groups-rings-fields

[dim widget]

what

[wild jasper]

I misread

then it's degree four

sorry I don't understand the question

@dim widget degree 2 extensions of F are square roots of elements of F

[south patrol]

Generated by square roots of elements yes

[dim widget]

okay so then if L/K has galois group Klein 4 then it contains two quadratic subextension K(sqrt(a_1)), K(sqrt(a_2))

and these have trivial interseciton since they are not the same

[south patrol]

(Note this requires characteristic not 2)

[dim widget]

so then L contains K(sqrt(a_1), sqrt(a_2))

but by degree considerations they are the same field

[wild jasper]

ahha

Sorry for being dense

thank you

[void cosmos]

if R is a PID

M and N are free R-modules of the same rank (??) f:M-->N and g:N-->M are morphisms such that gf is the identity on M show that f and g are isomorphisms?

i got lost with how there are so many to deal with this i just fucked up

i thought about exact sequences and that free modules are projectives

but also aren't literlaly M and N already isomorphic

this was a problem on a qual

[dim widget]

hmm I think most of this is general category theory and you only have to know one thing about rings

[void cosmos]

i see

[gusty swallow]

does anyone know

[void cosmos]

my gut says this has to do with the matrix representation

and determinants

[gusty swallow]

i mean i got some good progress considering cycles from Sn

[void cosmos]

thank you for the hint; i was so out of focus cuz the problem had so manay information and like paths to try out

[gusty swallow]

this did come from talkig abt the formula for the determinant of permutation matrices

[dim widget]

I think you should just prove that the function is multiplicative then show that it has the right value on transpositions (which is an easy calculation)

[gusty swallow]

oh wow thats way simpler than trying strong induction on n

[kind temple]

for a given integers n,m, what is the kernel of multiplication by m from Z/nZ --> Z/nZ?

i think that the image is Z/(n/d)Z where d = gcd(n,m)
but im not sure what the kernel is

can anyone verify this?

[long obsidian]

Can someone remind me why (S/p)_p= Frac(S/p) for S=k[x1,...,xn] and p a prime ideal? I guess I don't have a good feel of inverting elements

[dim widget]

m*d = 0 iff n|md so the minimal d = n/gcd(n, m), so that is the generator of the kernel

[kind temple]

sorry, what is the generator?

[dim widget]

in your notation n/d

[kind temple]

okay great

wait

[dim widget]

i am waiting

[kind temple]

so d + nZ generates in Z/nZ

[dim widget]

generates what?

[kind temple]

the kernel

[dim widget]

no d is unlikely to be in the kernel even

but n/d will be

[kind temple]

oh whoops

n/d + nZ generates

the kernel

[dim widget]

yeah

[kind temple]

what group is this

[dim widget]

the subgroup?

[kind temple]

<n/d + nZ>

yea

[dim widget]

it's Z/dZ

which it would have to be just by size considerations

[kind temple]

okay, thats what i thought

but

[dim widget]

since finite cyclic groups are determined uniquely by their order

[kind temple]

ig im just confused?

like, i thought that the image would be mZ/nZ

[dim widget]

yep

[kind temple]

so is mZ/nZ = Z/(n/d)Z in general?

[dim widget]

yep

[kind temple]

okay, im confused, since when m divides n, we have mZ/nZ = Z/(n/m)Z

[dim widget]

when m divides n d = m

so that checks out

[kind temple]

wait what

when m divides n, we have d = m?

[dim widget]

yeah

d = gcd(n, m) so if m|n then d = m

[kind temple]

oh my god

alr

thanks

i was having a massive malfunction

[dim widget]

it happens

[gusty swallow]

hey im back again, ive gotten through proving multiplicativity and that it does indeed give the correct value for transpositions, but im just not sure how to combine them. Idk if theres some property of finite products that im not aware of that allows me to combine them in a way that proves the fact

[quiet pelican]

You’re immediately done because then every σ \in S_n can be written as a product of transpositions
So the value a multiplicative function takes on the transpositions uniquely determines what that function is on all of S_n

[gusty swallow]

ik but im trying to prove that for all $\sigma\in S_n$
$$
\mathrm{sgn}(\sigma) = \prod_{1\leq i<j\leq n}\frac{\sigma(j)-\sigma(i)}{j-i}
$$

[cloud walrus]

BigBoyConst

[gusty swallow]

its not evident to me how this works if i split it up into transpositions

since i just get a product of products

namely
$$
\mathrm{sgn}(\sigma)=\mathrm{sgn}(\tau_1\cdots\tau_k) = \prod_{1\leq i<j\leq n}\frac{\tau_1(j)-\tau_1(i)}{j-i}\cdots\prod_{1\leq i<j\leq n}\frac{\tau_k(j)-\tau_k(i)}{j-i}
$$

[cloud walrus]

BigBoyConst

[quiet pelican]

So with multiplicativity you’re claiming to have proven something like
$$\prod_{1 \leq i < j \leq n} \frac{\sigma_1(j) - \sigma_1(i)}{j - i} \prod_{1 \leq i < j \leq n} \frac{\sigma_2(j) - \sigma_2(i)}{j - i} = \prod_{1 \leq i < j \leq n} \frac{\sigma_1 \sigma_2(j) - \sigma_1 \sigma_2(i)}{j - i}$$
Right?

[cloud walrus]

Micose

[quiet pelican]

Because that’s what’s meant by multiplicativity

[gusty swallow]

i thought by multiplicativity they meant the multiplicativity of sgn($\sigma$)

[cloud walrus]

BigBoyConst

[gusty swallow]

whoops

brb when i prove that

[quiet pelican]

Sure, you need both
But i assumed you could assume that

[gusty swallow]

now thinking abt it it makes so much more sense that i have to prove the multiplicativity for the product and not the actual definition of the function since its kinda trivial

[dim widget]

the point is to note that || \prod \frac{\tau \circ \sigma(j) - \tau \circ \sigma(i)}{\sigma(j) - \sigma(i)} = \prod \frac{\tau(j) - \tau(i)}{j - i} || and then it becomes easy.

[hidden wind]

i’m not tripping with the following equivalence right?:

[ R[x] \cong \bigoplus_{k\in\mathbb N} R ]

[cloud walrus]

rødbet

[rocky cloak]

As R-modules, that's true yes

[hidden wind]

oh the direct sum doesn’t make a ring?

or would it just be like clumsy to define the product

[rocky cloak]

I mean you can cook up several ways to make it a ring, but there's isn't really a canonical way to make it a (unital) ring

[hidden wind]

right

[rocky cloak]

Like you could do point-wise multiplication, but that isn't unital and is also very different from the polynomial ring

[hidden wind]

mhm, i can’t imagine why that’d show up

…i guess except if you were looking to equip that direct sum with ring structure

[hidden wind]

is this terminology in Clark standard? “a proper element of a ring,” namely, a non-zero and non-unital element of the ring

[gusty swallow]

and these are both transpositions or general n-cycles?

[dim widget]

it doesn't matter

so general n cycles I guess

[gusty swallow]

but wouldnt i have to consider both the cases where $\sigma(j)>\sigma(i)$ (which is what i'd want) as well as the case when $\sigma(j)<\sigma(i)$?

[cloud walrus]

BigBoyConst

[dim widget]

I don't think that

is useful

[quiet pelican]

I don’t believe it is

[hidden wind]

ok that makes sense then

i guess the idea is to take this equivalence as a definition: the principal ideal (a) is a proper ideal of R iff a is a “proper element” of R

[tardy hedge]

Im trying to prove 4th (lattice) isomorphism theorem for modules. Can I assume a priori that submodules of M/N are like A/N..?

or is that itself actually need to be proved

submodules of M/N are elements {m+N} that itself form a module

does my question even make sense

[tardy hedge]

Ok, how is this (maybe some details off but more or less is this a correct approach?):

Trying to show: Submodules of M/N are A/N for some A submodule of M containing N.

Take a proper subset of M/N, that has identity and say has more than 1 element. So this is {m+N | m in A}. Where necessarily N subset A subset M for this collection of cosets to be a proper subset of M/N (that contains identity in M/N). If this is to be a submodule then of M/N then by submodule criteria: (m1+N) + r(m2+N) = (m1 + rm2) + N, needs to be in our collection, so we need m1+rm2 to be in A, so A is a submodule of M containing N

I kind of like this cause its like starting generally and then seeing what needs to happen once u impose more structure

[tardy hedge]

Where da peeps at

[unkempt stream]

You can refine this using the fact that homomorphism preimages and images of modules are modules

For surjective maps like our quotient, image of preimage of a set is the original set

So it’s the quotient module of the preimage :3

[tardy hedge]

Hmm i dont get exactly how this works

[coral spindle]

Can you think of a map M -> M/N

[tardy hedge]

The projection map thing

[coral spindle]

Can you write it down

[tardy hedge]

pi(m) = m+N. oh ok so for surjective homs, preimage of a submodule is a submodule

[unkempt stream]

After this I implore you to show that the same idea works for the lattice isomorphism theorem of rings :3

[coral spindle]

Now show that taking preimages defines a function {submodules of M/N} -> {submodules of M containing N} and is bijective

[unkempt stream]

this is true for any homomorphism. Images and preimages send submodules to submodules.

[tardy hedge]

Oh the surjective thing is only for preimage of normal subgroups for groups ?

[unkempt stream]

(same for ring homomorphisms)

[tardy hedge]

U need surjective for normal subgroup to map to normal subgroup

[coral spindle]

Yes

[tardy hedge]

Not preimage tho?

[unkempt stream]

In a sense, we only care about the surjectivity

M/N is just a notation for the image of the module under the quotient

[coral spindle]

These are modules though, so we don't have the same issues as with groups.

[tardy hedge]

Yeah

[winged void]

i have small question

iin regard to the following notation

why do they say i. = i + nZ subset of Z

[unkempt stream]

If $\phi$ is a surjective homomorphism of $A \twoheadrightarrow B$, then for any submodule $M$ of $B$, $M = \phi(K)$ for some submodule $K$ of $A$. In fact, $K = \phi^{-1}(M)$, the preimage of $M$

[winged void]

and what does Z / n Z actuall means

i thought its modulo numbers

[cloud walrus]

Mizalign

[unkempt stream]

we just have $\phi(M) = M/N$ in the case of quotienting

[cloud walrus]

Mizalign

[unkempt stream]

Z/nZ is technically the set of congruence classes modulo nZ

So yes, i + nZ is a congruence class, a subset of Z

[coral spindle]

Do you know what a quotient group is? Have you seen G/N where N is normal in G?

[unkempt stream]

We just define the quotient group to be on the set of congruence classes

[winged void]

yes

[coral spindle]

So nZ is just {nz | z in Z}

And Z/nZ is just the quotient group of Z by nZ

If you know what a quotient group is, everything should be clear.

[tardy hedge]

So we havent showed that preimage of submodule in M\N is containing N tho right

[coral spindle]

M\N?

[winged void]

sure thanks alot

[coral spindle]

You mean M/N right

[tardy hedge]

Yeah

[coral spindle]

But yes I haven't shown that

[unkempt stream]

preimage of 0 is N

[coral spindle]

I asked you to show that

[unkempt stream]

every submodule contains 0

[tardy hedge]

Bruh u spoiled the answer but that makes sense haha

[unkempt stream]

oh sorry i thought you were asking

:c

[tardy hedge]

No worries

I appreciate the more high level description

Instead of going to nuts and bolts

[unkempt stream]

I mean that is the explicit reason why in a sense

it's easy to generalize it to like

most algebraic structures

[tardy hedge]

Yea

[unkempt stream]

The "tedium" is showing the image and preimages are modules for homomorphisms of whatever algebraic objects you're talking about

[coral spindle]

do the ruddy exercise!!

[tardy hedge]

Yea i will need to check if we proved rhat in class

[unkempt stream]

just show it yourself :3

thankfully modules don't have a lot of axioms to verify

[tardy hedge]

Yea i should do it for the exercise anyway

[unkempt stream]

yippee

[tardy hedge]

Btw for the group case, the subjectivity of the hom is only relevant for normal subgroups to map to normal subgroups

[coral spindle]

I know

[tardy hedge]

Did preimage of normal subgroup always map to normal even if hom is not surjective?

[unkempt stream]

no

[tardy hedge]

No ik sorry i was asking

[unkempt stream]

Try to think of an example :3

[tardy hedge]

I wasnt telling u i was asking lol

[unkempt stream]

oh okie

[coral spindle]

Just embed any group in another where it's not normal

Like

Z/2Z in Dih(6) or whatever

[unkempt stream]

sotrue

I should be working on my compeng project right now

[unkempt stream]

tldr lattice isomorphism theorem is due to the quotient projection being a surjective module map

I actually wonder why that preimage-image of homomorphisms preserving submodules or subobjects isn't covered that much

like, as much as it should be in intro AA classes from what I can tell

[tardy hedge]

Hmm right ok usually they talk about how the IMAGE is a submodule of codomain, but they dont say how images of submodules are submodules of codomain ?

Is that what ur referring to?

[unkempt stream]

yes

or preimages

[tardy hedge]

Yeah thats weird

[unkempt stream]

preimages of codomain submodules are submodules of the domain

even if that codomain submodule is outside of the image, 0 is in both

[tardy hedge]

Yeah thats kind of a good fact

Thanks for the insights

[unkempt stream]

you're welcome

[tardy hedge]

Usually when i do algebra things i go very low level but i want to try to think more in terms of these more fundamental ways

[unkempt stream]

once again, not only for modules. it works for like a vast majority of the algebraic things you'll see

[tardy hedge]

Instead of playing around dirty with coset stuff

Yea

[unkempt stream]

These ideas can make some cohomology sillies of modules make more sense, especially diagram chasing imo

I've seen some arguments use these DIRECTLY

[unkempt stream]

Like even then, the kernel is the preimage of the zero module

So you can describe the kernel of compositions of maps

If h = f \circ g, then the kernel of h is the g-preimage of the kernel of f

[tardy hedge]

Oh ya, bruh

[unkempt stream]

Btw the whole image preimage correspondence is similar to annihilators of submodules

[old sphinx]

What is cosets and isomorphism
Need help to determine its defintions

[tardy hedge]

Wat da flip bro

[drifting plover]

Hello peeps, I have a somewhat strange computer-sciencey question about polynomials (this is probably not the best channel to ask but I try).

Given a multivariate polynomial f, we can associate to it a (highly non-unique) formal expression e in terms of sums and products of monomials such that e is mathematically equivalent to f. For instance, if f is the multivariate polynomial 2xy+3yz, the formal expressions 2xy+3yz, y(2x+3z), and 2y(x+z)+yz are all equivalent to f.

Now define the size of the formal expression e to be the total number of variables that appear in it, counted with repetitions and multiplicity. For instance, the size of 2xy+3yz is 4, the size of y(2x+3z) is 3, and the size of 2y(x+z)+yz is 5. I call y(2x+3z) optimal for the polynomial 2xy+3yz because (I believe) it's the formal expression associated to 2xy+3yz with the smallest size.

Here's my question: is there a (heuristic/probabilistic) algorithm that, given any multivariate polynomial, outputs an optimal or close to optimal formal expression associated to it? Feels like (the decision version of) this problem is NP-complete, so I'm happy for a locally optimal solution.

[coral spindle]

You can find these, and all other definitions necessary to understand them, in any introductory book on groups. In fact you can find these definitions online if you google them.

[unkempt stream]

NP

Aight I’ll see myself out

[old sphinx]

Thank you then for that

[unkempt stream]

Plenty of textbooks (particularly older ones like Jacobson Basic Alg) have pdfs online

[unkempt stream]

Like how the inverse image map sends submodules of the codomain to submodules of the domain, the "annihilator map", Ann(M) = {r : rx = 0 for each x in M}, sends submodules of a module to left/right ideals (essentially the ring's submodules)

[tardy hedge]

ANNIHILATOR!!

[tardy hedge]

Hmm ok i have not actually parsed thru what u put here yet but i will

[unkempt stream]

The annihilator of a set is elements of the ring R that the module is over that send ever element of that set to 0

[tardy hedge]

Unfortunately my module theory lectures are so dry and i learn 100x more from u guys

My topology lectures are epic this term tho.

[unkempt stream]

jacobson is fun

[south patrol]

Are they monic

[south patrol]

Tbh yeah I feel like first courses in modules or whatever are bound to be a bit dry lol

[void cosmos]

do u have notes or like

anything to share

topology is like the most 'rewatachable" series ever

in a way

math way

[winged void]

can someon give me a hint regarding this question

Showing that the field of fractions of ℤ[\sqrt{d}] is ℚ[{sort{d}}].

is there an easy way to do it

[unkempt stream]

what are the units

of Z[\sqrt{d}]

actually probably just to use the 2-dim approach

[arctic trail]

The field of fractions is the 'minimum field' which contains it

since the field of fractions of Z is Q

and the field of fractions of that object is not Q since it doesn't contain sqrt(d) (assumind sqrt(d) is not rational, I'm assuming it's a non-square integer here, if that's not an assumption in the problem please say).
And Q[sqrt(d)] is a field and Q[sqrt(d)]:Q = 2 and Q[sqrt(d)] contains Z[sqrt(d)] we are finished

[unkempt stream]

1/(a^2 - db^2) exists in Q given d^2 is not rational

so for any field containing Z[\sqrt{d}], for any a + b\sqrt{d}, 1/(a^2 - db^2) is in Q (contained in that field since it's a field containing Z, thus Q = frac(Z) too), so (a - b\sqrt{d})/(a^2 - db^2) is its inverse

[arctic trail]

that seems like a lot of unnecessary work

[unkempt stream]

if d = (a/b)^2, then a + \sqrt{d} b is a zero divisor so there is no field of fractions for that element

[arctic trail]

That's why I added a disclaimer

[unkempt stream]

was just adding that to elaborate

[arctic trail]

this type of question seems to be one where you assume d to be a non-square integer

[topaz solar]

I am not aware of any, but how fast is the “obvious” method?

[arctic trail]

in one variable the algorithm should be* the same as the one used for evaluating polynomials

so representing a0 +a1x +a2x^2 + a3x^3 + ... anx^n as a0 + x(a1 + x(a2 + x(a3 +(...))

[topaz solar]

Well, what about e.g. (x+3)^3

[arctic trail]

that's 3 accounting for multiplicity

which is the same

(although I still changed it to should be)

[topaz solar]

True true, might just always be the same

[arctic trail]

this algorithm gives you n, which is a lower* bound

[topaz solar]

For multiple variables, there might be some weird possibilities that could be better?

Based on funny factoring being hard

[arctic trail]

wait

well first off we can remove any constants, since they're not important

[topaz solar]

Might as well

[arctic trail]

if the resulting is divisible by a variable you can just factor it

[south patrol]

Trivial lemma

[arctic trail]

lmao

[topaz solar]

I mean, it should just be like, total degree as a lower bound

But getting to that form

[arctic trail]

this is probably funny n > 1 means n = 2 moment, where everything's the same

if you have an expression of the form xp(x) + yq(y) + xy h(x,y) what's optimal?

[arctic trail]

if we reach the lower bound then we should have that for any evaluation on a variable we also reach the lower bound

[topaz solar]

I’d sure hope so

[arctic trail]

I mean it follows

[topaz solar]

That’s why I’d hope so

[arctic trail]

then like, how in which ways* can you optimally write an expression

also can you even reach the lower bound on x^2 + x^2y^2 + y^2

[topaz solar]

Is there much improvement possible on that one tbh

Other than like x^2(1+y^2) + y^2

[arctic trail]

you can apparently (1+x^2)(1+y^2) -1

[topaz solar]

True

Could pull out that 1

But would you be able to find that nicely algorithmically

[arctic trail]

e is invariant under translations on the variables, constants

transformations (x1,...,xn) -> (x1^k,...,xn^k) multiply by k the optimal

and x-> x^k preserves the optimal solution as well (although the value depends on the type of factorization)

x^2 + x^3y +y doesn't have any degree 4 factorization

[topaz solar]

Sad

[arctic trail]

I wonder if homogenization helps

[barren sierra]

Ok so this feels closely related to something that is already being studied that you may find interesting

there are these things called algebraic circuits

so these formally are directed acyclic graphs

the bottom (source) vertices are inputs and the top (sink) vertex is the output

so here is the same polynomial expressed as two different circuits

you can easily do this for multivariable polynomials as well

the size of the circuit is the number of nodes in the graph

and the depth of the circuit is the length of the longest path from any source to the sink vertex (both of these circuits are depth 3)

so one can fix a family of polynomials paramaterized by n, say the n x n determinant (which is a polynomial) and ask "what is the smallest circuit (in terms of big O notation) for the n x n determinant

or maybe the n-th elementary symmetric polynomials

or the n x n permanent

or the n, d Iterated Matrix Multiplication polynomial (multiply together d different n x n matrices, output the 1,1 entry of the resulting matrix)

and this type of computation model has some deep deep connections to the P vs NP problem

[sweet pendant]

For this question how do I get started for the second part? Do I just need to show that (a,b) times (x,y) = (1,0) and then just find the values of x and y?

[sharp ice]

yeah do exactly that. u can also notice that R is isomorphic to the ring of complex numbers. the addition and multiplication is exactly the same as addition and multiplication on the complex numbers, and u should be familiar already with finding inverses of complex numbers

[sweet pendant]

Okay thank you so much 🫡

[sweet pendant]

Another doubt related to the same question. When showing that (R,+) has a neutral element, I need to show that there exists some (c,d) in R such that (a,b) + (c,d) = (c,d) + (a,b) = (a,b). In this case the neutral element is (0,0) but do I need to prove it? Or can I just say (0,0)?

[barren sierra]

you should prove that (0, 0) satisfies the property you stated

but like it's kind of a one liner lol

[sweet pendant]

I just said that since R=R^2 (0,0) is in R and by the definition of addition in R (a,b) + (0,0) = (a,b).

[barren sierra]

Your explanation is inaccurate

This is not just the standard ring R^2

Yes the addition is the same but the multiplication is different

So it's not just R = ℝ^2 as a ring

Because when you just say ℝ^2 as a ring, you mean with coordinate wise addition and product

So you really should be arguing by the definition of R

So everything you said is right except for the R = ℝ^2 part

It's being really pedantic but it's good to be pedantic when just starting

[sweet pendant]

So instead can I say since R is a set of ordered pair of real numbers, (0,0) is in R etc,etc

[wintry sluice]

sorry i’m not too good with complex groups yet. i get that each element is a rotation on the complex plane but i don’t know how to demonstrate that. i also don’t know what the inverse for any element would be

[void cosmos]

use euler's theorem

that's how u would go to the geometric intuition

ie represent a complex number in polar form and then use re^itheta = r(cos(theta)+isin(theta))

so this is really just a circle

the set H i mean

now is anyone here up for a field theory qual problem proof check?

i wanna sleep and it's just a y/n proof verification

just from i --> ii the rest is normal

the hints would be to know what kind of structure an isomorphism respects really

it should be all of them

that is for example if two groups are isomorphic then u know they must have the same order

so the problem is asking u to find the order of the image of an element

surely ur intuition would say that somehow u should consider the order of the element first before sending it

and then on ur paper maybe try to find out how an isomorphism changes that

correct?

that is

if we have two groups say G and H and an isomorphism, if an element g in G has order n what is the order of phi(g)?

phi is ur isomorphism ofc

[void cosmos]

i) --> ii) consider the polynomial x^p-a and call a root of this polynomial y. if y is in F than we are done if not then y is in some algebraic closue and hence by assumption we must have min_y(x) be seperable. also min_y(x) must divide x^p-a and hence x^p-a is not irreducible. factor x^p-a into irreducibles and since (x-y) must be a factor then the minimal polynomial must be one of the irreducible factors. consider now x^p-a = min_y(x)*g(x) where g(x) is the product of the rest of the irreducible factors. we see that by taking derivatives we must have either min_y'(y) = 0 or g(y) = 0. as min_y is seperable we must have g(y) = 0, but this means min_y must divide g(x) and since min_y is irred and hence prime it must divide one of the irreducible factors which is a contradiction?

yeah

[void cosmos]

chatgpt says im right please prove chatgpt wrong ik this is ur fav thing to do

[chilly ocean]

Apologies if this doesn't belong here. In the first picture, T is assumed normal on a finite-dimensional real inner product space.

Why does the primary cyclic decomposition of (V, T) have the same number of summands as there are factors of the minpoly? (Is the primary decomposition already primary cyclic, and if so, why?) Assuming this to be the case (otherwise I'm not sure how the direct sum is orthogonal as indicated by the dotted circles), how do we get from the basis B_i to an orthonormal basis with the same matrix form - the statement of which I assume is missing in Theorem 10.10..?

[void cosmos]

do u know the fundamental theorem of f.g modules over PIDS?

[chilly ocean]

Yes

[void cosmos]

each summand annhilates the whole space no? cuz these are quotients

in a way

so the minimal polynomial must divide each one and each one must divide the other untill the last factor (invariant factor)

[void cosmos]

im sorry if what im saying is nonsense i was about to sleep after i threw my solution, i just said what my gut told me idk if this is right sorry if it isn't

good luck im off to sleep

[chilly ocean]

Hm I assumed we were working with the elementary divisor formulation

[void cosmos]

yeah this

mb

[chilly ocean]

Good night

[void cosmos]

but ig that was not the point

[chilly ocean]

So my understanding is this: the (unique) primary decomposition of (V, T) is of the form E_{l_1} (+) ... (+) E_{l_k} (+) W_1 (+) ... (+) W_m, where l_1, ..., l_k are the eigenvalues and W_j is the set of vectors killed by p_j(T). This decomposition is pairwise orthogonal because the orders of summands are pairwise relatively prime. But the first thing that I'm having trouble seeing is why the W_js must be cyclic, which I'm assuming Roman claims to be the case because otherwise the primary cyclic decomposition would involve more terms than is written (breaking up the primary summands into primary cyclic ones), in which case I can't use the relatively prime argument to claim that the direct sum is still orthogonal.

I'm also off to bed, will bump this when I get back ig lol

[cobalt heath]

Is rep theory through quivers supposed to be brutal? I feel it is somehow quite involved.

[lament flame]

From what I've seen rep theory in general is pretty brutal

[drifting plover]

I’ve heard of these things actually! I’ll look into them, thanks

Apparently they’re called arithmetic circuits

[drifting plover]

This is something called Horner form?

[arctic trail]

yeah I think I've heard that

probably that

[drifting plover]

I think Horner forms work for polynomials in any number of variables actually

http://www.cs.ru.nl/~freek/courses/tt-2014/read/10.1.1.61.3041.pdf

Would be funny if the optimal solution is just to write it in Horner form lol

[vast quiver]

I think reading Elements of the representation theory of associative algebras by Assem et al is really nice actually (at least through chapter 3 which is as far as I got haha). It's real surprising to me that any finite dimensional K-algebra (maybe K is perfect or alg closed or something) is morita equivalent to a quiver algebra modulo relations. I think it's very pretty to start with your algebra and "find" the quiver and the relations

[rocky cloak]

You need K algebraically closed, but there is a generalization of quivers called species that work for arbitrary fields. This also extends Gabriel's theorem to arbitrary Dynkin diagrams, not just ADE

[rocky cloak]

I don't think it's supposed to be brutal, but it is definitely supposed to be involved. It's a big field of study after all

[cobalt heath]

Thanks, I will look into the book

[cobalt heath]

I see, I felt that the Nakayama functor and related business is pretty intense.

[rocky cloak]

I mean, there is a lot of machinery underlying to make everything work. But when it comes down to it the Nakayama functor basically just map the projective cover of a simple to it's injective hull, that's it.

I guess one gets used to everything

[cobalt heath]

Hmm.. I am struggling to get the usage of it in proofs, and intuitive picture.

Maybe some of it is technical computations

[finite turtle]

has anyone read or looked at Paolo Aluffi Algebra: Chapter 0 and Algebra Notes from the Underground? wondering if i should get both or just Algebra Chapter 0

[glad osprey]

I've read a little bit of both, for a first introduction I would definitely go with Notes from the Underground. Chapter 0 focuses a lot on category theory, and I find that you end up spending more energy learning that than the actual algebra

[chilly sparrow]

I'm reading through notes from the underground & really enjoying it. Nice conversational style.

But I'm at the level of "re-learning math i vaguely did 8 years ago" so good chance it's too basic for a lot of folk

[south patrol]

Dostoevsky?

Jk

[cloud lynx]

Every Splitting field of a family of polynomials is normal but is every normal field extension also a splitting field?

and is every algebraic closure a splitting field?

[tardy hedge]

Yeah true but I had a great time reading about modules in dummit and foote tbh

So somehow they didnt make it sound dry to me at least

[tardy hedge]

Hm not really unfortunately, i mean i do have Prof’s notes but the notes themselves arent particularly special

His lectures are just really good. He’s passionate and really wants us to understand the ideas

[tardy hedge]

Lmao

[surreal dagger]

I think after the bit where you learn about categories at the very beginning youre not doing much category theory for a long time in Chapter 0

[rocky cloak]

Yes, every normal extension is the splitting field of the polynomials it splits. In particular the algebraic closure is the splitting field of all polynomials.

[cloud lynx]

also if the normal extension is infinite?

[rocky cloak]

Yes

[cloud lynx]

so nomal extension <=> splitting field <=> algebraic closure?

algebraic closure over K*

[rocky cloak]

First one yes, but you can be a splitting field without being algebraically closed

[cloud lynx]

oh because in lecture we had that splitting fields are exactly the algebraic closure

[cloud lynx]

do you have an example?

[rocky cloak]

Basically any example.

Like Q(sqrt(2)) is the splitting field of x^2 - 2

[rocky cloak]

I think you must be mixing something here

[cloud lynx]

but it is also the algebraic closure over Q because Q(sqrt(2) is algebraic closed and Q(sqrt(2))/Q is algebraic

[rocky cloak]

It is very much not algebraically closed

[coral spindle]

When you talk about the algebraic closure, what do you think that means?

I think you may be confusing it with a different idea

[cloud lynx]

because for example X^2+1 has no roots in Q(sqrt(2))?

[coral spindle]

Yes

[rocky cloak]

Yeah for example

[cloud lynx]

the algebraic closure T/K is algebraic closed (every polynomial over T has a roots in T ) and T/K is algebraic

thats the definition we have

[coral spindle]

So it should be clear that Q(sqrt(2)) is far from being the algebraic closure of Q.

[cloud lynx]

what do they mean with the splitting fields of F over K are exactly the algebraic closures of K?

[rocky cloak]

They mean if you take F (the set of all polynomials) and take it's splitting field, you get the algebraic closure

[cloud lynx]

AHH over ALL polynomials

my reading skills are bad sorry

so normal <=> splitting field <= algebraic closure, and splitting field <=> algebraic closure only if its the splitting field over all polynomials

thank you guys!

[rocky cloak]

Yes, that's right.

Though I guess it's worth noting that the algebraic closure can also be the splitting field of a smaller set of polynomials

[cloud lynx]

how?

[rocky cloak]

(different families of polynomials can have the same splitting field)

For example over R, the splitting field of x^2 + 1 is C

C is also the splitting field of the family of all polynomials

But you really just need one

[cloud lynx]

one polynomial?

[rocky cloak]

Yeah x^2 + 1

(for example)

[cloud lynx]

ah ok now i understand

[surreal dagger]

What do you guys cover in a module theory course?

[elfin prairie]

is there any criterion that shows that set wrt to two operations is a field?

besides proving all the field axioms

[rocky cloak]

I mean, the axioms holding is the definition. But in most cases you get most of them for free, because the operations you have aren't pulled from think air, but coming from something already with some structure

[rocky cloak]

If there was a criterion that was simpler than the axioms, then people would just use that as the axiom

[elfin prairie]

hmm okay

I googled it just now and it says that closure wrt to addition, subtraction, multiplication and division guarantees that it is a field

So closure wrt addition, multiplication and their respective inverses

is this valid?

[rocky cloak]

Well, that depends what you mean exactly.

If you take a subset of a field, that is closed under all that, then it is indeed true.

Otherwise, what exactly do you mean by closure?

[elfin prairie]

oh I see

so there is an assumption that it is a subfield of a known field

by closure I mean the set is closed with respect to the binary operation

I didn't mean it in any other context

[rocky cloak]

Well, that's just the definition of being a binary operation

[cloud lynx]

T=( subgroups of G) . If we have a galois group with Order n then the Order of T <= n+1?

[coral spindle]

The set Z with addition being defined as a + b = 13, a - b = 5, a * b = 32 and a / b = -1 is closed under all four operations

This is clearly not a field. You need to follow the field axioms. There's no way around this.

[rocky cloak]

If T is a subgroup of G, then the order of T is less than the order of G yeah. You don't even need to add 1

[elfin prairie]

Got it 👍

[cloud lynx]

no I meant T is the set of all subgroups of G

[coral spindle]

So this isn't a galois theory question, you're asking if the number of subgroups of a group G can exceed |G|+1

Groupprops tells me that S_4 has 30 subgroups

And indeed S_4 is the Galois group of some extension

@cloud lynx see above.

[cloud lynx]

i cooked

i am cooked

[coral spindle]

Culinary mathematics, great stuff.

[cloud lynx]

because my thoughts are the following

[rocky cloak]

Also, I think (Z/2)^3 has 16 subgroups

I think in general (Z/2)^n will give you a lot of bang for you buck if you're looking for many subgroups for a small order

[cloud lynx]

if I have a galoisgroup with order n then there exists n automorphisms which have maximal order n and every automorphism has therefore finite order so every automorphism is a generator of a subgroup of the galois group so there exists max n+1 subgroups of a galoisgroup

[coral spindle]

if I have a galoisgroup with order n then there exists n automorphisms which have maximal order n
This is false

In fact this never happens.

[cloud lynx]

if this comes in my algebra exam i am gonna quit math

[coral spindle]

and every automorphism has therefore finite order
Yes, this is trivial because the group is finite.

[rocky cloak]

Every automorphism generates some subgroup, but you can have subgroups generated by more than one element.

[coral spindle]

Have you studied group theory before tag?

[cloud lynx]

yes i did

[coral spindle]

OK

[cloud lynx]

do you have an example for a galoisgroup which has more subgroups than its order?

[coral spindle]

I just gave an example...

[cloud lynx]

why?

you mean S_4?

[coral spindle]

Yes

[cloud lynx]

mmh

then i am cooked for my exam

i dont know how to find all subgroups of S4 example

because by now i had only galois groups with order n which have only maximal n+1 subgroups thats why i had this wrong thoughts

[chilly ocean]

bump (text is above)

[rocky cloak]

Both me and Boytjie have examples

And (Z/2)^3 is pretty easy to cook up an extension for. Like Q(sqrt2, sqrt3, sqrt5) for example

[cloud lynx]

and (Z/2)^4 isomorh to Gal(Q(sqrt(2),sqrt(3),sqrt5,sqrt7)/Q))?

[rocky cloak]

Yup

And so on adjoining square roots of primes

[cloud lynx]

and Gal(Q(third root(prime1), sqrt(prime2))/Q) isomorph to Z3xZ2?

[rocky cloak]

Adjoining a third root of a prime, doesn't give you a Galois extension. You need the other third roots aswell, in which case the Galois group becomes S3. And then I think it depends on the prime wether it already contains the square root of the other prime

[cloud lynx]

what does adjoining square mean?

[rocky cloak]

Q(a) is Q adjoined a.

So Q(sqrt2) is what you get by adjoining the square root of 2

[cloud lynx]

ah ok

[cloud lynx]

i know but the extension does not have to be galois in order to have a galois group

[rocky cloak]

I see, but then the Galois group of Q(third root of prime) is the trivial group

[cloud lynx]

and Q(third root of prime2, sqrt(prime1))/Q?

[rocky cloak]

That would have the same Galois group as Q(sqrt prime 1)

(so Z/2)

[cloud lynx]

why

Z/2 has order 2

and this galois group has order 6

[coral spindle]

It does not have order 6.

It has order 2.

Because the extension is not Galois, the order of the group of automorphisms is not necessarily the degree of the extension.

This is an example of such a case.

[cloud lynx]

ah

[dim widget]

unless prime2 is -3

[rocky cloak]

In fact an extension is Galois if and only if it has degree equal to the order of its automorphism group

[cloud lynx]

ok tyy guyss

is there a trick or statements to see wether a galois group is cyclic or not? or abelian or not?

because u guys can just tell to which group a galois group is isomorphic to

[dim widget]

there's no trick

[cloud lynx]

for me its actually pretty hard unless i have determined all subgroups of the galois group

[dim widget]

some are just obvious, basically cyclic/abelian galois groups mean there is some way to write the extension where it is clear that there are no relations between the different roots of the minimal polynomial

so like Q(sqrt(p)) is quadratic so its galois with galois group z/2

then Q(sqrt(p), sqrt(q)) must have galois group the klein 4 group (intuitively, just because the two generators have no relation to each other, so the galois group should just be a product of the two galois groups)

[cloud lynx]

and how do i know wether they are cyclic/non abelian? or do i have to show it by hand

[dim widget]

certain extensions are cyclic because of general theory

e.g. if K contains an nth root of unity then K(nthroot(p)) is always cyclic and Galois

There is no silver bullet that tells you when something is abelian or nonabelian in general, you have to learn from experience/experiment around with automorphisms

But one example to think about is Q(sqrt(-3), cuberoot(p)) where p is a prime bigger than 3

this is galois over Q with galois group S_3 = Z/3 \rtimes Z/2

[cloud lynx]

nth root of unity is x^n=1?

[dim widget]

its a solution to the nth cyclotomic polynomial, which is the irreducible polynomial of largest possible degree which divides that polynomial

anyway in the case of Q(sqrt(-3), cuberoot(p)) the reason the galois group is not a product is because there is no galois degree 3 subextension

which there would be if the group was a product

if it were a product then taking the fixed field of complex conjugation (an element of order 2) would give a galois degree 3 subfield, but instead it gives Q(cuberoot(p)) which is not galois

so you see that the galois group must be nonabelian

[cloud lynx]

ok thank you!!!

[tardy hedge]

Starting definitions, module homs, Hom(M,N) as an r module, free modules, exact sequence, tensor product

I think we are also gonna do the modules over pid stuff

[finite turtle]

main question for me is how much overlap is there? actually i have dummit and foote already

[unkempt stream]

HELL

[arctic trail]

yeah that would be quite funny

what's the horner form of x + xy + y?

x(1+y) + y innit?

this isn't optimal as (1+x)(1+y) - 1 is better

[drifting plover]

Oh my thing tells me the Horner form is x + xy + y lol

Guess it's not the same Horner form as the one I had in mind

[arctic trail]

A good algorithm (as in to find better than horner) seems to be:
Decomposing a polynomial as something with a lot of factors + a remainder with low degree

[glad osprey]

If you already have dummit and foote then you won't need both chapter 0 and notes from the underground. If you think DF is difficult, get notes from the underground, otherwise get chapter 0

Personally I quite like Notes from the Underground. It has a very conversational style that is fun to read

[sweet pendant]

Can someone give me a hint for this question? So far I know I need to show that (1-sr)*t=1 which is essentially showing that (1-sr) has a multiplicative inverse. But I can’t figure out how to “construct” t?

[grave sedge]

How would you represent 1/(1-sr) as a series?

(think outside of abstract algebra, if you wish)

[sweet pendant]

Yes I figured it out. Thank you so much!

[main flower]

here’s a problem from an old qualifying exam archive but I’m not sure how to solve it. I have some facts that I know written down but everything I try leads me nowhere.

part 1b) is what i’m specifically struggling with.

[unkempt stream]

OOOOOH

[unkempt stream]

I am so close to flash banging you with cool radical (pun intended) shit

Naturally for any ideal (commutative), a in J or b in J implies ab is in J. This is two way if it's prime...

this applies to powers of a too lmao

[main flower]

Yes that makes sense.

I’ve done a lot of scratch work trying to get to the answer but I seem to just be spinning around.

[main flower]

.

stuff like this.

[unkempt stream]

do you mind zooming into the problem please :3

helps me a lot

Oh i think I see it.

Assume for each x in R there exists an m > 1 such that x^m = x. It's gonna be commutative even without unity then you want to show every prime ideal is maximal

[main flower]

yes

[unkempt stream]

let me ponder for a second, it's a new problem to me

[main flower]

that’s alright take all the time you need.

[unkempt stream]

Maybe assume J is an ideal containing prime ideal P strictly (i.e we have a z in J\P). Maybe we can show J = R

[unkempt stream]

Not sure if this is helpful, but if x^m = x, then x^(n(m-1) + 1) = x for each n

[main flower]

Yeah that’s what i’m trying with q. I’d like to use operations to show that two elements in the ideal containing q and the prime ideal will let us achieve a sum of 1

and if 1 is in it, then it’s equal to R

[unkempt stream]

OH

let p^a = p, x^b = x. If k = lcm(a - 1, b - 1), then (xp)^(k + 1) = xp

hm

[main flower]

I don’t think that’s true.

[heavy minnow]

Didnt check the details and i havent slept so take it with a grain of salt: Let P be a prime ideal. Then R/P is an integral domain. It suffices to show that it is also a field. Let a in R but not in P. It suffices to find b in R such that ab-1 is in P. There is m such that a^m=a. Then a(a^{m-1}-1)=0 in P, so either a in P or a^{m-1}-1 in P. The first case is already excluded previously, so a^{m-1}-1 is in P. Hence, take b = a^{m-2} (which is possible since m is at least 2) and we are done

[unkempt stream]

oh that works.

I am a moron

[main flower]

Ahhh that makes sense because it uses part a. I should’ve thought of something like that. Thank you!

[barren sierra]

arithmetic algebraic I've seen both around tbh, it's just a name

[arctic trail]

the set of 1-nilpotent elements forms a group, which is dope

[errant wedge]

So ik one can prove x and other degree >0 polynomials in R[x] have no inverses using a degree argument. R[x] embeds into R[[x]] by taking a_k = 0 for all k > some n in N, can I prove x in R[[x]] has no inverse by using such embeddings and counting degree, or is this flawed since degree isn't in general well defined? I want to use this to then show formal power series with constant 0 are not units, but I'm thinking this probably does not work

hrm wait nvm i think i got it

[alpine island]

It is trivial that aH = H implies a in H right?

[glad osprey]

Yes, a is in the LHS, so it must be in the RHS

[topaz solar]

Assuming H is a subgroup anyway

[signal thistle]

$f(X)^c = f(X^c)$ true or false yes or no

[cloud walrus]

bug free

[topaz solar]

What do you think

[signal thistle]

$f(X)^c = {x: x \notin X \vee f(x) \notin Y}$

[cloud walrus]

bug free

[signal thistle]

and $f(X^c) = {x: x \notin X^c \vee f(x) \notin \text{what goes here} }$

either way I think no

[topaz solar]

Are you sure about that

[cloud walrus]

bug free

[topaz solar]

Neither of the sets should be consisting of points in X or X complement

[signal thistle]

wow do i need to review some defnitions.

[topaz solar]

Assuming the domain and codomain disjoint

Anyway, it’s false

But you’re kinda doing these wrong I think

[signal thistle]

how so

[topaz solar]

Well, it shouldn’t be the set of x things whose f(x) has so and so property

So that’s already a bad sign

[rapid junco]

A transitive group action is one such that for all x, y \in X there is a g \in G so that gx = y

Is there a notion for when there is exactly one g?

like SO(n) on the sphere.

[void cosmos]

hi

this is easy right?

1. has (x-1) and (x+1) as zero divisors, and the constants are the units except 0

2 is a field ?

[south patrol]

Sure

[void cosmos]

or by 1 i mean R and 2 i mean S

yeah jusit wanted to check seemed suspectively easy

thank you

[south patrol]

Owo

[void cosmos]

u*

also

one more problem

i got stuck on

cuz i have no practice with this

[south patrol]

Wait I didn't read the units bit

[void cosmos]

yeah

okay 1 more problem that is very embarrasing but yeah whatever

the problem just stated ( and it recurred on multiple quals )

classify all semisimple rings of order n

[south patrol]

Lol

[void cosmos]

now since these are finite they are artinian

so i can use artin-wedderburn

but i can't figure out how the matrix dimensions and what are the division rings

cuz i remember one problem asking this + (how many are commutative?)

i know the commutative ones would have to be like the ones that have dimension 1 over their matrix rings

but i wasn't sure

how to like write this fully

any hints?

[void cosmos]

also another bad question

the shift up operator on the set of infinite sequences is a famous counterexample for like proving injectivity <--> surjectivity fails when the vec space has infinte dim

literally the same counterexample should work in the context of noetherian modules correct?

that is if a surjection on a noetherian module is an injection

[rocky cloak]

So I guess you want to keep in mind that all finite division rings are fields, and that you presumably know what the finite fields are.

From there I'm not sure how detailed you need the classification to be...

[arctic trail]

1. also has any linear polynomial x-a as a unit, for |a| != 1

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

[arctic trail]

oof, wouldn't that be annoying.

|GL(m,F)| = F^m
so like n has to be a prime power say n = p^n' and then you're essentially playing with the divisors of n'.
And your ring is of the form GL(m,F_p^m') with mm' = n'

(finite domains are fields by Wedderburns little theorem)

[arctic trail]

Depends on the domain and codomain

probably on their lengths if I had to guess or something

[true island]

I have to show the previous equality in a problem set on nilpotent groups and I'm not sure if they're integers and the left superscript operation means the tetration or if it's another operation defined on groups that I'm not aware of

Precisely, the exercise states 'Show that, in general: [above equality'

If anyone could enlighten me

[true island]

(first case seems absurd)

[arctic trail]

I've never seen the notation

then I guess using n = 1 we have ab = ^{a}ba

so ^{a}b = aba^(-1)

so does ^{h}g = hgh^(-1) in general?

[arctic trail]

by induction you have abab...ab = (ab)^n

so I guess it should've been (ab)^n in the problem and not ab^n

[arctic trail]

this makes sense if it's notation for the action of h on g (by conjugation)

[true island]

Right ! It does make sense thanks

[upbeat dirge]

Hey what can I say if there are two subgroups H,K of a group G such that they are conjugates (i.e there is a g s.t
gHg^-1 = K

[rocky cloak]

I'm not sure what kind of stuff you're looking for, but they will be isomorphic and of the same index at least.

[coral spindle]

You might be able to say worcestershire if you try hard

[cloud lynx]

how important is the theorie behind symmetric functions for galois theory?

or can i keep it out

lol

theory*

How does the field of the symmetric functions in X1,...Xn look like? I have no idea or intuition of this field

just K={f of K[X1,...Xn] | f(x1,...,xn)=f(x k(1),...x k(n)) for all k of Sn}?

[rotund aurora]

It's purely transcendental if I'm not mistaken

[coral spindle]

Isn't it a famous theorem that it's isomorphic to some polynomial ring

Like a classic result of invariant theory

[rotund aurora]

? it's due to Hilbert I think

and this is Galois theory

and field theory I guess

[cloud lynx]

are this answers to my questions?

these*

[rotund aurora]

K is isomorphic to K(x_1,..,x_n)

and you can take as the x_i the ith symmetric polynomial in the X_1,...,X_n

[wooden ember]

cursed notation

[rotund aurora]

yeah I just realized

[cloud lynx]

what do u mean by the c_i the ith symmetric polynomial?

x_i*

[rotund aurora]

Consider $\prod_{i=1}^n (T-X_i)=s_0+s_1T+\cdots+T^n$ where $s_i$ are polynomials in the $X_1,\dots, X_n$ (with integer coefficients). The $s_i$ are the symmetric polynomials (maybe there are some conventions as to whether you pick $s_i$ or $-s_i$, but this is irrelevant here)

[cloud walrus]

croqueta3385

[cloud lynx]

ahh okk tyy

[tender wharf]

what is equation 41 saying here? i dont think axa^(-1) makes any sense unless we we are considering the action of G on itself by conjugation

[south patrol]

It should be Stab ax

And then the statement just follows because g(ax) = ax <=> aga^-1(x) = x

@tender wharf

Where I have added unnecessary brackets for clarity hopefully lol

[tender wharf]

aoh I see

[south patrol]

It is a bit silly of them to say the proof is clear and get the statement wrong

[tender wharf]

So its a typo

I was trying to prove it myself and I was only able to do it when it was stab ax lol

So good to know

[south patrol]

Lol well nice

[tender wharf]

he makes another typo on the next page I think

he says equation 41' but there is no 41'

so its probably 42'

[void cosmos]

yes, thank you jagr and triviallemam

[tardy hedge]

If an infinite group with every element having finite order, can I show this group has infinite number of subgroups by just doing:
pick g in G, subgroup <g> is finite. Then pick g' in G that is not in <g>, and then <g'> is a different subgroup , just continue this wa

way

so G has infinite number of subgroups

The question I am trying to solve is: "A group with finite number of subgroups is finite"

So I was trying to show contrapositive

I dont know, maybe there is a better way or something, pls guide me with your smartness fellow math ppl

[rocky cloak]

You did great, no need for guidance

[tardy hedge]

Thanks a lot!

[arctic trail]

in fact a group has an infinite number of subgroups if and only if it's infinite

[tardy hedge]

Btw I will forever be grateful for this server

If you guys weren’t here i’d be way behind

Everything would just take longer to understand, have to go wait for prof office hours all the time etc

[glad osprey]

A group action of G on A is a group homomorphism $G \to Aut(A)$, right? Many sources also define it as a function $\sigma : G \to (A \to A)$ where $\sigma(e_G) = Id_A$ and $\sigma(gh) = \sigma(g) \circ \sigma(h)$. Isn't $\sigma(e_G) = Id_A$ redundant, as it just follows from the homomorphism property?

[cloud walrus]

sheddow

[prisma ibex]

yes

the property of preserving the identity is redundant if you are already assuming this is a group homomorphism sure, but you need this condition if you're spelling out what it means to be a homomorphism in the first place

[arctic trail]

shouldn't the bottom be Aut(A)?

or does it follow, or something

[void cosmos]

holy fuck guys

im having a crash

wtf are the roots of x^4+7

??

nvm

[glad osprey]

but f(e_G) = e_H is not part of the definition of a homomorphism

but it might have something to do with what Trivial says, when the codomain is A -> A instead of Aut(A), you can't assume sigma(f) is invertible, so you can't deduce the identity property from the homomorphism property

[prisma ibex]

it depends on how you define homomorphism

preserving the identity and inverses should be automatic from preserving products

whether you prefer this to be part of the definition or a consequence is a matter of taste

(it's basically the same as whether you consider the inverses and the identity to be a structure or a property)

[glad osprey]

So for the first definition you have a homomorphism G to Aut(A), and since Aut(A) is a group you automatically preserve the identity. For the second definition you only have a map G -> (A -> A), but you can use the axiom sigma(e) = Id_A to show that the range is actually Aut(A). Is that correct?

[south patrol]

Well the answer is basically a tautology

[void cosmos]

Wdym

[south patrol]

Well like

They are the fourth roots of -7

i.e. i^k 7^1/4

but that is like not a very amazing answer because i am just saying the 4th root of 7 is the 4th root of 7

[chilly radish]

If it respects composition then it mean that \sigma(g) is invertible for all g

If it's meant to be compatible with the operation on A it should be a map from G to End(A)

[arctic trail]

fair enough

[limber thorn]

I have a few questions on how surgery effects the homotopy groups of manifold. Assume we are examining the effect of $0$-surgery on a topological manifold of dimension $3$ which we denote $M$, and it's fundamental group. We know $\pi_1(\chi(M)) \cong \pi_1(M #(S^1 \times S^2))$. How do I get the final result $\pi_1(M) \star \mathbb{Z}$? And can we even follow this method for higher homotopy groups?

[cloud walrus]

Architect

[candid dove]

Quotient modules do have a universal property right?

[lone niche]

yes

[candid dove]

Cool

Can I say like the forgetful functor from mod to grp is adjoint to the Z module functor

So it happens?

Or is it needed to prove it separately

[lone niche]

idk tbh

[chilly ocean]

Let $A = (a_{ij}) \in \text{Mat}{m,n}(F)$. Then $A^t = (b{ij}) \in \text{Mat}{n,m}(F)$ with $b{ij} = a_{ji}$.

    \begin{proof}
        We have:
            \begin{equation*}
            \begin{split}
                T_A(e_i)&= \sum_{k=1}^m a_{ki}f_k \\
                T_A^\vee(f_j^\vee) &= \sum_{k=1}^n b_{kj}e_k^\vee.
            \end{split}
            \end{equation*}
        Applying $f_j^\vee$ to $T_A(e_i)$ yields:
            \begin{equation*}
            \begin{split}
                (f_j^\vee \circ T_A)(e_i) &= f_j^\vee \left(\sum_{k=1}^m a_{ki}f_k\right)\\
                & = \sum_{k=1}^m a_{ki}f_j^\vee(f_k) \\
                &= a_{ji}.
            \end{split}
            \end{equation*}
        Evaluating the $T_A^\vee(f_j^\vee)$ at $e_i$ gives:
            \begin{equation*}
            \begin{split}
                T_A^\vee(f_j^\vee)(e_i)
                & = \sum_{k=1}^n b_{kj}e_k^\vee(e_i) \\
                & = b_{ij}.
            \end{split}
            \end{equation*}
        We have $(f_j^\vee \circ T_A)(e_i) = T_A^\vee(f_j^\vee)(e_i)$. Hence $a_{ji} = b_{ij}$
    \end{proof}

[cloud walrus]

gian

[chilly ocean]

Can anyone explain why this proof is applying f_j^v to T_A

And also why we evaluate at T_A^v(f_j^v) at e_i

this is assuming E_n = {e1,...,en} is the standard basis for F^n and W_n = {f1,...,fm} is the standard basis for F^m

maybe i should ask this in linear algebra actually

[prisma ibex]

yes this is an easy application of Van Kampen

you should learn what the Van Kampen theorem is and how to use it, it's a standard thing that shows up quite early in most algebraic topology courses

if you know the statement of Van Kampen then it is immediately obvious, it's meant to be a pretty rapid thing

thankfully this is like, the standard result to use any time you try to compute \pi_1 of any sort of gluing or connect sum like this

so usually you do not have to make so many decisions around what results might be useful for problems like these

[warped bramble]

H is a cyclic subgroup of G with order p, where p is a prime.
a is in H iff aH=Ha

is this true?

[prisma ibex]

have you tried this in a basic example

[warped bramble]

oh it's true

Ok it's obvious

[jade flume]

Suppose that $G = H \rtimes K$. Previously proven $G$ may be identified as a set with
the Cartesian product $H \times K$, and that viewed this way, the multiplication on $G$ is given by
[
(h_1,k_1) \cdot (h_2,k_2) = \left(h_1 (k_1 h_2 k_1^{-1}), k_1 k_2\right) = (h_1 \phi_{k_1} (h_2), k_1 k_2),
]
where for any $k\in K$, $\phi_{k}: H \rightarrow H$ is the map given by $\phi_k (h) = khk^{-1}$.

(External) semi-direct products.
The previous problem motivates the following definition. Let $H$ and $K$ be abstract groups, and let
[
\phi : K \rightarrow Aut(H), \quad k \mapsto \phi_k
]
be a homomorphism. The (external) semi-direct product $H \rtimes_\phi K$ is defined as follows. As a set, it is the Cartesian product $H\times K$, with group law given by
[
(h_1, k_1) \cdot (h_2, k_2) = (h_1 \phi_{k_1} (h_2), k_1 k_2).
]

Status: I have begun but got stuck midway.

What is the relation between internal and external semi-direct products?

Claim:
For disjoint subgroups within the same larger group forming an internal semi-direct product of the larger group, there exists an isomorphism between their external semi-direct products.
Idea:
Show semi-direct internal products are isormophic to extern semi-direct products.
Then, show external semi direct products are equal to internal semi direct products.

[cloud walrus]

kaxi

[serene dune]

$|HK| = \frac {|H||K|}{|H \cap K|}$

[cloud walrus]

yeshua

[serene dune]

2nd isomorphism theorem should do it

[dim widget]

No its obviously false

a counterexample is p = 3, H = <(123)> and a = (12) in S_3

[obtuse bear]

oh wait, misread the original question, nvm

[void cosmos]

do i just go that |X| = |X^G| + number of elements in each orbit

but by orbit-stab theore mwe have number of elements in each orbit is equal to p^n/stab_g(a represnetative of that orbit)

but this goes to 0 mod p and we have the claim?

[rocky cloak]

Yup

[void cosmos]

thank you kng

king

[cloud lynx]

is the direct product of rings commutative if only if the components are commutative?

should be right?

if we have commutative Rings R1 and R2 commutative then

[arctic trail]

all parts must be commutative

R1 and R2 both embed in R1 x R2

[cloud lynx]

(a1,b1)(a2,b2)=(a1 a2, b1 b2)=(a2 a1, b2 b1)=(a2,b2)(a1,b1) because R1 and R2 are commutative

[arctic trail]

and if they're commutative then the product is commutative

[cloud lynx]

yes ok thank youuu

[cloud lynx]

if we have a a principal ideam domain

then (a1)+....+(an)=(a1,....,an)?

because we have a theorem that d gcd of a1,...,an if only if (d)=(a1)+...+(an) where as R is a PID

[rocky cloak]

This is true in any ring, just by definition.

[cloud lynx]

what is actually the thing about gcd and ideals? why do I need to know it?

[rocky cloak]

Like I + J is the smallest ideal that contains I and J, and (x, y) is the smallest ideal that contains x and y

That's just what this notation means

[cloud lynx]

ahh

[rocky cloak]

In a PID (x, y) = (gcd(x, y)) just like you said

(and similarly for more than 2 elements)

[cloud lynx]

but for what is it good for? only to simplify?

[inner steppe]

Let G be a group with the identity element 1 and a,b,c is also in G. Let x be in G which satisfies the equalities. Find G.

[rocky cloak]

Yeah, it's very useful to have an ideal generated by just one element.

Then every element in the ideal just looks like a multiple of the generator.

When you have more generators you have to think about relations between them and things get complicated.

[inner steppe]

I have been stuck on this a little too long. I filled in two tables but still cannot find x. I get like x^5 = c but don't know how to go one from there.

[cloud lynx]

ahh okk tyyyy

[rocky cloak]

Are we supposed to assume that a, b, c, 1, x are the only elements in G, or that G is generated by them? Are they necessarily distinct?

[inner steppe]

Let G be a group with the identity element 1, and a, b, c ∈ G. Given that x ∈ G satisfies:

This is the only thing on the question.

I guess they have to be distinct and that G only consists of only these 4 elements.

[rocky cloak]

Well, then no such G exists

If x^3 = 1, then either x=1 or the order of G is a multiple of 3

[inner steppe]

I forgot I have a solution lol but they did something like:

ax^2= bx <=> ax^3 = bx <=> [x^3 = 1] <=> a=bx

[arctic trail]

a LOT of things

[rocky cloak]

Yeah, sure (assuming you meant ax^2 = b at first there)

[arctic trail]

G = {1,x,x^2} a = x; b = 1; c = x^2

[rocky cloak]

Is there something about those equivalences you don't understand?

[inner steppe]

Yes, but why did you say that such no G exist.?I am asking because I actually wanna understand the topic fully.

[rocky cloak]

Like I said, the only groups in which x^3 = 1 is possible, are groups where the number of elements is a multiple of 3.

So if the problem is supposed to be interpreted as the group having 4 elements, then no such G exists

Because 4 is not a multiple of 3

[rocky cloak]

Of course it could be that the problem is supposed to be interpreted some other way.

Like the way Trivial did here for example

[arctic trail]

maybe it's a forall x in G satisfies?

[rocky cloak]

That seems even weirder to me. But at least then G would have to be the trivial group

Also c would be completely irrelevant

[arctic trail]

can be Z/3Z^n I guess

[rocky cloak]

Can't have ax^2 = b for all x

[arctic trail]

oh right

I was just thinking of the second one lol

[inner steppe]

(U(Z8), ×) are groups. U(Zn) consists of the invertible elements in Zn. Investigate whether the group is cyclic.

So U(Z8) = {1,3,5,7} then <1> = {1}, <3> = {1,3}

This is only what I sketched so far, how do I solve this?

The order of 1 is 1, then order of the rest in the set is 2. But how can this help me?

[coral spindle]

What does it mean to be a cyclic group.

This should be a one-sentence affair

[inner steppe]

From what I understand, the smallest sub group of a group that consists of x, like the smallest group made from x

[coral spindle]

OK so you don't know the definition

[inner steppe]

I am not sure that I totally grasp it tho.

[coral spindle]

A group $G$ is cyclic if there is some element $g \in G$ such that $\langle g \rangle = G$.

[cloud walrus]

Boytjie

[coral spindle]

So it should be clear to you know whether or not U(Z_8) is cyclic from your calculations.

[inner steppe]

Aha okay there needs to be an element g whose group generated from it is equal to the entire G

In that case no, there is no such element.

[inner steppe]

Did I understan it correctly?

[coral spindle]

Yes that is the definition

And indeed U(Z_8) is not cyclic.

[inner steppe]

Thank you so much for making this clear for me and making me realise. I sort of understood but like it was like not clear for me, but now everything I have done so far seems clear to me.

[coral spindle]

I think that next time you're struggling with a question, you should ensure that you know the definitions of all the terms involved

Because this was the only issue here

[cloud lynx]

AHHH

[inner steppe]

The group (Z13 \ {0}, ×) is cyclic. Determine all of its generators.

How do I even begin. In the solution they found the order of the elements and used Lagrange but I really cannot see it. Before I checked the answer I was finding the order by finding the cyclic group of each element and got exhausted and felt like I was doing something wrong. Is there antoehr way to find solve the equation 2^n mod 13 = 1 for example? Because if I am forced to find all the cyclic groups then do i even need Lagrange or whatever reason they used it for?

[rocky cloak]

So lots of different things you can do here:

For any specific element either it's a generator or the order divides 4 or 6.

You can compute x^4 as (x^2)^2 and x^6 as x^4 * x^2. So this uses Lagrange to verify if x is a generator slightly quicker.

If you find one generator, it's quick to find more. Since the order of x^n is o(x)/gcd(n, o(x)) where o(x) is the order of x. In particular x^n has the same order as x iff n is relatively prime to the order of x.

Otherwise, if you just start computing powers of various elements. Then any element which has a generator as some power is also a generator. And any element that is a power of a non-generator is not a generator. Then you can start crossing out and circling things quickly.

[limber sequoia]

Im reading algebra again and i've got this textbook which lays out all the definitions well but doesn't provide strong excersizes, so im feeling that im just reading without learning. Specifically for rings. It just feels like a construction after construction without it turning into anything in my mind.

Does anyone have a text to recommend, or an easyish research paper I can wrestle through? I don't want something like lang, that feels far too hard for me.

[inner steppe]

why 4 and 6

[rocky cloak]

By Lagrange the order divides 12

So you just look at what the divisors of 12 is

[chilly sparrow]

I hear dummit and foote has good exercises & examples

[inner steppe]

yes but like all 1,2,3,4,6 divide 12. I don't really get it.

[limber sequoia]

alright i can check it out

how hard would you say are the original 1920s papers of Noether?

came across them randomly in google

[cloud lynx]

arent the generators all elements of Z/nZ which are coprime to n?

so they are units and therefore generate all Elements of Z/nZ

[coral spindle]

No, that would generate Z/nZ and not U(Z/nZ)

We're looking at multiplication, not addition.

[cloud lynx]

ah

[rocky cloak]

Yeah, so if the order is 1, 2, 3 or 6 then x^6 = 1, and if the order is 1, 2 or 4 then x^4 = 1

[inner steppe]

Find a subgroup of S4 that is isomorphic to G□?

How do I even begin here?

[coral spindle]

What is G□

[inner steppe]

I am guessing they mean a square.

[coral spindle]

OK I don't want a guess though, I want to know the definition

[coral spindle]

Remember we talked about this up here

You cannot expect to answer a question whose meaning you don't know!

[inner steppe]

They have had multiple examples of square begin a group with operations of rotations etc... in my book. This is an exercise my prof made so it must be based on the book. Also when I checked solutions it says: "Number the corners of the square 1, 2, 3, 4 and write down which permutation corresponds to each element in G□" so I am pretty sure it is a square.

[coral spindle]

OK so you've checked the solutions as well

So that is a pretty good hint so I'd try it.

[inner steppe]

Honestly I have but it is not saying me anything nor leading me anywhere.

Like are they saying to write down all possible permutations of G?
(1)(2)(3)(4) -> G is unchanged. (2341) rotated 90 degree?

[cloud lynx]

isnt it D4? Dihedralgroup

[cloud lynx]

yes I think so

just rotate

then reflect

and do it again

then you should get 8 Permutations

so its isomorphic to D4

if I am not wrong

lol

[inner steppe]

Aren't we rotating and reflecting D4 to begin with if I understand you correctly?

[cloud lynx]

?

no I meant the square

every rotation and reflection is a permutation and there are only 8 ways to permute a square

so its isomorph to D4

[inner steppe]

Yeah but what is the isomorphism you are trying to show? I get that we can rotate and reflect but how does tell us that it is isomorphic to the subgroup of S4

[cloud lynx]

D4 is a subgroup S4

[inner steppe]

okay

[cloud lynx]

Sn is the set of all possible permutations

[mental venture]

Excuse me, May I ask a question about groups?

[inner steppe]

given that D4 is a square, why do I even need to show that it isomorphic to another square. I am confused , sorry.

[mental venture]

I wonder how to solve the first problem

[cloud lynx]

what do u mean

[mental venture]

I can just only prove that if g is in I, then the group generated by g is belonged to I. But I don’t know whether it is useful.

[cloud lynx]

thats isomorphism if two groups are isomorph than only the label is changing and sometimes its helpful to know to what group it is isomorphic to because we know all properties of D4

[inner steppe]

Sorry I forgot the question, I thought I was supposed to prove that the subgruop I find is isomorphic. How can I prove that D4 is isomorphic to square formally?

I get the relation f(ab) = f(a)f(b) but can someone make such function in this case?

[cloud lynx]

just change the labels

write down all permutations of G square

all permutations of D4

then map the right permutation of D4 to the right permutation of G square

for example if u map ,,rotate" to "reflect" it wont work

because it wont be a homomorphism

[mental venture]

So can I beseech you to have a glimpse into my question now?

[cloud lynx]

i think you can determine first the order of I

then put everything together

i think its good to think about the elements of G if G is finite

if g is element of G

then g^-1 must also be element of G

[mental venture]

Indeed, I have noticed that. But what can I do with it

[limber sequoia]

take two members from I. then phi(gh) = h^-1g^-1 = g^-1h^-1

so that gives you a lot of commutative multiplications

not sure how to continue

[mental venture]

Quite so, i have made numerous tries in the past three hours, but I didn’t find anything useful

[limber sequoia]

hm

how advanced is the material

whats the last thing you learned

[limber sequoia]

i wrote a silly thing

a subgroup contained in I would be abelian is what i should have wrote

[mental venture]

I’m sure the teacher didn’t teach anything about Automorphism. TA said students should learn more out of class

I think “3/4” is important to solve it. But I have no idea how can we get this number

Both problems are talking about 3/4

[limber sequoia]

Consider GXG the direct product. Consider the subset (g,phi(g)). This is a subgroup of GXG.
|GXG| = |G|^2. Now apply the constraints we know about the size of said subgroup

[mental venture]

I just checked my notes, the last thing I’ve learned is isomorphism

[mental venture]

This is a brand new method for me, I haven’t thought about it

[limber sequoia]

yes it is also new for me

absolutely devilish

im not sure it works lets do the details

so it is indeed a subgroup of GXG, check. it has 1,1 and is closed under multiplication and inverse

ach it doesn't work

you can have |G| = 20 for example and then a subgroup of size 16 of GXG

[mental venture]

May I ask why GXG has the order of 16 rather than 20^2 when |G|=20

I only learned the direct product in advanced linear algebra. I don’t know how it works in abstract algebra

[limber sequoia]

you're right, GXG will have order 20^2

what i was saying is that 16|400 so there could, theroetically, be a subgroup that size in GXG

[mental venture]

I got that

Then we may get some general regulations from it

[jade flume]

(External) semi-direct products.
Consider the following definition. Let $H$ and $K$ be abstract groups, and let
[
\phi : K \rightarrow Aut(H), \quad k \mapsto \phi_k
]
be a homomorphism. The (external) semi-direct product $H \rtimes_\phi K$ is defined as follows. As a set, it is the Cartesian product $H\times K$, with group law given by
[
(h_1, k_1) \cdot (h_2, k_2) = (h_1 \phi_{k_1} (h_2), k_1 k_2).
]

It is rather tricky to determine when two different choices of $\phi$ (for the same $H$ and $K$) give rise to isomorphic groups. However, at least we can give a {\it sufficient} criterion for two semi-direct products to be isomorphic, as follows.
The set of $\phi$ above, namely the set $Hom (K, Aut (H))$, carries an action of $Aut (H) \times Aut (K)$: if $(\psi,\mu) \in Aut (H) \times Aut(K)$, then
[
\left( (\psi,\mu) \cdot \phi \right) (k) =\psi \circ ( \phi ( \mu^{-1} (k))) \circ \psi^{-1}.
]
Show that if $\phi$ and $\phi'$ are in the same orbit for this action, then
[
H \rtimes_{\phi} K \simeq H \rtimes_{\phi'} K.
]

I've gotten to $\phi$ and $\phi^{\prime}$ have action $a \in Aut(H) \times Aut(K)$ such that $\phi = a \phi^{\prime}$. Not sure how to define the map moving forward or show isomorphism.

[cloud walrus]

kaxi

[limber sequoia]

i gotta go take a phone call if ill think of something ill return

[mental venture]

I want to go to bed since it’s midnight in my country. I have worked on it for too long

[tardy hedge]

In a (commutative) ring, multiplication by r in R on some a in R, a -> ra is not injective unless R is an integral domain?

What about surjectivity ?

Also, how can we show that the multiplicative inverse of a unit in R is unique?

So multiplication by r is not surjective

[tardy hedge]

Edited because i wrote the map wrong

[surreal dagger]

Whats the exact question?
under what conditions multiplication by r is surjective/injective in a (commtative) ring?

[tardy hedge]

Yea

Its injective iff the ring is an integral domain right

Oo wait

Maybe thats not an iff? If the ring is an ID then certainly injective yeah

[surreal dagger]

yeah

[tardy hedge]

But is it possible maybe that for this particular r you always have cancellation law, but not for others on the ring

In that case it wouod be inejctive but not OD

ID

[surreal dagger]

I think you dont have to require that your whole ring is an integral domain if youre only considering the multiplication function for a single element r

something weaker suffices

[surreal dagger]

regarding surjectivity: If multiplication by r is surjective, then 1 is in the image. What does this mean for r? And then show the converse of the satement also holds

I think then you have a cute way of showing that a finite commutative ring is a field iff its an integral domain

[tardy hedge]

Cool ok ill think about it

[south patrol]

In fact this generalises a lot lol

To "Artinian" rings

[wild jasper]

I have a subgroup $H\leq S_p$ for a prime $p$ where $|H|=p(p-1)$, I am trying to determine $H$ I have that $H$ has a single subgroup of order $p$. Does anyone have a hint?

[cloud walrus]

mh_le

[tardy hedge]

Did u read atiyah macdonald

I am supposed to work thru that book this semester

I am looking forward to understanding rings better because i dont think i really had a solid understanding of them

[south patrol]

I did indeed, would recommend

[tardy hedge]

Nice

[south patrol]

Yeah nice

Tbh I didn't know much about rings before starting atiyah macdonald

You learn a lot from it

[tardy hedge]

Awesome

I need to figure out what zorns lemma is now lol

[wild jasper]

Put another way, how do I determine the elements of the Galois group of $x^p-2$ for a prime $p$?

[cloud walrus]

mh_le

[tardy hedge]

“There exists rings with exactly one maximal ideal, for example fields”

Maximal ideal is just (0) in that case?

[vapid vale]

yes

[prisma garden]

https://media.discordapp.net/attachments/1012470948496167017/1041089767619690628/magik.png?ex=66efd7a7&is=66ee8627&hm=6803f6df5f366513c78e25c7d92e59e783cba7f8ba28b60f85fad5a08f0f6f8e&=&format=webp&quality=lossless&width=411&height=417

[wild jasper]

suppose that p is a prime that a group G is generated by two elements of order p and p-1 respectively, is the order of G p(p-1)?

[sharp ice]

consider the group of all words formed from x,y with the condition x^p = 1 and y^p-1 = 1. the group is infinite

[tardy hedge]

"every non unit of R is contained in a maximal ideal"

Is the reason just because, every ideal is contained in a maximal ideal, so take "a" a non unit and (a) is contained in maximal ideal so a is contained in maximal ideal

[dim widget]

^

[tardy hedge]

So I was trying to say that R/(a) has a maximal ideal (that was zorns lemma application) but then all I can say from that is that there is an ideal containing (a) in R, but i cant say its maximal can I?

like for pi: R -> R/(a), can I say preimages of maximal ideals are maximal?

It said for a ring hom A->B, preimages of prime ideals are prime but it said preimages of maximal ideals arent necessarily maximal ...

[surreal dagger]

In this case you can indeed say it

[tardy hedge]

Hmmm but for the general case no... is it something about surjectivity ?

[wild jasper]

Btw G is a subgroup of S_p

[tardy hedge]

If xy is a unit then x and y are units ??

Apparently so

Lol bruh OK im a noob i kno

but i am learning hooray!!

Hahahahahaha

Also i really like atiyah macdonald already

Its engaging