groups-rings-fields

[strong yacht]

For part (a), remember that the roots of any polynomial are related to its coefficients in a special way.

In particular, the coefficients of a polynomial can be expressed by a symmetric expression of the roots (think Vieta's formulas from school)

Therefore, since elements of the Galois group permute the roots, and symmetric expressions are not changed under permutations...

[tawdry crystal]

Commutative diagram is the name I'm looking for. Thanks!

[strong yacht]

This is quite a lovely question, the way I thought about this is that $D_8$ is generated by a reflection and a rotation. So a very natural polynomial to think about is $x^4-2$, whose roots are $\sqrt[4]{2}, i\sqrt[4]{2}, -\sqrt[4]{2}, -i\sqrt[4]{2}$. The Galois group of the splitting field of this polynomial acts transitively on these roots so is isomorphic to a transitive subgroup of $S_4$. There are two automorphisms that are interesting here: $i \mapsto -i, \sqrt[4]{2} \mapsto \sqrt[4]{2}$ (reflection) and $i \mapsto i, \sqrt[4]{2} \mapsto i\sqrt[4]{2}$ (rotation). Any permutation of the roots (automorphism) is entirely determined by these two maps, and these two maps behave exactly the same way as the generators of $D_8$ so it's completely obvious the Galois group is $D_8$. The construction used in the proof of the Primitive Element Theorem on $\mathbb{Q}(\sqrt[4]{2}, i)$ gives your $\eta$.

[cloud walrus]

Greenman

[median pawn]

I have a question about filters

Z(X) is the collection of zero sets of real-valued continuous functions on X

X is some topological space

So now you know what a z-filter means

A z-ultrafilter is a maximal z-filter

How does the stuff in blue follow?

(We know that every family of zero sets that has FIP is contained in a z-filter)

Hear me out: A z-ultrafilter F is a subfamily of Z(X) (as it's a z-filter). Suppose there is a zero set Z(f), such that F U {Z(f)} satisfies FIP. Then, F U {Z(f)} is contained in a z-filter F'. However, F' contains F, which is an ultrafilter; so F' = F. I guess I have figured it out (?)

A follow-up question: Suppose M is a maximal ideal in C(X), so that Z[M] is a z-ultrafilter. If Z(f) is in Z[M], can we say that f is in M? In general, we can't, but I think the fact that Z[M] is a z-ULTRAfilter may have something to offer

[latent anvil]

Yes, you can. Suppose f not in M. Then the ideal (f) + M strictly contains M, so (f) + M = (1) by maximality, so gf + h = 1 for h in M. For any point p in Z[M] we have g(p)f(p) = 1, so f(p) ≠ 0, so p not in Z(f). So Z[M] and Z(f) are disjoint.

So if Z(f) is contained in M and Z(f) is nonempty (ie f is not a unit) then Z(f) and M are not disjoint so f is in M

This is assuming Z[M] = intersection of Z(h) over all h in M, like the vanishing locus in algebraic geometry. Let me know if that's not right

[broken stirrup]

I want to prove that dual of finitely generated projective module is also finitely generated projective. I've already proven that dual P* is projective but not sure about how to prove that it's also finitely generated

Let's suppose P is generated by A={p_1,...,p_n}. What if i consider {f_p | p \in A} ?

would it form a basis?

where $f_p(x)=\delta_{xy}$

[cloud walrus]

backpack

[broken stirrup]

and delta denotes kronecker delta notation

[next obsidian]

I think it follows because P is a direct summons of a finite free module

So you have a split exact sequence

0 -> P -> F -> P’ -> 0

When you take duals this will remain split exact so that P* is surjected on by F* which is finite free

[pastel cliff]

does \leq apply to rings and subrings as well

[devout crow]

$S \leq R$ is often written to mean $S$ is a subring of $R$

[cloud walrus]

xdres

[devout crow]

or do you mean putting an order on the elements of a ring?

[pastel cliff]

no that's what i meant

ive only seen $H \leq G$ refer to groups/subgroups, was making sure

[cloud walrus]

μ₂ (46/47 🪲)

[devout crow]

I would say it's more common for vector spaces/groups than rings actually

replace "often" with "sometimes"

[broken stirrup]

yes that's how i saw that P* is also projective but I don't quite get how it imples that it's finitely generated

[next obsidian]

Because F* is isomorphic to F

This is giving you a surjevtion from a finite free module, which is practically definitionally equivalent to being finitely generated

[broken stirrup]

oh got it

thanks

[spiral wolf]

I'm beginning Galois Theory right now and have a simple assignment, to describe the Galois group and subfields of the 15th cyclotomic polynomial. I have the Galois group completely understood; it's Z4 x Z2, and so I have all the subgroups and everything. However, I'm not sure how to find the fixed subfields corresponding to each of those subgroups.

As an example, let $\alpha = e^\frac{2\pi i}{15}$ and $\sigma : \alpha \rightarrow \alpha^2$.
Then the subgroup ${e,\sigma, \sigma^2, \sigma^3}$ corresponds to the subfield $\mathbb{Q}(\alpha + \alpha ^2 + \alpha ^4 + \alpha^8)$

[cloud walrus]

Espio

[spiral wolf]

How can I find the subfields, starting with the corresponding subgroups?

[strong yacht]

They correspond with the fields that they fix

[spiral wolf]

Yes, definitely, I'm just not sure how to find those fixed fields

I'm sure there's some little obvious thing I'm missing

Should I just try to find a "large" element that the subgroup fixes?

[round jay]

Hi, I wasn't able to figure it out. Can you please help me? I have this homework due on Wednesday and I still feel like idk what I'm doing.

[spiral wolf]

To be more explicit, here's the chart my professor gave for the 15th cyclotomic field. I just don't know where he got the fixed subfields from; it seems random/patternless

[round jay]

All I got was that lcm(|gh|, |g|) = lcm(|gh|, |h|)

I don't know what to do from here, again I feel that this is a number theory problem but idk any number theory

[strong yacht]

Hmm yea I agree with you, struggling to see the pattern myself

It's easy to verify that the subgroups fixes those fields, then degree argument to say you're done, but where did those elements come from.....

https://en.wikipedia.org/wiki/Gaussian_period

This seems relevant @spiral wolf

[spiral wolf]

Thank you, I'll read that page

Yeah, these Gaussian periods keep popping up in this class but the professor has never explicitly named them or explained them, he just kinda takes it for granted that the algebra is correct and while I can verify his steps I never understand where it comes from

[strong yacht]

That doesn't seem very nice of him :/

Perhaps it's not so important that you come up with this stuff quite yet? Just that you're able to verify that it's correct maybe

[uncut girder]

To me its kind of obvious because I visualize the roots of unity on a unit circle and summing them up in various ways corresponds to adding unit vectors in R^2. The geometric symmetry combined with right angle trigonometry usually does the trick for computing explicit instances of these sums .

But yes, lecturers should make it clear when and where they use geometric intuition

[spiral wolf]

The assignment WAS to come up with it -- when all of us emailed him saying we couldn't he just posted the solutions and asked us to show proof of verification

[strong yacht]

Fair enough. I agree with you that it's not obvious!

[spiral wolf]

Yeah, I get the geometric intuition as far as adding the roots go. I'm not sure how to visualize what gets fixed by each subgroup

[uncut girder]

Also to find fixed subfields, a common method is to adjoin 1 element and then act on it by some element of the Galois group. Then form a sum over all those conjugates.

[spiral wolf]

Hm. So would that be how you find the first one, Q(a+a^2+a^4+a^8)? That was the only one that made any sense to me at least, lol

[pastel cliff]

another wew lad

[delicate orchid]

what permastudy does to a mofo

[pastel cliff]

https://tenor.com/view/cruelty-squad-gif-22461848

[strong yacht]

Maybe there is some degree of "guesswork". For example with the subgroup $<\sigma>$, you can spot that the automorphisms in this group permute elements of the set ${\alpha, \alpha^2, \alpha^4, \alpha^8}$, then use "geometric intuition" to see that the sum is $\sqrt{-15}$. Once you have that $\mathbb{Q}(\alpha + \alpha^2 + \alpha^4 + \alpha^8) = \mathbb{Q}(\sqrt{-15})$ you're done: since the group is size four so you know it fixes a field of degree at most two (and it certainly fixes this one).

[spiral wolf]

Yeah, I'm not as worried about simplifying it down to sqrt(-15), as long as I can come up with a subfield in terms of alphas

[cloud walrus]

Greenman

[strong yacht]

Basically the only non-obvious step to me here is why we should expect the sum to come out with anything nice at all, but I think the others have nice suggestions

[uncut girder]

Omay so you have the 15th cyclotomic field right? The field is obtained by adjoining zeta_15 to Q. A field automorphism is determined by where it sends zeta_15. The Galois group is isomorphic to the group of units mod 15. Now you just gotta pick a subgroup of the Galois group and find the elemtns invariant under that action. So let's say the first subgroup you picked was cyclic of order 4, generated by sending zeta_15 to zeta_15^2. (Check this has order 4 since 2*2=4, 2*2*2=8, 2*2*2*2=16 =1 mod 15).

Now do the sum of conjugates trick i suggested. So the conjugates of zeta_15 under the action of this automorphism is zeta_15, zeta_15^2, zeta_15^4, zeta_15^8. So the sum of these conjugates is indeed invariant under the action: zeta_15 + zeta_15^2 + zeta_15^4 + zeta_15^8 is invariant under the automorphism which sends zeta_15 to zeta_15^2. Now all the remains to to check that this has the right degree over Q, and you're done. Once you check the degree is the correct degree, you know the field obtained by adjoining this sum of conjugates to Q is both contained in the fixed field, and has the correct degree for the fixed field, so it must be the whole fixed field.

There i explained my whole thoughts process for obtaining the first subfield in that list

[spiral wolf]

Gotcha. So for the second subgroup in the chart, the conjugates of zeta_15 are: zeta_15, zeta_15^13, zeta_15^4, zeta_15^7. I then check that their sum is invariant under the automorphism that sends zeta_15 to zeta_15^-2 = zeta_15^ 13, and then check that this is degree 2 over Q.

Supposing this is correct, awesome. However, I'm not sure how the professor got zeta_15^5 - zeta_15^10, which isn't even composed of units mod 15

[uncut girder]

I'm not sure if those conjugates are correct, but I havnt carefully verified it.

[spiral wolf]

Perhaps as a simpler example, consider the automorphism sending zeta_15 to zeta_15^4 (order 2)

The conjugates would be very simple, zeta_15 and zeta_15^4, but the professor gives the field as Q(zeta_15 + zeta_15^2)

[uncut girder]

I see what you mean

Yeah thats weird

[spiral wolf]

Yeah, your logic makes a lot of sense to me and certainly could be right, but it doesn't seem to be the logic the professor is following, so I'm just confused

[uncut girder]

,w minimal polynomial of e^(2pii/15) + e^(22pi*i/15)

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by PTYamin#8224

[uncut girder]

@spiral wolf yeah its just wrong

Your prof made a typo

[spiral wolf]

Oh pog

I need to start using that function lol, I didn't know it existed

[uncut girder]

It should be zeta_15 + zeta_15^4 not zeta_15 + zeta_15^2

[spiral wolf]

Awesome

[uncut girder]

Wolframalpha.com was very helpful when I was learning galois theory

Also very helpful when I use galois theory

[spiral wolf]

Yeah, I have access to Mathematica and use it a bunch, there's always stuff I never knew though lol

,w minimal polynomial of e^(2pii/15) + e^(26pii/15) + e^(8pii/15) + e^(14pii/15)

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Espio#1653

[spiral wolf]

Nice!

[strong yacht]

Do you have a reference for the thing about summing the conjugates? I don't see why the sums should turn out to be nice. but it's a good trick I will remember.

[uncut girder]

The reason the sum of the conjugates is invariant under the action of the automorphism is because the conjugates are permuted by the automorphism. If you permute the terms in a finite sum, that doesn't change the value of the sum.@strong yacht

This is why its invariant.

[strong yacht]

Yes I understand the invariance, but why does it sum to nice things

Like how the first one sums to sqrt(-15)

[uncut girder]

Oh that. There are several approaches: one is algebraically, whats the minimal polynomial of this sum of conjugates? What's the degree of the simple extension? If its a degree 2 extension, then you know its a quadratic extension, so its Q(sqrt d) for some d. Now just figure out which.

There's also a geometric/trigonometric technique that works often with sums of roots of unity. If you plot the points on the unit circle, and add them up as vectors in R^2, you can draw some right triangles sometimes or use some symmetry of the figure to deduce what the value, or length, of the sum vector is.

This is a regular 15 gon

Let me mark the points that are involved in the sum

Actually I have no idea how to deduce this geometrically

[strong yacht]

It's not obvious, right?

[uncut girder]

Is it easy to see that this is a degree 2 extension?

I mean, from the fundamental theorem of galois theory, yes

[strong yacht]

Yes I agree with you

That's why you have to check the sum does sum to a quadratic thing right?

[uncut girder]

Well, knowing its a quadratic extension means that Q(sum)=Q(sqrt d) for some d.

[strong yacht]

Oh sorry, I did not know it was a quadratic extension

I thought we were checking that fact by computing the sum

[uncut girder]

If you know the fundamental theorem of galois theory, then you know that the fixed field of an index 2 subgroup has degree 2 over your ground field

The subfield we are considering rn is the fixed field of an index 2 subgroup of the galois group

Are you familiar with the fundamental theorem of galois theory?

[strong yacht]

I think I am - but we don't necessarily know that the fixed field is specifically Q(alpha+alpha^2+alpha^4+alpha^8) - this is certainly fixed by <sigma>, but not necessarily the fixed field (but contained in the fixed field)

Maybe I misunderstood - I thought the prof proves that this is the fixed field by doing the computation alpha+alpha^2+alpha^4+alpha^8 = sqrt(-15), which proves that this is a quadratic extension so it must be the fixed field

[uncut girder]

Okay good point

Its contained in the fixed field

But also, the fixed field is necessarily degree 2, do you agree with that?

By the Fundamental theorem

[strong yacht]

Yes I agree

[uncut girder]

Okay

[strong yacht]

Maybe there is a better way I could have phrased my question

[uncut girder]

So we have some field thats contained in the fixed field, which is degree 2, but also this thing we have is not the trivial extension of Q, so its degree must be >1. So this field has degree <=2 and >=2, so this field must have degree 2, and its the full fixed field of the automorphism.

This is a little degree argument trick

[strong yacht]

Right, i agree in this case maybe it is easy to see without the computation, in other words the prof did not need to do that computation at all (though I'd still like to know the trick for it)

What about other cases? In general, do we have Q(adjoin sum of conjugates) as the fixed field?

I see containment in the fixed field

[uncut girder]

Still the question remains, for what d is Q(this sum)=Q(sqrt d)?

[uncut girder]

You need to do a degree argument, either by computing minimal polynomial, or counting conjugates in the total field, or any other trick

[strong yacht]

Ah right i see

Ahhhh I see

Degree argument to get that it is the fixed field, but still we need to write it in a nice form right?

[uncut girder]

Yeah, degree arguments are a very powerful tool in galois theory

[uncut girder]

To write it in a nice form is kinda tricky because what counts as "nice"?

[strong yacht]

So in the original image, you see how the guy wrote sums and minuses of the conjugates of the Galois subgroups to obtain expressions?

[uncut girder]

If I were grading the problem, I would give full marks for just writing Q(sums)

[strong yacht]

Yeah I see where you're coming form

I think the written stuff in that image was a red herring all along, right?

[uncut girder]

But for my own sanity, I would try to reexpress the extension in a more familiar form, so that I can know for sure that didnt make a mistake

[strong yacht]

Ok I'm with you, the image made me think there was a reason for why the sums were so nice

[uncut girder]

I think its included to help you be more familiar with the subfields

[strong yacht]

Thank you, no doubts left about it now 😄

[uncut girder]

Its not strictly necessary to "evaluate the sums" in a nice way to answer the question

Also

[strong yacht]

Yes exactly, my mind thought that it was the argument the prof was using to show they were the fixed fields

[uncut girder]

Differnet people can answer the question in different ways, some people might answer the question by writing Q(sqrt -15) where as others would write Q(sum)

Both are valid answers, so the prof includes all likely answers in the answer sheet

[strong yacht]

That's another good point

But still, there is the fact that the sums actually are nice

[uncut girder]

Yeah

[terse crystal]

I have two questions about rational cohomology :
First given any group G, why we have
$dim_{Q}(G_{ab} \otimes Q)=dim_{Q}(H_{1}(G,Q))$?
Second given an exact sequence of groups $1 \rightarrow A \rightarrow G \rightarrow H \rightarrow 1$ where A is an abelian group, why we have
$H_{1}(A,Q)$ being isomorphic to $H_{1}(A,Q)_{H}$ ?

[cloud walrus]

Cogwheels of the mind

[terse crystal]

G_ab is the abelianization of G, =G/[G,G]

[median pawn]

Ah

Z[M] is just the collection of all Z(f), where f is in M

[latent anvil]

Oh, my bad!

[chilly radish]

To recall, you know that the order of g^N divides the order of g, but we also know that (h^-1)^N^|h|=e, so the order of (h^-1)^N=g^N divides the order of h, but |g|,|h| are coprime. If a number divides two numbers, what does it also divide?

Wait sorry I said it wrong

Let me edit

Ok now it's fine

Feel free to @ me when u see this

[broken stirrup]

We know that if P is a projective space then there exists a free module F and submodule K such that F= K + P. But why if P is finitely generated then F must be finitely generated as well?

[harsh anvil]

Does the Cayley graph of any finite symmetric group always have a Eulerian path?

[terse crystal]

P is finitely generated therefore there exists a free module of finite rank F and an epic homomorphism F—>P. Let K be the kernel of this homomorphism we therefore have an exact sequence 0—>K—>F—>P—>0. Since P is projective this exact sequence splits, therefore F is isomorphic to the direct sum of K and P.

[broken stirrup]

oh got it

now the only thing i have to do is proving that Kernel is finitely generated since it's not trivially finitely generated as a submodule

[wooden ember]

I’m having trouble showing that y^4 +xy^3 + xy^2 + x^2y +3x^2 -2x is irreducible over Q[x,y]

I’m missing the smart use of a reduction morphism or substitution

I think I figured it out

Evaluate x at 1-y then reduce mod 2

[terse crystal]

Of course K is finitely generated, because that sequence splits, you have K—>F—>K being 1_K for some F—>K. K is finitely generated as a quotient of F

[opal osprey]

I will call a group $G$ extensive if for any two subgroups $H$ and $K$ of G and an isomorphism $\psi : H \rightarrow K$ there exists an automorphim $\tilde{\psi} \in \text{Aut}(G)$ such that $\tilde{\psi}|_{H} \equiv \psi$.
\
\
Could someone give me examples of non-simple and non-trivial groups which are extensive (in the abovementioned way)?

[cloud walrus]

MISTERSYSTEM

[wooden ember]

Q_8?

Seems like a functioning example

[wet glacier]

Pls help me in this problem "A ’maximal’ subgroup H of G is one that is maximal among proper subgroups of G, i.e., H is proper
and it is not properly contained in another proper subgroup. Prove that if a finite group G has exactly
one maximal subgroup, then it must be cyclic."

[wooden ember]

Take an element of the maximal subgroup M, call it x. Suppose G isn’t cyclic. Then <x> is a subgroup of M so you can pick y outside of M and hence outside of <x>. What can you say about <y>

[south patrol]

I'm not sure the point of you introducing x there? "Suppose y is any element not in M. What can you say about <y>" is how I'd think of it

[wet glacier]

I'm confused what should I do with y

[wooden ember]

Yeah fair I wrote as I thought so my argument shifted midway

Which I suppose makes it confusing sorry

[wooden ember]

Every proper subgroup is in some maximal subgroup

[south patrol]

Oh dw

cool just wanted to clarify

[wet glacier]

Okay ı understood that but now how can ı proof ıt, G has only one maximal subgroup

prove*

[vagrant star]

well, <y> isn't in M right?

[lethal cipher]

Okay, so I am currently learning about constructible numbers. Technically it is geometry, but I am learning this in Galois theory, so I'll ask here. I was wondering if someone can help me understand why the y-axis in the complex plane is constructible.

[delicate bloom]

I think you start with Q[i] in that approach

[lethal cipher]

Sadly no

Btw, Exercise 1 doesn't tell us any tips

Oh wait, I got it. By drawing a circle of radius 1 centered at 0, and creating a line through 0 and 1, we get that -1 is constructible. Then we can create a circle of radius 1 for both 1 and -1. Their intersection creates a point on the imaginary axis.

Then we draw a line from 0 and that point, and that line is our y-axis

Out of curiosity, what would be a number in the complex plane that isn't constructible?

[strong yacht]

Quite famously cube root of 2 is not constructible (it was an unsolved problem for a long time before Galois theory, the Greeks considered it)

[mint seal]

pi is not constructible either

no transcendental number is constructible

[strong yacht]

Manifold is right and here is the more general theorem from my notes

You can only construct numbers if their degree is 2^k for some k

[lethal cipher]

I may sound silly, but what does 1 step constructible mean?

[lethal cipher]

Can it be an intersection of a line and a circle. Or do the two shapes have to be the same?

[strong yacht]

Shapes can be different I believe

[chilly ocean]

Wow, so you are a firefighter, a teacher, an astronaut, a doctor, a police officer AND a mathematician? Respect man

[lethal cipher]

I'm suprised you didn't get that comment sooner

[wary verge]

Do anyone know how i can show that the groups Sp (2) and SL (2, R) contain exactly the same real 2x2 matrices, ie that they are trivially isomorphic with each other.

[kind jungle]

Why is the union of a nested set of ideals (as in the ascending chain condition) an ideal?

[thorn delta]

if you take two elements of the union, they are both contained in some common sufficiently large ideal in the union

[kind jungle]

thank you 🙂

[thorn delta]

np

[broken stirrup]

Any recommendations about textbook on Character theory of finite groups

We use Irvin Martin Isaac's book but I don't think it's for beginners especially chapter 1

[opal osprey]

I have studied character theory mostly in the context of representation theory. If that's your main motivation, then Fulton-Harris has a first chapter covering some basic (complex) representation theory of finite groups and it has a nice introductory presentation to character theory. If you want something a bit more advanced, you may check Serre's "Linear Representations of Finite Groups".

Fulton-Harris and Serre respectively.

[next obsidian]

Fulton-Harris is such a weird book

Because it’s just two algebraic geometers that made a rep theory book

[opal osprey]

Tbh, I have only used it to study representation theory of finite groups, idk how it deals with rep theory of lie groups and lie algebras.

And I guess it is fine for rep theory of finite groups

But can you notice a bit of influences from algebraic geometry in the later chapters?

[shell brook]

Is this true? $|\bQ(\sqrt[n]{2}):\bQ| = n$

[cloud walrus]

jesse

[shell brook]

or is only for certain n

n > 1 sry

this is a dumb question god

yeah so its minimal polynomial is $x^n - 2$ and so it has basis ${1, \sqrt[n]{2})^2, ..., \sqrt[n]{2})^{n-1}}$ and so yeah it has degree n i think

Latex.

[cloud walrus]

jesse

[shell brook]

okay so maybe I'm missing some condiiton but I think its right ikd

[thorn delta]

right, and in general [F(u) : F] is the degree of u's minimal polynomial

[shell brook]

Okay the thing I'm confused about is that

In the book they write "If a is algebraic over F and it's minimal polynomial is degree n then |F(a):F| = n". But isn't \sqrt2 not algebraic over Q? Am I misunderstanding something

Isn't a number algebraic in F only if it's a root of a polynomial in F[x]?

[thorn delta]

sqrt(2) is the root of x^2 - 2 in Q[x] so its algebraic over Q

[shell brook]

oh fuck lol

Okay sorry I am confusing myself yes of course

this makes sense now

thank you

[thorn delta]

npnp

[round jay]

Their product, right?

@chilly radish

[chilly ocean]

7

what is the question am I high?

do i have to show that a finite union of irreducible closed spaces implies set of irreducible components is finite?

because isnt this straight from definition?

is 7 a question or just a statement?

[vestal snow]

Do we require n and r to be positive here?

[golden pasture]

basically yea lol

[golden pasture]

er

not rlly

you basically use "noetherian induction" as given in the hint

[lavish nexus]

let Z be the center of a division algebra over F

how to show Z is a finite dimensional extension of F?

[shell brook]

can someone give me an example of a relatively simple application of the primitive element theorem? (if F char 0 and a, b algebraic in F then there is c in F(a,b) such that F(a,b) = F(c))

it doesnt seem very useful

what becomes easier when you only have one element in your field extension? I can't imagine this is at all nice if a, b are actually given, for instance

or even a really awful complicated one I guess

i just want to see why this even deserves a name

also i dont have a proof for this on hand but why does char 0 matter

the silence is proof that this theorem is dogwater

steinitz should stick to proving cool ag theorems.

[south patrol]

Just looked at my notes again and basically uh so our notes didn't use char 0 but char 0 means ur extensions are separable and the field is infinite which simplifies matters

[shell brook]

ah okay

I looked at another version that wanted subfields of C lol

but that makes sense

[south patrol]

Like our proof basically revolved around saying if K this infinite field and L/K a finite extension, say L=K(a, b), then there are only finitely many intermediate fields

And then u find distinct c, d such tbat K(a+cb) = K(K+db)

which lets you find ur alpha

But yes the thm is more general

[shell brook]

curious

is there a use for it

[south patrol]

Lol idk the use tbh

idk much field theory

Although apparently it was used more in like old galois theory before artin idk

[round light]

One to one correspondence & injective are exactly same thing right ?

I don't understand why it isn't required to show phi (U) is unique for each U

[lethal cipher]

I may sound silly, but I'm not too sure of the 'high school geometry' specifics to get the parallel line l'

Can someone help clarify for me

Topic: constructible numbers

[small bison]

basically it's saying that you can construct a parallel line through another point using a compass and ruler

https://www.mathopenref.com/constparallel.html

[terse crystal]

How to prove that f(x)=x^7+7x^2+2 is irreducible over Q[x]?

[shell brook]

I think this is a bad way to state this theorem. Notice that since every U in S contains the kernel, when we do p(U), we are sorta "losing" everything that the kernel contained (b/c it's getting sent to 0). So when you actually get that p is a map between R and R'/ker(p), because p(U) = {a + ker(p) : a \in U}. Does that make sense?

So then for U' in S', we have that p^{-1}(U') = {a : a + ker(p) \in U'}.

[inland sandal]

could anyone help me with this problem

i think i should use the first isomorphism theorem but I dont know what map i should define that has kernel I

i was thinking maybe phi(f(x,y)) = xyf(x,y) but idk if this is right

[shell brook]

So now you can symbol bash to show that:
${\varphi(\varphi^{-1}(U')) = {a + ker(\varphi) \colon a \in \varphi^{-1}(U')} = ... = U'}$, and also
${\varphi^{-1}(\varphi(U)) = {a : a \in b + ker(\varphi), b\in U}}$.
And so you can show that the second one is equal to $U$ also. Thus we have a bijection.

[cloud walrus]

jesse

[shell brook]

@round light

this is awful + ugly etc but

idk i dont get the proof in the book you posted at all lol

[shell brook]

you got mat401 tmrw too?

[inland sandal]

@shell brook yes lmaoo

your in this class?

[shell brook]

yep

not looking forward to it lol

[inland sandal]

i just started studying a few hours ago im screwed lol

[shell brook]

i skipped that problem b/c i dont know nayhting about rings in two variables and didnt want to bother lol

oh shit

[inland sandal]

i skipped the last 2 homeworks also so i still need to study galios stuff

[shell brook]

holy fuck

hw9 was killer

[chilly ocean]

good luck to both of you

[shell brook]

have u taken a class w kudla tteppa

[chilly ocean]

nope

[shell brook]

get him 2 teach u langlands

[chilly ocean]

langlands

[shell brook]

the whole thing

[chilly ocean]

all of it?

why do that when i can just intuit it?

[shell brook]

also i was talking to someone about 488 and they said that it was fucked

why do u love it so much

[chilly ocean]

448?

[shell brook]

ya sry

[chilly ocean]

why was it fucked

[shell brook]

488 is the racist version

[chilly ocean]

i liked it

[shell brook]

idk they just said it was hard

[chilly ocean]

skill issue

[shell brook]

very true

i told them the person i know that took it is painfully geometry brained

so tahts hwy u liked it

[chilly ocean]

to be fair it was a little tricky at some points

for example during all the sheaf stuff

and some of the stuff on integral extensions was hard

it was a bierstone course so of course it's gonna be hard

[shell brook]

i will never take a bierstone course

i cant handle this kudla course

[terse crystal]

R^2

You can show that any f(x,y),f(x,y)+I=ax^m+by^n+I for some a,b,m,n

[west violet]

How woiuld you find the class equation of an abelian group?

[inland sandal]

is xy in R[x,y]?

[terse crystal]

And ax^m+by^n+I=ax^m’+by^n’+I so you map f+I to (a,b)

Yes

[west violet]

$$|G| = |Z(G)| + \sum_{i=1}^r |G:C_G(g_i)|$$

[cloud walrus]

suck2015

[west violet]

Thus if G was Z_5? then

$$5= 5 + \sum_{i=1}^r |G:C_G(g_i)|$$

[cloud walrus]

suck2015

[west violet]

which meeans the sum term would be 0

however most sites I've seen list the equation as 1+1+1+1+1?

[inland sandal]

@terse crystal would it be easier to use the first isomorphism theorem?

[terse crystal]

It’s a concrete question, specific calculation is needed

[inland sandal]

okay thanks

[west violet]

@terse crystal sorry for the ping but would you have any insight on my q?

[glass grail]

why not just f(x,y) -> (f(0,1), f(1,0))

nvm im not ok in the head

wait i thinkit works

[inland sandal]

yes i think this works

because the kernel will be when f(0,1) = 0 and f(1,0) = 0 so its I

[glass grail]

yeah

i had a brainfart and thought homomorphism was f(x1+x2,y1+y2)=f(x1,y1)+f(x2,y2) for some reason

[inland sandal]

f(x,y) = bx + ay gets mapped to (a,b) so its surjective

then R[x,y]/I ~= R^2 by the first isomorphism theorem

[terse crystal]

It should be 5=5

[glass grail]

yeah alr

[inland sandal]

thanks @glass grail

[chilly radish]

Well, yes, but better than that, it divides their gcd. By definition, it is a divisor of both numbers, so it divides the greatest common divisor

[round jay]

Yes so it is 1

[chilly radish]

Yep

So the order of g^N is 1

As we wanted

[round jay]

Lit

[round light]

Sorry, got busy I'll go through what you have said

[west violet]

no i mean the sum

[round jay]

Do you have any tips for proving that the rational under addition are not isomorphic to the product of two groups

Two non-trivial groups

[west violet]

just find elements wth different orders

or an element of a certain order in one group that's not in the other

oh wait isn't that a ring? sorry

[chilly radish]

This almost finishes the proof, no you just gotta say smth about the fact that g^N=e=h^N, and how this implies |g||h| divides |gh|

[terse crystal]

There is no sum. That sum is taken for all conjugacy classes that have more than 1 elements

An abelian group doesn’t have such conjugacy classes

[west violet]

yeah i understand , since the abelian group = it's own center

but why is the equation written as 1+1+1+1+1?

[terse crystal]

Because it counted |Z(G)| as part of the sum

c(g)=G for any g in the center

[west violet]

so.. it's doing 5/5 (essentially by group index) for each element in the group

thus 1+1+1+1+1

since [G : C(g)] = 1 for each element in Z_5

[terse crystal]

Yeah

[chilly radish]

Try proving that any two nontrivial subgroups intersect nontrivially. This is not true for a direct product

[west violet]

but then Z(G) would be 0? which is wrong

|G| = Z(G) + sum([G:c(g)])

5 = Z(G) +( 1 + 1+ 1 + 1 + 1)

[terse crystal]

….

That’s just two different sum

One is over conjugacy classes containing more than 1 element, one is over all conjugacy classes

Why make it a big deal

[west violet]

oh ok

sorry ive just done s_4 and i got terms for both Z(G) and the sum

i guess what you've said here makes sense, sorry for any trouble

[round jay]

Cool idea

I think I got the order problem

We know that N<= lcm(|g|,|h|) = |g|*|h|

Suppose N is less than this number

Then N cannot be a common multiple of both |g| and |h|

Thus, either g^N cant equal e or h^N cant equal e, a contradiction

[round jay]

The question comes before the subgroup section of the book so I'm not sure if I can use that

[chilly radish]

That's... Weird

Sorry I'm not sure I see how to do this otherwise

You don't really need contradiction here

g^N=e so |g| divides N. Similarly |h| divides N, so their lcm also does, but they're coprime, so |g||h| divides N

You already proved the other direction so we're done

This is basically what you said without the contradiction

[round jay]

Yeah we're using Aluffi so its a bit different to most treatments

I was thinking about using the universal properties for products/coproducts somehow

Thank you for all the help

[chilly radish]

Np

Lemme think

But you know the universal property and isomorphism before knowing what a subgroup is?

[round jay]

yeah lol

first chapter of aluffi goes into category theory, even before defining groups

[chilly radish]

[mint seal]

sweet

how does it compare to D&F or hungerford?

[round jay]

Not sure, I am a noob to this stuff

[shell brook]

@maiden ocean can u let me back into ivory

@uneven folio 🥺

:(

im going 2 cry

[chilly radish]

Why are u not in ivory

[shell brook]

i asked to be kicked out for a bit

but im studied as much as i can for this godforsaken exam

[chilly radish]

I c

[shell brook]

and i cant keep talking to first years in their mathcord

[chilly radish]

[shell brook]

whos ijn ivory rn

tell them to get in here and save me

[maiden ocean]

done

[shell brook]

thx

[chilly radish]

You're welcome

[median pawn]

I know this:

but I'm not able to make sense of this proof

Could someone help?

[teal iron]

aha you're reading gillman and jerison! I love that book

[median pawn]

YES!

[teal iron]

the proof is just that if I is the z-ideal corresponding to some zero-set A, then I = {f : f|A = 0}. The corollary says that the intersection of prime ideals containing I is equal to { f : exists n with f^n in I }. So the Theorem is saying these two are equal. Clearly I is a subset of the latter, since we can take n = 1. And if there is some n such that f^n is in I, then Z(f^n) = Z(f) so in particular, f^n(x) = 0 for all x in Z(f), and Z(f) contains A, so f^n(x) = 0 for all x in A. So f^n(x)|A = 0, so f^n is in I. Hence the reverse inclusion also holds.

[median pawn]

what do you mean by "I is the z-ideal corresponding to some zero-set A"

[cloud walrus]

Hausdorff

[teal iron]

oh shoot, I've been only working with compact spaces for the last while so I kinda skipped ahead and only in that setting. More generally, Suppose I is a z-ideal. Let f^n be in I for some n. Then Z(f^n) = Z(f), so because f^n is in I, it follows that Z(f^n) is in Z[I], so Z(f) = Z(f^n) is in I, so f is in I.

[median pawn]

okay so my question basically is, why do you start the proof with " Let f^n be in I for some n."

i understand the rest of the proof

[teal iron]

ah okay, because you want to prove it is the intersection of all prime ideals containing I, and by the Corollary, that's the same as showing its equal to the set {f : f^n in I for some n}

Clearly I is included in that set, so you just need to show the converse

[median pawn]

ahhhhh right brilliant

sorry this was trivial

thanks a lot

[desert dome]

Hi, I'm reading through this proof. Could anyone help explain why the highlighted line holds? Thank you!

[proud bear]

This is a fact about primes ideals in general. If $P$ is a prime ideal and $P\supseteq IJ$ for ideals $I,J$, then $P\supseteq I$ or $P\supseteq J$. This is because if there is some $x\in I$ that is not in $P$, since $xy\in P$ for any $y\in J$ we must have $y\in P$ as well. Then you can extend this to any finite product of ideals by induction.

[desert dome]

I see. Thank you!!

[median pawn]

@teal iron Sorry for the tag, but I have another question - could you help me show that every maximal ideal in C(X) is a z-ideal?

My approach is as follows: suppose Z(f) is in Z[I]. If f is not in I, then (I,f) = C(X) by maximality of I

So that the constant function 1 = h + sf, for h in the ideal I, and s in C(X)

[median pawn]

This answer has a similar starting point, except they assume Z[I] to be something else, so it doesn't work

[broken stirrup]

thank you

[west violet]

Whats the difference between $S_n$ and $S_{(n-1)}$

Essentially I'm trying to find the difference in elements

[cloud walrus]

suck2015

[west violet]

i know $|S_n| = n! $ and $|S_{(n-1)}| = (n-1)!$

[cloud walrus]

suck2015

[west violet]

so there are n times more elements in S_n

Never mind i thin i've got it

[sharp sonnet]

i wouldnt even know how to answer this, im sure you will find more differences...

well, one is a subgroup of the other

[west violet]

it's a little unclear

what i was trying to say is if you could show S_{n-1} being a subgroup of S_n

or atleast isomorphic to a subgroup of S_n

my reasoning is that if we fix 1 element of a permutation in S_n, there are exactly (n-1)! permutations that do this (including the identity)

[west violet]

ah yeah how do you show that?

[sharp sonnet]

basically what you said

[west violet]

well what i said shows that there's a subgroup of order (n-1)! containing the identity

oh wait, if 1 eelement is fixed, then it's the group of permutations on (n-1) elements ?

[sharp sonnet]

any permutation on {1, ..., n-1} is just a permutation on {1, ..., n} that fixes n

and vice versa

[west violet]

hmm ok, thanks

[wary verge]

Do anyone know how i can show that the groups Sp (2) and SL (2, R) contain exactly the same real 2x2 matrices, ie that they are trivially isomorphic with each other.

[broken stirrup]

I suppose {f_x} forms a linearly independent subset but doesn't generate every element right?

I also see why F* is isomorphic to direct product but don't understand why last sentence leads to contradiction

[cloud walrus]

Hausdorff

[median pawn]

What's next?

[wooden ember]

Take p of smallest degree in x, suppose it isn’t divisible by y^2 - x and perform euclidian division or something around those lines

That’s usually how you go about showing these kernel identities

[chilly radish]

Idt that works here

F[x,y] is not euclidean

Only noetherian

Not even a PID

[wooden ember]

It’s fine though cause coefficients are invertible here

So you can still perform division

[chilly radish]

[wooden ember]

You can do euclidian division with y^2-x because its leading coefficient is invertible

As a polynomial in x over F[y]

[chilly radish]

Hmm okay yea that might work

[wooden ember]

In general if you quotient by something with a linear term for a polynomial ring you tend to be able to apply this argument I’m pretty sure

Cause you can do division with remainder and be left with a polynomial that’s constant in your isolated variable

In this case x

[chilly radish]

Yea you're right

Cuz you can scale the leading term of both polynomials and the euclidean division algorithm works without a hitch

[pastel cliff]

does bilinearity of the lie bracket give me that [x,cy] = c[x,y]

where c is a constant ofc

[delicate orchid]

it may just do so

it might just be the case

[broken stirrup]

oh so every element of {f_x | x \in X} is from the direct product but since X is infinite, direct sum and direct products are different thus {f_x | x\in X} may not be a subset of F, right?

[thorn delta]

Im not sure im understanding your reasoning, but the isomorphism would give $\bZ^{\oplus |X|} \cong \bZ^{\Pi |X|}$ and that is a contradiction

[cloud walrus]

kxrider

[broken stirrup]

im sorry but I don't see why F* should be isomorphic to $\bZ^{\oplus |X|}$

[cloud walrus]

backpack

[lethal dune]

F* is a Z-module?

[thorn delta]

Suppose ${f_x : x \in X}$ forms a basis for $F^$. The isomorphism $F^ \to \prod \bZ x$ given in the hint sends ${f_x : x \in X}$ to a basis which generates $\bZ^{\oplus X} \subset \bZ^{\Pi X}$.

[cloud walrus]

kxrider

[thorn delta]

but since the image of the basis would generate the image, this gives $\bZ^{\oplus X} = \bZ^{\Pi X}$

[cloud walrus]

kxrider

[thorn delta]

i mean basically you have $\operatorname{span}{f_x : x\in X} = \bZ^{\oplus X}$ while $F^* \cong \bZ^{\Pi X}$

and this is bad

[cloud walrus]

kxrider

[broken stirrup]

F* is dual of F and is Z-module so yes

thank you kxrider

I'll inspect your solution thanks for your time

[thorn delta]

np. I think maybe the important takeaway is that $span{f_x : x \in X} \hookrightarrow \bZ^{\Pi X}$ just generates $\bZ^{\oplus X}$ in $\bZ^{\Pi X}$ but we know this inclusion must be proper

[coarse stag]

can someone explain to me what the {1,\tau} means?

[cloud walrus]

kxrider

[tall jay]

Say $$f(x) = ([x]_m,[x]_n)$$ where $x\in\mathbb{Z}$. Why does m and n have to be coprime for f to be onto?

[cloud walrus]

beeswax

[carmine fossil]

Take m=4 and n=6 for example you will never hit something like (3,4)

[tall jay]

I see. To prove that it's onto, would the best approach be contradiction?

[carmine fossil]

It will be explicit construction

Consider m' such that mm'=1 mod n and n' such that nn'=1 mod m
Then
Consider x=bmm'+ann' . This will hit (a,b) for any a,b in the appropriate domain

(m' and n' exist because of coprimality)

[tall jay]

ah

[broken stirrup]

how do I prove part (b)?

[thorn delta]

Think $Hom(F^{\oplus X}, F) \cong \prod_{x\in X} Hom(F,F) \cong F^{\Pi X}$. What are elements of $F^{\Pi X}$?

[cloud walrus]

kxrider

[normal oasis]

I can see from closure that all these new cyclic groups generated by elements under the hyper operation series will include the latter groups, but I have trouble understanding how these groups have specific equivalence relations between each other. I only get a sense of congruence and that there exists a mapping, can someone show me how to understand subgroup and supergroup equivalence?

[broken stirrup]

so it's copies of F indexed by the basis set

[thorn delta]

yes, so like, what do individual elements of this product look like?

F^{prod X}, not prod Hom(F,F)

by sequences of maps, do you mean like maps Z --> {some set of functions}?

[broken stirrup]

sequence of elements from the field in which even all coordinates can be nonzero

[thorn delta]

to avoid causing more confusion, the point is that elements of F^{prod X} are functions X --> F

that's what a "sequence indexed by X" is

but using sequence is perhaps confusing since we think of sequences being countable

[broken stirrup]

oh you mean the function notation of a direct product

yes

[thorn delta]

anyway that's it

[broken stirrup]

it's clear now thanks

[thorn delta]

npnp

and ye bianca

[teal iron]

Suppose Z(f) is in Z[I]. Then there exists g in I such that Z(f) = Z(g). If f is not in I, then as you said, (I,f) = C(X) by maximality, so 1 = h + sf for some h in I and s in C(X). hence Z(h) and Z(sf) = Z(s) cap Z(f) are disjoint. Since Z(f) = Z(g), it follows that Z(h) and Z(sg) = Z(s) cap Z(g) = Z(s) cap Z(f) = Z(sf) are disjoint. But g is in I, so sg is in I. Hence sg and h are elements of I with disjoint-zero sets. So Z((sg)^2 + h^2) is empty, so (sg)^2 + h^2 is invertible. So I contains a unit, contradicting that I is not C(X).

Alternatively you can use the theorem in the book that the pullback of Z[I] is always an ideal that contains I. Then by maximality of I, these are equal, and hence I is a z-ideal.

[thorn delta]

its nice to have an alternative way to refer to elements of the dual rather than elements of the space Hom(oplus_x \in X F, F)

[simple mulch]

Can anyone explain the little group method? please

[pastel cliff]

idk how trivial this is but can i usually state without proof that the units of a subring are also units of the "super"ring it's in

[thorn delta]

probably

[pastel cliff]

actually wait no

Z/2Z is a subring of Z right

[south patrol]

No

[pastel cliff]

fuck no i mean 2Z

[south patrol]

Depends on whether subrings need to share the identity of the ring in your convention

[devout crow]

yes, if something has a multiplicative inverse in the subring then it has the same inverse in the superring

it's obvious enough to state imo

[pastel cliff]

mildly related but what's special about commutative rings

besides being commutative ofc

[thorn delta]

“Good” definition of prime ideals.
Free modules over commutative rings have well defined rank

Free modules are modules with a basis (in the same sense as for vector spaces). This definitely comes up in AT (like homology theory)

But with noncommutative rings, you can have free modules with a basis containing any finite number of elements for example

Free groups are the same kind of thing. They all have the same universal property

[devout crow]

I think the most immediate thing is that the ideal generated by some element x \in R is xR ("everything with a factor of r")

noncommutative abstract algebra seems sort of gross to me because you can't do so many things

[thorn delta]

Ah yea right, the structure of ideals suck in noncommutative rings

[chilly ocean]

Can someone explain adjouin notation to me? I.e, Z[x]

What’s the mathematical definition? The wiki article is using different notation

[open pilot]

Commutative algebra isn't equated with the study of ideals for no reason

[next obsidian]

Comm alg is surely more than just that

Although its roots do lie in ideal theory

[lavish nexus]

\

what is meant by "universal with respect to this property"?

[tribal moss]

In this case it must mean that whenever there is a k: X -> B such that fk=0, then there is a unique m: X -> A such that k=im.

Hmm, or is that even right?

[latent anvil]

that's right

How so?

[tribal moss]

I'm not sure "universal with respect to this property" is unambiguous enough to pinpoint that expansion if you don't already know what a kernel is supposed to be, though.

[south patrol]

Misread sorry sleepy

Swapped two letters round

[lavish nexus]

looks like universal properties of tensor products and localizations

[tribal moss]

Universal properties look like that.

[lavish nexus]

so for cokernel it would be
whenever there is a k: C-> X s.t. kf =0, there exists unique m: D ->X s.t. k = me

[latent anvil]

I don't think so

nevermind, I was being silly. This looks right

[lavish nexus]

if there is g: A' -> A such that ig = 0

ig is a map from A' to B such that fig = 0
so ig is a kernel of f

but i is also a kernel of f, so fi =0, and by universal property there exists g' : A -> A' such that igg' = i

yet ig = 0 so igg' = 0 and i = 0?

[toxic zephyr]

in my textbook, an ideal of a ring is defined by being

1. nonempty
2. closed under subtraction
3. closed under multiplication by the ring
   is the second requirement a standard definition? i havent found any other sources that do it like that. the way i keep seeing is that its a subgroup under addition. our textbook is dumb and the chapters go like Ch 2. Rings and Fields and Ch 4. Groups so im wondering if this is because of that stupid ordering

[next obsidian]

This is equivalent to being a subgroup

Let’s prove it, but let’s even not assume it’s abelian

If H is a subset of G that’s nonempty and closed under division

take arbitrary g in H, then gg^-1 = e so e is in H

Then, eg^-1 = g^-1 is in H so it’s closed under inverses

Then, g(h^-1)^-1 = gh so it’s closed under products

[toxic zephyr]

that makes sense. im just wondering if its standard, rather than questioning if it's valid

[next obsidian]

Ahhh

Well, normally you might say “a subgroup such that…”

But idk, definitions are whatever

Idk if there’s really a standard because everything is ewuicalent so ppl just do whatever

I would say it’s probably not that common tho

[prisma shuttle]

can someone explain why ht(p)=dim(A_p), where A is a ring and p is an ideal of A

[prisma shuttle]

when every element has its inverse in it(which it must as an ideal), then addition and subtraction are analogous. So if you were to just say its closed under addition, you would have the additional requirement that all elements in the ideal must be invertible

[next obsidian]

There’s a bijection between primes of A_p and primes contained in p

I suggest you pick up a book on commutative algebra and read the first few sections, localization will surely be covered in it

[upper cape]

Can someone tell me if this is true?

I think it is, since if k[A] is semi-simple, then it is Artinian, and if A is not of finite order, then we can come up with an infinite descending chain of ideals, hence contradiction

And otherwise if A is of finite order, then the only maximal ideal is {0} so k[A] is semi-simple

[wicked zephyr]

Are in general, automorphisms of an object, a measure for the complexity of the object itself?

I know this is vague, but I'm trying to motivate galois theory a bit

Ambiguity is a better word maybe, but I'm also interested in the answer since I've heard in AG seminars that automorphisms complicate things in general

[frail zealot]

automorphisms are a symmetry of the object so in a sense yes the more symmetries the more complex unless the automorphism group is of prime order or something and a really easy infinite class of groups

[next obsidian]

In algebraic geometry automorphisms screw up things being an actual scheme, and force you to use stacks and stuff

One idea in this direction is that you can glue stuff if you have a cocycle condition on things which says certain pairs need to be equal

You know that the two things differ by an automorphism, so if it has no automorphisms then you trivially know they’re equal

In the case that automorphisms exist there might not really be a way to do things up to isomorphism so you’re forced to consider things not up to isomorphism while remembering they’re isomorphic

And this leads to needing groupoids and crap, and you just keep going and then you have a stack

That’s my 1 minute shit explanation

[strong yacht]

If you're trying to motivate Galois theory, check out this vid, really helped me motivate it a few months ago: https://youtu.be/CwvuZ8aHyH4

[cloud walrus]

gristality of the cardinuum

[echo orbit]

I proved this, call it proposition A

the part i'm having trouble with is proposition B:

[cloud walrus]

gristality of the cardinuum

[echo orbit]

the book says that, by proposition A, it suffices to prove proposition B for the case where I = 0

but i just don't see how

book is d&f

[terse crystal]

Prime ideal P of R containing I correspond to prime idea P/I of R/I.

(Intersection of P containing I)/I=intersection of P/I=nilrad(R/I)=nilrad(I)/I , and nilrad(I) is contained in (intersection of P containing I) so those two equal

[echo orbit]

two things there, at least

what is A?

and why is that first equals sign true?

[terse crystal]

Sorry R not A edited

?

There exists a one-to-one correspondence between ideals J containing I to ideals J/I of R/I

You can easily prove that J is prime iff J/I is prime in R/I

[echo orbit]

oh

intersection of P containing I = intersection of P/I

[terse crystal]

Yeah

By definition

[echo orbit]

(P/I)/I = P/I?

[terse crystal]

No

I wrote over I

[terse crystal]

Here

[echo orbit]

i'm looking at the first equals sign

[echo orbit]

mod both sides of this by I

to get what you got on the lhs

[terse crystal]

I wrote over I you missed it

[echo orbit]

?

what do you mean

wrote what over I

(intersection of P/I)/I = intersection of P/I?

yes, i'm totally lost here. it's 5am. i think i'll ask my prof tomorrow

[terse crystal]

Here

Make sure you didn’t misread anything

[wicked zephyr]

@frail zealot @next obsidian @strong yacht thank you for your answers

[echo orbit]

i didn't misread it, i just didn't understand how you got that

[chilly ocean]

How do you count the subgroups of index two of $\left(\frac{\mathbb{Z}}{2\mathbb{Z}}\right)^n$?

[cloud walrus]

Gio

[delicate bloom]

start with small examples like the klein 4 group and see if you can work out a pattern

[chilly ocean]

it should be 2^n-1 but I can't prove it

[delicate bloom]

how many elements of order 2 are ther ein (Z/2Z)^n

[chilly ocean]

2^n-1

[delicate bloom]

right, so you can quotient out by the subgroups generated by those elements

and so you get one subgroup of index 2 for each element that way

and that's it, you're done

[chilly ocean]

how are they of index two if the order is 2?

[delicate bloom]

you're right what I said is total nonsense wtf

oh we might be saved if we can then show that we have a direct product of groups there

[ember field]

I know groups actions on a set give a homomorphism into the symmetric group on the set, but what if I have a group action on a group? Does that give a homomorphism into the automorphism group of the group?

[lavish nexus]

how do you show they satisfy the universal property

[next obsidian]

I’m pretty sure you just do it component-wise IIRC

[latent anvil]

Let $f(x) = x^4 + 2x^2 + x + 3 \in \Q[x]$. How can I compute the galois group of $f$ in a manner that's practical on a qualifying exam?

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

Even calculating the discriminant is kind of hairy here bc I would need to memorize the formula for it and successfully calculate it by hand under time pressure

[strong yacht]

@latent anvil it is S_4, since it is irreducible mod 2 and it factorises as a linear * a cubic mod 3

And also because of the following theorem

[latent anvil]

how do you deduce it contains a 3 cycle using the factorization over F3?

There's this theorem, but that relies on committing the discriminant, no?

And that's a huge time sink and is error prone

[strong yacht]

No I don't think you need to compute the discriminant at all

[latent anvil]

So how do you get information about the galois group from the factorization over F3?

[strong yacht]

This is what I used, doesn't say anything about discriminant here

If you do need the discriminant, I didn't know that

[latent anvil]

Oh, maybe the repeated roots condition is equivalent to the discriminant not being divisible by p

That would be sick

Do you have a reference for that theorem?

[strong yacht]

https://dec41.user.srcf.net/notes/II_M/galois_theory.pdf

[latent anvil]

Ty

[strong yacht]

Page 62

[latent anvil]

Right so if you have a monic polynomial the discriminant is some sign times Π (ri - rj) where r1,...,rn are the roots

Then p divides the discriminant iff p divides ri - rj for some i, j

Iff the polynomial has repeated roots over Fp

And that's much easier to check

[strong yacht]

Yes, very easy to check, can compute GCD with formal derivative yes?

[latent anvil]

Yep

[rough escarp]

quick question which should be obvious but I cant figure out
how can I show that all elements of a ring with unity will be of the form n*1?(this should be true right?)

[chilly ocean]

n being? if you want it to be an integer, this is false.

[south patrol]

note n*1 =n, so your question is equivalent to asking if elements are of the form n

[rough escarp]

really?? that was unexpected

[chilly ocean]

it should be expected

there are rational numbers that aren't integers

there are real numbers that aren't integers, complex numbers, etc.

[rough escarp]

hmm for some reason i was thinking that 1 should generate the whole ring through addition

[chilly ocean]

the only such rings are quotients of Z

[delicate orchid]

that is true in commutative rings I believe

oh only Z quotients

ah yes, R^2 fails

[chilly ocean]

unity

[south patrol]

Well Z is the initial object in Ring anyway

So 1 under addition generates a ring iff that map is a surjection i.e. quotients of Z

[chilly ocean]

if R is generated additively by 1, then Z -> R, n -> n*1 is a surjective homomorphism. apply the first isomorphism theorem to get R = Z/kernel

i was sniped

[south patrol]

lol

[grand sigil]

I have trouble seeing the implications in the proof. How does (M_n) being an a-filtration imply that M’_n=a^nM is subset of M_n. Similarly How does (M_n) a-stable imply a^nM_n0 a subset of M’_n=a^nM?

[grand sigil]

source: Atiyah Macdonald. Introduction to Commutative Algebra

[chilly ocean]

for your first question, repeatedly applying the definition of an $\mathfrak{a}$-filtration gives $$\mathfrak{a}^nM = \mathfrak{a}^n M_0 \subseteq \mathfrak{a}^{n-1}M_1 \subseteq \cdots \subseteq \mathfrak{a}M_{n-1} \subseteq M_n.$$

[cloud walrus]

TTerra

[chilly ocean]

the other question is similar. $$\mathfrak{a}^nM_{n_0} = \mathfrak{a}^{n-1}M_{n_0+1} = \cdots = \mathfrak{a}M_{n_0+n-1} = M_{n+n_0},$$ and this is a subset of $\mathfrak{a}^nM$ since $M_{n_0} \subseteq M$ implies $\mathfrak{a}^nM_{n_0} \subseteq \mathfrak{a}^nM$

[cloud walrus]

TTerra

[chilly ocean]

@grand sigil does this answer your questions?

[echo orbit]

is this rly true

[lavish nexus]

3 seems pointless

[echo orbit]

that's dope af if true

[lavish nexus]

if the composite A -> C is zero then ofc it's a complex and any two rows exact gives the other row exact if they are complexes

[echo orbit]

o yea its obviously true

bc free group 1 generator

[grand sigil]

Sort of but I think I misunderstand. I see how we are defining (M’_n) to be a^nM. However I do not see how a-filtration gives us a^nM_i subset of a^n-1M_i+1. I thought it only means aM_n subset of M_n+1. Im going to spend more time later clarifying my misunderstanding.

Thanks for the help

[chilly ocean]

$$\mathfrak{a}^nM_k = \mathfrak{a}^{n-1}(\mathfrak{a}M_k) \subseteq \mathfrak{a}^{n-1}M_{k+1}$$

[cloud walrus]

TTerra

[chilly ocean]

using only the definition of an a-filtration

Why isn't [Q(gamma):Q]=2

[teal iron]

it could be, it depends on gamma

it just says its 2^r for some r >= 0

it could be higher than 2, for example, by the tower law. So like [Q(fourth root 2) : Q] = [Q(fourth root 2) : Q(sqrt 2) ][Q(sqrt 2) : Q] = (2)(2)=4

[grand sigil]

Thanks a lot that clarified what you did.

[lilac trench]

can you have polynomials of infinite degree in a polynomial ring

[chilly ocean]

no

but you can talk about (formal) power series rings

[grand sigil]

No worries if no responses because I will try and read about later, but does anyone know what prime/maximal ideals correspond to on rings of smooth functions on a manifold. And if so what do quotients and localizations correspond to?

[chilly ocean]

for compact manifolds, maximal ideals are in one-to-one correspondence with points. i don't think that's true for non-compact manifolds. i don't know if in general you can say anything nice about the prime ideals there

by "one-to-one correspondence with points" i mean that every maximal ideal in C^\infty(M) is of the form {f : f(p) = 0} for some (unique) p in M. the quotient in this case is the field of real numbers, and the localization is the set {f/g : g(p) \neq 0}, which is just C^\infty(M) again

that last part being because we can take reciprocals of smooth functions and stay a smooth function

which you might want to compare to the following statement: the reciprocal of a polynomial function is not a polynomial function

whitney has a paper on ideals of smooth functions that you can find somewhere on google that might give you what you want

[shell brook]

this is cool

[teal iron]

i think it might be true more generally that every closed prime ideal is maximal, and corresponds to a point

[median pawn]

Thanks a lot

There is a small error in your proof, just replace \cap by \cup everywhere

All good otherwise, thanks!

[south patrol]

Yes, send 1 to 1 and extend in the unique fashion

[grand sigil]

I do not see how 1+a a unit implies \cap a^nM = 0

Does this come from 1+x unit implies x nilpotent?

[south patrol]

Look at the theorem this is a corrolary of

Namely, that intersection is the set of x in M annihilated by some element of 1 + I, i.e. {0} since all of those elements are units.

[grand sigil]

I was thinking of this, but then I couldnt find a reason for why units can’t be annihilate x in M. All I would know is that Ann(x)=(1) for all x in M

[south patrol]

If Ann(x) = A then x = 1.x = 0

[grand sigil]

That is very clear. Thank you for the clarification

[south patrol]

Np

[lethal dune]

@grand sigil how do you define topology using a?

[south patrol]

It's the I-adic topology, where the base at 0 is given by M, IM, I^2 M, I^3 M, ... for an ideal I

[next obsidian]

You have to extend it linearly tho

Otherwise you haven’t said much about the other points’ neighborhoods

[south patrol]

Sorry i meant for that to be implied lol

because aren't topological rings like topologically homogeneous

[next obsidian]

Yeah

You want the mult and addition map to be continuous and since addition is invertible you can always just look at nbds of 0

[long obsidian]

Is there an intuitive idea of what a character of a representation is?

[lavish nexus]

how exactly do you lift the cycle c to b in B_n

C is just a chain complex not an exact sequence

there can be stuff in Z_n that are not in B_n

[thorn delta]

the context is you have a SES 0 -> A -> B -> C -> 0 of chain complexes, right? The chain map B -> C is surjective

[tranquil parcel]

For part a), it seems sufficient to show there is an automorphism which takes a to a+1. Any hints on how to find this?

[thorn delta]

there is a theorem about extending isomorphisms of fields to isomorphisms on simple extensions of those fields. does that ring any bells?

[tranquil parcel]

No I don't think it does.

[lavish nexus]

I was dumb I thought it was the B_n(C)

[thorn delta]

ah yea okay i figured, that's kinda confusing lmao

[barren sierra]

How do I find a galois group

[cloud walrus]

Spamakin🎷

[barren sierra]

Not quite sure how to do this

[tranquil parcel]

What ways can you 'conjugate' that expression you set alpha to?

[barren sierra]

wdym

[tranquil parcel]

as an example, i can change that interior + to a minus... plug it in, and get another root

[barren sierra]

yea

[tranquil parcel]

so that is an automorphism. Really you're relying on the idea that automorphisms take roots to roots.

[barren sierra]

Hm

so I have 4 choices for where to map alpha

and i must go to some version of i I guess

i and -i?

[thorn delta]

btw this is the result that should help you. nothing super complicated

[tranquil parcel]

So, I need to take an automorphism and push F(a) through it, and show that a+1 is a root to that polynomial, so that I may invoke theorem 1.8 to produce an isomorphism of F(a) onto F(a+1)? @thorn delta

that polynomial meaning the image of the polynomial under sigma

[thorn delta]

yes, you start with the identity F --> F. It maps f to f, so you can construct an automorphism sending a root of f, say a, to another root of f, say a + 1

[tranquil parcel]

I see this makes a lot of sense, thanks @thorn delta.

[thorn delta]

npnp

[dull root]

I'm having a bit of trouble showing the following:

Let K be a subfield of C. If we had an polynomial f in K[x] such that

.f has a root in some radical extension of K.

Then it is true that f has some irreducible factor p such that the Splitting field of p over K has a solvable Galois group.

It is obvious that I need to apply Galois's theorem there, which states that

1. Some polynomial g \in K[x] has all its roots contained in some radical extension of K \iff the Galois group, (ie the galois group corresponding to the splitting field of g over K) is solvable.

So we know that our poly f has some root in a radical ext of K, say alpha. My thought is to look at the minimal poly of alpha, which must be an irreducible factor of f, and call this p. Now if this radical ext has all roots of p we are good, but the issue is that might not be true. If this radical extension was normal, we would be good since it must have all roots of p, but it is not true that all radical extensions are normal....

The only thing that I have which relates radical extensions to normal extensions is that the normal closure of a radical extension is radical. But I don't think this is useful here.

Opps nvm, it is clear this is exactly what I need. This statement is basically trivial using the final line xD

[chilly ocean]

Why is this true

[delicate bloom]

Since E is finite, it has positive characteristic, call it p. So we can just take 1 and add it to itself to get {0,1,...,p-1} and they multiply just like they would in Z_p because they're integers mod p.

idk if that's satisfying enough

[south patrol]

One way of viewing it: the map Z -> E has kernel pZ, p the characteristic of E (ig this is how I'd define characteristic in fact), and so we have an embedding Z/pZ -> E by the first iso thm

[dull root]

If I have an primtiive n-th root of unity (not necessarily n prime) then the degree of its simple extension is counted by the Euler totient function at n right? IE that extension has degree equal to counting the integers less than n coprime to n

[barren sierra]

Is this Galois Group (it is a Galois Group right?) this a Klein 4 group?

I think it is because the 4 roots are +- sqrt(2) +- sqrt(3). Thus permuting sqrt(2) with -sqrt(2) (and sim for sqrt(3)) gives the 4 mappings of one root to another. However looking at it in this angle gives a group of C_2 x C_2 = Klein 4 group

[barren sierra]

Dumb question isn't it primative iff n is prime? if n isn't prime we just say n-th root of unity I thought.

But yes I believe that's correct

[dull root]

yea, i meant just n-th root. not primtitive mb

[sharp sonnet]

huh

an n-th root of unity is primitive if it is not an m-th root of unity for m < n

n doesnt have to be prime

the statement is true though

[strong yacht]

Looks good

[strong yacht]

If the degree of the splitting field degree is 8 then I think the Galois group is D_8 (for quartics), at least I think this is always true

But the question asks you to find it out using what you have

The roots can be written as (x-a)(x+a)(x-ia^-1)(x+ia^-1) where a is what you said it was, if you label the roots as {1, 2, 3, 4} instead, complex conjugation gives you the aut {1, 2, 3, 4} -> {1, 2, 4, 3} i.e. the transposition (3, 4)

Your target is D_8 which is <(1, 2, 3, 4), (1, 3)>

Hmmm

[barren sierra]

Yea I found some automorphisms that generate a Klein 4 group and a cyclic group

So it must be D_8

[strong yacht]

Ah ok nice

[next obsidian]

Wait what?

This is a degree 4 extension is it not?

How can you have D8 for the Galois group

[strong yacht]

I thought the question said it is degree 8 in the first part

[next obsidian]

Oh different question lmao

[strong yacht]

Ahh

[next obsidian]

I was looking at the Q(sqrt(2),sqrt(3))

[strong yacht]

How do you find more auts to this actually

Galois group acts transitively so e.g. there must be an aut that sends 1 to 2, but I know nothing else about this aut

[chilly ocean]

Every finite field E is a finite extension of a prime field isomorphic to the field Zp, where p is the characteristic of E.
Why is this true

[sharp sonnet]

what part of the answer 10 mins ago didnt you like

[next obsidian]

Wtf lmao

[tardy yacht]

I was trying to show that cube root 2 is not in Q(sqrt(2),sqrt(3),..sqrt(p)...) and I was initially under the impression that simply showing that it wasn't rational or a square root would suffice, but now I realized that I must make sure it is not a linear combination of square roots over the rationals. I tried to suppose it was a linear combination and square it meany times to get a contradiction but didn't get anywhere. How should I proceed?

[strong yacht]

Does the Tower Law help? If it was a (finite) linear combination then Q(cuberoot(2)) contained in Q(sqrt(a_1), ..., sqrt(a_n))

[tardy yacht]

By tower law do you mean [M:K]=[M:L][L:K]?

[strong yacht]

Indeed

[tardy yacht]

Thank you lol

[strong yacht]

no problem

[lethal cipher]

Question: given a group of order 2^m, is it guaranteed to have subgroups of order 2^i for all i?

[next obsidian]

Yes by induction

Any p-group has a nontrivial center so you can mod out by the center then use induction

I guess you need to handle the case that the center is everything, but then you can use the classification of finite abelian groups

[lethal cipher]

Hmm, so I see we can use Lagrange's theorem for order p^2. But I'm not sure how you'd use induction for p^n?

[elfin furnace]

After you have a nontrivial center you get can an element x of order p in the center so you can just quotient by (x) which has a subgroup isomorphic to a subgroup (of the original group) of order p^{n-1}

Yeah i think a subgroup of order p^{i-1} in G/(x) is isomorphic to a subgroup or order p^i in G?

@lethal cipher hope that helps

[cloud walrus]

limbostar

[long obsidian]

I mean \chi(e)=\chi(g)

[delicate orchid]

Every normal subgroup is the intersection of the kernels of some characters

If all of these kernels are trivial then the only possible intersection is trivial (or the entire group if you just take the kernel of the trivial character), thus the group is normal

[delicate orchid]

This follows from the fact that you have the kernel of the character equal to the kernel of the representation for representations into GL(n, C), and the kernel of any homomorphism is a normal subgroup - the other direction is a bit harder to prove but I do think you use lifted characters in it

I forgor

[long obsidian]

That helps so much thank you

[terse crystal]

Any n=(p^r)m, where p doesn’t divide m, there exists subgroups of order p^s where s<=r. The number of those subgroups is also =1 mod p. I don’t know why most proofs only prove the case where s=r, but the proof of sylow theorem by bourbaki doesn’t say anything about r=s

[deep sky]

If a group G is acting on a set of n elements by permutation, does G have to be a subgroup of S_n?

[latent vector]

hey i need some help with galois theory

i'm supposed to draw the lattice of intermediate fields of the extension E/Q, where E is the splitting field of x^4-6x^2+6

so far, i've shown Gal(E/Q) = D4

but idk how to find the generators of the group

E = Q(sqrt(3 - sqrt(3), sqrt(3 + sqrt(3))

[prisma ibex]

yes

[lethal cipher]

I'm not entirely sure why the center is non-trivial and how we know there is an element of order p in there. Other than that, I think I understand how that fact gets me to the one I want.

[inner holly]

hello, i am a bit confused on how to determine the elements the symmetric groups S_n. why is S_3

and not the set of all permutations of (1, 2, 3)?

[smoky cypress]

Those are all the permutations of 1 2 3

[lethal cipher]

Crown, are you familiar with cycle notation?

[inner holly]

yes

not too sure how to word it, but now i am guessing instead of having {(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)}. S_3 is the set where if i do (1, 2, 3) times any the elements in S_3 i will get an element from the set above?

[lethal cipher]

Ah, I see. So let's take a step back for a small moment. (12) sends 1 to 2 and 2 to 1,while fixing 3.
So 123 maps to 213

And 213 is indeed an element of this set

Each cycle represents one of these permutations you've mentioned

So the two sets you have listed, the one in cycle notation, and the permutations of 123, are indeed the same set.

But the problem with writing S_3 like {(1,2,3),(1,3,2),...(3,2,1)} is it can be hard to keep track of what each permutation does. This is especially true when you are looking at a larger symmetric group S_n.
Cycle notation is a very clean way to tell us exactly what the permutation is doing

[inner holly]

ahhhh okay i understand now

[lethal cipher]

Glad to hear it

[inner holly]

but just for clarity

the permutation {3, 2, 1} would be the element (1, 3) in S_3?

[lethal cipher]

Precisely

If the number isnt mentioned in the cycle, 2 in this case, then the permutation fixes that number

[inner holly]

thank you very much

[smoky cypress]

Plus the benefit of cycle notation is that it tells us what the permutation is doing essentially

So you can write (13)(24) for the permutation that makes 1234 to 3412, so you can multiple the cycles, and it turns out you can write any permutation this way

And the order doesn't matter

[lethal cipher]

When the cycle is disjoint it doesn't matter. I just want to make that clear

[inner holly]

that reminds me, how exactly do you multiply disjoint cycles?

i was searching around on google, but all the examples were not disjoint

[smoky cypress]

If you want to write it in cycle notation you just put them right next to each other

[lethal cipher]

If they are disjoint, ya kinda don't. (13)(24) is exactly that: (13)(24)

When they aren't disjoint is when it gets a bit trickier

[inner holly]

so i can go from 3412 -> (13)(24) but not the other way around?

[lethal cipher]

Actually, (3412) and (13)(24) are completely different permutations

3412 sends 1234 to 2341 and (13)(24) sends 1234 to 3412

They are not the same

[inner holly]

ah okay

thank you both

[lethal cipher]

No problem!

One more thing to keep in mind, cycles are functions, and they are a group under composition. So when composing functions, you combine cycles together from right to left

Example: (12)(23) in the right 2 cycle, 2 gets sent to 3, and 3 gets sent to 2.
In the left 2 cycle, 1 gets sent to 2 and 2 gets sent to 1.
So composing the functions gives the following images for 1,2, and 3:
1->1->2
2->3->3
3->2->1

So (12)(23)=(231)

[inner holly]

(1 2 3)(1 2) = (1 3)?

[lethal cipher]

1->2->3
3->3->1
2->1->2

Yep, that checks out

[inner holly]

very cool

[lethal cipher]

S_n is highly NOT abelian. Of course, S2 is the only exception

[smoky cypress]

Wait (12)(34) and (34)(21) are the same

[lethal cipher]

Wait...disregard

I meant (12)(23)=/=(23)(21)

Grabbed the wrong example. Thanks for catching that

[inner holly]

ahhh okay

[lethal cipher]

Iirc, the identity element is the only element that is in the center (abelian with all elements).

[smoky cypress]

Except when n=2

[lethal cipher]

Lol, yes of course

[echo orbit]

I don't at all understand rschwieb's answer here. I know about Lattice Isomorphism Theorem. But this does not seem to clear anything up.

https://math.stackexchange.com/questions/1591177/shortcut-to-the-radical-of-a-proper-ideal-i-is-the-intersection-of-all-prime

I'm trying to show that $\sqrt{I} = \cap P_I$

[cloud walrus]

gristality of the cardinuum

gristality of the cardinuum

gristality of the cardinuum

[echo orbit]

What I can tell we're really doing is a little clearer to me when I rename things a bit

(I already was able to verify that nilrad$(R') = \cap P'$. And I'd like to let $R' = R/I$.)

[cloud walrus]

gristality of the cardinuum

[echo orbit]

But then, when I do that, I get...

OH

[chilly ocean]

if you have the result for the nilradical in the quotient, take the preimage under the quotient map

[echo orbit]

i needed to do the following to the LHS

[cloud walrus]

gristality of the cardinuum

[echo orbit]

by LIT this corresponds bijectively to

[cloud walrus]

gristality of the cardinuum

[echo orbit]

(ignoring the middle part)

ok, ok

[echo orbit]

the above is what you're talking about?

ah

yeah, it has to be

all you really need is that prime ideals have only prime ideals under the preimage under the quotient map

ok, ok

[grand sigil]

Does anyone know how extended elements of Galois groups work?

So if you have x in Gal(L/K)

how can you extend this to some y in Gal(F/L)

or do there need to be special conditions on the type of extension?

[terse crystal]

Not Gal(F/L) but Gal(F/K) I think:
F is a splitting field of f over K, F is also a splitting field of f over L.you just extend it like when you extend an automorphism to an automorphism on the splitting field

[grand sigil]

Thanks for catching the mistake

[elfin furnace]

Nontrivial center just comes from the class equation. I also misspoke, subgroups of order p^{i-1} in G/(x) correspond with subgroups of order p^i in G that contain (x)

And if the center has an element x with order p^r then x^p^{r-1} should have order p

[quartz perch]

What's the difference between a Weyl chamber and an alcove?

[chilly ocean]

do you want a different kind of answer for each of the four channels you posted it in?

[quartz perch]

ahhh, sorry, it's just I don't know if different people go to different channels

[woeful flint]

I'm having the following problem. I have a finite group and an integer n such that x^n annhilates any element x of the group i.e. yields the identity. Can I conclude that the order of the group divides n?

[quartz perch]

any answer that explains the difference will do. It seemed like it could likely be relevant in the channels I posted, and I didn't know if there would be someone who knew what they were in each channel.

didn't mean to spam

[woeful flint]

I tried showing it using the division algorithm but it tells me that for any element there is an m strictly less than the order of the group which annihilates the group. But I'm not sure whether this is possible or not?

[cloud walrus]

Kraft Macaroni

[delicate bloom]

it's possible, look at the klein 4 group

[deep sky]

So apparently S_4 can act on three elements?

[woeful flint]

Ah of course

Is there anything I can say then?

Surely the order of the group is less than or equal to this n no?

and if I take O to be the smallest annhilating power in G then this indeed does divide n

[woeful flint]

wait now that I think about it this is wrong too

[long narwhal]

[on the topic of rings and ideals] if <2> is an ideal of Z how does that ideal generate say, 3, or 9? i'm having this issue from trying to understand why <2+3> of Z is equal to <2>.
<2+3> includes every term in Z, but <2> seems to only include 2Z which doesn't.

[simple mulch]

Is Z/1Z = {0} or {1} ?

[long narwhal]

{0}

[south patrol]

Neither or both or one depending on what you mean

[lethal dune]

[elfin furnace]

well 1 = 3 - 2 so once you have that you just generate all of Z

[long narwhal]

i get that you can do that from <3+2>

but from <2> you don't have 3 in the ideal to do that, right?

[elfin furnace]

yeah bc its just 2Z