groups-rings-fields

[south patrol]

It's ok

[spice wagon]

ah it’s a good way to remember that thanks

[thorn jay]

0 is convenient because it's the neutral element of the underlying group, and this means that kernels of homomorphisms (as equivalence relations) are uniquely determined by what they send to 0

In the Netherlands and in Germany they call fields "bodies"

[south patrol]

Kernels are very much a additive thing. Like when you have map A -> B of rings you can view it as a map of A-modules and then the kernel is the kernel of this map of modules

[spice wagon]

deepl translated it like that but I remembered it was field in english :))

[thorn jay]

[south patrol]

Körper

[thorn jay]

Yurr

[south patrol]

What is it in Nederlands

Korps

[spice wagon]

I’m not into module theory yet 😩

[south patrol]

Idk

[south patrol]

Dww

Well in your case you can do it as vector spaces over K aha

[spice wagon]

i have the book of Serge Lang about algebra but never read it lol

[thorn jay]

Lichaam

[south patrol]

Ah I remember seeing that

I wonder if it is related to uhhhh

[thorn jay]

No clue Abt the etymology

We use like "hemellichaam" (celestial body) too

[south patrol]

Yeah so in german there's Leiche for like corpse and stuff lol

And Leichnam

I'm sure it is the same origin

[thorn jay]

NO trace of either soma or Körper

[south patrol]

Nice

[thorn jay]

Mm right

[south patrol]

R u Dutch I assume lol

[thorn jay]

Yis

[south patrol]

Nice

[thorn jay]

I love vector spaces because they're actually just numbers with weird morphisms between them

[south patrol]

Real

Finite dimensional ones you mean

😏

[thorn jay]

no, infinite dimensional vector spaces do NOT exist THEY ARE NOT REAL

Yes I am SCARED of them

Like wdym they can have injective nonbijective endomorphisms

[south patrol]

Lol

Z scares u

[thorn jay]

Yes

Well, not as a ring

[jade mason]

What about as a Z-vector space

[spice wagon]

what about as a F_p vector space

[thorn jay]

"vector space"

[thorn jay]

That's not Z anymore

F_p vector spaces have torsion

[south patrol]

What about as a S-module

[jade mason]

p=0

[spice wagon]

wtf is a torsion..

[south patrol]

Where S is the sphere spectrum

[thorn jay]

I have heard that name before..

No clue what it is

[thorn jay]

F_0, the field with one element??? No way

Wait no that would be F_1

[jade mason]

No, F_p = Z/(p)

F_0 is Z already

[thorn jay]

.

[jade mason]

also known as the field of zero elements

[thorn jay]

[thorn jay]

Every element of an F_p vector space must have finite order

I.e. add it to itself p times and you get 0

[tardy hedge]

VECTOR SPACE!? HOW COULD HE!!!

[south patrol]

An Fp vector space is equivalently an abelian group A with pA = 0

Kinda funny when like

"Structure becomes a property"

[spice wagon]

ah

[tardy hedge]

I like i like

[south patrol]

Similarly with Q-vector spaces

[thorn jay]

It's interesting

There's a universal algebraic reason, naturally

(being that quotients correspond to the satisfying of equations)

[tardy hedge]

Man those equations love being satisfied

[thorn jay]

(hello, algebraic geometry department?)

[south patrol]

Lol

I was gonna say something but it is just redundant basically lol

[thorn jay]

Me all the time

[cloud walrus]

tm

tm

[spice wagon]

is one definition used more often than the other?

[velvet hull]

they're equivalent

functionally it doesn't really matter, (left) modules are (left) modules are (left) modules

[thorn jay]

They're the same definition

[spice wagon]

ah lol

alright

[south patrol]

An R-module is an abelian group M equipped with a ring map R -> End(M)

[velvet hull]

an R-module is an R-action

[mighty kiln]

An R-module is an Ab-presheaf of R^op

[knotty badger]

Oh so a functor from R^op to Ab…?

What’s R here though

[rocky cloak]

A ring is a preadditive category with one object

[knotty badger]

Oh so this is just “an R-module is a ring acting on an abelian group”

[rocky cloak]

Yup

[knotty badger]

Cool

[rocky cloak]

You can also replace R by the category of fg projective modules if you like your categories additive

[knotty badger]

I’m good thanks

[spice wagon]

if R is a commutative field no?

[thorn jay]

Fields are always commutative

[mighty kiln]

Le corps commutatif

[spice wagon]

not in my book

[mighty kiln]

What book is it

[elfin wraith]

French moment

[spice wagon]

Arnaudies/Bertin-Groupes algèbres et géométrie Tome 1 , 1998

page 43

[mighty kiln]

"Corps" should translate to "division ring"

So "corps commutatif" is "field"

[spice wagon]

ah

[mighty kiln]

It's like how "positif" is "nonnegative"

[spice wagon]

its written division algebra when it’s not commutative

[thorn jay]

That's only when it's an algebra over a commutative ring, usually a field

Unless that's different too

[little shadow]

why is it necessary to use induction here? If p = p(x) is a zero-divisor then $\exists \text{ non-zero } q = q(x) \in R[x]$ such that $pq = p\sum_{i=0}^m b_i x^i =\sum_{i=0}^m (pb_i) x^i = 0$. Since q is non-zero there is an $i \leq m$ s.t $(pb_i)x^i = 0 \Rightarrow pb_i = 0$

[cloud walrus]

dompa

[rocky cloak]

Just because a sum of stuff equals 0 doesn't mean each summand is 0, so you would need some argument for that

[little shadow]

ah okay i see the error now

or actually i dont, $a(x) + b(x) = 0 \Rightarrow a(x) = -b(x), deg(-b(x)) = deg(b(x))$ and $pb_i x^i$, $pb_j x^j$ have distinct orders unless both are equal to 0 (which in itself would prove the only if part)

[cloud walrus]

dompa

[delicate orchid]

I believe jagr is looking for the statement "x^j, x^i for i != j are R-linearly independent"

[rocky cloak]

That's not true though.

The degree of pb does not have to equal the degree of p.

For example 2(2x+1) = 2 in Z/4[x]

[rocky cloak]

I mean that would be helpful if we were doing a combination with coefficients in R.

But these are combinations with coefficients in R[x]

[delicate orchid]

my case is that the i = 0 term of the above sum is enough. You're not going to get any constant terms from any of the other terms, thus pb_0 = 0?

am I smoking that good gas station zaza here

[rocky cloak]

Well, that argument shows that
p0 * b0 = 0.

Then you can use induction to get the rest

[rocky cloak]

Which is what happens here (except they start with the leading term)

[delicate orchid]

right yeah I've just done the full thing on paper. I basically did just do the argument we were trying to avoid backwards

whoopsie doodles!

[crystal vale]

So, the automorphism of the group preserve the maximality of the subgroup, right?

I mean if H is a maximal subgroup of G then f(H) is also maximal subgroup of G, f is an automorphism of G

[south patrol]

For any two subgroups H and K, H is contained in k iff f(H) is contained in f(K)

[crystal vale]

I see

So I can use this fact to show that the Frattini subgroup is a characteristic subgroup, right?

[woeful sage]

This is from a CMI question paper? I remember seeing something about the Frattini subgroup

[crystal vale]

No

[candid patrol]

Yes

[crystal vale]

Okay thank you

[ivory ore]

need to find all the rings with identity of 4 elements

[crystal vale]

There are 4 rings up to isomorphisn

[ivory ore]

so, we have 0, 1, a, b

yeah there are 4

[crystal vale]

First one Z/4Z, Z/2Z × Z/2Z

Now think about Z_2[x]/< ax^2+bx + c >

[ivory ore]

i was thinking about setting up the multiplication of these elements

a^2 = a gives Z/4

oh yeah ping me in that server

[ivory ore]

would setting up a^2 to be different elements (0,1,a,b) just work?

[rocky cloak]

Whether or not you get Z/4 is mostly dependent on what 1+1 equals in your ring

[ivory ore]

i mean there are two abelian groups of summation

[rocky cloak]

Yes, but a^2 = a doesn't really tell you about the additive group

Also 0 and 1 are the only elements of Z/4 with a^2 = a

[ivory ore]

yeah, right, i was just thinking about multi

[ivory ore]

would you encourage to check this with two groups

[rocky cloak]

I would check the two different groups seperately

In general to classify finite rings it's good to split into cases based on what the characteristic is

Then it usually becomes easier to classify

[crystal vale]

Great idea

[ivory ore]

alr

[crystal vale]

But say, it has characteristic 2, I know there is a field of order 4, but what if someone doesn't know so how he/she come up with it?

[rocky cloak]

At that point just writing down
{0, 1, a, a+1} and asking what a^2 can be will get you everything

[crystal vale]

I see

But what if our ring size is bigger?

[rocky cloak]

Or really the same as you did before think about
F2[x]/(x^2 + ax + b)

[crystal vale]

I mean there is some ring, which is not coming from F2[x]/ (x^2 + ax + b ), right?

[rocky cloak]

there is Z/4

[crystal vale]

Yes

Because characteristic 4

[rocky cloak]

Like if you're characteristic n and commutative, then you're equal to
Z/n[x1, x2, ...] modulo some ideal

[crystal vale]

Oh

[rocky cloak]

You can bound how many xs you need based on the structure of the additive group

Like for Z/2 x Z/2 you just need one extra, because then 1 and x already generate the additive group

[crystal vale]

I see

[wispy light]

In a commutative ring, a non-unit generates a proper ideal. In a non-commutative ring, is it still true that the two-sided ideal generated by a non-unit is properly contained in the ring?

[tough raven]

No - for example, any non-zero two-sided ideal of M_n(k), where k is a field (or a division ring), is improper.

Rather x is a left (resp., right, two-sided) unit iff Rx (resp., xR, both Rx and xR) is equal to R.

[wispy light]

Ah, I see, thanks. I have zero intuition for noncommutative rings.

So matrix rings have exactly two two-sided ideals, and yet they aren't division rings. Very much unlike the commutative case, where every commutative ring with exactly two ideals has to be a field.

How complicated can a noncommutative ring with exactly two two-sided ideals be?

Consider, for a noncommutative ring R, the following three statements:

1. Every ascending chain of left ideals is eventually constant.
2. Every ascending chain of right ideals is eventually constant.
3. Every ascending chain of two-sided ideals is eventually constant.
   Clearly, 1 and 2 together imply 3. But 3 doesn't imply 1 and 2, right?

[rocky cloak]

So if a ring is a division ring iff it has exactly two left ideals iff it has exactly to right ideals.

Rings with exactly two two-sided ideals are called simple rings. The only artinian simple rings are Mn(D) for a division ring D, but non-artinian simple rings can get wacky. The Weyl algebra is one example.

[rocky cloak]

This is correct yes, either 1 or 2 will imply 3, but there are no other implications amongst them

[wispy light]

Thanks to you too!

[keen badge]

Can someone help me find a polynomial in Q(π) that vanishes at x=sqrt(i)+sqrt(π)?

[tribal moss]

Start with x = sqrt(i) + sqrt(pi).
Get rid of the square root around the pi by isolating sqrt(pi) on one side and squaring both sides.
Then get rid of the square root around the i by isolating sqrt(i) on one side and squaring both sides.
Then get rid of the i by collecting all terms with i in them on one side and squaring both sides.
You'll get lots of terms, plenty of opportunity for silly sign errors, etc., but at the end I think you ought to end up with no worse than degree 8.

[keen badge]

Great thx

[twilit wraith]

hello, i am wondering if i could have some help understanding how i should begin to think about proving this

im thinking about just representing the subgroups in terms of their generators but im not sure

maybe i should instead think of H_i like a set and find the smallest construction that both contains all the H_i and is a subgroup

[wraith cargo]

Do you know of the concept of a subgroup generated by a subset of the group

[glad osprey]

Hint: think about intersecting a collection of certain subgroups. This same construction is used generally when you want to find the smallest X satisfying P, assuming P is preserved by intersections, like the composite of subfields or the closure of a set in topology

[twilit wraith]

i think thats what i started doing

[twilit wraith]

i think i understand what that means

like theres a subgroup generated by some elements from the group

[twilit wraith]

This is what i have so far

[arctic trail]

degree 8 is optimal by the tower rule

[arctic trail]

f(x) = 0

[glad osprey]

Nice, you're definitely on the right track 👍 it's a bit hard to parse the definition of S, but I think it's the set of all subgroups H such that every Hi is a subgroup of H, right?

[twilit wraith]

yes

is there a way i should rewrite that?

i was wondering if my notation was correct

[arctic trail]

tip do a pagebreak

[glad osprey]

No I think it's fine, it's just the subset vs subgroup notation, plus lack of universal quantifier, but it's no big deal

[twilit wraith]

i see

[arctic trail]

you're starting really low, so might as well do a pagebreak

[twilit wraith]

yeah ur right

im not fully sure how to show closure

this is only day 3 of my abstract algebra class so im still pretty new to all this

[arctic trail]

That's fine

What you do is the following

take a,b in H_0

by definition this means a,b are in H_i for all i in I

but if each H_i is a group then ab^(-1) is in H_i for all i in I

[twilit wraith]

wait doesnt that cover every requirement?

i remember seeing the ab^(-1) trick before

[arctic trail]

the ab^(-1) is basically a trick for simple proofs

in general it's best practice to do

• ab is in H
• a^(-1) is in H

[twilit wraith]

i see

[arctic trail]

it's just that in this case it follows immediately for the trick

in general the easiest way to show ab^(-1) is in H usually is to show ab is in H for a,b in H and showing b^(-1) is in H lol

[glad osprey]

I need to go to bed, but I think you got this, just show closure, then the fact that it's the smallest and unique should be relatively straightforward. Good luck!

[arctic trail]

have you had topology?

[twilit wraith]

i see

not yet

but i am taking it in the fall

[arctic trail]

damn

[twilit wraith]

technically my current research is on topology but they have us avoiding the technical stuff for now

[arctic trail]

there is a way to connect this to topology

[arctic trail]

but it's not really useful

just a neat fact

so I'll refrain from saying it

[glad osprey]

No, say it

I can't go to sleep on a cliffhanger like that

[arctic trail]

Oh uhh, call the groups that contain all of the H_i closed

[twilit wraith]

(just real quick, H_0 is unique because if there were another H_1 that does everything H_0 does, we just have both of these sets being the smallest)

right?

[arctic trail]

this defines a topology

not a useful one but still one

then H_0 is quite literally the closure of the union of all H_i for this topology

[twilit wraith]

oh wait ive seen the definition of a topology

isnt it just like some sort of finite intersection

at least thats one of the requirements

[arctic trail]

finite intersection is for open sets

[glad osprey]

I see, nice

[twilit wraith]

i see

[arctic trail]

yeah, and then basically the minimum follows from existence of closure in a topological space

Multiplication by an element in H_0 (or its inverse) defines a homeomorphism

And all closed sets are fixed under that homeomorphism

as the groups that contain H_0 form a basis for the closed sets and are fixed by this action

so H_0 is a group

So I guess you can actually do this from just topology @glad osprey

[twilit wraith]

ok heres the proof done

at least i think its done

i dont see an error in it so i think im in the clear

[arctic trail]

You're missing H_0 being not empty

You show S is not empty by G in S

[twilit wraith]

i have that

oh i see

i just didnt say it clearly enough

[arctic trail]

why so many lines?

[twilit wraith]

well if G is in S then we have H_0 just being the union of all H_i right

[arctic trail]

no

[twilit wraith]

im not sure

[twilit wraith]

sorry i should say

if G is the only element of S

[arctic trail]

still no

you just need to see H_0 has an element

[arctic trail]

I mean, you're doing paragraphs, isn't this LaTeX?

[twilit wraith]

yes

but i mean i feel like having paragraphs with the lines so close to each other makes it kinda hard to read

[arctic trail]

I'll show you how I'd write it and you can see if you like it

[twilit wraith]

sounds good

oh wait im being silly

G being the only thing in S would make H_0 equal to G

[twilit wraith]

alright

[arctic trail]

oops I wrote "Notice that all since"

[arctic trail]

that happens sometimes when you're rewriting stuff

[twilit wraith]

btw how do u do that thing where the math is in the middle

[twilit wraith]

yeah that happens to me

[cloud walrus]

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

[twilit wraith]

thank you

btw i was luckily able to get the not empty part myself

since G is in S its never the case that all elements in S are disjoint and so H_0 is not empty

[arctic trail]

sure

You just need to make sure to write it

[twilit wraith]

alright

thanks for the help

hopefully the next 3 problems arent as bad

[arctic trail]

The reason as well that I did the math in the middle

wasn't really about showcasing that

it was more so I would avoid lengthy walls of just text

[twilit wraith]

oh i also have to add in the uniqueness part

[arctic trail]

I'm deleting this from my pc, I'll send the file itself as well

What was the guy

ring with p elements?

[ivory ore]

yeah i thought i might be interupting the convo which is why i deleted

so rings with p element

if 1 \in R

then it is the field

[arctic trail]

sure

[ivory ore]

for any other case(s) we have to look without identity

any hint for this?

[arctic trail]

you want to classify non-unital rings with p elements?

[ivory ore]

yeah

i mean the exercise mentions all the rings with p elements

but with identity it's straightforward

[arctic trail]

Hint: Additive structure

Hint2: ||Group theory||

Hint3: ||Lagrange's theorem||

@ivory ore do you want more hints?

[ivory ore]

i'm trying to think

[ivory ore]

well let's take all the non-zero element and take multipliction in whatever order, it should not give back any non zero element

in non-zero elements again product of all the elements but a specific a_i should give a_i or 0

implying (a_i)^2 = 0

again if we do the same by taking 2 elements out of the string say a_j and a_i

the product should give a_i or a_j or 0

well i have exhausted too much time and couln't figure out any multiplication that works

[crystal vale]

If g^m ≠ e and g^m has order k, can I say order of g is mk?

[spice wagon]

the order of g must divide mk but it is not necessarily equal to mk

take Z/12Z and g=2

[crystal vale]

And what should I take m here?

[spice wagon]

m=2

[crystal vale]

g^2 = 4, 4 has order 3, and g has order 6, so 2 × 3

[spice wagon]

ah

take m=4 then lol

g^4=8, the order of g^4 is 3 but the order of g is 6≠12

[crystal vale]

Oh g^4 = 8, 8 has order 3, but g^2 has not order 12

Got it thanks

I have to prove if G is a finite group and N is a normal subgroup, if G/N has an element of order n, then G has also an element of order n. Any hint?

So assume gN has an order n in G/N, then g^n in N.

If g^n = e, then we can conclude that |g| = n.

If g^n ≠ e, then g^n = n1 for some n1 in N.

Since G is a finite group so g^n has a finite order

[chilly ocean]

isnt it trivial

[crystal vale]

How?

[chilly ocean]

well sorry i was thinking something else

[chilly ocean]

you should try the case when n is prime

[crystal vale]

I think then we can do it by using Sylow

[chilly ocean]

when n is prime that case i found intresting

i am getting some infinite descent for contradiction

[crystal vale]

Oh

[chilly ocean]

without sylow

[crystal vale]

Yeah I know

[crystal vale]

I think we can do it by contradiction, taking the corresponding mapping G -> G/H, it is onto mapping

Suppose there is no g in G such that |g| = nk, for some k in N.

Since the mapping is onto, |gH| divides the |g|, so if there is some element of order n then n must be divide |g|, so we get the contradiction.

We can show directly

Take |gN| has order n, then |gN| divides |g|, so n divides |g|, |g| = nk, so |g^k| = n.

[queen quarry]

Elements in N have finite order, and since g^n \in N, we have k such that (g^(n))^k = 1
the element g^k has order n

oops thats what you did

[crystal vale]

So g^n has order k how it implies g^k has order n?

Without using group homomorphism how can I see |gN | divides |g|?

[queen quarry]

g^(nk) = g^(kn) by commutativity of integer multiplication

[crystal vale]

Yes but how it implies g^k has order n?

[queen quarry]

(g^k)^n = g^(kn) = g^(nk) = (g^n)^k

wait

[crystal vale]

It implies |g^k| divides n

[queen quarry]

ur right let me think

[rocky cloak]

Say g has order m=nk, then g^k has order n

To see this note that g^kd =/= e for d<n

[queen quarry]

ok in general its a false statement, as in Z/pZ k*1 (k<p) has order p, but p*1 has order 1

[crystal vale]

Yes

[crystal vale]

This one?

[rocky cloak]

|gN| is the smallest d with g^d in N, that should be clear.

Say g has order n, then certainly g^n is in N.

Let ad + bn = gcd(d, n)
then g^ad * g^bn = (g^d)^a is in N.

So d <= gcd(d, n)

[crystal vale]

I got it

Thank you, jagr

[lyric berry]

Hello! I have a question about semi simple actions. Let’s say a group A acts on a group V, and that this action is semisimple. In their book, Kurzweil and Stellmacher wrote that the restriction of the action of A tu a subgroup that is not normal may not remain semisimple. And for that they say that the exercise 1 in the picture justifies it. Can someone help me with that?

[rocky cloak]

What is pi and CA(V) here?

As for your semisimplicity statement:

Let ||S3 act on C2xC2 by permuting the 3 non-identity elements. Then this is a simple S3-module||
||If you restrict to a subgroup of order 2, then the resulting module is not semisimple||

[lyric berry]

It is the set of elements of A that fixes every element of V

[rocky cloak]

And what is pi?

[lyric berry]

Prime divisors

Prime divisors of the order of the group*

[rocky cloak]

So replacing A by A/C(V) you can assume C(V) is trivial.

If p divides the order of A, then A has a subgroup of order p. Then the question is how can Cp act on an elementary p-group?

The answer should be trivially or non-semisimply

I don't know what tools you have, but how I would approach it would be that
||ZG acting on an elementary p-group would factor through ZG/(p) = FpG.||
||FpCp = Fp[x]/(x^p), and you can use the structure theorem for PIDs to tell you what all the modules are. The only simple one being the trivial module||

[lyric berry]

What are ZG and Fp for you?

[rocky cloak]

ZG group ring, Fp field with p elements

I guess I changed A to G

[crystal vale]

So we can make a field of p^n, n in N, just by taking the irreducibile polynomial of degree n+1 in Z_p[x], right?

But how can I always generate irreducibile polynomial of n in Z_p[x]?

[glad osprey]

Irreducible of degree n, not n+1. You can prove that there is an irreducible of degree n for every n in Z_p[x], but all the proofs I've seen have been non-constructive. I think it's pretty non-trivial to find irreducibles of arbitrary degree

[cobalt heath]

What is Z_p for you guys

😈

[crystal vale]

Z/pZ

[crystal vale]

Yeah

[thorn jay]

Its whatever I need it to be

[cobalt heath]

Good answer

[thorn jay]

If I'm referring to the ring Z/pZ without prior context then I'd use F_p though

[cobalt heath]

The convenience to shorten Z/(q) is worth angering number theorists

[thorn jay]

Everything is worth angering number theorists

[velvet hull]

If your goal is to just construct a field of order p^k, you let it be a splitting field of x^p^k-x over Fp

[crystal vale]

Oh

[velvet hull]

Prove that the collection of all roots in the extension are closed under field operations

And then Fp adjoin all the roots is a field

[crystal vale]

I don't know field theory

[velvet hull]

Hmmmmm I can’t think of a clean way of doing it without field theory

[velvet hull]

I guess for an extremely janky way to do it is to consider the set of roots of x^p^k-x in the algebraic closure of Fp

And you can prove that the algebraic closure of a field always exists without needing any field theory

[south patrol]

P adics

[rocky cloak]

I wouldn't call that janky, it's pretty natural.

Just sort of requires knowing field theory

[south patrol]

Of course also no need to construct the algebraic closure there

Just adjoin roots of that polynomial

[velvet hull]

I said that but this is their response

[rocky cloak]

If you don't know how to adjoin roots to split a polynomial, then you surely don't know how to adjoin roots to split every polynomial

[white oxide]

Does this work for showing that $\text{Spec } A$ irreducible implies $N(A)$ prime? Suppose that $\text{Spec } A$ is irreducible and let $N(A)$ be the nilradical of $A$. Suppose there was $ab \in N(A)$ with $a$, $b \notin N(A)$. Then there exists prime ideals $\mathfrak{a}$ and $\mathfrak{b}$ with $a \notin \mathfrak{a}$, $b \notin \mathfrak{b}$. Thus, $\mathfrak{a} \notin V(a)$ and $\mathfrak{b} \notin V(b)$. Therefore, $X \setminus V(a)$ and $X \setminus V(b)$ are nonempty open sets. Let $J \in X \setminus V(a) \cap X \setminus V(b)$. Now, $ab \in N(A) \subseteq J$, but $a$, $b \notin J$. This contradicts the fact that $J$ is prime.

[cloud walrus]

okeyokay

[white oxide]

Oops I should mention here that X is Spec A

[fringe marten]

Hello, i have a question about proving the existence of irreductible polynomials of degree n over a finite field F_q.
The problem is that most of the literature on the internet, i either don't understand or it uses the existence of the field F_q^n.
Do any of you know if it is possible to first prove the existence of irreductible polynomials?

I mean to prove the existence of F_q^n using the irreductible polynomials

[glad osprey]

Once you have an irreducible p(x) of degree n over F_q, then F_q^n is just F_q[x]/(p(x)). Did you mean something else, like proving the existence of an irreducible of degree n in the first place?

[fringe marten]

Yes exactly, i wanted to prove the existence of an irreductible of degree n first and then construct F_p^n because:
-idk how to construct F_p^n any other way
-it makes more sense to do it this way in my project

[glad osprey]

The usual proofs first prove the existence of F_q^n as the splitting field of x^q^n - x, then prove that every finite field F_q^n is isomorphic to F_q[x]/(p(x)) for an irreduible p(x), thereby indirectly proving the existence of an irreducible of any degree. I'm not familiar with a more direct way to prove this, but maybe there is

[fringe marten]

i know the proof of the existence of p(x) from the field but i don't like it

[glad osprey]

Keith Conrad's notes are a really nice reference for this btw: https://kconrad.math.uconn.edu/blurbs/galoistheory/finitefields.pdf

[fringe marten]

i'll check that

i think the key for my understanding is here: how do we get this splitting field in the theorem if we don't use irreductible polynomials?

do you see what i mean?

the lemma uses irreductible polynomials to prove the existence of the splitting field

and then: "general theorems guarantee the existence of such a field" doesn't help me much

[dim widget]

if you want I think you can prove the existence of an irreducible polynomial by counting irreducible polynomials of degree d with an inclusion-exclusion argument

[dim widget]

the point is you take the splitting field of t^{p^n} - t. This gives L/F_p some field extension. Then L^* is equal to all of the (p^n - 1)th roots of unity thus L has size p^n -1 + 1 = p^n

[fringe marten]

the problem is that idk how to properly define field extensions without those irreductible polynomials

[glad osprey]

Neither lemma 2.1 or 2.2 assumes the existence of any irreducible polynomials. You can construct the splitting field by just adjoining roots, and this always works regardless of what irreducibles exists. The crux of theorem 2.2 is showing that the splitting field of x^q^n - x has exactly q^n elements. When adjoining roots you might be forced to add "too many" roots: the splitting field of a polynomial of degree n may be an extension of degree n!.

[dim widget]

well if I have any polynomial p(x) it has irreducible factors \prod_i p_i(x) and each of those forms generates a field F_i (one could either take the splitting field, or just adjoint a root of a p_i), then I can take the field generated by all of the elements of the F_i and I get F/F_p

[glad osprey]

Maybe this can help: https://kconrad.math.uconn.edu/blurbs/galoistheory/splittingfields.pdf

[fringe marten]

how do you extend the field structure to the roots we make up?

[dim widget]

What do you mean?

[fringe marten]

if i got it well, we make up roots that we add to our field to create a new one, but for it to be a field we need to extend addition (shouln't be hard) but also multiplication to the new elements

[dim widget]

the ring you are getting is F[x]/(p_i(x)) where p_i(x) is irreducible

it turns out that if p_i(x) is irreducible that ring is a field

[fringe marten]

yesyesyes

[dim widget]

and it's isomorphic to F(a_i) where a_i is any root of p_i(x)

[fringe marten]

lemme explain my problem

-i know how to construct F_q^n knowing an irreductible polynomial of degree n exists over F_q

[dim widget]

Yes I know, but you don't need that

[fringe marten]

-i know to provethe existence of irreductibles knowing F_p^n exists

[dim widget]

you can construct fields just by taking a compositum of these fields F_i

[fringe marten]

but idk how to prove the existence of any of them independently

[dim widget]

you take X^{p^n} - X

[fringe marten]

yes

[dim widget]

factor into irreducibles

take the splitting fields

take the field generated by all of those fields

[fringe marten]

ohhhhh

[dim widget]

within a fixed algebraic closure

[fringe marten]

ok

[dim widget]

then every element of that field satisfies X^{p^n} - X

[fringe marten]

how didn't i think of that

[dim widget]

so you can show that it can only have p^n elements

[fringe marten]

it is very clear now that you say it

[dim widget]

and then you can show that it has at least p^n elements

[fringe marten]

yeeeee

really

ty

[dim widget]

np

[lyric berry]

Hello! Im having a doubt about what I have overlined here (A is a group acting on a group V). My problem is that for a normal subgroup B of the semi direct product AV with B\subset V, we have that B is A-invariant AND normal in V. But this seems then to imply that every minimal A-invariant subgroup of V is automatically normal in V…

[fringe marten]

i think it's one of those things i really struggled to understand but won't be able to hold back from calling it trivial from now on

[dim widget]

I think V is abelian here

so every subgroup is normal

[lyric berry]

Yeah that’d make sense

[glad osprey]

Every mathematician's journey "wtf is this, this makes no sense" - spends 3 weeks trying to understand - "oh it's trivial"

[potent condor]

Wow I’ve completely mastered the first step, I’m halfway to achieving mathematical greatness

[rocky cloak]

"After carefully considered your question I have concluded it is in fact trivial"
-professor getting back to you about your question from last week

[true bolt]

bonus points if it also took the prof a week to solve it

[cloud walrus]

esca

[coral spindle]

No, because all the ideals listed are equal to Z(1/2)

1/2 is a unit in Z(1/2)

[quartz wind]

oh right whoops

thanks boytjie

[coral spindle]

Np

I think you can argue that any localisation of a Noetherian ring is Noetherian

(at a multiplicative set, I mean -- not just an ideal)

Yeah ok yes

This is true

So we immediately get that Z(1/2) is Noetherian.

[warm dove]

I don't quite get this argument

namely, I don't see where separability is proved in 3 -> 1, and I also don't see why each of the automorphisms produces distinct \alpha_i

[elfin wraith]

Is this true for all rings or all commutative rings? I know it’s true for left Noetherian semiprime rings, but we only covered localisation for domains in my noncom class so I’m not sure of the general theory

[coral spindle]

I have no idea how noncomm localisation works

I'm assuming commutative here

I remember reading a chapter in a book that showed how completely fucked noncomm localisation is but I simply do not remember how it works

[next obsidian]

“I know it’s true for left Noetherian semiprime rings”

Bro why tf do you know that

[next obsidian]

So I think here we don’t say every automorphism produces a distinct root

Consider the orbit of alpha under G, this is a set of elements, all of which are roots of f, maybe some automorphisms send alpha to the same alpha_i but idc

[warm dove]

How do we know deg g < deg f?

I undertand everything but that I think

[next obsidian]

Because g is just products of x - alpha_i where alpha_i are roots of f

[warm dove]

oh wait im fucking stupid

[next obsidian]

Yeah

😘

[warm dove]

thanks lmao

[next obsidian]

🫶

[elfin wraith]

Even the “nice” special case I learned in class was just

[thorn jay]

how does it work

what property of localisation is kept

[elfin wraith]

It’s a theorem of Goldie lol I took a course on non com Noetherian rings

[elfin wraith]

You just kinda construct the localisation via a universal property, at least for domains, by building what you want out of the free ring, and what you get is sort of a mess, but under nice enough conditions (usually the Ore condition and some other adjectives) things behave somewhat ok

[thorn jay]

gotcha

[elfin wraith]

Generally it just is horrible though, like in general you can write things as simple fractions, it’s basically awful

I loved that class but the time we spent on localisation was just painful, and I’m still not sure what the point of it is

Like I’m yet to come across somewhere it’s actually useful

[next obsidian]

It gives you a model to base the construction of the derived category as a localization of a category

[thorn jay]

I tried to define localisation for certain types of general algebraic structures once, for some algebraic geometry thingy

I fear I hath angered the gods

[next obsidian]

That’s the only time I’ve ever seen localization of noncomm rings mentioned

[elfin wraith]

Model as in model theory? I understand essentially none of that sentence beyond knowing derived categories are a thing that exist lol

[next obsidian]

Like

When you think about how you would do it

You take a note from how the ore condition for noncomm rings is and how they do it

Or something

Idk

[rocky cloak]

A ring is a preadditive category with one object goes brrr

[elfin wraith]

I need to learn category theory properly soon I know such a strange collection of things about it and absolutely do not see the big picture

[knotty badger]

do it

[elfin wraith]

Ideally in my masters!

[thorn jay]

heh

ideal

[elfin wraith]

Heh

[knotty badger]

same time I learnt about it!

[elfin wraith]

I’ve learned about abelian and monoidal categories for other classes before ever properly having done the basics, so I kinda get it all but no where near as well as I’d like

More so when it comes to limits and stuff, I’ve just never had to deal with them

[knotty badger]

Universal properties are a great place to start I’d say

[elfin wraith]

To be fair looking at Leinsters basic category theory I’d say it’s only really the last 3 chapters I don’t know about so that’s probably not too bad, my foundations are potentially a little less spotty than I thought

[tough raven]

What it does show is that ℤ[1/2] is not a Noetherian (or finitely generated) ℤ-module.

[next obsidian]

Wut

[tough raven]

ℤ[1/2] is not a finitely generated ℤ-module.

[next obsidian]

Yeah but how does what Esca wrote prove that

[dim widget]

using the same notation to refer to modules that would be an ascending chain that doesn't stabilize

[tough raven]

I meant more that that's what formalising the intuition of what was written leads to, IG

[next obsidian]

Oh yeh

Nvm I gets it

[dim widget]

I guess you have to accept Z[1/n] as good notation for \frac{1}{n} \cdot Z

for it to work

[next obsidian]

Yeh

The benefit is

I can use my mind to imagine that

[quartz wind]

yeah and it would be nice to say thats what i meant all along but the exercise is finding a commutative ring thats non noetherian lol

[next obsidian]

Just pick a really big ahh ring

[keen badge]

If I have a group G=<x1,x2,dots,xn: r1,r2,dots,rm> where r1,dots,rm are relations.
Is there a way to find the derived group G'?

[wraith cargo]

G' is just the abelianization right?

so you'd just add in the relations [xi,xj] for all i!=j

[keen badge]

If G^ab is the abeliazation, then G=G/G'

[keen badge]

so it is enough to check for generators

ok, cool, thx

[wraith cargo]

ah okay

actually wait

yeah okay I had it backwards

[keen badge]

so what we said is true, yeah?
G'=<[x,y]: x,y different generators of G>

[wraith cargo]

I believe this is true
I can't immediately see what happens to the relations

but I think nothing really substantial happens

[keen badge]

yeah I also think it works. thx

[next obsidian]

Is it clear that it’s normal tho?

[dim widget]

yes

[next obsidian]

If it’s normal then I think you win

When you quotient by those relations you can see the quotient is abelian

Cuz this makes all generators commute

And this is also includdd inside the commutator subgroup

[quartz wind]

does the set of all countably infinite sequences with sum and prod defined index wise work

im really new to all this so apologies if this sounds dumb

the chain being $\langle (1,0,0,\dots)\rangle \subset \langle (1,0,0,0,\dots), (0,1,0,0\dots)\rangle\subset\cdots$

[cloud walrus]

esca

[tardy hedge]

Hahahahahhaha

A big ahh ring

[next obsidian]

You’re onto something

Sequences of what tho?

[quartz wind]

$\mathbb{Z}^\bbN$

[cloud walrus]

esca

[quartz wind]

thanks chmonkey

[next obsidian]

It actually doesn’t matter

You could take a product of an infinite number of whatever nonzero rings

And this works

[twilit wraith]

once again i am coming here to ask for help

seems like everyday im introduced to a new problem that i have no idea how to solve

i know that every element of G has an order that divides 35 by the last equation

[dim widget]

the second hypothesis isn't even necessary

actually all the hypotheses aren't necessary

except for being of order 35

[dim widget]

what are the possible orders of an element of G then?

[elfin wraith]

This is an odd question tbh. I think I would just argue that there is an element of order 5 and an element of order 7 by Cauchy, they’re coprime hence the order of their product is just the LCM

As TTEG points out there’s no reason to use the last hypothesis, and even assuming abelian feels odd since it makes it quite a bit easier

Though I guess if you don’t know sylow you might be stuck

35 is a pretty small and simple number though so maybe there’s a nice counting argument without appealing to Sylow if you don’t assume abelian

[twilit wraith]

well i will say

this has to be done without lagrange or sylow

[dim widget]

sylow isn't useful, but without lagrange explains the weird hypothesis

[elfin wraith]

Even without it you don’t need that assumption, I’m not sure how you even could use it

[elfin wraith]

Oh wait am I being dumb?

[dim widget]

isn't lagrange's theorem that the order of a subgroup divides the order of the group?

so the second hypothesis "gives you lagrange" even if you didn't already know it was true

[elfin wraith]

Yeah but don’t we just need Cauchy? Though I suppose Cauchy tends to come after Lagrange

[dim widget]

Yes I guess you could just use Cauchy, strange to even call them different theorems to be honest

I actually can't think of a good way to prove cauchy without lagrange

[elfin wraith]

In any case, I do think Sylow is useful, you can even drop the abelian hypothesis I think and just make a counting argument on the size of the p groups

[dim widget]

but I think Sylow isn't necessary since the group's order is squarefree

[elfin wraith]

Yeah potentially you use it in the proof, I genuinely can’t remember it’s been a while lol, something something orbit stabiliser

[dim widget]

you just need the thing about p||G| thus there is an element of order p

[elfin wraith]

Oh yeah definitely not necessary

Finite groups are simply a nail, and I have a very large Sylow shaped hammer

[quartz wind]

proving the forward direction, we have $\pi \colon M\to R/\mathcal{I}\colon (r+\mathcal{I})u\mapsto r+\mathcal{I}$, which is an epimorphism. then $ker(\pi) = \mathcal{I}u\cong \mathcal{I}$ and $M\cong \mathcal{I}\boxplus R/\mathcal{I}\cong R$. but $R$ isnt assumed to have identity so im confused, the statement just seems false to me? what am i missing?

[cloud walrus]

esca

[twilit wraith]

oh i dont have cauchy either

this is day 4 of a summer abstract algebra course

tomorrow is isomorphisms

[elfin wraith]

Yeah ok that’s fair then, but essentially the last condition is the part of Cauchy/lagrange that you need

[twilit wraith]

i see

i mean without those tricks, i cant think of anything else besides assuming an element that isnt generated by some other element is in the group

and then show it doesnt work somehow

[next obsidian]

There’s a bunch of proofs of Cauchy

I can think of one which is pretty standard you do without Lagrange

Or well, you can do without it

Because it’s an orbit argument but you can manually do it to not appeal to orbit stabilizer and therefore don’t need Lagrange

But in some sense the proof you do ends up being so similar to how you prove Lagrange

But I’ve seen like 4 proofs of Cauchy’s theorem and I imagine one of them doesn’t use it

[twilit wraith]

i dont even have orbit stabilizer

we essentially have no tools rn

[next obsidian]

Yeah you don’t need it

Okay I looked at the problem

Rip lol

I can only see how to do this via Cauchy which would involve just proving it lmao

[dim widget]

you just count elements based on their orders

knowing that there are only 35 elements and one of them has order 1

[twilit wraith]

wait is this it?

[dim widget]

yah

[quartz wind]

.

[twilit wraith]

im not seeing it tho

i mean i know theres 1 element of order 1, 4 elements of order 5, 6 elements of order 7, and 24 elements of order 35

but why couldnt there exist more than one group up to isomorphism that does this

[dim widget]

if there is a single element of order 35 you win by definition

[twilit wraith]

why

[dim widget]

that's the definition of a cyclic group

[twilit wraith]

oh duh

ive done too much math today my mind is blown out

[dim widget]

so you just have to rule out all elements having order 5 or 7

which is pretty easy

[twilit wraith]

also i just realized this only applies to cyclic groups

[twilit wraith]

oh i see

i suppose this would be done by virtue of G being abelian

or i could show that suppose that there are elements of order 5 and 7 in the group

then create an element that must be order 35

[dim widget]

yep

so you just have to rule out every element having the same order except for Id

[next obsidian]

Okay so you already need to know that |g| | |G| tho right?

[dim widget]

that's part of the hypotheses

[next obsidian]

It is

?

[twilit wraith]

uhh im not fully sure weve proven this

[next obsidian]

Oh lol

Yeah x^35 = e

[twilit wraith]

so if every element had order 5 then there are smth like 7 generators i think

[next obsidian]

You can’t solve 1 + 4n = 35

[twilit wraith]

oh i see

likewise for 1 + 6n = 35

[next obsidian]

Yeah. Altho you have to argue that <x>\cap <y> = 1 or <x> (= <y>) if x and y are both of order 5 or 7

This would be trivial with Lagrange, but still easy af without

I mean I think it’s kinda trivial either way

[twilit wraith]

this would be easy af if only i had remotely any intuition with group theory

its only 4 days in tho so ig i cant be too hard on myself

[next obsidian]

What’s the order of a non identity element of <x>

[twilit wraith]

something that divides the order of x

[next obsidian]

Which is?

In this specific case

[glad osprey]

I haven't looked very closely at this, but is pi an isomorphism of M and R/I as R/I-modules? Aren't you supposed to treat R/I as an R-module?

[twilit wraith]

5,7,35

[next obsidian]

I was assuming |x| is 5 or 7

[twilit wraith]

oh i see

in the case its not cyclic

its only 5 or 7

[quartz wind]

ah good catch, thanks. but treating R and M as an R module and constructing an epimorphism that way also gets us M iso R i think

[next obsidian]

I don’t think you understand what I’m asking

But I think it’s better if you just try to make your own argument atp

[next obsidian]

That’s hella false

[twilit wraith]

fair enough

[next obsidian]

Ru is not isomorphic to R

It’s R/ann(u)

This isn’t a vector space so you can have torsion

[quartz wind]

ann u is {r : ru= 0}?

right of course

thanks chmonkey

[next obsidian]

And the rest of your proof is hella over complicating it

If you’re generated by m youre R/ann(m) by the map x -> mx

And if you’re an R/I you’re generated by 1’s image

And that’s the end

[quartz wind]

well R isnt guaranteed to have id

[next obsidian]

Wat

Is R even a cyclic module if you don’t have an identity lol

[next obsidian]

I have seen this image

[quartz wind]

does R have to be cyclic for M to be cyclic?

[next obsidian]

I mean R is R/0 lol

So this follows from the if

[quartz wind]

right

thanks

[next obsidian]

I am dubious of this being true if you don’t assume R has a unit

But I don’t know anything about these sorts of rings so

Idk

I just don’t see how you can show it has a generator otherwise lol

[quartz wind]

nw ill take it on faith for now, thanks

[tough raven]

Consider any one-sided ideal I of a ring R with unit. Then I is a rng. Moreover, the I-submodule of I generated by any subset of I is included in the R-submodule generated by the subset (I is also an R-module and its I-module structure is the restriction of its R-module structure, so any R-submodule is an I-submodule; this proves the claim). Thus, if I is not principal as a one-sided ideal of R, it is not cyclic (generated by one element) as an I-module.

[small salmon]

Could anyone recommend a book or website that tabulates the classic identifications of homogeneous spaces—something like a lookup table? For example,
SU(2)/U(1)  ≅  SO(3)/SO(2)  ≅  S2,
CPn  ≅  SU(n+1)/S(U(n)×U(1)).

[fringe marten]

Well, i just realized i don't understand why it is true

[rocky cloak]

The nonzero elements of a field form a group under multiplication

So this follows from Lagrange

[fringe marten]

But how do i know that the splitting field of X^{q^n}-X has less than q^n elements first?

[rocky cloak]

A polynomial of degree n has at most n roots

And the roots of x^q^n - x forms a field

[fringe marten]

The thing is that i want to prove the existenc eof such fields, and i understood how to construct the splitting field of x^q^n-x

But how do i know that there's only its roots in it?

Wait

I think i can do it by induction or something...

Idk

Have to write it

Maybe not

No idk

[rocky cloak]

Consider the set of roots for this polynomial.

Show that is closed under addition, multiplication, subtraction and division.

Conclude it is a field

[fringe marten]

How do you define the operations on the roots if you don't know F_q^n yet?

[velvet hull]

you don't have to know what they look like

as jagr said, you just verify that the roots are closed under field operations

for instance if a^q^n-a = 0 and b^q^n-b = 0 then (ab)^q^n - (ab) = 0 is true, and you don't need to know anything about a and b for it to be true

you treat them as some element of some field

[fringe marten]

I see

It's indeed stable

[rocky cloak]

You said you understood how to construct the splitting field. So that's the operation

[fringe marten]

This way you show that the roots is a subfield of the splitting field of x^q^n-x

[fringe marten]

Ye i just didn't make the links quite right

Ty for your patience everyone

[rare cipher]

hey sorry i have a quick question

can the complex numbers be given the structure of a lattice-ordered field? my professor mentioned this question in class

[rocky cloak]

What's the definition of lattice ordered field?

Partially ordered ring which is a field and where the order is a lattice?

[chilly ocean]

$\alpha= \sqrt{22+2\sqrt{5}}$ and $\beta= \sqrt{22-2\sqrt{5}}$, then is it true that $\mathbb{Q}({\alpha} +\beta)$ and $\mathbb{Q}({\alpha})$ are isomorphic?

[cloud walrus]

Akhi Mishra(Riemman's Cat)

[rocky cloak]

Shouldn't be no

[chilly ocean]

what about $Q(\alpha)$ AND $Q(\beta)$

[cloud walrus]

Akhi Mishra(Riemman's Cat)

[rocky cloak]

They're isomorphic yes

[velvet hull]

they're conjugate as subfields of the splitting field of Q(alpha, beta), so yes

[chilly ocean]

okay then i am right

[velvet hull]

presumably jagr made a typo, but it's no because alpha and alpha + beta have non isomorphic minimal polynomials

[chilly ocean]

yeah thanks for confirming

[rocky cloak]

I made a typo?

[velvet hull]

shouldn't be no sounds like a double negation to me

[rocky cloak]

Ah, the quarrels of the English language

No, they should not be isomorphic

[chilly ocean]

today prof said i should know french also for good maths , how it is related

[elfin wraith]

A lot of maths was historically written in French, you can definitely avoid it though

[elfin wraith]

“Shouldn’t be no” is really “shouldn’t be, no” which reads fine. English is weird

[rocky cloak]

I'm not sure it's so relevant anymore. I had to read a Russian paper and Google translate really takes you a long way these days.

[chilly ocean]

so we need french only to live in france now

[rocky cloak]

You might need it to live in parts of Canada and a few African countries aswell

[chilly ocean]

why

[velvet hull]

immigration and colonialism

[rocky cloak]

Cuz they speak French there

[knotty badger]

✨colonialism✨

[keen badge]

A commutative algebra is a field iff its a division algebra, right?

[mighty kiln]

Yes

[long obsidian]

How would I go about getting an basis of the quotient of J=<x^2+x^3,y^2> in k[x,y]?

[velvet hull]

it's just the quotient of a basis for k[x,y], no?

[rocky cloak]

Like Gröbner basis? Or what do you mean?

[long obsidian]

k[x,y] is infinite dimensional as a vector space but k[x,y]/<x^2+x^3,y^2> is finite dimensional as a vector space. I mean a basis as a vector space

[velvet hull]

As a vector space it’s 1,x,x^2,y,yx,yx^2

[spice wagon]

yes exactly

we have very good books in france from Bourbaki

[long obsidian]

Is <1,x,y,xy,x^2y,x^3y> also a basis?

[velvet hull]

x^2y and x^3y are linearly dependent

[velvet hull]

You view the ideal as telling you what relations you are imposing on k[x,y]

So you add the requirement that y^2=0 and x^3=-x^2

And then the basis is all the lower degree terms

[long obsidian]

I see thanks ok

[velvet hull]

Well all the lower degree terms in terms of the degree of each variable

So for instance y^2=0 tells us that every polynomial f(x,y), when viewed as a polynomial in (k[x])[y] has degree at most 1

[chilly ocean]

$\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}....)$ and $\mathbb{Q}(\sqrt{3},\sqrt{5}....)$ not isomorphic how to show?

[cloud walrus]

Akhi Mishra(Riemman's Cat)

[rocky cloak]

One contains a square root of 2, the other doesn't

[chilly ocean]

that we have to show

[velvet hull]

Any Q-linear map preserves 2

[rocky cloak]

Indeed, just write down an element and square it

[chilly ocean]

my prof messed my mind with simplis

it is all because of that radical question

please give an example of field which is isomorphic to its proper subfield

[crystal vale]

If I want to find all subgroups of R* such that it has index 2, then how can I do that?

And in S_n, I want to prove that A_n is the only subgroup of index 2, without using that A_n is simple for n≥5.

So let H be the subgroup of index 2, then a^2 in H for all a in S_n. And since we can write every 3 cycle into a^2 so we get A_n \subset H, so we get H = A_n, right?

[rocky cloak]

Q(x)

[chilly ocean]

iso morphic to Q(x^2)

[chilly ocean]

positive reals

[crystal vale]

I know

But how can I show that there is no other?

[dim widget]

Do you mean closed subgroups? Otherwise it’s probably not true

that there is only 1

[crystal vale]

Closed subgroups?

[dim widget]

Closed subgroups.

[crystal vale]

How can I show there is only one?

[dim widget]

Only one?

[crystal vale]

.

Forget it, how can I do that question?

[rocky cloak]

Should be true either way.
R^* is like C2 x Q^N right

[dim widget]

ah true i was being silly

[crystal vale]

I don't get it what does it mean by C2 ×Q^N?

[rocky cloak]

Think about a map R^* -> C2

What happens to elements that have a square root? Which elements is that?

[crystal vale]

Okay

[crystal vale]

So are we finding the total number of homomorphisms R* -> C2?

[crystal vale]

Every positive real number has square root

Negative have not square root

[crystal vale]

Onto homomorphism

[rocky cloak]

Yes, an index 2 subgroup would correspond to an onto homomorphism to C2

[crystal vale]

Yes

[crystal vale]

I got it so every number who has square root will be map to 0, thank you jagr ❤️

[chilly ocean]

every subgroup is either containing full even elements or half even or half odd so case 1 gives A_n second case gives a^2 is even but a^2 cant be odd so a^2 is only only turning out to be even

[near tapir]

Hey, could anyone share some feedback on my proof of 2.17?

[cloud walrus]

Aguacate

[chilly ocean]

Ring is abelian group

[echo vine]

oh

thanks haha I forgot about that

[crystal vale]

So a^2 is even, so it contradicts the second case?

And what about my idea, is it correct?

[chilly ocean]

i didnt understand the meaning of writing every 3 cycle into a^2

[crystal vale]

Let (ijk) = (ikj)^2

So that means (ijk) in H

We don't need to be let 😅

A_n is generated by all three cycles

[chilly ocean]

its good

[chilly ocean]

yes

[crystal vale]

So a^2 is even, how does it help me here?

[chilly ocean]

we cant get -1 signature

[chilly ocean]

S_N/H= C_2

[crystal vale]

Yes

[chilly ocean]

implies there is a homomorphism S_N to C_2 with kerner H

[crystal vale]

Yes

Onto homomorphism

[chilly ocean]

YES

look at 3-cycles

[chilly ocean]

also H has odd cycles inside use that also

[chilly ocean]

ok let me finish, H cant contain all 3 cycles so we must have one 3 cycles out side H say a with phi(a)=-1 then 1=phi(a)phi(a^2) and phi(a^2)=-1, which cant be, so phi(a^2)=1 and gives 1=phI(a^3)=-1,contradiction

[chilly ocean]

i was observing not prooving

[crystal vale]

How 1 = phi(a)phi(a^2)?

[chilly ocean]

a is of order 3

[crystal vale]

I see

But why are we doing this? I mean we already get the idea that all 3 cycles in H so A_n \subset H, then H = A_n.

[crystal vale]

Thank you

[chilly ocean]

alternate

[crystal vale]

Yes thank you

[crystal vale]

I have to prove that if G is a group of order 2^n and G is abelian then the number of elements of order 2 is odd.

Can I say, let H = {x | x^2 = e } then H is a subgroup and it has order 2^k, so it implies that the number of elements of order 2 is odd.

Is it correct?

[candid patrol]

Good

But explains

[crystal vale]

So H is subgroup, it is clear, right?

[candid patrol]

The number of elements of order 2 is 2^k - 1 btw

[crystal vale]

Yes

Because we count e only in H which does not order 2, and every element who has order 2 will come in H,

And x in H, x≠e implies x has order 2

[candid patrol]

Yes

Then why H has 2^k elements ?

[chilly ocean]

USE FUNdamental theorem

i think we should once look at it

[dim widget]

the condition on the order of G is unnecessary. If G is a finite abelian group then G[2] = { g \in G|g^2 = e} is a finite abelian group of order 2^n, and every element except for e has order 2...

same argument works if you just assume that G has finite 2-torsion

[candid patrol]

A better exercice would be to proove that for G finite, if x^2 = e for every x € G, then G is abelian and G ~ (Z/2Z)^n

I mean it’s the same exercice, but it’s clearer

[crystal vale]

Lagrange theorem

[crystal vale]

Yes but I found this one easier

[crystal vale]

I see

[crystal vale]

What is a finite 2-torsion group?

[dim widget]

a finite group where every element satisfies x^2 = e

[candid patrol]

Yo what ?

[crystal vale]

Oh I see

[crystal vale]

It is given that G has order 2^n, so H has to be order 2^k

And every even order group has at least one element of order 2

[crystal vale]

Oh I see

[dim widget]

so you keep saying!

[candid patrol]

Ok but I was talking about my exercice

[crystal vale]

It is easy to see G is abelian

[glad osprey]

He's called Notknow, not Notsee

[crystal vale]

abab = aabb => ab = ba

[candid patrol]

?!

I don’t get u

[candid patrol]

^

[crystal vale]

We can use the fundamental theorem, so it will imply G isomorphic to (Z/2Z)^n

[candid patrol]

You can use it if you proove that G is abelien yes

[crystal vale]

.

[dim widget]

that's lucky since there are a lot of those on discord

[candid patrol]

U don’t get it

But it’s okay

[crystal vale]

I proved that G is abelian

[spice wagon]

i mean it’s pretty trivial no? ab=(ab)^-1=b^-1 a^-1=ba

[candid patrol]

Gg OMG

[candid patrol]

You didn’t proove anything

[spice wagon]

mdrr

[glad osprey]

He did, he just skipped a few steps. (ab)^2 = e = e*e = a^2 b^2

Don't sleep on Notknow, he's not called Notprove

[next obsidian]

Oh I see

[crystal vale]

?

[crystal vale]

[tardy hedge]

❤️❤️

[twilit wraith]

just to check i have something right, I am computing Aut(U(10)), with U(10) being the multiplicative group mod 10

i found that there are two automorphisms because there are two generators, 3 and 7

so theres one automorphism that sends the identity to itself and the other automorphism just flips them

am i right on this?

when i look it up it seems i dont have the right idea

[fresh bison]

Other than sending the identity to itself, what does that first automorphism do?

[digital notch]

what does it mean to "classify subgroups up to automorphisms" and what is this point of this

[fresh bison]

I’m still learning abstract algebra, so you shouldn’t trust me that much, but swapping 3 and 7 while 1 and 9 are kept the same seems to be an automorphism

[next obsidian]

You’re right that any automorphism needs to send a generator to a generator, and that to show automorphisms are equal you only care where the generator goes (as this determines everything else), but you don’t immediately know that swapping the two actually gives a homomorphism

You should verify that it does

(Note: in these simple cases you can show it will work anyway, but you need to do an argument to show that)

[twilit wraith]

i see

am i right though that these are the only two homomorphisms?

as in i have to send things of certain orders to other elements of that same order in the automorphism

in which case, every permutation of the elements in U(10) that fix 1 and 9 are automorphisms

[next obsidian]

Okay sorry, it’s clear the map will be a homomorphism, you need to show its well defined

[glad osprey]

You can show that the automorphism group has order two because U(10) is iso to Z/4Z, then Aut(Z/4Z) is iso to Z/2Z. This is basically two applications of the euler totient function

[next obsidian]

My man bro is like 1 week into group theory

He needs to toil in the mines

And figure stuff out by hand

[glad osprey]

Lol, sorry 😅

The children yearn for the mines

[twilit wraith]

yeah i was thinking it was Z/4Z

there is a hw problem were i need to find two groups that arent isomorphic but their automorphism groups are

[velvet hull]

My first immediate thought is to find ||groups with no nontrivial automorphisms|| but your idea should work as well

[twilit wraith]

i mean i just thought of Z at first

since its automorphism group is very simple

[velvet hull]

Z doesn’t work (as a group with no nontrivial automorphism)

1 -> -1

But it does work for your problem

[twilit wraith]

oh yeah sorry i was meaning i was gonna like Z and U(10) together

i actually dont know enough about automorphisms yet to have the intuition about a group that has no nontrivial automorphisms

wouldnt that have to be one where every element has different orders?

oh wait so Z/2Z works

and then D_1

well no but those are isomorphic

[velvet hull]

Actually, now that I think about it, my approach is more annoying than I thought. Just do what you were doing originally

Also using Z as your first group is a good approach

[twilit wraith]

thank you

i just took by far the simplest group we have

infinite cyclic group with 2 generators

i should say the simplest that isnt almost useless

[next obsidian]

What is even simpler than Z_2…

[twilit wraith]

well im trying to think of a nontrivial group with a trivial automorphism group

[next obsidian]

What

All groups have a trivial subgroup lol

[twilit wraith]

sorry i mistyped

[tardy hedge]

Yuh

[next obsidian]

I mean u identified one

Z_2

[twilit wraith]

well yeah

hey so i did the homework problem twice

look at me go

[next obsidian]

And then what’s another group with a trivial automorphism group

Yah

Okay

[twilit wraith]

uhh

D_1

[next obsidian]

What’s D_1

Cuz this depends on ur conventions

[twilit wraith]

dihedral group of order 2

[next obsidian]

That’s the same thing

As Z_2

[twilit wraith]

oh yeah i cant find something thats isomorphic

uhh

[next obsidian]

So hence I said

What’s even sikpler than Z_2

[twilit wraith]

{e}

[next obsidian]

And what’s that automorphism group

[twilit wraith]

e \mapsto e

[next obsidian]

So it’s…

[twilit wraith]

trivial

well yeah i know

[next obsidian]

😧

so that’s ur pair

Yeah?

[twilit wraith]

i guess that would probably be a better example

since {e} might be a better place to start

but i quickly associated Z and U(10) in my head

[next obsidian]

I mean that’s good cuz that’s also right

And working through it is instructive

[twilit wraith]

yeah

[next obsidian]

But U(10) is just Z/4Z so ¯_(ツ)_/¯

[twilit wraith]

yeah

i probably couldve used that instead

[next obsidian]

Idk why U(10) came to ur head before Z/4Z

But all the same

[twilit wraith]

i ended last semester with primitive roots in elementary NT class

so its still fresh in my head

[next obsidian]

Lmao

I see

Here’s a maybe fun thing you can try too

[south patrol]

Chmonkey uses Z_2 to mean Z/2?

[next obsidian]

No but I was matching notation

[south patrol]

Yeah I assumed so, just teasing

[next obsidian]

For this let’s just say G and H are abelian cuz I forget if that’s necessary

If |G| and |H| are coprime, show that Aut(G x H) ≈ Aut(G) X Aut(H)

I think you probably can’t do this yet without Lagrange

[next obsidian]

But if you take this as given for a moment

[twilit wraith]

we havent got to direct products yet

but i know what it is

[next obsidian]

If G is any odd order abelian group (I don’t think you need abelian btw) then G x Z/2Z and G have the same automorphism group

[twilit wraith]

oh thats pretty cool

and wait

is the trivial group and Z/2Z the only groups with trivial automorphism groups?

[next obsidian]

Nah

Probably

Actually idk

[dim widget]

S_n for n >> 0

[next obsidian]

Oh yah

[twilit wraith]

i guess i would just have to cook up a group where all the elements have different orders