groups-rings-fields

[fading field]

do you mean ring when you say field?

[grizzled crane]

obvio, i mean a field

[fading field]

okay, Z is not an example of what you’re talking about then

[grizzled crane]

I was giving a example i had with rings

[fading field]

oh okay

oh i see what you’re trying to say

[grizzled crane]

I was asking how does it go, when i have fields

english isn’t my first language, sorry

[fading field]

it’s okay

it’s not a language issue at all, your english is good. i just misunderstood on my end

[grizzled crane]

I asked, what about fields in general

that could imply Z[x] is some specific field im looking

that could be misleading

[vivid tiger]

Surely the answer must be no.

Isn't this basically asking "How does the ring of algebraic integers differ from the field of algebraic elements?"

And, the answer is "more than by nothing"

[crystal vale]

G->G by x->gx is homomorphism when g=e, right ?

[chilly ocean]

the identity map is indeed a homomorphism

[crystal vale]

for what value of g it will be homomorphism , g must be e, right?

[chilly ocean]

yes. you can see this right away by plugging in e for x

if it's a homomorphism, then it sends e to e. but it sends e to g, so...

[crystal vale]

Okay, thank you

[crystal vale]

I want homomorphism mapping between D_4 and S_3 so if I define D_4 as {1,r,r^2,r^3,s,sr,sr^(2), sr^(3)} and r^4=1, s^2=1 ,sr=r^3s.

If 1 maps to 1 and r maps to (12) and s maps to 12 then is it correct?

[tawny magnet]

I want to show that the Gaussian period is only fixed by the maps that it should- Is there an obvious way to show this?

want to show this is only fixed by sigma^km and nothing else

[winged void]

I have small question regarding when we say that homomorphism preserve the group structure

what do mean actually with the word group structure

i mean those are two groups with two diffrent operation

so what does it exactly mean that the group structure is preserved

[ivory trail]

a group is made of three things: a set of elements, a "structure" of operations (nullary identity, unary inverse, binary multiplication), and some properties (associativity, left inverse law, right inverse law, left unit law, right unit law). preserving the structure means that you can apply the homomorphism before or after the operations and it's the same

for groups in particular, you can just check that a map f preserves multiplication (f(ab) = f(a)f(b)), which implies that it preserves the identity (f(1) = 1) and inverses (f(a^-1) = f(a)^-1)

in these equations, take the identity/inverses/multiplications to be the appropriate ones from either group in question

[winged void]

sure i agree about that

but i mean are the groups like not diffrent

like G for example here can be integers and say for example H is rational or something but not integers

could you give an example if its possilbe

@amber wraith do you have maybe intuitive way of seeing it ?

[tawny magnet]

Bump 😭

[rocky cloak]

In a group you have a bunch of relations like a*b = c. These together form the structure of the group. Now the names a, b, c or the exact nature of the operation is not really the important part, it's this relationship that forms the structure of the group.

And a homomorphism preserves these relationships.

[winged void]

so its actually sort of preserving the closure

of a group

kind of

like if a * b = c then c must be in that group or somethign

am i right

[rocky cloak]

I mean, c must be in the group sure, but it's saying more than that. Like I couldn't just replace c with any element of the group. It's a special thing that a*b = c

[winged void]

i see. its a really difficult thing

because i have see a lot of example and i can solve the questions

but the intuition behind it kind of vague

like i was solving this

its homorphism

[rocky cloak]

Then I think you might be overthinking it.

It's just that in your first group the operation is + and in the second it's *, then you want the map to take things involving + to things involving *. That's all there really is to it. You take instances of +, replace them with *, and check that things still work out.

[winged void]

I see but then what is here the group structure

I guess at the end

That’s not so important

[coral prawn]

Was messing around with quotient groups and I made this - is this a lie group? (Also whiteboard flex 😎)

[rocky cloak]

These are indeed lie groups. In general (R/Z)^n is called the n-torus.

[coral prawn]

Oh awesome

Wait is that the same as the donut shape? That’s how I imagined a torus

[last spoke]

Imagine taking the sides of the square you’ve equated and gluing them together

You get a donut

[coral prawn]

Okay now that’s sick

[lime junco]

the way I look at group homorphisms is when I talk about a group - the operations like *, +, x, whatever are really just window dressing or language, What I really have is a bunch of objects, that follow those 3 important properties. A homorphism is in a way, letting me translate the current language into a new language. For example, the exponential map takes these symbols, real numbers, and the operation +, and translates them into the language of x

so while in my first group,

i get a + b

when i do exp, i can trnaslate that structure

so I now get

exp(a + b) = exp(a) x exp(b)

so I've basically just translated the objects into a new "language" (group)

It helps more when you know about for example isomorphisms

which basically mean that

these two groups are the exact same, just described in different languages

[winged void]

I see

[amber wraith]

so lets say I have a group like Z mod 2

this is just {0, 1} where 1 + 1 = 0

now lets say I have another group like {1, a} where a^2 = 1

so in the first group, I have 1 + 1 = 0

and in the second I have a * a = 1

in both cases the group has two elements, one of which is the identity, and the other one goes to the identity when applied to itself

we say these groups are isomorphic, because the only difference between them is that you're using the symbol + vs *, and differnt symbols for the identity

but you could say "well you're just calling it * instead of +, and using 1 instead of 0, and the identity 0 in the first group is acting like the identity 1 in the second, and instead of calling it a you're calling it 1, but other than that it's the same group"

[crystal vale]

The set of all g such that mapping G to G x->gxg^(-1) is identity mapping is centre of G, right?

[delicate orchid]

yes

[crystal vale]

Okay, thank you

[sly rain]

For Ideals I,J and Ring R, does I+J=R imply that R is commutative?
Context: This is the assumption of an exercise, I+J=R to show that IJ=I /cap J

Unless its follows extremely directly, feel free to give me a nudge into the right direction

[delicate orchid]

I don't think this result holds for non-commutative rings

[boreal inlet]

Let L/K be finitely generated field extension. How do we show any intermediate subfield E is also finitely generated?

(Problem is, finitely generated field extensions can be infinite. If it wasn't, it's similar to how vector spaces work)

[sly rain]

I see thanks, maybe we are assuming commutative Rings now but its weird because usually it was stated If its commutative/or Just a Ring

[dull ginkgo]

If you are actually fucking insane you can probably use that every intermediate subring is thus a field, and use valuation maps from the polynomial ring

[boreal inlet]

I don't know what valuation maps are..

[dull ginkgo]

Then don’t use that route lmao

Uhh let me think of another way

[boreal inlet]

But I don't know what to do 😭

[dull ginkgo]

Almost all the field shit I have done uses absurd bullshit with polynomial rings that comes to the same conclusions but makes me look psychotic

[boreal inlet]

Are these left ideals or right, or both sided

[dull ginkgo]

Usually when left/right is omitted I assume two sided

[delicate orchid]

I think the counter example is when R is a two by two matrix ring, we take one ideal to be generated by one of the columns, and the other to be generated by one of the rows. Then they add to R, have non-trivial intersection (one of the corners), but multiply to give 0

[dull ginkgo]

I’m trying to think of ways that don’t use borderline insane methods

[delicate orchid]

L and E are both K-vector spaces

E is a subspace of L and is thus of strictly smaller or equal dimension, hence is also finitely generated

does that work?

[sly rain]

Cool counter example, thanks

[dull ginkgo]

Would you need to show that its basis must be of E-elements?

[delicate orchid]

basis of what

of E?

[dull ginkgo]

Yes

[delicate orchid]

well yes... the basis of a vector space is a subset of that space

[dull ginkgo]

Yeah I guess that takes a bit of work but it can be ignored because I’m lazy

[delicate orchid]

how does that take work? b in B, 1b = b in V?

[dull ginkgo]

I always get mixed up between module finite generation or algebra finite generation but for fields you don’t need to give a shit due to Noether normalization lmao

[dull ginkgo]

Fair

[delicate orchid]

oh yeah

oh well!

[dull ginkgo]

I unironically love the proof of Noether normalization

[delicate orchid]

I don't recall the statement

nor do I care because it's alg geo

[dull ginkgo]

True!

[dull ginkgo]

It’s rather long so fair

[boreal inlet]

Thing is, Q(x)/Q is finitely generated but has infinite dimension

[delicate orchid]

right so you want algebraically generated

[dull ginkgo]

Isn’t that easier

That’s just going through the polynomial ring

[boreal inlet]

When I ask finitely generated I mean there is a finite generating set S= {a_1,a_2,...a_n}

[delicate orchid]

yeah

noether normalisation should do that but lets not just throw that result at the problem

ugh but it's just so easy to jsut go "Krull dimension"

[delicate orchid]

elaborate

[dull ginkgo]

”nor do I care because it’s alg geo”
krill dimension

[dull ginkgo]

Realized it doesn’t work as nicely as I thought

The superfield is of the form F[X_1…X_N]/M for maximal ideal M

[delicate orchid]

yup

[dull ginkgo]

Then every subring is of the form R/M

For R subring of F[X_1… X_N] containing F

[delicate orchid]

and if that subring does not contain M?

I presume you mean R/M cap R, that I can see

or do I

R/RM

[dull ginkgo]

By the isomorphism theorem of rings

Every subring of A/I is if the form R/I for R between I and A

[delicate orchid]

honestly, try this

seems like the easiest way at this point

there's no alg geo involved at least

[dull ginkgo]

I need to go soon and my family is pissy because I am procrastinating

[delicate orchid]

f^5 = (12)(354)

[coral spindle]

There was a mistake here, indeed f^5 was calculated wrongly. All other things written in the screenshot-within-a-screenshot are correct.

[scenic saffron]

Oh right

Poor from my professor

Thanks

[coral spindle]

Professors are not magical

They make mistakes

As do you.

Note that they got the correct answer, everything else was just a typo.

[delicate orchid]

yeah but... cmon now...

[scenic saffron]

I will be seeing the disciplinary committee about this

[delicate orchid]

and you will be summarily ignored

[coral spindle]

Oh I'm sorry do you never copy-paste and forget to change something

[delicate orchid]

no I type things up by hand each time

[boreal inlet]

I honestly don't know what is a Valuation map actually

My only background is Ring Theory and the Galois theory

[sly rain]

Whats the usual group theoretic progression after a basic abstract algebra course?
Like usually afterwards you can do commutative algebra and then algebraic geometry (atleast at our uni).
What would be the group theory focussed counter part?

Everyone can answer

[dull ginkgo]

It’s the name for the map from a polynomial ring when you “set the indeterminates”

I.e give them a value, valuation

[delicate orchid]

commutative algebra then representation theory

branching off into lie group nonsense if you want or stuff like algebraic topology

or you can do things like combinatorial group theory

these are all intimately connected but have very different flavours

that being said, a basic abstract algebra course does not cover much group theory

[sly rain]

Yeah we kinda did a little bit of everything from groups to galois theory in the abstract algebra course
Thanks for the answer

[scenic saffron]

is there an algorithmic approach to finding the order of an element in the multiplicative congruence classes or do i have to keep on multiplying till i get e?

Ofc reducing the congruence class back down each time

[sonic coral]

Let K/Q be a Galois extension. Let L be the set of elements α ∈ K that are expressible in
terms of radicals. Show that L/Q is Galois and Gal(L/Q) is a solvable group.

I think i need to use the fact that quotients of a solvable group are solvable, but i’m not sure about much more than that. Any hints?

[fading basin]

I would first show that L is a field

This follows because sums, differences, products and quotients of elements expressible by radicals are also expressible by radicals

Then show L/Q is a normal extension, i.e. every irreducible polynomial in Q[x] that has a root in L splits completely in L[x]

This holds because if α∈L, then all conjugates of α are also in L, since they differ from α only by roots of unity which are expressible by radicals

So the minimal polynomial of α splits in L[x]

L/Q is separable since char Q = 0

By the Fundamental Theorem of Galois Theory, Gal(K/L) is a normal subgroup of Gal(K/G) with quotient group isomorphic to Gal(L/Q)

Since K/L is abelian, Gal(G/L) is solvable

Quotients of solvable groups are solvable, so Gal(L/Q) ≅ Gal(K/Q)/Gal(K/L) is solvable too

[sonic coral]

why is K/L abelian?

[fading basin]

Well, it's not necessarily an abelian extension

A solvable group can have non-abelian subgroups and quotients too

Only if all the factor groups Gi/Gi+1 are cyclic in the group called supersolvable, in which case all subgroups and quotients are cyclic hence abelian

[sonic coral]

so are you saying that the galois group with these properties is super solvable?

[fading basin]

Nope

The problem states that K/Q is a Galois extension, and L is the subfield of K consisting of elements expressible by radicals over Q

The goal is to show L/Q is Galois with solvable (not necessarily supersolvable) Galois group Gal(L/Q)

A group G is called supersolvable if it has a normal series 1 = G0 ⊴ G1 ⊴ ··· ⊴ Gn = G such that each factor group Gi/Gi-1 is cyclic

Every supersolvable group is solvable, but the converse is not true in general

[sonic coral]

i figured that, thank you

i’ll get to work on this and see what happens, thanks

[sonic coral]

do we need K to contain roots of unity for this to work?

i’m not convinced that K needs to have the roots of unity in it

[rocky cloak]

If K contains the roots of x^n - a it will also contain the nth roots of unity.

[sonic coral]

perfect that’s what i needed

[dire bramble]

how does this determine an isomorphism of real algebras?

is the basis for C + C not 4 dimensional?

[south patrol]

Both sides are 2D C vector spaces and this is a C-linear isomorphism

[dire bramble]

if this is C-linear, why can't I do $(i, 0) \mapsto \frac{1}{2} (1 \otimes i - 1 \otimes i) = 0$?

[cloud walrus]

bacono

[dire bramble]

then it's not an iso

[coral spindle]

Potat, hows C (x) C 2d as a C-space? I don't see what's happening here from this alone

[dire bramble]

as a real tensor, C otimes C = C

yeah

hmm

[coral spindle]

So as a C-vector space it's still 1d

[dire bramble]

so I think this can only be the case over R

[coral spindle]

Yeah

So this is a tensor product over R, with the obvious left action of C, right?

That should work?

Yeah indeed, so they're both 2D C-spaces as potat says

[dire bramble]

okay perhaps i'm just having a bit of a morning

[chilly ocean]

,ti bacono

[cloud walrus]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

[dire bramble]

its 3pm

[barren sierra]

what are you looking for

[dire bramble]

leave me alone

[barren sierra]

ah

[dire bramble]

okay but back to my point

[dire bramble]

like did we not just establish that C (x) C = C, so C + C cannot be C (x) C over C

one is dim 2 over C, the other is dim 1

i'm sorry if this is stupid i'm just not seeing it still

[coral spindle]

Nono as in, tensor product over R, not over C

So C (x)_R C rather than C (x)_C C

The former is 4d over R, and the latter is 1d over C

So if the former has a C-vector space structure, it is 2d.

(And indeed it has such a structure)

[south patrol]

I nearly mentioned in my response that the tensor product must be over R oops should've done that

[dire bramble]

Oh I see

Thank you for clarifying, I believe it makes sense now

[delicate orchid]

2+2 = 2*2

I won?

[south patrol]

this is true

I guess you can just argue via Wedderburn's theorem as a meme proof

I learnt a new proof of Wedderburn's theorem like yesterday

Well "learnt" = i read it and have partially forgotten

[delicate orchid]

I never remember anything so that's ok

[mortal folio]

sums up my degree

[hollow mica]

So basically here you use the following fact:
If F(x_1, ..., x_n) and G(x_1, ..., x_n) are homogenous polynomials in n variables and are both of degree k < n, then if the equality of polynomials F(x_1, ..., x_n) = G(x_1, ..., x_n) holds for all selections of n-k variables x_i as 0, then F = G.

You said this can be proven by induction but I already don't see how to deal with n = 2, k = 1. In that case
F(x_1, 0) = G(x_1, 0) and F(0, x_2) = G(0, x_2).

[glossy crag]

What was it?

[dull ginkgo]

How the fuck am I supposed to interpret this?

$D$ is a domain such that for each pair $(a,b) \in {D^\times}^2$ there is a pair $(x,y) \in {D^\times}^2$ such that $ax = by$
\
Define relation $\sim$ on $D \times D^\times$ where $(a,b) \sim (c,d) \Leftrightarrow \forall (x,y) \in {D^\times}^2 \left[ ax = by \Leftrightarrow cx = dy \right]$

Is that what it's referring to?

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[hollow mica]

Ok I guess we are fine with just the base case of n = 1. Now assuming the statement holds for all n less than m, in the case n = m we have that at least one of the equalities holds: F(0, x_2, ..., x_m) = G(0, x_2, ..., x_m), F(x_1, 0, x_3, ..., x_m) = G(x_1, 0, x_3, ..., x_m), ..., F(x_1, ..., x_{m-1}, 0) = G(x_1, ..., x_{m-1}, 0) by the inductive hypothesis, but I don't see how to get F = G from here. @dim widget

[dull ginkgo]

then idk what it's talking about in "independent of the choice of b_1, d_1"?

[hollow mica]

Nvm I figured it out 🙏

[dull ginkgo]

idk how to go about the first part

assuming sqrt(3) is in Q[sqrt(2)], then u^2 - 3 = 0 for some u in Q[sqrt(2)]. Every element of Q[sqrt(2)] is of the form a + sqrt(2)b, thus (a + sqrt(2))^2 - 3 = (a^2 + b^2 - 3) + 2ab sqrt(2) = 0, i.e sqrt(2) is rational. a contradiction

[hollow mica]

You can do it without using the first fact too: after a = sqrt(2)b + sqrt(3) you can just square both sides and isolate sqrt(6)

Or wait I guess this indirectly uses that

[dull ginkgo]

The next part uses something similar

these both seem like they are of the same form.

Let $R$ be a ring, $S$ a semigroup, and $I$ an ideal of $R$, then $R[S]/I[S] \cong (R/I)[S]$

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[dull ginkgo]

A bit finnicky because R[S] usually doesn't have a multiplicative identity

[hollow mica]

If we have a ring R and a subring F that also happens to be a field, isn't it true that R is a F-vector space?

[dull ginkgo]

Yes, it’s actually an associative, unital F-algebra

[hollow mica]

For a commutative unital ring R, in general R[x1, ..., xn] is not an integral domain.
But some polynomials are cancellable. Is there a characterization for them?

[crystal turtle]

Any time you have a ring homomorphism f: R --> S, any S module M becomes an R module by restriction of scalars: r•m = f(r)•m. This includes when you take M=S, which gives your example

[crystal turtle]

I don't think I've ever seen people use "nonassociative algebras" lol I'm just a naive idiot

[rocky cloak]

Well, lie algebras are quite common

[vivid tiger]

Jordan!

Malcev!

Octonions!

Composition algebras over other fields!

and indeed I've seen "nonassociative algebras" used to describe them

[crystal turtle]

I stand corrected then

[vivid tiger]

at the terminology or at if anyone uses nonassociative algebras?

The octonions are probably the most common of the things I mentioned. But I only had occasion to learn about them (and Jordan algebras) due to Lie, so perhaps general composition algebras are more common

of course Lie is king

Lie is life

Lie is god

Lie is daddy.

[fading field]

ok

[rocky cloak]

That being said in my ear "algebra = associative algebra" and if you mean nonassociative algebra you would say what (or specify further)

[crystal turtle]

I'll be honest I just didn't think people used them. But I'm also very tired and not thinking straight

[crystal vale]

Let F be the additive group of all functions from R to R. Let H be the subgroup of constant functions. Then I need to find the subgroup of F which is isomorphic to F/H.

So if I map f->f -f(c) , where c is a real number then its kernel is H so the image set will be the set of all functions which have at least one root, right?

[cobalt heath]

Nonassociative algebra is not cannon ime

Fite me

[rocky cloak]

Should be the set of functions that vanish at c.

If you just consider having an arbitrary root it won't be closed under addition.

[hollow mica]

Is there a more intuitive proof of the following fact: x_1^k + ... + x_n^k a non zero-divisor / cancellable element in R[x_1, ..., x_n], where R is a commutative unital ring.

The proof I have in mind uses that

• Monic polynomials in R[x] are cancellable
• R[x_1, ..., x_n] = (R[x_1, ..., x_{n-1}])[x_n]

[rocky cloak]

An element is cancelable if it's not a zero divisor.

So just directly calculating the multiplication with another polynomial and noticing it's not zero should be enough.

[crystal vale]

Yes, but it works for all c, right

[hollow mica]

For some reason I can't manually show that last part.

To me it looks like there's room for cancellation

Like in (x1 + x2)(x1*x2^2 - x1^2*x2), a cancellation occurs

(thought the product is still nonzero)

[rocky cloak]

Say wlog xn is one of the variables that appear in f. Then f has a term of the form xn^r times something, where r is of largest degree. Then the product has a term xn^r+k times something. Nothing else can cancel that since r was the largest degree of xn in f.

(I guess this is just writing out the proof that monic polynomials are cancelable)

[hollow mica]

Ohhh I see

Nice

If R is the ring of symmetric polynomials over Z (i.e. a subring of Z[x1, ..., xn]), then how can I show that the polynomials f_i = x1^i + ... + xn^i don't generate R ?

so like I wanna show {f_1, f_2, f_3, ...} is not a spanning set

[crystal vale]

Got it, thank you

[topaz solar]

Model theory

[chilly radish]

Don't forget poisson Algebras also!

Well technically they are part associated

[dim widget]

I think for you to understand i just have to rephrase what you're saying. So you have two linear polynomials $P(X, Y), Q(X, Y)$ which are homogenous (so they have no constant term). You know that $P(0, Y) = Q(0, Y), P(X, 0) = Q(X, 0)$. If you write P(X, Y) = aX + bY, Q(X, Y) = cX + dY do you see why they are equal?

[cloud walrus]

missa inte chansen

[hollow mica]

Yea

I also dmed you my solution earlier lol

[crystal vale]

Let U be the multiplicative group { z in C | | z | = 1 } , I need a hint for which group is isomorphic to U/<-1> ?

[summer notch]

is one-sided normality (gN subset Ng for all g in G) a weaker condition than normality? clearly they're equivalent for finite groups

i don't really know any nonabelian infinite groups besides the free group and i can't think of any counterexamples there

[topaz solar]

Well, try sketching it out

[hollow mica]

if gN ⊂ Ng then gNg^-1 ⊂ N and N ⊂ g^-1Ng. these inclusions hold for all g so ye

[summer notch]

oh my god that's literally trivial

ok i have a little egg on my face hehe

i'll blame it on it being 5 am

thanks for your help!

[hollow mica]

no problem!

it is also 5 am for me lol

almost 6

[crystal vale]

I don't know how to do it

[winged void]

I have question regarding proving that lcm(a,b)*gcd(a,b) = ab im learining some new notations to writing proves so i proved it using ord_p(a) notation

but im not really sure if its a true prove

though

[dire siren]

use the first isomorphism theorem

[winged void]

i said if a or b = 0 then its true

simple

and then i did the following

here is ggd = gcd

and kgv = lcm

but it seems a little bit not true except if its all the time {ord_{p}(a)} < ord_{p}(b)}

[dire siren]

it's fine

[winged void]

do you mean that the prove is correct

[dire siren]

yes

[winged void]

but i have feeling that its only true if all the time ord{p}{a} < ord{p}{b}

for every single prime number

[dire siren]

min{x,y}+max{x,y} is always x+y

[winged void]

how is that true can you give a concret example

[dire siren]

I mean, if x<y, then it is x+y
and if x>=y, then it is y+x

[winged void]

that is true

[dire siren]

let me just rant about the notation ord_p

that's normally the multiplicative order modulo p

and the p-adic valuation is v_p

[winged void]

that is true btw

well we did not study anything yet about the p_adic numbers

i tried to take the min or the max and i came at the same number

[dire siren]

that was not about p-adic numbers

the thing that you used in that exercise is called p-adic valuation

[winged void]

well im not really sure what does p-adic valulation means

[winged void]

i have one comment regarding this

well here the result is true only if all x > y or x < y or x = y that is true but here again its only if all x or y

but i mean for example the example i provided above well we see that the number on left and right = 6 sure

but if i pick like one time min order randomly like one time from right and then left i will get another number

[crystal vale]

I am not sure about what is G_1 such that U-> G_1 surjective with kernel <-1>

[coral spindle]

Does anyone know of a universal property that characterises the matrix ring functor? I can't put my finger on what it might be.

I don't think it's left adjoint to anything particularly obvious, which would be my first instinct.

[delicate orchid]

you can force one by joining the universal property of polynomial rings with the universal property of the quotient by describing the matrix ring as such a quotient but that seems really lame

[coral spindle]

Yeah it just feels cheap

[delicate orchid]

maybe if you actually work out which maps factor it's nice but I am not paid enough to do so

[coral spindle]

Well, if anything neat comes to mind feel free to ping me. Honestly I wouldn't be surprised if there's nothing super nifty

[dire siren]

No, it doesn't have to be always v_p(a) < v_p(b), or always v_p(a) = v_p(b), or always v_p(a) > v_p(b)
the point is that min{x,y} + max{x,y} = x+y is always true, regardless of the relation between x and y
so yeah, it might happen that v_2(a)<v_2(b), while v_3(a)>v_3(b), but that's not relevant

[dire siren]

you want to find a morphism f with kernel <-1>
then U/ker(f) is isomorphic to Im(f)

[winged void]

i see but can you give me a concret example because i really do not see how its not relevant

[dire siren]

try a = 2^3 * 3^8 * 5 * 7^6 and b = 2^5 * 3 * 7^6

[south patrol]

The proof you had in mind seems best anyway

[south patrol]

It's on the Stacks project

[winged void]

sure then min{a,b} = {3+1+0+6} max_{a,b} = 5 + 8 + 3 + 6 , v_p(a) = 3 + 8 + 1 + 6 v_p(b) = 5 + 1 + 6

[dire siren]

@winged void why are you adding them?

[winged void]

because min{x,y} + max{x,y} = x+ y

for each number

[dire siren]

I mean, "min{a,b} = {3+1+0+6}"

I don't get this

[winged void]

like i took the least power of prim between the two

2^3 is in a and 2^5 is in b so the least is 3

[dire siren]

yeah, but why are you adding them?

also, you want min{v_p(a),v_p(b)}, not min{a,b}

[winged void]

this is the definition right this is what i used to prove the question about gcd * lcm

[winged void]

that is actually what i did but i wrote not good notation indeed

[dire siren]

in the example I gave above you have
min{v_2(a),v_2(b)}=min{3,5}=3
max{v_2(a),v_2(b)}=max{3,5}=5

min{v_3(a),v_3(b)}=min{8,1}=1
max{v_3(a),v_3(b)}=max{8,1}=8

[winged void]

yes sure i found that

but how is that going to prove that min{a,b} + max{a,b} = a+b

i do not see it

[dire siren]

that was not supposed to be a proof for that

the proof is merely done by inspecting cases a<=b and a>b, as we did before with x and y

the proof uses the identity for x=v_p(a) and y=v_p(b)

[winged void]

i do not get it im not really sure

[dire siren]

if x<=y, then min{x,y}=x and max{x,y}=y
if x>y, then min{x,y}=y and max{x,y}=x

[winged void]

No, it doesn't have to be always v_p(a) < v_p(b), or always v_p(a) = v_p(b), or always v_p(a) > v_p(b)
the point is that min{x,y} + max{x,y} = x+y is always true, regardless of the relation between x and y
so yeah, it might happen that v_2(a)<v_2(b), while v_3(a)>v_3(b), but that's not relevant here you said this.

like you have mentioned that

[winged void]

exactly but then the v_2(a) < v_2(b) while v_3(a) > v_3(b) is relevant

right

[dire siren]

it's not

[winged void]

i do not know im not seeing how its not relevant

[dire siren]

because you first choose
x=v_2(a) and y=v_2(b) and apply min{x,y}+max{x,y}=x+y
and then choose
x=v_3(a) and y=v_3(b) and apply min{x,y}+max{x,y}=x+y

and the identity is true regardless of x<y, x=y or x>y... that's why I say it's not relevant

[winged void]

oh i see right now

i do not know what i was thinking

sorry

both of them are going to be choosen

either way

so order does not matter

[crystal vale]

Yes, but that's a question what is im(f)

[dire siren]

@crystal vale did you find f? it is easy to find Im(f) afterwards

[crystal vale]

No I didn't find f

[dire siren]

you want a function that sends 1 and -1 to 1

[crystal vale]

But to which group? Like what will be the structure of the Group

[dire siren]

the group was U={ z in C : |z|=1 } with multiplication

and you search f:U-->U a homomorphism

with ker(f)={-1,1}

[crystal vale]

But how do we know that f: U -> G is homomorphism with kernel<-1> then G = U ?

[dire siren]

it's not necessary for G to be U

[crystal vale]

Yes

[dire siren]

in my example I have in mind it is

[crystal vale]

So how do I find that G ?

[dire siren]

choose f(z)=z^2 and see its image

in fact it is better to think about a mapping first, and then determine the image

so f:U-->unknown

[crystal vale]

But how do I guess an unknown group?

[crystal vale]

Got it, thank you ❤️

[dire siren]

in this particular example, since U contains complex numbers, my idea was to have something like f:U-->C

[crystal vale]

Okay, thank you

[hollow mica]

Ye I was just sure a “direct” proof along the lines of jagr’s existed, I just couldn’t finish it

[dull ginkgo]

Is it easier to prove that every automorphism of a polynomial ring fixes the subring it is a poly ring over through generating sets or through the universal property

[delicate orchid]

I'm not sure I believe this. R[x_1, x_2], the map swapping x_1 and x_2

[dull ginkgo]

i MISPOKE

Fixes the ring it's a poly ring over

[delicate orchid]

try the universal property on this one

[cobalt heath]

Wdym 'fixes indeterminates'?

[dull ginkgo]

Assume $R[X_1...X_N] = P$. Let $i$ be the injection of $R \hookrightarrow P$, and $\alpha$ be an automorphism of $P$. Thus $\alpha \circ i = \beta$ is a monomorphism of $R$ into $P$. Thus there is a unique morphism $\beta^\prime$ such that $\beta^\prime(X_n) = \beta(X_n) = \alpha(X_n)$, and $\beta^\prime \circ i = \beta \circ i$

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[dull ginkgo]

idk how to show beta \circ i = i

wait

[south patrol]

I don't believe this.

[dull ginkgo]

i misstated it

and didn't feel like correcting the message

[south patrol]

yes but i mean with your updated statement

there's an automorphism of the polynomial ring R[x] over R = Z[y] given by permuting x,y

which clearly doesn't fix R

[dull ginkgo]

ah, i see

[south patrol]

the universal property is about R-algebra maps. But then your qustion would become a tautology

[delicate orchid]

yeah actually good point

[dull ginkgo]

ah true

[delicate orchid]

are these ring automorphisms or R-algebra automorphisms

[dull ginkgo]

jacobson states it for rings

There's like two different statements aaaaa

[delicate orchid]

then uhhh I don't buy it either

[dull ginkgo]

it's incorrect

ignore it

[delicate orchid]

extend any automorphism of R

[south patrol]

yeah for ring autos it's wrong, for R-alg autos it's a tautology

[dull ginkgo]

The universal property has like, two variants lmao

For rings and r-algebras

[delicate orchid]

I just think "evaluation maps"

no thoughts. head empty.

[south patrol]

uwu

[dull ginkgo]

is there like a generalized version that encapsulates both. One has associativity and unitality as a requirement, the other just needs to fix R

AAAAAAAAAAAAAAAAA

[south patrol]

one of them is about normal rings, and the other

Ugh I am confused by a statement in a paper and not even convinced this statement is true after having talked to others

[delicate orchid]

wait can I even draw the triangle from memory?

ok I can

[tranquil musk]

I wont post the full question but generally, If I have a field F and a ring R, and I need to show that there is no ring isomorphism between the two would it suffice to show that the number of units in F and R are different? I'm trying to find a proof that ring isomorphisms preserve the number of units and I cant find one

[dull marsh]

1. Yes, it does suffice
2. Have you tried proving it yourself?

[tranquil musk]

I havent give it a red hot crack yet, but I am super busy so was trying to find a source for a proof I can study. My main issue was, I am using the lack of invertibility of all elements in R to prove that there exists no ring isomorphism. 'RING isomorphism' means that the isomorphism doesnt have to preserve the properties of a field ie invertibility of all elements so using the lack of invertibility is not correct because all rings lack invertibility. Do you know what I'm saying. Perhaps I need to review the definition of ring isomorphism.

[fading basin]

Yes, because the key fact is that ring isomorphisms preserve units. If φ: A → B is a ring isomorphism between rings A and B, then φ restricts to a group isomorphism between the group of units A* of A and the group of units B* of B

[dull marsh]

I don't see how invertibility is not a part of the structure of a ring

[tranquil musk]

Yeah actually my brain was malfunctioning lol

But my main problem is now constructing/finding a proof that a ring isomorphism preserves units. Anyone know a good source

[dire bramble]

You should be able to write down a 1-liner

Consider the image of aa^-1 under a ring hom

[fading basin]

I could give you an example proof, but then I would have to write so much, and on the phone it's crap

https://dept.math.lsa.umich.edu/~kesmith/RingHomomorphism2018-ANSWERS.pdf

Maybe this will help you

[rocky cloak]

If a is a unit, then
1 = ||f(1)|| = ||f(a a^-1)|| = ||f(a) f(a^-1)||
Hence f(a) is a unit.

[tranquil musk]

what I ended up with is: consider a field F and ring R, if there exists a ring isomorphism between the two then for any b in R there exists an a in F such that f(a)=b.

Then f(1) = 1 -----> f(a a^-1) = 1 ------> f(a a^-1) = f(a) f(a^-1) -------> b b^-1 = 1. Therefore the ring isomorphism must preserve invertibility

Is this legit?

[barren sierra]

Very legit

[tranquil musk]

Thanks for the help everyone

[south patrol]

Lengthen the arrows more to improve the proof

[cloud solar]

What are the finite groups G with the property that the sum of orders of all elements ≠ e equals to the order of G

[next obsidian]

Uhhhh quite literally none other than Z/2Z

The former is bounded below by 2(|G|-1) > |G| for every value of |G| not equal to 2

[cloud solar]

Can you give me a proof please?

I was thinking about class equation

[next obsidian]

I literally gave a proof

In the next message

[celest furnace]

just beacuse every non-identity element has order at least 2, so $\sum_{g \in G \setminus e} |g| \geq \sum_{g \in G \setminus e} 2 = 2(|G| - 1)$

[cloud walrus]

Ultimate Chad

[celest furnace]

So the tightness condition is you need 2(|G|-1) = |G| so |G| = 2 and every element to have order 2, but the first condition already tells you just Z/2

[rapid junco]

a poly irreducible over Z_p for some p is irreducible over Z
is there an analogy for prime ideals

[crystal vale]

I need an example where gof is monomorphism but g not.

So if I take f: Z_3 -> Z_6 such that x-> 4x and g: Z_6 -> Z_6 such that x -> 4x then gof=f and f is injective but g is not injective.

Is it correct?

[topaz solar]

if g is injective on the image of f, the g o f is injective too

[crystal vale]

Yes, but here I need an example for g is not injective but gof is injective so is it correct?

[rapid junco]

every finitely generated and algebraic extension is necessarily finite

is this true?

can't think of a c/e

[rocky cloak]

Try thinking about the simple case first. If an extension is generated by a single algebraic element, is it then finite?

Then note that ||F(x, y) = F(x)(y)||

[rapid junco]

so this is true?

[rocky cloak]

It's true

[rapid junco]

oh duhh

degree of minimal polynomial is precisely the dimension

was overthinking

ah but calculating the dimension can get wonky, since the minimal polynomial can change at every extension

is this correct thinking @rocky cloak

[rocky cloak]

The minimal polynomial can change, but it can only get smaller. (And either way it's definitely finite)

But yes, you're thinking correctly

[long obsidian]

Hey just wanna verify if I have a ring S which is a subring of a ring R then does it follow that there is a ring of the form S/J which is a subring of R/J

When can this go wrong

[rocky cloak]

There is indeed such a ring, though you might want to describe it as S/(J \cap S), since J isn't necessarily contained in S.

You can think of it like
S -> R is a ring homomorphism and R -> R/J is a ring homomorphism, so S -> R/J is a ring homomorphism, and then the image will be a subring.

[long obsidian]

Hmmm what if I had something like k[aaa] as a subring of k[a,b,c] then considered the ideal J=<bc=1> in k[a,b,c] then does the ring k[aaa]/J exist as a subring of k[a,b,c]/J?

This is a fine subring?

[rocky cloak]

Yes, or rather J is not contain in k[a^3] but if you take the intersection with J and mod that out, you get a subring.

k[a, b, c]/J is just k[a, b^±], so the resulting subring is just k[a^3] again (because J cap k[a^3] is 0)

[summer brook]

I want to find the minimal polynomials over Q of e^{2 i pi/p} and e^{i pi / p} for primes p and odd primes p respectively. my approach would be using cyclotomic polynomials, but we haven't covered those so I'm wondering, is there another way using field theory (without galois as well)? or would it just boil down to re-proving cyclotomic polynomial properties without calling it that?

[crystal vale]

If D and D_1 are division rings such that there is a non - zero homomorphism of rings f: D->D_1 then char D = Char D_1.

So since mapping is non-zero therefore it is injective mapping.

Let Char D_1 = p then for any d in D f(p•d) = p•f(d) =0 and since it is injective p•d=0 so char D =p

Is it correct when Char D_1 is prime ?

What does it mean by the epimorphic image of a prime ideal, does it if here R to S homomorphism then that mapping is surjective?

[dull ginkgo]

The image of a prime ideal under an epimorphism

[dull ginkgo]

lol there’s epimorphisms of rings that aren’t surjective

[cobalt heath]

You can say it’s essentially surjective tho

[dull ginkgo]

I like putting the fear of god into people studying ring theory

[cobalt heath]

What was the map, again?

Z -> Q?

[dull ginkgo]

Yeah usually the prime ring maps

[crystal vale]

{0,2} is the prime ideal in Z_4, right?

[dull ginkgo]

{0,2} = (2) in Z/4Z
So (Z/4Z)/(2Z/4Z) ~= Z/2Z so yeah it’s prime

[crystal vale]

If I define mapping Z to Z/4Z then image of prime ideal of Z , im(3Z) is not prime ideal in Z/4Z, right, and also 3Z is maximal ideal so it's image is not maximal in Z/4Z, right?

[rocky cloak]

Well, the first one is a root of (x^p - 1)/(x-1), which is a cyclotomic polynomial. You can directly prove it's irreducible using ||chance of variables|| and ||Eisenstein||.

The second one is similar, being a root of (x^p + 1)/(x + 1).

[rocky cloak]

Any localization would do, which makes sense, since a localization of rings is an embedding of affine schemes.

[cobalt heath]

Yeah. Oh

[rocky cloak]

The image is everything, so yes. Not prime, not maximal.

[crystal vale]

Thank you

[long obsidian]

Thank you so much!

[coral steeple]

Are euclidean domains important in number theory and AG (the only follow-up subjects to ring theory I know of)? My introductory book doesn't define them which seems suspicious

[coral spindle]

Not really, but ig they're part of the story of algebraic number theory

[rocky cloak]

I don't think so. Euclidean domains are pretty rare, so it's not that often you need to talk about them in general.

But I guess the fact that k[x] is euclidean is important for Galois theory.

[coral steeple]

It sounds like they're at least worth defining, then. Heh

Thanks

[crystal vale]

Let f : R -> S be a monomorphism of commutative domains (here domains means no zero divisors not necessary unit exits). Let Q(R) and Q(S) be the field of fractions of R and S respectively. Show that induced map Q(f) : Q(R) -> Q(S) by a/x -> f(a)/f(x) is monomorphism. And if Q(f) is an isomorphism then f is isomorphism.

I have shown that Q(f) is well defined and monomorphism but I have not an idea how to show the isomorphism part, any idea?

I have one idea, let s≠0 in S then there exists an element in Q(R) such that f(a/x)= s^2/s, it imply that f(a)/f(x) = s^2/s which gives us sf(x) = f(a) but it doesn't give any useful results

What terminology is used for the set {a/b such that a and b are in Z and p does not b, where p is a prime number}

Is this set the same as the localisation of Z at pZ ?

[coral spindle]

Yes

[crystal vale]

Thank you, any hint for the monomorphism question?

[delicate orchid]

you've done the hard part of the monomorphism question

like, what's the obvious extension of f - that one works

maybe think about the field of fractions as the localisation away from zero if you know about the universal property of that functor

[crystal vale]

If Q is the prime ideal of R then if QR_P = R_P implies Q is not a subset of P, where R_P is localisation of R at prime ideal P.

Let Q a subset of P then since QR_P is subset of R_P but for r which is not in P then r/xr is in R_P but is not in QR_P, it contradiction.

Is it correct?

[crystal vale]

No I don't know about functor

And for Q is not a subset of P implies QR_P = R_P. Since QR_P is a subset of R_P. Now let a/x in R_P then since Q is not a subset of P there exists q in Q not in P then aq/xq in QR_P which is same as a/x so R_P is subset of QR_P.
Hence QR_P = R_P.

Is it correct?

[long obsidian]

I'm confused what is the sign of a permutation s of a finite set in Sn.

(132)=(12)(23) so I'm comfortable calling this permutation even. Is there a convention where it makes sense to assign sign(132)=1?

[delicate orchid]

literally every convention will say sign(132) = 1

[untold garnet]

There’s no convention for the signature, it’s completely canonical.

[south patrol]

How are you defining the sign of a permutation

All (standard lol) definitions agree

[rapid junco]

What questions might a rings and modules exam ask for Jordan normal form

[rocky cloak]

I guess "prove that Jordan normal form exists" might be a reasonable one

[rapid junco]

Besides computing it

Prof said he was giving us a 3x3

So I dunno what would be entailed besides computing it

[crystal vale]

?

[long obsidian]

You're exactly right. I mixed up sign with the two equivalence classes of permutations lol

[last spoke]

Trying an exercise from Rotman about group actions.

let p be a prime and X a finite G-set with |G| = p^n and p not dividing |X|. Then there is an element of x fixed by every element of G.

Proof: say $x \in X$. Then we know $|O(x)| = [G:G_x]$ and G is finite, so $|O(x)| \cdot |G_x| = p^n$ . This means the orbit and stabilizer of x both have orders being powers of p. Since x was arbitrary and the set of orbits ${O_j}$ partitions X, we can write $|X| = \sum O_j$. However, by assumption p doesn’t divide the order of X so one of these orbits must have order 1, and the claim is satisfied.

Right?

[cloud walrus]

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

[last spoke]

What’s causing the italics 🤔

[dull ginkgo]

RAAAAAHHHHH SYLOW

[dull ginkgo]

accidental mathmode

[last spoke]

Sylow theorems haven’t been introduced yet

[dull ginkgo]

it's related

actually a part of sylow

anyway

yeah that proof's right

[last spoke]

💪

[dull ginkgo]

This I think is a part of proving Sylow 2?

well a related notion using basically the exact same proof

[stark helm]

I am wondering if |Gal(Q(2^1/4 i)/Q)|=4, I find it by irreducible polynomial x^4-2 over Q, but can not find exact 4 elements in this galois group, I only figure out identity and 2^1/4 i ->-2^1/4 i, what about the other two?

[dull ginkgo]

2^1/4 and -2^1/4 i think

Cyclic extension

2^1/4 {1, i, -1, -i}

[topaz solar]

Yes

Italics from the p^n in your quote

[stark helm]

2^1/4 is not in Q( 2^1/4 i) right, why it can be mapped?

[stark helm]

and also, do you mean 2^1/4 i map to these four elements?

[whole basalt]

Hello! I'm planning an activity for my undergrad abstract algebra students and I'd like to make sure I have some interesting examples.

I have this thing I introduce students to called the "algebra game", which is where you take a given number and your goal is to get to zero using the following operations:

• Add, subtract, or multiply by an integer other than zero
• Raise to a positive integer power
• Store or retrieve a result

The idea is that doing this generates a polynomial in Z[x] that has that number as a root, hence showing that the number is algebraic. For example, if they're given √2 + √3, they could square it, subtract 5, square it again, and subtract 24 to get to 0, which shows that √2 + √3 is a root of (x² - 5)² - 24.

However, I've previously had trouble when teaching this lesson with numbers with mixed roots such as √2 + ³√2. How might one generate a polynomial with that as a root? Do I need to add rules to my game to make that work?

[delicate orchid]

nope, that'll have a minimal polynomial over Q - and thus over Z by just multiplying out denominators

I'll check what it is

x^6-6x^4-4x^3+12x^2-24x-4 apparently

[topaz solar]

Is that reachable with those operations though?

[whole basalt]

If you have store and retrieve yeah I guess

But it would be nicer if it was more akin to what I did with √2 + √3

[topaz solar]

The store and retrieve are very powerful

[whole basalt]

Do this, do that, get things to cancel out

I would prefer to not have to use store and retrieve

But I don't know if I can do without it

[topaz solar]

But also that’ll make a disgusting expression because degree 6

[delicate orchid]

yeah I'm not sure how you'd do it with a single value

the fact that we have coprime exponents on the x's makes me doubt it can be done

[topaz solar]

Yeah idk if it’s possible without memory

maybe starting with x^6 stuff could get you somewhere, but that’s also abhorrent and you’ll certainly not get the minimal poly

[whole basalt]

Any suggestions for other modifications to the "game" that could make it work?

Actually, come to think of it, Horner's method should show that technically it IS possible with just a single number

[delicate orchid]

which one's that

[whole basalt]

So perhaps multiplying by the original number could be allowed as an operation

Synthetic division

[delicate orchid]

I'll google

[whole basalt]

Take a look at the very end!

[delicate orchid]

ahhh those geezers

[topaz solar]

Disgusting

[delicate orchid]

you'd need to store the x though wouldn't you?

[topaz solar]

This does use remembering x yes

[whole basalt]

Yeah, but just the x

[delicate orchid]

true, a big improvement

[whole basalt]

Which is just restrictive enough to be interesting

[delicate orchid]

it's actually really computationally efficient if you need to evalaute some fixed polynomial loads of times

[topaz solar]

That is a significant weakening but still

[topaz solar]

I do not base my aesthetic standards on mere efficiency

[delicate orchid]

it is aesthetic lil nerd

[topaz solar]

Ok fusion system enjoyer

[delicate orchid]

:itsmid:... just take the L lil bro...

[south patrol]

Gm

[whole basalt]

Okay that's probably good enough :V
Thanks for letting me hash some of this out

[topaz solar]

Troll them with a hard to recognize transcendental

[whole basalt]

Oh I'm gonna give them π at the end. XD

[topaz solar]

Nah that’s too common

[whole basalt]

For you maybe!

[topaz solar]

Root 2 ^ root 2 tho

That “looks” algebraic

[whole basalt]

I doubt most of my students know what a transcendental number is

It's not commonly taught before this point tbh

[topaz solar]

Ok fair, but numberphile and such online things often talk about pi trivia

[delicate orchid]

it is in any... saved by the bell 🔔

[whole basalt]

Or at least if it is, it's like a footnote of a footnote

[topaz solar]

Ye, I wouldn’t expect it to be taught, but I would expect them to have heard about it

[south patrol]

uwu

Tbh like

I've not encountered transcendental numbers at all in ug curriculum lol

Besides by omission in talking about algebraic numbers

[delicate orchid]

did you not do galois theory

[topaz solar]

That’s because we know like 4 theorems about them

[delicate orchid]

saved by the bell 🔔

[south patrol]

Lol

Algebraic numbers of course have come up tons

But like not a single proved example of a transcendental number has come up lol

[whole basalt]

We're gonna try to get a glimpse of Galois theory at the end

[south patrol]

Though it is very cool

[topaz solar]

Nice

[whole basalt]

I'm trying to work up to just like a baby version of it

[topaz solar]

Throw them into the fire

We invoking the Kim-Pillay Galois group

[delicate orchid]

moderators he is making things up again

[south patrol]

Witt vectors and Artin-Schreier-Witt theory

Introduce Grothendieck galois theory first

[topaz solar]

the nG self react

[delicate orchid]

introduce galois cohomology first

[south patrol]

Yeah

[south patrol]

That's a special case if u do étale cohomology

[delicate orchid]

yes I know it is

but I don't know what etale cohomology IS

[coral steeple]

Can I get a hint on how to generalise CRT for arbitrary commutative rings from two ideals to n? I can't show that for ideals I,J,K, pairwise coprime, we have IJK is contained in contains IJ cap K. I need this to show R/(I cap J cap K)=R/(IJK). (I used n=3 here to be concise)

[whole basalt]

Also just to make sure... $\bQ(\sqrt2+\sqrt3)=\bQ(\sqrt2,\sqrt3)$ right?

[cloud walrus]

DM Ashura

[whole basalt]

Clearly including both √2 and √3 gives you √2+√3

[delicate orchid]

I think this is correct

you can get a sqrt(6) by just squaring sqrt(2)+sqrt(3) and subtracting some rational junk

then do sqrt(2)+sqrt(3)-sqrt(6) to get a sqrt(2)

ah but there's a root(3) still in there

ah wait, sqrt(6)/(sqrt(2)+sqrt(3)) = sqrt(6)*(sqrt(2)-sqrt(3))/ -1 = uhhh

3sqrt(2)-2sqrt(3)?

[whole basalt]

Oh riiiight I forgot about dividing!

Dividing will give you a √2 - √3 with some coefficient

That does it!

[delicate orchid]

then add sqrt(2)+sqrt(3) until u only get one of the mfs

[whole basalt]

Yup, perfect

[delicate orchid]

nice that this should generalise to Q(sqrt(n+m)) for coprime m, n - I think

I can't do the bezout's lemma in my head

[topaz solar]

Alternatively, Q(sqrt 2 + sqrt 3) is a subfield of Q(root 2, root 3), and is degree 4

And the whole [A:C]=[A:B][B:C] thing

[delicate orchid]

it's not obvious to me why it's degree 4 so I ignore such statements

[topaz solar]

Well, minimal poly is degree 4

[delicate orchid]

no shit sherlock?

I'm saying I can't conjure such a polynomial in my head

https://tenor.com/view/fila-shoes-funny-gif-20527988

[topaz solar]

True

However

Proof by WA said so

[delicate orchid]

true!

[whole basalt]

CONGRATS BOYS

WE'VE PROVEN π IS ALGEBRAIC

[south patrol]

yeah

I wonder why it is so close to π

[rapid junco]

is this def wrong

cuz that means every element is independent

[boreal inlet]

LMAO

[topaz solar]

no

how does it mean that?

[rapid junco]

nvmnvm

[topaz solar]

ye, do note that {y} is independent as long as ry = 0 iff r = 0

but that's kinda degenerate obviously

[rapid junco]

wait why

[topaz solar]

well, ry = 0 -> r = 0

[rapid junco]

what if R is not an integral domain and you look at R as a R module over itself

[topaz solar]

therefore

[topaz solar]

then some single elements will suck in this regard

having n independent elements means you have a submodule of R^n, does that make sense?

(assuming unital whatever idc, if your ring isn't unital seek professional help)

[rapid junco]

lmao

[topaz solar]

do you see why

[rapid junco]

no

your saying those n elements form a submodule?

of R^n?

I don't really follow the statement itself

[topaz solar]

I'm saying they generate a submodule of M that's just R^n

[rapid junco]

Oh I see

uhh lemme think

R / Ann(m) iso Rm

[fading field]

shrubmodule

[rapid junco]

but Ann(m) is 0

nope nvm

oh wait yeah it is

Is this the start?

[fading field]

take Z/4 as a Z-module, 1 + 3 = 0, so 1 and 3 are not independent

also 1 is not independent since $4 \cdot 1$ is 0 (ignore, i misread the message and texit won't let me remove)

[cloud walrus]

|smay⟩

[fading field]

for example

unrelated i like ur banner

oh wait

oh my god i misread your definition, and then thought it was wrong, and then read the rest of your conversation

[fading field]

wait sharp what are you saying here

this doesn't match with the definition

r and y can be nonzero

and ry can be zero

in M an R-module

[topaz solar]

yeah it does? if ry = 0, then r needs to be 0?

[fading field]

no it doesn't

[topaz solar]

ah wait

they said ry = 0 not the coefficients 0

mfw reading comprehension

[fading field]

ack

[topaz solar]

They say a_i y_i = 0 in the image not a_i = 0

[topaz solar]

if it's just a_i = 0 then r = 0 iff ry = 0 tho yeah?

[fading field]

OKAY, i think i get where the def is coming from.

this is kind of dirty, yes it suggests that every singleton is independent

[topaz solar]

the evil vile clutches of zero divisors

Which since I have realized I'm blind, ry = 0 does indeed imply ry = 0 so {y} is independent for y =/= 0

[rapid junco]

on a domain right

not in a general ring

[fading field]

note that this doesn't recover linear independence for vector spaces since if you have a field, and M is a vector space, then {0} is independent but not LI

[rapid junco]

oh wtf lol

[topaz solar]

no it says y_i =/= 0

[fading field]

oh okay

then it does

[topaz solar]

yep

[fading field]

gerat

okay, then actually this definition is fine. you can think of examples, just use your intuition about FG abelian groups and vector spaces

where if you have two things in the same summand you won't get independence

[topaz solar]

I am blind for not noticing the a_i y_i = 0 rather than a_i = 0, but yeah it's just counting generators (So <y_1, .. y_n> = R^n/something qotienting each coordinate separately). If you require exactly a_i = 0 and R is unital, this asks for a copy of R^n rather than merely a quotient

[fading field]

you really do recover linear independence from this definition when R is a field

[fading field]

you can do this for general rings

[rapid junco]

yeah i guess it is a tautology

if y_ia_i = 0 then y_ia_i = 0

[topaz solar]

indeed

[fading field]

yeah and you should note that singletons really are independent

and you want this

[topaz solar]

if you have something like A x B, then a in A and b in B nonzero has {a, b} independent

[rapid junco]

how do you see then it generated R^n

[topaz solar]

well, the idea is that if \sum a_i y_i = 0 implies a_i = 0, \sum Ry_i = \bigoplus Ry_i, and each Ry_i ~= R as an R-module

If we just know each a_i y_i = 0, then the submodule generated by the y_i is still \bigoplus Ry_i

and, as you said earlier, is R/Ann(m) or wtv

[crystal vale]

They define local homomorphism between local rings such as non-unit maps to non-unit.

I want an example, homomorphism between local rings which is not local homomorphism.
Any hint?

If I define mapping from Z to Z such that x maps to 2x then it is not homomorphism but it is ring homomorphism when we consider Z as Lie ring, right?

[rapid junco]

is there a nice way to classify all prime ideals in Z[x]?

using projection map Z[x] to Z[x]/(x)

and using the fact for a surjective ring hom the preimage of an ideal is an ideal.

prime ideal is prime ideal

preimage^

[crystal vale]

For preimage of an ideal is ideal, surjective is not necessary right?

[rapid junco]

for a surjectvie map there is a bijection between those prime ideals in the domain which contain the kernel and those prime ideals in the image

[sterile garden]

Given integers $a,b$ not necessarily coprime, I'm trying to construct the formula $\varphi(ab) = \varphi(a)\varphi(b) \frac{\gcd(a,b)}{\varphi(\gcd(a,b))}$ by constructing an exact sequence
$$
0 \to \mathbb{Z}/\gcd(a,b)\mathbb{Z} \to U_{ab} \to U_a \times U_b \to U_{\gcd(a,b)} \to 0.
$$
I've got the map $f: U_{ab} \to U_a \times U_b$ by $x \mapsto (x \mod a, x \mod b)$ and I've shown that the kernel of this map is indeed the non-trivial group on the left. I'm trying to show the cokernel is indeed the non-trivial group on the right. I'm kind of stuck though. I was trying to show the image of $f$ ends up in the kernel of the map to $U_{\gcd(a,b)}$, but I'm not sure how to do that.

[cloud walrus]

IAmDerek

[sterile garden]

Worth mentioning, the map I found on the left is $x \mapsto 1 + lcm(a,b)x$

[cloud walrus]

IAmDerek

[sterile garden]

And I think the map on the right is just (x,y) -> xy mod gcd(a,b)

[rocky cloak]

I think the usual way is to think of preimages over the inclusion
Z -> Z[x]

For any prime P, P \cap Z is a prime ideal. If it's equal to (p) for some p, then this corresponds to an ideal in (Z/p)[x]. So this is either just the ideal (p) or (p) plus a polynomial irreducible modulo p.

If the intersection with Z is (0), then the ideal doesn't contain Z\{0}, so corresponds to an ideal in the localization Q[x], so is either 0, or an irreducible polynomial in Q[x].

[dim widget]

This only sees some of them

For any UFD R the point is that prime ideals in R[X] are (like jagr has said) either (p) for p a prime of R (you could call these horizontal ideals) (q(x)) an irreducible polynomial in R[X] (vertical prime ideals), or a combination (p, q(x)) with q(x) irreducible mod p (the closed points of Spec(R[X]))

That all maximal ideals containing a nontrivial prime ideal of R are of the third type is obvious since R/p[X] is a Euclidean domain

Then one can classify the rest by looking at the prime ideals of Frac(R)[X] using that R[X] is a UFD

[crystal vale]

Inverse image of a prime ideal is a prime ideal in R if both R and S are commutative, how can I show that inverse image of prime ideal is proper ideal of R?

[coral spindle]

If you tell me your rings aren't unital I'm gonna have an aneurysm

Anyway, just look at 1.

[crystal vale]

No it's not given that it has unity

[coral spindle]

AGgaghaghga the pain... the pain in my brian ahdhdad

Anyway, then it's obviously false. Look at the inclusion 2Z into Z.

Wow it's almost as if nonunital rings suck

[delicate orchid]

Switch textbooks

[crystal vale]

I can't

[delicate orchid]

Wasn’t a question.

[crystal vale]

Yes 2Z is prime in Z but not in 2Z, right?

[coral spindle]

Last I checked

[languid trellis]

2Z isn't a ring so we can't have ideals surely

[analog sedge]

Do you think abstract algebra is weirdly placed? Give 5-yo kids some tokens and ask them to make addition/multiplication table with those tokens, then maybe add some more rules over time, then I think they could make groups/fields/etc.

[delicate orchid]

it's a non-unital ring

[rocky cloak]

I'm sure they could, but you're framing this as though the main goal of the education system is to teach people abstract algebra.

As an early university course you reach the people that want/need to learn it, and you still reach them quite early.

[coral spindle]

It’s a nonunital ring, which notknow is somehow pursuing.

And it is increasing human suffering across the globe

[analog sedge]

It's not about teaching people AA, it's about how the subject seems to be quite orthogonal to subjects taught in grade schools and so on... but well... you could argue quite the same about algorithm courses.

[rocky cloak]

Yeah, there are lots of subjects that are fairly independent of what you learn in grade school, but I'm not sure that's really counts as an argument for introducing that topic in grade school.

[analog sedge]

Perhaps the argument might be that learning them can kinda improve abstract reasoning skill?

[languid trellis]

Well the issue is that I don't believe in nonunital rings.

[languid trellis]

They don't exist

[coral spindle]

You and me both

[rocky cloak]

nonunital rings exist, they're just not rings

I'm on the fence about rings with enough idempotents.

[delicate orchid]

u can never have enough

[rapid junco]

lol

[untold garnet]

We tried this a few decades ago. Apparently, it worked for very few people.

[south patrol]

what does this mean

[rocky cloak]

Like I'm not sure if rings with enough idempotents quite qualify to being called rings. But they're up there.

And a ring has enough idempotents if there are orthogonal idempotents with R = directsum Re

[south patrol]

Ah okay thanks that was going to be my question

so for a commutative ring is this just the same as having a non-trivial idempotent

since like R = Re (+) R(1-e)

[rocky cloak]

No, like it's a weaker version of being unital

[south patrol]

Oh wait orthogonal

[rocky cloak]

If you have 1, then R = R1

[south patrol]

Oh lol

i assumed you meant non-trivial idempotents

fair enough

[next obsidian]

I’m a chmonkeyyyyyyyy

:3

I just… don’t think about what channel I’m in when I type tbh

[crystal turtle]

:3

[summer path]

:3

is that how chmonkey sent the uwu gif in adv lounge then insta deleted it

[dim widget]

in French and some English schools basic abstract algebra is part of the grade school curriculum if you specialize in math.

[low wyvern]

I wasn't aware that we could take aa at A level 😅

[dim widget]

I thought some people did! maybe im wrong

[low wyvern]

People in this discord are doing aa at GCSE

[delicate orchid]

there are three main exam boards in the UK, and it's only in one (maybe two?) of them

so it's not exactly common

[dim widget]

ah okay I don't actually know that much about the education system in the sceptered isle

I went to the airport in manchester once

one of the darkest times in my life

[delicate orchid]

I'm very sorry to hear that

[coral spindle]

Shite airport tbh

[low wyvern]

I'd say it is one of our better ones tbh.

[delicate orchid]

glorious newcastle airport

[dull ginkgo]

You guys are able to leave that shithole?

[delicate orchid]

newcastle is the shining jewel of europe

[coral steeple]

Polynomials over a PID don't necessarily form a PID because, for example, (2,x) is not principal in Z[x], right?

[dull ginkgo]

Yes

[coral steeple]

Thanks

I have my final later today so I will likely be asking lots of stupid questions here till then out of nervousness

[delicate orchid]

you say stupid questions, I say easy answers to boost my social credit score in Jagr2808's algebraic utopia

[dim widget]

they are never a PID unless the PID you start with is a field

[coral steeple]

This is good to know

[south patrol]

The same proof works

If you replace 2 by any non-unit

[languid trellis]

newcastle has, unfortunately, a very nice airport

[dim widget]

Newcastle has become surprisingly pretty

a lot of leveling up money

[delicate orchid]

there's a wetherspoons past security iirc

[dim widget]

or just public money in general

[languid trellis]

as long as you stay inside the city centre

[south patrol]

Pog

[languid trellis]

least alcoholic britbonger

[south patrol]

Not sure if I've ever been to Newcastle lol

[languid trellis]

don't bother

[dim widget]

even just more generally compared to the stories I've heard from many years ago

[delicate orchid]

even some places outside of the centre are nice, but nothing south of the tyne lol

[languid trellis]

You're probably right, but there is a stark contrast between the city centre and everywhere else, especially gateshead lmfao

[coral steeple]

The non-unit condition is necessary because otherwise you could divide with remainder to get a gcd, which would be a generator for the ideal, right?

[south patrol]

Well if it is a unit you get the whole ring

[languid trellis]

NOTHING south of the tyne is pretty huh?

[delicate orchid]

YUP

[languid trellis]

you're about to get so owned. what about bath huh?

[delicate orchid]

"baf" more like barf hahahah

[south patrol]

Like if u is a unit of R then (u,x) contains R and x hence R[x]

[languid trellis]

let's not forget that you pronounce bath like a southerner. you are an impostor

[coral steeple]

Sorry yes my message wasn't any good

[south patrol]

Dw

[delicate orchid]

I don't, where have you gotten this idea from

[languid trellis]

i've literally heard you say it

[south patrol]

There's also the fun fact that if R is noetherian then dim R[x] = 1 + dim R

[coral steeple]

What I meant was that if you take (f,g) with f,g of degree >= 1 you can divide with remainder to get a gcd and the ideal is principal

[delicate orchid]

I say all of my "-ath"s northern

[south patrol]

Hm not sure what you mean

Why would you be able to divide like that

[dull ginkgo]

Is this Krull dim?

[delicate orchid]

yur

[south patrol]

Yh

[languid trellis]

afaik we only have unique division with remainder in the case F[x], f a field

[delicate orchid]

I think it holds for R[x_1, ..., x_n] as well u just replace 1 with n

although that might just be fields

[south patrol]

Well that follows from what we're doing

[coral steeple]

Yeah I may be confused

[south patrol]

Sure but that immediately follows by induction

[coral steeple]

I probably was thinking of F[x]

[south patrol]

Dw

[delicate orchid]

oh yeah lol

R[x_1,x_2] = R[x_1][x_2] moment

[south patrol]

Stuck on some AG rn ugh

[languid trellis]

what a surprise

[delicate orchid]

all euclidean domains are principal ideal domains by your exact reasoning

ironically, your assertion works if a, b are degree 1

[delicate orchid]

same but alg top

[coral steeple]

Wait isn't this true

[languid trellis]

yes

[delicate orchid]

yes, it's true - that's why I said it

[coral steeple]

But I thought my reasoning was flawed

[delicate orchid]

it is

because Z[x] isn't a PID

[barren sierra]

Society if it was:

[delicate orchid]

actually @south patrol while we're here, why is $H^p(S^{2n+1}, H^q(U(n); \bZ)) \cong H^p(S^{2n+1}; \bZ) \otimes H^q(U(n); \bZ)$? My source just says it's because $S^{2n+1}$ is simply connected but I don't get it!

[cloud walrus]

Wew Lads Tbh

[south patrol]

Universal coefficient theorem

[delicate orchid]

oh

duh

[south patrol]

I don't really know why simply connected is relevant there, rather it is because the stuff is free

[delicate orchid]

I knew it was going to be something obvious

[south patrol]

Are you doing a serre spectral sequence

For the fibration with the U(n)

[delicate orchid]

there's a H inside of a H

yeah it's an exercise in this book

[south patrol]

Nice

[delicate orchid]

and that was the hint

[south patrol]

Yeah like this is always what u want to be the case for the spectral sequence I'd say lol

Fortunately I mean

Cohomology of the sphere is simple enough that you can just do this by hand though without UCT

[dim widget]

what is U(n)

[south patrol]

the topological group

[dim widget]

the relevance of simply connectedness is that you want to know that the cohomology of the fibers of what you are considering is a constant local system on the base

it has nothing to do with the isomorphism you are asking about

[delicate orchid]

oh ok

[dim widget]

You are presumably trying to compute the cohomology of Gln

by induction

[delicate orchid]

yur

not GL though just U(n)

[south patrol]

They are homotopy equivalent

[delicate orchid]

I don't believe it!!!!

oh wait yeah I do

[dim widget]

mfs when KAN

[south patrol]

As witnessed by uh

[dim widget]

you KAN understand this

[south patrol]

Gram Schmidt

[delicate orchid]

as witnessed by uhhh something something equivalence of representations something something homotopy class of maps out of BU(n) something something yoneda

[south patrol]

Lol

No it is just that Gram schmidt is a continuous ting

Gives u a deformation retraction

[delicate orchid]

polynomials are NOT continous... they would never...

[dim widget]

The point is that if you take the subgroup $B_+$ of matrices which are upper triangular with $R_{> 0}$ on the diagonal then $U(n) \cdot B_+$ is all of $Gl_n$

as Potato is saying this is a fancy rephrasing of Graham Schmidt

[rocky cloak]

Are you implying that this isn't true if R is not Noetherian? And do you have an example of that? Sounds pretty sick.

[cloud walrus]

Math_Discord_Final_Girl

[south patrol]

Uhh not off the top of my head but tteg will knowi m sure

[delicate orchid]

it's going to be something stupid with infinite krull dimension

[dim widget]

yes indeed

[south patrol]

Well that'd be pointless right lol

[delicate orchid]

counter example nevertheless

[south patrol]

Not under the convention oo + 1 = oo

[delicate orchid]

unless there's noetherian rings with infinite krull dimension

[south patrol]

P sure there are fin dim counterexamples

[topaz solar]

Counterexampletheless

[rocky cloak]

But if you have infinite Krull dimension then it's obviously true

[south patrol]

There are

[delicate orchid]

there better not be or I'm muting this channel

ohhhhhhhhh

it's over

[dim widget]

Im actyually struggling to come up with a counterexample but I know this is a known issue

[topaz solar]

Ordinal valued krull dimension

[dim widget]

probably it has to be non-excellent

[south patrol]

Oh yeah this is fun

So I think u always have a bound like

dim A + 1 <= dim A[x] <= 1 + 2 dim A

[delicate orchid]

harsh...

[rocky cloak]

There's one here
https://math.stackexchange.com/q/1267419/306319

[south patrol]

And it's been proven you can get examples with any number in between

[south patrol]

Yeah

[dim widget]

I see

quite ugly

[delicate orchid]

the counter example is k(t)[[y]][x]

I'm gonna do PDEs now

[coral steeple]

Is the content of the third iso theorem for rings the fact that I can write ideals of R/I as {j+I: j in some ideal J of R which contains I}?

What I should ask is, "is that what ideals of R/I look like"

[languid trellis]

My understanding is no author agrees on which theorem is which number iso theorem. But yes, you're right. Consider a surjective ring hom R-> R'. By the first iso theorem, there is an iso between R/K and R' (K the kernel). By what you call the third iso theorem (some call it the correspondence theorem), there is a bijection between the set of ideals of R/K and R' given by the induced map from the first iso theorem. In fact, there is an isomorphism between R/K and R'/K' (K' is the image of K under the induced map). We can extend this result to any ideal I of R containing K.

[rocky cloak]

Yes, all the ideals are like that.

[delicate orchid]

bro was SO eager to use the sweet jacobson knowledge

except that's not the third iso theorem, that's the corrispondence theorem! better luck next time sweaty...

[languid trellis]

I was only doing it in the egoistic pursuit of reinforcing my own understanding

[languid trellis]

um. someone can't read

[coral steeple]

This makes sense mostly, thanks. In fact I was trying to grok this yesterday out of Jacobson, but I don't have as much time as I want to really be comfortable with it

[languid trellis]

I don't think Jacobson's exposition of this theorem is too amazing to be honest, I only really understood the result in the rings section after spending an hour reviewing the correspondence theorem for groups lol

[coral steeple]

That's what I did 😭

But I didn't properly study the group section, just skimmed it to make sense of when he called back to it

(yay, rings-first algebra courses)

[languid trellis]

I'm sure your uni knows what its doing

maybe

[coral steeple]

It's apparently because a lot of the students in the course are cs majors who are unused to proofs/abstraction, and need lots of concrete examples (hence, not groups)