groups-rings-fields

[delicate orchid]

alternatively just like, (a,b) in R x S is invertible if and only if a is invertible in R and b is invertible in S

[tardy hedge]

Yea thats how i found it

[delicate orchid]

yurrr

[tardy hedge]

My final is on saturday

Just doing random questions

[delicate orchid]

so (R x S)^{\ast} = R^\ast x S^\ast

[tardy hedge]

Tbh studying for exams is weird

I look at questions in text and i just freeze cuz im like wtf questions should i even try

I wanna choose “good” ones

But then i realize trying any question is better than just sitting here staring at the page like an idiot

[rocky cloak]

Another way to see it is that to be isomorphic as rings they would have to be isomorphic as groups, which they're not.

[cloud solar]

Let G be a finite group. s(G)= the sum of all orders of elements in G. For a fixed n natural mumber find min s(G), max s(G), when G cover the set of all abelian groups with order n.

The max is obtained when G is cyclic right?

So max s(G) = sum when d|n of d×phi(d)

But how about the minimum

[tardy hedge]

Yes i was considering looking at it that way too!

[dire siren]

@cloud solar here's the solution: https://archive.org/details/gazeta-matematica-b/page/204/mode/2up
also, add the link to your resources list

[cloud solar]

Mersi frumos din nou

[cloud walrus]

Leno

[frosty tartan]

Is there any general way to prove the distributive property to show that something is a field?

[uncut girder]

Any polynomial in Z[x] that has alpha = sqrt(2) + sqrt(3) as a root must also have -alpha, beta=sqrt(2) - sqrt(3), and -beta as roots. Thus,the kernel of that ring homomorphism must be included in the ideal generated by the monic, irreducible, integer polynomial with \pm alpha, \pm beta as roots. Clearly, this ideal is also contained in the kernel. So this ideal equals the kernel.

Here's the proof: let alpha be a root of an integer polynomial f(x) and let sigma be a field automorphism C->C.
Claim: Then sigma(alpha) is also a root of f(x).
Proof: f(sigma(alpha)) = sigma(f(alpha)) = sigma(0) = 0.
The first equality uses the definition of field homomorphism.

Alpha is a complex number

I just did tho

What do you want me to explain

Do you know what a field homomorphism is?

Whats confusing

Do you know how if w is a root of a polynomial with real coefficients then the complex conjugate of w is also a root?

Its the same proof here

So maybe look at that proof again

Here's the general fact:
If K/F is a field extension, f(x) is a polynomial in F[x], alpha is a root of f(x) in K, and sigma is an automorphism K->K that fixes every element of F, then sigma(alpha) is also a root of f(x).

[rocky cloak]

If you're not a fan of field automorphisms I think you might be able to get an elementary argument with just a little brute force.

So it's enough to show that it's not a root of any polynomial of smaller degree, and by computing the (sqrt(2)+sqrt(3))^k for k=0,1,2,3 we see that it's enough to prove that 1, sqrt(2), sqrt(3), and sqrt(6) are linearly independent.

Assume
a + bsqrt(2) = csqrt(3) + dsqrt(6)

We may assume c and d are not both 0. Multiply both sides by c - dsqrt(2) yields

ac - 2bd + (bc - ad)sqrt(2) = (c^2 - 2d^2)sqrt(3)

Dividing by c^2 - 2d^2 and renaming constants gives

e + fsqrt(2) = sqrt(3)

Squaring both sides gives

e^2 + 2f^2 + 2efsqrt(2) = 3

So ef=0, which means either e^2 = 3 or 2f^2 = 3, neither of which have solutions.

So they must be linearly independent.

@marsh fulcrum

[south patrol]

hm weird q: does anyone have a particularly nice proof that if V is a k-vector space with dim V infinite then V^* is bigger than V ?

It's funny cause whenever I look up a proof I just find people making up their own arguments on stack exchange or smth lol

[south patrol]

Wait I'm now confused cause the proof I cooked up seems simpler than ones i've seen before, wonder if I messed up somewhere lol

[rocky cloak]

If k is countable or finite I guess it's pretty easy, since if V = k^(I) then the dual is k^I. So V has cardinality I while V* has cardinality 2^I.

I guess in general the argument would be similar, you just need a little extra work when the cardinality of k is bigger than I

[south patrol]

Yeah you can go from that to the general case by tensoring up and extending functionals (on stack exchange that's a top-voted argument)

The way I cooked up was like

Oh hm

Yeah so are the cardinalities of $V*$ and $V$ the same for when |F| >= |2^S|? Idk

[chilly ocean]

can someone give an explicit example of IJ where I and J are ideals?

don't know what they mean by all finite sums of products

[south patrol]

This certainly holds for, say, S = N and F = R (reals) i guess, so yeah something more algebraic is needed more generally

[chilly ocean]

the definition makes it looks like the size of I and J are the same which makes no sense

like what does a1.b1+a2.b2+ ... an.bn mean

[south patrol]

just a sum

So like

$IJ$ is the set of all $\sum_{i=1}^{n} a_i b_i$ over all tuples $(a_1,\dots,a_n)$ of elements of $I$ and $(b_1,\dots,b_n)$ of elements of $J$ and over all natural $n$

[cloud walrus]

homotopy coherent potato

[chilly ocean]

Can a1 and a2 be any element from I?

[south patrol]

Really though the important thing is that we take the set $S := { ab \mid a \in I, b \in J}$ and want to form the ideal generated by that. Now $S$ is clearly closed under multiplication y any element of our original ring and closed under multiplication, but $S$ isn't closed under addition, so we just chuck in all finite sums of elements of $S$

[south patrol]

Yes, a_1,...,a_n are just arbitrary elements

[chilly ocean]

So they can be equal too?

[cloud walrus]

homotopy coherent potato

[south patrol]

Yeah

[chilly ocean]

The set must be huge then

[south patrol]

Well it's a subset of R

(your ring)

it's also a subset of I

and a subset of J

so it's really not that big

(relatively speaking at least)

[chilly ocean]

If you have a1.b1 + a2.b2 there's |I| choices for a1 and a2, |J| choices for b1 and b2

[south patrol]

Yes but some choices coincide

[chilly ocean]

it's even bigger for a1.b1 + a2.b2 + a3.b3

[south patrol]

and anyway, it's a subset of I

and a subset of J

[chilly ocean]

Hmm repetition probably fixes this

I can't visualize an example of IJ

a summation of length 2 will take forever to check

let alone length n

[south patrol]

Hm I mean ideals can be complicated, yes

But I, J may already be complicated

[chilly ocean]

ring theory trips me up

At this point i was like

Maybe this pure math stuff isnt for me

I don't understand why things are defined the way they are

Like integral domains came out of nowhere

[tardy hedge]

Im new to it too but i think u can think of integral domains as like the ring being “one step closer” to being a field

Cause u have cancellation law in integral domains ab = ac a nonzero gives b = c

And then i also just learned today about quotient fields. For any integral domain , you can extend it and construct a field that contains the integral domain as a subring

[chilly ocean]

quotient algebras trip me up

There's many conditions that define whether R mod something is a quotient algebra or not

For quotient groups you need a normal subgroup, for rings you need prime or maximal ideals

for fields idk the general result but if Fp is a polynomial field the thing you're dividing by is an irreducible polynomial

Not sure how integral domains relate with quotient algebras

[south patrol]

No you don't for rings

You can quotient by any ideal

[rocky cloak]

To make a quotient ring you just need an ideal.
Quotienting by a prime ideal gives you an integral domain and quotienting by a maximal ideal gives you a field.

For the polynomial ring F[x], every ideal is generated by a single polynomial, and the ideal is maximal iff said polynomial is irreducible.

[chilly ocean]

aren't prime and maximal ideals equivalent?

[rocky cloak]

Maximal ideals are prime, but not all prime ideal are maximal

In just the same way that all fields are integral domains, but not vica versa

[chilly ocean]

Ohh

[white oxide]

why does $0 \otimes b = 0 \otimes 0$?

[cloud walrus]

okeyokay

[rustic crown]

what's (0+0)⊗b

[white oxide]

oh $0 \otimes b + 0 \otimes b = 0 \otimes b$

[cloud walrus]

okeyokay

[white oxide]

which implies that $0 \otimes b$ is the additive identity ig

[cloud walrus]

okeyokay

[white oxide]

yeah because quotient group

ok makes sense thanks

[rustic crown]

its the same way you show 0 * b = 0 in any ring :p

[white oxide]

yee makes sense i was too focused on the homogeneity condition in trying to see why lmao

[novel spruce]

Tensor products are bilinear, universally.

[rustic crown]

you would still need to do some form of that argument right?

but ig yea, it's sort of absorbed in the language

f : M x N --> P bilinear would mean f(-, b) : M --> P is linear, so f(0, b) = 0 always

hmm

ig another way would be this
0⊗b = (0*0)⊗b = 0⊗(0*b) = 0⊗0

since you're also allowed to juggle scalars (from the base right)

[white oxide]

so theorem 7.10 says that universal objects (and initial objects) are unique up to isomorphism in categories, but wouldn't that say that the map i is unique up to isomorphism in the category of all middle linear maps on A x B, rather than A (x) B is unique up to isomorphism?

[rustic crown]

yea common abuse of language

[white oxide]

ye but how would that translate to A (x) B being uniquely determined up to isomorphism?

[rustic crown]

that's just false

when people talk about tensor product, the structure map is also in their head

[white oxide]

oh so when they say A (x)_R B is uniquely determined up to isomorphism by this property they're referring to the canonical middle linear map?

[rustic crown]

same as kernerls/cokernels, you just refer to the object but there is always a map attached to it

[white oxide]

i see okay

oh wait yeah

[white oxide]

this doesn't make any sense at all lmfao

[rustic crown]

to show this, one would need to show taht it has no non-trivial automorphisms

but of course one can just look at multiplication by -1

[white oxide]

ah ok cool

thanks det

[rustic crown]

[south patrol]

it's the ideal generated by X^2

which is an ideal

all multiples of X^2

[rustic crown]

ideal generated by a set of elements S in a ring R is the collection
{r_1 * s_1 + ... + r_n * s_n | r_i in R, s_i in S}

[south patrol]

i suggest you look at ideals generated by a set of things

it is analogous to that yes

it's the smallest ideal containing S

i.e. intersection of them

analogously to having the smallest subgroup containing the generators

and so forth in other algebraic situations

[rocky cloak]

In general 'thing' generated by 'stuff' usually means the intersecting of all 'thing's that contain 'stuff'.

[tribal moss]

That is, if the stuff is already known to be a subset of an appropriate thing.
(If it's not, we need to talk about free things instead, but sometimes the "generated by" wording is used in that case too).

[white oxide]

can somebody explain to me how exactly tensor products make bilinear maps out of a product U x V "linear" (now considered with codomain U (x) V) - my guess is that it's linear in the sense that the induced mapping f has the property that f(a (x) b1 + a (x) b2) = f(a (x) b1) + f(a (x) b2), but when we call things linear, don't we usually want it to be true for all elements in the codomain? i,e, for all a1 (x) a2 and b1 (x) b2, we would want f(a1 (x) a2 + b1 (x) b2) = f(a1 (x) a2) + f(b1 (x) b2) rather than fixing the first element or the second element

idk if that makes any sense lol

[delicate orchid]

Call it middle linear one more time and I will scream at 240db

[delicate orchid]

It makes linear maps out of bilinears because there’s an isomorphism BiLin(AxB, M) = Hom(A (x) B, M) by universal property

[white oxide]

bro i'm just going by hungy 😹

[rustic crown]

what do you prefer? R-balanced/R-midlinear

[delicate orchid]

Multi linear. I prefer multilinear

[rustic crown]

but it's not that here right

A is a right module, B is a left module, C is an abelian group, A x B --> C

[delicate orchid]

Oh it’s non-commutative. I’m even more enraged now

[rustic crown]

[white oxide]

my genuine, honest to god reaction to that information:

[delicate orchid]

RAAAARRRRGGHHHH

[tribal moss]

You mean "domain" each time you say "codomain", right?

i,e, for all a1 (x) a2 and b1 (x) b2, we would want f(a1 (x) a2 + b1 (x) b2) = f(a1 (x) a2) + f(b1 (x) b2)
Yes, that is true. Your screenshot uses a special case of that to show how this agrees with bilinearity-to-the-right of the original map U×V -> W.

[white oxide]

oh yeah sorry oops

oh wait so

for all a1 (x) a2 and b1 (x) b2, we have f(a1 (x) a2 + b1 (x) b2) = f(a1 (x) a2) + f(b1 (x) b2)

[rustic crown]

mid-linear sounds cute tho ><

[tribal moss]

Yes.

[white oxide]

something's not clicking here for me then ig

[tribal moss]

I think the point of your screenshot is that whenever we have a linear map f: U otimes V -> W, the map g: U×V -> W defined by g(u,v) = f(u otimes v) will automatically be bilinear.
Conversely, if we start with a bilinear g: U×V -> W, then there is exactly one linear f: U otimes V -> W that satisfies f(u otimes v) = g(u,v) for all u and v.

(In the latter direction it is clear that can be at most one such linear map, because every element of U otimes V is a finite sum of things of the form u otimes v. The content of the claim is bilinearity of g is enough to guarantee that there will be such a map).

[tribal moss]

Looks like a prototypical example might be the evaluation map V* × V -> k when we view V* and V as right and left End(V)-modules.

[delicate orchid]

Gimme a min

Yeah I see it

[tribal moss]

I suspect physicists need to speak of such things in connection with operator algebras.

[crystal turtle]

det sounds cute >.<

[still ledge]

i need some help for this problem, here V is a finite dimensional
vector space over some field F and A(V) is the collection of homomorphisms
of V into V, i'm stuck at proving $vE \neq 0$ implies $vE = v$.

[cloud walrus]

pzh

[rapid junco]

if G iso H

and G_1 iso H_1

is G/G_1 iso H/H_1

this seems true but formally where does this follow

[novel spruce]

I'm not surprised you're unable to prove $vE = v$, for it isn't true in general. Hint for problem in screenshot: if $E$ is idempotent, so is $I - E$.

[cloud walrus]

swallowfisheye

[naive whale]

Counterexample, take G, H, and G_1 to be Z, H_1 to be 2Z. Z is isomorphic to 2Z but the quotients (cyclic group of order 2, and the trivial group) are not isomorphic

You can try to construct an iso or get one from universal property of the quotient but you will see the implication is that G_1 has to be the preimage of H_1 under the iso G -> H and H_1 the preimage of G_1 under the inverse

And all we know is that G -> H sends normal subgroups to normal subgroups, but we don't know which normal subgroup

[summer path]

rotman has an abstract algebra book that's quite good

[dim widget]

Why does it matter if there is other material in the book

You can just not read it

[dim widget]

I don’t know what you’re expecting. If you’ve read atiyah Macdonald that is basically the equivalent for commutative rings. Fields don’t really have very much structure just “as fields” so a book on number theory or Galois theory is what you want if you want to pursue this the way you’re thinking about…

In terms of noncommutative rings there are things you can read but it’s probably much better to have a more specific goal

I think this is more an issue of groups having a lot more which is known about their structure than say “all rings”

Prerequisite for Atiyah Macdonald is ring theory at the level of a general text on algebra

Galois theory is covered well in Lang’s algebra or Dummit and Foote if you do lots of exercises

[dim widget]

Dummit and Foote or Lang

there really isn’t a good one but reading the relevant sections of those will allow you to understand everything

that you want to understand

[mighty kiln]

Aluffi go brrr

[glossy crag]

Was there some elementary result that neatly classifies subgroups of Z/a_1Zx...xZ/a_nZ or does one have to do it ad hoc? Or, rather, does a subgroup of such a product necessarily a product of subgroups of each factor?

[mighty kiln]

No, e.g. {(n, n)} ⊂ G×G

But yes if a_i pairwise coprime then ye

And ideals of product of rings are products of ideals of rings

[glossy crag]

I am aware, I meant for the a_i general, e.g. stuff like Z/4ZxZ/2Z (this one's easy and is not the problem, my question was do I have to do it by hand every time or not)

[mighty kiln]

I think this is iff actually

[glossy crag]

I don't understand what this is supposed to mean, is this the diagonal (because that's not a subgroup) or the trivial subgroup (because that's the product of trivial subgroups)?

[mighty kiln]

Ye the diagonal

[glossy crag]

That's not a subgroup, dude.

[mighty kiln]

?

[glossy crag]

Oh right, of GxG, not general products, my bad.

[mighty kiln]

Actually what if you factor into products of (Z/mZ)^n with coprime m using CRT or sth

[glossy crag]

Hm, I'll think about that.

[mighty kiln]

Oh wait that doesn't work

[still ledge]

Yep, my previous attempt was wrong, but I don't know how to use I-E is idempotent. However, I do realized v = v(I-E) + vE, and v(I-E)E = 0, (vE)E = vE.

so only thing left is to prove the intersection of the image of I-E and the image of E is trivial.

[rotund aurora]

I think it might be better to consider the decomposition Z/d_nZx...xZ/d_1Z with d_i|d_i+1 for each 1<=i<n

then what Ocean said is true, in terms of isomorphism

[mighty kiln]

How do you classify the subgroups

[rotund aurora]

I didn't

I said that they are isomorphic to a product of subgroups of each part

[mighty kiln]

Are they

Say <(1, 2)> ∈ Z/2Z × Z/4Z

[rotund aurora]

Yes, because the order of every element divides d_n and these are abelian groups

that has order 2?

[rotund aurora]

if you just want to see this, it is enough to show that it is a direct product of subgroups of Z/d_nZ

[cloud walrus]

pzh

[mighty kiln]

<(1, 2)> is not <1> × <2> tho

[rotund aurora]

it is isomorphic to Z/2Z

right?

[mighty kiln]

Ye

[rotund aurora]

Z/2Z is a subgroup of Z/4Z, so there you go

[mighty kiln]

??

[rotund aurora]

I said in terms of isomorphism, not a classification

[mighty kiln]

It's not a product of subgroups tho

[rotund aurora]

Z/2Z iso Z/2Z x {1}

{1} is a subgroup of Z/2Z

but

or of Z/4Z even

[coral shale]

💀 what the

[mighty kiln]

Oh any isomorphism

I thought you mean a canonical one

[rotund aurora]

that's why I said I didn't classify them, in terms of a nice description of how they lie inside the group

[mighty kiln]

Tho this actually gave me an idea

[teal vessel]

so I have to show that the union of all subgroups in H is a subgroup of G, right? specifically each subgroup is an element of H, not a subset of H?

[void cosmos]

u have a chain of subgroups of G

u want to prove that the union of this chain of subgroups is in S basicaly

hence being a maximal element in the chain

[teal vessel]

I'll get there. Right now I'm having to read the definition of "chain" very carefully. The definition I had was slightly ambiguous but I think I'm getting it

a chain is an ordered subset, right?

[void cosmos]

yes

its just any subset of a poset

if its totaly ordered

[teal vessel]

poset being a partially ordered set?

[void cosmos]

ofc

yes

[teal vessel]

ok. the definitions used the word "in" when talking about both subsets and elements in the same sentence, so it threw me for a loop

[void cosmos]

no ur elements are the subgroups themselves

they are ordered by inclusion

is the union of the subgroups an element aswell? ( is it a subgroup )?

do not be mislead into thinking thats the case in general tho

the union of any two subgroups may not be a subgroup

but this is a chain

so they are ordered by inclusion

[teal vessel]

in a well ordered set by inclusion, union just acts as effectively a max function

[untold basalt]

still field theory

[void cosmos]

it would be a maximal element yes

the question is asking ifts even an element

if its*

[teal vessel]

S is the set of all subgroups of G, which is finitely generated. The subgroups of a finitely generated set are finitely generated (by a subset of the generating set of the supergroup). Given that S is treating each subgroup as a single element, we can bound the cardinality of S to be, at most, equinumerous to the power set of the generating set for G, which is finite. Any subset of a finite set must be finite, so C is finite.

looking at the chain C, because it is well ordered by inclusion, the union of any two elements returns one of the elements (the greater one, but this is unnecessary information for this part). Because there are finitely many elements of C, we can define the union across C as a collection of finitely many pairwise unions. Union is both associative and commutative on finite sets, therefore we can simply define the union across C as c_1Uc_2U...Uc_n. Since each pairwise union will return a member of C, we can collapse this down like a single-elim tournament where at the end we have an (largest) element of C. C is a subset of S, which is defined as having subgroups of G as elements, therefore the union across C must also be a subgroup.

overexplained? yes, but thorough.

it seems easier to prove given that C is a subset of S which only contains proper subgroups. I'm not sure what contradiction would arise. I suppose if every element of the generating set is in the chain, then C=S, and technically the generating set for G would have superfluous elements in it. I'm not sure how this is a contradiction tho

final part, though: Since C was an arbitrary chain in S, the argument that H is an element of C (and subsequently S) applies to all chains in S, therefore, by Zorn's lemma (if the union across any chain in A produces an element of A, then A has a maximal element), S must have a maximal element, which is the maximal subgroup by the construction of S.

[untold basalt]

Is this proof correct?

[glossy crag]

This is dumb, but what's the smallest equivalence relation generated by a subset S<= X x X? According to the internet it's "xRy if and only if there is a sequence z_1,...,z_n such that xSz_1 and z_jSz_j+1 and z_nSy", but this definition doesn't even seem to be reflexive and symmetric. Shouldn't the answer be this united with {(x,x)} and {(y,x)} for every xSy?

[wraith cargo]

Smallest in what sense

[tribal moss]

Yeah, it needs a correction along those lines.

You can get {(x,x)} by allowing the case n=1.

[wraith cargo]

Wouldn't the smallest one just be the diagonal?

[tribal moss]

Wait, not exactly since it doesn't say x=z1 but xSz1..

Usually the diagonal will not contain S.

[wraith cargo]

Ahh

You pick out a random subset

I see

I thought we were looking for the smallest S lol

[tribal moss]

It's probably most economic to say

xRy iff there is a sequence z1,...,zn with n>=1 such that x=z1 and zn=y and for each applicable j we have zjSz(j+1) or z(j+1)Szj.

[mighty kiln]

Oh wait that's covered

[celest furnace]

Can anyone give me a hint to showing the nilradical of a ring is the intersection of all prime ideals

[wraith cargo]

You'll need to use Zorn's lemma

To show there doesn't exist a prime outside the nilradical

[rocky cloak]

It's not true that a subgroup of a finitely generated group is finitely generated.

For example the free group on two generators has the free group on countably many generators as a subgroup.

So maybe you meant for G to be a finite group...

[teal vessel]

I don't know that example, can you clarify?

[rocky cloak]

If G is the free group generated by a and b, then the subgroup generated by a^n b a^-n for all n in Z is freely generated by those generators.

(which is not finitely generated)

[teal vessel]

I don't know what a free group is.

[rocky cloak]

The free group is just the group of words under concatenation.

So the free group generated by a and b consists of things like
a, b, ab, aab, abab, a^(-1)bab^(-1), etc
Where the only restriction is that a a^(-1) is the identity and vica versa + same for b.

https://en.m.wikipedia.org/wiki/Free_group

[teal vessel]

Dammit

[languid trellis]

swifteeee's foray through artin

[teal vessel]

I suppose, because I know that the chain C is necessarily well ordered (by definition of being a chain), I can still hold the same properties without supposing that C has finite cardinality. I need to look over my set theory notes for this one.

my brain really wants to set up induction.

by the union axiom, I know that H=Uc exists, which is reassuring, at least. I also know that the bottom of the chain c has to be {e}.

[celest furnace]

This was crazy

[white oxide]

a bit of a silly question ig, but when we talk about bilinear maps factoring through the tensor product, the induced homomorphism from A (x)_R B --> C (say given the initial bilinear map A x B --> C) is simply considered a group homomorphism and not a R-module homomorphism or Z-module homomorphism, right? because C is an abelian group, and A (x)_R B is a module over any arbitrary ring R

[tribal moss]

(Never trust a book that writes \in as \varepsilon).

[white oxide]

LOL

yeah hungerford's font is weird

[tribal moss]

One needs to be pretty careful here. For a concrete example, let R = Z[i]. Then we can define a Z-bilinear map f: R × R -> Z by f(1,1) = 1, f(i,1) = 5, f(1,i) = 42, f(i,i) = 0. This map doesn't factor through R otimes_R R, though -- because i otimes 1 is the same element of the tensor product as 1 otimes i, and that single element cannot map to both 5 and 42.

(It factors through R otimes_Z R, of course).

As your quote points out, a tensor product over one ring can inherit a module structure over a different ring that only one of the factors has a scalar multiplication for. However, the usual universal property of tensor products will not even make sense for that module structure. The theorem can be thought of as stating what you can know about this inherited module structure (all while making things maximally confusing by doing everything for the non-commutative case).

[white oxide]

huhhh okay

oh wait yeah i think that makes sense

yeah it's mad confusing

thank you

here do we even need the universal property? for showing that A (x)_R R is isomorphic to A why can't we just define maps f: A (x)_R B --> A given by a (x) r |-> ar and g: A --> A (x)_R R given by a |-> a (x) r and show that they're inverses? is it because if we use the universal property, we are immediately guaranteed that these are well-defined and also homomorphisms?

[rustic crown]

why can't we just define maps f: A (x)_R R --> A given by a (x) r |-> ar
how will you show this is well-defined?

[white oxide]

yeah okay makes sense

[white oxide]

so basically the "universal property of tensor products" is a way to ensure that the map you're considering is well-defined

i.e. if i wanted to construct a map f A (x) B --> C and i have some idea, i can just express it as a bilinear map f': A x B --> C, get my factorization through the tensor product which is exactly f, and boom it's well-defined and shit

because it seems like it's very tedious to check things are well-defined in the tensor product since you have lots of collapsing

[remote nymph]

this problem is giving me a lot of trouble.

I've seen people online write $u \otimes v = 1/2(u \otimes v + v \otimes u) + 1/2(u \otimes v - v \otimes u)$

[cloud walrus]

srhoosteen

[remote nymph]

But what does multiplication by 1/2 even mean in an arbitrary vector space?

[delicate orchid]

Multiply by the inverse of 2

Hence why F cannot be characteristic 2 in your question

[remote nymph]

Let R be an integral domain, let M be a nonprincipal ideal. why is it the case that if M is free, then M is isomorphic to R?

[brave agate]

Been staring at this way too long, any ideas?

so since $\phi$ is a surjective homomorphism, by the First Isomorphism Theorem

$G/K \cong I$ where K is the kernel of $\phi$

[cloud walrus]

BLONK

[wraith cargo]

Now show that G/K is exactly H

Or eeer isomorphic to it

[brave agate]

I noticed $h_1 h_2 K = h_2 h_1 K$ since $G/K$ is abelian, but went nowhere from there
Bijection seemed impossible to construct

[cloud walrus]

BLONK

[brave agate]

Showing G/K is isomorphic to H would require a bijective homomorphism, which isn't guaranteed here (only surjective), so I wasn't sure where to go

the correspondence theorem seems like it could be useful here but H isn't normal, but K is

[next obsidian]

A non principal ideal cannot be free because if it were it would be isomorphic to R, and thus principal.

To see that a free ideal is isomorphic to R, it suffices to show that you cannot have a free module F and a submodule F’ free of a higher rank, then the statement above is a special case when F = R.

This is true over any ring, but the proof is easier over an integral domain. If F is rank n and F’ is rank m, then letting S = R\{0} you have S^-1F’ < S^-1F, and these are vector spaces over S^-1R of rank m and n respectively. Thus, m <= n

[brave agate]

also we know G has an abelian subgroup of order p

[celest furnace]

Do you know about semi-direct products?

[wraith cargo]

Do you want him to use schur zassenhaus here lol

[wraith cargo]

Notice that K is of order p so it's also abelian

Tho that probably isn't useful at all lol

[brave agate]

yes the kernel is abelian (order p)
and there is a group we know that is abelian of order n (G/K)
not sure how we can introduce H into this idea though

thought maybe i could apply these

[celest furnace]

||If you write K = ker phi then G is iso to K semidirect H and then K semidirect H mod K is iso to I which is just H||

BLONK DO NOT READ THAT MEZZAGE

Hint : try showing H cap K = 1

[celest furnace]

I actually didn't know this had a name thats awesome

[rapid junco]

If G has a subgroup of size k then does it also have a subgroup of size g/k?

[wraith cargo]

Yeah
If you map G into the subgroup of order k
Then the kernel would have order g/k

[rapid junco]

Just the projection map?

Does this not guarantee then that every group has a non trivial normal subgroup which is false?

[wraith cargo]

Okay wait I've been awake for at least 24h lemme think about it more

Hm okay my bad this isn't true

The map needn't always exist

[rapid junco]

bump

[mighty kiln]

S5 has a subgroup of size 2, but no subgroup of size 60

Oh wait it does

Well it has a subgroup of size 3, but no subgroup of size 40

[rapid junco]

mm okay

what if we require our subgroup to be normal, then is this true

then G/H is defined and by the lattice theorem its in bijection with the subgroups of G that contain H

do those subgroups have opposite order?

[mighty kiln]

A subgroup of size k in G/H corresponds to a subgroup of size k|H| in G

[rapid junco]

mm okay

[granite topaz]

Is there any characterisation of a subset S such that the normaliser of S is perfect?

i.e. $[N_G(S), N_G(S)] = N_G(S)$

[cloud walrus]

Megumi_Tadokoro

[granite topaz]

this cannot hold for any subsets for sure, let S = G, then it is equivalent to G^{ab} being trivial, and it's definitely not true for all groups G

[cloud walrus]

Seagull

[untold basalt]

Is this correct?

[chilly ocean]

can someone explain the following proposition to me?:

Let $\cdot$ denote the action of a group $G$ on a set $X$. For $g \in G$, define the map $\phi_g : X \to X$ by $\phi_g(x) = g \cdot x$ for all $x \in X$. Then $\phi_g \in \text{Sym}(X)$ and the map $\phi : G \to \text{Sym}(X)$ defined by $\phi(g) = \phi_g$ is a homomorphism.

[cloud walrus]

whwwi

[coral spindle]

Be specific about what part you're struggling with.

[chilly ocean]

so

I don't understand the significance of $\phi(g)$ being a homomorphism

[cloud walrus]

whwwi

[chilly ocean]

are we trying to say that every group action can be viewed as a permutation?

[coral spindle]

I don't see what you mean by that. Wym 'significance'. It is a homomorphism.

[chilly ocean]

how so

[coral spindle]

Try proving it.

[chilly ocean]

because of how group actions are defined?

oh yes

ok

so i have to prove that $\phi_g(x) \phi_h(x) = \phi_{gh}(x)$?

[cloud walrus]

whwwi

[coral spindle]

Yup

[chilly ocean]

i kind of don't understand why $\phi_g(x) \in Sym(X)$.

[cloud walrus]

whwwi

[coral spindle]

Then prove that too.

Oh wait no

phi_g is in Sym(X).

phi_g(x) is in X.

Got that?

[chilly ocean]

oh

OHHH

phi_g is a mapping from X to X so it IS in Sym(X) because it is a bijection?

[coral spindle]

Have you proved it's a bijection?

[chilly ocean]

I have not

[coral spindle]

Then that's all you need to prove in order to convince yourself it's an element of Sym(X).

[chilly ocean]

by proving that phi_g is a bijection will I have proven that each group action is actually a permutation?

[coral spindle]

You've answered your own question, yes

[chilly ocean]

oh...

lmao thanks so much

[coral spindle]

I still have no idea what the confusion was

[chilly ocean]

so I can think of any group action as an element of Sym(X)?

[coral spindle]

Yes

Well

[chilly ocean]

is that related to Cayley's theorem?

[coral spindle]

more accurately, a homomorphism G → Sym(X).

[coral spindle]

Think about this yourself.

You should see it quite quickly.

[chilly ocean]

one sec... i learned about cayley's theorem a while ago. will read a wiki page and remember

oh

every group is isomorphic to a permutation group

[mighty kiln]

Can you find the isomorphism and the permutation group

[chilly ocean]

but will it be OK to view a group action, say D_6 acting on P_6, as an element of Sym(P_n)?

as P_n is just the set {1, 2, 3, 4, 5, 6}

[chilly ocean]

good question

[chilly ocean]

and so this would be a homomorphism $D_6 \to Sym(X)$?

[cloud walrus]

whwwi

[mighty kiln]

The action of one element in the group can be thought of as an element of Sym

But the whole group action is a homomorphism from the group to Sym

[chilly ocean]

OH

this was exactly what I needed. thank you @mighty kiln and @coral spindle

much appreciated

[tribal moss]

It that is true, it can only be vacuously -- if M is isomorphic to R, then it would be free by definition.

(Whoops, missed several pages of scroll, and this was resolved already, sorry).

[glossy crag]

Barbeau and Prasolov both have books titled "Polynomials", Prasolov's in particular goes pretty hard, and IIRC they have some problems, but I don't expect there to be a whole book dedicated to what is essentially basic algebra. That said, there was also a problem book on fields/Galois theory, it might have what you're looking for.

I remember only the first chapter or two, but I read them a long time ago. It starts off at high-school level and I think more or less stays there (early uni at most).

[glossy crag]

The book I mentioned is "Galois Theory through Exercises" Brzezinski

[scenic torrent]

Hi guys! Could someone explain me, which areas use group theory?

[glossy crag]

There might be something useful to you here @marsh fulcrum

[glossy crag]

For a question this broad you're better off reading the relevant Wikipedia article https://en.wikipedia.org/wiki/Group_theory#Applications_of_group_theory.

[scenic torrent]

omg yes) I don't know why I missed this option, thank you)

[languid trellis]

I'm struggling quite a lot to even get started on this exercise

Could someone give a hint to help out please?

i've got 2 pages of stuff that I wrote down

i suppose it suffices to show that xH \cap yK and some coset of the subgroup H \cap K share an element right?

wait maybe not

im weirdly lost on this exercise

[cloud walrus]

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

[delicate orchid]

have you overloaded the use of "y" here or is this actually want you want

I don't see how you've shown y is in the intersection

or rather, what happens if it isn't?

[languid trellis]

i got to the stage where I convinced myself $y^{-1}x \in y^{-1}x(H \cap K)$

[cloud walrus]

swifteeee

[languid trellis]

from xh = yk \in xH \cap yK

and I'm lost

[coral spindle]

Hint: if $k \in xH$ then $xH = kH$. Now imagine the intersection is nonempty, and see what that gets you.

[cloud walrus]

Boyt

[delicate orchid]

assume intersection is non empty, then we have some $g \in xH \cap yK \iff g = xh = yk \Rightarrow h = x^{-1}g, k = y^{-1}g$, it's clear from this that $gH = xH$ and $yK = gK$, hence $xH \cap yK = gH \cap gK = g(H \cap K)$

[cloud walrus]

Wew The Lads Tbh

[coral spindle]

Gah

[delicate orchid]

sorry kinda spoilt the entire thing but you know what. I'm the boss around here

[coral spindle]

UM MODS???

[languid trellis]

didn't even think about how representation of cosets isnt unique

[delicate orchid]

the way I approached this is that I knew we had to get some kind of "x = y" situation going on because then it's obvious

and an element of the intersection struck me as a good choice to try

[languid trellis]

that's so unbelievably obvious now

[coral spindle]

Dontchya wish you kept trying huh

[delicate orchid]

it's always obvious once you understand it

[languid trellis]

I was trying for like an hour boyt

who made algebra so god damn hard man

[coral spindle]

Dw it gets harder

[delicate orchid]

Grothendieck

Grothendieck is the one to blame

[languid trellis]

i can't believe i'm having less isssue with rudin than this book

[mighty kiln]

Algebra is like linear algebra

[languid trellis]

maybe i should just ask more here

[delicate orchid]

except without a basis. Or a nice invariant like dimension.

[mighty kiln]

A lot of it becomes easy once you can intuit

At least for undergrad algebra

[tardy hedge]

For factor rings, we are quotienting the ring based on its group structure right?

And the multplication is just kind of tacked on it seems

[languid trellis]

wait how did you conclude gH = xH

[delicate orchid]

they differ by an element of h

the specific equivalence relation that gives you cosets is x ~ y <=> xy^-1 \in H

and since we have h = x^-1g...

[languid trellis]

oh right

yeah ofc

[coral spindle]

Yes, the equivalence relation comes from the additive group structure. For a general algebraic structure (such as a semigroup), when we quotient we need to consider special equivalence relations called congruences, which can be very hard to do. However, groups are somehow 'rigid' enough that these congruences correspond to normal subgroups, so we can consider quotients by normal subgroups as opposed to quotients by congruences. Ring quotients also exploit this, since they are – in particular – Abelian groups.

[delicate orchid]

I don't actually know about congurences of monoids

do they corrispond to subobjects

[coral spindle]

No unfortunately not

We really need inverses

[delicate orchid]

very sad

what about loops then

[coral spindle]

But this isn't too hard to intuit: every monoid is a semigroup, and every semigroup can naturally be made a monoid, so clearly they have equal grossness

no idea for loops

[delicate orchid]

and every monoid can be naturally (literally) made into a group so I legit don't see this argument

[delicate orchid]

you passed the sanity check

[delicate orchid]

although if you want to be heuristic, yes - adjoining formal inverses is a "bigger" operation than adjoining a single identity

[coral spindle]

Well not really

not every monoid embeds in a group

They can in a natural way, yeah, but this will often cause collapse.

A bit like how noncommutative localisation fucking SUCKS ASS

[delicate orchid]

well yeah that's precisely why it sucks ass lmao

[coral spindle]

Now for commutative monoids, you can do this though. Hm.

So

[delicate orchid]

I forgot grothendieck completion was only for commutative monoids OK

I ADMIT IT

YOU GOT ME

[coral spindle]

Every day I wake up and think of ways to one-up wew

Commutative, cancellative monoid. I would imagine there's a nice correspondence here. I may spend some time thinking about it... later.

[delicate orchid]

I spend as much time as possible avoiding thought

[coral spindle]

Holy shit me too... don't tell my supervisor

[languid trellis]

quick question, here, we're just observing that the upper bound of [G : K \cap H] is [G:H] * [G:K} by the previous exercise?

[delicate orchid]

yeah that works

[languid trellis]

thanks mr angloid

[delicate orchid]

WHAT

MODDDDSSS

[languid trellis]

its okay i can't be racist against angloids because im (supposedly) an angloid

[mighty kiln]

What's an angloid

Is it like an anglerfish but humanoid

[languid trellis]

no its a derogatory term for an english person

(but actually thanks wew for saving me)

[chilly radish]

......ore not

[brave agate]

H cap K = 1 because H cap K is a subgroup, and its order must divide both H and K, and since their orders are coprime, it must be trivial

therefore G is a semidirect product of H and K

therefore H and G/K are isomorphic, and G/K is isomorphic to I, done

it seems I could also not use semidirect products at all and just apply second isomorphism theorem and prove HK = G

[untold basalt]

I was recommended Jacobson by my teacher

Basic algebra 1

also not the right channel for this

[celest furnace]

Yep that works too

[tardy hedge]

Dihedral group Dn = {a^k, a^kb} with k going from 0 to n right

a is just one rotation and b is a flip?

[coral spindle]

Yup

[tardy hedge]

And it doesnt matter what i say b and a is right

[coral spindle]

A rose by any other name

[tardy hedge]

Could be any flip and rotation then all the rest of the elements come oit from rhat

[coral spindle]

No, only some rotations will do that

If you look at D_10000000 or whatever, and you look at the subgroup generated by the 180-degree rotation and flip stabilising that, this will not generate the whole group.

[tardy hedge]

Yeah right

Needs to be a rotation that generates everything

Generator of cyclic group

[coral spindle]

Yup

[brave agate]

The other problem I was having trouble with: it's easy to show the order 5 subgroup with Sylow's theorem (of the subgroup of order 15, let's call it H), but normality isn't transitive. I thought maybe a counting argument could work but the possible number of 5-Sylow subgroups in the original subgroup is 1, 6, or 21, none of which create a contradiction.

wait this is trivial isn't it

no nvm abelian subgroups aren't necessarily normal

[gentle pendant]

A bit rusty at this kind of thing, but if you let H be the normal subgroup of G of order 15, and N the normal (in H) subgroup of H of order 5, then conjugation of N by any element of G gives another subgroup of G of order 5, but by normality of H in G, this subgroup must be contained in H.

By uniqueness of the 5-group N in H it follows that this conjugated subgroup must always be N itself, so N is normal in G.

[white oxide]

lorenz

[white oxide]

can somebody explain to me how this is not a priori reasoning? (proof of division algorithim in A[x] where A is commutative) because f has degree n

namely the last math display part

[delicate orchid]

all of the degree >= n terms are taken care of by the g(X) term so we can safely set f_1(X) to be degree less than n

we're shifting g(X) up to be a degree n thing and adjusting the coefficients to equal f(X), then sorting out the lower degrees by sticking an f_1 on the end

[white oxide]

i see, okay

okay got it, i think that makes sense

thank you

[white oxide]

why do we replace $X_i^{v_i}$ with $X_i^{\mu + 1}$ instead of $X_i^{\mu}$? For $a^{v_i} = a^{q + \mu} = a^q \cdot a^{\mu} = a^{\mu}$

[cloud walrus]

okeyokay

[rose prism]

why is a^q = 1?

i dont think that is right

[white oxide]

am i trippin

i thought that |k| = q => a^q = 1 for all a in k

[rose prism]

no

[white oxide]

damn

[rose prism]

i am almost certain that is always false

if q = prime then it is just fermat little theorem

which disagrees with what you wrote

[white oxide]

damn guess i gotta learn algebra again

[rose prism]

and i am almost certain that the conclusion of fermat little theorem applies for q = p^n

recall that the multiplicative group of a finite field is cyclic

also by lagrange theorem the elements all have order dividing q-1

[coral spindle]

Yes you're just forgetting that the group of units has order |k|-1, not |k|.

I mean for goddness' sake how could it be that 0^q = 1???

Anyway

Back to timewasting

[delicate orchid]

watch this

[coral spindle]

I'm watching

[long obsidian]

Can someone help me with this basic homology computation

If I have ...->Z_8->Z_8->Z_8->0.... With the last differential 0 and the others d(x)=4x then H2={0,2,4,6}/{0,4} since any even number will be sent to 0 and {0,4} is clearly the image of d.

What is H2 here?? I can't do algebra right now. Do I need to think about the orders of everything here. So for instance 4/2=2 so H2=Z_2?

[south patrol]

that notation seems a little messy

[south storm]

Yes, one has that |G/N||N|=|G| for any group G and normal subgroup N of G. Consequently your second homology group will have order 2 and thus will be cyclic of order 2.

[delicate orchid]

it's also just a quotient of a cyclic group?

[celest furnace]

Need help with this one--determined that <A, B> is iso to S_3, and of course C multiples the top row by i (when on the left), so the group is iso to some sort of weigthed combination of S_3 and (Z_4)^3

[chilly ocean]

finding the orders of A, B and C might help

since by Lagrange's theorem o(g) for g \in G divides o(G)

[tardy hedge]

What is <2,x> in Z[x]?

Is it the set {2p(x)+xq(x)}

For p and q in Zx

If not i fucked ip my exam lol

[rocky cloak]

The group is indeed the semidirect product of (Z/4)^3 and S3, which you can see by conjugating C with A and B.

Since the max order of an element of S3 is 3 and in (Z/4)^3 it's 4, you can't have any element of order > 12. So if you can find an element of order 12 you're good.

[rocky cloak]

That is indeed what it is.

[tardy hedge]

Oh cool. The question said show that <2,x> is not a principle ideal. I said basically if it was then there is a polynomial that divides everything in <2,x>, but thats not possible since 2 and x are in <2,x> and 2,x have no common divisors

Is that good?

[rocky cloak]

Yeah, that sounds pretty good

[tardy hedge]

Cool thanks

[rustic crown]

to find the number of elements of order 12, ig its easier to work with the semi direct product description where S3 acts on (Z/4Z)^3 by permuting the factors. so in (Z/4Z)^3 ⋊ S3, order of s will divide (a, s), therefore you want s to be order 3 permutation, and then the first factor in its cube will look something like a + s(a) + s^2(a), and you would want this to have order 4. in other words if a = (x, y, z) then you want x+y+z to be not even as s(a) and s^2(a) are the two cyclic permutations of (x, y, z). half of the elements will have this quantity even, while others half will have it odd, so 4^3/2 choices for each s, and there are 2 permutations of order 3. so 4^3?

[celest furnace]

I was visualizing it like this too--"weighted permutations of {1, 2, 3}"

[delicate orchid]

oh my golly gosh it's a wreath producterino

[tardy hedge]

Thank u for help this term

I finished exam today

And thanks other ppl too

[celest furnace]

W

[delicate orchid]

I don't recall helping you but I recognise you lol so yw

[tardy hedge]

Hahahaha

[median rock]

if i have a group G with |G|=8 then i have only one 2-sylow subgroup right?and if i am right then is it right to say that G is isomorphic to Z8?

[delicate orchid]

the entire group is the sylow two subgroup

and no, the group can be Z8, Z2xZ4, Z2xZ2xZ2, D8 or Q8

[median rock]

fu** you are right ,i am trying to prove that there is only one element with order 2 and i thought an isomorphism would help

[delicate orchid]

do you know anything else about G

because at the moment that just isn't true

[median rock]

ye

i forgot

G=<a,b> ,a^4=e,a^2=b^2 and bab^(-1)=a^-1

i said that there is at least one from cauchys theorem

[delicate orchid]

that looks like Q_8 to me

so b^2 = a^2 will be your element of order 2

not sure if there's a faster way for this particular group than just... checking the orders?

[celest furnace]

Sorry to bother you but i was wondering if the second fact was true here since (Z/4)^3 is abelian I cant seem to prove it otherwise

[median rock]

ye i thought that i could show it with a more 'elegant way ' if you know what i mean

[rocky cloak]

Which fact do you mean?

[delicate orchid]

the order argument

[celest furnace]

If every element in G has order dividing n and every element in H has order dividing m then every element in G semidirect H has order dividing m*n (This was false! What is true is every element has order <= m*n, like jagr said before!)

Was able to show it for G abelian

[delicate orchid]

well that's just true by lagrange lol

oh they're not the orders of the groups

rip

[celest furnace]

Yeah not necessarily

[rocky cloak]

It doesn't really have anything to do with being abelian

[celest furnace]

I got like: (x,y)^3 = (xy*xy^2*x, y^3)$ for example and you can see if you take this to the mn power you can commute all the elements to get (y^alpha x)^n = 1 for some alpha (For abelian in the first coordinate)

[rocky cloak]

It's just if you have a normal subgroup N < G and you consider an element g in G. Then if n is the order of gN in G/N, then g^n will be in N.

[rocky cloak]

Right, so if y^n = e, then (x, y)^n will have order at most m.

[celest furnace]

Aaaah

I see thats awesome

[warm saffron]

Hey there, so I'm working on a problem that asks me to show that there can't be a surjective homomorphism from a group of order 10 to a group of order 6. So I suppose f:G->H is an onto homomorphism, where |G|=10=2(5) and |H|=6=2(3). And then I was looking at a given h in H. I know there is some g in G for which f(g)=h, and e_H=f(g) f(g)^(-1), but I'm not getting anywhere....

[charred iris]

do you know the first isomorphism theorem (might have been introduced without a number idk)?

[warm saffron]

yea something about G/ker f being isomorphic to f(G) ?

[charred iris]

yes

[warm saffron]

so f(G) is the image and ker f is all the things in G that map to e_H, right?

[charred iris]

yes

[warm saffron]

hmm I'm lost on how to apply that here

I know we must also have |f(g)| dividing |g| as well for all g in G

[prisma ibex]

you know f(G) has size 6, so G/ker(f) has size 6, but this can't happen since...

[warm saffron]

so since we assumed f was surjective f(G)=H.

and

since it's isomorphic to G/ker f, that must mean G/ker f has size 6 as well

I'm just connecting the dots sorry guys

oh

wait G/ker f= {a in G: a ker f}

yea I'm lost again

[coral spindle]

Hint: let N be a normal subgroup of G. What is |G/N| in terms of |N| and |G|?

[warm saffron]

6 doesn't divide 10, so that's a problem

oh

my

gosh

[coral spindle]

(:

[warm saffron]

wait

so I should explain why ker f must be of size 6

?

[coral spindle]

Gl

[delicate orchid]

No because that’s completely unrelated

[prisma ibex]

ker(f) would have to be of size 6/10

[warm saffron]

my prof just threw 20 theroems at us about homomorphisms and I'm so lost on how they all connect here

I know ker f is a normal subgroup of G

[delicate orchid]

|img(f)| = 6 = |G|/|ker(f)| = 10/|ker(f)| by first iso and everything here must be an integer

Immediate contradiction

You’ve said all of these things yourself but didn’t seem to realise that you had proven it

[warm saffron]

and img(f)=H, there cuz surjective. Ohhh

thank u all for helping me connect the dots that were floating around in my head for some reason

am I correct in thinking G/T in the context of group theory, implies T is normal. My prof stated a theorem as Let T be normal in G. Then G/T={gT:g g in G}. If he stated "definition" instead of "theorem", I wouldn't think to consider the case where H is a subgroup of G but not normal in G and be looking at G/H

[coral spindle]

We can define G/H where H is any subgroup, but it will only have the structure of a group when H is normal.

Otherwise, it is just a G-set.

[warm saffron]

okay gotcha, thank u.

[celest furnace]

By the way this was awesome!!!

U can interpret that group as the automorphism group of the following tree: where C4 acts on the bottom row by rotating them (90 degrees) and S3 acts by permuting the upper part

Then you can get elements of order 12 by rotating at least 1 of the bottom row by 90 or 270 degrees, and then picking a permutation of the top 3 of order 3 (of which there are 2) so the answer ends up being something like 2(3^3 - 1)

So basically we interpreted a matrix problem as an automorphism group of this tree problem which is the coolest thing i think ive done al lweek

[delicate orchid]

Yeah it is super cool, this is how wreath product behave generally

They act on “tree like” graphs in a way that preserves the height of each branch

The cooler part is that this is how the Sylow subgroups of GL itself look

Implying a deeper connection than this one isomorphism

[celest furnace]

Crazy stuff

[cloud walrus]

pzh

[coral spindle]

b = 1

[still ledge]

it may not have a unit element.

[coral spindle]

Take any Abelian Lie algebra as an example then.

If you're about to say "but it's associative" I will shout "context" at you

Not that it matters – Abelian Lie algebras are associative after all.

[still ledge]

i'm sorry, but this is the first time I come across Lie algebra, let alone the abelian ones

[coral spindle]

Take any K-vector space and define xy = 0. Tadah, that's an associative algebra.

[still ledge]

emm, i understand now, but if my original statement is false when the K-algebra is not unitary, how do i prove this lemma when when A is not unitary?

Here A(V) is Hom(V, V).

[sand cradle]

Unitialization probably (I'm not an algebraist, so take my words with a grain of salt)

[still ledge]

that's interesting, sadly i don't know how to embed A into an algebra with unit element

[rose prism]

you just add one, see eg https://ncatlab.org/nlab/show/unitalization

[still ledge]

thanks for the link!

[remote nymph]

How could I prove the ideal $(3, x+1, x^2-x+1)$ is prime in $\mathbb{Z}[x]$? (I'm pretty sure it's prime)

[cloud walrus]

srhoosteen

[remote nymph]

I know it is prime iff the quotient is an integral domain but that seems tricky here

[next obsidian]

Mod out by x+1 first, then you end up with Z where you replace x by -1

So now you mod out by 3 and by (-1)^2 - (-1) + 1 = 3

So you mod out by 3

So the result is F_3 which is a field

[remote nymph]

Crazy stuff

You’re a goat

[next obsidian]

Nah, once you see how to do it once you know how to do it kekw

[remote nymph]

We get Z because all the variables get replaced with constants so only the first coefficient really matters?

[next obsidian]

Just prove it

Take the map where you evaluate at x = -1 and show the kernel is (x+1)

It’s true over any ring but the fact that Z[x] is a UFD makes it even easier

[remote nymph]

Ah ok makes sense

[faint fractal]

wow the chat has been dead 😔

this is from a HW I've already submitted, but I had no idea how to do it. Could anyone help me with it?

[dim widget]

So did you try computing f mod 2 and mod 3? How does it factor?

[faint fractal]

So for 2 I had like x(x^3+x+1)? I wasn't too sure about 3, I know you get some form like x^4 + 1 which can be factored from sophie germaine, but I wasn't sure if that was the right approach

[dim widget]

should be x(x^3 + x^2 -1) but yes. How does that cubic factor mod 2? Does it have any roots?

Mod 3 you get x^4 - 2, is that irreducible? What would be its factors?

[faint fractal]

wait sohuldn't it be x^3+x+1? Either way both of them are irreducible right? Since it must have a linear factor and x+1 doesn't factor it right?

like by polynomial division or whatever

[dim widget]

+1 = -1 mod 2 so it doesn’t matter

[dim widget]

Yes

[faint fractal]

ah ok

and then for mod 3 I was kind of stuck trying to find the reducible factors because it could factor into qudartaics

[dim widget]

Yes, what would a root of the quadratics look like?

[faint fractal]

well in R I know that it can be written as (x^2-sqrt(2))(x^2+sqrt(2)) so does this mean that it's irreducible in mod 3? Or is that bad reasoning

well it's not irreducible since sophie germaine? but that also would give irrational factors?

iirc

[dim widget]

Considering the quadratic would have to divide x^4 - 2 you can say that over any field a root of the quadratic would have to be a 4th root of 2, so over the algebraic closure of F_3 the only options for roots of quadratic factors are zeta_4^i * \sqrt{4}(2)

[faint fractal]

ok right

sure

[dim widget]

Then each quadratic factor would have to be a product of two linear polynomials with those roots. Are any of those polynomials with coefficients in F_3?

You can just brute force it

[faint fractal]

ah ok

(x^2+2x+2)(x^2+x+2) works right?

[dim widget]

Er, not exactly

[faint fractal]

really?

[dim widget]

that doesn’t have the right constant term

[faint fractal]

well it gives you 4 which is 1 mod 3 which is -2 mod 3 right?

[dim widget]

Ah no it does

Okay

I had written it differently but it’s equivalent mod 3

[faint fractal]

ah right right haha

[dim widget]

Anyway yeah

[faint fractal]

ok cool so we have those two now. Is there a way to use that to show that f is irreducible? I remember I just used the ratinoal root test, but I figure there's something specific I was supposed to do that i didn't

[dim widget]

now you have to use the fact that if f factors over Q or Z it has to factor mod p for every p

in a way which refines the factorization over Z

or you can use Galois theory

[faint fractal]

wait why is this true? is this a named theorem

[dim widget]

Because of f = hg over Z[X] implying the same mod p

[faint fractal]

ah ok

[dim widget]

So if f factors into a product of quadratics, say, then it factors at least into quadratics mod p

[faint fractal]

but 2 and 3 contradict each other?

[dim widget]

to go from Q to Z it’s called gauss’s lemma

[dim widget]

Yep

Now you have to do Galois theory to show the galois group is S_4

[faint fractal]

ok, would you be able to help me with that as well? I don't feel super comfortable with it still. So like we have some roots and since it's irreducible none of them are in Q and so that means our splitting field would have to adjoin all 4 roots?

or is that not correct

[dim widget]

The tool to use here is that if f factors into polynomials of degrees (d_1, d_2) mod p then the galois group of f contains a (d_1,d_2) cycle in S_n where n = d_1 + d_2 = deg(f)

The same would be true if it factored into more than two factors but I don’t want to write it out

[faint fractal]

wait but why wouldn't that always just be the degree of the polynomial?

like is sum d_i not always equal to deg(f)?

[dim widget]

Like in this notation a (2,2) cycle is (12)(34)

[faint fractal]

oh oh

I see

my bad

AH OK

I remember talking about this in class

[dim widget]

yep

[faint fractal]

ok so we have a (2,2) cycle and (1,3) cycle right?

[dim widget]

Yep!

[faint fractal]

hmm

[untold basalt]

Can't find a proof

[faint fractal]

is the order of a (2,2) cycle 4 or 2?

[dim widget]

So now you know by group theory that those generate some subgroup of S_4

[dim widget]

It’s 2

But the two two cycle and the 3 cycle cannot possibly commute

[faint fractal]

ah

I see

right right

wait

why is the order greater than 12 then?

[dim widget]

So you have to rule out a nonabelian group of order 6 (there is only 1)

[faint fractal]

ab

ah

[dim widget]

This part is just group theory

[faint fractal]

ok cool

so once you have that the order is greater than 12, how do you use that to extrapolate it is just S_4 itself? Like some galois correspondance thing

?

and does the discriminant ever come into play?

[dim widget]

Anyway you will get that they generate A_4 and then you have to use the discriminant to show that Galois is S_4

[faint fractal]

what does discriminant do in this problem? I don't remember talking about it

or maybe I do

ahhhhh

[dim widget]

This is a nice and important result: if f is of degree n then Disc(f) is a square if and only if the galois group is contained in A_n

I guess f has to be separable

[faint fractal]

interesting

do you hvae a link to a proof of that? or is it like provable for someone learning the content

[dim widget]

This just comes from the fact that the discriminant is \prod_{i < j} (r_i - r_j), which some people take to be the definition

Where r_i are the roots

[faint fractal]

oh ok I see

[dim widget]

So a galois element acts on this by the sign of it as a permutation of the roots

[faint fractal]

so we know that the discriminant isn't a square which means that the galois group isn't continaed in A_n? And so looking at like the lattice diagrams of subgroups means that it must be S_n?

or like A_4 and S_4?

[dim widget]

So if all galois elements act trivially (if it is a square) then galois is in A_4

[dim widget]

Yep! There are no subgroups strictly containing A_n except all of S_n

Because it has index 2

[faint fractal]

ah cool

ok that makes a lot more sense

thank you so much! I'll try to resolve this problem later knowing this now

[dim widget]

Great!

[faint fractal]

do you have any good advice for getting better at Galois theory stuff? Our final is soon and I still just feel very wish washy about it all

I plan on rereading the sections in DF but I don't know how much help that'll be

[dim widget]

I think do calculations and read solutions to Galois theory questions online

[faint fractal]

cool thanks

[dim widget]

I learned a lot from reading on mathstackexchange about how clever people would solve elementary Galois theory problems. When it was more of a centerpiece of math research before other aspects of number theory took off a lot of people came up with clever ways to do it.

[untold basalt]

Let $E/F$ be an algebraic field extension. Suppose $\Omega$ is an algebraic closure of $E$. Show that $E/F$ is normal if and only if every $F$-homomorphism $\varphi: E \to \Omega$ is $E$-invariant.

\textbf{Solution:} Let $\varphi: E \to \Omega$ be a $F$-homomorphism, and suppose $r_1, \dots, r_n$ is a basis of $E/F$, so that $E = F(r_1, \dots, r_n)$. If $f_1, \dots, f_n$ are the minimal polynomials of $r_1, \dots, r_n$, then it is easy to verify that $E$ is a splitting field of $f := f_1 \cdots f_n$. Observe now that since $E/F$ is normal, $f$ has all roots in $E$, and $\varphi$ maps roots into roots, so $r_1, \dots, r_n \in E$, hence $\varphi(E) \subset E$.

[cloud walrus]

Seagull

[untold basalt]

How do I prove the converse?

[glossy crag]

You want to show every irreducible f that has a root in E splits. Fix such an f and let a be its root in E, f has a root in Omega too (alg-closed), so there's an F-homomorphism F(a)->Omega. See if you can take it from there.

[blissful zinc]

Hello, I re-did an exercise (prove that Q(sqrt(2)) is a field).
The first thing I thought of, is to do it with the field's definition.
taking a non-zero element of Q(sqrt(2)) and see if 1 / (that element) is in Q(sqrt(2))
And the 1!=0 is clear.

And after that, I rechecked my teacher's proof: he computated two elements of Q(sqrt(2)),
( a+b sqrt(2) ) / ( c+d sqrt(2) )
I forgot the reason he did that. Can someone explain me? Please

[cloud walrus]

Seagull

[glossy crag]

Q(sqrt2) is a subset of R, a field, so I guess he decided to show the quotient of any two elements of Q(sqrt2) is again in Q(sqrt2), which is the same thing as showing every element is invertible. Or he wanted to demonstrate to you how to divide in Q(sqrt2) (in particular, this shows that 1/a+bsqrt2 is again in Q(sqrt2)).

[glossy crag]

Sure.

[blissful zinc]

I might be dumb (Now, i remember that part confused me):
Why is doing the quotient of two elements same as showing every element is invertible?

[glossy crag]

If you show a+bsqrt2/c+dsqrt2 is an element of Q(sqrt2) for any a,b,c,d (c,d!=0), then in particular 1/c+dsqrt2 is an element of Q(sqrt2) (a=1,b=0). Conversely, if you show 1/c+dsqrt2 is an element, then so is a+bsqrt2/c+dsqrt2, since that's the product of a+bsqrt2 and 1/c+dsqrt2. These are equivalent problems.

[blissful zinc]

I understand now. Ty

[untold basalt]

Show that an algebraic field extension $E/F$ is normal if and only if $E$ is the splitting field of a polynomial $f \in F[x]$.

\vspace{1ex}

\textbf{Solution:} If $E/F$ is normal it is easy to construct $f \in F[x]$ such that $E$ is its splitting field. Conversely, suppose that $E$ is the splitting field of a polynomial $f \in F[x]$; then $E=F(r_1,\dots,r_n)$ where $r_1,\dots,r_n \in E$ are the roots of $f$ . Suppose $\Omega$ is an algebraic closure of $E$ and let $\varphi: E \to \Omega$ be a $F$-homomorphism. Since a basis of $E/F$ can be constructed from powers and products only of $r_1,\dots,r_n$ we see that $\varphi$ is completely determined by $\varphi(r_1),\dots,\varphi(r_n)$, which must be roots of $f$ too, which means $\varphi({r_1,\dots,r_n})={r_1,\dots,r_n}$, hence $\varphi$ is $E$-invariant. $\square$

[cloud walrus]

Seagull

[untold basalt]

Actually I think it should be from a subset or r_1,...r_n that we construct the basis.

But except for that does this sound right?

[flint crater]

Bit confused about finding all isomorphism classes of Z-modules of order 5. Can I use the Fundamental Theorem of FInite Abelian Groups for this in any way?

[south patrol]

Z-modules are exactly abelian groups so yes

[flint crater]

Right right but I am not sure how exactly I would use it. Would every direct product I can construct of cyclic groups be one isomorphism class?

[south patrol]

Hm I mean if you understand isomorphism classes of abelian groups then those correspond to isomorphism classes of Z-modules

So it sounds like you're more asking about what the fundamenta ltheorem of finite abelian groups says about isomorphism classes or smth

[flint crater]

I guess I kind of am yes

[south patrol]

(I should say as well that classifying finite groups of order 5 doesn't need anything as strong as this!)

[south patrol]

Well the full form of the theorem does actually describe all isomorphism classes

Like each group is a direct sum of groups Z/p^k across some primes p, and that decomposition is unique up to reordering

[flint crater]

Sorry Z/p^k being Z/p^kZ or is this something else?

[south patrol]

Yeah

(this is common alternative notation)

[flint crater]

Ah my bad. Hadn't seen it before in the lectures

But in the case we have order 5 anyway so we just end up with Z/5Z?

[south patrol]

Yes exactly

[flint crater]

But in the case of something like order 8 we end up with 3 possibilities no? Z/2Z x Z/2Z x Z/2Z, Z/2Z x Z/4Z and Z/8Z?

[south patrol]

Exactly yeah

[flint crater]

Awesome think I got it

[south patrol]

Coolio

!

[tardy hedge]

Why do ring homomorphisms out of Fields need to be one-to-one or trivial?

I did the proof for it once but i forget it

[rustic crown]

because kernels are ideals

and ideals in a field look super nice

[tardy hedge]

Oh ok yeah

Thats a more abstract way of looking at it

Fields have no proper nontriv ideals

But a level deeper has something to do with fields having inverses doesnt it

I think thats how i did the proof

[rustic crown]

well you'll essentially be proving that only ideals in a field are (0) and (1)

[tardy hedge]

Thanjs for reminding me of this tho!

Ring homomorphisms preserves units right?

[rustic crown]

yee

and a unit = 0 iff the whole ring is 0

want me to spell it out?

<

[tardy hedge]

Maybe lol

Also trying to understand the point of this thm

[rustic crown]

say F --> R sends a non-zero element to 0, then since F is a field, that non-zero element is a unit, which must go to a unit in R which is 0, this forces the whole ring to be zero and so the whole map is 0.

so the map has either trivial kernel or is 0

[tardy hedge]

I see

Thank yoy

[rustic crown]

it's telling you how construct maps out of Q(D) (to a field)

[tardy hedge]

Thank YOY!

[rustic crown]

to construct a map Q(D) --> F

you just have to give an injective map D --> F

(in general, a map Q(D) --> R is given by an injective map D --> R which sends non-zero elements to units in R)

you check it once that this gives you a ring hom, and never again later :3

[wary lantern]

I have a question regarding a problem from an old exam. Two profs claimed in their solutions to 3c that F[X]/(X^2) is a PID and then used the structure theorem for finitely generated modules over PIDs to claim that a F[X]/(X^2)-module is isomorphic to (F[X]/(X^2) )^s ⊕ F^u, for some nonnegative integers s and u. What I don't understand is how F[X]/(X^2) can be a PID if (X + (X^2))( X + (X^2)) =(X^2 + (X^2)) = (0) ?

[south patrol]

F[X]/X^2 isn't a PID

They're wrong lol

[tardy hedge]

To construct a 1-1 map from Q(D) to field F, you can just take some 1-1 map from D to F and “extend it” in the manner described in the thm

[south patrol]

It is a principal ideal ring in that every ideal is generated by one element

But im pretty sure the classification requires domains

I think the key bit though if you can view any R/I module as an R module

Then use the classification of those

[tardy hedge]

And , there is only one way to extend it in this manner (the uniqueness part)?

[rustic crown]

any map between two fields in 1-1

[tardy hedge]

Ok

Besides me specifying that its 1-1 is what i said the right way to think about it

[rustic crown]

yep

[tardy hedge]

Thanks. Now im just trying to figure out why the map needs to be theta(a)theta(b)^-1

In an intuitive way

[rustic crown]

theta' = hat(theta)
needs to satisfy the identity
theta'(a/b) * theta'(b/1) = theta'(a/1)

because we want it to be a ring hom

now what does the commutativity of the triangle say?

(oh btw maybe i should have written [a, b] instead of a/b)

theta'([a,b]) * theta'([b,1]) = theta'([a,1])

[tardy hedge]

It says that theta’([b,1]) = theta(b)

And then i see how it comes about

[rustic crown]

[tardy hedge]

But i still dont rlly feel super satisfied with it. But maybe i will after seeing more examples etc

[rustic crown]

basically you know where it sends a and b to

so it should send a/b to (where it sends a)/(where it sends b)

[tardy hedge]

Oo

That was good

[rustic crown]

as Q(D) is trying to construct fractions in stuff from D

[tardy hedge]

Yeah

[rustic crown]

and to make sense of the denominator, you want it to be non-zero which is why we require theta to be injective

[tardy hedge]

So , you could maybe have completely different maps from Q(D) to F correct? But this theorem is saying if you want to make a map you can do it easily by using any map from D to F

And extending it

[wary lantern]

Thank you!

[rocky cloak]

I think any principal ideal ring is the product of quotients of PIDs, so then the classification should translate.

[rustic crown]

well i would say the easier part is to show that any map has to look like that... the "harder" and more annoying part is to check that this indeed gives a ring homomorphism which nobody likes to do. checking the axioms is boring and if i were to do this everytime i was working with Q(D) i'll die.

[south patrol]

Sure

[rustic crown]

these things are called universal properties, they tell you a recipe to construct maps to or maps from some object in terms of other easier to do things

[tardy hedge]

I see

Thanks so much for ur help btw

This server is crazy useful for self study

[rustic crown]

[chilly ocean]

How can II' be contained in <I intersection I'>
If i is in I and not in I', then i=i1 is in II' and not in I intersection with I'

[tardy hedge]

Im a bit confused by 5.4.7, every element of F has the form ab^-1 for some a b in D, so we are requiring elements in D to be invertible?

[next obsidian]

What

II’ is elements which are sums of elements ij with i in I, j in I’

ij is in I and ij is in I’

So it’s a sum of things in I\cap I’

[rustic crown]

chmuwu

[tardy hedge]

Does this not supposed to say every element in F needs to have the form ab^-1 for a b in F, not a b in D?

[next obsidian]

No

It’s correct as written

Your proposed change is trivial

Every element is of the form a(1)^-1 always

[tardy hedge]

Ok so im missing some understanding again

[chilly ocean]

@next obsidian
If i is in I and not in I', then i=i1 is in II' and not in I intersection with I'

[next obsidian]

That makes no sense

Why is i1 in II’?

[rustic crown]

the statement is saying if every element in F looks like a quotient of two things in D, then F is actually Q(D).

[chilly ocean]

1 is unity

[next obsidian]

So?

Why is it in II’

[chilly ocean]

In case 1 is in I'

[tardy hedge]

A “quotient” in D being ab^-1?

[next obsidian]

Then I’ is all of the ring lol

And then you can’t have i in I and not in I’

[next obsidian]

In F

[rustic crown]

two "things in D" and not "quotient in D"

[next obsidian]

/b for b in D makes sense in F

Because F is a field and b in D < F

Think about Z and Q

[tardy hedge]

Yea i was getting confused cause i was thinking we are saying these elements need inverses in D

[next obsidian]

No

[tardy hedge]

Ok thats where my confusion was

We are identifying the things in D but we are computing the quotient in F

[chilly ocean]

Why

[next obsidian]

Because r1 is in I’ for all r

[tardy hedge]

So its saying if every element in F we can identify with being a quotient in F from elements in D then F isomorphic to Q(D)

[chilly ocean]

So

[next obsidian]

???

If 1 is in an ideal I

Then for all r in R, r•1 = r is in I

Because of I’s absorptive property

An ideal is equal to the ring if and only if it contains 1

[tardy hedge]

Satisfactory?

[next obsidian]

There is no “identify”

This is literal equality

[tardy hedge]

Well im trying to not get confused by inverses in D and inverses in F

[next obsidian]

You aren’t doing anything fancy, given an integral domain D and a field F containing D, you can form the set {a/b| a,b in D} < F

If this is an equality then F is isomorphic to Q(D)

[next obsidian]

It quite literally doesn’t matter

[tardy hedge]

Interesting

Im probably getting lost in the weeds

[next obsidian]

If you can make sense of all a/b then D is already a field

And then this is just saying if D = F then F is isomorphic to Q(D)

[tardy hedge]

Oh what yeah this is like “obvious” then right

[next obsidian]

Yes

[tardy hedge]

But the field of fractions is not literally a sub thing right

Its like isomorphic or smth isnt it

[next obsidian]

This is

Nebulous

[tardy hedge]

Cause elements of F is not like [a,b]

[next obsidian]

You have seen a model of the field of fractions

[south patrol]

Field of fractions is a universal object and a useful model is this

[next obsidian]

To me the field of fractions is something satisfying a universal property and it virtually makes no sense or no purpose to really even ask the question

[south patrol]

Lol

[next obsidian]

And for everything I care about (except for when it’s different which is hard to describe) they just are all the same

[tardy hedge]

Yeah im just beginning to learn this so its hard to see the big picture sometimes

[next obsidian]

I mean like

Is Q the field of fractions of Z?

In one sense “no” but if you try to

Ummm ackshually 🤓 me about that

[tardy hedge]

Ohh ok

[next obsidian]

I’m gonna punch you

[blissful zinc]

Let A, a field (and commutative, Idk it has a role here)
And B, a subdomain (subring if you wish) of A

Why can we say that B is also a field?
I can't convince myself from this.

[tardy hedge]

Yea so ur saying im being “umm ackshualy” about it

[next obsidian]

I mean not to your fault

[tardy hedge]

Yes but this helps me understand ur point

[next obsidian]

When you first learn this stuff you care about that kind of thing

[tardy hedge]

Yeah

[next obsidian]

But the point is you shouldn’t

[tardy hedge]

Haha yeah but tbh i think it takes time to get to ur POV

I can see where ur coming from though

[next obsidian]

It isn’t

Consider Z in Q

[tardy hedge]

A a field and B subring of A then B is an integral domain right

[next obsidian]

Yes

A subring of a domain is a domain

[tardy hedge]

Ok

[blissful zinc]

ah an example ty!

[rapid junco]

Is there anything useful to be said about G acting on Z(G)

[coral spindle]

Acting how

The only way I can think of is by conjugation, and that is literally the trivial action

[rapid junco]

By left multiplication or conjugation

[coral spindle]

The left multiplication action is not well-defined