groups-rings-fields

[vocal pebble]

Becauze if its smaller than An then the intersection must be equal to An

This is what we are trying to do here

[candid patrol]

Ker phi intersection An <= n so it's smaller than An ok

[cloud walrus]

UGOBEL

[lusty marlin]

Ker φ ∩ A_n is the kernel of the restriction of φ to A_n

Making it normal in A_n

[candid patrol]

Ohhhh, and so, An is simple so ker phi intersection An is An or {Id}, but it's impossible because |ker phi| = n

[lusty marlin]

Yes

[vocal pebble]

Alternatively ker phi intersection An is a subgroup of An and is a normal subgroup of Sn so a normal subgroup of An<=Sn (what was said earlier)

[candid patrol]

Si the intersection of two normals subgroups is always a subgroup ?

[vocal pebble]

It is infact a normal subgroup

[candid patrol]

Yeah a normal subgroup i wanted to say

Thank guys 🙏

[keen terrace]

Anyone?

[candid patrol]

Did you translate with google lens ? Because it has any sense

Try with GPT

[tardy hedge]

For 5, im trying to show it by using exactness of formation of fractions

I got S-1R = 0, (as an S-1A module), how can I conclude R = 0?

[keen terrace]

My answer or the exercise?

[next obsidian]

You can’t

What you are really asking about is faithful flatness of the ring Prod_p\in Spec A A_p

You should just prove by hand that if x/1 = 0 in A_p over all primes p it’s 0

[keen terrace]

I'll try to write both of them clearly:

In the Z[X] ring, consider the principal ideals I=(x-1) and J=(5), respectively generated by x-1 and 5, and the subset K={f(X)∈Z[x] | f(1) ≡0 mod 5}.
a)...
b) Prove that φ:Z[X]-->Z, f(x)-->f(1), is a surjective morphism of rings, and deduce that Z[X]/I is isomorphic to Z

Can I say something like "since phi is the canonical projection (or at least, I think(?)), I know that it is surjective, therefore its image is Z, and since I is a bilateral ideal of Z[X], I know that it is a kernel of the canonical projection. Using the first homomorphism theorem, I can say that Z[X]/I is isomorphic to Z"?

[next obsidian]

If it’s asking you show it’s surjective I don’t think saying “it’s the canonical map which is surjective” is the point

And just because I is an ideal doesn’t mean it’s the kernel of any specific map, you need to show it’s the kernel

[keen terrace]

The teacher wrote this during class, it says "Every I bilateral ideal of a ring R is kernel of the morphism pi:R--->R/I canonical projection". Do i still need to prove it?

Same with the surjectivity: isn't it enough to point out that the function is the canonical projection, thus it has to be surjective?

[crystal sundial]

For this example aren't we more using this corrolary of Gauss's lemma?

The corrolary is this:

[delicate bloom]

I would say using that corollary would be more like reducing x^3-3x-1 = 1 mod 2 since in F_2 it has no roots (a cubic has a linear term if it's factorable)

I also won't pretend to not already be drinking on xmas if I've missed the mark

[next obsidian]

I mean if you appeal to this you’re already implicitly using the first iso and stuff. This method of proof basically assumes you already know everything this problem is asking of you

[crystal sundial]

In Dummit and Foote he says: "A polynomial of degree 4, for example, may be
the product of two irreducible quadratics, hence be reducible but have no linear factor." I am curious as to whether this is true for any n>4

[delicate bloom]

take the product of an irreducible degree n-k and degree k polynomial

for instnace take x^{n-k} + p and x^k + p for p some prime are irreducible by eisenstein

[crystal sundial]

I see, thanks a lot

[crystal sundial]

I am confused as to the highlighted section: why do we only need to consder in F[x]? . The 2nd image is our version of Gauss's Lemma. Shouldn't we need something stronger like : irreducible in F[x] iff irreducible in R[x]?

[tiny jolt]

If you've figured out how to reduce things in F[x], then you know whether things reduce in R[x] as well by Gauss' lemma

On the other hand is it clear to you why a polynomial factoring in R[x] implies that it factors in F[x]?

[crystal sundial]

I think I understand what you mean.

What I was trying to say is that if we are irreducible in R[x], Im not sure that that would imply we are also irreducible in F[x]. For example, 2 is irreducible in Z[x], but it isnt in Q[x].

Because of this counterexample I am confused when D and F says it suffices to check if we are irreducible in F[x]. (This was my interpretation of the highlighted part in my own words)

I presume he really means this corollary to Gauss’s lemma instead, or that my interpretation is completely wrong

[lunar spindle]

Help

[sharp sonnet]

this isnt the place for memes

[next obsidian]

Yes

The point is you can just first check if the gcd is 1 (which by the way is called a primitive polynomial)

And if it is you apply that

[crystal sundial]

Ah I see

[next obsidian]

Cuz like

If the gcd isn’t 1

then it clearly is reducible lol

So the only thing of actual content is dealing with primitive polynomials at which point you can just work over F

[crystal sundial]

Right

I see, thanks so much

That clears it up

[sly crescent]

I don’t think he realized it was a meme

[glad osprey]

Is O(n) isomorphic to SO(n) x {±I}? I think the answer is yes, using phi : SO(n) x {±I} -> O(n), phi(x, I) = x, phi(x, -I) = -x

[rocky cloak]

If n is even, then -I is already in SO(n)

For n odd this should be correct though

[glad osprey]

I see, thanks

[rocky cloak]

But yeah O(2) is not abelian, but SO(2)x{±I} is

[glad osprey]

Right, good point

[somber sleet]

Hey guys I was reading through this proof and I do not understand a little detail. In the second part, where we assume that both N and M/N have finite length, how do we get the composition series (*)? I understand the whole construction, but I am not entirely sure how to get (**)

[coral spindle]

allegedly by Exercise 3.19(b).

It says so in the text

[somber sleet]

I mean in the second part though, wher eit starts with conversely

[coral spindle]

This is just the correspondence theorem. There is a 1-to-1 correspondence between submodules of M/N and submodules of M containing N. It is literally just taking preimages of the M_i.

[cedar condor]

anyone know videos online that teach groups rings and fields? maybe using Abstract Algebra A First Course by Dan Saracino

[icy totem]

Why are irreducible elements in a ring defined only on a domain and not on arbitrary rings?

[rocky cloak]

If xy = 0, then x = x(1 - y). So the concept of irreducibility becomes a little more muddled. You could still define it I guess, but it's not as useful.

[icy totem]

Ah ok, basically it's just not really useful

Cause I was thinking about how a lot of theorems about irreducible stuff dont hold if A is not a domain

Tyanks

[next obsidian]

= x(1-y)^n

Power series

Swag

Ok this isn’t a series but whatever

Power

[barren sierra]

So I am learning about syzygy modules right now

Working on the last sentence where it says to check as an exercise this if-and-only-if condition

[cloud walrus]

Spamakin🎷

Spamakin🎷

[tough raven]

∑_i (c_i - d_i) f_i = 0 in M, but not necessarily in R^t. Your argument is not valid because you do not yet know that phi is well-defined. phi(∑_i c_i f_i) is defined to be Bc = ∑_i c_i (i^th column of B), and phi(∑_i d_i f_i) as Bd. So we need to know that B(c-d) = 0 for the two definitions to match up.

Thus phi is well-defined iff Bc = 0 for all c in the syzygy module.

[serene dune]

https://math.stackexchange.com/a/121971/1015164

But $(2)\cap (3)=(6)\neq (0)$, yet clearly $(2)(3)=(0)$.

[cloud walrus]

nastasya

[serene dune]

i dont get the multiplication here

how is it (0)

[scarlet estuary]

2 * 3 = 0 in the rng they defined

indeed, any product of ideals in that rng must be (0)

[serene dune]

oh, so it is not the natural ring right ?

it just shows such a ring can be constructed

now i need to prove it is indeed a ring

[scarlet estuary]

the answer you linked explicitly defines the rng

and yes, when proving a statement false you only need to find one counterexample

[serene dune]

yeah but i didnt notice that, so im saying i should rather focus on how this is ring now

[scarlet estuary]

yes you'd need to verify that it's a rng in order for it to work as a counterexample

[serene dune]

skipping details, im cooked

[scarlet estuary]

clearly ℤ-addition satisfies the addition axioms and the multiplication axioms are all fairly trivial though

as is distributivity (all relevant products are just 0)

if i was a grader in a ring theory class and a student submitted this example without proving that the rng they constructed is actually a rng, i'd still accept it as a correct counterexample

since it's "obviously" a rng

your class might differ tho

[serene dune]

brain got too comfortable to assume 1 would exist, why should i care

i can see it now(hopefully) tho

[feral knot]

Can someone help me to prove this ?

[rocky cloak]

So one inclusion is hopefully clear. Conversely say
x^n is in I + (a) and x^m is in I + b.

Then x^n = i + ra and x^m = j + sb for i and j in I.

What can you say about x^n+m ?

[feral knot]

i can say x^n+m is in i+(a) and also in i+(b)

(x^n)^m = (i + ra)^m = (j + sb)^n

[rocky cloak]

That's true, but you can say a little more

x^n+m = x^n * x^m
what does that equal from the expressions above?

[feral knot]

I don't see it

[rocky cloak]

So it's
(i + ra)(j + sb)

[feral knot]

ok

[rocky cloak]

Then you can distribute that out, and what do you notice about each term?

[feral knot]

i see it now

all in I

thanks

[crystal vale]

if A is a set, then a subgroup H of S_A is transitive on A if for each a,b in A there exist f in H such that f(a) = b. Show that if A is non-empty finite set, then there exists a finite cyclic subgroup H of S_A with | H | = | A | that is transitive on A.

any hint?

[somber sleet]

Thank you very much for answering and sorry for such a late reply, your insight really helped me a lot! ✨

[tardy hedge]

the equivalence A-p is multiplicatively closed <-> p is prime assumes p is already an ideal does it?

[somber sleet]

Does someone know, why it must follow that "in particular, every maximal ideal of R_P must be contained in P^e"

the correspondence is this

[next obsidian]

The previous line says “every prime is contained in P^e” and maximal ideals are prime

[somber sleet]

this confuses me since two years, which implication holds?
max ->prime or prime->max?

I would have said this first one

[next obsidian]

Maximal -> prime

[somber sleet]

I still don't see why this should follow though.... (I know this should be easy, pls don't judge)

[tardy hedge]

yoo i learning about dis shit rn tooooooooooooo

im using atiyah macdonald though

[next obsidian]

Take a maximal ideal. It’s prime. Thus by the previous line it’s contained in P^e

[somber sleet]

THANK YOU ❤️

yeah sorry, my brain doesn't work full time

I just didn't see it

[somber sleet]

thanks again

[next obsidian]

[somber sleet]

My prof completely follows Gathmann script, so it makes more sense to me to study with this

[rocky cloak]

Yes

[vagrant zinc]

Is this true??

[quiet pelican]

It’s true if you don’t require (sub)rings to have identity

[tardy hedge]

A-S is a minimal prime ideal in what sense? Every prime ideal p with p subset of A-S has p = A-S?

[next obsidian]

Yeah

[tardy hedge]

ty

[glad osprey]

[e^(iθ) 0; 0 e^(-iθ)] is a maximal torus of SU(2), right?

[elder stump]

This isn't from any book or problem. I was just messing around with the idea of harmonic arithmetic.\
\
I defined the set $G=(\bR\setminus{0})\cup{\infty}$, where $\infty$ is just some non-real value. It is defined to have the arithmetic properties $\infty^{-1}=0$ and $0^{-1}=\infty$.\
\
I then defined the binary operation $\bigstar :G^2\to G$ by
$$a\bigstar b=(a^{-1}+b^{-1})^{-1}$$
I've already proven this is an Abelian group.\
\
My question is the following: Is there a known name for this group? Does it have any applications anywhere?

[cloud walrus]

SWR

[digital seal]

you removed 0?

[elder stump]

I forget which group operation gets broken with 0, but one of them absolutely does

[stark pebble]

But what is infinity inverse

If zero is gone

[vagrant zinc]

That is to say that if I have an ideal without identity, then it is a sub ring, I had read that it is not, as far as I remember.

[elder stump]

Infinity is the identity element of this group. It is its own inverse

[stark pebble]

I mean infinity to the power of negative one, not the inverse under star

[elder stump]

Oops my notation may have been misleading. I meant to say $\frac{1}{\infty}=0$ and $\frac10=\infty$

[cloud walrus]

SWR

[stark pebble]

Oh

Right

Similarly in the definition of star i guess?

[elder stump]

And this is strictly in defining $\bigstar$

[cloud walrus]

SWR

[digital seal]

yeah it's fine it's just kinda funky that you remove 0 but then say infinity^-1 = 0, necessitating it in the calculation of the star operation

[stark pebble]

This is what i was trying to point out

[elder stump]

I could write it more tediously

I'll define $a\bigstar\infty=\infty\bigstar a=a$ for all $a\in G$

[quiet pelican]

The reason you’ll have seen that is that almost everywhere does require a ring to have identity

[quiet pelican]

But the source that said that seemingly doesn’t

[cloud walrus]

SWR

[elder stump]

And then I'll define $a\bigstar b=\frac1{(1/a)+(1/b)}$ for all $a,b\in G\setminus{\infty}$

[digital seal]

in either case this is isomorphic to R under the usual addition

[cloud walrus]

SWR

[elder stump]

Oh that's neat at least

Is there a known name or application for this group though other than a simple isomorphism? I was just playing with this idea, so i know nothing about it

[digital seal]

[stark pebble]

Seems like the name is "the additive group of real numbers"

[elder stump]

[tardy hedge]

OMG!

[crystal sundial]

just a small question: it should be that the f_i are pairwise nonassociate instead of pairwise distinct right? i am assuming Dummit and Foote means distinct as in distinct under the equivalence relation: "is associate to"

[cloud solar]

G finite group and the only automorphisms are the identity one and x^7. Show the order of G is a power of 2

[next obsidian]

I guess so, yeah

[cloud solar]

If a is an element of order p prime > 2 i took the inner automorphism a^-1xa . If this is the identity one then a is in Z(G) so p divides Z(G). If this is x^7 then xa=ax^7 for every x in G. So a^2=a^8 so a^6 = e so p=3. Any ideas?

[vagrant zinc]

Pairwise?

[velvet hull]

this reasoning in the first case isn't correct. just because a ∈ Z(G) has order p does not mean that p divides |Z(G)|

[serene dune]

now assume |G| =2^n p^m (p is some odd prime)
now in H7 it looks like x\mapsto x^7 is gonna be the trivial automorphism [if, p=7 we are done, otherwise (7,p)=1] so both cases collapses into one to produce the contradiction ?

[velvet hull]

I'll share my progress on the problem so far-
since Φ(x) = x^7 defines an automorphism of G, then Φ(Φ(x)) = x^49 must also be an automorphism of G. So Φ^2 = Φ or Φ^2 = id, and checking both cases we can get that x^42 = 1 or x^48 = 1 for all x in G.
You can further reduce the x^42 = 1 case to get that x^6 = 1 for all x in G, because x^7 is an automorphism
The prime factors of 6 and 48 are just 2 and 3, and so by Sylow 1 |G| = 2^n3^m for some n,m
(note - this result isn't needed)

[lusty marlin]

The order of any element of a finite group must divide the order of the group

[lusty marlin]

If |Aut(G)|=2, then |Inn(G)|≤2, forcing it to be cyclic. But Inn(G)≅G/Z(G), which means it cannot be both cyclic and non-trivial. Hence Inn(G) is trivial, which means G is abelian.

Not sure how to proceed

[velvet hull]

why can't it be cyclic and nontrivial?

wait no that's not even needed, Aut(G) being cyclic already forces G to be cyclic abelian**

[lusty marlin]

G/Z(G) being cyclic implies G is abelian

[velvet hull]

But then |Aut(Zn x Zm)| >= |Aut(Zn)| + |Aut(Zm)| = Φ(n) + Φ(m), and from there I'm pretty sure you can prove that G is uniquely isomorphic to Z4 by considering its cyclic factors

[tall latch]

What does he mean by various identifications can be made?

Reading Ian Stewart Galios Theory

[kind temple]

K and i(K), L and j(L) are technically different as sets, but if you view them as being identical or equal, then the commutativity condition can be relaxed from the whole commutativity square to just the restriction formulation. the author then considers what happens if you make a more extreme identification of the fields and says that you should also make use of these identifications

i.e., the various levels of identifications

[hidden cairn]

is it simply because N(q) = q^2 ?

(q), 1 + (q), ..., q^2 - 1 + (q) ?

but where do i use q = 3 mod 4?

or q prime

[digital seal]

(q) = q + (q)

[hidden cairn]

true

[digital seal]

q = 3 mod 4 prime is required for field

[hidden cairn]

(q) is maximal in Z[i] so the quotient ring is a field

i don't see how q = 3 mod 4 is used there

[digital seal]

only if q = 3 mod 4

otherwise q is not prime in Z[i]

[crystal vale]

but why q is maximal the reason is q = 3mod 4

[hidden cairn]

ah i see

what about N(q) = q^2?

is that the reason why it has q^2 elements?

[crystal vale]

do you know why prime p, p =1 mod 4 is not prime in Z[i]?

[hidden cairn]

because the norm for when D = 1 mod 4 in Z[D] is different?

[crystal vale]

no, here D = -1, and i am asking about element of Z[i]

[hidden cairn]

ah right

ah because p can be written as sum of two squares

and then a + bi and a - bi are divisors of p

where p = a^2 + b^2

[crystal vale]

right

[hidden cairn]

gotta read the proof of this again tho

was too complicated when i first read it

[crystal vale]

which book do you following now?

[hidden cairn]

dummit and foote

[crystal vale]

oh

[crystal vale]

you can refer Herstein

so now Z[i]/ (q) has structure a + ib, where a and b in Z/qZ

[rocky cloak]

I wouldn't really say it's the reason, though it's vaugely related.

Z[i]/(q) = Z[x](x^2 + 1, q) = (Z/q)[x]/(x^2 + 1)

Which as an abelian group is just (Z/q)^2, this has q^2 elements if q is any positive integer. But only a field when q is a prime over which x^2 + 1 is irreducible

[hidden cairn]

i see, then the field has q^2 elements of course. thanks!

[hidden cairn]

sorry i dont get how x^2 + 1 appeared out of nowhere

are generalizing the Z[i] case to the polynomial ring case?

[rocky cloak]

Z[i] = Z[x]/(x^2 + 1)

But yeah a similar argument would work for Z[x]/(f) for any monic polynomial f

[hidden cairn]

ah i see

[hidden cairn]

this makes sense now

thanks

[somber sleet]

I am confused about the proof of the theorem that R is artinian iff R is Noeth and all its prime ideals are maximal. I was reading the proof in Gathmann, but I can't see why this proof makes sense. First of all, for me it doesn't really make sense that Q_n=R, since it should hold that P_1 ... P_n =0. Also, why is Q_i/Q_i-1 an Artinian R-module? is it because the Q_i are "submodules" of R and hence also their quotient must be?

[rocky cloak]

Yes, that's correct. Submodules and quotient modules of artinian modules are artinian

[somber sleet]

Okay, that makes sense then. Do you maybe know, why we have that Q_n=R? Why should be the empty product 1?

[rocky cloak]

So for any ideal J you want
J^n+1 = J^n * J

And this only works if J^0 = R.

But for the purposes of this you can just define Qn to be R regardless

[somber sleet]

Also I have a question in the last part of this other direction. Why would P being maximal among all other ideals I st R/I is not artinian imply that quotienting S =R/P by a non-zero ideal must yield an artinian ring?

[somber sleet]

Oh okay that's clear, I wasn't really sure why they didn't explicitely write down that they define Qn=R, but it makes sense now, thanks! 🙂

Sorry for spaming with so many questions

[rocky cloak]

Because P is the largest ideal where is not artinian, so if you mod out a larger ideal it will be artinian. That's just what maximal means

Remember ideals of R/P are just ideals of R that contain P

[somber sleet]

Oh yes, I see that now, thank you!

[tardy hedge]

show r(p^n) = p, where p a prime ideal:

if x in r(p^n), then x^k in p^n so x^k is in p, so either x or x^k-1 is in p. If x is not in p then u keep doing this process until u get x is in p

Is that the idea?

r(p^n) is radical of p^n

[somber sleet]

I kind of find it funyy that we are doing the same stuff

[tardy hedge]

Lol yea

Im working thru atiyah macdonald

[lusty marlin]

Yes

[somber sleet]

As far as I know, it also generally holds that if you have and ideal I in R, then r(I^n)=r(I) for all n > 0

[tardy hedge]

Oh yes

[somber sleet]

Then you can use the reasoning you wrote above

Then it might be easier because you only need to show that r(P)=P, reducing anyway to showing that r(P) \subseteq P, since the other direction is "trivial"

[somber sleet]

I have kind of a really easy question, if an ideal doesn't have a primary decomposition, then this ideal can not be primary, right?

Also, in this picture my professor was writing down the proof of the existence of a primary decomposition for every proper ideal in a Noetherian ring and a big question came to my mind. Why should 0=I/I be primary?? This doesn't really make any sense to me since we actually assume that I/I is the only ideal without a primary decomposition, how can it be suddenly be primary?

(pls guys tell me he's wrong )

[tardy hedge]

Your professor records their lecture? Lucky you!

[somber sleet]

Yeah, almost all of my prof do because there are some lectures that intersect. I am really grateful because I can not capture everything from the lecture, so when I go through the chapters again it really helps me, whenever I do not understand what's written, they often add some useful vocal explanation

[tardy hedge]

"The primary ideals of Z are (0) and (p^n), for these are the only ideals with prime radical, and it is immediately checked that they are primary"

I am missing what the relevance of those ideals having "prime radical" have to do with it

Oh nevermind, I think its because it is necessary for primary ideals to have a radical that is a prime ideal

But an ideal having a prime radical isnt sufficient for it to be primary right, since we know that r(p^n) = p for a prime ideal, but p^n need not be primary

[young thorn]

What is an easy way to show isomorphisms between non cyclic groups

[rotund dragon]

do you have a particular problem in mind? that is not at all an easy problem in general (see, for example, the group isomorphism problem)

[long nebula]

The easiest way to show two groups are isomorphic is to write an isomorphism down

The easiest way to show they're not isomorphic is to find a property of groups that holds in one but not the other

[young thorn]

I would but I’m out rn 😭

[wicked patio]

map generators to generators and check hoping the map works out?

[lusty marlin]

There is no general method

[potent condor]

Yeah for one thing, any method will only work when the groups are isomorphic

[abstract rock]

technically if given a graph isomorphism oracle, for finite groups atleast you can resolve the group isomorphism problem

[lusty marlin]

What are the complexity bounds on this?

[abstract rock]

its actually an open problem in computer science if there exists a P-time algorithm for the graph isomorphism problem

but we know its also not NP-complete, and that it is at most NP-intermediate (so current tries are still exponential time)

[lusty marlin]

Oh ok

[lusty marlin]

Yeah

[wicked patio]

If P = NP then it is automatically NP-complete, so this is a strange statement

[abstract rock]

should have just said we know its not NP-hard

[wicked patio]

If it's NP-complete then it is also NP-hard

[abstract rock]

but it is not np hard

that's the point?

[wicked patio]

So we've found a problem in P which is not NP-hard?

[abstract rock]

pedantics aside this isnt the channel for this kind of discussion

[wicked patio]

You should publish that as a proof for P=NP then

[young thorn]

how do u guys work out the group table for the group of rotations of a tetrahedron

im very much struggling to find what happens when u combine rotations with different directions

[acoustic igloo]

it's a 12x12 table right?

it might be easier if you convert everything to permutations on 4 elements

[clever dagger]

Not sure if lower-level question are on-topic here. I'm starting abstract algebra for the first time and wanted to know if I'm doing things right.

For showing that the Direct Product is satisfied by GR 1, 2, and 3, is it sufficient to say:

GR 1. is satisfied since we've defined the Direct Product to be component-wise, then e.g., for (x,x') * (y, y') * (z, z'), we have that (xy)z = x(yz). Since x,y,z are members of the group G, and thus associative. (Don't want to type the rest of the component-wise for the primes).

GR 2. is satisfied since we've defined the DP to be component-wise, then our identity element will just be an ordered pair (e, e') where e is the identity of group G and e' is the identity of G'.

GR 3. is satisfied since we've defined the DP to be component-wise, then our inverse will just be an ordered pair (z, z') where z is the inverse of xy such that xy(z) = e, where e is the identity of G. And z' is the inverse of x'y' such that (x'y')(z')=e', where e' is the identity of G'.

[kind temple]

your proofs for GR 1 and GR 1 are fine but could be phrased more clearly and precisely.

your proof for GR 3 seems to be off a bit. you need to show that for any (x,x’) in the product, there is an ordered pair (y,y’) such that
(x,x’)*(y,y’) = (e,e’) = (y,y’)*(x,x’)

[clever dagger]

I think I'm underthinking the problem. Is it not immediate (again, since DP is component-wise) that if y is the inverse of x, and that y' is the inverse of x', then we have:

(x,y)(y,y') = (xy,x'y') = (e,e')?

Good! Thanks a bunch. This was very helpful. I'm sure I'll get better at actually writing them as I do more exercises.

[rocky cloak]

It is not known whether it's NP-hard or whether it's in P or neither (or both).

But it's in NP, that part is for sure.

[rotund aurora]

Not sure what you are trying to do but the symmetry group of the tetrahedron is the same as the one of the cube, maybe that makes it easier

[regal hornet]

Hey, i just had an exercise were we found all the subrings of Z^2, and i was wondering if there was also a way to list all the sub-groups of Z^2. After a quick search, I cant’ find anything. Would any of you know ?

I presume they are either Z2 whole, or either (x, y)Z, but im not sure

Nah that cant be true cause that doesnt include things like every point so that the sum of coordinates is even

[wicked patio]

Should pretty much be all the ways to choose 2 elements to be generators (ofc you will get many duplicates this way)

or 1 element as a generator, or the trivial subgroup ofc

hmm

[rocky cloak]

There are quite a lot of them, but like Dreyuk is aluding
Any subgroup of Z^2 is free of rank at most 2. So you can think of this as the subgroup generated by the rows of a 2x2-matrix.

So it will be the same as the set of 2x2 matrices up to (integer) row operations.

Such matrices can always be reduced to the form
[x, y; 0, z]
with x, z non-negative
and either z=0 or 0 <= y < z
and x=0 implying z=0

I think that should cover all the redundancy

[somber sleet]

why does it directly follow that if an element is integral over R, then it also is over R[x]? I suppose definition, since R \subseteq R[x] but I'm not sure

[lone niche]

well if an element b satisfies some monic* polynomial p(t) with coefficients in R, that exact same polynomial works for R[x]

[empty kernel]

R,+,• is a ring

r•r=r for every element in R

prove that r+r=0 for every r

Idk what to do i feel like it is some simple thingy but i dont see it

[mighty kiln]

Show 2 = 0

[empty kernel]

Like its prob some smart use of distributivity or sum

[somber sleet]

Makes sense, thanks, was just not sure anymore

[empty kernel]

Ait will try

[somber sleet]

I have kind of a stupid question regarding minimal polynomials. Why is the minimal polynomial of a + b sqrt(d) over Q(sqrt d) \ Q what's written in the last line and not (x - a - b \sqrt(d))? I do not remember a lot of minimal polynomials, so if it's trivial I'm sorry

Oh wait, it's over Q right? Not Q(\sqrt d), so we can't have any roots, right?

[young thorn]

This is more of a mathematical writing qu but when u answer qus and u use x^(-1) as an inverse to x where x is an element of a group let’s say

Do I need to specify x^(-1) is the inverse to x

Or can I just use it directly

Before defining it

[young thorn]

cz im always terrified of losing marks by making an assumption

[young thorn]

just realised rotations by pi that preserve the polygon also act as lines of symmetry

made it infinitely easier to visualise

lol

[empty kernel]

What does the notation I+J mean

I and J are both ideals of a ring R

[mighty kiln]

{i+j: i ∈ I, j ∈ J}

This is also an ideal

It is a subideal of (I, J)

[empty kernel]

Yuh thats what i have to prove but i doubt that will be very hard

Thanks for helping me

[lusty marlin]

No

[empty kernel]

Z[x]/(2)

My friend says its isomorphic with Z_2 but i dont see how

[lusty marlin]

It isn't

It clearly has more than 2 elements

[empty kernel]

I KNEW it

[empty kernel]

Isnt it isomorphic with Z

thats what i was thinking

Wait no dont spoil

Im thinking

[lusty marlin]

No. It has characteristic 2

[empty kernel]

Hmmmm

[rocky cloak]

Maybe try to list out what some of the elements are, and how they relate

[south patrol]

Z_2 is infinite anyway

[empty kernel]

OH WAIT OMG

[tardy hedge]

Whats the context for this

[empty kernel]

IS IT Z_2[x]

[lusty marlin]

Yes

[rocky cloak]

Z_2 usually denotes the 2-adic integers

[tardy hedge]

Oh ok

[empty kernel]

Yipie when i said out load all the polynomials with coefficients 1 and 0 i was like hol up

[empty kernel]

Oh ye we learned that

But our prof was like

Z_2 is easier notation so we do that anyway screw the p-adic integers

[rocky cloak]

I don't feel like Z/2 is that hard, but if you're not gonna talk about p-adics anyway, then whatever

[empty kernel]

Ye we didnt rlly talk abt it lol
He just said
Yo u should ask the analysis prof abt the p-adic int

[empty kernel]

here i am again

[tardy hedge]

Sup Obama

[empty kernel]

Z_2[x]/(x³+x+1)
bc of division algoritm we can show that u can write every f+(x³+x+1) with a unique representing element of degree 2 or less
and then we know that Z_2[x]/(x³+x+1) only has 8 elements cuz only 8 polynomials with degree 2 or less in Z_2[x]

is this logic right?

i tried translating it to english as best as i could

[empty kernel]

thanks jagr

i think im starting to understand it

yipie

[tardy hedge]

Yassss

Good job Obama

[south patrol]

ye this lol

In particular since everything i do is mod p or p-adic lol

[lucid abyss]

hi, some people willing to share their galois theory exam?

[empty kernel]

how do u show that a polynomial is irreducable

[void cosmos]

alot of ways

[empty kernel]

like uhm i now have a third degree polynomial over Q so i know that deg f + deg g = 3 if my polynomial is reducable

[void cosmos]

if you have a third degree polynomial over Q then it is enough to show that it has no roots

[empty kernel]

so either f or g has degree one which means that it is enough to show that the polynoial has no roots in Q right?

[void cosmos]

yes

u can do that by using the rational root theorem

[empty kernel]

df is that

[void cosmos]

google it

[empty kernel]

yup was doing so

ok but what are more general ways?

lets say i got a dgree 4 polynomial

ok im back

the rational root theorem is quite powerful i believe

[tough raven]

Then it's not enough to check for roots because it could factorise into two quadratic factors.

[empty kernel]

ye i figured thats why i want to know a mroe general way

[tough raven]

I don't think it's more general but Eisenstein's criterion can be used to show irreducibility for polynomials of any degree (although it's only sufficient, not necessary).

[empty kernel]

oh wait we might have learned that theorem

it appears i made a mistake while making a summary i skipped the page where we learned all the criteria lol good thaat i noticed this now

thanks both of you

[potent condor]

np

[young thorn]

when doign groups of rotations do we normally assume anticlockwise rotations or clockwise

[lusty marlin]

It doesn't matter

[tardy hedge]

new faces around here these days

[ashen heron]

obama has been around here longer than you

[tardy hedge]

i mean like

i guess

ive been very active here for like 1.5 years though

obama doesnt even have an active tag

[tardy hedge]

oh wait i forgot i was actually kicked at one point lmao

im not sure when i joined this server but it was pre 2024

i was crashing out and being a troll

this server is much too valuable for me to ever be silly again though

[noble adder]

What's the difference between a direct product and a direct sum?

[velvet hull]

For abelian groups they mean the same thing

There exists two names because they’re special cases of the more general product and coproduct, from category theory

[noble adder]

is there anything about the number of nonzero elements

[noble adder]

ok i thought this had something to do with it

[velvet hull]

What do you mean?

IIRC direct sum actually isn’t defined for nonabelian groups

[noble adder]

hungerford is saying something about how in a direct sum there can only be a finite number of nonzero elements

well he's using a big sigma so im assuming it's a direct sum

in context this is for the group ring

[static mauve]

If the left and right cosets of a subgroup are the same, that doesnt mean that the "parent" group of the subgroup is necessarily Abelian, right?

[vapid vale]

correct

[coral spindle]

Consider the dihedral group with 6 elements, D_6, which has a normal subgroup isomorphic to C_3, the cyclic group with three elements. You can check easily that the left and right cosets of the subgroup are the same, but D_6 is not Abelian. In fact you can find examples of subgroups and overgroups for which this holds, but neither are Abelian.

[rocky cloak]

The condition of left and right cosets being the same is called being a normal subgroup.

The quaternion 8-group is an example of a non-abelian group in which every subgroup is normal.

[acoustic igloo]

things got a little awkward while showing that the set of automorphisms of a group G form a group under function composition

but i guess you have to do something like this?

show that the inner automorphisms of a group G form a normal subgroup of the group of all automorphisms of G
bro this is getting too meta

[lusty marlin]

You also need to show:

1. The identity map from G to G is an automorphism.
2. The composition of two arbitrary automorphisms of G is an automorphism of G.
   You have only shown that the set-theoretic inverse of an automorphism is also an automorphism.

[acoustic igloo]

i defined $\iota(g)=g$ earlier

[cloud walrus]

Axe

[somber sleet]

Is the quotient of a local ring by an ideal always local?

[next obsidian]

Apply the correspondence

[somber sleet]

oh yes, thanks!

An other question if I have the multiplication of two ideals I,J in an integral domain st IJ=0, does it follow that one of them has to be 0?

[lusty marlin]

Yes

[coral spindle]

Well if both of them have some nonzero element, then what happens in IJ?

[somber sleet]

then it's obv not zero since the multiplications generate IJ

What happens if one of the ideals is the localization R_Q at a prime ideal Q in R

are these idealsnever 0?

Okay no, They can't be zero if R is not zero, I guess

[rocky cloak]

What does "the ideal is the localization" mean?

[coral spindle]

Yeah I don’t understand what that means either

[somber sleet]

I was actually trying to fully understand a proof, Krull's principal ideal thm. And the last sentence states that Q^nR_Q=0, where R is an integral domain and Q a prime ideal in R. After stating this, they directly follow that Q must be 0, since R is an integral domain

[coral spindle]

And what is n?

Does it matter?

[somber sleet]

I was confused, because I didn't see why they didn't at all consider that $R_Q=0$, but if $R_Q = {\frac{a}{s} \mid a \in R, s \in Q^C}$, then the whole ring R must be 0

[cloud walrus]

damn_guuurl

[rocky cloak]

The localization at a prime ideal is never 0 anyway.

But R_Q is not an ideal

[somber sleet]

I thought no, but now you're giving me a doubt

[somber sleet]

they are in general only rings, right?

[rocky cloak]

They are rings yeah.

They're also R-algebras.

They're local rings.

[somber sleet]

does the fact that they are local rings help me? Because I don't see it.

[rocky cloak]

No, none of those facts help you.

The most helpful thing is just to write out the definition of Q^n R_Q

And what it would mean for something to be 0

[somber sleet]

Isn't it equal to Q^n?

[rocky cloak]

Not really no

[somber sleet]

I mean, I see that taking an element $q^n \frac{b}{c}$ from $Q^nR_Q$ is 0 iff qb is 0, but I don't recognize if the set should fulfill some explicit definition

[cloud walrus]

damn_guuurl

[spare orbit]

are these things even used

[next obsidian]

Not really

[candid tulip]

If a and b are arbitrary elements of (G, •), does it mean that a•b is arbitrary too?

[lone niche]

ig it depends on what you mean by arbitrary, once you fix a and b, a dot b is also fixed. But if you mean that there is no way of knowing what it is without knowing a and b (except in degenerate case ig), then yeah.

[candid tulip]

I meant the second sentence

[candid tulip]

Wdym by “degenerate case?”

[lone niche]

trivial group lol

[candid tulip]

Oh

[serene dune]

Let $K$ be a field and $S$ a set. Let $x_0 \in S$. Let $F(S, K)$ be the ring of all mappings from $S$ into $K$, and let $J$ be the set of maps $f \in F(S, K)$ such that $f(x_0) = 0$.

1. Show that $J$ is a maximal ideal of $F(S, K)$. \

2. Show that $F(S, K) / J$ is isomorphic to $K$.

[cloud walrus]

nastasya

[serene dune]

this is fairly straight forward right ?

we are gonna just focus on the 2nd claim, take the map, well-def and the kernel is J
now for all a \in K there is some map f \in F(S,K) such that f(s) = a

is there anything more to it ?

[lusty marlin]

If by s (in the message right above) you meant x₀, then no, you're done

[serene dune]

s is arbitrary element of S for which it's not mapped to 0 ig

also without considering the field xan we show the first claim by brute force

i hope so, i will try that

[lusty marlin]

The homomorphism you consider to prove claim 2 should be the map ϕ: F(S,K) → K defined by f ↦f(x₀)

This has kernel J and is clearly surjective

[thorn jay]

No, but they do exist to make the picture of "group-like algebraic structures" complete, in a way.
This is because not necessarily associative binary operations arise all over math, so it's convenient to describe them using "a magma with so and so property", and because mathematicians like naming things.

It's always mathematicians who like naming things.

[crystal vale]

let G be finite group such that there exists a non-identity element which conjugate to its inverse. Then what can we say about cardinality of G, even or not?

i think it is even and we have to show existence of non-identity element which has order 2.
any hint?

[abstract rock]

if an element is conjugate to its inverse, what does that relation look like?

[crystal vale]

they are conjugate to each other, they are in same orbit with conjugation action

[south patrol]

Actually yeah what I had was fine I think

Try to show that the conjugacy class of any element c9njugate to its inverse is closed under inverse

From there it is straightforward

[crystal vale]

oh

done

[crystal vale]

how?

[south patrol]

So what can you say about the order of that conjugacy group

[crystal vale]

even

[south patrol]

Yeah

I guess technically there's a special case if g has order 2 but then we are done anyway by Lagrange

Otherwise yeah we just pair up elements with their inverses

[crystal vale]

and if it is not then some element also follows x = x^-1

[crystal vale]

yes

[south patrol]

And now can you see what to do next?

[crystal vale]

how did you get this idea?

[south patrol]

Idk lol

[crystal vale]

yes its order divides | G | so | G | is even

[crystal vale]

sad

[south patrol]

I guess I knew of the idea of pairing up w inverses

[crystal vale]

thank you @south patrol

[south patrol]

And it sort of made sense that if one element was conjugate to inverse then all the others in the class will be too

[crystal vale]

yes

[south patrol]

But hm I wonder if there is another way to do it

[loud herald]

Maths be like

[south patrol]

Well you can consider the image in the abelianisation but what happens if g is in the kernel of the abelianisation map

[loud herald]

Fr fr

[rocky cloak]

So you have
gxg^- = x^-
that means
g^2 x g^-2 = x
so what does that say about the order of g?

[crystal vale]

Abelianisation?

[lusty marlin]

The abelianisation of a group is the quotient of the group by its commutator subgroup

[crystal vale]

Thanks

[crystal vale]

I don't know

[rocky cloak]

Well, can the order be 3 for example?

[somber sleet]

are field of fractions of integral domains always integral domains?

[crystal vale]

Yes

[rocky cloak]

Even more, they are in fact fields. Hence the name

[crystal vale]

Yes, the ring of fractions and the field of fractions are somehow different

[crystal vale]

No then g = e

[rocky cloak]

There is something called the total ring of fractions, which for an integral domain coincides with the field of fractions. But it will not be a field of you start with something not an integral domain

[crystal vale]

Yes

[rocky cloak]

Indeed, and can you generalize that to orders other than 3?

[crystal vale]

Total ring of fractions and rings of fraction, are different? No right?

[rocky cloak]

Should be different names for the same thing yeah

[somber sleet]

thank you, was not sure anymore (Could have just thought about Q....)

[crystal vale]

I think the point is g^4 = e

[rocky cloak]

Well, not quite. But you're onto something.

What made you conclude it would be 4 in particular?

[static mauve]

So Im confident in my answer to #28 in these 2 questions, but Ive been struggling with #29. Is #29 truly the converse of #28? To me, Im not convince its really the converse. Every left coset equaling its associated right coset (gH = Hg where g is a fixed number) doesnt seem like the same thing as 2 partitions being the same

[velvet hull]

so let's start with the assumption. for every left coset aH, there exists a b in G such that aH = Hb

since the left/ right cosets of any subgroup partition the group, let's choose a g in aH

then, there must exist a h in H, such that g = ah (by def. of a coset)

but since aH = Hb, there must also exist a h' such that g = h'b

so g = ah = h'b, and by rearranging we can get that h'^-1h = ba^-1

but since the left side is in H, and a and b each cannot be in H individually

this means that a=b

i.e. aH = Ha for all a in G

[velvet hull]

from this it's easy(ier) to show that ghg^-1 is in H, can you figure out how? :)

[crystal vale]

I am dumb

[static mauve]

I dont see how you can get h'^-1h = ba^-1... maybe im missing something very trivial here. Inverting h' will sandwich a like -> h'-1 a h = b

[velvet hull]

oh wait, you're right lol. let me rethink this real quick

[crystal vale]

Which book?

[static mauve]

Fraleigh (A first course in Abstract Algebra)

[crystal vale]

Yeah I did that exercise

[crystal vale]

So let it give us gH = Hg, do you know why?

Don't let

It give us gH = Hg

Because g in gH left coset, since they making same partition so gH = Hg, because g also in Hg

[static mauve]

Oh, and thats the case because since e is an element of H then ge = g (where g is the identity) right?

[crystal vale]

If g^3 =e, then xg = gx so xg = x^-1g give us x^2 = e

[crystal vale]

Yes

[crystal vale]

Here I think your point is if g has order odd then x has even order, right? Because g can have an odd order

[static mauve]

Hmmm. For the general case Im not sure what you are suggesting helps me with the aH = Hb case... I need to think about this/convince myself of this.

[velvet hull]

if g has odd order, then x = g^(o(g)) x g^(-(o(g)) = -x, i.e. this forces x to be equal to its inverse, which is not necessarily true

[velvet hull]

his proof is a lot cleaner than my aH = Hb proof lol

you can ignore that

[static mauve]

I guess I dont understand this particular detail specifically

[velvet hull]

you can still prove it real quick: if aH = Hb, if we know that aH = Ha then that implies Hb = Ha as the right cosets partition the group

[static mauve]

Im not sure that I know that aH = Ha in the general case.

[velvet hull]

so, left and right cosets partition the group G

we are trying to investigate how the partition looks like

if we know that the partitions are in fact the same, then if I can show that a section from the first partition contains a common element with a section from the second partition

then the two sections are necessarily the same section

[static mauve]

If the 2 partitions are the same, then I know that 1 left coset equals 1 right coset... but that doesnt necessarily mean that each cell of the partition is generated from the same element in G (though maybe that is indeed the case)

[velvet hull]

right, that's good thinking!

[velvet hull]

but can you agree with this at least

okay, now let's pick some random a in G

since H must contain 1, we know that a1 = a must be in aH and also 1a = a must also be in Ha

but then that means aH and Ha share at least a

[static mauve]

Riiiiiiight

So then we can make the further conclusions about g-1 h g is an element of H for all h and g....

Ok, I think this illuminates all of the details 😄

Thanks for walking me through this. I appreciate yall!

[crystal vale]

But if g has an odd order then x has an even order

[velvet hull]

where did you get that from?

[crystal vale]

What?

[velvet hull]

this

[crystal vale]

.

[velvet hull]

oh right

yeah it's true, I'm just pointing out you know more than that

x doesn't just have even order

but that's not the general case, and I think it's not the point jagr wanted you to figure out

[crystal vale]

I think it is on g, if g has even order we are done

If g has odd order then we are done by x

[velvet hull]

ok yes it works, just not what I had in mind

but that's fine lol

[rocky cloak]

Yes, if g has odd order then x has order 2. So if x doesn't have order 2, g has even order.

In any case you get an element with even order, which is what you wanted to prove.

[velvet hull]

oh, so that was the intended idea, since you pointed out g^2xg^-2 = x I thought you wanted that o(g) = 2*o(x)

[tardy hedge]

Every ideal in A/a can be mapped back to an ideal of A containing a. But there can be ideals in A that neither are contained in or contain a, so what happens to these ideals in A/a?

these are not ideals in A/a?

Not sure why A/a has only one prime ideal

I know that nilradical of A/a is intersection of all prime ideals of A/a

Is it because if there was another prime ideal, then it would contain the nilradical, but since r(a) is maximal in A, it is maximal in A/a?

[south patrol]

The image of an ideal under any surjection is still an ideal. But note that if I is an ideal of A (and J another fixed ideal) then the image of I in A/J is the same as the image of I + J.

[velvet hull]

the "hence A/a has only one prime ideal by (1.8)" is likely referring to the theorem that the nilradical ideal is the intersection of all prime (in fact prime minimal) ideals

[south patrol]

This is why one must restrict to primes containing J to get a bijection

[velvet hull]

and so since the nilradical ideal of A/a (i.e. the image of r(a) under quotient) is maximal by assumption, the only way for the intersection of prime ideals to be prime is for it to be r(a)/a only

[south patrol]

Yeah lemma is that if the nilradical is prime then that is the unique miminal prime ideal

[tardy hedge]

So why do we need r(a) to be maximal in that theorem?

[velvet hull]

r(a) is maximal implies r(a)/a is maximal

but then since r(a) = intersection of prime ideals

r(a)/q must be the unique prime ideal, and be the intersection of itself with, well.. itself

if r(a)/a is no longer maximal then it could be the intersection of two distinct prime ideals

[tardy hedge]

unique minimal prime ideal?

[velvet hull]

sure, but that's not needed

[tardy hedge]

Oh, A also has only one prime ideal?

[velvet hull]

A/a does

not sure about A

A might contain prime ideals inside r(a), but none otherwise of course

[tardy hedge]

not sure what u mean here

I think im getting myself confused here

[velvet hull]

actually, I don't think you even need the fact that r(a)/a is the unique prime ideal

r(a) is maximal, so by considering the quotient map A/a -> A/r(a) which is a field, every element in A/a has the form u + r for some unit u and r in r(a)/a (i.e. nilpotent)

so every zero divisor in A/a is nilpotent, and you get the statement

[long geyser]

no, unique prime ideal period

there is only one prime ideal in A containing a, therefore there is only one prime ideal in A/a, by prime ideal correspondence

[crystal vale]

Yes thank you

[wooden rain]

In real analysis, it's common to identify linear maps R^m -> R^n with vectors R^{mn}. Is there a geometric intuition behind that isomorphism?

The best I can think of is:

1. we fix the bases of the domain and codomain of a linear map,
2. we represent the map as a matrix,
3. we embed the columns of the matrix in R^{mn} and "combine" them into one big vector

But well, it's quite abstract and maybe there's a nicer way of looking at it that can lead to some non-trivial conclusions or simply be useful

[crystal vale]

i think M_mn(R) is equivalent to R^(mn), but i am using equivalent in sense of metric space

[wooden rain]

Now that you say that, I think it is an isometry indeed

[viscid pewter]

yeah, matrices form a vector space

[velvet hull]

R^m -> R^n is isomorphic to M_mn(R), which is a mn-dimensional vs over R so is isomorphic to R^mn, as everyone above me said

but really the isomorphism between Hom(R^m, R^n) and M_mn(R) is a lot more natural

for example if n=m, then function composition can be represented nicely as matrix multiplication, whereas you will have to define some crazy operation on the nm-dimensional vectors

[wooden rain]

Yeah, that's true. Both isomorphisms are still clear, but it's very hard to try to imagine something in higher dimensions, which is why M_mn(R) -> R^mn is tricky. But I should probably also provide more context of my question

I'm trying to understand the second total derivative of a function f : R^m -> R. Using the Taylor expansion, I can associate D^2f(x) with the best "quadratic" approximation (in the sense of a "coefficient") of f in the neighbourhood of x, which can give some intuition why that object may be related to whether a function is convex or not, etc. But that requires some knowledge about Df(x) in the first place + I don't think it fully encompasses the meaning of D^2f(x) yet.
The way D^2f(x) is usually defined is we consider Df(x), so a linear map, and use the isomorphism phi between L(R^m, R) and R^m (for higher-order derivatives, it's naturally a more complex object, which is why I can't simply waive it off as trivial) and consider the total derivative of the map phi composed with Df. While it makes sense to do that and I get it algebraically, I don't really see what is going on geometrically, which brought me to my question

[young thorn]

if u have permutations that share the same cycle type are they conjugate?

[lusty marlin]

Yes

In fact two permutations have the same cycle type if and only if they are conjugate in S_n

[young thorn]

but then

when i was working with the dihedral group D_4

and i expressed R^2, V and H (V, H being reflections in the vertical axis and horizontal axis respectively)

i got R^2, V and H have the same cycle types

but only V and H are conjugate to each other

whereas R^2 existed in its own conjugacy class

R^2 = (13) (24) H = (14) (23) and V = (12) (34)

[lusty marlin]

I meant conjugate in S_n

[young thorn]

(where 1 is the top left vertex, 2 is top right, 3 is bottom right, 4 is bottom left)

[tardy hedge]

Why is it that if an ideal is primary, it has only one associated prime ideal?

[young thorn]

but if u express it in terms of permutations then dont u get elements of S_4 only

[tardy hedge]

This is soo groups rings and fields

[viscid pewter]

(13)(24) is conjugate to (14)(23) in S_4, but R^2 is not conjugate to V in D_4

the group you're working with matters

in S_4, there are just more elements you can use for conjugation

[young thorn]

why tho arent both elements of S_4

[viscid pewter]

so it makes sense that more things will be conjugate

ok so x and y are in the same conjugacy class if there's g such that gxg^-1 = y right

there exists such a g in S_4

but not in D_4

S_4 is larger than D_4

[young thorn]

yes

[lusty marlin]

If two elements are conjugate in a group, that doesn't mean they are conjugate in a given subgroup of that group

[young thorn]

but in S_4 wouldnt (13)(24) and (12)(34) still be conjugate to each other

[viscid pewter]

yes

[lusty marlin]

They are

But not in D_4

[lusty marlin]

As said here

[young thorn]

so essentially if i wanted to show conjugacy in dihedral groups using permutations would be problematic?

[viscid pewter]

yes

very

[young thorn]

unless it's d_3

[young thorn]

so the only way to really do it is just go through every element

[viscid pewter]

no

there are tricks iirc

do you know presentations?

[young thorn]

no

[viscid pewter]

well they help ok

[young thorn]

ive finished a

introductory groups module

which covered all types of groups up till basic conjugacy

[viscid pewter]

do you know generators and relations? they're really useful

[young thorn]

yes

i do

[viscid pewter]

oh ok that was what i meant by presentations

[young thorn]

ah ok

[viscid pewter]

ok so you know that all elements in D_n are either of the form r^k or sr^k for some integer k?

and rs = sr^-1

it works out

[young thorn]

i normally use rs

is that wrong

[viscid pewter]

same thing

[young thorn]

aight

[viscid pewter]

ok so suppose x is r^k. then first let g = r^l and compute gxg^-1. then let g = (r^l)s and compute gxg^-1

then consider if x = (r^k)s and do the same thing

[young thorn]

ah i see what u mean

[viscid pewter]

it makes it a lot easier lol

[young thorn]

the question which told me to find the conjugacy classes explicitly gives 4 different reflections however

as in expressed as D, D', H, V

[viscid pewter]

i mean ok

let whichever one you want be s

then all the other ones are just rs, r^2s, r^3s

in some order

[young thorn]

wont i need to figure out which one is equal to rs and then r^2s

for some s

[viscid pewter]

sure

that's not hard

[young thorn]

ye

aight thxthxhthx

wait

[young thorn]

but if ur working with D_4 and D_4 is isomorphic to some subgroup in S_4

wait

no it wouldnt

nvm

wait

yea it wouldnt

D_4 is isomorhphic to some subgroup in S_8 i believe

by cayley's theorem

[lusty marlin]

D_4 is isomorphic to a subgroup of S_4 because its elements are permutations of 4 vertices, but D_4 is also isomorphic to a subgroup of S_8 by Cayley's theorem because it is a group of order 8

[young thorn]

but then if they are isomorphic to a subgroup of S4

and we showed that both H and V can be represented as permutations of that subgroup

then if those permutations are conjugate why are H and V not conjugate...

[unkempt stream]

Does the ring theoretic form of Chinese Remainder Theorem for coprime ideals still hold when there doesn't exist a unit?

My proof works fine for the simple case of two coprime ideals, as if A + B = r, and we have some u and v in R:
We want an x in R such that:
x = u (mod A) implies x = a + u
x = v (mod B) implies x = b + v, then we have a - b = u - v, so we can choose our (a,b) due to A + B = R, and set x = a + u = b + v

But the issue is the inductive step for more (pairwise) coprime ideals I can't really formulate without the existence of a unit

doom.

[south patrol]

The usual proof uses 1 very explicitly

[unkempt stream]

i was wondering if it didn't require the existence of a unit for the C being coprime to A \cap B

i guess it does

[long geyser]

doom.
also known as non-unital ring

[proud hedge]

hey guys quick question
what books did you read for an undergraduate abstract algebra or group theory course?

[young thorn]

i started topics in algebra by herstein in the summer

but didnt get that far cz uni started lul

but from the little i did i thought it was very good

[proud hedge]

i have a little problem understanding cosets and i thought about looking to other books but couldn't find a good one

do you think herstein could help with that
i am studying fromFundamentals of Abstract Algebra
Textbook by D. S. Malik, John M. Mordeson, and M. K. Sen

and it explains cosets before quotient groups

[coral spindle]

There is no such perfect book or perfect explanation. I would suggest just choosing a book and sticking with it.

If you are having trouble understanding some specific part of a book, why not ask for clarification here

In reality, no matter how well explained something is, it is only your thoughts and your work which can lead you to actually understanding and being able to use a concept.

[boreal flame]

I'm going through abstract algebra and in the section about field degrees there's a theorem: If a, b ∈ K where K is an extension field of a field F, with [F(a) : F] = m, [F(b) : F] = n, and gcd(m, n) = 1, then [F(a,b) : F] = mn and F(a) /\ F(b) = F.

My initial idea is to prove that it's false that a∈F(b) and go from there (from there it seems fairly obvious what to do) so I say assume (for the sake of contradiction) a∈F(b). Then a = k_0 + k_1b + ... + k_n-1b^n-1 for some k_0, ..., k_n-1 elements of F. so a - k_0 = b(k_1 + k_2b + ... + k_n-1b^n-2.

Am I on the right track here ?

[coral spindle]

Why do you think it's false?

[boreal flame]

If it was true that a∈F(b) then the theorem I'm trying to prove would be false.

[coral spindle]

...?

But you say you're trying to prove that it's false

[boreal flame]

oh

[coral spindle]

OK, well in any case

[boreal flame]

I'm trying to prove that a∈F(b) is false

by contradiction

[coral spindle]

It's not a great approach I think. Perhaps there is a way, but I think the most straightforward way is just number theoretic

Hint: Use the tower law (a few times).

Hint: show that [F(a,b) : F(b)] <= [F(a) : F].

[boreal flame]

Got it, I'll try that. Thank you

[ripe crest]

I'm reading Category Theory by Awodey and I'm confused about what he means by the 2nd condition

[rocky gorge]

this is a classic awodey statement

what he means is that two words w and w' are equal then there is a finite sequence w = w1 = ... = wn = w' where each step is an application of a monoid axiom

[ripe crest]

Hmm, I'm afraid I don't understand this either. For example, let's look at N as a monoid - but let's define the map, 1 -> a, 0 -> z, + -> (string concatenation). Then we have 5 = aaaaa. And we also have aaaaa = aaaaazzzzz. So let w = aaaaa, w' = aaaaazzzzz. What would the finite sequence look like? like this: aaaaa, aaaaaz, aaaaazz, aaaaazzz, aaaaazzzz, aaaaazzzzz?

[rocky gorge]

yeah exactly

it looks like you're sweeping the associativity axiom of monoids under the rug by ignoring parens (which is fine), so the only nontrivial monoid axiom is xz = x for all x (and zx = x)

what point 2 of your screenshot is trying to say is that applying this rule over and over is the only way two elements can be equal in the free monoid: so for example, if you had a free monoid with two generators a and b, then a /= b because there's no way to get from a to b by adding and removing zs

[ripe crest]

ahhh i see. so if M = (A, +) is a monad, the free monad on A will not contain any of the structure of M because we forget all that information because we have to go to Set before we can go back to Monoid

[rocky gorge]

I assume you mean monoid instead of monad, but yes

[ripe crest]

ah yeah, whoops

[tidal torrent]

random but what does the statement "up to isomorphism" mean in this context

i might have misinterpreted this in all my proofs

[ripe crest]

same in terms of structure

[long nebula]

e.g. "There is only one group, up to isomorphism, satisfying X property" means that technically there are infinitely many groups satisfying X property, but since they're all isomorphic, we just think of them as the same group

so there's only really one group with X property

[tidal torrent]

that being said, does there exist a maximal algebraic extension of a field F, meaning there is an algebraic extension of F that has no proper algebraic extensions?

[unkempt stream]

The algebraic closure

[tidal torrent]

trying to prove this with no sufficient set theory background is kind of pain

seems trivial idk

[unkempt stream]

Constructing the algebraic closure is a pain in the nuts

[tidal torrent]

for such E to exist, does this mean every polynomial in E[x] split in E?

[long nebula]

Here's a fake proof: order the algebraic extensions of F by inclusion. Every chain has an upper bound (the union). Therefore, by Zorn's lemma, there exists a maximal algebraic extension, which is the algebraic closure.

[long nebula]

yes

think of how the complex numbers are the algebraic closure of the real numbers (or rational numbers)

[tidal torrent]

meaning every field F must contain a unique algebraic extension im assuming

up to isomorphism idk

[long nebula]

all of the algebraic extensions are isomorphic, yes

[tidal torrent]

oh ok

[long nebula]

er

algebraic closures* not extensions

[tidal torrent]

otherwise if E does not exist, then i believe it follows kronecker's theorem in which f(x) in F[x] has a zero

which means it would have proper algebraic extensions

[unkempt stream]

If you tried to naively construct it, if you're given a field F, you want to "add" an indeterminate for every root of an irreducible. For each irreducible polynomial f(X), assign it the set {f_1 ... f_n} for it's n many roots. If we take the (disjoint) union over all of these sets of indeterminate roots, call that set S.

Now we want to add all these roots to the field, i.e we want to find F(S) [adjoining all of S]. But when we just consider the polynomial ring F[S], it's essentially "free", we have no relations between the roots themselves, even though we'd definitely have relations between them in the actual closure.

What i mean by this is if we just have the rationals Q, and we might want to append the roots of f(x) = x^2 - 2 to them, and say f_1 = sqrt(2), f_2 = -sqrt(2). In the poly ring F[f_1, f_2], we just have f_1 + f_2, evidently nonzero... but really it's 0 in the "actual" Q(sqrt(2)).

So we need to quotient/factor out the all relators, formal sums and products of the roots that "truly" equal to 0. But we need to somehow "encapsulate" all of these relators in some maximal ideal disjoint from F so when we quotient them out, it's a field containing F. That's usually why we need choice (or something like it, e.g. boolean prime ideals), to actually get that maximal ideal containing all of our relators

[tidal torrent]

not sure if im getting the right idea

[unkempt stream]

It's almost like if we wanted to create a field with a given "relation" on it. Like if we wanted to find a field extension of R with two elements X and Y where XY = -1. In F[X,Y] (the poly ring in two variables), XY + 1 is definitely nonzero, but if we find some MAXIMAL IDEAL M containing XY + 1 that only trivially intersects our original field R, then we can quotient out by M to have a field where XY + 1 = 0 mod M, so the relation is satisfied

We're doing that with an ungodly amount of indeterminates

sorta like if we wanted to find a simple group extending a given group with relations on some elements we append to it

[unkempt stream]

So uhm. Gonna see if this is the correct proof for part a:

Lemma (rank-nullity theorem): If F is a map from D^N to D^M, then D^N ~= ker(f) (+) im(f)

By Smith Normal Form calculations, F is similar to a diagonal map U, where U(e_n) = d_n e_n for n <= N, 0 for N < n <= M, and where the d_n form an ascending divisor chain for the first r <= N terms, and are 0 after.

Thus U(e_i) is nonzero iff 0 <= i <= r, implying the kernel of U has the basis {e_i, r < i <= N} and the image has the basis {d_i e_i, 0 <= i <= r}.

By similarity, F = AUB^-1, where A and B are invertible. The image of F takes the form of A \circ U \circ B^-1 (D^N) = A(U(D^N)), so is thus has basis {d_i A(e_i), 0 <= i <= r}. likewise the kernel takes the form of (AUB^-1)^-1(0) = B( U^-1 A^-1(0)) = B(U^-1(0)), so thus has the basis {B(e_i), r < i <= N}. These maps give pairwise disjoint embeddings into D^N.

Rank-Nullity theorem follows from the basises of ker(F) and im(F) forming one total basis of rank r + (N - r) = N, along with invariant basis number theorem.

[empty kernel]

What is F(X)?

F is a field

And this notation popped up when we were talking abt field expansions
If F a sub field of E and u in E is a transcendent number then F(u) is isomorphic with F(X)

[next obsidian]

Rational functions in X

It’s the field of fractions of F[X]

So just ratios of polynomials

[wanton crescent]

I want to study the last 6 chapters from Artin's Algebra in order to prepare for my universities Algebra 2 class next semester. I have already started, revising chapters 1 and 2, stuff I saw in our Linear Algebra and Group Theory class (our version of Algebra 1 I guess).

How dependent are those last 6 chapters on the other chapters like Group Representations and Billinear Forms? How necessary are they for the problems seen in these chapters? I'm nervous If I skip over them I may be f'd later on, yet I want to maximize my coverage of directly relevant/assessed content.

In the preface Artin says that those chapters are largely logically independent of many of the preceding ones, but I'm not sure how well reflected this is in the problems, as for example Chapter 2 problems do make usage of Chapter 1 ideas, though I would say Chapter 1 could be treated as pre-requisite knowledge for the book?

[velvet hull]

which last 6 chapters?

[wanton crescent]

I'm using second edition let me just find it rq

[empty kernel]

Ah so all rational polynomial functions?

[wanton crescent]

11. Rings, 12. Factoring, 13. Quadratic Number Fields 14. Linear Algebra in a Ring, 15. Fields, 16. Galois Theory.

[velvet hull]

yes it's all independent

[velvet hull]

ch 11- 14 is ring theory, so they're all codependent

ch 15-16 is field theory, which builds upon ring theory and group theory

[wanton crescent]

I see, just nervous I might see some previous content show up in problems or be secretly necessary for those problems haha

[velvet hull]

nope, you won't

there will be too much new content anyways 😈

[wanton crescent]

What do you mean haha

By new content that is

[velvet hull]

there's a lot of new stuff

asking questions about previous chapters is a waste of time

[wanton crescent]

Ah I see thank you!

[crystal vale]

the number of element of order 15 in Z/60Z times Z/50Z is 16, right?

[velvet hull]

what do you mean "16"? Z60 x Z50 is not cyclic, so that doesn't tell me which element you're talking about

[crystal vale]

number of element of order 15 is 16

elements

[rocky cloak]

I think there is a bit more. What's your thoughts process?

[crystal vale]

i take (a,b), there are cases,

i. a has order 15 and b has order 1
ii. a has order 3 and b has order 5
iii, a has order 5 and b has order 3 this is not possible

but now i see other options a has order 15 and b has order 5

thank you jagr

[south rain]

I'm trying to prove that if $R$ is a primitive ring (without unity) then a non-trivial ideal $I$ is also primitive (as a ring). Now I thought this is intuitive but I'm missing something.

If $M$ is a $R$-Module then it naturally is also an $I$ module since $I$ is also a ring. Since $(0: M) = {0}$ as a $R$-module it must also holds as a $I$-module. Now it is left to prove that $IM \neq {0}$ and that it has no trivial submodules. But $IM \neq {0}$ must be true since $I \neq {0}$ and $(0: M) = {0}$.

I now also argue that $M$ (as a $I$-module) can't have any non-trivial submodules since if it would, this would also be a non-trivial submodules (in the $R$ sense). What I'm now missing is where I need $I$ to be an ideal in the first place, thus I suspect something is wrong with my argument.

[cloud walrus]

dellinger

[rocky cloak]

I mean "this would also be a submodule in the R sense" needs a justification. Because you don't expect it to be true in general.

[south rain]

Alright so there's where I need for I to be an ideal. But if I calrify that I'm good on my other arguments? Thanks!

[rocky cloak]

Yeah I think so

[south rain]

Yeah I'm pretty much stuck there now. If $N$ is a submodule of $\prescript{}{I}{M}$ then $(N, +)$ is a subgroup of $(M, +)$ (this shouldn't depend on the ring I use right?) and $IN \subset N$. Now since $I$ is an ideal $IRN \subset IN$ so $RN \subset N$, but I only used "right ideal".

[cloud walrus]

dellinger

[rocky cloak]

The left ideal part is really the one you want to use

[south rain]

I don't see how $RI \subset I$ helps me here. I would get $RIN \subset IN \subset N$

[cloud walrus]

dellinger

[rocky cloak]

And what is R(IN) ?

[south rain]

I'm not sure, are you looking for an equality here? Do you mean that IN is a submodule of R?

[rocky cloak]

Let me ask another way.

IN is a subset of M. What does
RIN < IN tell you about what kind of subset it is?

[south rain]

I think I'm just too dense. I mean that would make it a submodule of R right?

Or that it is again an ideal?

[rocky cloak]

Submodules of R need to be subsets of R at the very least

Which it isn't, it's a subset of M.

But what do you can a subset U of M such that RU < U?

(that is closed under addition)

It's the definition of a ||submodule||, so since ||M is simple|| it's either ||0 or M||.

Then you just need to show that ||IN is non-zero||, where you might use that ||I is a right ideal||

[south rain]

Alright so I just confused everything. Of course it can't be a submodule of R because R is the ring, my god. I think I'm going for a break.

I need to think about that line of argument a bit longer, but even so, I'm only using right ideality and not left right?

[rocky cloak]

You will need to use both. But the initial step would be using the left ideal structure

Not sure if you read my spoilers

But the main point is that
I being a left ideal ensures
RIN < IN
and I being a right ideal ensures
I(RN + N) < IN

[south rain]

Yea thanks, I read them, helped a lot!

Say $N$ is a submodule of $\prescript{}{I}{M}$. Then $IN \subset N$. By $I$ being a left ideal we have he $RIN \subset IN$ so $IN$ is a submodule of $\prescript{}{R}{M}$. So either $IN = 0$ or $IN = M$ since $M$ is simple. Because its a right ideal we also have that $IRN \subset IN$ (why the "+N" ?)

[cloud walrus]

dellinger

[south rain]

I'm not sure when I can conclude that $N$ is also a submodule of $\prescript{}{R}{M}$.

[cloud walrus]

dellinger

[rocky cloak]

The +N is because you didn't assume R has 1.

So the submodule generated by N is RN + N as opposed to just RN

[south rain]

That I don't quite understand. Because I start of with N being a submodule of M w.r.t. I. Then I know that IN is a submodule of M w.r.t. R. Now I somehow need to show that N is also a submodule. And that I do by using that IN < N and with the usage of I being a right ideal IRN < IN < N. But that +N comes out of nowhere for me right now, in that chain of arguments. I thought that I can conclude that RN < N, but I don't see that happening with a +N.

[rocky cloak]

Instead of trying to prove that N is a submodule, just prove that IN = M

Then it would follow that N=M

[south rain]

Oh now I see the goal with this reasoning

That makes much more sense

[rocky cloak]

Like you have to think about what tools you have at hand.

M is simple so we need to use that somehowe. Okay, what does that mean, well it implies the submodule generated by N is all of M. Okay, let's use that.

RN + N = M

Then we loop this back to IN since we already proved stuff about that
IRN + IN = IM

-> IN = IM

[south rain]

Alright, I think I've got and understood the whole argument now. I'm trying to show that any submodule is eiher 0 or M directly.

Finally, thanks for your patience and all the help

[tropic obsidian]

I am reading ravi vakil's intuition on short exact sequences, and here is one of the frames. I think this is supposed to represent the splitting of W -> V -> V/W in vector spaces, in which case this makes sense

but then he says this in the next frame, which I dont quite get

the sequence H -> G -> G/H doesn't split even for abelian groups?

aren't the category of abelian groups an abelian category?

maybe I am missing something

taken from: https://www.3blue1brown.com/blog/exact-sequence-picturebook

[rocky cloak]

The diagram isn't representing the splitting of the short exact sequence.

It's simply a visual representation of the short exact sequence.

You think of A as a kind of glued together blob, and then parts of that blob would be subquotients.

Like they explain a submodule is supposed to be "bulging out" so A/B is not a submodule in that picture

[lone niche]

I think at some point this might have been a video, and he's saying: "This works more generally for abelian groups", and then says it holds in an even more general case and says "This works more generally for modules over a given ring" and crosses out the previous part because it is now redundant. And then once again says "This works more generally for objects in an abelian category", and crosses out the previous part (again because it's redundant).

[serene dune]

i have asked my concern in the comment just a while ago so yeah, if you take a look over there

[crystal vale]

i think option a is correct, they saying 5 is not in S, but if we take S_24 then 5 is in S_24, right?

[rocky cloak]

You can drop unital if you like, it doesn't make a big difference.

But commutativity is somewhat crucial, as without it R would not be an R-algebra.

[serene dune]

which one is identified by id_M ?

[serene dune]

yeah i'd like to get specific details to see why its not an algebra any more which i thought would be a good exercise

[rocky cloak]

Why R is not an R-algebra if it's not commutative?

It's simply because one of the requirements is that
(rx)y = x(ry)

[serene dune]

so the bilinearity ?

im not sure if im using right terminology here

[rocky cloak]

Yes bilinearity.

As you can see if rx doesn't equal xr this would get you into trouble

[serene dune]

ok yeah it gives a flavour of vector spaces now, lol

my lifealgebra
my rules

[serene dune]

in anyways, this is another thing i'd like to get some help!

how stupid of me, it has to be 1(m) = m, so trivial

[crystal vale]

Which book? Blyth?

[serene dune]

yeah!

as it has to be 😝

[crystal vale]

anyone?

[kind temple]

i am trying to reason about why Sn can be generated by any set of transpositions whose corresponding graph is connected (V = {1,…, n} and edges are drawn if the corresponding transposition is in the given set).

i have noticed that chains let you generate cycles and you can generate (m k) for any vertex k adjacent to m.
but i am not sure how to argue that connectedness lets you generate everything

for example, if T = {(1 i) : 1 <= i <= n}, you get a star graph

and if T = {(i i+1) : 1 <= i < n}, you get a stick

[south patrol]

Pretty sure you can reduce to those two cases

Unless it is just a star, consider the subgraph of vertices it's immediately connected to and then those transpositions (by induction) generate the corresponding subgroup of Sn

I think that works

[kind temple]

how do you plan on stitching together the subgroups

oh

i think i get it

so

if you want to generate (m k) for any 1 <= m < k <= n, pick any path from m to k, say v1, …, vr

the permutation v1 v2 … vr vr-1 … v1 should be (m k)

[south patrol]

Tbh maybe it's easier. Take a maximal tree T, take out some "end" a and then consider the subgroups corresponding to stuff immediately connected to a and stuff corresponding to T \ {a}

I imagine working w that will work

Like it gives you big subgroups of Sn

[kind temple]

you can’t just take any sub-tree, you want an MST, right?

[south patrol]

Yeah so you get the subgroup of stuff fixing a vertex a and then like it's straightforward

Like given a permutation s sending a to b, compose with any permutation sending b to a (which is in our subgroup by connectedness)

[kind temple]

need to fix this. v1 = m, vr = k
the permutation is (v1 v2) (v2 v3) … (vr-1 vr) (vr-1 vr-2) … (v2 v1) which should be (m k)

[south patrol]

Yeah okay so basically this is just the usual induction but I'm being more careful with ensuring the graph remains connected after removing a vertex

So yeah I guess you just need to take your graph and remove any such vertex for the induction

:)

[kind temple]

im not sure i get your solution atm, but i think my solution should work as well?

[south patrol]

Ah yes i agree with this. Especially since you can take a minimal path and then the calculation is easy

[kind temple]

yea

i think its hard to see the subtree argument, but i don’t doubt that it works

[kind temple]

yea, this works

this is really neat wtf

[south patrol]

I'm basically saying: take your connected graph X, pick a vertex (wlog n) such that X \ {n} is still connected. Given a permutation s, you can compose with a permutation from the subgroup corresponding to X to assume s(n) = n. Then by induction on the size of the graph, s is in the subgroup corresponding to X \ {n}

(Trees aren't necessary, but i found them useful psychologically when coming up w the proof cause it simplifies the graph a little)

[kind temple]

you are saying that you get the subgroup of permutations fixing n?

3 — 1 — 2 — 4 and i want s = (3 4), i can’t compose with edges to fix 4. what should i compose with to fix 4?

[south patrol]

If I write permutations using function composition ordering, then consider (24)(12)(31)

I.e. pick a path from 3 = s(4) to 4

And then (24)(12)(31)(34) fixes 4

[kind temple]

i see

i think the converse of this should also be true, that if the set of transpositions generates, then its graph must be connected

[south patrol]

Certainly, since you can take a path in this way

[kind temple]

how do you argue it though

i intuitively get it

but making it precise is a bit difficult

[south patrol]

Damn

Okay yeah hm

[kind temple]

a product of transpositions for (m k) from T may need to be refined. and even afterwards, i see no reason for this to be a path going through m and k

at least not immediately

it should be though

maybe the contrapositive is easier? the group elements generated by connected components are only disjoint cycles. so you can’t possibly get the transpositions corresponding to connections between components. but just a sketch

[south patrol]

Exactly

[kind temple]

anyways. this makes some proofs about presentations of Sn a bit more streamlined

[south patrol]

I do wonder if there is any funny more topological proof

But I guess topological thinking is more useful for infinite groups generally

[kind temple]

do the transpositions generate infinite symmetric groups?

all of them, yes

but there was nothing about the arguments above that required the graph to be finite

[south patrol]

Well you need the infinite symmetric group in the sense of finitely supported things

And then yes that follows quickly from the finite case

[kind temple]

but you generate not finitely supported bijections from certain collections of transpositions

what goes wrong if its not finitely supported? i can still generate any transposition if G(T) is connected since the path is of finite length

oh

oh nvm

the issue is really similar to hamel vs shauder basis

[south patrol]

Yeah like any product of finitely supported tings is still finitely supported

[kind temple]

i wonder if there is any funny condition on the graph to generate An

[empty kernel]

How does one prove that Q(sqrt(3)) isnt isomorphic with Q(sqrt(2))

[rocky cloak]

The easiest way might be to show that it doesn't contain a square root of 2

[empty kernel]

But that doesnt show that they arent isomorphic though right?

[rocky cloak]

If one structure has a property the other doesn't, then they can't be isomorphic

[empty kernel]

Is containing a certain element a property?

[rocky cloak]

Having an element with a specific property is something that counts as a property of the structure yes

[empty kernel]

Bc like what if their is an isomorphism that just directs sqrt(2) at sqrt(3)

[empty kernel]

Yes i understand that but what property does root2 have that root3 hasnt

Or vice versa

[rocky cloak]

Well sqrt(2) satisfies
x^2 = 1 + 1
And sqrt(3) does not satisfy this. Hence no isomorphism can map one to the other

[rocky cloak]

That it is a square root of 2

[empty kernel]

Wut this is making me so confused

[lone niche]

I think a key part of the argument is that an isomorphism fixes the rational numbers

so if phi is an isomorphism, phi(3)=3

[empty kernel]

So all the Q(sqrt(something)) arent isomorphic?
If sqrt(something) isnt in Q obv

[empty kernel]

Yes that makes sense

[rocky cloak]

Well Q(sqrt(8)) is isomorphic to Q(sqrt(2))

[empty kernel]

Oh ye i already proved that

Its cuz sqrt(8) is 2sqrt(2)

[lone niche]

so if there is a sqrt of 3 on the left side of the isomorphism, phi(sqrt(3)sqrt(3))=3, but also phi(sqrt(3))phi(sqrt(3)) =3 by definition of homomorphism.

So this implies that 3 has also a sqrt on the other side of the isomorphism

[empty kernel]

Oh ye that makes a lot of sense

But obv such this does not exist i already proved that in an earlier exercise

Yipie

Thanks

Ok uhm i have one more question

Is Q(root3)(root2)=Q(root2,root3)

I think yes but my intuition tends te be weong a lot of times in algebra

[lone niche]

yeah, and jagr had a cool proof iirc

[empty kernel]

@rocky cloak u got a cool proof for this?

[rocky cloak]

Yes, just by definition.

Q(x, y) just means the smallest field that contains Q, x and y.

So Q(x)(y) is the smallest field that contains Q(x) and y.

[empty kernel]

Oh ye that makes a lot of sense lol

[lone niche]

oh I misread that, i was thinking of the root3,root2 and root3+root2

[empty kernel]

This makes it a lot easier to find the primitive element

[empty kernel]

Nah thats what im trying to find now

[empty kernel]

Algebra is hard

[kind temple]

i think a direct argument does actually work. a composition of transpositions which is equal to (m k) must contain m in at least one transposition in the composition. otherwise, it fixes m. same goes for k

[empty kernel]

Im struggling to prove this

Now idk what to do with this [F(a,b):F(a)]

,rotate

[lone niche]

Hmmm I think have a proof idea: We would like to show [F(a,b):F] <= [F(a):F][F(b):F] AND [F(a):F][F(b):F] divides [F(a,b):F].

I think this would establish the equality

The right hand side of the AND follows by a number theory argument that if a,b divides c then lcm(a,b) divides c.

Oh I think you reached the same conclusion as I did

And you are stuck on the left hand side of the AND, mainly you have the equation [F(a,b):F]=[F(b,a):F(a)][F(a):F] and want to see how [F(b,a):F(a)] relates to [F(b):F].

Which you should be able to relate by arguing with the minimal polynomial for b with coefficients in F, and with coefficients in F(a).

[tardy hedge]

Is a decomposable in A/a because pi(a) = pi( cap qi) = cap pi(qi) = 0 and each pi(qi) is primary in A/a?

Why is the image of qi in A/a primary?

[trim wind]

a lies inside of the q_i, and then just use this definition of primary applied to the image of q_i inside of A/a

[south patrol]

Ye

You can remove transpositions in the composition until it becomes a path anyway

[tardy hedge]

Idk im still stuck on this.