groups-rings-fields

[delicate bloom]

now as for actually getting solutions, I dunno if there's any kind of pattern to them in (Z/nZ)* from the perspective of trying to take the 1 and p^k - 1 and combining them with CRT

[delicate orchid]

ah I see, they were asking about which ones are self inverse - not when all of them are self inverse, no clue how I made that mix-up in my mind

Think I have an idea though, ||The group (Z/p^kZ)^\times acting on Z/p^kZ^+ by multiplication is exactly Aut(C_p^k) \cong C_{p^{k-1}(p-1)} acting on C_p^k, so we just need to count the number of elements in C_{p^{k-1}(p-1)}, of even order - which is related to the size of the sylow 2-subgroups in some way|| could be nonsense but it's an idea

[glossy crag]

If L/K is the splitting field of f, in order to show "f irreducible iff Galois group acts transitively on roots of f" do we at any point need separability of f? For => definitely not I'm thinking, for the converse I don't see it either (as long as f is not a power of an irreducible).

[formal ermine]

I'm assuming you also mean to assume that f is separable

you need to use that f is separable for <=

[glossy crag]

How so? If f has two distinct irreducible factors, then the action cannot be transitive. Like I said, only reason I can fathom separability of f playing a role is to show f is not a power if the action is transitive, but I can't see how to do it.

[formal ermine]

the proof I have in mind is like

if f is separable and f = gh where g,h are nonconstant then the roots of g and h are disjunct

then go for a contradiction

yk?

[chilly radish]

galois group doesn't make sense if f isn't separable

the splitting field is only galois if f has no repeated roots

[glossy crag]

Don't split hairs, that's a catch-all term for Aut(L/K) regardless of whether the extension is Galois.

[glossy crag]

Yeah, that'd probably work. But it's enough for my purposes to show f has no distinct factors, so I think I'm fine without assuming f is separable.

Actually nvm, you're using a different definition of separability of f.

[chilly radish]

you use the fact that f is separable when you want to show that the aut group is transitive. In the proof I know, you take any two roots a,b, then you want to extend the isomorphism K(a)->K(b) into an automorphism of L

this can fail if f has repeated roots

[formal ermine]

what is your definition

[glossy crag]

Not "no repeated roots", but all irreducible factors don't have repeated roots.

[glossy crag]

Yeah but for the converse you're a priori assuming the action is transitive.

[chilly radish]

for the converse yea, but i'm talking about showing that irreducible implies transitive

[glossy crag]

You don't need separability for that.

[chilly radish]

what's your argument then

[glossy crag]

If a and b are two roots of f, then K(a) and K(b) are K-isomorphic, that extends to all of L since L is a splitting field.

[chilly radish]

if i'm not mistaken, this can fail if L is a splitting field of a non-separable polynomial

that's my point

[glossy crag]

It shouldn't

That's just the splitting field isomorphism extension theorem

If K and K' are iso, then L and L' are iso, extending the iso.

[chilly radish]

can you show me your proof of this theorem. Maybe it's different from the one I know

but in the one I know the fact that f has no repeated roots is featured quite heavily

[glossy crag]

I don't have a source literally on hand, but look in Lang or something else.

[glossy crag]

Number of roots comes into play when you want to show the number of distinct extensions is the degree. That's when you use separability.

[glossy crag]

Here

https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Abstract_Algebra%3A_Theory_and_Applications_(Judson)/21%3A_Fields/21.02%3A_Splitting_Fields

It works the same way for infinite splitting fields by Zorn.

[chilly radish]

hm yea I think you're right then

[wet zodiac]

Cat Bread dies to Artin

[untold cloud]

Hi, guys, suppose i am given a field K, and a natural number n, is it possible that we can always find a field L containing K such that [L : K]=n ?

[formal ermine]

yes

[glossy crag]

No, R doesn't have such extensions

[glossy crag]

It depends on the field

[formal ermine]

yeah except when n exceeds [alg closure of K : K]

[glossy crag]

I think there might possibly be pathological cases even if the alg closure has infinite degree, but I may be wrong.

[formal ermine]

hmmm we just need to find an irreducible polynomial of degree n

[glossy crag]

Exactly, I don't know if it holds for absolutely every field.

[formal ermine]

yeah

[glossy crag]

It does for finite fields and for Q definitely, don't know about other ones.

[untold cloud]

Thank you!

[white yoke]

What is an induced module?

[rustic crown]

if R --> S is a ring map, and M is an R-module then S ⊗_R M is naturally an S-module, this is called the induced module.

Hom_R(S, M) is also naturally an S-module. this one is called coinduced module

[main needle]

I am trying to find the commutative unital rings of characteristic $p$. What i have right now that we have a natural map from $F_p \to R$ and if we pick $x$ outside it's image $R$ takes a structure of a vector space spanned by 1 and $x$. So we have a surjective map from $F_P[x] \to R$. How to find the kernel of this map ?

[cloud walrus]

ru0xffian

[rotund aurora]

cant you just look at the sign?

like if x is smaller than 1, all the products will be negative. If x is between 1 and 2, all the products except one will be negative (so sign changes), and so on. And you want m to be big so that the sign of the whole expression does change (ie, so that whenever the product in the middle is negative, then the whole expression is negative. And when its positive, the whole expression is positive).

[fading basin]

The kernel of the map is the set of polynomials in $F_p[x]$ that map to 0 in $R$. To find this set, you can use the Euclidean Algorithm. For example, suppose you want to know if $f(x) \in F_p[x]$ is in the kernel. Then compute the greatest common divisor of f(x) and $g(x)= x^n+a_{n-1}x^{n−1}+\cdots+a_1x+a_0$. If it is a nonzero polynomial, then f(x) is not in the kernel. Otherwise, it is in the kernel

[rotund aurora]

I don't know, not really thinking about it. I assume that the factor of p and the x^p are to show that you have a cycle or something, and is irrelevant for showing you have p-2 roots, otherwise I'm saying nonsense.

[main needle]

I've seen somewhere that the kernel is just the ideal generated by $x^2-bx-a$ for some $a,b \in F_p$. I see why this ideal should be inside the kernel by considering $x^2$ as it must be $a + bx$ for some $a,b$ but why is it the whole kernel ?

[cloud walrus]

ru0xffian

Nats

[fading basin]

Because it captures all polynomials in $F_p[x]$ that map to 0 when evaluated at $R$. This is due to the fact that any polynomial in the kernel can be written as a combination of powers of $x^2-bx-a$, so all we need to do is show that these polynomials map to 0

[cloud walrus]

Nats

[main needle]

can you elaborate more on why it captures all of them ?

[fading basin]

Sure, consider a simple example. We want to know whether $f(x)=x^2+2x+3 \in F_p[x]$ is in the kernel, so we can rewrite this polynomial using the form of the kernel: $$f(x)=(x+1)^2-2x-4,$$ where $b=-2$ and $a=-4$. And since any polynomial in $F_p[x]$ that maps to 0 when evaluated at $R$ can be written as a linear combination of powers of $(x+1)^2-2x-4$, the ideal generated by $x^2-bx-a$ captures all such polynomials

[cloud walrus]

Nats

[white yoke]

Thank you! Do you have a reference that mentions this definition? And by ring map do you just mean a map between rings, or do you mean a ring homomorphism?

[rustic crown]

ring homomorphism >.<

you should find that is most algebra books

[white yoke]

I don't really know many algebra books 😅 any you could recommend?

[rustic crown]

maybe someone else knows a nice algebra book, i don't have any at the top of my head

[white yoke]

No worries, thanks

[rustic crown]

i saw it in aluffi, but he didn't use the word "induction" or "coinduction"

[chilly radish]

(Tbh I never heard the term induced module)

[rustic crown]

it was extension of scalars

[chilly radish]

I've always heard it referred to as extension of scalars

[rustic crown]

i think

[chilly radish]

Yea

[white yoke]

ive never heard that haha

I was looking at a book by John Milnor and he mentioned an induced module without defining it which is why I wanted to know

[rustic crown]

if you understand tensor-hom adjunction, you should be good

[white yoke]

Literally never heard of that lmao

[rustic crown]

(the word "induced" could be used loosely as well... like a map of modules induces this other morphism, somethign something)

[main needle]

Yeah but why any polynomial in the kernel can be written as a combination using $x^2 - bx - a$? sorry if i am missing something obvious.

[cloud walrus]

ru0xffian

[white yoke]

Ah I see

[rotund aurora]

So I wanted to show I is prime, but maybe I'm saying nonsense. So (1,1+sqrt-3=z) form a basis Z-basis for the whole ring and (2,z) forms a Z-basis for the ideal. Suppose (a+zb)(c+zd) is in the ideal, with a,b,c,d integers. Then multiplying everything out you will see that ac is even, therefore one of a or c is even, and we are done.

Is this correct?

[rotund aurora]

In this case I'd say its easier to calculate [L : Q] by first calculating the size of the Galois group

Here is a solution for x^p-2, p any prime. The group is actually pretty simple.

@willow geyser

(maybe you solved it already idk)

yeah and all roots (except 1) are primitive

||try finding two generators of the Galois group||

and like I think that was it

see what can you send 5rt 2 to

I don't think complex conjugation is too useful in this situation

[rotund aurora]

Yeah

And the other roots of unity are z^k

So you can fix one z and that automorphism composed k times will give 5rt 2 -> z^k 5rt 2

Now you know z is sent to a root of unity, so z -> z^k, and you want to ask if this automorphism produces all automorphisms of the type z -> z^a

Yes, because 5rt 2 and z are generators

Of the whole field

There are 5 fifth roots of unity, but only 4 primitive ones

One is just one

And you want to send z to a primitive root of unity, definitely not one

So 20

And you can check all of them do work

You dont really have to do a lot of work tho

So you look at the automorphisms that act on z and fix 5rt 2, and vice versa. And you can compose them like fg where f acts on 5rt 2 and g acts on z. Then fg(z)=g(z) and fg(5rt2)=f(5rt 2), I think

The two automorphisms I was talking about were z -> z^g (and fix everything else) and 5rt 2 -> z 5rt 2, where g is a primitive root of 5

And I think you can realize this as a matrix group by sending the first automorphism to the matrix ((g,0),(0,1)) and the second to ((1,1),(0,1))

Everyone knows matrix multiplication

So you wouldnt need to write down the relation needed to specify the group (given the two generators of order p and p-1)

But in that case, the group is pretty simple, so I guess its fine to not put it as a matrix group. But I guess its always nice to try to look at matrix groups

What do you mean?

Matrix groups do contain every galois group

But you get shit matrix groups, so there is no point. In this case it was all nice

Idk lmao

Like in general probably not, but maybe there are other funny families of polynomials like this one. I dont know more examples tbh

[south patrol]

Lol we had this problem and yeah using matrices is cute

[rotund aurora]

It should be noticed that we invoked a primitive root modulo p, so this is a very specific thing

[meager jolt]

Galois sheet moment

Well I'm about to ask a rep sheet 4 qs so dw

[meager jolt]

I'm trying to show that the Haar measure on SU(2) is translation invariant and is normalised by (1/2pi^2)

But I don't know how to paramaterise over the 3-sphere

[meager jolt]

Could anyone help me with the parameterisation or this part?

[next obsidian]

Use the first part of i), then ask what (x + x^-1)^p is

[meager jolt]

Btw Joe I can give you my sheet if you'd like

which 1 is it?

ok I'll send both lol

any chance you could find rep sheet 4 from somewhere?

[next obsidian]

By the first part, this is equivalent

[meager jolt]

lmao

I know of alimos haha my gf is on it

[next obsidian]

But here you need to use some fact that is particular to this situation to figure out what (x + x^-1)^p is

[meager jolt]

woah

any help would be appreciated 🙏

[next obsidian]

My guy you’re in char p

[meager jolt]

alao bump

[sonic coral]

i’m gonna ask a dumb question probably but it’s fine since i’m learning

but showing that a group G is abelian, we need to show xy=yx for all x,y in G

is it not enough to show that x’y’=y’x’, since x and y have unique inverses?

[chilly ocean]

all elements in G are inverses of something (of what?), so it would suffice

but why do you want to do this?

[sonic coral]

just trying to fill in some gaps

[next rain]

Does AATA have a more comprehensive solution booklet by any chance?

[chilly ocean]

what is AATA?

[meager jolt]

bump again

[next rain]

Abstract Algebra Theory and Applications

by Judson

[tender wharf]

you might be able to find it on certain

websites

[vagrant zinc]

What do you think of the demonstration as to whether it is a group or not?

[next rain]

Would you be able to point me to one?

[vivid tiger]

(i think that was an allusion to piracy)

[white oxide]

this was in my professor's lecture notes, but how do we know that Map(X, S) has a binary operatoin induced from (S, *)? Is it just defined to be that way?

[chilly ocean]

the operation is being defined at the bottom

[white oxide]

ohhh okay oops

im stupid

thanks!

[next rain]

AATA is freely available online, there is no need for piracy.

[junior vale]

ninja about to drop some heat

[ripe glacier]

hey can someone check my proof for 4.8 b
I just wanna know if it's right, if theres anything important I've skipped or just any general bad practices I've done
(matrices of the first type are those that represent adding a multiple of one row to another)

[ripe glacier]

hot enough for ya?

[junior vale]

oo yea

[bleak abyss]

Alright I'm here

So at a glance Ninja

Do you know globally what the idea is? There's a term for this

[ripe glacier]

wdym?

[bleak abyss]

Row reduction

Every invertible matrix row reduces to the identity

[ripe glacier]

i guess my argument is the opposite of row reduction

i try to generate the arbitrary matrix from the identity

[bleak abyss]

Well you can undo row operations

So it's the same thing really

[ripe glacier]

im not really sure what the point you're getting at is?

[bleak abyss]

I just wanted to make sure you "knew what was going on" in summary lol

My thing about undoing row operations is that if you show "every matrix row reduces to the identity"

That statement would auto imply this anyway

Because you'd say oh Id = E_1 ... E_n A

Now invert gg

[ripe glacier]

yeah I did consider that, but I thought of this idea first and wasn't 100% sure how to show that i can reduce them using just matrices of the first type

[bleak abyss]

I'm guessing second type means scale a row right?

[ripe glacier]

3rd type is scale a row

2nd type is swap rows

[bleak abyss]

Ah okay that requires more thought on how to avoid

[ripe glacier]

btw this is a proof for b, so I can't use 2nd or 3rd type

[bleak abyss]

So honestly I feel like this can be made a lot simpler?

To elaborate

Actually

Nah maybe this is it

[ripe glacier]

Oh me too but I'm not nearly smart enough to do it

[chilly ocean]

b ii) need help please

[lavish spoke]

it's easy enough to just find the square and cube of that matrix for general a, b, c by hand

write down what it means for the matrix to not have order 1 or 2 but have order 3 in terms of a, b, c and argue why this means that no such order 3 matrix exists

[chilly ocean]

I tried this

I got $\begin{pmatrix} a & 2b \ 0 & c \end{pmatrix}^3 = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$

[cloud walrus]

isomorphism

[chilly ocean]

expanding the LHS

allows me to select values of a, b, c such that this is true

[lavish spoke]

the diagonal of the cube should be a^3, c^3 so you can write down the value of those immediately, then it should be straightforward

[chilly ocean]

how so? I can just choose b = 0

and then a = c = 1

and this gives us the identity matrix

[lavish spoke]

that's the correct answer but you've got it the wrong way around

you've got to show that the equation holds if and only if a = c = 1 and b = 0

[chilly ocean]

oh, wait

[lavish spoke]

you already know the identity matrix will be a solution

you've just got to show it's the only one

[chilly ocean]

the identity matrix is a solution with order 1

[lavish spoke]

yeah

[chilly ocean]

we want something of order 3

I see

thank you

that makes sense

[lavish spoke]

I said this instead of saying solve A^n = I because you need to get into the habit of checking that the A you've found is not order 1 or 2 or etc.

which is an easy mistake to make

[chilly ocean]

yeah...

[lavish spoke]

obvs here because of Lagrange's theorem any solution only has order 1 or 3 but ya

[chilly ocean]

how come

because of Lagrange's theorem?

[lavish spoke]

you can also prove it with division by remainder, if g^k = e then the order of g divides k. so if k is prime you just need to establish that g != e

for the proof by division with remainder, let ||n be the order of g and write k = an + r for 0 <= r < n||, then ||show r = 0||.

[chilly ocean]

I see

that's clever

I still need help with this though (showing the index [Z^3 : H] is infinite)

[lavish spoke]

for each n you can prove that {(x, y, z) \in Z^3 : x + y + z = n} is a coset of H

[chilly ocean]

how so?

and how does that help me with showing the index is infinite

[lavish spoke]

well then you've got infinitely many cosets

[lavish spoke]

try looking at (1, 0, 0) + H first

[chilly ocean]

x + y + z = -1

[lavish spoke]

you should get x + y + z = 1

eg. that set should contain (1, 0, 0)

[chilly ocean]

how do you show that $\frac{1}{2} |S_n| = |A_n|$ using Lagrange's theorem?

[cloud walrus]

isomorphism

[chilly ocean]

I tried leveraging that $|S_n| = [S_n : A_n] |A_n|$

[cloud walrus]

isomorphism

[chilly ocean]

I don't know how to show the number of left cosets of $A_n$ in $S_n$ is 2 though

[cloud walrus]

isomorphism

[tender wharf]

we don't need lagrange's theorem for this actually

are you asked to use it

[chilly ocean]

I'm told I can use it

how would I show it otherwise?

[smoky ivy]

I mean you know the cardinality of S_n and A_n, no?

[tender wharf]

I guess you meant order

I think they want to prove that the order of A_n is actually half

of the order of S_n

[chilly ocean]

yes, the order of S_n is n!

[smoky ivy]

order, yeh

[tender wharf]

well first if a is any odd permutation then (12)a is even

so we have at least as many even permutations as odd ones yeah

[smoky ivy]

yeah, pretty much

[tender wharf]

similar argument if a is an even permutation

the lagrange theorem argument would be very similar actually

(12) A_n produces all your odd permutations

[smoky ivy]

you can consider the signum map from S_n -> {+-1}

the kernel is A_n

[tender wharf]

isomorphism have you learnt about homomorphisms yet

[chilly ocean]

yes

[tender wharf]

then this method would work too

[smoky ivy]

and using the first isomorphism theorem yields S_n/A_n isomorphic to {+-1}

[chilly ocean]

is there any way I can use lagranges theorem tho

[smoky ivy]

and considering the order yields the result

[tender wharf]

^

well I mean

this doesn't actually use lagrange tbh

it just tells us we have 2 left cosets of A_n

[smoky ivy]

yeh

anyway

does anyone have ideas for this haha?

i’m a bit stuck on this question

[chilly ocean]

is b) ii) $\mathbb{R}^\star = {A \in R : \text{det}(A) = 1}$?

[cloud walrus]

isomorphism

[chilly radish]

First make sure it's obvious that this is an ideal. Next, try showing that every element outside of this ideal is invertible, this is the same as saying it's unique maximal, but it might be easier to see. You will need to explicitly use some analytic arguments about continuity here

[rotund aurora]

Exercise: Show that the 8th cyclotomic field contains sqrt 2.

So I simply showed that it contains cos(pi/4) and that cos(pi/4)=1/sqrt 2

[smoky ivy]

how can i know that there is no other maximal ideal

[rotund aurora]

but idk if that's the intended way to do it. The discriminant is just 256

[chilly radish]

If you prove every element not in m is invertible, that shows that m is maximal

[smoky ivy]

oh wait, i see

[smoky ivy]

okay, but hmm, you only have f(0) not equal to 0 given

not sure how i would try to get an invertible element out of this

[chilly radish]

Ok if f of 0 is not equal to 0, then there's some small nbhd on which f is not 0

[smoky ivy]

yeah

[chilly radish]

And you don't care what happens outside of small nbhds of 0 because of quotienting out by I

So in this small nbhd f is invertible

Outside of this nbhd you can do whatever it doesn't matter since they will identify in the quotient

[smoky ivy]

okay, yeah, this is what i was stuck on, why does it not matter what happens outside of the neighborhood

[chilly radish]

Ok so

If f and g agree on some nbhd of 0

Then f-g|U =0

So they agree kn the quotient

Essentially this is capturing the local behaviour of functions near 0

So if you define a new function h such that h=f on the nbhd of 0 where it doesn't vanish, and constant elsewhere (so that it's still continuous but nonzero everywhere)

Then you get that h+I=f+I

But h is an invertible representative, since it never vanishes

So you can take its reciprocal

[smoky ivy]

ohh, okay, i see

that makes sense

okay got it, thank you

[chilly radish]

No problem!

Feel free if you have more questions about this

[smoky ivy]

sure :)

[lavish spoke]

no, a matrix being invertible is equivalent to nonzero determinant

[chilly ocean]

yes, but we have to preserve the structure of the ring

we must have integer entries

which is only achieveable when a^2 + b^2 = 1

[lavish spoke]

Oh ya you are right soz

[rotund aurora]

Any hints? 11(c) is that if all of the Galois conjugates of an algebraic integer have absolute value 1 then that algebraic integer is a root of unity. So I think the intended path is to show that all of the Galois conjugates of u/\overline u have absolute value 1, but this is equivalent to |sigma(u)|=|u| for every complex embedding sigma, which is most likely false, so I'm pretty lost.

[rotund aurora]

ok now I realize I didn't use the fact that u was a unit nowhere lmao

uh wait but you use it to invert \overline u so idk

[south patrol]

(Also which book is this?)

[rotund aurora]

marcus

pretty nice, and many exercises

[south patrol]

Number fields?

[rotund aurora]

yes

[south patrol]

Nice cheers, but don't wanna distract from your question lol

[pliant forge]

Let \Gamma = 1 \to 2 \ot 3 with a: 1 \to 2, b: 3 \to 2 be a quiver with the representation V = k^2 \to k^2 \ot k^2 with f_a and f_b being a 2x2 matrix with all entries 1, and k some field.
Based on the picture below, I am struggling to find some maps from the representation V to the direct sum of the 4 representations given, say (W, g).
Based on my intuition, the a map for the direct sum would be f_{1,a} \oplus f_{2,a} \oplus f_{3,a} \oplus f_{4,a} = 0 \oplus 0 \oplus 0 \oplus 1, given that V_i = (V_i, f_i) for i = 1,2,3,4.
But if I want to construct an isomorphism to show that V \cong W as representations, I have to find a map, \phi = (\phi_1, \phi_2, \phi_3), from V to W, where \phi_i: V(i) \to W(i) for i = 1,2,3, where V(i) and W(i) are the vector spaces at position i in the quiver \Gamma.
These \phi_i's have to satisfy a commuting diagram such that \phi_j f_a = g_a \phi_i for all a: i \to j \in \Gamma_1 (the arrows in Gamma)
Say, we want to find the maps such that the diagram commutes for the path a: 1 \to 2. Then we have \phi_2 f_a = g_a \phi_1 = (0 \oplus 0 \oplus 0 \oplus 1) \phi_1.

Let me just draw out the diagram

[smoky ivy]

i’m stuck on b) haha

so i tried providing an explicit function that would be in the intersection, but i can’t seem to find one in this way

[lament dawn]

You wrote your question out in TeX, then did not call the texbot

[pliant forge]

yeah but mainly my issue is I cant understand how you can have isomorphism maps \phi_1, \phi_2 such that the diagram commutes

[pliant forge]

Also I usually do this, but I suppose this question was a bit too long for doing it this way.

[pliant forge]

like it would always end with some value (0,0,x,y) in W(2) if I attempt to go via phi_2 whilst it would have (0,0,0,z) in W(2) via phi_1

if that makes sense

[oblique river]

I think you might be overthinking it a little. For any complex number z, |z/conj(z)| = 1. Now for u our unit, we want |sigma(u/conj(u))| = 1 for all sigma. But this is the same as |sigma(u) / sigma(conj(u))|, which is the same as |sigma(u) / conj(sigma(u))| because conj is in tbe galois group which is abelian

Now apply the first fact to z = sigma(u)

[rotund aurora]

oh the Galois group is abelian

right, that was what I was missing

thanks!

[oblique river]

Np

Note that this works equally well in any abelian number field! But also any abelian number field is a subfield of a cyclotomic field, so it’s also just true by that fact

[rotund aurora]

yup

but it should be shown that u/conj(u) is an algebraic integer, no?

well I guess its just that things in Z[omega] are algebraic integers

lul

So we have |sigma(a)|=|a| if and only if conj(sigma(a))=sigma(conj(a)), which is fun

[pliant forge]

diagonalize matrix. got it

[oblique river]

I dont think this is true

The second statement is always true (in an abelian number field) but the first is not

[chilly radish]

Oh this is a tricky one. There is an explicit function. The idea is you want a function that decays fast enough so that the limit of it divided by x^n is 0, so it's still continuous at 0 when divided by x^n for all n, then you could multiply by x^n to get that your function is in m^n (Since x is in m)

[rotund aurora]

Maybe I'm wrong, but this is what I did: Suppose sigma(conj a)=conj(sigma(a)). Then multiplying both sides by sigma(a), we see that |sigma(a)|^2=sigma(a*conj a)=sigma(|a|^2)=|a|^2, since |a|^2 is rational. Now, suppose |sigma(a)|=|a|, then |sigma(a)|^2=|a|^2 and so sigma(a) x conj sigma(a)=|a^2|=sigma(|a|^2)=sigma(a)sigma(conj a), and so conj sigma(a)=sigma(conj a)

[smoky ivy]

yeah, i was thinking abt that too, but i can’t seem to find it explicitly

[rotund aurora]

idk if it would be better if I latexed it lmao

[oblique river]

Latex or not I dont understand the first half of the proof. I dont contest tbe other half, because again, sigma(conj(a)) = conj(sigma(a)) is just true for all a in an abelian number field

[rotund aurora]

Maybe I'm wrong, but this is what I did: Suppose $\sigma(\overline a)=\overline{\sigma(a)}$. Then multiplying both sides by $\sigma(a)$, we see that $|\sigma(a)|^2=\sigma(a\cdot\overline a)=\sigma(|a|^2)=|a|^2$, since $|a|^2$ is rational. Now, suppose $|\sigma(a)|=|a|$, then $|\sigma(a)|^2=|a|^2$ and so $\sigma(a) \cdot \overline{\sigma(a)}=|a^2|=\sigma(|a|^2)=\sigma(a)\sigma(\overline a)$, and so $\overline{\sigma(a)}=\sigma(\overline a)$

[cloud walrus]

Croqueta

[rotund aurora]

well I'm saying in any number field |sigma(a)|=|a| if and only if sigma and complex conjugation commute

[oblique river]

Take a very large prime p (not necessary, jsut for demonstration purposes) and consider a = 1 + zeta_p. Zeta_p is very close to 1, so a (and |a|) are very close to 2. However, zeta_p has some conjugate very close to -1. For that corresponding sigma, sigma(a) = 1 + sigma(zeta_p) is very close to 0, as is its magnitude

[oblique river]

This isnt really a reasonable statement though because one half depends on a and the other doesnt.

[rotund aurora]

yes

[oblique river]

Unless you mean “for all a” in which case, it’s not true

By my example

[rotund aurora]

I actually meant sigma(conj a)=conj(sigma a), which is what I wrote at first, sorry

[oblique river]

Okay, still, it’s wrong lol

[rotund aurora]

mmh

[oblique river]

Please see my counterexample

Again, sigma(conj(a)) = conj(sigma(a)) is always true in an abelian field, but |a| = |sigma(a)| is not

[rotund aurora]

yeah okay your example definitely makes sense

Suppose $\sigma(\overline a)=\overline{\sigma(a)}$, multiplying both sides by $\sigma(a)$ we obtain $\sigma(a)\sigma(\overline a)=|\sigma(a)|^2$, and the left hand side is $\sigma(a\overline a)=\sigma(|a|^2)$.

[cloud walrus]

Croqueta

[rotund aurora]

This is correct I think

now I think i concluded that |a|^2 should be fixed

but it need not be rational actually, so there's no reason for that

I think that was the mistake

[vagrant zinc]

Guys for this exercise
I have this idea
Let be the binary operation ab=ab for all a,b belonging to R.
now let's fulfill the associativity of a group then for all a,b,c that belongs to R* then
(ab)c=a(bc)
therefore we have that
(ab)c=(ab)c
=(ab)c
=abc
=a(bc)
=a(bc)
=a(bc)
therefore is the binary operation is associative

Now for the inverse we have

Let a belongs to R* then there must exist an a^(-1) that belongs to R* therefore it follows that
ab=ab
aa^-1=aa^-1
=e
so the identity e belongs to R* a which is equal to
aa^-1=1

Now let e belong to R then let ae=ae be the binary operation ae=ae
=a(1)
=a

[rotund aurora]

[oblique river]

I agree, that is the mistake. All you can say is that it’s real.

[vagrant zinc]

Guys, how about the demo?
I'm not showing it with "*" but it doesn't appear here in discord.

[rotund aurora]

Thanks for the help

[smoky ivy]

actually, can’t i just take x for x >= 0 and 0 else

[rotund aurora]

This is actually interesting

consider x^2+4ax+2 (irreducible by Eisenstein), then the two roots of this polynomial differ by sqrt(4a^2-2) lmao

and both are real, so conjugation commutes with the automorphisms

[formal ermine]

this feels like "R* is a group because it is a group"

why must a^-1 belong to R*? you're saying "the inverse exists because it exists"

[smoky ivy]

is m^n/I isomorphic to (m/I)^n

[analog zephyr]

Así es

[analog zephyr]

What does "m^n/I" denotes?

[smoky ivy]

m is an ideal and I is an ideal as well

m/I is just = {a + I | a \in m}

[analog zephyr]

I haven't studied ideals yet

[analog zephyr]

Like coset of groups

Now I see. It is the same idea of group factor

But for Ring

[vagrant zinc]

Could you tell me the correct way to define it please?

[tender wharf]

recall the definition of a group

and understand exactly what it is saying

by the way the binary operation for R* is multiplication

[smoky ivy]

yeah

you can prove it for groups as well i think

[tender wharf]

you don't say ae=ae is a binary operation

[smoky ivy]

just considering normal groups

is it isomorphic to anything special?

so my try was

[south patrol]

Lol

R^2/Z^2 being S^2 is very different

[formal ermine]

oh wait this is a group

lmao

[south patrol]

to what is being discussed

lol

[smoky ivy]

yeah

take the canonical epimorphism m^n -> (m/I)^n

[formal ermine]

yeah nvm me then lol

[smoky ivy]

show that the kernel is I

but i’m not sure if the kernel is actually I

one inclusion seems fairly easy

[south patrol]

i think the kernel must be I just from the fact the isomorphism holds for abelian groups

But ofc gotta actually check that

[smoky ivy]

yeah, that’s what i thought

[south patrol]

Wait no I got confused because like

When you talk about m^2 you mean actual squares of elements

right?

[smoky ivy]

yeah

[south patrol]

Oh sure oopsies

[smoky ivy]

or well, no

the product

so like

[south patrol]

But then how are you quotienting by I

[smoky ivy]

i mean the ideal product

[south patrol]

Wait sorry I mean you mean the product of ideals

ye

[smoky ivy]

it’s an ideal, no?

[south patrol]

Yeah i just meant the other interpretation was false

[smoky ivy]

yeah

[south patrol]

It's all fine and dandy now lol

[chilly ocean]

need help please

[south patrol]

You can actually give a nice description of the quotient group

Try giving that a go

[chilly ocean]

can you elaborate?

[south patrol]

Okay so I'm not sure if you've done quotient groups yet so maybe I can go a bit slower

But basically what you should think about is: given a,b in C^x, when are they in the same coset of S?

[i.e. when can we write a = bc for some c in S]

The idea is of course that you wanna show there are infinitely many such cosets, so understanding what the cosets actually are is nice

[analog zephyr]

U was right

[chilly ocean]

I have done quotient groups with additive abelian groups

so this would be the cardinality of the quotient group $\mbb{C}^\star/\mbb{S}$

[cloud walrus]

n!

[chilly ocean]

we have to show the cardinality is infinite

I just have no idea how to go about this

[analog zephyr]

If G/H is a factor group factor from a homomorphism then H is the kernell

[chilly ocean]

idk what a kernel is

[formal ermine]

"null space"

all elements that get send to the identity

[analog zephyr]

like in linear algebra

The inverse imagen of 0

[formal ermine]

this is worded a bit weirdly btw

[chilly ocean]

still not sure how to approach this

[coral spindle]

Try finding an infinite group G and a surjective map C* -> G with S being the kernel.

n.b. this map must be a homomorphism

[chilly ocean]

I don't think we are supposed to go about it that way since we haven't studied kernels yet

is there an alternativ emethod?

[coral spindle]

Sure.

Find infinitely many cosets of S in C*.

[analog zephyr]

I couldn't explain you now cuz I'm at the job right now

[chilly ocean]

.....

[coral spindle]

Try drawing a single coset, diagrammatically. Draw a few. You will discover, in fact, that you know what all the cosets look like.

[analog zephyr]

Coset from what?

I'm lost in the chat

[formal ermine]

are you the one being helped or are you trying to help

[analog zephyr]

Trying

[formal ermine]

a coset is an element of G/H

G/H are the left cosets of H

H\G are the right cosets

[analog zephyr]

No no wait

aH denotes the left coset of H in G

a is any element of G

Ha denotes the right coset of H in G

[formal ermine]

"the left coset" implies there is only one, which just isn't how it works

[analog zephyr]

The factor Group G/H turn out when the left coset are the same as the right coset

When aH=Ha

[formal ermine]

G/H need not have a group structure

[analog zephyr]

No

[delicate orchid]

but it is a group when aH = Ha, is what they're saying

[analog zephyr]

But in order to form a group, it must

[analog zephyr]

Yeah

[formal ermine]

what does this have to do with the above

[delicate orchid]

lol

[analog zephyr]

But is when aH=Ha that we use G/H and names it as factor group

Or I'm wrong?

[analog zephyr]

Well, G could have many proper subgroups it will implies many set of cosets

[formal ermine]

What

[formal ermine]

you're wrong

we use G/H even when H isn't normal

[analog zephyr]

No

Is when H is normal too

[delicate orchid]

literally what illuminator just said

[analog zephyr]

Ok, but i ask before

"I'm i wrong"

:/

[formal ermine]

I said yes

also fwiw subgroups have nothing to do with the amount of cosets, like I already said above

[analog zephyr]

Cuz I was confusing about if we can call G/H when it doesn't even form a group

[analog zephyr]

I know

[formal ermine]

Well, G could have many proper subgroups it will implies many set of cosets

[delicate orchid]

illuminator you have to work with the language barrier here

they meant proper subgroups corrispond to many different sets of cosets

they didn't mention the size of the sets of cosets

[formal ermine]

ah

[analog zephyr]

Show me an counterexample please

[formal ermine]

my bad

[analog zephyr]

That

[delicate orchid]

anyway WTF ARE YOU TLAKING ABOUT

this is irrelevant to C^\times/S

[analog zephyr]

I mean, if G have D and U as subgroup aD and aU could be different set of coset

[delicate orchid]

they will absolutely be different cosets unless U = D

[analog zephyr]

So there's no one single left cose

Could be many

[formal ermine]

what are you saying

[analog zephyr]

Idk wtf about C^\times/S bruh

I don't even what it means

I don't even know

[delicate orchid]

the conversation you interrupted

[analog zephyr]

Maybe I was responding to someone more

Sorry

[celest cairn]

I have a question, whenever you guys are done with your conversation.

[analog zephyr]

?

[formal ermine]

we are done

what's your question

[celest cairn]

Is the Galois Group of $p(x) = x^4+3x^3-13x^2+15x-90 \cong S_{2}$?

[cloud walrus]

Sapphire

[chilly radish]

How would that work

[formal ermine]

yeah

[smoky ivy]

why not?

[rotund aurora]

[delicate orchid]

how tf did u do that so quickly - I'm impressed

[chilly radish]

Why is it in m^2?

[formal ermine]

I looked at its factorization

[smoky ivy]

cant you just take the root

[formal ermine]

two linear terms

one quadratic one

[smoky ivy]

so sqrt(x), x > 0 and 0 else

[delicate orchid]

yeah I was working on the factorisation - did u CHEAT??!!

[smoky ivy]

so it’s the product of that

[formal ermine]

no

obviously not

I'm a human calculator

[chilly radish]

Hmm, yea that works

You're right

[smoky ivy]

in fact, i think you could take almost any continuous map and take the root

[chilly radish]

I was thinking of a slightly different problem where you look at smooth functions

[smoky ivy]

i was overthinking it i think

[chilly radish]

That's a bit trickier

[smoky ivy]

yeah, pretty much

that would be harder

what did you have in mind v

[chilly radish]

Yea you're right

[smoky ivy]

some exponential function?

i was thinking too much abt it being smooth as well

[chilly radish]

In the case of smooth functions you take e^(-1/x^2)

[smoky ivy]

yeah, i had something like that in mind

[chilly radish]

Then this vanishes faster to 0 than any polynomial, so dividing it by x^n it remains smooth

[smoky ivy]

yup, when you said that earlier, i immediately thought of exponential

[chilly radish]

But yea in this case just taking the n-th root totally works

[smoky ivy]

but that case gives even more

[celest cairn]

Cool, thanks guys!
For fun I was just determining if p(x) was solvable by radicals. (It’s obvious it is.) So I just wanted to confirm the Galois Group was iso to S2

[formal ermine]

why don't you do some actual galois theory instead of just plain calculations btw?

because then you could see that every polynomial with degree less than or equal to 4 is solvable

the galois group embeds into S4 (or S3,2,1)

subgroups of solvable groups are solvable

S4,3,2,1 are solvable

[delicate orchid]

therefore by induction S_n is always solvable

[south patrol]

S_n is divisible by more than two primes so it is solvable by Burnside

[rotund aurora]

If all the roots of P(x) all lie in the unit circle, what can yous ay about P? Trivially, you can see that the coefficient of x^r will be less than or equal to nCr (in absolute value)

[coral shale]

what polynomial ring is this

[rotund aurora]

real of course

I think?

I think I assumed the polynomial should be monic

oh yeah it also works for complex coefficients

its just vieta and triangle inequality

I guess you can also do a less explicit reasoning

and the polynomial monic of course, because otherwise you can just multiply by stupid constants and the roots don't change

So I was looking at this problem: Suppose all the roots of P have absolute value 1. Show that all the roots of 2xP'(x)-nP(x) also have absolute value 1, where n is the degree of P

[formal ermine]

can't you maybe do something like uh factorize it into irreducible, monic polynomials, then those have to be the cyclotomic polynomials?

[rotund aurora]

this is fake

consider 3/5+i*4/5, this is not a root of unity, yet it has absolute value 1

(pythagoras)

[formal ermine]

yeah but like

if the polynomial is over R or Q

then you can factorize it into irreducibles that are monic

F_n,0 is irreducible

[rotund aurora]

yes, but they need not be cyclotomic polynomials

[formal ermine]

so the minimal polynomial has to divide them

[rotund aurora]

What do you mean here?

[formal ermine]

nth cyclotomic polynomial over Q

[rotund aurora]

x^2-6/5*x+1 is a counter example to what you are saying

[formal ermine]

ah I see

I was thinking something else I think

f has to divide some F_n,0

[rotund aurora]

why?

[formal ermine]

wait not F_n,0

some x^n + 1

[rotund aurora]

no

[formal ermine]

why not? if you look at it over C it will contain some of the roots

like not divide in Q[x] but in C[x]

yeah this is pretty stupid

[rotund aurora]

the roots of the problem need not even be algebraic

[formal ermine]

oh yeah

[rotund aurora]

but this is false even in the algebraic case, the example above is a counter example

tbh, idk if this problem is analysis or algebra

[formal ermine]

I should probably not help with stuff I can't help with

[rotund aurora]

nah don't worry, I don't know what to do either

I think I could show they cannot be strictly less than 1

[flint crater]

If I have two R-modules M and N and I take the group homomorphism f: M -> N. If the R modules are over a field, is the group homomorphism then just a linear map like in linear algebra?

Group homomorphism satisfying f(rm) = rf(m) with r in R and m in M

[formal ermine]

do you mean module homomorphism

[flint crater]

Idk, they use the name R-homomorphism or R-linear map too

[formal ermine]

but yeah this is true

if R is a field then it's a linear transformation

[rustic crown]

yea, you call it one of these things
R-module homomorphism/map
R-linear homomorphism/map

[flint crater]

Alright ty for confirming :)

[flint crater]

Got it!

[rustic crown]

(btw be a little careful if you your abelian groups can be considered modules over multiple rings, for example the conjugation map C --> C is R-linear by not C-linear)

[formal ermine]

what does $\bL(\bK)$ mean?

[formal ermine]

is it just like all stuff from K added to L?

ok another question

let K be char 0 and f in K[x] with splitting field L

why does Gal(L/K) is solvable => L(U_n)/K is a radical extension?

don't we need L/K to be finite?

[chilly radish]

Splitting fields are always finite

You get them by adding finitely many algebraic elements

[formal ermine]

ah right

ok thanks

[chilly ocean]

has anyone read the chapter on generalized free products in Robinson's "A course in the theory of groups"?

there's a theorem I've been reading for days but still can't make sense of it

[rotund aurora]

I haven't, but just ask the question

[chilly ocean]

I don't understand the whole proof

I know this sounds like I didn't even try but it's not the case

can someone tell me the rough idea what he's trying to do here?

we're trying to prove the uniqueness of normal form in generalized free products of groups

here's the definition also

[agile burrow]

Ok so the outline of this proof is to define an action of G on the set of normal forms. This can be done by specifying an action of each G_{\lambda} on the set of normal forms, and then verifying that N acts trivially. Do you see what this action is?

Once we've defined the action, we can specify a map G -> M that maps an element g to the image of its action on the normal form of the identity element. Conversely, we have a map M -> G as follows: given a normal form in M, we get an element in G by looking at what element the normal form specifies.

Finally, the composition M -> G -> M is the identity, which implies that the map sending a normal form to an element of G is injective, hence normal forms are unique.

There's a neat geometric interpretation of this in terms of groups acting on trees, which I can sketch out if you're interested

[chilly ocean]

yea kinda makes sense now, thank u very much

in the middle part when we make 2 cases

and he says that the other case can be handled similarly

its still not clear to me what's the difference between those cases

aight I'll figure it out I think (I'll try)

such a tedious and cumbersome proof

[chilly ocean]

I'm not sure if I'll understand but would be interesting

[agile burrow]

It's a bit hard to grasp, yeah. But it's a cool idea

Sure, I'll finish writing it up in a sec

Actually I'll just talk about it anyway because it's kind of fun. This is kind of specific to an amalgam of two groups, but just induct of something idk. I'm pretty sure there's a way to generalize it to more, but I haven't thought about it too much yet.

So if G is a group acting on a graph X by automorphisms, we call an edge e with vertices v and w a fundamental domain for the action of G on X if the quotient graph X/G is isomorphic to the subgraph consisting of v, w, and e.

A few observations: the stabilizer of e, say G_e, is the intersection of the stabilizers of v and w, G_v and G_w respectively. Also, two edges g*e and h*e have a vertex in common iff g*h^(-1) is in G_v or G_w.

Now I claim that X is a tree if and only if the map induced by inclusions G_v *_{G_e} G_w -> G is an isomorphism.

[chilly ocean]

that definition of a fundamental domain I dont understand (is it wrong?)

one edge is a fundamental domain if the quotient group is isomorphic to that one edge?

[agile burrow]

the edge e and its vertices v and w form a subgraph of X, right?

we say that it's a fundamental domain if the quotient graph X / G is isomorphic to this subgraph

The idea is that any edge in X can be written as g * e where e is the edge in our fundamental domain, and every vertex can be written as g * v or g * w, but not both

[chilly ocean]

so either all edges (with their vertices) are fundamental domains or none of them is?

[agile burrow]

I don't want to prove one direction, but I'll sketch the other. Using our observation on common vertices, you should be able to note that paths in X correspond to reduced words. Like you start at the vertex v, go to w via e, go to g*v via g*e (where g is in G_w), go to gh*w via gh*e (where h is in G_v), etc.

If X is a tree, then it has no loops, so these reduced words are not equal to the identity. Then the unique path from v to any vertex in X corresponds to a unique reduced word. Conversely, given an element g in the group, we can associate the reduced word from v to g*v where the first edge is e.

[agile burrow]

I don't think this is true

[chilly ocean]

I'm having hard time understanding the definition of fundamental domain:
we say that if the quotient graph is isomorphic to a single edge with its vertices then call v-e-w a fundamental domain, but in that case it's also isomorphic to any other subgraph with a single edge?

and on the other hand if the quotient group isnt a single edge then there are no fundamental domains?

[agile burrow]

I mean there's a more general definition of a fundamental domain which is just a subset of the space which contains one point from each orbit

But in this case I'm just restricting to the case where the fundamental domain has one edge and two vertices

It's not necessarily the case that every edge with its vertices are such that the two vertices are in distinct orbits

[chilly ocean]

oh, okay I didn't get the orbits part

[agile burrow]

Ah yeah, I probably should've said that explicitly.

I'm not explaining this too well either lol

[chilly ocean]

it's fine u don't have to, I'm trying to figure out tho

[agile burrow]

It's not a huge deal if you don't get it, but if you want some more exposition on this I think it's laid out in more detail in Serre's book on Trees

But yeah, at the very least I hope you can see the big picture of the proof of the uniqueness of normal form in amalgamated products

[chilly ocean]

yea I'm slowly coming to the point where this should connect to it

thx for ur time, and also, are you undergraduate?

[agile burrow]

I'm an undergrad right now, yeah

[chilly ocean]

focusing on group theory?

[agile burrow]

I like thinking about groups and I keep a collection of group theory facts, but I don't want to commit myself to anything right now. I like lots of different parts of math

[chilly ocean]

what kind of facts?

[agile burrow]

Well, for example, let G be a torsion-free subgroup of SL(2, Z) of finite index k. Then G is free on 1 + k/12 generators

[coral spindle]

black magic

[prisma ibex]

modular curves :^)

[agile burrow]

Here's another cool one, let d be the gcd of all [G : H] where H is a torsion-free finite index subgroup of G. Then the prime divisors of d are exactly the primes p such that G has p-torsion

[chilly ocean]

do u also know the proofs to these?

[agile burrow]

I do (though I might have to review some bits). The proofs I'm most comfortable with involve Euler characteristics of groups, but the first one can be proven by studying the action of SL(2, Z) on the Farey tree (which is also how you can deduce its structure as an amalgamated product, as we were talking about earlier)

[delicate orchid]

what the fuck

it makes sense that it's free but k/TWELVE? wtf...

[agile burrow]

it's related to the funny number -1/12

[delicate orchid]

the time has come

[agile burrow]

and also the fact that the abelianization of SL(2, Z) is Z/12Z

[delicate orchid]

ok that makes more sense

infinite groups are crAzy

read the proof, not particularly enlightening

[agile burrow]

Here's another fun one. I posted this last night, but I'll rephrase it to be even more group theoretic. Let G be a finitely presented group; let <S|R> be a presentation and let F = F(S) be the free group on the generators. The group (R \cap [F, F]) / [R, F] is a finitely generated abelian group, let r be the minimal number of generators.

Then any presentation of G with n generators has at least n + r relations.

[chilly ocean]

this one is more interesting by far

[elder wave]

One day I will learn group theory properly

One day

[delicate orchid]

presentation theory my beloved

[chilly ocean]

since we're all here

is combinatorial/geometric group theory worth studying full time?

it it active branch with perspective in it?

[delicate orchid]

geometric group theory absolutely has a lot going on

[agile burrow]

It is definitely active

[south patrol]

Get it out of my head

[chilly ocean]

I've started doing stuff related to amalgam free products with my undergrad advisor (studying still)

and I'll probably keep going this way

[agile burrow]

it's fun stuff for sure

and there's a lot to explore

[chilly ocean]

and since I'm not a math beast like some of u

to do grad lvl math as an undergrad

I'll probably just focus on this one thing

I'm doomed

[delicate orchid]

u r not doomed...

[chilly ocean]

I just need to be lucky with my future research in group theory

this should work out

no idea how other undergrads who are not focusing on group theory know more group theory than me lol

[grand cliff]

just went to a math camp and geometric group theory was one of the topics

something about group presentations

[long nebula]

Why is the map from k(x1,...,xn) to functions : k^n->k injective if k is infinite?

I see that it's clearly not injective if k is finite due to cardinality considerations

k being a field

[agile burrow]

hmm

so if f and g get mapped to the same function, then f - g is identically zero on k^n

[barren sierra]

It may be easier to show the kernel is trivial

[delicate orchid]

can you not view k(x1,...,xn) as a vector space iso to k^n and then the space of functions is just the dual of this vector space? Then nice things happen cause k is infinite and the vector space is finite dim

unless you want a field map I suppose

[barren sierra]

It's not a field map cause polynomials may not invert

[delicate orchid]

true!

wait

[barren sierra]

Wait they wrote k()

Not k[]

[delicate orchid]

yeah

[barren sierra]

So it is a field, neat

[delicate orchid]

take them as algebras

[barren sierra]

Wait field homomorphisms are automatically injective

We did it team

Wait no the space of functions k^n -> k isn't a field

Right?

[obsidian sleet]

Pointwise operations

it might be a field then

maybe

product and sums are ok

commuting ok

division is (1/f)(x,...) = 1/f(x,..)

[barren sierra]

Ah

Ok yea we did it team

[agile burrow]

my mfw face when the function has a zero

[obsidian sleet]

pain

[barren sierra]

Skill issue

[delicate orchid]

simply do not

yeah it's not quite as simple as "hurr vector space"

[long nebula]

wait how

the book I'm reading takes it to mean a polynomial ring in n variables

[agile burrow]

oh you're talking about polynomial ring??

[barren sierra]

💀💀💀💀💀

[delicate orchid]

ring
then isn't QFERVSTBYNH K**[x_1, ..., x_n]**

CMON PEOPLE

[barren sierra]

k[] != k()

[long nebula]

it says () 😭

[agile burrow]

lol yeah I've always seen [] to denote the polynomial ring and () to denote the field of rational functions

[delicate orchid]

reeeeeeee

[long nebula]

pain

[agile burrow]

i should've asked for clarification anyways

[delicate orchid]

nah it's like asking for clarification about if "G" is a group

[barren sierra]

Yea

[agile burrow]

But yeah, if f and g are mapped to the same function then f - g is identically zero. A polynomial over an integral domain has finitely many roots, but k is infinite, so f - g = 0

[delicate orchid]

we use le universal property

no we le don't

yeah can't improve on walter's solution

[long nebula]

ohhh gotcha

okay that makes total sense idk why I didn't think of that thanks!

[agile burrow]

happy to help

[barren sierra]

Yea important part is infinite field

[obsidian sleet]

alright

ive been waiting ooga

i am not good at this projective injective flat stuff

time to post my problem

If $A\otimes_R I\to A\otimes_R R$ is injective for every finitely generated ideal $I$, prove that $A\otimes_R I\to A\otimes_R R$ is injective for every ideal $I$. Show that if $K$ is any submodule of a finitely generated free module $F$ then $A\otimes_R K\to A\otimes_R F$ is injective. Show that the same is true for any free module $F$.

[cloud walrus]

metal cofe bad kernel oog

[obsidian sleet]

this has three parts

i did the first part for just ideals I of R

i have no clue how to do the second and third parts

more importantly the third part

i havent thought of the third part yet but i imagine its similar to how i did the first part

but like

im taking F as R^n and distributing the tensor product thru

and i have some element in the kernel of the map A otimes K to A otimes F

im struggling to wrap how im supposed to work with K here i dont really know what it looks like

its just any submodule

[delicate orchid]

finitely generated free R modules must be isomorphic to R^n, and ideals are submodules of R = R^1 (and conversely all submodules of R as an R-module are ideals), so it's a generalisation of the ideas in the previous question

[obsidian sleet]

right its like

the non rank 1 case

maybe i can use M being finitely generated somehoiw

[delicate orchid]

my idea is you take $R^n = R \oplus R \oplus ... \oplus R$ and then write a submodule as $K = I_1 \oplus I_2 \oplus ... \oplus I_n$ with $I_j \trianglelefteq R$

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

does this enumerate all the submodules? probably not

[obsidian sleet]

Do the submodules actually come out that way

i see

[delicate orchid]

I know that for groups all subgroups of a free group are free by nielsen-schreier but these aren't groups

[obsidian sleet]

yeah u can def have nonfree submodules

[delicate orchid]

counter example: take a field and k and look at submodules of k^2

oh wait no still works

hmm

do we know anything else about R or is it just le ring

[obsidian sleet]

le ring

[delicate orchid]

all rings are PIDs -> structure theorem

[smoky ivy]

im stuck on b), i've already shown that it's a local ring

[next obsidian]

By induction it suffices to let k = 1

Compute how much the dimension of the ring drops, and then show that the # of generators for m drops by the same amount

You will almost certainly use Nakayama’s lemma

@smoky ivy

[next obsidian]

Show that the generators are contained in the other ideal

For 2 this is… well pretty easy

So youre reduced to showing that 1 + sqrt(-5) is in the other ideal, and likewise

That’s just some algebra, idk off the top of my hand, but I’m sure you can figure it out

Actually I didn’t just figure it out off the top of my head, but I need to leave some work for you

xD

You… write it down

As an element of the other ideal

I mean

You tell me

Well you’ve now shown one ideal is contained in the other

So now you need to do the other one

[toxic zephyr]

could something like $\bZ/2\bZ\times\bZ/4\bZ$ be a $\bZ$ module?

[cloud walrus]

nilpotent nix

[lethal dune]

Yea

[next obsidian]

A Z-module is just an abelian group

[lethal dune]

Z mods are abelian group

[toxic zephyr]

okay sure. then, is it even possible to use matrices for it?

[next obsidian]

It’s not free so no

I mean idk what “use matrices for it” means

[toxic zephyr]

free means it has a basis right?

[next obsidian]

You can’t just write down every map as a matrix and do linear algebra like you can for vector spaces

[toxic zephyr]

hmm. then why isn't (1,0),(0,1) a basis?

[next obsidian]

That isn’t linearly independent

4(1,0) = 0

A free module never has torsion

[toxic zephyr]

ah that makes sense, but i didn't know that.

[next obsidian]

Free is equivalent to being isomorphic to a direct sum of the ring

So being Z^(+)alpha for some cardinal alpha

[toxic zephyr]

so Z/nZ can never be a free Z module?

[next obsidian]

Yes

You know the sturxfure of finite abelian groups right?

Or I guess finitely generated

[toxic zephyr]

a little bit. they are all isomorphic to a direct sum of cyclic groups, right?

[next obsidian]

Yeah

And for finitely generated it’s this and then a part which is just Z^r

r is the rank which is the maximal # of lin ind elements

So it splits into a free part, Z^r

And a torsion part, made up of cyclic groups

So from this you see a fg abelian group is free iff it is torsion free

[toxic zephyr]

ah okay. yeah that makes sense

[toxic zephyr]

I'm trying to figure out a way to characterize automorphisms on this group. does anyone know if there's some sort of well known result that characterizes the in possible automorphisms on finite abelian groups such as that?

[next obsidian]

Automorphisms commute with products

And you know the automorphisms group of every cyclic group

“Know”

[toxic zephyr]

haha yeah probably should know. are the automorphisms of a cyclic group something relatively simple? something like phi(1) has to be coprime to the order?

[next obsidian]

I mean

You know their size

Actually you don’t know specifically what they are in general

Also I misspoke, I think this is actually hard sorry

It’s true that Aut(G x H) = Aut(G) x Aut(H) when the order of G and H are coprime

But then you can only break apart your group into parts based on primes

So you have to deal with an arbitrary product of Z/p^nZ for various n inside each block

And eg Aut(Z/2Z x Z/2X) = S_3

Given by permutinf the 3 nonidentity elements

But Aut(Z/2Z) is trivial

In general though Aut(Z/nZ) has size phi(n) since it’s determined by sending 1 to some element coprime to n, so it’s just multiplication by any of the phi(n) elements coprime to n

In “a lot” of cases this is cyclic, for example for a prime number

But in general you’d do some number theory to figure out what when you admit a primitive root mod n

https://brilliant.org/wiki/primitive-roots/

But you get screwed once you deal with products of Z/p^nZ for different n, for one fixed p

[chilly ocean]

need help with part iii)

I tried saying the following when checking for the subring axioms:
$0 \in S$ because $\text{det}\begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} \leq 10^{50}$

[cloud walrus]

isomorphism

[chilly ocean]

$1 \in S$ for the same reason - because $det(I_2) = 1$ which is less than or equal to $10^{50}$

[cloud walrus]

isomorphism

[chilly ocean]

but we meet an issue when checking for additive closure, I think

if $a, b \in S$ it is not always true that $a + b \in S$

[cloud walrus]

isomorphism

[chilly ocean]

right??

i.e. if we take $a$ to be $\begin{pmatrix} 10^{50} & 0 \ 0 & 1 \end{pmatrix}$ and $b$ to be $\begin{pmatrix} 10^{49} & 0 \ 0 & 1 \end{pmatrix}$

[cloud walrus]

isomorphism

[chilly ocean]

the determinant of a + b will be 2(10^50 + 10^49)

which is clearly greater than 10^50

[toxic zephyr]

tysm for all of this! I'll have to read it more carefully, take it in, and mess around with it to fully understand it, bit thank you.

[formal ermine]

how do we show this

[next obsidian]

Exercise

[formal ermine]

someone remind me in 7 hours

to do this

[next obsidian]

But

Clearly if phi is an automorphism of G and psi one of H

[chilly ocean]

anyone

[next obsidian]

Then the formula (phi x psi)(g,h) = (phi(g),psi(h)) defined an auto morphism

You will end up showing everything is of that form

[formal ermine]

G and H have a trivial intersection

[smoky ivy]

i honestly don’t see how to show this by induction

[next obsidian]

Quotient by just one thing first, then by k-1

Assuming you can handle k = 1, after step 1 you have a regular local ring still

And the other k-1 things are still linearly independent

[smoky ivy]

yeah

so i.e. R/(a_1) regular local ring

how would i follow that R/(a_1, a_2) is a regular local ring

[next obsidian]

By induction…

The image of a_2 through a_k satisfy the same hypotheses

If you look at what (m/(a_1))/(m/(a_1))^2 is

This will be the same as m/(m^2 + a_1)

The images of a_2 through a_k will be linearly independent in here because a_1 through a_k were assumed to be linearly independent in m/m^2

[chilly ocean]

I proved that it is indeed a subring of C

as for the second part, showing it's part of the unit group I did the following

If $a = 8 + 3\sqrt{7}$ then $a^{-1} = 8 - 3\sqrt{7}$ which clearly follows the structure of $\mbb{Z}[\sqrt{7}]$

[cloud walrus]

isomorphism

[smoky ivy]

i don’t rlly get it, could you expand?

[chilly ocean]

from first sight it looks that it does since 8^2 = 64 and 9*7 = 63

[chilly ocean]

yeah good

[rotund aurora]

In an arbitrary ring $R$ if $I$ and $J$ are two ideals such that $I\supseteq J$ does there exist an ideal $K$ such that $J=IK$ ?

[cloud walrus]

Croqueta

[rotund aurora]

R commutative and unital

I tried looking at the set of ideals ${K \colon J\subseteq IK}$ and this satisfies Zorn I think, because you can take the intersection

[cloud walrus]

Croqueta

[rustic crown]

i would guess not. don't have a counter example at the top of my head, but i think that's true when R is a dedekind domain.

[rotund aurora]

how would you prove it? Using UFD?

"containment-division ring" wtf

[rustic crown]

you prove it for PIDs, and then this thing behaves well with localizations so you also get it for dedekind domains

i looked at these things when i didn't under localizations properly. ig it's a good time to revisit it

like there is a notion of ideal quotient (J:I) = {r in R | rI contained in J}

and by definition it satisfies (J:I) * I contained in J

the other inclusion isn't always true

(i think)

but for dedekind, you just localize and reduce this to DVRs, which are in particular PIDs so all was good

so if you're looking for a counter example, maybe start with a dimension 2 ring like k[x, y] or Z[x]

the ideal (x, y) in k[x, y] should definitely be involved lol

let's see if that works then

(x, y) contains the ideal (x)

so we want to look for (x, y) * I = (x)

so y * I is contained in (x)

but x and y are coprime

so I is contained in (x)

[rotund aurora]

tru

[rustic crown]

yea should work lol 😂

(x) = (x, y) * I <= (x, y) * (x)

dividing by (x) this gives (1) <= (x, y)

that's def false

[rotund aurora]

okay this is neat

thanks

[rustic crown]

ah, so it's equiv to dedekind, nice

[rotund aurora]

assuming integral domain

wait can a Dedekind domain have zero divisors?

[elder wave]

dedekind DOMAIN

[lament dawn]

Wiki says no

[rotund aurora]

[lament dawn]

Therefor no, wiki has never been wrong

But I mean your screenshot says R is an integral domain so this shouldn't be very surprising?

[rotund aurora]

No its not surprising lmao

[left estuary]

So I was searching for extra practice questions on homomorphisms when I found this document and I haven’t read through all of the questions but it seems to go in an order that would be similar to how you would learn ring theory but what do you guys think? http://www.math.kent.edu/~white/qual/list/ring.pdf

[rustic crown]

det should read some commie algebruh >.<

but me get bored easily >.<

[lament dawn]

This doesn't seem to go in a very particular order

Also these are qual exam questions lol

Maybe they are loosely ordered within the subjects

[elder wave]

i think you should refer to some book/script for the order you're gonna learn stuff in instead of a collection of qual exam questions

[rotund aurora]

I have doubts on ideal arithmetic, what is the algebraic structure formed by the ideals with the usual operations of addition and multiplication of ideals? Does it have a name?

I think its just like a ring except there are no additive inverses ?

[delicate orchid]

It’s a semi ring

[rotund aurora]

ok nice

[delicate orchid]

Might even be a sub-semi ring of the semi-ring of modules over a ring but I’d actually have to think about it which I’m not gonna do

[lethal dune]

what's the additive identity?

[rotund aurora]

(0) ?

[lethal dune]

then like what's the additive inverse of ideal (a)?

[delicate orchid]

There isn’t one

Hence semi ring

[rotund aurora]

ye

[lethal dune]

what's a semiring

[rotund aurora]

in a way, when you "idealize" your ring, you forget about units

[lethal dune]

well you can always Grothendify it

[rotund aurora]

hence no additive inverses at all (no -1)

[delicate orchid]

Yur

I’m doing a lot of stuff with semi rings as we speaK

[lethal dune]

cool

like?

[rotund aurora]

you can talk about ideals in a semiring just fine right?

[delicate orchid]

The semi-ring of modules over a group algebra that are fixed under a certain set of actions

[chilly radish]

Sounds lamw

[delicate orchid]

Think so? - they just won’t be groups under multiplication

[delicate orchid]

Bit like you then

[lethal dune]

[rotund aurora]

you meant addition?

[delicate orchid]

I did

[rotund aurora]

Because I was thinking, you could start with a semiring and "idealize" it, and then you can ask wether the original semiring and this idealization are isomorphic as semirings

[delicate orchid]

Hmm

Ok let’s try this

I’m taking Z because why tf would I take anything else