can think about this as a module.

Hmmm

[sturdy marsh]

a module over k[x]

[opal osprey]

So you consider a ring of polynomials of one variable to be the scalar ring.

Looks cool

[sturdy marsh]

k[x] module structures correspond to ring homomorphisms k[x] ---> End(V)

[opal osprey]

But how is the operation defined?

[sturdy marsh]

it might be instructive to think about it yourself

[opal osprey]

You take two elements in your vector space and sum it in the usual way?

[sturdy marsh]

the abelian group structure on V is the same

[opal osprey]

Yeah, I don't really see how the endomorphism plays a bigger role here.

[sturdy marsh]

What does it take to get a map k[x] ---> End(V)

what is the 'data' of the map

[opal osprey]

Let me think it straight

End(V) is a vector space

And for each polynomial over k

You are associating an endomorphism T : V --> V

Which respects the algebraic structure of both sets.

So a homomorphism.

Is that correct?