groups-rings-fields

[tough raven]

Distributive congruence lattice?

[swift root]

congruence permutable

[tough raven]

And what's that?

[frail shoal]

this one is slightly annoying cuz the easy to remember version is also only a one-way implication, i.e. if there exists a term M such that M(x, x, y) = M(x, y, x) = M(y, x, x) = x, then congruences distribute. but not conversely

there is a way to fix this. that i forgot

[swift root]

Two binary relations R, S on a set commute iff R ∘ S = S ∘ R, ∘ being the relational product.

V is congruence-permutable if for every A ∈ V, every pair of congruences on A commute. The Malcev condition for this is a term m(x, y, z) satisfying m(x, x, z) = z and m(x, z, z) = x

this is the first one discovered, by Malcev

[tough raven]

Oh yes.

[swift root]

and it is by far one of the most powerful ones

[tough raven]

OK, now explain this.

[swift root]

it is also equivalent to every tolerance (internal symmetric reflective relation) being a congruence

[swift root]

do you know about interpretations of varieties?

[swift root]

Jones terms

you can also characterize it as the commutator being equal to the intersection

[swift root]

Basically, think of how you can interpret groups as magmas in multiple different ways simply by choosing different binary terms.

For example, choosing the term c(x, y) = xyx^-1 interprets groups into the variety of left-distributive magmas

[tough raven]

Oh

OK sure

[swift root]

k-algebras can be interpreted into lie algebras by choosing for [x, y] the commutator

[tough raven]

Literally any binary term

[swift root]

yes because magmas require no condition to be satisfied

[tough raven]

Sure, sure.

[tough raven]

Wait that's kind of crazy

[swift root]

it is! and there higher dimensional analogues of this that pop up in commutator theory

[tough raven]

OK but semigroups

[tough raven]

Man I want to ask more

[swift root]

higher dimensional commutators

[swift root]

So, if W satisfies some Malcev condition, i.e. has a collection of terms { ti } that satisfy some equations, and V interprets into W, then by definition V also has a collection of terms { ti } satisfying those same equations.

A concrete example: groups have the term m(x, y, z) = xy^-1z which satisfies the identities for a congruence permutable variety. Consequently, any variety that interprets into the variety of groups must also be congruence permutable

even more general is for example quasigroups

[frail shoal]

i guess a malcev condition is essentially "V interprets at least one of a family of varieties"

and intepretation is transitive

[swift root]

Clearly every variety interprets (uniquely) into Set; this just forgets all structure. Thus, any Malcev condition that doesnt hold in every variety doesnt hold in Set.
So if Set interprets into a variety, by transitivity it follows that this variety cannot satisfy any nontrivial Malcev conditions

[frail shoal]

i.e. there is a variety of sets equipped w/ a malcev term

[tough raven]

Ahhhh

OK very strange that Set interprets into Semigroup now

But not Monoid hmmmm

[swift root]

Now note, that this doesnt mean the variety has no structure, its just that there often will be a Set-like behavior

[swift root]

its the fact you need a nullary operation

[frail shoal]

yea actually Set can literally only interpret n-ary operations in n distinct ways

[swift root]

Set does not have nullary operations, but pointed sets do interpret into monoids

[tough raven]

Of course.

I suppose one could enumerate all term equations that hold of the projections

And get the universal variety for which there is a natural structure on all sets.

Amusing.

[frail shoal]

wait hold on that sounds really funny

[tough raven]

smth smth operads

[frail shoal]

it's kinda funny that we can view interpretation as putting a co-W structure on F_V(1)

where V and W are varieties and F_V is the free functor

[tough raven]

Oh lol

This is why k[t] is a coring object

[frail shoal]

(i doubt this perspective is particularly useful but i like collecting funny facts like this)

[swift root]

interpretations the cleanest way to see them is using clones

[tough raven]

Category of interpretations hmmm

Compose co-W structure on Free_V and co-V structure on Free_U into co-W structure on Free_U

[frail shoal]

hm, so for instance, x(yz) = (xy)z is valid, because if you only look at the leftmost or rightmost variables, they're equivalent

[tough raven]

Oh well

Consider the variety with an n-ary p_{n,i} for all i in [n]

[frail shoal]

ohhhh

wait that makes more sense than what was in my head

i get ur idea better now

[tough raven]

and equations <composition of p_n,i's> = p_{do the right thing}

[frail shoal]

wait no i think that becomes trivial

[tough raven]

and that's all possible operations and equations you could ever naturally put on all sets

[frail shoal]

cuz you can just write down that these operations are projections

[tough raven]

How?

Ah OK

[frail shoal]

p_{2, 0}(x, y) = x

ah you realize

[tough raven]

IG first-order language implicitly has projections

[swift root]

yes

[tough raven]

But restrict to equations with the same variables on both sides

You'll get a more interesting variety

[frail shoal]

the idea i had was, we take a binary operation, and demand that its equations satisfy both the theory of the left projection AND the right projection

yea

it's kinda arbitrary admittedly

but it's funny at least

[tough raven]

I think if we include this constraint on the variety then the interpretation of Set is an equivalence

i.e. it is "Morita-equivalent" to Set

[frail shoal]

yea

[tough raven]

This shouldn't be though.

[frail shoal]

hm we get xyz = xwz

actually we might get something equivalent to Set^2

these look like rectangular bands

[tough raven]

Which variety?

[frail shoal]

a band is an idempotent semigroup

[tough raven]

OK but

what did you start with?

[frail shoal]

one binary operation

[tough raven]

Oh this

Ah, you're intersecting the axioms.

[frail shoal]

yea

[swift root]

does this have to do with factor bands?

[frail shoal]

idk

[next obsidian]

Can u factor into your calculation how many bands I got?

https://tenor.com/view/tana-money-spread-gif-11512930832744462476

[quiet pelican]

0 doesn’t affect the calculation

[next obsidian]

It does if u multiply….

[swift root]

copium

[next obsidian]

U r just denying facts and logic to protect ur feelings

[swift root]

i would never

[maiden crater]

Determine the minimal polynomial over Q for the element 1+ i .

I was thinking x^4+4=0, so it now remains to show its irreducibility

[quiet pelican]

Is this meant to be 1+i?

[quiet pelican]

This polynomial is not irreducible

I don’t think

[maiden crater]

right I just realised

wait

[quiet pelican]

I suggest adding 4x^2 and subtracting 4x^2, then applying difference of squares

[maiden crater]

I'm just going through various tests in my head

it fails Eisenstein

RRT is useless here

hmm

is there any test I'm forgetting

[maiden crater]

mhm

[quiet pelican]

(I’m kinda cheating here because I know enough Galois theory to just immediately know what the min poly should be)

[maiden crater]

using basic ring/field theory is there a standard way

this may help

[quiet pelican]

So x^4 + 4 = x^4 + 4x^2 + 4 - 4x^2 = (x^2 + 2)^2 - (2x)^2 = (x^2 + 2x + 2)(x^2 - 2x +2)
At one point you see this factorisation trick, and remember how to abuse it forever

[maiden crater]

right

got it

[quiet pelican]

But also if I were trying to manually compute the min poly of 1+i what I’d do is like
x = 1 + i
x - 1 = i (get the rational parts to one side)
(x - 1)^2 = -1
and then rearrange

[elfin wraith]

Complete the square and difference of squares in one go? Somewhere out there a GCSE teacher just got really excited

[maiden crater]

Here, would it be wrong to say both are of degree 2

Lemme supply formal proofs

$F(2+\sqrt{3}) \cong F[x]/(m_{\alpha}(x))$
\ From which it follows $[F(\alpha):F] = deg(m_{\alpha}(x)) = deg(\alpha)$.
\ We propose $1, \sqrt{3}$ to be a spanning set. It's also a basis.

[cloud walrus]

Wai

[maiden crater]

Thus the degree is $2$

[cloud walrus]

Wai

[maiden crater]

As for $1+ \sqrt[3]{2}+\sqrt[3]{4}$ We propose $1, \sqrt[3]{2}, $\sqrt[3]{4}$ to be a basis thus making it an extension of degree $3$

[cloud walrus]

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

[rocky cloak]

Do you know about how degree of field extensions is multiplicative?

[maiden crater]

As in if one is a subfield of other

that theorm?

I have not yet reached there in my revision

[rocky cloak]

Yes

In particular here is relevant that
1 + cbrt2 + cbrt4 is an element of Q(cbrt2)

[maiden crater]

so it's an extension of degree 2?

[elfin wraith]

No

whats the minimal polynomial of cbrt 2

[maiden crater]

x^3-2

right

got ot

thanks

[elfin wraith]

(This counts as revision for my galois exam, im not just procrastinating)

[quiet pelican]

I’m definitely not procrastinating revising by making notes for our topics in infinite groups course

(Note: said course is non-examinable)

[elfin wraith]

Ooh, what did you cover?

[quiet pelican]

Sigma invariants mostly

They let you determine when a metabelian group is finitely presented

He’s been promising to cover some of Dawid Kielak’s work for a couple lectures so I’m waiting for that lol

[elfin wraith]

Classic group theory example of working out all situations when semi-schleebs with the floob property are XYZ

[quiet pelican]

😠

Also: *algebra

[elfin wraith]

Group theory does feel particularly like making adjective soup and seeing what it does

Im joking im sure this is an iteresting property lol, finitely presented groups are nice and I guess a lot of groups are metabelian

[delicate orchid]

that's ring theory buddy

well, alg geo

[swift root]

same thing

[delicate orchid]

I can pull some insaneballs adjectives out that make "meta-abelian" look natural. "Q_d(p)-free" and "standard component" come to mind

[swift root]

lol

[delicate orchid]

u get most of these through highly technical conditions that come up in the cfsg

[quiet pelican]

it’s not like I can say anything when my paper is literally “when are Xs Y?”

[quiet pelican]

Do not the finite groups

[delicate orchid]

should've told that to Aschbacher in the 80s 💔

[swift root]

why not

no Galois groups? :(

[quiet pelican]

No

[tough raven]

linear algebraic groups

[delicate orchid]

oh like PSU_n(p^k) yeah I know those ones

[swift root]

you mean commutative Hopf algebras?

yeah i know those

[latent solstice]

How do you find the conjugacy classes in A4?

There's no solution in my book for this example and doing this by hand is taking forever... there has to be a better way right?

[rocky cloak]

Worth noting, if you write a permutation in cycle notation, then conjugating by another permutation just permutes the elements.

E.g. conjugating (1 2 3) by (12)(34) gives (2 1 4)

[latent solstice]

I wrote out all the elements of S4, determined which ones were even, and then tried conjugating every element of A4 with every other element of A4 to see which ones are in the same conjugacy classes

Its just a big confusing mess tbh

I found this

https://groupprops.subwiki.org/wiki/Splitting_criterion_for_conjugacy_classes_in_the_alternating_group

[delicate orchid]

yeah this would help but you can also just do what jagr said

so if you have two permutations x, y written out in cycle notation, one is conjugate to the other by permuting the elements in the cycles. Call this permutation z. S4 and A4 are centreless so z is the only element satisfying zxz^-1 = y, so they are conjugate in A4 if and only if z is even

[vestal jay]

How do I show that a relation is compatible with a group structure?

In particular I got this. I could show that the relation is indeed well-defined but I couldn't advance any further

Oh I got it

[tribal moss]

Are you asking for a formal definition of "compatible with the group structure"?

(I'm not sure that is unambiguous for relations that are not equivalence relations, or at least reflexive).

[quiet pelican]

This should be an equivalence relation

Because everything’s commutative

[tribal moss]

That would depend a lot on what g is.

Sure it's an equivalence relation when 2g=0, but as far as I can see not otherwise.

(Hmm, or perhaps "for some g" is supposed to be parsed as part of the rhs of the ==>)

[swift root]

sure there is

substructure of G × G

[tribal moss]

(But how does it even define a particular relation when it says ==> instead of <=> ?)

[swift root]

its meant to be <=> probably

[quiet pelican]

This

[tribal moss]

This, children, is why quantifiers go before the formula they quantify (and are not written with words in the middle of a symbolic formula!).

[tribal moss]

That is one possibility. But in the general case that is not equivalent to "if a ~ b then ac ~ bc and ca ~ cb", which is how it's formulated when we ask whether a (strict) ordering relation is compatible with the group structure.

[swift root]

compatibility means a ~ b and c ~ d => ac ~ bd though

maybe im too ua pilled

yeah what youre describing essentially is Malcev chains lol

[tribal moss]

I don't think that's strong enough to work for ordering relations, though.

(That is, when we say that an ordered field requires that the order relation is compatible with addition).

[swift root]

hm

[swift root]

(and a ~ b => a^-1 ~ b^-1)

well groups are Malcev so reflexive and compatible with my definition implies synmetry and transitivity anyways lol

[tribal moss]

Huh? Your definition is satisfied for < on (R,+), but that is definitely not a symmetric relation.

[swift root]

am i dumb

yeah im dumb

[tribal moss]

Oh, I see you wanted compatibility with inverses too.

[swift root]

yes

suppose x ~ y
=> y = m(x, x, y) ~ m(x, y, y) = x
where m(x, y, z) = xy^-1z
Then suppose x ~ y ~ z.
=> x = m(x, y, y) ~ m(x, x, z) = z

[tulip otter]

let $f(x)=x^4+ax^2+b\in\mathbb Q[x]$ be irreducible over $\mathbb Q$ with roots $\pm\alpha$ and $\pm\beta$ and let $L=\mathbb Q(\alpha,\beta)$ be the splitting field of $f$. I want to prove that Gal$(L/\mathbb Q)$ is one of the following groups: $C_4, C_2^2$ or $D_4$ ($D_n$ here denotes the dihedral group of order $2n$). I am thinking about doing it as follows:
\First if $[L:\mathbb Q]=4,$ then |Gal$(L/\mathbb Q)|=4$ and hence Gal$(L/\mathbb Q)$ is either $C_4$ or $C_2^2$. \\Otherwise, we know that $[L:\mathbb Q]=[\mathbb Q(\alpha,\beta):\mathbb Q]=[\mathbb Q(\beta)(\alpha):\mathbb Q(\alpha)][\mathbb Q(\alpha):\mathbb Q]=4[\mathbb Q(\beta)(\alpha):\mathbb Q(\alpha)]\mid 24$ so that $[\mathbb Q(\beta)(\alpha):\mathbb Q(\alpha)]\mid 6$. Now $[\mathbb Q(\beta)(\alpha):\mathbb Q(\alpha)]\leq [\mathbb Q(\beta):\mathbb Q]=4$ so that $[\mathbb Q(\beta)(\alpha):\mathbb Q(\alpha)]=2$ and $[L:\mathbb Q]=8$.\\Now I can look at where the $\alpha$ and $\beta$ can be sent by the $8$ elements of Gal$(L/\mathbb Q)$, $\alpha$ can be sent to $\pm\alpha,\pm\beta$ and $\beta$ can be sent to $\beta$ or the other root of Irr$(\beta,\mathbb Q(\alpha))$ (is that necessarily $-\beta$? but if so then arent we saying that Irr$(\beta,\mathbb Q(\alpha))=x^2-\beta^2$?). But from here I dont think that there is an element of Gal$(L/\mathbb Q)$ of order $4$ so where am i going wrong here

[cloud walrus]

ali yassine

[rocky cloak]

beta would not be sendt to a root of Irr(beta, Q(alpha)), but to a root of the image of this polynomial under the map
Q(alpha) -> Q(alpha, beta) you have defined thus far

Like, for example, say
beta was a root of
x^2 - alpha

a homomorphism taking alpha to gamma would need to take beta to a root of
x^2 - gamma

[tulip otter]

but isnt this just inclusion? So what is being done here is veiwing Irr(beta, Q(alpha)) as a polynomial in Q(alpha,beta)[x] which factors completely there?

and then looking at automorphisms of Q(alpha,beta) to see where alpha and beta can be sent to?

[rocky cloak]

What is just inclusion?

[tulip otter]

I meant the homomorphism Q(alpha)->Q(alpha,beta) whose restriction to Q(alpha) is the identity map on Q(alpha)

[rocky cloak]

Like if the automorphism takes alpha to alpha it will be the inclusion.

If it takes it to for example -alpha or beta, then it won't be

[tulip otter]

right

[tulip otter]

ah wait so what you meant by Q(alpha)->Q(alpha,beta) was the restriction of an automorphism Q(alpha,beta)->Q(alpha,beta)?

[rocky cloak]

Yeah, that what you're trying to do right?

Describe the automorphisms

So you first determine where alpha should go, then where beta should go

[tulip otter]

yea

but shouldnt i know what Irr(beta,Q(alpha)) is first?

[rocky cloak]

You can compute that if you want.

But you can also just use the fact that an automorphism will permute the four roots, so there aren't that many options

[tulip otter]

so the other root would be -beta since the roots of f are \pm alpha and \pm beta but beta is not in Q(alpha) so alpha*beta isnt there too?

[rocky cloak]

So indeed Irr(beta, Q(alpha)) is x^2 - beta^2

(Not that that is very helpful without relating it to alpha, but that's what it is)

But like what are you actually trying to do?

Because if you're trying to determine the Galois group in the case it has order 8, then I would just consider how the roots are permuted.

[tulip otter]

so when alpha is mapped to beta, beta cant be mapped to itself and if alpha is mapped to -beta, beta cant mapped to -beta

[rocky cloak]

If alpha maps to beta, then -alpha maps to -beta, so the only options left are for beta to map to ±alpha

[tulip otter]

yea I am looking at where alpha and beta can be mapped to because Q is fixed and so knowing where these 2 go determines the automorphisms

[rocky cloak]

Yeah, so 4 options for alpha, then only 2 options for beta. 4*2 = 8, so that must be everything

[tulip otter]

alright so let σ be the automorphism that maps alpha to -beta and beta to alpha and let τ be the automorphism that takes alpha to -alpha and fixes beta, then σ^4=τ^2=id and τστ=σ^{-1} and hence Gal(L/Q)=D_4

is that correct?

[tulip otter]

tysm jagr, have a great day/night!

[chrome verge]

A friend and I were discussing things and she came up with an interesting question

[velvet hull]

it's not clear to me that this limit would even be well-defined

because there are generally more than one morphism between two finite groups

but limits only make sense over a fixed diagram where all the morphisms are necessarily compatible

so you would either need to duplicate some groups, or remove some morphisms

[noble nexus]

Finitary infinite symmetric group is a good candidate

it's the group of all permutations of N that move only finitely many points

contains every finite group as a subgroup by Cayley

it's also more or less minimal because it's the union of S_n

making that precise seems a bit tricky

if you have any group G containing each S_n in such a way that S_n\subseteq S_{n+1} for each n (the obvious embedding) then at the least you get an inclusion from the finitary infinite symmetric group into G

[quiet pelican]

More precisely, such a group must contain each S_n, and we can fix representatives to form an increasing chain S_1 \to S_2 \to …
The union of this chain is the group you mentioned
(I’m interpreting this to mean “take the colimit of a skeleton of the category of finite groups with morphisms the injective homomorphisms”, which feels like a reasonable interpretation of the question)
Every group has a representative in this chain, because every group injects into one of these S_n

[noble nexus]

That said it doesn't seem particularly well defined since for instance you could also take the infinite general linear group and it also contains every finite group and is a union of an increasing sequence of finite groups

[velvet hull]

as blake said, injective homomorphisms still isn't enough to guarantee compatibility

but it is a step in the right direction, I think

[noble nexus]

but yeah I think the issue is as HChan said you have multiple injective maps

so you can't really take a colimit

or if you can it's going to be pretty ugly

or it might be zero

[velvet hull]

either make a choice of representative from the set of injectives to make it work, or you let your diagram consists of the ascending chain S1 -> S2 -> S3 -> S4 -> .... and every group of size n to have one embedding into Sn

that would be a way of making Sinfinity be a colimit of all the finite groups, in some sense

[rocky cloak]

I mean you can consider the category of all finite groups and all injective homomorphisms and take the colimit of its inclusion into Grp.

The colimit would be the trivial group though.

[fleet cairn]

whats the reason the definition of initial/final objects to want uniqueness? it looks equally "universal" if its simply there exist a morphism with A to everything (and converse)

[velvet hull]

well, if you simply require existence then it does lose universality actually

in Grp, for instance, by that definition every group would be initial and terminal

since the trivial group homomorphism always exists

[velvet hull]

or maybe here is a better answer, turns out the uniqueness requirement will enforce any initial / terminal object to be unique up to (unique) isomorphism (prove it!), which is not something you get with just existence

[swift root]

without uniqueness you cannot conclude anything because the existence of a morphism is in general not a hard condition to satisfy

[frail shoal]

i genuinely cant tell if im crazy or not rn, isnt the limit of any diagram with an initial object, that object

no one else has said this, so i think im misunderstanding the question

[swift root]

yes

colimit is more interesting

which i assume is what theyre acc talking about

[frail shoal]

well then it's the final object of Grp, i.e. also 0

[swift root]

oh lol

[frail shoal]

ok yea we can consider only injective maps, i do wonder why it gets trivialized then

oh yea cuz of stuff like

Z/2 embeds into (Z/2)^2 3 different ways

in general G embeds into G^2 in at least 3 ways

so we get (g, 1) ~ (g, g) ~ (1, g) and thus G^2 is trivialized

[lilac mango]

I am trying to see whether in k[x1, x2,...] theres an ideal without a primary decomp and I thought of <x_ix_j: i \neq j>

So that ideal is radical, so a minimal primary decomp would have to be an intersection of minimal primes containing it

And the minimal primes containing it SHOULD be <x_i: i\neq j> for fixed j

Is it correct to conclude from there that the primary decomp should contain all of them?

I mean I know that should be true in this specific instance by symmetry

But my question is in general do I have to use all minimal primes in the primary decomposition?

[lilac mango]

Or not even by symmetry, just showing that if theres one of them you dont use (say the one with no x_k), the intersection has x_k and thus cant be <xixj: i \neq j>

[fleet cairn]

allufi is using the word terminal to refer to both initial and final objects, weird

[elfin wraith]

I think his point is more that terminal means at the end so it could mean either initial or final, but in fairness ive never seen terminal used that way, ive only seen it be synonymous for final, but maybe this is just dated

[swift root]

i use terminal for final too lol

[elfin wraith]

As does every book I have to hand (I checked out of curiosity), im unsure if it is dated terminology or just a weird comment

[swift root]

i think its dated

[elfin wraith]

Even Maclane uses terminal in the normal way though

[swift root]

wtf

aluffi just freaky

[pliant forge]

what would be the minimal requirements for a module M so that Hom(M,-) to reflect isos?

[noble nexus]

what do you mean by reflect isos

[pliant forge]

hom(M,f) iso implies f iso

I guess its more commonly called a conservative functor

[noble nexus]

seems quite restrictive, I can't think of any examples aside from free modules

[pliant forge]

well i think for instance this is true when there is some n such that M^n has R as summand

[rocky cloak]

M being a generator is equivalent

[pliant forge]

how would you define a generator?

[rocky cloak]

For any pair of parallel map Hom(M, -) takes them to different maps (if they're different).

[rocky cloak]

This is also equivalent

[swift root]

seems remniscent of being projective

but like, the other way around

[pliant forge]

is this also known as a separator?

[rocky cloak]

Yes
https://en.wikipedia.org/wiki/Generator_(category_theory)

[pliant forge]

oh woops got stuck reading nlab instead

[rocky cloak]

I mean they're both conditions on Hom(M, -)

Projective being that it's exact and generator that it is faithful.

[small yacht]

How do we know that phi preserves addition? I don't believe that it's simply a requirement we can set

[wraith cargo]

I'm not sure what you mean

The fact that it preserves addition is a consequence of the group operation on Z

And by definition of phi

[rocky cloak]

Well
(m + n)*1 = m*1 + n*1
So it preserves addition

I guess it comes from addition in R being associative if you want to blame it on something

[small yacht]

Oh, buh😭

In retrospect that was obvious

I’m half asleep so that’s probably why I missed it

[next obsidian]

Maya the jolli

[small yacht]

Oh, i failed to consider associativity

So not used to rings

[small yacht]

Challenging monkey

[next obsidian]

Challenging what

[small yacht]

The challengers

[swift root]

social norms

or wtv

gender norms

[next obsidian]

I don’t feel I particularly strongly challenge gender norma

Maybe once these Hakama pants I ordered off mercari get here tho

[next obsidian]

Makes sense

[swift root]

those are cool

[vivid kestrel]

https://groupprops.subwiki.org/wiki/Splitting_field is this definition of splitting field of a group equivalent to saying that the endomorphism ring for every simple KG module is just K * id

asking because that is how splitting field for a group was defined in lecture notes when i took rep theory and that was sufficient to show that KG is isomorphic to a direct sum of matrix over K and then we showed the important stuff

hm lets say we just require char K = 0

[rocky cloak]

Yes, its equivalent.

[vivid kestrel]

thanks

is it trivial or do you have a reference?

[rocky cloak]

So for a KG-module M and a field extension F you have End(F(x)M) = F(x)End(M), so if End(M) = K extending scalars take simple modules to simple modules. Since FG and KG are the sum of all the simple modules and FG = F(x)KG every simple FG module is extended from a KG module.

Conversely if End(S) is not K you can extend scalars by a subfield of End(S) so that F(x)S is not simple. Then a summand of F(x)S will give you a summand of S using that every FG module is extended from a KG module

[vivid kestrel]

(x) is tensoring right

so thats with G acting trivially on F in the first line right

[rocky cloak]

Well it's not really about G acting on F, but that
F(x)M becomes an FG module, by
ag * b(x)m = ab(x)gm

[vivid kestrel]

ah okay i see

but i fail to see the link between what you typed up and the statement that any representation over a field extension is equivalent to a representation over K

[rocky cloak]

A representation M of FG being equivalent to one over K is exactly saying that M = F(x)N

[vivid kestrel]

oh ill have to think about that, thanks a lot!!

[rocky cloak]

The basic idea is if V is an F-vector space and you pick a basis, if you let V' be the K-span of that basis then V = F(x)V'.

So the question is when does V' become a KG module

And that's when the matrices describing the action of G has coefficients in K

[small yacht]

what is alpha1|Z = alpha2|Z in this context?

[lusty marlin]

They mean the restrictions of α_1 and α_2 to ℤ

[fleet cairn]

can someone give the full proof for this (Does this category have initial objects)
I havent had any luck trying myself

[mint seal]

in this situation, an initial object (if it exists) is of the form f: A —> P for some particular set P, with the property that given any other object g: A —> Z, there is a unique function h: P —> A making the triangle commute, i.e. g = h \circ f

now there aren’t very many sets around that are naturally related to A, to try as a candidate for what I called P

This also has to be about the equivalence relation ~ on A

so a natural thing to try is the map A -> A/~ taking a in A to its equivalence class [a]

now suppose we have a function g: A —> Z for any set Z, with the property a ~ b implies g(a) = g(b)

Define a function h: A/~ —> Z by h([x]) = g(x)

this is well defined, because if a ~ b, so [a] = [b], we have g(a) = g(b) by hypothesis of g

and it makes the triangle commute

to make the triangle commute forces that definition of h, so it’s unique as well

[high hazel]

Hi, I'm struggling a bit with problem b here.
I assume it's the same route of application as problem a, but I don't exactly know how to generalize this in the same way due to it being a polynomial ring.
Translation: 'Let p be a prime number and n>=1. Show that every irreducible polynomial f(x) in Zp[x] of degree n divides the polynomial x^(p^n) - x in Zp[x]'

[rocky cloak]

All you need to show is that the roots of f are elements of a field with order p^n, then it follows from a

[high hazel]

I think I got it, thanks! :D

[small yacht]

??

[next obsidian]

You have a function on Q

This gives you a function on Z

[small yacht]

Ohhh ok

[next obsidian]

By just saying it’s value on an integer is the value of the function when you pretend that integer is a rational

[small yacht]

I searched it up 😭

[next obsidian]

You’re just restricting the domain

[small yacht]

Choice of axiom monkey

[next obsidian]

Zorn’s monkey

[frail shoal]

it means alpha1 ○ iota = alpha2 ○ iota

but outside of category theory we often think of post-composing by an injective map as "restricting" the domain

[frigid epoch]

isn't that precomposition

[tribal moss]

One map's precomposition is another map's postcomposition.

[frail shoal]

fuck

○ notation is confusing

[frigid epoch]

petition to write (x)f○g for applying f to x, then g to the result of that

[proud vigil]

yes!! i support this

[delicate orchid]

Wait this gif is terrible let me get a better one

[delicate orchid]

https://tenor.com/view/son-im-crine-gif-13090932573727637773

[frail shoal]

i say "○" out loud as "after"

like "f ○ g" is "f after g"

[azure cairn]

i think i say "circ"

f serk g

[frail shoal]

the issue is that i genuinely lose 50% of my reading comprehension when i can't say things in english

[azure cairn]

lmfao

sometimes i say f "composed with" g

[frail shoal]

maybe i should say "f of g" like in middle school

[south patrol]

I just write and say fg

[vapid vale]

my preferred convention is f ○ g is f after g but fg is f then g

[elder hollow]

[south patrol]

Ah okay so your preferred convention is just being wrong

[elfin wraith]

I say f of g, o for of. But I’m also not 12 so I just write fg

[vapid vale]

One is function composition the other is multiplication in an algebraic structure

[frail shoal]

i will admit that the use of a symbol that takes 2 arguments to denote an operation that acts on lists is pretty annoying at times

[hollow moss]

it is opposite multiplication in an algebraic structure

[kind temple]

do you mean like, taking a group G, and defininig G^op where x *' y = y * x?

[vapid vale]

i think its in response to my message

[somber goblet]

herstein w

[maiden crater]

maybe I'm missing something

but how's this any different from saying exactly one of a,b is a unit

because if a,b are both units, r is a unit

this has been confusing me for a while because if we blindly apply this defn, yeah, this proof make sense

but a could be invertible too, could it not

[rocky cloak]

It's not

Like theyre equivalent, so it's just picking the simpler statement.

[rocky cloak]

It couldn't because one of them needs to be in p

[maiden crater]

I don't follow

[rocky cloak]

Following the proof a = pr, hence a is not a unit

[maiden crater]

ah , I see as primes aren' invertible

[balmy python]

any tips on determining whether a quadratic form has solutions?

[maiden crater]

Kind of weird way to go about shwoing (x). is prime, but how does this look

suppose there were a polynomial in $(x)$ such that$p(x,y)=q(x,y)\cdot r(x,y)$ where $q$or $r$ have a term of the form $a \cdot y^n$, then and only then would $(x)$ not be prime. However, $p(x,y) \in (x)$ and thus every term of the polynomial has $x$ as a factor. But if our assumption were true, there would be a term without $x$ as a factor , violating that $p$ is in (x)$.

[cloud walrus]

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

[quiet pelican]

That works (assuming you meant “q and r”, not or)
The quicker way would be to observe that Q[x, y]/(x) = Q[y] is an integral domain

[maiden crater]

right, I wrote that at first and then backtracked 😭

The quicker way would be to observe that Q[x, y]/(x) = Q[y] is an integral domain
right,makes senbse

[maiden crater]

well, to show $(x,y)$ is prime I can just use $Q$ is an ID

[cloud walrus]

Wai

[elfin wraith]

You can do even better than that, Q is a bit more than an ID

[maiden crater]

it's maximal

[elfin wraith]

Yur

[maiden crater]

as for the first to show it's not maximal, it's contained in (x,y)≠Q[x,y]

Let $R= R'[x_1,x_2,\dots,x_n]$. It's then sufficient to show $(x_1,x_2)$ isn't principal. Let $(x_1,x_2)=(p(x_1,x_2))$. We then consider $r(x_1,x_2)=q(x_1,x_2)\cdot p(x_1,x_2)+x_1$. We note all these elements lie in $(x_1,x_2)$. However $x_1$ isn't a multiple of $p(x_1,x_2)$. We thus have a contradiction.

[cloud walrus]

Wai

[south patrol]

A fun exercise is that if R is a commutative ring (with unity) and R[x] is a PID, then R is a field

(This implies what ur told to prove)

[maiden crater]

will do it once I finish this RA exercise 🙏

[twilit wraith]

hey everyone, i have my algebra 2 final later today, and im wondering if i could discuss this practice final with anyone here to help prep

its a bit long but ofc you can help with as much as youre willing to

the first part is pretty easy

1(i) is just a routine checking of Q, Q(sqrt[5]{2}), Q(zeta) as not being splitting fields of x^5 - 2

well after showing that each alpha_i is a root of x^5 - 2 which shows that we have a complete list of complex roots by FTA

then as sqrt[5]{2} and zeta induce simple extensions of Q of degree 5 and 4 respectively, and 5 and 4 are coprime, it follows that Q adjoin both is the product

and since that extension is Galois the Galois group will have the order that is the degree of the extension

for 1(ii), we know that elements in Gal(K/Q) send elements to other elements that satisfy the same minimal polynomials

and this is necessary and sufficient pretty sure

sigma just sends a root of x^5 - 2 to another root

and tau just permutes the 5th roots of unity

uniqueness might be a little annoying, unless theres an elegant way to do it that im forgetting

1(iii) is kinda confusing, i mean its pretty obvious that sigma and tau do these things so maybe its just supposed to be an easy question

1(iv) is kinda tough

i mean sigma and tau have to generate Gal(K/Q) since the group they generate is at least order 20 by appealing to element orders

nonabelian is just showing that (12345)(2354) isnt the same as (2354)(12345)

oh and then

Sylow p-subgroups are cyclic because 20 = 2^2 * 5 so that the Sylow 2-subgroups are the only ones to check

but (2354) generates a cyclic Sylow 2-subgroup and all the other Sylow 2-subgroups being conjugate to it means theyre isomorphic

i.e. theyre also cyclic

seemed tough but it wasnt bad at all

alright question 2 is definitely testing my knowledge of the fundamental theorem of galois theory much more heavily

actually the first one is Sylow theory

mmm its kinda both

well ok it suffices to show by FTGT that there are five subgroups with index 5 in Gal(K/Q)

i.e. order 4

by Sylow theory we know that there is either 1 or 5 subgroups of order 4

there isnt 1 subgroup otherwise Gal(K/Q) would be abelian

alright and that works

ok finding the elements explicitly

i think its just Q(alpha_i) is it not

i think you can show that if these werent distinct among each other then you can find that theyre equal to Q(sqrt[5]{2})

but thats not possible since Q(sqrt[5]{2}) is in R

maybe the ones were zeta's degree is opposed by 2 is less easy but meh

ok 2(ii) is invoking the fact that subfields are isomorphic if and only if their corresponding subgroups are conjugate

by construction the Fi's correspond to subgroups that are conjugate by Sylow theory

and none of them are Galois since their subgroups arent normal in Gal(K/Q)

[twilit wraith]

ok sorry i was gone for a sec

alright next i have to show that there are unique degree 10 extensions of each Fi

i think this amounts to showing that each subgroup corresponding to Fi contains a unique subgroup of order 2

and this follows from the fact that each Sylow 2-subgroup is cyclic

[elfin wraith]

I think sigma is unique by Sylow, it generates a cyclic group of order 5. I’m trying to think of why tau is unique but I imagine you can do something similar (you can probably entirely avoid Sylow though if you’re brighter than me)

[twilit wraith]

tau being unique by Sylow might require a later argument

the fact that the Sylow 2-subgroups are cyclic

hmm no

since there are multiple Sylow 2-subgroups

[elfin wraith]

Yeah you probably have to say something about the roots (and I should really know what lol, lemme ponder)

[twilit wraith]

i guess doing it directly could be not too bad

like the cleanest way to represent elements constructively in K is by $\sum_{j = 0}^4 \sum_{i = 0}^4 c_{i,j} (\sqrt[5]{2})^i (\zeta)^j$

[cloud walrus]

hiidostuff

[twilit wraith]

i think this is right

mm the indexing is a bit off

[elfin wraith]

Yeah I guess actually thinking about it is actually “clear”

Which sounds like a cop out but if you just write down what you’re trying to do it’s forced

But yeah sigma is easy by Sylow at least

[twilit wraith]

yeah

like even just constructively its clear

the only reason i wanted to avoid that direction is because id have to show that 20 terms get sent to the same thing but even thats not too bad

ok ok

the reason its not too bad is because those 20 terms are just products of two things

though if i were to do it that way my intimidating british prof would comment on the fact that i shouldve done it more elegantly lol

[elfin wraith]

For this, just remember how you labeled the alpha I guess

Sigma sends alpha_i->alpha_i+1

[twilit wraith]

yeah i mean thats what i meant by it was confusingly easy

[elfin wraith]

Oh sorry I misread but yeah I think it’s just that

(I actually have a Galois exam tomorrow so I’m working through this and comparing against what you got btw which is why I’m just replying at random points lol)

[twilit wraith]

fair enough lol

good luck to you as well then

[elfin wraith]

Thanks! I think it should be pretty chill tbh, I’m not too worried

Also it seems like the new lecture is much more likely to ask us to reprove like every lemma from the notes rather than do any sort of computations so

[twilit wraith]

im somewhat worried since my prof really implied that theres gonna be a lot of group theory

im a tad rusty on it

Sylow theory is fine but if i have to do a bunch of group action stuff i might be in trouble

[elfin wraith]

I mean if it’s mostly Galois the main things to know id think are Sylow, Lagrange, S_5 is generated by a 2 cycle and 5 cycle and that A_5 is simple non Abelian

[twilit wraith]

yeah fair enough

these are pretty easy

[twilit wraith]

on another note though, do you remember the argument that cyclic Sylow 2-subgroups have unique elements of order 2?

i know the visualization of how these subgroups are related is like a flower and thats a "visual proof"

[elfin wraith]

I don’t remeber the exact details but I think it’s an induction argument on 2^n, and basically a case of writing down what needs to be shown

[twilit wraith]

ill just have to think about it some more i suppose

for now ill move on though

alright 2(iv) is also not too bad

pretty sure theres a property of Sylow p-subgroups in that every subgroup of order p^k is found in some Sylow p-subgroup

so if we have some subgroup of order 2 in Gal(K/Q) its found in some Sylow 2-subgroup but then that must be the unique subgroup of order 2 in that Sylow 2-subgroup

meaning by pigeonhole theres only five subgroups of order 2

alright 2(v) is a direct result of this

since the Sylow 2-subgroups are conjugate among each other and conjugation preserves order, it follows that these conjugations restrict to conjugations of the subgroups of order 2

[chilly radish]

Cyclic groups have a unique element of each order dividing their order

[twilit wraith]

true

i just cant remember why we cant have something like two Sylow 2-subgroups sharing the element (12) or smth

[chilly radish]

This isn't a priori true

[twilit wraith]

yes i should specify in the case where the Sylow 2-subgroups are cyclic

[chilly radish]

I'm not sure that actually helps with having elements in the intersection of two sylow subgroups. In this case you can argue though that if E_i contains F_i AND F_j, then it would be equal to K, so each E_i contains a unique F_i and thus there are 5 unique ones (corresponding to the order 2 elements)

[twilit wraith]

i just realized i interpreted the problem incorrectly

well i suppose 2(iv) is where we need that no element of order 2 is found in more than one Fi

[chilly radish]

Each subgroup being cyclic (make sure you argued why that is) implies that there is some field containing F_i of degree 10. For uniqueness you'll probably need to make the kind of argument I did

[twilit wraith]

actually wait a minute

no we never need to know that there are five Ei's

i think there does end up being five

but 2(iii) just establishes that there are subgroups of order 2 that correspond to each Fi

and 2(iv) says that these subgroups of order 2 are the only such ones

[chilly radish]

The reasoning here is that group theory alone doesn't imply the intersection is trivial, but the fixed field of an intersection is the composite of the fixed fields (fun fact), so trivial intersection is equivalent to composite being K

[twilit wraith]

but nothing is stopping us from having only one Ei right

[twilit wraith]

i see

oh wait 2(v) is where we need that theres more than one

if there isnt then that subgroup corresponding to Ei is normal in Gal(K/Q)

[chilly radish]

It's enough to say the intersection of two sylows is trivial, because then the subgroup corresponding to E_i can't be fixed by conjugation

[twilit wraith]

makes sense

though im not fully sure why Ei containing Fi and Fj implies that Ei is K

[chilly radish]

The subgroup they generate is at most order 20. It can be smaller if the intersection of the subgroups generated by them isn't trivial

[twilit wraith]

oh

well the intersection has to be trivial

bc order of elements in <sigma> and <tau>

[chilly radish]

Each F_i contains a root of x^5-2, which are given by primitive 5th root of unity * 2^(1/4). Once you have two of these in one field, you get a field containing 2^(1/4) and a primitive 5th root of unity, so it's all of K

[twilit wraith]

thats what i was vaguely referencing when i said "appealing to element orders"

[chilly radish]

Oh, yea in this case that's true because the orders are coprime

[twilit wraith]

oh yeah by 2(i)

very nice

alright

then finishing 2(v) no subgroup corresponding to an Ei in Gal(K/Q) is normal since the Sylow 2-subgroups are conjugate and theres more than 1

[chilly radish]

Exactly

[twilit wraith]

alright 3 is certainly the hardest

not surprising

honestly i find it easier to first express complex conjugation as a word in the generators sigma and tau

itll just end up being tau^2 anyways

i guess the explanation as to why we can be certain complex conjugation is in Gal(K/Q) in the first place is that it fixes R and thus Q and sqrt[5]{2}, and that complex conjugates of primitive 5th roots of unity are still primitive 5th roots of unity

[chilly radish]

Yea, that's all you need

[twilit wraith]

next up i have to show that L = Q(zeta) is Galois and is also the unique subfield of K that is degree 4 over Q

ah i see what to do here

well L is Galois because its the splitting field of the 5th cyclotomic polynomial

its degree 4 because that polynomial is degree 4

[next obsidian]

No you won’t

You’re gonna go make a grilled cheese and eat it

https://tenor.com/view/jedi-starwars-mind-trick-obiwan-gif-5199308

[twilit wraith]

and its unique as any subfield of K that is degree 4 over Q corresponds to a subgroup of order 5 in Gal(K/Q) and Sylow theory shows theres only one of those

[twilit wraith]

too late 😎

[next obsidian]

STOP RESISTING

https://tenor.com/view/adultswim-loitersquad-balloonman-bully-cops-gif-4890669

[twilit wraith]

oh boy 3(iii) is an A5 moment

[chilly radish]

Wdym

Determine whether L/Q is Galois

Is hysterical. My guy, it's a degree 2 extension

[twilit wraith]

lol

i mean you can also use the uniqueness

L' corresponds to A5 which is normal in S5

[chilly radish]

But the galois group is not S5

Not even close

[twilit wraith]

oh uhh oops

i got ahead of myself

uhh well L' is fixed by complex conjugation that might be important

wait ive done a hw problem about this

[twilit wraith]

wait wait

since Gal(K/Q) is a subgroup of S5 cant i pass to S5

[chilly radish]

S5 has order 120

[chilly radish]

Then you wanna show that it's the intersection with A5?

[twilit wraith]

like if the subgroup corresponding to L' is index 2 in S5 its restriction is index 2 in Gal(K/Q) right

[twilit wraith]

yeah exactly

[chilly radish]

That still doesn't prove it's a unique index 2 subgroup

[twilit wraith]

yeah uniqueness isnt there yet

but thats probably the nicest way to get thats its index 2 at all

[chilly radish]

If you just wanna show existence, you can just take the subgroup generated by Gal(K/Q(\zeta)) and an order 2 element

[chilly radish]

It won't be index 2. It would contain a subgroup of order 5 (since a 5-cycle is even) but it won't contain any element of order 2 cuz those are odd

So it can't be order 10

[twilit wraith]

oh i see

[chilly radish]

So here's one thing you can do:
Any order 10 (index 2) subgroup is generated by Gal(K/Q(\zeta)) and an element of order 2.
Show all order 2 elements are conjugate via an element of Gal(K/K(\zeta))

[chilly radish]

In fact, every order 2 element must be contained in a sylow 2-subgroup, so they're all conjugate by question 2, so every index 2 subgroup is conjugate (Must contain Gal(K/Q(\zeta)) since it contains an element of order 5, and an element of order 2)
and any index 2 subgroup is normal, so every subgroup generated by an elt of order 2 and order 5 is is the same.

[twilit wraith]

hmm

this makes sense

ok wait so every subgroup generated by an element of order 2 and an element of order 5 is the same, doesnt that mean theres a unique subgroup of order 10?

[chilly radish]

That's my conclusion, yes

Every order 10 subgroup is generated by g,h where g is of order 5 and h is of order 2.

Conjugating g doesn't change anything since Gal(K/Q(\zeta)) is normal and every element is a generator of it, and conjugating h gives another element of order 2. You can conjugate any 2 subgroups of order 10 by conjugating one order 2 element to another.

Order 10 (index 2) subgroups are normal, thus since they're all conjugate there's just one

[fleet cairn]

what the helly do i do here

the chapter already gives us the lemma that two final objects are iso to each other

[next obsidian]

What do you know

[fleet cairn]

this proof would just be they are iso because they are iso

[next obsidian]

Lmao

Okay uh, prove the stronger thing that

They’re unique up to unique isomorphism

[fleet cairn]

this is a play on words

what does that mean

[next obsidian]

It isn’t actually

As an example “group generated by an element of order 5” are unique

But not uniquely, there’s many ways to define an isomorphism between them

By mapping any generator to any other generator

[fleet cairn]

allufi

[next obsidian]

Here for final objects, there’s a single isomorphism between them

[twilit wraith]

Oh yeah true

[next obsidian]

So they’re uniquely unique

I promise you

This deadass matters

[fleet cairn]

i have allufi so i dont know anything abt groups

next chapter is groups tho

[next obsidian]

Okay well whatever

[next obsidian]

Look my point is

[fleet cairn]

this lowkey makes sense

[next obsidian]

If you have two things that you knew were Z because idk

God told you they were

You can make an isomorphism between them

[twilit wraith]

I also realized that L is a degree two extension of L' and L is degree 4 so ofc L' is degree 2 over Q

[next obsidian]

Errr

[twilit wraith]

So thats existence ig

[next obsidian]

I’m talking about Z with just +

[fleet cairn]

what do you mean were Z? named Z or like property of Z

[next obsidian]

Then you can’t distinguish +1 and -1

[next obsidian]

Like are isomorphic to Z

Look man

Just put the unique isomorphism in the bag

[fleet cairn]

Ohh

[next obsidian]

And when you get to groups you’ll get what I was saying

With the cyclic group thing

[chilly radish]

Once you prove it's not rational or all of L yea. Since L is Galois inside of L it's just the fixed field of tau^2 (at least it's contained in the fixed field)

[chilly radish]

But not fixed by, e.g. tau, so it can't be L and it can't be Q, so it must be a degree 2 extension (and is in fact equal to the fixed field)

[fleet cairn]

i think this is an example

[twilit wraith]

actually, how would i go about showing that zeta + zeta^-1 isnt rational?

its easy to see that its real

wait wait no i dont need to show it isnt rational'

we just have [L : L'][L' : Q] = 4

and [L : L'] = 2 since L' is a real field and zeta satisfies a degree 2 polynomial over L'

[chilly radish]

This is circular since you are assuming it's not rational

What I'm saying is this element is fixed by some of the galois group, but not all of it, so it can't be Q or L

[twilit wraith]

well zeta cant possibly be in L'

so its extension is at least degree 2

[chilly radish]

Oh sorry you said L/L'

[twilit wraith]

yeah my bad

[chilly radish]

Yes that also works

[twilit wraith]

3(iv) part 1 isnt that bad

part 2 might be kinda troublesome but maybe not

well its easy to see that E' = Q(sqrt[2]{5}, zeta + zeta^-1) is degree 10 over Q

since 5 and 2 are coprime

so by problem 2 its one of the Ei

well then the subgroup that fixes it is order 2

we know that complex conjugation fixes E' since its extension elements are real

and so complex conjugation must generate the subgroup that fixes E'

(well more formally the restriction of complex conj but whos counting)

ok 3(v) is showing that theres exactly 14 subfields which kinda feels like its coming out of left field

not counting the obvious ones theres 12

5 of these are the Fi

id like to say that another 5 of these are the Ei but we didnt show that these are in bijection with the Fi

actually yeah we totally did nvm

ok and then theres L and L'

itll end up being the case that Q, L', L, and K are the only Galois subfields

well from here its just using the previous classifications weve done

if F is some subfield of K then it corresponds to a subgroup of Gal(K/Q)

that subgroup must have order 1,2,4,5,10,20

[chilly radish]

You just count subgroups. You already know there's a unique order 10 one, and a unique order 5 one. The only other options are order 4 (The F_i) and order 2 (The E_i)

[twilit wraith]

1 or 20 corresponds to K or Q respectively

if subgroup is order 2 then field is order 10 hence one of the Ei

if subgroup is order 4 then field is order 5 hence one of the Fi

if subgroup is order 5 then field is order 5 hence L

[chilly radish]

Ye

[twilit wraith]

if subgroup is order 10 then field is order 2 hence L'

cool

alright that solves that

[chilly radish]

4 is cute but kind of unfair if you don't remember your trig

[twilit wraith]

besides getting stuck on 3(iii) i feel pretty confident

lets see if i do remember my trig

well Q(sqrt 5) being a subfield is gonna mean that the real part of zeta is a rational expression of sqrt 5

i think i constructed a pentagon this semester so i should know this

well cos(2pi/5) is a root of the polynomial 16x^5 -20x^3 + 5x - 1

yeah i gotta be more clever about this

i mean cos has a period of 2pi so if x = 2pi/5 then cos(2x) = cos(3x)

then $\cos{2x} = 2\cos^2{x} - 1$ and $\cos{3x} = 4\cos^3{x} - 3 \cos{x}$

[cloud walrus]

hiidostuff

[twilit wraith]

so that cos(2pi/5) satisfies the polynomial $4x^3 - 2x^2 - 3x + 1$

[cloud walrus]

hiidostuff

[twilit wraith]

ok is this reducible

4-2-3+1 = 0 so (x-1) is a factor

4x^2 + 2x - 1 is the quotient

so its finding the roots of this

yeah its $\frac{-1 + \sqrt 5}{4}$

[cloud walrus]

hiidostuff

[twilit wraith]

so Q(sqrt 5) is indeed a subfield

[chilly radish]

Epic

[twilit wraith]

the group Gal(K/Q(sqrt 5)) is gonna be generated by only a power of tau

[chilly radish]

No

[twilit wraith]

yeah nvm thats wrong

[chilly radish]

It's gonna be the unique order 10 subgroup found before. Take your pick at generators

[twilit wraith]

every power of sigma is in this group

well i mean i know sigma is in here

tau^2 is too since zeta and its conjugate share a real part

uhh so that actually means sigma, tau^2 generate this Galois group

and im pretty sure this group is isomorphic to D10

since the roots of x^5 - 2 are permuted freely

and tau^2 is reflection across the real axis

[chilly radish]

cos(2x)=cos(4pi/5) = -cos(pi/5) = -cos(3x)

[twilit wraith]

oh yeah that too

[chilly radish]

Sorry typo, my point is it's equal to minus each other

[twilit wraith]

yeah after checking relations its definitely D10

sigma corresponds to r and tau^2 to s

[chilly radish]

Yea, so it's generated by an order 5 and order 2 subgroup, one of them is normal and they intersect trivially, so it's a semidirect product of these two subgroups

Then it's just about checking the action of conjugation, and it's exactly sending sigma to sigma^(-1), which is the same as the action for the semidirect product presentation of D10

[twilit wraith]

ah thats a better way

i forgot a bit about semidirect products so i resorted to checking relations though i think thats insufficient iirc

though i think two groups that have the same order and whose generators satisfy the same relations are isomorphic

maybe not

[twilit wraith]

isnt only the order 5 subgroup normal?

otherwise D10 would be abelian

[chilly radish]

Normal in Gal(K/Q(sqrt(5))

[twilit wraith]

wait thats certainly not true

the order 2 one is

[chilly radish]

Er wait

Sorry I meant only the order 5 one is normal* yea

[twilit wraith]

yeah <s> is the center of D10 right

wait am i losing it

[chilly radish]

That would make it normal

[twilit wraith]

no no the center of symmetries of odd polygons is trivial

yeah only the order 5 one is

[chilly radish]

Yea, so the order 5 subgroup is normal and is acted on by conjugation by the order 2 element

[twilit wraith]

yep

alright cool im remembering now

semidirect products are determined by how the non-normal subgroup acts on the normal one

awesome, since my exam is only about galois theory, im feeling pretty confident, i was more ready to do this practice exam than i thought

[twilit wraith]

is there anything of note that i seemed lacking in or do things seem fine

[chilly radish]

Maybe just being a little more comfortable with finite group theory

Other than that you did well, but some arguments seem to elude you since they lie entirely within group theory (like the uniqueness of the order 10 subgroup)

[twilit wraith]

yeah i agree my group theory is pretty rusty

thanks for the help!

[chilly radish]

Happy to help!

[maiden crater]

I suppose here I simpy say $F[x]$ is an ED and thus a PID. Then $(f(x))$ is prime and thus maximal

[cloud walrus]

Wai

[maiden crater]

another problem

Uh, here would $x^p-(p-1)$ work

[cloud walrus]

Wai

[maiden crater]

where p is the order of F

[next obsidian]

That’s just a single polynomial

[maiden crater]

well I suppose not, it just makes sure that there's no factor in the field

[next obsidian]

My hint is that Euclid could have proved this

[maiden crater]

mhm, div algo then

[maiden crater]

how

[next obsidian]

That’s up to you

[maiden crater]

F[x] is a PID and UFD

I suppose I'd need that

[maiden crater]

oh, got it

$x^{np}-(p-1)$

for n in N

[cloud walrus]

Wai

[maiden crater]

or ofcourse just use euclids proof of infinite primes

[maiden crater]

I suppose this is what I was supposed to use

[rocky cloak]

This polynomial isn't irreducible.

[maiden crater]

so this yea

I remember doing this earlier too

[fleet cairn]

is there a real difference between defined up to isomorphism and unique up to isomorphism

here's how i understood the first one, and ill be precise
Let C be a well defined category. And apply a definition P(x) on all elements of ObjC. And for all x, y: P(x) and P(y) => x and y are iso. You could also think of it as set of objects satisying the definition, and as a result all its possible pairs are iso.

[solar shore]

the iso part is correct. the unique up to iso just specifies that only one exists of whatever property you say is unique but there may be different ways of expressing it

theres also unique up to unique iso

[fleet cairn]

i still dont get the difference

can you explain in more detail

[solar shore]

sure

do you agree that Z and nZ are iso?

[fleet cairn]

Ok assume they are

[solar shore]

well they are

[fleet cairn]

unless its notation for smt idk

[solar shore]

as in Z is set of integers, and nZ is just the set of integers scaled by some integer n

sorry probably shouldve led w that lol

[fleet cairn]

Ohh yes they are iso

[solar shore]

anyways, those two are examples of infinite cyclic groups right

and they are all isomorphic to each other, but they arent unique

[fleet cairn]

im not familiar with groups sorry but i think i can follow

yes they arent, in terms of set theory atleast

[solar shore]

oh oops

hm okay sets

one sec

oh actually i dont think there is a real difference

two sets are isomorphic if you can just create a bijection between them i.e satisfy the same properties.

unique up to isomorphism just classified such sets that satisfy some set of properties but may be written differently

[rocky cloak]

I would say that the only difference is that the former is about a definition while the latter could be more general.

I.e. something is defined up to isomorphism if the definition specifies something unique up to isomorphism.

But for example I could say that the category of abelian groups has a unique indecomposable projective up to isomorphism. But I wouldn't say that "indecomposable projective" is defined up to isomorphism.

[tardy hedge]

I don't get what defined up to isomorphism would mean

Sorry i guess u said it there

[rocky cloak]

A definition that determines something up to isomorphism. Like the definition of projective cover for example

[tardy hedge]

Oh huh

[rocky cloak]

which is defined up to isomorphism (but interestingly not up to unique isomorphism)

[tardy hedge]

Cool I haven't seen projective covers before

ive seen injective envelope briefly tho

[rocky cloak]

It's just the dual

[tardy hedge]

yea

[rocky cloak]

But yeah injective envelopes come up a little more, since they exist for every module over every ring

Whereas projective covers are a little rarer

[kind temple]

i don’t understand what you are saying here. something seems wrong in the sentence “And for all x,y: …”

[fleet cairn]

it just means all pairs that satisifies the definition are iso

[kind temple]

okay, thanks

[kind temple]

just trying to understand this one. so a projective cover for an object X is a pair (P,p) where P is a projective object and p : P —> X is a superfluous epimorphism (whatever that means).

if i take another pair (Q,q), the claim that there is an isomorphism of P and Q that is compatible with the choice of superfluous epimorphism?

[rocky cloak]

Yes, specifically there is an isomorphism f: P -> Q with
p = q f

For example consider R-modules for R = k[[x]], then the projective cover of k = R/(x) is the canonical projection R -> k.

But there are several automorphisms R -> R fixing this. For example multiplication by 1+x

[kind temple]

okay, sorry if i’m just completely dropping the ball on this one, but how does this apply to the situation @rocky cloak ? projective objects are supposed to be defined up to isomorphism, but i’m struggling to see how to apply JFK’s reasoning in this case.

[rocky cloak]

If P is the property "x is the projective cover of k", then any two objects with property P are isomorphic (to R)

So the projective cover of k is defined up to isomorphism

[kind temple]

ah okay. thank you!

[torpid grotto]

Are flash cards a smart choice for memorizing important definitions and theorems in algebra?

I've got a final exam in abstract algebra coming up. It's an oral exam, and I'm pretty shaky when it comes to proving things. Like, it takes me a very long time to gather the necessary definitions and theorems. I have heard that it's better to just do a larger volume of problems. However, at the speed I do problems it's unlikely that I can do more than 2 or 3 a day and actually understand them. Does anyone have thoughts or advice on this?

[south patrol]

Idk I thought the usual advice is not to be memorising definitions and theorems but rather to understand stuff

[next obsidian]

I feel like you shoulda been memorizing this over the semester

If you really are at a spot you just gotta cram make a quizlet or some shit but you’re supposed to have memorized these definitions by using them in proofs throughout the class

[karmic moat]

study the examples you learned in class

and ideally do problems

[verbal valley]

i have always thought this a little silly advice to give to an undergrad -- if it's stuff they're using a lot ofc they will just naturally internalize the important ideas, but ofc we also just take classes which we don't really care about and just need to pas the exam for and memorizing stuff here is totally reasonable

honestly memory is also useful for our field of interest it's just not something we usually need to worry about since we will remember stuff anyway

[verbal valley]

but yeah memorizing definitions is a perfectly good study method

esp for an oral exam where you will be expected to know the key definitions well

and to be able to relay them quickly

[noble nexus]

I find that I memorize things much easier by not actively trying to memorize them

Doing examples especially is one of the best ways to accidentally memorize things

[quiet pelican]

For me personally, the processes of memorising a proof and condensing the proof to its core ideas are essentially the same thing

Try condensing a proof 3-4 times and I’ll literally remember it for years lol

[noble nexus]

because then instead of having to remember the definition I'm like "oh it's like this one example" then all the details just come to me

[noble nexus]

yeah exactly this is big

I try to get proofs to a point where I could explain the proof in a few sentences

[elfin wraith]

Must be nice, in my Galois exam today I forgot 2 proofs I had done an hour before

[past fjord]

can anyone recommend me a groups, rings and fields indepth yt video so i can learn and understand better, since my lecture notes that my prof gave are abit too hard to understand

[rotund aurora]

Benedict Gross has a nice lecture course on it. Also Richard Borcherds, but I preffer the former for an introduction

[past fjord]

how hard is groups rings and fields bc im a 1st year uni student doing maths and its my first time studying this

[verbal valley]

it's not inherently very hard, but it does take time to get used to the ideas involved and the style of problem solving -- especially depending on how much experience you have with proof-based courses

[past fjord]

i basically have no experience w proof based courses, abstract algebra is one of my first courses😭

[verbal valley]

this is not a problem but will mean you'll need to put more time into it than you might otherwise

it's just a matter of practice really

[rotund aurora]

Pretty much any math topic can be arbitrarily hard. But the basic abstract algebra sequence is great, a lot of fun and provides new ways of thinking. With patience and enough interest, it should come pretty smoothly

[past fjord]

i was struggling on the first part of algebra, relations for like a day or two, idk why it was so hard for me to grasp the concept...............

but groups, rings and fields do seem fun icl

[rotund aurora]

why the sob emoji? The division proof vs non-proof courses shouldn't exist, if it exists it's because there's people making programs who don't know what they are doing. Abstract algebra as a first course with proofs is fine (in this context)

[past fjord]

yeah thats true yk

ty

[tough raven]

I think the precaution exists to say, "don't memorise without understanding" (based on the assumption that if you say memorising is OK, it might be effectively interpreted as "memorising without actively building understanding/familiarity is OK"). If it's possible to effectively add the caveat (or you're talking to someone whom you think already knows it), then I think it's fine to suggest memorisation.

[azure cairn]

why are all intro algebra books so Big.

[mint seal]

groups, rings, and fields is a lot of material to cover

Herstein isn't too big though

[quiet pelican]

Also a lot of those books tend to cover more than a first course in groups rings and fields will

[mint seal]

indeed

[azure cairn]

Fair

[novel star]

algebra is big

[azure cairn]

Fair

[verbal valley]

also by virtue of being introductory books there is very little material they're skipping -- if you cut out all the routine stuff and just focused on key ideas they'd be a lot shorter, but also a lot less useful as introductions

[quiet pelican]

Why is “(., .)_M defines an inner product” needed here
I can’t see where it’s used

[verbal valley]

to even define the reflections you need to be in an inner product space

[rocky cloak]

I'm assuming it's part of their definition of root system

[quiet pelican]

There’s a comment right above this that you don’t actually need that much
You only need for (a, a) =/= 0

[verbal valley]

oh hmm

well yeah the point is that root systems generally live inside a real inner product space

[quiet pelican]

This is it
They just happen to separate the assumption “V is an inner product space” from their definition of a root system by about 2 pages

[verbal valley]

oof

[maiden crater]

I'm kind of confused as to why $char(F)$ being odd matters here

[cloud walrus]

Wai

[quiet pelican]

Why do you think it doesn’t?

[maiden crater]

fair

Could I have a hint

[quiet pelican]

Do you know that F^* is always cyclic?

[maiden crater]

Yes

oh right

[quiet pelican]

Use that

[maiden crater]

🤦‍♂️

yeah, that tells me at once the cardinality is 2

but char F being odd wasn't used at all

[quiet pelican]

Prove this

(Hint: you will need that the order of F^* is even to prove it)

[maiden crater]

looks like I missed some thoery

will be back in a bit

[candid patrol]

You can show that G/G^2 is an elementary abelian 2-group

for any group G

So [G : G^2] = 2^n for some n in the finite case

[quiet pelican]

G^2 isn’t necessarily a subgroup

(If G is abelian then this all holds, but not if it’s non abelian)

[maiden crater]

here why is alpha = t+ (p(t)) , it surely should be f(t)?

oh jus an assumption

[quiet pelican]

alpha is the coset in E_0 represented by t

[maiden crater]

and how is f(alpha)=((p(t))

because f(t)=0?

[quiet pelican]

Yes

[candid patrol]

It is, the notation G^2 is common, it denotes the subgroup generated by the squares of G. If G is abelian, then G^2 more simply refers to the group of squares of elements of G.

[maiden crater]

sorry for the follow up, but here are we assuming t is a root

[rocky cloak]

f(alpha) = f(t) + (p(t))
But f(t) is contained in (p(t)) by assumption, so
f(t) + (p(t)) = (p(t))

[maiden crater]

okay, so lemme try that method here

We start off by constructing an extension $F_3[x]/((x^3-x+1))$

[cloud walrus]

Wai

[maiden crater]

in this extension alpha is a root, so I divide x^3-x+1 by x-a ?

[rocky cloak]

Probably good to use t or some other symbol in the construction of your field, so you don't mix it with the indeterminate x.

Anyway, you mean to find the other roots of x^3 - x + 1? Then sure

[maiden crater]

I'm kind of confused

so I performed the long division

to get x^2-ax+a^2-1, with a reminder of -a+1

which makes no sense as that would force -a+1=0 or a=1

[rocky cloak]

If you got a remainder then you must have messed up a step

[maiden crater]

well, I will ,as a is just some element right. I'm just assuming that element is a factor

okay got it

[rocky cloak]

Well when I do the division I get a remainder of
a^3 - a + 1

Which was 0 by assumption

[maiden crater]

yup, just got that too

now I suppose I should see if a^2-ax+(a^2-1) has any factors in F_3[x]/(x^3-x+1)

before making another field extension?

[rocky cloak]

Think the should be
x^2 - 2ax + a^2 - 1

[maiden crater]

idts

lemme redo it

infact it should be +ax not -ax

[rocky cloak]

Yes -2 = 1

So that would be the same thing

[maiden crater]

okay so all roots are in $F_3[x]/(x^3-x+1)$, and similarly for $F_3[x]/(x^3-x-1)$

[cloud walrus]

Wai

[maiden crater]

now to get an explicit iso b/w them

okay,got it

thanks!

I suppose I have to just brute force (a)?

well, I could use the einiestein criteria ?

(sorry for butchering the spelling)

Well, so any polynomial with 2 as its constant coeff is irreducible

this doesnt give an exhaustive list though

[quiet pelican]

There are 18 possibilities
And you only need to check if +- 1 are roots

(I’ve ignored 0 as a root, and all the polynomials with constant term 0, because they’re trivially not irred)

[maiden crater]

±1 as -1=2 right

[maiden crater]

okay, but I still have to brute force

[quiet pelican]

Yes but it’s not hard brute force

[tribal moss]

No, consider (x+1)(x²+2).

[maiden crater]

uh, (n) is confusing me

what do they mean construct fields of order 27

F_3[x]/(x^3-x+1) is of order 27 and is a field, would that be sufficient

[tribal moss]

Yes, that's one of them.

[maiden crater]

they're all isomorphic though, are they not

so that's enough

[tribal moss]

Yes.

[maiden crater]

cool

thanks

[tribal moss]

You can also be systematic about it, though I doubt it will actually be faster:
A cubic polynomial over a field is irreducible iff it has no root in the field. That means there are 2^3 possibilities for what your polynomial's values on 0, 1, and 2 are. For each such set of values, Lagrange interpolation gives you a quadratic that takes those values; then add x³-x to make it monic of degree 3.
(Conversely, for any monic irreducible cubic, subtracting x³-x gives you something of degree at most 2 with the same values on F3, and we've just constructed all of those).

[maiden crater]

How does this sound.
\
Let $M$ be the set of common multiples of $a,b$. As $ab$ is a common mutliple we know $M$ is non-empty.
\
By WOP $M'={N(x) \mid x\ in M}$ has a least element. Let some $x$ corresponding to this least element be denoted by $m$.
\
Let $m'$ be another multiple of $m$. Then by the Euclidian algorithm $m'=mq+r$
\
\textbf{ I'm now wondering how to show r is zero, I think it will have to do with minimality of N(m) but unsure}

[cloud walrus]

Wai

[rocky cloak]

I guess you mean that m' is another common multiple of a and b.

Now remember that euclidean division tells you that N(r) < N(m) or r=0

[maiden crater]

Sure, but to use that I now need to show r is a multiple of a and b or even better a multiple of m

oh, r= m'-mq both of which are multiple of a,b

so r is a multiple of a,b

but N(r)<N(m)

so contradiiction

The last time I tried proving this I remember taking ≥1 hr

☠️

[quiet pelican]

Me when revising Coxeter groups

[maiden crater]

Prove that a quotient of a PID by a prime ideal is again a PID.

So just to be clear, I need to show

R/(p) is a PID?

[karmic moat]

yeah

[maiden crater]

hmm

[wraith cargo]

Hmm

[karmic moat]

hmm

[rocky cloak]

hmm

[maiden crater]

Well, as $(p)$ is prime, $R/(p)$ is an integral domain

[cloud walrus]

Wai

[maiden crater]

but idt that helps in proving it's a PID

[karmic moat]

PIDs are necessarily integral domains, by definition

[maiden crater]

PIDs are integral domains yes

but all integral domains arne't PIDs

[wraith cargo]

Wha

[maiden crater]

oops

wait

[south patrol]

I guess you mean "not all integral domains are PIDs" but yes

[wraith cargo]

Prime ideal domains

[karmic moat]

integral domain where every ideal is prime

what a wonderful idea

[maiden crater]

If only such objects existed

[karmic moat]

fields

[wraith cargo]

Are there non fields with this property hm

[maiden crater]

Every ideal of R/(p) is of the form $((a_1)/(p), (a_2)/(p),\dots,(a_n)/(p))$ But as $R$ is a PID, $(a_1,\dots,a_n)=(r)$ , so $((a_1)/(p), (a_2)/(p),\dots,(a_n)/(p))= (r)/(p)$. Thus it's indeed a PID

[cloud walrus]

Wai

[wraith cargo]

How do you know (r)/(p) is necessarily generated by a single element

[maiden crater]

as (r) is gnerated by one element

[wraith cargo]

Yeah but why can you conclude the quotient is as well

[maiden crater]

i don't follow

[wraith cargo]

You're making a jump in concluding it's principal without telling me why can't it happen that (r)/(p) = (a,b)

Like you haven't given me a good reason to think that (r)/(p) isn't generated by two elements

[maiden crater]

Becaiuse R is a PID every idea is pincipal

so (r) must be generated by 1 element

[wraith cargo]

Well you know (r) is principal