groups-rings-fields

[distant sleet]

Thanks again

[thorn jay]

No problem :3

[glad osprey]

I'm reading this proof that x^p - x - 1 is irreducible over Z_p. I'm not quite sure why [Z_p(a) : Z_p] = p, is it just because adjoining a automatically gives you all p roots? What's the significance of it being a splitting field?

[tardy hedge]

Hello sheddow hello jagr

Hello enpeace

[rocky cloak]

I don't know if this is exactly what the proof is trying to say, but the fact that
Zp(alpha) = Zp(beta) for any two roots, means they all have the same degree minimal polynomial.

So if you write f as a product of minimal polynomials, they all have the same degree.

And p is prime, so that means they have degree either 1 or p

[tardy hedge]

One of my homeworks was all about this polynomial

[glad osprey]

That makes sense, thanks tbh, if the book was trying to convey this, then it failed hard 💀

[thorn jay]

aeaeaeae

[rocky cloak]

An alternative could be that the Frobenius homomorphism has order p on Zp(alpha), showing it has degree p

But yeah, I don't really see how they're justifying the claim that the degree is p. But it can be justified

[glad osprey]

uh, stupid question, but since a being a root implies that a + 1 is a root, why doesn't this imply that every element of Z_p(a) is a root of x^p - x - 1? There would be p^k roots, but it can only have p roots

[thorn jay]

There wouldnt, a is already a root of the polynomial

a is not an element of Z_p, so a + k will only reach elements in a + Z_p

[glad osprey]

ooh, I was (subconsciously maybe) thinking of GF(p^k) as cyclic, but yeah, I get it now

you can't reach every element of GF(p^k) by just adding 1

[thorn jay]

Nope

The characteristic of an R-algebra must divide the characteristic of R :3

[glad osprey]

the characteristic of GF(p^k) must equal the characteristic of GF(p)

[thorn jay]

Prime field moment

[crystal vale]

I know how to prove the converse part, but in forward direction, I think we just need to use the converse part

Say it has $((1, 2 \dots n_1 )(n_{1} +1,\dots 2n_1)\dots ((k-1))n_1 + 1\dots kn_1))$

[cloud walrus]

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

[crystal vale]

So every cycle has n_1 size and there are k disjoint cycle

So I think p is v^k for some cycle v

and v has length kn_1

[worthy solar]

Is there any easy way of determing the intermediate fields between the extension Q to Q(sqrt(2+i), sqrt(2-i)) (this should be a degree 8 extension if my diagram below is correct)?

I am working with the polynomial x^4-4x^2+5 over Q and am having trouble identifying all the field extension in between.

right now i have something like

Q -> Q(i) -> Q(sqrt(2-i) -> Q(sqrt(2+i), sqrt(2-i))
| ^
-> Q(sqrt(2+i) - |

the original problem was making finding the minimal polyniomal for for sqrt(2+i) over Q. I know all the roots should be +/- sqrt(2+i) and +/- sqrt(2-i) but I can't prove that say sqrt(2-i) is not contained in Q(sqrt(2+i)).

[worthy solar]

Im getting conflicting definitions of the ideal. Are these equivalent defintions of an ideal. Because the latter doesnt mention anything about it needing to be a subring of R.

Is it just because in DF they require a subring to have the unity of R whereas in gallian it it is only required that (S, +, x) is a subset of R which forms a ring with the same operations as that of R

Because based on these definitions

If I consider Z then 2Z is a subring (if I follow gallians definitions) but 2Z isnt not a subring of Z because it doesnt contain the unity of Z (per DF).

[charred iris]

do gallian rings have multiplicative identity

the df one talking about "the" unity of R tells me that dummit foote ones do

[next obsidian]

Yes

[thorn jay]

I like the convention of calling rings with unity rings and rings without unity rngs

[next obsidian]

I don’t consider rings without unity

[thorn jay]

No same, but the once in a blue moon that i do it's nice to have a name

[next obsidian]

I refuse to recognize them

We must keep rings pure

[thorn jay]

Nonunital associative Z-algebras

[glad osprey]

Aren't sqrt(2 + i) and sqrt(2 - i) both roots of x^4 - 4x^2 + 5?

if so, Q(sqrt(2 + i), sqrt(2 - i)) can't have degree 8 over Q

[worthy solar]

Yes

:/

I'm struggling with this

[worthy solar]

The roots are just +/- those 2 terms

[glad osprey]

Not 100% sure what the best way to solve it is, but I think the idea is that if F is an intermediate field between Q and Q(sqrt(2-i)) then F is algebraic and simple, and the minimal polynomial of F divides the minimal polynomial of Q(sqrt(2-i))

[tough raven]

Let a be algebraic over Q (probably any field of characteristic other than 2 works equally well) with minimal polynomial p. Is it true that the minimal polynomial of a^2 is p(sqrt(x)) if that is a polynomial in x and p(sqrt(x)) p(-sqrt(x)) otherwise?

[tough raven]

You can show that the quadratic extensions ℚ(sqrt(2+i)), ℚ(sqrt(2-i)) of the field ℚ(i) are equal iff (2+i)/(2-i) (or equivalently (2+i)(2-i)) is a square in ℚ(i). (This is a special case of what is called Kummer theory.) So is (2+i)(2-i) a square in ℚ(i)?

[tough raven]

If deg(p) = deg(a) is odd, then usual divisibility arguments show that deg(a^2) = deg(a) and p' st p(x) p(-x) = p'(x^2) has the right degree. More generally, if deg(a) over ℚ(a^2) is 2 then a, -a are conjugate over it hence over ℚ, which implies p(x) ∣ p(-x) ⇔ p(-x) = ±p(x) ⇔ p is a polynomial in x^2 or p(x) = x; and in the latter case deg = [ℚ : ℚ] = 1. Conversely, if p(x) = p'(x^2), then deg(a^2) ≤ deg(p') = deg(a)/2. Thus, we see that deg(a^2) = deg(p)/2 if p is a polynomial in x^2 and deg(a^2) = deg(p) otherwise. Hence our candidate minimal polynomial has the right degree; we are done.

[tardy hedge]

YIIIIKES!

[tough raven]

Hello to you too.

[thorn jay]

w e a k

[minor fulcrum]

is there a type of product that one can use to express C_p^2 as the product of C_p and C_p

clearly it isn't the direct or semidirect product

[tough raven]

C_{p^2} is an extension of C_p by C_p, i.e., there is a subgroup of C_{p^2} isomorphic to C_p such that the quotient is C_p.

However, extensions are not unique (e.g., C_p ⨯ C_p is also such an extension), so there is not any "binary operation" called "extension" that takes in two groups and returns another one.

[tardy hedge]

I need dumb stuff to study again

Dis is too much thinking 4 me 😊

[minor fulcrum]

right but like, C_p x C_p is the direct product, there are other kinds of "binary operations" that are a "type of extension" like the wreathed product or the semidirect product

is there something that generalized the kind of way that C_p^2 is an extension of C_p by C_p

[glad osprey]

I love that you write the solution instead of just "nvm, figured it out"

[tough raven]

I didn't know whether I had figured it out until I finished typing and at that point it seemed a waste not to send it.

[thorn jay]

if you wanna do dumb stuff then go do a first course in proofs or smt

[tardy hedge]

Lolol

[tough raven]

Now someone do a+1/a for me 🥺

It should be the same replacing p(-x) with p(1/x) (i.e., p with the order of the coefficients reversed) but I don't want to prove it.

[thorn jay]

I love kind of winging how much preliminaries I put into my paper

I've never written a proper paper before, but from what I've gathered from reading it's really just to make sure everyone knows what conventions and notation you're using, right?

[delicate orchid]

depends on how mean you want to be

[thorn jay]

you get NO preliminaries and have to FIGURE IT OUT YOURSELF BY CONTEXT

[tough raven]

There's a subtelty where it's not as easy to rule out p being "skew-symmetric" as it was in the even case.

Ah, got it: if x^{deg p} p(1/x) = -p(x), then the constant coefficient of p is -1, which forces 1 to be a root of p, so p(x) = x-1 by irreducibility, and that's a separate case.

This is very reminiscent of every skew-symmetric polynomial being Vandermonde times a symmetric polynomial, honestly.

[elfin wraith]

Theres also rig categories, as another fun play on the name

[thorn jay]

no additive inverses right?

lol

[elfin wraith]

Yep, ring without negatives lol

Its the categorification of N

[thorn jay]

what about a rg

[thorn jay]

oidification?

[tardy hedge]

Im so tired dude

[elfin wraith]

Its a bimonoid, unsure what you mean by oidification

[thorn jay]

ah

monoid -> category
group -> groupoid

the like

[thorn jay]

same ngl

[elfin wraith]

Ahh

Yeah its a monoid under 2 operations, i dont know a huge ammount about them ive just heard a talk about building quantum computing out of reverseable classical computing and rig categories where the kinda key object

[thorn jay]

math is so weird

[elfin wraith]

They were talking about finding a similar notion to like algebraic closure, such that you can recover quantum computing from classical

Was pretty interesting

[thorn jay]

closure operators making a comeback 2025?????

I might have a job!!

[elfin wraith]

Get into the categorical quantum information sphere

[thorn jay]

why do those words make sense together

[tough raven]

Precisely: let K be a field, c ∈ K \ {0} and a be non-zero and algebraic over K with minimal polynomial p. Assume that char(K) ≠ 2 or a^2 ≠ c. Then a has degree 2 over K(a+c/a) ⇔ p(x) = x^{deg p/2} p0(x + c/x) for a unique monic polynomial p0 (of degree deg p/2). If this happens, p0 is the minimal polynomial of a+c/a. If this doesn't happen, then the polynomial q(x) = p(x) x^{deg p} p(c/x) is of the form x^{deg p} q0(x+c/x) and q0 is the minimal polynomial of a+c/a. (Finally, if char(K) = 2 and a^2 = c, then a+c/a = 0 has minimal polynomial X, while p(X) = X^2 - c = X^2 + c = X(X + c/X) or p(X) = X - a and p(X) X p(c/X) = X^2 - c. In both cases the previous algorithm is still correct.)

This is probably all a special case of elimination theory using resultants. Hmmm.

Now we can use this to find the minimal polynomial of 2 cos(rational multiple of pi) and 4 cos^2(rational multiple of pi).

[tough raven]

Suppose a, b, k are elements of a ring such that ab-ba = k(a - b) (e.g., take the free ring on 3 generators with that relation). Must there exist some x such that aba-bab = x(a-b)?

I do have an example of this where x = k^2 + k + (a+b)-1 although I can't fathom why that works. There should be roughly no extra relations in my example beyond k^2 being central. Assuming a+b+k to be central may also simplify things somehow, although I'm not sure how.

[tough raven]

Turns out it's really simple: a(a-b) = a^2 - ab, k(a-b) = ab-ba so (a+k)(a-b) = (a-b)a. Similarly (b+k)(a-b) = (a-b)b. Hence the left ideal generated by a-b is closed under right-multiplication by a and b. Hence difference of any two words in a, b can be obtained. For example, aba - bab = (ab-ba)a + ba(a-b) = k(a-b)a + ba(a-b) = (k(a+k) + ba)(a-b), etc.

[rocky cloak]

I'm not sure the product itself has a name, but extensions with abelian kernel are classified group cohomology, and the construction is fairly similar to the semidirect product:

For a group G and a G-module M, you consider a 2-cocycle s: GxG -> M, then you put a group structure on MxG by
(m, g)(n, h) = (m + g(n) + s(g,h), gh)

In your example G=Cp and M=Cp with trivial action.

Also note that the semidirect product happens when s=0.

Also in your example s can be interpreted as a "carry". I.e. you simply choose s to be the carried digit of the sum g+h

[sturdy spear]

i think the product is written in above form that is (a1...am)(a{m+1}...an) instead of

$(a_1*\cdots *a_{n-1})a_n$

[cloud walrus]

Abstract Afzal

[sturdy spear]

cuz of this definition isn't it?

[tardy hedge]

For L/K Galois, can I say a sub extension L/E is Galois because it is still normal and separable?

I understand why its separable but why is it normal? Normal means what again, the splitting field of a set of separable polys?

[rocky cloak]

Being normal is equivalent to being a splitting field.

L is still the splitting field for the same polynomials that show L/K is Galois

[tardy hedge]

So the fact that L/E is Galois is kind of an obvious fact?

[rocky cloak]

Sure

[amber burrow]

jagr im still kinda struggling with that finite index problem

how did the automorphisms thing work?

im trying to do it purely by considering the actions itself, as im struggling to understand the homomorphism approach

the index m subgroups would be given by distinct stabilizers given by the action of G on {1,...,m}

i can probably just look at stabilizers of 1 across every action

although, it might be nice to use that every other stabilizer is conjugate to one another

regardless, the possible transitive actions for m =3 are

b(1) = 2, b(2) = 3, b(3) = 1

or b(1) = 3, b(3) = 2, b(2) = 1

a(1) = 1, a(2) = 2, a(3) = 3, or a is some transposition

i anticipate that the stabilizer of 1 given by one choice of b is the same as another choice of b by choosing a different a

so i can probably just focus on the first action for b

idk its hard to think about because its an infinite group

so its hard to think about when two stabilizers are distinct

[rocky cloak]

If you first determine the kernels of maps G -> Sn, then you can work with G/N which is a finite group.

[amber burrow]

but those would all be normal

[rocky cloak]

Yes exactly

[amber burrow]

but i want to count index m subgroups in general

not necessarily normal

[rocky cloak]

Well they're not of index m, they are the normal cores of subgroups of index m

Like you have H of index 3 for example.

Then the action of G on G/H induces a map G -> S3. Then there are two cases, either the image has size 3 or is all of S3.

In the first case H is normal. In the second case S3 = G/N for some normal subgroup contained in H.

Then you know that the index 3 subgroups that contain N correspond to indeed 3 subgroups of S3, hence there are 3 of them.

[amber burrow]

is there like a super obvious way its either 3 or all?

or is it because, for it to be transitive, b has to have an orbit of size 3

so either you have 1, (123),(132)

[rocky cloak]

Yeah it's just because S3 is so small. The only transitive subgroups are A3 and S3, so it must be one of them

For m=4, you have Klein 4, C4, A4, D4, and S4. But it should be easy to see that the first two cannot be quotients of G

[amber burrow]

Ok this is making more sense

Ok for m=3, once you have the two possible transitive subgroups A3 and S3

How did u get H is normal for A3?

So I mean using first iso thm, G/ker ~= A3

Oh and the kernel is precisely the intersection of the stabilizers or smth

Wait I saw this in dummit and Foote lemme check

[rocky cloak]

Yes, but also the kernel is an index 3 subgroup contained in H, so must equal H

[amber burrow]

Oh because |G: ker| = |G/ker| = |A_3| = 3

That makes a lot of sense

But in the other case G/ker ~= S_3

So the index is 6

Oh and |G:ker| = |G:H||H:ker| = 3|H:ker| = 6 so |H:ker| = 2

Oh yeah I just check dummit and Foote

Kernel of this action would be intersection of conjugates of H, and kernel is largest normal subgroup of G contained in H

Oh and the last thing you said is the fourth iso thm/correspondence/lattice

yo this is actually making me so happy

so many different things used in a problem

i really gotta do more exercises so i can actually do these problems

I think i actually understand the m=3 case 🙂

[rocky cloak]

I think the m=4 and m=5 case should go through pretty similarly. But m=6 seems like pain

[amber burrow]

yeah google tells me the answer is 22 for 6

[rocky cloak]

Feels like there should be some clever tricks beyond this brute force approach

Maybe there's a solution sheet for this problem with something clever

[amber burrow]

no solution out yet

i wouldn't be surprised if its brute force, you are given a month to work on the problem set

it looks like there is a lot of theory on counting finite index subgroups of PSL(2,Z), but its beyond my understanding

like this series of blogposts: https://qchu.wordpress.com/2015/11/15/finite-index-subgroups-of-the-modular-group/
counts them, but like i have no idea what they are doing

[rocky cloak]

I guess by not just consider transitive actions you can think of any subgroup of index <= n as a map G -> Sn. Then you just have to sort of what you're double counting

[amber burrow]

Hmm, when looking at the previous post where he got the formula he uses what I think is category theory, so I stopped really reading

[amber burrow]

just to confirm why:
to get a homomorphism, we need to map generators to generators. For C4, we would be forced to have b |-> e, and phi(a) is the generator of C4, but obviously that isnt surjective

for V4 (<x,y | x^2,y^2>), we have to map b to e and a to x, which is also not surjective

then for the rest its a bit repetitive?

so im just counting index 4 subgroups of A4, D4, and S4?

that doesnt feel quite right

[rocky cloak]

Yup, though D4 you can also throw out.

[amber burrow]

alr nice

but im already going over

for A4

If H is index 4, |A4:H| = |A4|/|H| = 12/|H| = 4, so |H| = 3

and vice versa

it would suffice to consider cyclic subgroups because so small

so lcm of sizes of disjoint cycles is 3

so its just all the possible 3 cyces

which is 432/3 =8

but the answer overall is 8

and for S4, |H| = 6, so it would cycles of type 2,3 and type 6

which is obviously a lot more

actually maybe there is no surj homo to S4

thats also def not true, because google says S4 ~= PSL(2,Z/4Z)

oh wait nvm S4 doesnt work

[rocky cloak]

A group of order 3 contains two 3-cycles. So you've double counted

So A4 has 4 index 4 subgroups and same with S4

[amber burrow]

wait actualy why is it 4 for S4

since you need order 6 subgroup

it cant be cylic, since there are no 2,3 or 6 type cycles in S4 (cuz theres only 4 numbers)

[rocky cloak]

Yeah, there are 4 such subgroups, just count them

[amber burrow]

alr

[rocky cloak]

S{123}, S{124}, S{134}, S{234}

[acoustic igloo]

my book is asking if this is an acceptable definition

a prime field is a field that has no proper subfields
and apparently it is, but i don't understand why, when the book defines a prime field as Z_p and Q. it provides a theorem:
a field F is either of prime characteristic p and contains a subfield isomorphic to Z_p or of characteristic 0 and contains a subfield isomorphic to Q
but how can i rule out the possibility, for example, that Q contains a proper subgroup isomorphic to Q?

and actually something seems wrong, because there could be a field isomorphic to Z_p which has no proper subfields

[glad osprey]

Is that Fraleigh? He describes prime fields extremely poorly in my opinion. The basic idea is that there is a unique ring hom from Z to any field F, which must have kernel (0) or (p). In the first case, F contains Z and therefore Q, otherwise F contains Z/(p) by the first isomorphism theorem

[acoustic igloo]

yes it's fraleigh

[glad osprey]

Consider looking up the definition in another book, Aluffi's Notes from the Underground for example

[acoustic igloo]

thanks 🙏

oh fraleigh does explain the homomorphism idea

and actually if a field has no proper subfields, it must still have a subfield isomorphic to either Z_p or Q, so the field itself must be isomorphic to Z_p or Q. Conversely, Z_p contains no proper subfields. i just need to see that Q contains no proper subfields

[thorn jay]

Notice that Z must be contained in every subfield of Q, and as its a subfield, every inverse of elements in Z too

[acoustic igloo]

oh

right

but how do we know unity in the subfield is unity in Q?

[thorn jay]

Unity? Do you mean inverses?

[acoustic igloo]

i mean the multiplicative identity

how do we know it's the same

[thorn jay]

Because thats how subfields work

No?

[acoustic igloo]

by definition?

[thorn jay]

I believe so? Do your rings not have a unit element

[acoustic igloo]

no

[thorn jay]

Shame on your book (kidding)

[acoustic igloo]

lol

[thorn jay]

Any nonzero idempotent element in a field must uniquely be 1

Because of properties of groups

Anyhow, I believe it silly to not include this in the definition itself

[acoustic igloo]

ok fair enough

[glad osprey]

if R < S and S has a multiplicative identity 1, then 1 acts as a multiplicative identity in R too, so R can't have a different one because of uniqueness as enpeace said

[acoustic igloo]

if R < S and S has a multiplicative identity 1, then 1 acts as a multiplicative identity in R too
that assumes 1 is in R

oh maybe i misunderstood

i think i get it

[tardy hedge]

Galois group of (x^2-x+1)(x^3+x^2-1) over Q. When doing problems such as these, is it worthwhile to try first explicitly computing the splitting field?

For (x^3-2)(x^3-3), i know I had the splitting field be Q(3rd root 1, cr(2), cr(3))

So Gal group for that should be order 18?

[glad osprey]

Uh yeah, that might not be correct, noggin is not working today

[tardy hedge]

In any case, you know that for a polynomial factored into irreducibles pq with p deg n q deg m, the galois group is subgroup of Sn x Sm right

[thorn jay]

Not necessarily

[acoustic igloo]

i found an example for rings with unity
Z_6 has unity 1, and the subgroup {0,2,4} has unity 4

but for fields it seems unity must be shared

[thorn jay]

Uniqueness follows either locally from group properties, or uniqueness of identity is only applicable for two elements from the same monoid

[acoustic igloo]

so a subfield of Q contains 1 and thus Z and thus Q

[tardy hedge]

Ye

[thorn jay]

[tardy hedge]

Enpeace learn galois theory rn so u can help me

U could learn it in like idk 30 mins

Hehe

[thorn jay]

Could I qwq

[tardy hedge]

You could learn it very quickly tbh

its just groups and fields

[thorn jay]

Universal algebra is just sets and functions

[tardy hedge]

lol

have you learn field theory stuff?

[thorn jay]

Some, yes

[alpine plank]

should be possible until the fundamental theorem atleast, I think we learnt (basic) galois theory for number fields in one lecture in my number fields course

[thorn jay]

Though I am busy with a research project atm

[thorn jay]

Thats true, but time yk

[tardy hedge]

i need so much time

to learn alla dis shi

AllaDisShi

🔥

hey, im happy though because my presentation today went well

on infinite Galois theory

Keith Conrad W

[alpine plank]

I just learnt about infinite galois theory recently too! I was using Milne's book

[tardy hedge]

Nice, I just learned the beginnings of it from keith conrads notes, and i presented the galois correspondence for subextensions and closed subgroups

Milne's book is something on number theory is it?

[alpine plank]

nope his Fields and Galois theory book

[tardy hedge]

oh

is it good? i just reading from dumit foote mostly

[alpine plank]

although he speedruns some basic facts in one paragraph in his elliptic curves book, i havent read any other stuff of his

[alpine plank]

it was good for me but I was already familiar with a little galois theory ( But I didnt do a course in it, I didnt know much about separable extensions for example)

It has a 2 hour examination at the end which I thought was pretty cool (for self study), and the last chapters have some advanced stuff

its free on his website

[tardy hedge]

I wasnt sure about the discussion on the splitting field being given by adjoining the square root of the discriminant and all that … where is that coming from?

I know root(D) is always contained in the splitting field

So does that just follow from that? In that case where D is not a square in F, the degree of the splitting field is 6, so adjoining root D gives a quadratic extension, and adjoining some root of f would multiply to degree 6 extension

[alpine plank]

yep that seems correct

[tardy hedge]

What elements in F(theta, root(D)) are roots of the irreducible cubic?

Theta is one, what are the others ?

Random: the only subgroup of S4 divisible by 6 that isnt A4 is S4?

[alpine plank]

Well you have the polynomial as one algebraic relation for the other roots and root(D) is the product of their differences, ig you can get it from these

by manipulating it

[dense root]

whats the point of the charastirstic of a Ring

[tardy hedge]

Ok makes sense

[alpine plank]

but you might be rederiving the cubic formula in the process (someone fact check me here)

[tardy hedge]

Holy Molay

Lol

[alpine plank]

use the sum of roots product of roots stuff too

[rocky cloak]

S3 is also a subgroup, otherwise yes

You can find all subgroups here
https://groupprops.subwiki.org/wiki/Subgroup_structure_of_symmetric_group:S4

[alpine plank]

Is there a database of websites like this? I know there's one for counterexamples in topology too. My bookmark folder is already overloaded enough

oh wait https://mathbases.org/

[tardy hedge]

S3 being a subgroup of S4 in the sense of, leave say 4 fixed and then take the 6 permutations of the other 3 elements?

So its not a transitive subgroup

[rocky cloak]

Idk about complete list, but
https://ringtheory.herokuapp.com/
https://groupprops.subwiki.org/wiki/Main_Page
https://www.math.uni-bielefeld.de/~jgeuenich/string-applet/

Are good

[rocky cloak]

Any transitive subgroup will have order a multiple of 4.

(So a multiple of 12 if it's also divisible by 6)

[tardy hedge]

Its been hard for me to understand “transitive subgroup” properly

Obviously it keeps coming up with Galois theory

[tardy hedge]

Is it because any transitive subgroup would need to contain a 4 cycle? Im not convinced on that im just throwing it out there

Prob not

[rocky cloak]

It doesn't need to contain a 4-cycle (see Klein 4 for example), but it needs to contain an index 4 subgroup as per the orbit stabilizer theorem

[tardy hedge]

Ay yai yai!

I have been needing to learn this stuff

Ty for klein 4 example

What is orbit stabilizer theorem? Is that just the class equation for G acting on a set S?

Oh so, if G acts transitively on S then there is one orbit, meaning there needs to be a subgroup of G with index size of S

By the partition equation

[thorn jay]

Its essentially the kernel of the natural map Z -> R

Also, for example, fields with certain characteristics have nice properties with respect to for example representation theory

Its just some property of a ring that one needs to consider a lot (and it behaves especially nice in integral domains, as then n • r = 0 <=> n = char R), so we've given it a name

[tardy hedge]

Lol i was about to ask can the characteristic of a ring not be prime (cuz i been studying fields so much recently)

Z/4Z be like

[thorn jay]

The humble Z/nZ

[tardy hedge]

Ikr!

We love Z/nZ

[thorn jay]

If you take the the natural numbers (including 0) as a posetal category where n -> m iff n | m, noting that every number divides 0 as 0 * x = 0, char is a contravariant functor from Ring to this category lol

This is not unique to rings though, just that the lattice of ideals of Z is opposite to the lattice of integers ordered by divisibility

[rocky cloak]

For a group action G on X and an element x, the index of the stabilizer (all g such that g(x) = x) is the size of the orbit of x.

More explicitly the map
G -> X
g |-> gx
Gives a bijection between G/H and the orbit of x, where H is the stabilizer

[rocky cloak]

It's not really by convention.

0 = 0*n, so every number does indeed divide 0

[thorn jay]

Yeah, true, bad wording
Fixed

Every variety of algebras has a such a characteristic functor to the congruence lattice of the initial algebra, its just that the congruence lattice of Z has a convenient representation :3

[acoustic igloo]

true or false

A ring with zero divisors may contain one of the prime fields as a subring

QxQ contains Qx{0} which is isomorphic to Q, does that count?

i'm gonna say true

[eager willow]

Does not count because Q x {0} is not a subring. It doesn't contain 1 = (1, 1)

However, the same example of your ring with zero divisors does have Q in it. Just somewhere else.

[acoustic igloo]

oh

we can use {(x,x) | x in Q}

[thorn jay]

The humble diagonal map X -> X x X

[south patrol]

Also more obvious examples like Q[x]/x^2

[thorn jay]

Oh ofc, just adjoin a zero divisor

Lmaoo

[south patrol]

Yes

[thorn jay]

Thats smart

[south patrol]

I mean really like these two are isomorphic anyway right

But I find it more easy psychologically lol

I mean also like

Any ring map out of a field is injective lol (except a map to 0)

And every ring admits a map from a prime field

[thorn jay]

No?
QxQ is reduced, Q[x]/x^2 isnt

[south patrol]

Sorry yes I am dumb

Good catch

LOL

[thorn jay]

:P

[south patrol]

Yes the second is a square zero extension of Q whilst the first isn't lol

[thorn jay]

QxQ ≈ Q[x] / (x - x^2)
Though, right?

[south patrol]

Ye

[thorn jay]

Just pulling it out of my ass

[south patrol]

Chinese remainder

[thorn jay]

Yeah ok

[south patrol]

Freely adjoining an idempotent

[thorn jay]

I will NOT be calling it chinise remaejinder this is the lemma used in proving that an algebra is subdirectly irreducible iff it has a unique minimal nontrivial congruence >:((((

Lol

[thorn jay]

I hate this fact

It looks so unclean

[tardy hedge]

What does a reduced ring mean?

[tardy hedge]

Also, how?

[wraith cargo]

No nilpotents

[tardy hedge]

Oki

[thorn jay]

Equivalently, some subring of a direct product of fields

Because reduced rings are quotients by radicals, radicals are intersections of primes, and primes are ideals where the quotient ring can be embedded into some field

[wraith cargo]

Isn't it enough to require it to be a direct product of integral domains?

Or even reduced rings lol

[thorn jay]

Only subdirect product

I believe

Yeah

Reduced rings are subdirect products of integral domains

I guess one could say that ISP(K) = IP_s(IS(K))
But I'd look like a crackpot

[warm ember]

let homomorphism $f\colon A\to B$

[cloud walrus]

IHaveABoner12

[tardy hedge]

So true

[velvet hull]

$\to$

[cloud walrus]

HChan

[velvet hull]

ain't no waaay lmao

[tardy hedge]

Lmaooooo wtf man

That is a pp

[vagrant zinc]

xde

[amber burrow]

$\to$

[cloud walrus]

donut123

[amber burrow]

$\rightarrow$

[cloud walrus]

donut123

[amber burrow]

why is weird for you all

[potent condor]

$\to$

[cloud walrus]

Homogeneous spaces

[chilly ocean]

, \to

, to

$\to$

[cloud walrus]

NAT Enthusiast

[sturdy spear]

isnt is a bit confusing

one defined symmetry interms of rigid motion and other defined rigid motion in terms of symmetry

[velvet hull]

they're both just essentially saying that you have some map that preserves the structure of the object

[sturdy spear]

"structure of the objective"

wdym by this

[velvet hull]

a rearrangement of the figure preserving the arrangement of its sides and vertices as well as its distances and angles

[sturdy spear]

oh i see, so rigid motion is a map that preservers structure of the objective and symmetry is also the same thing

so rigid motion isn't same as symmetry

[velvet hull]

well depends on what context you're using the words in

but practically its not something to worry about, just semantics

[weak ridge]

In mathematical terms, you define a plane symmetry as an isometry of R^2, that is a map f from R^2 to itself that is 1-1 and onto, and preserves distances. For a geometric figure, say M in R^2, you want M to be invariant under f in the sense that f(M) = M, and in that case you say f is a symmetry of M. Now plane symmetries are equally termed to as rigid motions.

[sturdy spear]

i see, so plane symmetries are rigid motion in R^3 are like same?

[tough raven]

Well, not ℤ.

[weak ridge]

You extend the notion naturally to R^3 and R^n, n >= 3, and futhermore to abstract metric spaces, wherever there's a notion of distance.

[acoustic igloo]

i think a rigid motion in R^3 would be an isometry that preserves orientation

are reflections rigid motions?

i guess for the n-gon you can make a reflection with a rigid motion in R^3

[sturdy spear]

i see

[velvet hull]

well reflections are orientation reversing isometries

[sturdy spear]

why these terms are so frustrating

[acoustic igloo]

the idea of "symmetries of a shape" became clearer to me when i understood isometries

[weak ridge]

Well, mathematics can be so frustrating in any case but you just need to get used to all this kind of terminology coming around. Afterall it specifies the language we speak of in order to gain precision in what we are trying to say, formulate and prove. As frustrating as it may be, notice that the same term can be also used by others in various places with a slightly different meaning, if not equivalent, and in some sense it's part of your understanding to try to make things clear. For some it's also fun, a conceptual one!

[acoustic igloo]

sorry, i feel like my comments were more confusing than helpful

[thorn jay]

Me tbh

[acoustic igloo]

is this the principal ideal generated by a+M ?

[sturdy spear]

yeah i hope so i will be able to understand these terms soon, after having some experience with algebra

thank you guys for help
HChan, Flexibile , axe

[weak ridge]

Yes.

[acoustic igloo]

thanks 🙏

[weak ridge]

In general, for R a commutative ring and x in R, Rx is an alternative notation for the ideal (x) of R generated by x. If R is not commutative, Rx is merely a left ideal.

[thorn jay]

Rx looks so cursed

[weak ridge]

It just signifies we multiply all elements of R with x from the left.

[thorn jay]

I know, I guess I havent seen it much so im not used to it

[weak ridge]

Yes, I don't use it either much. I prefer the more conventional (x).

[thorn jay]

It does make ideal notation more homogenous

[sturdy spear]

i have written this def

[weak ridge]

Sometimes it's preferable where there are more than one ring involved, so we can keep track which ideal in which ring we are referring to.

[weak ridge]

It stands but it can become better!

[sturdy spear]

oh, for now i hope it will work, maybe will edit it more once i read about this stuff more

[acoustic igloo]

i do wonder why they defined "rigid motion" as preserving symmetry
because i think it's any map preserving distance and orientation

[weak ridge]

It comes as a theorem that a symmetry of an n-gon preserves the arrangement of sides and vertices.

[sturdy spear]

idk

[weak ridge]

Also that "n" in n-gon is generally different from the other "n" for the dimension of the ambient space.

For the case of n-gons, it's better to stick in R^2 even though things can be also done in higher dimensional spaces.

[sturdy spear]

make sense

[weak ridge]

I should try this: A map f: R^d -> R^d, d >= 1, that preserves distances is called a rigid motion. A symmetry of an n-gon P, n >= 3, is a rigid motion in R^2 that maps P to itself.

[sturdy spear]

oh its clear one

[green fox]

Why is $\langle g \rangle H \cap K = H$

[cloud walrus]

Luke

[boreal inlet]

What's K

[green fox]

K is a subgroub of G/H

[boreal inlet]

Ok so let me ask just to clarify, we have G a group, with H a normal subgroup, and K is a subgroup of G/H. Right?

Also, is anything said about small g?

[green fox]

G is a finite abelian p-group and g is an element of maximal order

[boreal inlet]

I don't know.. maybe you can do an induction? A finite abelian p-group is of always order p^n for some n. Maybe induction on n should work

[rocky cloak]

If G/H = <g> x K, then they have trivial intersection per def .

But it's a little unclear what information is given to you

[green fox]

What definition? This is a random proof from a video I am wathcing, I have never seen this defined anywhere

[rocky cloak]

A group A is a product BxC if B and C are normal subgroups that don't intersect and generate A. There are many equivalent definitions.

Protip: when asking for something it's a good idea to identify what you don't understand and give the relevant context.

Just throwing out some symbols you don't understand and ask "why?" Won't lead to helpful answers

[green fox]

Except your helpful answer. checkmate

[balmy vector]

bit of a stupid question here uhhh

is there a name for a structure consisting of a set X with addition and multiplication
where X, + is an abelian group
and * distributes over +
and * isn't necessarily commutative or associative?

[rocky cloak]

People usually call that a (non-associative) algebra.

[balmy vector]

thank you!

[boreal inlet]

Lie Algebras

[glad osprey]

Don't you need scalar multiplication for it to be an algebra?

[thorn jay]

Any abelian group is vacuously a Z-module, as Z is the initial ring

So nonassociative rings are nonassociative Z-algebras

[glad osprey]

I see

[thorn jay]

Module, not algebra, oops

[hidden wind]

it feels very wrong to me to say that this vacuous, there is something to show here

[thorn jay]

For me, R-modules with underlying abelian group G are ring homomorphisms R -> End_Z(G)

So, as Z is initial, the Z-module structure on G is unique and essentially by definition

Though vacuous isnt really the right word i suppose, as Z-module doesnt add any new information

[hidden wind]

yeah, but vacuous rather means there is no information

[thorn jay]

What word would you use?

[glad osprey]

this sounds intriguing btw, do you have an example other than rings and Z?

[rocky cloak]

Any abelian group is canonically a Z-module, or uniquely a Z-module, or trivially a Z-module

[thorn jay]

Yeah thats better, thanks jagr

[thorn jay]

Groups lol, although as the trivial group has a one element normal subgroup lattice, i cant say that the functor is very interesting :P

for something more interesting, take the comma category (G \downarrow Grp) for some group G. This is the category of group homomorphisms from G and maps making a triangle commute. The initial object here is G, and this gives a functor from this category to the normal subgroup lattice

Also, yes, coslice categories are also varieties

What you do, for the coslice category of A, is adjoin nullary operations for all elements in A and gives those the structure of A with respect to the fundamental relations

I dont think any of this is particularly useful, it only is for Z as the ideals are conveniently numbers

I could be wrong but im not sure how much the characteristic says about your algebraic structure or vice versa

[glad osprey]

interesting category theory is crazy

[hidden wind]

what does variety mean here

[thorn jay]

Variety of algebraic structures, in the universal algebra sense

A bit of unfortunate naming

They are classes of algebraic structures of the same type which satisfy some axioms

[thorn jay]

(Although in a sense they are very alike)

[hidden wind]

i still haven’t had a look at universal algebra, just seen it mentioned here and there

do you know of some nice expository article

[thorn jay]

Its really awesome, although it is built more on lattice theory and closure operators than category theory. This meant that, with the rise of Bourbaki, it kind of faded into obscurity, but didnt disappear completely, luckily. What comes close is the study of monads or operad theory, but I like the classical UA

[hidden wind]

iirc it predates category theory so that’s not so surprising

[thorn jay]

Hmm, I dont know of expository articles, I'd have to look for you, but I do believe that Burris and Sankappanavar's book is a great introduction

[hidden wind]

i think Ore published something universal algebra -like

[thorn jay]

😔 category theory is making its way into UA though

[hidden wind]

noted, thanku

[thorn jay]

Like, direct limits, pushout and pullbacks, and free objects are all important objects

Free algebras especially, as they are essentially the algebras of expressions over some set of variables

So that gives the link between algebraic logic and category theory

Furthermore, I am working on linking (universal) algebraic geometry to algebraic logic, which is also cool

Anyhow

Yes, universal algebra, very cool

[hidden wind]

lovely

they don’t seem to have the book at my uni library, which is rare; on another note i see this burris seems more than healthily obsessed with george boole

[thorn jay]

The boolean algebra chapter is longer than the "elements of universal algebra" chapter in his book

[hidden wind]

in case you’ve never had the displeasure of reading early 19th century logic: boole was of tremenduous influence, not only on mathematical logic, but also on spiritualism and parapsychology

[thorn jay]

Seems like a lovely man..

At least boolean algebras are cool

[hidden wind]

[thorn jay]

Stone duality gives rise to cool constructions

[chilly ocean]

is $\frac{\mathbb{R}[X,Y]}{<X^2+Y^2-1>}$ a ufd and what if We replace $\mathbb{R}$ by $\mathbb{C}$.

[cloud walrus]

Ricci 10

[sharp ice]

consider X in A = R[X,Y]/(X^2 + Y^2 - 1). you can show that X is not prime but irreducible so A cannot be a UFD
if you replace R with C, you can show that its isomorphic to C[t,t^-1] which is a localisation of C[t], which is a UFD, at t so it must be a UFD

[chilly ocean]

what is the map for isomorphism

[sharp ice]

C[t,t^-1] -> C[X,Y]/(X^2 + Y^2 - 1) via t -> X+iY

[glad osprey]

Trying to make sure I understand this right: for any field extension k <= F (not necessarily algebraic) there is a correspondence between intermediate fields k <= E <= F and subgroups of Gal_k(F), which is actually an adjoint functor between their preorders. When k <= F is Galois this correspondence is bijective, and the preorders are isomorphic. Is this correct?

[chilly ocean]

isnt x is not prime also in C

[languid trellis]

Not sure about the question about adjoint functors, but the first part I'm pretty sure yes, because given an intermediate field E, k \subset E subset F, we can construct a subgroup of Gal_k(F) by considering Gal_E(F). The same goes the other way, given a subgroup of Gal_k(F) (a bunch of automorphisms of F which fix k), we can construct an intermediate subfield by considering all elements fixed by your subgroup.

https://en.wikipedia.org/wiki/Adjoint_functors#Posets There is a section on the wikipedia page here that you may find useful

[chilly ocean]

ISNT THERE any elementary answer without localization

[glad osprey]

thanks I wanna learn more about lattices and posets, I bet it's possible to understand this part 2 of the fundamental theorem in a lattice-theoretic way, atleast if it's possible to define normal subgroups in terms of lattices

[woven trout]

Im trying to understand the construction of rational canonical form in D&F and I'm confused. Suppose we have a pair (V,T), where V a vector space and T a linear map of V. Our goal is to find a basis for V which gives a matrix rep for T as close to diagonal as possible. To this end, first pass the pair (V,T) to its corresponding F[x]-module. Then the fundamental theorem gives us this isomorphism:

Ok, so now let's focus on one of these F[x]/(a(x)) in particular.
we know F[x] is an F vector space, and (a(x)) a subsapce, so F[x]/(a(x)) is an F vector space
It has the basis described here:

that is, if a(x)=x^k+b_{k-1}x^{k-1}+...+b_0, then a basis of F[x]/(a(x)) is the images of x^{k-1},...,1 under the projection map to the quotient.
Now, we have a linear map T:F[x]/(a(x)) to F[x]/(a(x)): [p(x)] mapsto [xp(x)]

[woven trout]

Its matrix represnetation relative to that basis is the comapnion matrix described here

thats fine, all of it makes sense to here

but then Im confused here:

what is B_i exactly?

The basis for F[x]/(a_i(x)), with say a_i(x)= x^k+b_{k-1}x^{k-1}+...+b_0 is the equivalence classes of x^{k-1},...,1
what are the elements of V corresponding to this basis?
would it just be for example, the element corresponding to [x^{k-1}] is (0,0,...x^{k-1},0,...0), were x^{k-1} occurs at the ith component?

[thorn jay]

This is something called a Galois connection

Its the same as an adjoint functor between categories, except for the special case that the categories are thin (there exists at most one morphism between two objects)

[glad osprey]

Yep, that's pretty much what I pieced together; in this case the categories are preorders of field extensions and subgroups respectively. I read on wikipedia that every Galois connection gives rise to a closure operator. What is the closure operator in this case?

[thorn jay]

So you've got
E -> Gal_E(F)
H -> field fixed by H
Right?

I dont know enough field theory to know whether or not this is an actual inverse mapping and such

I'd have to look into it

[glad osprey]

In general it's not, but they are inverses when the field extension is Galois

[thorn jay]

The closure operator though is given by the composition
E -> Gal_E(F) -> field fixed by Gal_E(F)

So the "closed" fields are all fields which are fixed by some subgroup of automorphisms of F

[glad osprey]

ahh, so the last two lines of this proposition is actually describing the closure properties of the closure operator?

[thorn jay]

Yes

[gilded fulcrum]

a)
We need to show that if a = b in Zn, then f(a) = f(b) in Zm. Since a = b in Zn, then a - b = 0 in Zn. Then we can apply f. So, f(a) - f(b) = f(a-b) = f(0) = 0 in Zm. Since f(a) - f(b) = 0 in Zm, then f(a) = f(b) in Zm. f is well defined.

Is my proof correct?

[glad osprey]

wow, cool thanks

[thorn jay]

Are Galois extensions the fields which are fixed by some subgroup of the galois group?

Because if so, then the closure operator on the subgroup lattice of Gal_k(F) is the normal closure

[languid trellis]

"Galois field" typically refers to a finite field of order p^n

[thorn jay]

Sorry, i meant extensions

[glad osprey]

hmm, so the question is if G <= Gal_k(F), then is k <= F^G a Galois extension? I think so, let me see if I find can it

[languid trellis]

only if G is normal in Gal_k(F)

[glad osprey]

Yeah, that seems correct. I only found that F^G <= F is Galois in general

[acoustic igloo]

i think i was able to prove this, but i never used the fact that p,q are prime. only that p,q > 1

[thorn jay]

p, q prime is a requirement

You can use the fact that integral domains can be embedded into a field, and then that Z_p must be the prime subfield of that field
Think about the characteristic

Maybe

[languid trellis]

haha

[thorn jay]

Much more simple :P

[acoustic igloo]

i believe a subring isomorphic to Z_n means the integral domain has characteristic n

[thorn jay]

Yes

[acoustic igloo]

even for composite n

[thorn jay]

Integral domains cannot have a subring isomorphic to Z_n for some nonprime n

[acoustic igloo]

true

then it's vacuously true

[thorn jay]

Yes

[acoustic igloo]

thanks 🙏

[gilded fulcrum]

Can someone take a look at this please

[glad osprey]

hmm, if Z_n <= R does R have to have characteristic n? Or is it just that n divides char(R)?

[languid trellis]

Looks good

[acoustic igloo]

the subring must contain unity of the integral domain, right?

that's what i wrote

any "a" in the integral domain added to itself p times can be written a(1+1+1+...) = a(p dot 1) = a(0)=0

that's Theorem 19.15

[glad osprey]

yep, that looks correct

[acoustic igloo]

🙏

[glad osprey]

yeah, the argument you gave works for any ring AFAICT, so the characteristic should be equal

I was thinking of the result that if R -> S is a ring hom then char(S) divides char(R), if I remember correctly

[acoustic igloo]

oh, interesting

[glad osprey]

which ties in to the big brain stuff enpeace was talking about, where the characteristic is a functor from the category of rings to the divisibility lattice of integers

[dense root]

i just started studying again after a break i kinda forget the idea behind of an order of a ring or the order of an element , if anyone could tell me the idea behind it

[acoustic igloo]

the order of an element "a" of a group is the order of the cyclic subgroup generated by "a", or equivalently, the smallest positive integer n such that a^n is the identity of the group

[tardy hedge]

A W explanation, Axe

[dense root]

thanks big man

[thorn jay]

The order of an algebraic structure is usually the number of elements in the underlying set

(Sometimes also called power >~>)

[arctic trail]

my brain is bugging for some reason. Can anyone help find what I overlooked?

If I have a left ideal I, I can consider the left R-module R/I.
The annihilator is a bilateral ideal, however it should also be equal to I?

[tough raven]

Divides if ℤ/nℤ maps to R, equal if it maps to R injectivedy.

[arctic trail]

nvm

poor sleep I see my mistake

[thorn jay]

I guess you could say it was trivial.. (lemma)

[worthy solar]

im reading the product of two ideals A and B is an ideal where we define AB is the finite sums of product ab. But i am a bit confused on where in the proof that we require both of them to be ideals

i must be missing something obvious

like if AB = { sum a_i*b_i | ai in A, bi in B}

the abosring can all be done using the ideal A

and likewiise the inverse requirement I believe

and i think by defintion of AB then any s1 + s2 where s1 and s2 are finite sums of products aibi then it must be in the set AB

[arctic trail]

is rA an ideal?

[rocky cloak]

It's not an if and only if.

If A and B are ideals, then AB is an ideal, but that will also happen if only A is an ideal for example

[tough raven]

For any sets A, B, AB is an additive submonoid (by construction). For any r, (rA)B = r(AB) and A(Br) = (AB)r (more generally, (AB)C = A(BC) = ABC for any sets A, B, C), so if A (resp. B) is closed under multiplication on the left (resp. right), e.g. because it is a left (resp.) right ideal, then AB is a left (resp. right) ideal.

[minor fulcrum]

is there a name for the subgroup of S1 of only the elements of finite order

[gilded fulcrum]

hey, where do i learn about the ring of taylor polynomials?

[south patrol]

This is the subgroup $\mathbf Q/\mathbf Z \subset \mathbf R/\mathbf Z$

[cloud walrus]

Prismatic Potato

[minor fulcrum]

ok cool thanks

[worthy solar]

Okay 👌.

[tough raven]

It's p-torsion is called the Prufer group (Prufer p-group? p-Prufer group? this is there somewhere on Wikipedia) sometimes.

[worthy solar]

Yes (I assume here R was commutative)

[south patrol]

And Prüfer pedantically aha

But yes this is a good point

And then those I guess are equivalently like Z[1/p]/Z

[tough raven]

I want an easy way to type those characters on phone :/

[lean sail]

Question here for anyone who has used Fraleigh 8th edition for an abstract algebra course. Wondering if you would be willing to share a list of chapters and homework exercises that you had to do for the course. Not the solutions, but just the problem numbers.

[south patrol]

Lol yeah

I guess I should say like typically the German convention is to spell ü etc as ue if you don't have the character

[tardy hedge]

Been a while since I said hi to the Potato

Hello Potato

[south patrol]

Hi Kian

Heh

[arctic trail]

ah lol

that's important

what name do you use for the arithmetic of subsets of a ring?
Or even when you have a subset of a ring with a subset of a module

this type of arithmetic on sets is constantly used but I don't know if it has a name

it's really useful for saving time cause I don't have to grab elements from the set when using it

[thorn jay]

Arbitrary subsets?

[arctic trail]

just the algebra in general though usually ideals

and submodules I suppose

though sometimes we consider small sets

like singletons

[thorn jay]

Well, thats just ring/module theory no?

For ideals you can make the case that its commutator theory but besides that

Lol

[arctic trail]

Sure but in this argument I wanted to refer to it directly

R = I1 + ... + It (direct sums, R is semisimple these are minimal left ideals)
RM = (I1 + ... + It)M = I1M + ... + ItM
Rm = (I1 + ... + It)m = I1m + ... + Itm = I1m

[thorn jay]

I guess "by commutative algebra"

[arctic trail]

these are non-commutative rings

[thorn jay]

This is about modules, which is commutative algebra no?

Nvm

That just seems like module theory

[arctic trail]

yeah it's module theory

[thorn jay]

I meant nothing else specifically, like arithmetic of ideals

Its just what you do

:P

[arctic trail]

I was solving all modules are semisimple iff all finitely modules are semisimple iff regular R module is semisimple

[south patrol]

By regular R-module is this like R as an R-module

[arctic trail]

yeah

[glad osprey]

For our first course we did

• Section 0
• Part I: Sections 1–6
• Part II: Sections 8–10
• Part III: Sections 12–15
• Part IV: Sections 16, 17, 27
• Part V: Sections 22, 23, 24
• Part VI: Sections 27, 28, 30, 31
• Part VII: Sections 33
• Part VIII: Sections 42
  For exercises, just pick and choose. I recommend a couple of computations, a few from the concepts section and a few from the theory part. (I particularly like the true/false questions and questions about correcting definitions)

[lean sail]

Nice! Thanks for putting that together, you have no idea how helpful that is.

[grizzled crow]

Which group am I looking at here?

my thought was that it must have 2 generators cus it has two types of lines

[coral shale]

what diagram is this

[grizzled crow]

This is a Cayley Diagram

From Pinter

[thorn jay]

Looks like S_3

Generated by two transpositions, but not commutative

[coral shale]

well i guess the expectation is u write the group
as <a, b | ababab> ?

and then figure what it is?

or am i being silly

[thorn jay]

Also a^2, b^2

[coral shale]

ah right

[thorn jay]

So that gives the relation aba = bab
I.e. its a coxeter group, D_6

Or D_3, whatever your convention

[grizzled crow]

Isn't s_3 supposed to look more like a triangle inside a triangle?

hmm, apparently it can look different depending on the generating set

[thorn jay]

Yee

Cayley graphs are great for certain things but not great for determining stuff up to isomorphism

[coral shale]

til ngl

never seen that cayley of s3

[thorn jay]

Same

Wait no actually

I have!!

[worthy solar]

Does the notion of two ideals being comaximal/co-prime have some analog to that of two numbers who have a gcd of 1 so we can find a linear combination of those elements?

Reading Dummit and foote generalization of Chinese remainder theorem and trying to draw connections to the typical way it is viewed about some integers being coprime

[void cosmos]

yes

[south patrol]

Yes, this is the motivation for it. You can define the gcd of two integers $a,b$ as the (positive) generator of the ideal $(a,b) = a\mathbf Z + a \mathbf Z$ and check that this matches the useful definition, so in particular if $a,b$ are coprime then the corresponding ideals are coprime/comaximal. Note though this doesn't really work well in general, which is a reason to use the term comaximal here

[cloud walrus]

Prismatic Potato

[next obsidian]

Bro said aZ + aZ 😂😂😂

[nocturne sun]

If $\exists n\in Z$ such that $n \cdot a=0,\forall a \in R$ then we call $n$ the characteristic of the ring R.\
For example, in $\mathbb{Z}_6={0,1,2,3,4,5}$, we have\
$6\cdot1=0$,$3\cdot2=0$,$2\cdot3=0$,$3\cdot4=0$,$6\cdot5=0$, \so the character would be lcm ${6,3,2,3,6}=6$

I noticed that in some places, they just check what smallest $n$ satisfies $n\cdot1=0$ and then say characteristic of R is $n$.
\
I was wondering if there was a ring where the unit element's corresponding $n$ is not the chracterstic of R.

Does there exist a ring where some of the elements (includes identity) have $m$ as the smallest integer and the oher elements have $l$ as the smallest integer.\
eventually making the characterstic of the ring , lcm$(m,l) \neq m$

[cloud walrus]

HandT

[nocturne sun]

I would appreciate some directions on this

[velvet hull]

the characteristic of a ring can equivalently be defined as the smallest positive integer n such that n*1 = 0 (try and prove this)

[nocturne sun]

I see,n.a= (n.1).a=0.a=0

[chilly ocean]

What does class number of cyclotomic field mean

[tough raven]

As for any number field, the cardinality (which is finite) of the ideal class group of its ring of integers.

[rocky cloak]

I'm using G-board, you just hold a letter and accents/variations show up. Is that not how all phone keyboards work more or less / is that not an easy way?

[tough raven]

Oh. I use a different keyboard, but I can indeed do that if I switch back. 🤦

[potent condor]

ai is getting scary smart 🤯

[thorn jay]

Lobotomy

[candid tulip]

Is a. basically asking to prove that there exists no permutation with n elements can be written as a product of more than n – 1 transpositions? I’m confused since there is a counterexample to this i believe

[languid trellis]

What's your counterexample?

[candid tulip]

This permutation has only 4 elements but can be expressed as a product of 5 transpositions

[languid trellis]

Yes, but it can also be written as (14)(24)(34), which is 3 transpositions. The point is that there exists such a decomposition, not that "there are no products of more than n transpositions in S_n"; there are, because S_n is a group

For example, the identity can be written as (12)(12)(12)(12)(12)(12)(12)(12) \in S_3, but it can also be (12)(12) or the "empty" product.

[chilly ocean]

Any ideas on Q8

Let D be a division ring prove that every finite commutative subgroup of the multiplication group of D is cyclic

[candid tulip]

So…it says that all permutations S_n can be expressed as a product of n – 1 transpositions?

[languid trellis]

at most n-1 transpositions

not exactly n-1 transpositions

[chilly ocean]

I have very interesting problems on algebra please DM if you are interested we can discuss

[crystal vale]

This one is interesting

[rocky cloak]

You can use the classification of finite abelian groups.

[crystal vale]

|| you can make that commutative subgroup into the field, just add 0 and then use that x^n = 1 has at most n roots ||

[crystal vale]

Is this one correct? Jagr?

[chilly ocean]

is this that simple

[crystal vale]

Maybe wrong

[languid trellis]

There is an argument that goes along those lines

[chilly ocean]

that we do exactly when D is field and yes the previous problem assumed D to be field

[rocky cloak]

The technique is exactly the same. You can even explicitly reduce to a field like notknow said

[tardy sage]

does this #groups-rings-fields topic exist as a subject on itself or is it always part of another subject?

[chilly ocean]

yes correct

[rocky cloak]

What do you mean?

[tardy sage]

like do u get a subject called fields for example. Like the study of fields

[languid trellis]

Yes. Field theory

[tardy sage]

ohh i see

[languid trellis]

And it's hard

[rocky cloak]

Group theory, field theory and ring theory are all fields in their own right yeah.

[tardy sage]

oh i see. thank you

[chilly ocean]

i was reviewing the proof (dummit n foote)for existence of algebraic closure can any one tell why the resultant field is algebraic over the base field

[rocky cloak]

The reason to group them together is just because they're usually introduce at the same time (in an academic setting)

[crystal vale]

I think I have to do more, because adding 0 is not enough to make into the field, what about closure under addition?

[chilly ocean]

take field generated by set , if you are desperate

[rocky cloak]

You need to take the division ring generated by the elements, so close under addition subtraction multiplication and division

[crystal vale]

Then its multiplicative set needs the same?

[crystal vale]

I see

[chilly ocean]

theorem for finitely generated group is precise way to do this

[crystal vale]

Oh so its multiplicative set is cyclic and our original multiplicative set is subgroup so it is cyclic

[rocky cloak]

I guess it's not completely obvious that the division ring generated by commuting elements is commutative, but it's true

[languid trellis]

Shout out wedderburn

[crystal vale]

But why do I have to make a division ring, why not a field?

[chilly ocean]

why to overkill

[rocky cloak]

What would it mean to "make a field"

[chilly ocean]

wedderburn theorem is outlined in excersises and involves many steps to proove it

[crystal vale]

I mean make a field of fraction

[candid tulip]

wait what does “at most” mean

[crystal vale]

What does it mean by division ring generated by the elements?

You mean make a subring generated by the elements?

[rocky cloak]

The smallest subdivision ring that contains all the elements

[crystal vale]

Which will be the division ring because our ring is division ring

[languid trellis]

There exists a way of writing an element of S_n that is the product of < n transpositions, so at most n-1 transpositions.

[crystal vale]

Yes I got it

[candid tulip]

Oh thanks
I thought it meant it has always to be less than or equal to n – 1

[thorn jay]

No

The subring of Q generated by 1 is Z

[crystal vale]

Yes I got the smallest subdivision ring

Containing all those elements

[rocky cloak]

I guess you can also split it into first looking at the ring generated, which is clearly commutative.

Then if two elements commute they also commute with their inverses (simple calculation)

So adjoining inverses gives the field of fractions for this commutative ring

[crystal vale]

By adjoining you mean add inverse such that it will be closure under addition and closed under multiplication, right?

[rocky cloak]

It's enough to adjoin all the inverses, and closure under addition follows

So it's closing under multiplication I guess

What I mean it's consider everything of the form a*b^-1

[crystal vale]

I think it closed under multiplication but I have to show it is closed under addition

Say x^-1 and y^-1 , then I have to x^-1 + y^-1 is an inverse of S, where S is ring generated by that set

x and y in S

[rocky cloak]

It's common denominator, just like in primary school.

x^-1 + y^-1 = y(xy)^-1 + x(xy)^-1 = (x+y)(xy)^-1

[crystal vale]

Oh it makes sense

So here they are assuming a positive norm?

[rocky cloak]

Not necessarily no. For example if R is the ring of polynomials then you can use the degree as a norm, so N(1) = 0

It follows from the division algorithm that N(a) = 0 only if a is a unit or 0 though

[crystal vale]

I am not sure how it comes from a dummit's division algorithm definition

I know other authors use that N(xy) ≤ N(x)N(y) but dummit didn't introduce yet

[rocky cloak]

Consider
1 = qb + r.

If N(b) = 0, then N(r) < N(b) is impossible, so you have r=0 and b a unit

[crystal vale]

I see

[tardy hedge]

how is Klein 4 a transitive subgroup of S4?

I mean I guess writing it out I can see that any element in {1,2,3,4} can go to any other element by applying some element of Klein 4

but, I dont see that with the galois group of (x^2-5)(x^2+5)

galois group of that is Klein 4 isnt it?

But you cant make sqrt(5) map to sqrt(-5) ... so i dont get it

unless the galois group is not klein 4 and im just wrong there

Ok I guess it shouldnt be because the galois group acting on the roots should have two orbits not one

Not sure where im going wrong

Do you guys see my confusion?

The galois group of that should be C2 x C2

Ok I think gpt helped me

Klein 4 is a subgroup of S4 in different ways ... ech

{(), (12), (34), (12)(34)} is the one im looking at

but then you have {(), (12)(34), (13)(24), (14)(23)} which is the transitive one

I've never seen this stuff before, can someone please elucidate whats going on here?

My group theory isnt the best

[tough raven]

Is this the subgroup {1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}?

[tardy hedge]

Yea

[tough raven]

Yeah.

[tough raven]

I think this is what you have here though.

[tardy hedge]

Yeah that makes sense

But for some reason I just havent seen this before

Like, I usually would just think "its a subgroup" full stop

not like its the transitive one

[tough raven]

But if you find a primitive element for the field extension the Galois group should act transitively on that.

[tardy hedge]

makes sense

[tough raven]

So that would give you a different embedding of the Galois group (which is still the Klein-4) into S4 (corresponding to its action on a different set of 4 elements) which is a transitive action, i.e., has image a "transitive" subgroup.

[tardy hedge]

the Galois group should act transitively on the roots of the minimal polynomial for that primitive element

is what it means fully explained?

[tough raven]

IG then it would be the (12)(34) (13)(24) (14)(23) subgroup.

[tardy hedge]

Ok, weird / interesting .. (to me). It has been kind of tricky thinking about Galois group acting on the stuff

Prob because i havent worked with group actions that much before in general

[tough raven]

Well, everything comes with experience, I guess.

[tardy hedge]

Haha yeah

So how do those two Klein 4 subgroups of S4 relate? Are they related by conjugation or just something like that?

is there a way of describing the difference?

They are isomorphic groups but not the "same"?

i just read in dummit and foote say "... is a biquadratic extension hence the Galois group is isomorphic to the klein 4 subgroup of S4"

i dont like how they said the

[tough raven]

Hmm. Conjugation by an element of S4 corresponds to relabelling the elements the group acts on, which shouldn't be able to change whether the group acts transitively or not.

Indeed, if we look at the cycle types one subgroup has 3 "product of 2 2-cycles" and the other has only 1. So they're not conjugate.

[tough raven]

Well, that's bad language.

I think the "transitive" subgroup is the unique subgroup of A_4 isomorphic to the Klein-4 though.

Maybe Dummit & Foote meant to write A_4?

[tardy hedge]

Oh ok and the transitive one {1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} is in A4 because each element is an even permutation (even number of transpositions?)

and the other one is not in A4 because it has odd permutations

[rocky cloak]

Yeah, different subgroups can be isomorphic.

So here being transitive is a property of the subgroup, not of the group.

[tardy hedge]

And being transitive or not requires us to tecnically specify a set S and define a group action ... ?

Is it just with elements / subgroups of Sn its natural to think of it already as a group acting on a set cause its just acting on {1,2,...n}

I know this is kind of like fundamental group theory knowledge but i havent thought about this too much

[rocky cloak]

Yeah, I mean transitive is really a word about group actions, which has been translated to subgroups of Sn

[tardy hedge]

thanks im learning

[thorn jay]

Absorbing the knowledge, even

[tardy hedge]

So true

[tough raven]

Yep. In the other direction, an action of G on a set with n elements is the same as (after choosing a labelling by 1, ..., n) a homomorphism G → S_n and whether or not it's transitive only depends on the image of this homomorphism. So both definitions are the same thing really.

[tardy hedge]

I think I understand it up until "both definitions are the same thing really"

[tough raven]

I mean that you can define either in terms of the other: H ⊆ S_n is transitive iff H on {1, ..., n} is transitive, and G on {1, ..., n} iff im(rho) ⊆ S_n is transitive, where rho: G → S_n is the action map.

[chilly ocean]

$\frac{\mathbb{C}[x,y]}{<x^2-y^3>}$ is UFD?

[cloud walrus]

Curvature

[rocky cloak]

Notice that it's isomorphic to C[t^2, t^3], is this a UFD?

[worthy solar]

I am / was also confused on this because I was introduced to the idea then given a specific ring and they gave me a norm which was multiplicative to do the proof. My book called this the measure and it was never specified that this measure had such multiplicative properties.

So like I don't know if given a euclidean domain I'm unsure if I can find a euclidean function which is multiplicative.

If that is always possible or not. Or how that works

[tardy hedge]

Ok ye so im still tryna figure out the Galois group of x^5+x-1 = (x^2-x+1)(x^3+x^2-1) over Q. So I know its a subgroup of C2 x S3. The discriminant of the quadratic and cubic are -3 and -23, not squares in Q, so the splitting field is divisible by 2 and 6. They are not related by a square so Q(sqrt(-3), sqrt(-23)) is contained in the splitting field so the splitting field is also divisible by 4. So split field is divisible by 12 so the galois group must then be C2 x S3?

[arctic trail]

I'm curious

solvable groups are usually(?) defined as groups that have a subnormal series whose factors are abelian

but the derived series will be a normal series, and its factors will be abelian. So why even say subnormal series?

[tardy hedge]

Shii i gotta learn that^

My exam next friday lol

I have to learn those radical extension solvable extensions stuff

With the solvable groups and allat

[rocky cloak]

Why use a seemingly stronger definition when a weaker will do I guess

[arctic trail]

fair

[tardy hedge]

Y use big word when lil word do trick

Or whatever he said

[rocky cloak]

https://youtu.be/2p_8gx-XHJo?si=EISDPAonOtysCs-C

[round portal]

hey guys

can you guys help me with the following statement

Let G be infinite cyclic. There exists exactly two possible generators for G

how to approach this?

[south patrol]

wlog G = Z

Now just think about the subgroups generated by different elements

[round portal]

ok...

we could have like <1> = Z... <2> = 2Z and so on

but Z can only be generated entirely by <1> and <n>? idk

wait

this kinda makes sense

either it makes sense or i got it completely wrong

wait

fuck

I think I got it

the possible generators are exactly <g> and <g^{-1}>

duh

RIGHT?

apologize the spam

[paper sonnet]

what if g = g^{-1}

[crystal vale]

I got it, but in that case I think we are not sure that the Euclidean algorithm will stop at some point, right?

No it will

[tough raven]

Then g^2 = 1, so it's not infinite cyclic.

[paper sonnet]

I knew the answer tho

cause that question was on my exam and I lost a point cause I missed that lol

[chilly ocean]

You can see G as a Z module here so G is isomorphic to Z^n but G is cyclic so it is enough to check for Z

[chilly ocean]

No this is not

[worthy solar]

Is there a way to avoid unity argument? I can't figure out how I would extend this or show it is surjective if R was without unity.

i know given A + B = R then for all r in R there exist an a and b such that r = a + b. But I don't know how I can work this backwards from saying

given (r1 + A, r2 + B) there exist an r in R such that r -> (r1+A, r2+B)

[rocky cloak]

r1 = a1 + b1
r2 = a2 + b2
What's the image of
a2 + b1

I think you'll have problems with the induction step though

For example consider R = Z/2 x Z/2 where all multiplications are 0.

Then A1 = ((1,0)), A2 = ((0, 1)), A3 = ((1, 1)). Then they are pairwise comaximal and R/Ai = Z/2.

But R is of course not (Z/2)^3

What you need is that A1\cap A2 is comaximal with A3, and this doesn't follow in the non-unital case

[worthy solar]

a2 = (r2-b2)

b1 = (r1-a1)

f(a2+b1) = (a2+b1+A, a2+b1+B) = (a2+r1-a1+A, r2-b2+b1+B) = (r1+A, r2+B)

b1 , b2 in B so b1 + B = 0 + B = b2 + B
a1, a2 in A so a1 + A = 0 + A = a2 + A

correct?

[proven swift]

from what book is this? if i may ask

[chilly ocean]

looks like dummit foote

[sturdy spear]

True

[chilly ocean]

If F is a field and algebraic closure of F has finite dimension over F, is F isomorphic to Reals

[rocky cloak]

Not isomorphic, but it will be a real closed field
https://en.m.wikipedia.org/wiki/Real_closed_field

[chilly ocean]

Truncated exponential polynomials are irreducible for every n

Means n terms of tayler series of e^x

[hidden wind]

whew i remember elementary equivalence being a personal highlight of the model theory course i took

[weak ridge]

As mentioned by @south patrol you can assume without loss of generality that G = Z. Now if n is a cyclic generator of Z, then nZ = Z which implies that n | 1. That forces n = 1 or n = -1.

[weak ridge]

Knowing in advance that any infinite cyclic group G is isomorphic to Z, you cannot have g = g^(-1) unless g is of course the identity element of G. In additive notation, that would be translated in (Z, +) as g = -g => g + g = 0 which automatically yields g = 0 (the additive identity of Z). Moreover any element of G has infinite order apart from the identity that has always order 1.

[crystal vale]

Can I get a hint?

[rocky cloak]

Consider the smallest possible norm of an element in R-tildeR

[crystal vale]

I got it, thank you

But how did you get the idea that we have to take smallest norm of an element in R-tildeR?

[rocky cloak]

Well, I looked at the equation
x = qu + z with N(z) < N(u)

How can I force z to be a unit? Well if everything with norm less than N(u) was a unit that would do it. Okay then just do that

[worthy solar]

Dummit and foote chapter 7 section 6

[crystal vale]

Okay

[tardy hedge]

Galois group of (x^3-3)(x^3-2). So Galois group is a subgroup of S3 x S3. The splitting field is degree 18 over Q so the Galois group is not going to be the full S3 x S3... so can I just say its A3 x S3 then?

Do subgroups of G1 x G2 have to be like H1 x H2?

[rocky cloak]

Maybe try to figure out what all the subgroups of C2xC2 is

[tardy hedge]

oh right

[tardy hedge]

So I guess I can't really say A3 x S3 right away

[rocky cloak]

Yeah, it's not that if you think of it as a subgroup of S3xS3, but I think it will be that up to isomorphism

[delicate orchid]

https://en.wikipedia.org/wiki/Goursat's_lemma if you're interested

[rocky cloak]

I guess enpeace would love to mentioned that the Galois group would need to be a subdirect product of the Galois groups of x^3 - 3 and x^3 - 2

[languid trellis]

The first part seems clear to me from orbit stabiliser, we have a map H-> Orb(i) which induces a bijection H/Stab(i) -> Orb(i). I'm unsure how to prove the second part, and also where to use the hypothesis that H is normal in G. Any help would be appreciated

Actually I think I'm going to take a break from galois theory and come back to it in a few months. It's becoming a real slog honestly

Hopefully by then my brain will have matured and understood what is going on

It's becoming pretty hard to carry on studying this subject

[delicate orchid]

All H orbits are the same size. So if n = p what are the possible orbit sizes

You’ve done the hard part lol

[tardy hedge]

What are you using to study it? Dummit and foote?

[languid trellis]

I feel like the possible orbit sizes are 1 or p but i dont know dawg im falling apart

[languid trellis]

jacobson, basic algebra 1

[delicate orchid]

U literally just cited orbit stabiliser og so yes it’s 1 or p. If it’s p it’s transitive so u just have to reason why H != 1 doesn’t act trivially

[languid trellis]

im literaly falling apart

im going to go play league

[tardy hedge]

i feel u

[woven trout]

Totally stuck on this one

My first idea was to get a basis for ker(phi) and extrnd it to a basis for M

But i don't think you can do that in PID modules like u can in vector spaces

[tough raven]

The hint is supposed to be applied to ||the image of phi||.

[woven trout]

Ok let me think

[woven trout]

Maybe if phi(V) has a basis so does M/ker(phi), take coset representatives of this basis of M/ker(phi) in M?

The submodule they generate might direct sum with the kernel to M

[desert verge]

if you have 2 vector spaces and a homomorphism of groups between them, must it be a homomorphism of vector spaces?

[prisma ibex]

no

[tardy hedge]

[glad osprey]

I seem to remember some deduction where you can get phi(ax) = a phi(x) from phi(x + y) = phi(x) + phi(y), but I'm guessing that only works in a field? I think in a vector space you can only deduce phi(a x) = a phi(x) for integers a

[woven trout]

For example think of complex conjugation

Over C as a C space

[woven trout]

Thanks

[thorn jay]

I guess any nonidentity field automorphism haha

[south patrol]

This doesn't work in a field either

Well I guess kerchooboi's example works for that lol

[tardy hedge]

Brb

[glad osprey]

I think I was thinking of the proof that if f : R -> R is additive and continuous then it's linear. Does additive imply linear for maps between vector spaces over prime fields?

[tardy hedge]

Oh so the size of homF(Es, Fs) being equal to [Es : F] is clear right

[Es : F] is like some [F(a) : F] so every embedding of Es is determined by where it sends a and you have all [F(a) : F] distinct choices because Fs contains all roots

[tardy hedge]

Im not sure how dummit and foote justifies that separable degrees are multiplicative

In the corollary there

[worthy solar]

for the product of ideals

How is something like A(BC) defined if A,B,C are ideals from a ring R

my confusion is

BC = {b1c1 + .... + bkck | k in N and bi in B and ci in C)

so I was unsure of it was some nested sum

[thorn jay]

BC is an ideal, and as the product is defined for any two ideals, A(BC) is also defined
Im not sure what the problem is? Do you want an explicit description?

That involves just some distributivity

[worthy solar]

I guess I want to know if saying A(BC) has elements of the form

$\sum_i a_i (\sum_j (b_jc_j))$
was incorrect

[cloud walrus]

Brandon7716

[thorn jay]

Its the definition, so it is correct

[worthy solar]

but I cant figure out how to work with a basic example and say rewrite

a1b1c1 + a2b2c2

into something of the form a1'(x)+a2'(x)

where x would just be whatever the value of that finite sum was iterated over j

like i have to find a choice of a1' and a2' that gives back the original sum

[thorn jay]

Oh wait no this isnt correct

[worthy solar]

or maybe im misunderstanding something

[thorn jay]

It has to b_ij and c_ij

[sweet echo]

^ your inner sum depends on i

[thorn jay]

Also, the max value of j depends on i too

Apologies, i didnt look very well

[worthy solar]

oh...

[sweet echo]

Just like how in BC it wasn't the same c for each b in the sum

[acoustic igloo]

just write it as a1(b1c1) + a2(b2c2)
b1c1 and b2c2 are in BC

in this case the indeces of the inner sums run from j=1 to j=1

[thorn jay]

The summation is doing heavy work there 🔥

[acoustic igloo]

or i guess i should say the first one runs from 1 to 1 and the second one runs from 2 to 2

[sweet echo]

Could be worth trying out a specific example with 3 ideals in your favourite ring (as long as you don't end up making it trivial)

[thorn jay]

A = B = C = (0)

[worthy solar]

would this be a better description?

$A(BC) := {\sum_{k=1}^i a_kx_k | a_k \in A, x_k \in BC, i \in \mathbb{N}}$

and then I guess x_i does have dependencies on the inner sum of BC stuff so another index?

[sweet echo]

Yeah you just know what an x_i looks like because of the defn of BC

[cloud walrus]

Brandon7716

[tardy hedge]

if a field is separable over K, is it also separable over F if F subset of K?

prob not

[south patrol]

Let the field = K.

Now all field extensions are separable

One thing that is true is uh
Suppose you have extensions L/K/F. Then tfae; 1) L/K and K/F are separable 2) L/F is separable

[heady acorn]

"We note that f is separable" why?

[heady acorn]

Any counterexample in the case where we consider proper extensions?

[south patrol]

Well you can just compose with any separable extension

As in let L/K be separable and K/F inseparable

Then L/F inseparable

[heady acorn]

Ah, i see, cool

[south patrol]

The roots are distinct by construction

[heady acorn]

Oh whoops

Right

Thanks haha

[south patrol]

Npp

[tardy hedge]

yes im really struggling to see why dummit and foote says separability degree is multiplicative

like based on their discussion there

was so bad i hit my vape again after quitting for like 1.5 months

lol

[tough raven]

IG we need to show that if F/E is purely insep and K/F is sep, then when we take K^sep the separable closure of E in K, [K^sep : E] = [K : F] and [K : K^sep] = [F : E].

[tardy hedge]

they never even defined separable closure so idk what they are getting at

[tardy hedge]

also every time i ask for help on this question here, ppl mention something like Q7

but Q6 (showing sep degree multiplicative) is before Q7

so idek

[tough raven]

I mean the field E_sep in the second proposition.

[tardy hedge]

oh ok

[tough raven]

IG I should have written K_sep but 🤷

[tardy hedge]

idk why this topic in particular is so confusing to me

Its just strange

[tough raven]

Anyway, IDK how to show this off the top of my head.

The proofs about this stuff that I can actually remember are in terms of counting embeddings.

[tardy hedge]

Do you think I should think about Q7 before looking at Q6 then?

That Q7 in my hw was stuff about counting embeddings

[tough raven]

NGL I don't understand this notation.

Let me just give some statements to develop this in terms of embeddings.

[proven swift]

tysm

[tough raven]

1. For E/F a field extension, define S_E(K) := Hom_F(E, K) for all extensions K/F. Obviously for K ⊆ L, S_E(K) ⊆ S_E(L) (if you want to be categorical, replace ⊆ with → and call S_E a functor). Define [E : F]_sep = max S_E(K) over all extensions K/F, as an element of ℕ ∪ {∞}.

2. If E/F is finite, |S_E(K)| ≤ [E : F] is bounded above. Therefore it has a finite maximum, i.e., [E : F]_sep is finite.

3. [E : F]_sep is multiplicative in towers (the actual degrees don't need to be finite as long as the _sep degrees are).

4. If a is algebraic over F, write its minimal polynomial p(x) as q(x^i) where q is separable irreducible and i is a power of l in characteristic l and 1 in characteristic 0. Then [F(a) : F]_sep = deg(q); in particular, [F(a) : F]_sep divides [F(a) : F] (their ratio being i).

5. [E : F]_sep divides [E : F] for all finite E/F so we can define [E : F]_insep as their ratio; it is also multiplicative in towers.

6. The composite of finite separable (meaning [:]_sep = [:]) extensions is still separable. Hence for any extension E/F, the union of all finite separable subextensions is a subfield of E; call it E_sep.

From now on, assume E/F is algebraic (these definitions aren't that useful for transcendental anyway).

7. Redefine [E : F]_sep as [E_sep : F]; check that this is the same as before. Define [E : F]_insep = [E : E_sep]; check that this is the same as before if [E : F] is finite (but now makes sense slightly more generally, namely when E_sep/F is infinite but E/E_sep is finite).

8. For E/F algebraic, define E/F to be separable if E_sep = E i.e. [E : F]_insep = 1 and purely inseparable if E_sep = F i.e. [E : F]_sep = 1. By 4., this can be phrased in terms of the minimal polynomials p(x) = q(x^i) of a in E: E/F is separable ⇔ p is separable (i.e., i = 1) for all a in E; E/F is purely inseparable ⇔ p(x) = x^i - b for some b in F (because deg(q) = 1 ⇔ q(x) = x-b for some b in F) for all a in E.

9. If p(a) = 0 for any separable p, then the minimal polynomial of a is separable. If a^i = b in F for any i a power of p in characteristic p or 1 in characteristic 0, then the minimal polynomial of a is also x^i' - b' for some i' a power... and b' in F (in fact, you can get there by taking p^th roots of b' as many times as possible). So you can check separability/pure inseparability of extensions by using any polynomial killed by a whether or not it's minimal.

[tardy hedge]

Oh shoot, thanks. I will read this when i have the energy to

[rocky cloak]

I feel like the main question of the exercise is pretty straightforward. But then they say "conclude that [...]". and then you have to actually do something

[amber burrow]

just wanted to ask whats a good philosophy for doing exercises?