groups-rings-fields

[quaint ivy]

$\rotatebox{90}{\text{nice}}$

[cloud walrus]

derivada.schwarziana

[chilly ocean]

what do i have to check if something is norm or not?
https://gyazo.com/0917c09058ec8f02fb83eebbfc5fea8f
like this

Lmgtfy

?

"let me Google that for you"

But actually not, because it is simple enough for you to Google it as well

?

[quaint ivy]

you should follow your course/book's definition, but check that it's positive, =0 iff the function phi here is the 0 function, that it spits scalars out in abs. value and triangle inequality

also #advanced-analysis

[chilly ocean]

i moved to #linear-algebra , but ty ^^

[quaint ivy]

right, hadn't thought of that

[glad juniper]

Not sure if this is the best channel, but does anyone know how I could show that, say $x^n-r(x^{n-1}+x^{n-2}+\cdots+x+1)$ never has repeated roots when $r>0$? Calculating with small $n$ suggests that the discriminant is probably always a polynomial in $r$ with no sign changes, but I don't know a good way to approach it generally.

[cloud walrus]

dirib

[chilly ocean]

This sounds vaguely like rouches theorem

But I guess rouche was about roots at all, not about repeated roots

Oh

Is it that one criterion?

You take a derivative or something

[glad juniper]

Certainly it suffices to show that the derivative and it don't share any zeros. Is there a clever way to apply Rouche to show that?

[glad juniper]

(for my problem) I think there's an ugly proof out there like "here is explicitly the pattern in the resultants and an induction proof about it", but if anyone knows of a better approach, just message or ping me.

For giovanni, what's giving you trouble?

[placid falcon]

i have no clue how to start

[pallid ember]

which part

[glad juniper]

Forgetting all the details of the problem, you can try to start every "if [something about variables] then [something about variables]" problem the same way:

[pallid ember]

in a your just showing closure under addition

[glad juniper]

Let all the variables be given. Assume the first thing (the "hypothesis"), and then try to do a manipulation to get to the conclusion

[chilly ocean]

could someone check if my last gyazo from #linear-algebra is alright pls?

[old hollow]

Is a field extension written L/K distinct from L itself? Sorta like for quotient groups how G, H, and G/H are completely distinct objects

This is what wikipedia is saying

If K is a subfield of L, then L is an extension field or simply extension of K, and this pair of fields is a field extension. Such a field extension is denoted L / K (read as "L over K").

So when I write L and L/K, are these two things completely different objects?

[chilly ocean]

to me, this is like saying "is X a topological space, or is it (X, T)?"

[old hollow]

oh I see

so L/K is like being specific that L is a field extension of K?

or

[chilly ocean]

obviously in some cases it is importat to note, like Gal(L/K)

[old hollow]

ye

[mild laurel]

This is more like asking if G and H \subseteq G are different things

[old hollow]

yeah my main thing was that quotient groups are written exactly the same way and it doesn't work like that

[mild laurel]

Except there's no quotienting going on here, It's just K \subseteq L

[old hollow]

yeah

[tall jay]

If two of the primes I picked were equal, would I still be able to come up with 3 consecutive integers using some different method/theorem?

Bc I know that the gcd of each mod needs to be 1

[ivory cosmos]

It obviously fails

E. G. Find an x = 1 mod p and = 2 mod p

[tall jay]

I see. What is the reason that it fails, aside from fulfilling the conditions of the CRT?

[ivory cosmos]

I gave an example

[tall jay]

Last question of the early morning: how am I able to check if there are other solutions to this?

[carmine fossil]

let x=[a] and y=[b]
then [a+2b-4]=[0] and [4a+3b-4]=[0]
implying a+2b-4=7m and
4a+3b-4=7k
implying
(2a-8)+4a-4)=7p i.e,
7 divides 6a-12, which implies
7 divides a-2, similarly you get 7 divides b-1

Which implies a has to be 7k+2 for some k which implies
[a]=[2] and [b]=[1]

[molten silo]

Why is this not a field?

[carmine fossil]

What's the inverse of x?

[chilly ocean]

same proof different words: it has a non-trivial ideal that isn't the entire ring, (x)

[chilly ocean]

yo can someone expalin this: in this book they say A_3 has index 2 in S_3 therefore S_3/A_3 is abelian. why is that?

I was thinking sth like if H is normal then G/H is abelian but couldn't quite prove it

they don't seem to use the fact A_3 is abelian

And actually I don't get what they say after: (There was the theorem saying that if H is normal in G then G/H is abelian iff [G,G] subgroup of H)* If S_3/A_3 is abelian then by theorem* [S_3,S_3] \subset A_3. Therefore [S_3,S_3] = A_3 (Why the reverse inclusion?)

[golden pasture]

groups of size 2 are abelian

[chilly ocean]

ye

[rustic crown]

that's not true in general... it just means A3 is normal and the quotient is S3/A3 which has order 2 hence abelian

[golden pasture]

G/H is in general not abelian

[chilly ocean]

oh ok now I see why they talk about the index 2, its just to conclude S_3/A_3 has size 2

actually im not sure how they got it has index 2, they don't elaborate

cant we say if |S_3| =6 and |A_3| = 3 then S_3/A_3 has 2 elements?

[rustic crown]

S3 has order 3! = 6 and A3 has order 3!/2 = 3

yep

[chilly ocean]

is that true always if the quotient is normal?

[rustic crown]

what do you mean "quotient is normal"?

[chilly ocean]

|G/H|=|G|/|H| if H normal

right

[golden pasture]

G/H only exists iff H is normal

[rustic crown]

ye, that's even true if H is not normal

[golden pasture]

as the set of all cosets

[rustic crown]

but then G/H is just a set and not a group

[golden pasture]

^ this

ye

[chilly ocean]

ye

but why do they even talk about index

[golden pasture]

wdym

[chilly ocean]

like why do they state A_3 has index 2 to conclude S_3/A_3 is abelian

[rustic crown]

|G/H| is called index of H in G or number of left cosets

[chilly ocean]

seems like you dont need to use that

[golden pasture]

index 2 -> S3/A3 is a group of order 2

[chilly ocean]

oh okay yeah obviously

sry brainfart

and is that always true that if H is normal then H is a subset of [G,G]?

[golden pasture]

no

H=G

[chilly ocean]

but if H is also abelian?

then its true right

wait no

if H abelian but G not abelian maybe

(I just don't see why [S_3,S_3]=A_3 )

[rustic crown]

[S3, S3] is a subgroup of S3 and can only have even permutations in it

this gives [S3, S3] is a subgroup of A3

just show there is a non-trivial commutator

[chilly ocean]

why

[rustic crown]

and that follows easily because S3 is not abelian

oh so a particular commutator is [g, h] = ghg^-1h^-1... what can you say about its sign?

[chilly ocean]

sign?

as in odd or even number of transpositions?

[rustic crown]

ye

g and g^-1 will have same parity of (number of) transpositions

[chilly ocean]

hmm okay so always even

[rustic crown]

yep!

[chilly ocean]

ohhHHHH

I SEE NOW

thanks det

[rustic crown]

If we use an argument like this, are we sure if the isomorphism is canonical or not?

maybe i'm just mixing up 2 things

also could someone tell me more about natural isomorphisms concretely? i have seen the categorical definition, but like can someone tell me when we say "Consider the natural projection G --> G/N"... where exactly are the 2 functors hiding? can we really say "the" natural projection?

[oblique river]

here's how I think about it: suppose you have some construction of a homomorphism M --> M' which you can do for any object M. Now if you have two objects M and N and some map between them M --> N, we can do the construction on M and N separately to get M --> M' and N --> N'. Does the map M --> N induce a map on M' --> N' making the square commute?

normal is also sometimes used to just mean "canonical"

for your group quotient example, I think when people say "the natural projection G --> G/N" they just mean canonical, but it also is natural under my definition. If you have two pairs of a group and a normal subgroup (G, N) and (H, M), and if you have a map G --> H which takes N inside of M, then you do get an induced map G/N --> H/M

for an example which isn't natural, we know that every finite-dimensional vector space is isomorphic to its dual

but if you choose isomorphisms V --> V* and W --> W*, there's no way to get a map V* --> W* from a map V --> W

[rustic crown]

in what sense is there not a way? I can just say use the map V* --> V --> W --> W*

[oblique river]

uh hrmmm let me try to phrase it differently then

so I guess that wasn't actually a good example because "taking duals" isn't even a functor from Vect to Vect

[rustic crown]

its a contravariant functor, right?

[oblique river]

yeah

[rustic crown]

so do we only talk about natural transformations between 2 covariant or 2 contravariant functors? can we not mix them up?

[oblique river]

yeah

maybe a better way to phrase it is the following: given every map V --> W you get a map from W* to V*. It's not possible to choose isomorphisms V --> V* and W --> W* which make all of these diagrams commute for all maps V --> W simultaneously

(I think I kind of had my quantifiers backward earlier)

[amber charm]

anyone able to help me with a query?

[oblique river]

[amber charm]

Q (Group Theory): I have been asked to find permutations of two elements a and b, both of which has order 2 such that the order of ab is 5, this to me seems like it is impossible to be able to do, i do not want a solution but want to make sure the question is right before continuing.
i guess what i am asking is if this question makes sense or that it cannot be done

[oblique river]

yes it is possible. For any positive integers p, q, and r, you can find a group G and two elements a, b in G where the order of a is p, the order of b is q, and the order of ab is r.

[amber charm]

and p, q and r can be of any value? hmm never knew that

[oblique river]

yeah it's kind of an obscure theorem

[amber charm]

so could you have two permutations c and d of order 1 and then have cd have order 100 or something?

or is there a limit to what you are allowed?

[oblique river]

actually sorry what I wrote isn't correct, you need p, q, r > 1. otherwise it's false haha.

no you're right -- I was being sloppy

if c has order 1, then it must be the identity, so c*d = d

so the order of cd is just equal to the order of d

my bad

but it is true for all p, q, r > 1

so for example you can find a group G and two elements of order 2 whose product has order 12361234

or two elements of orders 23515633 and 235092362 whose product has order 19

[amber charm]

wow! this is nuts haha

[oblique river]

yeah :)

[amber charm]

best go try figure it out now lol

[oblique river]

don't try to prove that general statement haha

[solemn rain]

can u prove this

[amber charm]

i would never haha

[carmine fossil]

Is that super hard to prove?

[solemn rain]

it looks like

[oblique river]

I remember that you do it in some matrix ring

like SL_2(F_q) or something

[golden pasture]

woaaa

[oblique river]

I don't remember the exact construction

[solemn rain]

col

[oblique river]

found it https://www.jmilne.org/math/CourseNotes/GT.pdf theorem 1.64

it's in the group SL_2(F_q)/{I, -I}

(at least, that's how this construction works)

it actually looks relatively straightforward

at least if you're familiar with some basic group theory and matrix algebra

[solemn rain]

ty

[chilly ocean]

@solemn rain fun little problem: show if |G|=p^i q^j for p,q prime then G is solvable

[solemn rain]

i forgot about these things

but ty for the problem

looks hard

haha

@chilly ocean

[chilly ocean]

jk don't try it

it's super hard

in general case

[worthy haven]

hey this is a problem in my galois theory class!

[chilly ocean]

lol no way

my prof told us this generalization is too hard and it uses a lot of rep theory

[worthy haven]

sorry full screenshot

[chilly ocean]

damn gl then

[unique juniper]

Let N be a normal subgroup of G, Let H be set of elements of G such that $hn = nh \forall n \in N$. Show H is a normal subgroup of G

[cloud walrus]

Yes

[unique juniper]

can someone give me a hint on this?

what ive tried is

$(ghg^{-1})(gNg^{-1})(gh^{-1}g^{-1}) = N$

[cloud walrus]

Yes

[chilly ocean]

maybe you can use different definition of normal group?

for any gin G for any h in H g^-1 h g in H

this should follow, maybe you need to multiply by 1 somewhere

[unique juniper]

ok

[unique juniper]

this is what we want to end up with tho

hm

[chilly ocean]

oh this is the channel for vector spaces

so... idk how to translate it but, it sais something like: Considering M 2x2 the vector space made by 2x2 matrixes. Guess its dimension and a base

ye i delete from there

[maiden ocean]

#linear-algebra is where that question belongs

lol

[chilly ocean]

no

this channel is for vector spaces

[maiden ocean]

uh

no...?

[chilly ocean]

and this is regarding to a vector spaces

homological algebra, commutative algebra```

yes...?

lmao

[maiden ocean]

this level of question does not belong here

[chilly ocean]

like, idc honestly where to ask, but i read vector sapces here and thats what i am studying right now, so i posted here

[maiden ocean]

in #linear-algebra

yea it should

its a bit confusing

@rose axle can you remove the "vector spaces" from the channel description here

its a lil confusing so it might be better to leave it out

[chilly ocean]

i move to linear then

[maiden ocean]

yea you should do that, more ppl will see ur question

[chilly ocean]

why does it say vector spaces twice?

[maiden ocean]

xd

by that metric it says it more than twice

[chilly ocean]

I don't exactly understand why it is necessary to be so anal about correct channel

[maiden ocean]

i mean if nothing else there are far more people ready and willing to answer questions like that in that channel than here

it gets a bit annoying for everyone else if questions like that are consistently asked here anyway

[unique juniper]

[past temple]

i'm currently trying to figure out this problem

consider the rational function field over one variable F_p(x)

and consider the F_p-automorphism phi: F_p(x) -> F_p(x)

such that phi(x) = x + 1

i have to show that phi has finite order in the Galois group Gal(F_p(x)/F_p)

which I was able to do because if you just compose phi on itself p times

then u get x + p = x giving u the identity function

and now i have to find a u in F_p(x)

such that F_p(u) is equal to the fixed field of phi

any help would be appreciated!

[mild laurel]

is your automorphism here phi(f(x)) = f(x + 1)? Where f is an element of F_p(x)?

okay, I think that for odd primes, the polynomial f(x) = x^(2p) - x^p is fixed by this automorphism

[oblique river]

you can find an example of degree p

hint: if a is a root of f(x), then so is a + 1 because f(a+1) = f(a) = 0. in particular, if your polynomial has a single root in F_p, then everything in F_p must be a root. Can you write down such a polynomial and verify that f(x+1) = f(x)?

[past temple]

okay yeah

i made a lot of progress

using the freshman's dream

i found that the polynomial

u = x^p - x

is fixed by the automorphism phi

so cosidering F_p(u)

we know that this automorphisms of this field are uniquely determined

by their action on u

so then we have that

F_p(u) is a subset of the fixed field of F_p(x)

from there

it suffices to show that

[F_p(x):F_p(u)] = [F_p(x): Fixed field of phi]

i used some results in my book

where

[F_p(x):F_p(u)] = max{deg (numerator of u), deg (demoninator of u)} = p

and a result about

[K: Fixed field of H] = |H|

to show that their dimensions are both p, so they're equal

is this reasoning correct?

[oblique river]

sounds good! @past temple

[south temple]

Hey guys, I was wondering if my proof of the claim that ``The only proper normal subgroups of $S_n$ is $A_n$ for $n\neq 4$" is correct?

[cloud walrus]

vov&sons

[south temple]

just wanted to make sure my contradiction for H actually makes sense

[mild laurel]

No this doesn't actually make sense

Simple doesn't mean no normal subgroups, it means no trivial normal subgroups

So it's possible that H intersect A_n is the identity subgroup

[stray rivet]

What was your thinking behind "H may not contain any elements of A_n?"

[old hollow]

I see some authors use capital $X$’s for in determinates and some use lowercase $x$’s, like in $\mathbb{Q}[X]$ or $$Y^2 = X^3 + AX + B$$

[cloud walrus]

abs_0

[old hollow]

Does it mean anything to use capital vs lowercase or is this just a convention

[south temple]

this is the definition in my book

also how can any group have no trivial normal subgroups, if every subgroups has a trivial subgroup, namely the identity subgroup

[sturdy marsh]

no non-trivial normal subgroups

[south temple]

then I don't see why my argument fails though

which part is the issue exactly?

ah right

the intersection is never empty because of the identity

ok but

if it does contain a non-identity element, then $H\cap A_n$ is a nontrivial normal subgroup of $A_n$

[cloud walrus]

vov&sons

[south temple]

contradicting that A_n is simple

right?

well I want to show that this element cannot exist

since its existence leads to a contradiction

[south temple]

yeah but I believe just rewording my argument from earlier accordingly gets the job done

[past temple]

can anyone help with this problem

consider the rational function field k(x)

and consider sigma, tau to be automorpisms of k(x)

where sigma(x) = 1/x

tau(x) = 1-x

how do i find the fixed field of {tau,sigma}

basically ive gotten to the point where

im trying to prove that

|<tau,sigma>| = 6

and if i find some u such that

[k(x):k(u)] = 6

then k(u) is the fixed field

but idk how to show that|<tau,sigma>| = 6 or what u is

i know so far that |tau| = 2, |sigma| = 2, |tau*sigma| = 3

[gritty adder]

This doesn't have anything to do with the specific group. If you know that a group is generated by x and y, and they have relatively prime orders p and q, then the order of the group is pq

[oblique river]

@gritty adder that is absolutely false. The group SL_2(Z) is generated by an element of order 2 and an element of order 3 but it's an infinite group

[gritty adder]

oh i'm being stupid my bad

[chilly ocean]

You should have a "| dumbass" next to your name 🙂

[gritty adder]

good point

[oblique river]

@past temple if you want to show that the order is 6 you can just list the elements. I'll write t and s for tau and sigma. every element must be of the form "tsts...." or "stst...." since if you had any squares they would cancel out. that leaves you with: e, s, t, st, ts, sts, tst, stst, tsts, ststs, tstst. (You know that ststst = tststs = e.) now you just need to figure out which of those are equal. you know that ststs*t = ststst = e, so ststs = t^(-1) = t so we can get rid of that. using that kind of thinknig you should be able to narrow it down to e, s, t, st, ts, sts

now you can prove that those are all different by just writing them down as functions and observing that they're not the same

[chilly ocean]

Hah tststststs

https://c.tenor.com/bvsFM3G1oiAAAAAM/pspspsps-neature-walk.gif

[jagged dune]

How can you tell if a group operation is preserved in terms of isomorphisms?

[chilly ocean]

I am not sure what you mean. Certainly it is always true if you mean phi(x*y)=phi(x)*phi(y)

[jagged dune]

Alright I'll rephrase: I'm trying to see if two groups are isomorphic.

How do I tell if the group operation is preserved?

I see that it's bijective.

How can I tell if it is preserved

a little fuzzy

we did go over it

Gotcha.

Thank you.

[restive star]

Hey can someone double-check my work on finding the splitting field of $X^4 + 5$?

[cloud walrus]

randleS2003

[restive star]

I think it's Q(i,forth root of 5), but when I find similar problems online they all do different weird things to find it

[chilly ocean]

That looks right to me, yeah

[restive star]

Is the right way to do it just to notice that the roots are the forth roots of negative 5?

cause there is some really weird stuff some people do for similar problems that I don't quite get the point of

[chilly ocean]

Oh I messed it up (what you wrote would be correct for x^4-5=0)

[chilly ocean]

Yeah, and what do you get for them?

[restive star]

sorry this is the part I suck at lol. I'm assuming I use the primitive 4th roots of unity?

[chilly ocean]

Something like that (maybe 8th roots of unity)

[restive star]

how would I tell which to use?

[chilly ocean]

I imagine the way you are thinking about it is that you let y=x/fourth root of 5, then y^4=-1 right? But then y is an 8th root of unity

I think

[restive star]

x divided by the forth root of 5?

Im sorry Im not following what u are asking

[chilly ocean]

Well

How did you come to Q(i, fourth root of 5)?

[restive star]

I'll be honest, I was stupid and forgot about the 4th roots of unity and just eyeballed it lol

[fathom delta]

how did you get 8th root of unity? the solutions to y^4 = -1 are the 4th roots of unity

[carmine fossil]

*8th roots

[chilly ocean]

fourth root of unity means y^4=1 right?

[fathom delta]

oh wait, it's not quite 4th root of unity

[carmine fossil]

Yes

[chilly ocean]

So but eg you can find at least one root, x^2 = i sqrt(5), then x = sqrt(i) * 4rt(5)

I guess we are talking about 8th roots of unity, but only 4 of them

[restive star]

why only 4 and not all?

[chilly ocean]

Because y^4=-1 has 4 solutions

[fathom delta]

^fundamental theorem of algebra

[chilly ocean]

So these are the 8th roots that are not 4th roots

[restive star]

ah I see

So do we find the 8th roots and compare them to the 4th roots?

[fathom delta]

that's not necessary

it would be just sqrt(i)*5^1/4

for the roots

[restive star]

isnt that just 1 root though?

[fathom delta]

yeah. so you have to go through all sign combinations. x^2 = (plus or minus)i sqrt(5), so you get four possibilities for x, x=sqrt(i)*5^1/4, x=- sqrt(i) * 5^1/4, x=-\sqrt(-i) 5^1/4. x=\sqrt(-i) 5^1/4

[chilly ocean]

Ah, I guess you can say it is about 4th roots of unity if you take one solution to the equation and rotate them by the primitive 4th root

[fathom delta]

the terminology is confusing, but it's just the solutions to x^4 = -1. it's the solutions to the 8th cyclotomic polynomial whatever they are called

[restive star]

sorry for the confusion, but I'm still not completely sure what to do. So I find the roots, then i just match a field to it?

[fathom delta]

yes, mostly

[chilly ocean]

You could literally write it like Q(r1, r2, r3, r4), but I think this is bad style

[restive star]

so then it'll end up being something like Q(sqrt(2),i,4throot(5))?

[fathom delta]

but you have to find the minimum number of generators to generate all the algebraic combinations

[chilly ocean]

But this field is too big

[fathom delta]

that's right I think

[restive star]

wdym too big?

[chilly ocean]

I don't think sqrt2 should be in there, for instance

The splitting field is contained in that field, but is not equal

[restive star]

Wont I need sqrt2 tho?

[fathom delta]

sqrt(2) has to be in there. otherwise, how do you get e^{pi * i/4}

[chilly ocean]

You need it, but it should be mixed in with the 4th root of 5 or something

[fathom delta]

I don't think so

[chilly ocean]

Ok here is another argument, what is the degree of the extension over Q of that field? It should divide 4! Right?

[restive star]

I think so

[chilly ocean]

But it is degree 16

[restive star]

huh

RIP lol

[chilly ocean]

At least, you can write the splitting field as Q(i, sqrt(i)*4rt(5))

I think there are standard ways to write the generators, but I have not done any field theory in too long to formalize what it should be, probably something like each generator is "full degree" or something

[restive star]

but does that have sqrt(2)?

[chilly ocean]

It "has" it in sqrt(i)

[fathom delta]

that works. but I wonder if you can write in terms of integer roots

[restive star]

The final goal is to find the degree of x^4+5 over Q, so I do def need int roots haha

[chilly ocean]

I'm not sure I understand this thing about integer roots

[restive star]

I just thought it'd be easier

sorry Im V new to this lol

[chilly ocean]

But you can show that it is degree 8 by showing Q subset Q(sqrt(i)*4rt(5)) subset Q(i, sqrt(i)*4rt(5)) has strict inclusions

[restive star]

wait why does this show it?

Is it since Q(sqrt(i),sqrt(i)4rt(5)) has degree 16) and the degree of Q(sqrt(i)*4rt(5)) must then be 8?

[chilly ocean]

Q(sqrt(i)*4rt(5)) should be degree 4 over Q

[restive star]

oooooh and then q(i) has degree 2 over q

[chilly ocean]

Yeah

[restive star]

sorry I'm dumb lol

working through this stuff

[chilly ocean]

Also you should carefully check whatever I have said, I have a disclaimer in my username that I may say incorrect things

[thorn delta]

any ideas on how to show that for an abelian group $A$, $$A \otimes_\bZ (\bZ/m\bZ) \cong A/mA?$$

[cloud walrus]

kxrider

[thorn delta]

there is an obvious balanced map $(a, \bar k) \mapsto k\bar a \in A/mA$, but is this on the right track?

[cloud walrus]

kxrider

[thorn delta]

because, if so, I don't see why the map it induces on the tensor product is injective

ive also been trying to think of something fancier, like tensoring two exact sequences, but nothing's panned out so far

[sturdy marsh]

look at $0 \rightarrow m\mathbb{Z} \rightarrow \mathbb{Z} \rightarrow \mathbb{Z}/m\mathbb{Z} \rightarrow 0$

[cloud walrus]

Brofibration

[sturdy marsh]

tensor with A

[thorn delta]

🤦‍♂️ thanks

[vestal snow]

Let R be a rng (ring without 1). Then does R[x] contain x?

I can see both answers to be true

Depending on the definition of R[x].

If we define it as "adjoin x and ringify" then it must contain x

If we define it as the polynomials with coefficients in R, then it would not

[chilly ocean]

rngs show up in things like analysis, and polynomial rings show up in algebra, so this should never happen, right? 🙂

[vestal snow]

algebraic analysis

or analytic algebra

Lmao algebraic analysis is actually a thing

[golden pasture]

you would probably define R[x] as the smallest rng containing R and x

and in that case no

[sturdy marsh]

no

a definition of poly ring is the ring of sequences of elements of R with finite support

[vestal snow]

Wait it should in that case

[sturdy marsh]

with multiplication defined appropriately

[latent anvil]

sure but we could also define it by a kind of free construction

right?

like it's true that if you define R[x] in your fashion it will not contain x

[latent anvil]

which is essentially this statement repackaged

[vestal snow]

I can see why most algebraists require 1 in rings

[sturdy marsh]

do you mean a free R-algebra on 1 generator

[vestal snow]

I think Shamrock's definition is "better"

[sturdy marsh]

I'm pretty sure people always ask for the ring to be unital when talking about algebras

[latent anvil]

I meant you take words in the language {elements of R union x} with formal negatives

and then take a big set-quotient

which is to say, "formally adjoin a new element x"

[golden pasture]

"smallest rng"

the rng that doesnt contain 1 is smaller than the one that contains one

[vestal snow]

If we define R[x] to be the smallest rng containing R and x

then x is in R[x]

[next obsidian]

How can you define something to be the smallest rng containing something without even embedding it into something lol

[latent anvil]

I didn't want to say it

[golden pasture]

x is in R[x] but 1 may not be in it

hm but good point tho

maybe it is better to define it as

{sum rx^n|r\in R}

ah right

yea i get what you mean we kinda need to define an action of Z on R

[summer geyser]

https://math.stackexchange.com/questions/62762/the-degree-of-a-splitting-field-of-a-polynomial

For the first part of the answer on this, they say that f/p is in L[x] however isn't it also in F[x]? Or are they just saying L[x] so that the tower law argument follows?

Then in the second part they say f/(x-a) is in L[x], but I can't really see how we know that the resulting polynomial will be in L[x].

[chilly ocean]

f and x-alpha are in L[x], so f/(x-alpha) is just a normal polynomial division that divides exactly, so it should be in L[x], right?

[summer geyser]

Is there no concern for if we divide out x-alpha that the coefficients of the result will not be in L?

[chilly ocean]

I'm not sure how this can happen, L is a field

I don't remember all the details, but division works in whatever kind of algebraic environment, so f(x)=q(x)(x-alpha)+r(x), and r is actually 0

[summer geyser]

Oh right the division algorithm states that q(x) is in L[x]

Then for the first part where f is reducible, they state that K is a splitting field for f/p over L, but how do we know that there isn't a smaller field for which f/p splits fully?

[chilly ocean]

If f/p splits in a smaller field M with L<M<K, then M contains L so f splits in M right? But this is a contradiction

[summer geyser]

Ohh okay I see, thanks for the help!

[cloud walrus]

mirzathecutiepie

[viscid pewter]

h in is G and x is in G, so you can use the structure of G?

i don't think i understand the problem

[mild laurel]

writing h_1x = h_2x doesn't even make sense if you're only considering x as an element of O, since you can't multiply an element of G with an element of O if you're considering O as just a set

[viscid pewter]

yeah, you're using the group operation

[cloud walrus]

mirzathecutiepie

[viscid pewter]

we just use the group structure

that's fine, right?

but the group action is defined by using the group operation

left multiplication

i think by using the group operation on the set we have all the group structure

all the group properties, including inverses

i don't see why we can't do it

why can't we just consider them as elements in G

the effect of the group action is just elements of G being operated on by the operation of G

too rigorous for me, lol, i've irretrievably conflated sets and groups already i think

it works

so far

[mild laurel]

I think you already have to implicitly use this bijection to even define the action

h_1 acting on x is defined by h_1x where we use the group operation of G and so will give us an element of G. To get to the set underlying G again, you need to use this bijection

[viscid pewter]

??

elements of the underlying set are just elements of the group

[cloud walrus]

mirzathecutiepie

[viscid pewter]

ok but like

we know it's a group, that's a given

so elements of the underlying set are just elements of the group

it's equivalent

so i still don't get the problem

oh

you want a bijection that isn't just a bijection.

soooo

do you want like. structure on the set? in a group-like way? that's preserved by the bijection? that's an isomorphism

what do you mean by 'isn't just a bijection between the sets'

...

what.

??

this entire thing just seems entirely incoherent to me

[stiff tundra]

the shorthand G is used when talking about a group so we don't have say (G,*) every time the group is referred to

[viscid pewter]

i mean generally there's no real reason to distinguish the set and the group

i think

[oblique river]

I'm not sure if this is what you're trying to say, but if you have any set X and any group G, and a bijection f from X to G, then you can use that bijection to turn X into a group, too

even if X didn't have any group structure

for x, y in X, define x*y = f^(-1)(f(x)*f(y))

basically use the bijection to push x and y to G where we know how to multiply them since G is a group, and then use the bijection to pull the result back to X

I see -- I have a meeting now so I can't chat more sorry :(

[viscid pewter]

they're both in the group G

that's it

[shut halo]

If I have the equation: $x^2 - 23xz + 16z^2 = 0$ what is the factorisation over $F_11$ (field of integers mod 11). I found the answer to be $(x+2z)(x+8z)=0$ but my notes mention that it is $(x-3z)(x-9z)=0$.

[cloud walrus]

snypehype

[mild laurel]

Those are the same since

2 = -9 and 8 = -3 (mod 11)

[shut halo]

oh lol I didn't even notice

thanks

too much coffee is bad

[cloud walrus]

Mecejide

[glad juniper]

It took me more work than I'd hoped, but I solved this in the end. Maybe I should suggest this as a challenge problem?

[chilly ocean]

Oh nice, what was the solution?

[brittle hemlock]

anyone know how to solve this/what this means?

[thorn delta]

yes and yes

[brittle hemlock]

alright, what's C?

I completely forgot

[thorn delta]

the complex numbers

the complex numbers are an abelian group under addition, and the reals are a subgroup of the complex numbers, and their quotient is isomorphic to R

[brittle hemlock]

my bad I went afk I didn't see this

ok so we're trying to show that their quotient is isomorphic to R

so do we have to find a mapping between Real over Complex Numbers to Real Numbers?

I'm kinda confused what the quotient group really means

if it were just normal division, then wouldn't a Real over a complex be a complex and therefore not an isomorphism to Real numbers?

I'm kinda confused

[thorn delta]

do you know about the first isomorphism theorem?

[brittle hemlock]

nah I don't

[thorn delta]

alright, so, it might help to make sense of G/N first. We know that two elements of G/N are congruent if their difference is a real number, i.e. their imaginary part is the same.

[brittle hemlock]

alright that makes sense

since we're trying to set it equal to an isomorphism to only the Real numbers right?

I'm kinda confused abt what G/N really denotes to begin with

really new to this notation stuff

[thorn delta]

G/N is the set of "cosets" of N by elements of G. A coset is denoted by gN = {gn : n \in N} for each g in G. Is that familiar?

[brittle hemlock]

my bad I keep going afk you can leave

don't wanna take up your time

but I see ok that makes sense

yeah I learned abt cosets earlier

ok so that makes sense, it seems like more of a multiplicative group than a quotient group lmao

[thorn delta]

what do you mean "it?"

[brittle hemlock]

G/N

since you're taking every g and applying it to N (to the left, so left cosets)

[thorn delta]

the quotient group G/N inherits its operation from G. When N is normal, the operation (aN)*(bN) := (ab)N is well defined, and makes G/N = {gN : g \in G} into a group.

in the case of C/R, it would make sense to write things like cosets as z + R for z \in C for example.

[brittle hemlock]

a and b are in G right

and yeah ok that makes sense

[thorn delta]

yup

[brittle hemlock]

but a complex + a real number would be a complex number so how is this gonna be isomorphic to the set of real numbers?

unless the complex part is 0

or am I misunderstanding the question

[thorn delta]

right, remember that z + R = w + R (think of z+R and w+R as elements of C/R), if z-w \in R.

consequently, their imaginary part is the same, so you can think of C/R as the set of "horizontal lines" in the complex plane (intuitively, we can identify this with R).

[brittle hemlock]

ohh ok I see

so z-w is always in R if their imaginary part is the same obviously, so z + R = w + R

so how do we concretely show that this is isomorphic to R? Say that for every element z there exists another element w such that the imaginary part of w is the same as z?

[thorn delta]

First, use your intuition to try to write down a map. Sure, there are many elements z,w which have the same imaginary part. The map you define should map such elements to the same thing in R. This is what it means for the map to be "well defined"

[brittle hemlock]

uhh can I just make any map?

like a map of only the real part?

[thorn delta]

not quite sure what you mean by that. What do you have in mind?

[brittle hemlock]

uhh idk can I just define a map on z + R and w + R such that we get (z-w) in R?

this would be dependent on having two elements with the same imaginary part

I'm confused on what an isomorphism is really asking for

like do we have to have 2 elements that satisfy a certain property?

and ignore what I said earlier, what I had in mind was just dropping the imaginary part in the map

[thorn delta]

yea, we'll get to the isomorphism part. The first step is to write down a map. The most natural one is going to be the correct one (that defines a isomorphism).

and ignore what I said earlier, what I had in mind was just dropping the imaginary part in the map
👀 that sounds possibly correct depending on what you mean by dropping the imaginary part.

[brittle hemlock]

I mean we can just map a number a + bi to a no?

by subtracting by bi

idk

[thorn delta]

ok, so suppose that we defined $f(a + bi + \bR) = a.$ We have that $a + bi + \bR = c + bi + \bR$ while $f(a+bi + \bR) = a \neq c = f(c+bi+\bR)$ so $f$ is not well defined here.

[cloud walrus]

kxrider

[thorn delta]

but you're close. right idea, wrong term

[brittle hemlock]

oh, so we just do the same thing except with the real term?

I'm so confused idk what's going on

could you just show me an example map so I can figure out why it would work?

[thorn delta]

are you familiar with the integers mod n?

[brittle hemlock]

wdym

like how n+1 mod n = 1?

and e.t.c, we have multiples of the sequences 0..n-1 in a row?

[thorn delta]

yea. So Z/nZ are the integers mod n as a quotient group. a = b mod n means a - b = kn \in nZ for some k \in Z.
Anyway, you can define a map f : Z/4Z to Z/2Z such that f(k ) = k. For this map to be well defined means that if k = n mod 4 then k = n mod 2.

Consider the map f : Z/3Z to Z/2Z defined by f(k) = k. I claim that this map is not well defined. in other words, there is k, n such that k = n mod 3 while k != n mod 2.

[brittle hemlock]

isn't the Z/4Z to Z/2Z map also not well-defined?

since k can be 3 for mod 4 when it's 1 for mod 2

[thorn delta]

that is fine. what you don't want is for f to "split up" the coset of 3 in 4Z. For example, its okay that 3 = 1 mod 2 as long as 7 = 1 mod 2 and 11 = 1 mod 2... and 3 + 4k = 1 mod 2 for any k!

[brittle hemlock]

oh ok I see, thanks

I still don't really see how we can directly apply that to this problem

I get that we need to drop the complex part to keep it a real number, but what would a map look like

[thorn delta]

f(a+bi + R) = b is the map we want. f(a + bi + R) = a doesn't work because it is not well defined. You'll have to show why the first one is well defined. The basic intuition about congruence modulo some subgroup carries over from the digression on the integers. If you think you got the hang of it, we can move on proving that f is an isomorphism though.

[brittle hemlock]

alright, I think I got it

you're just keeping dropping the i of the complex part of the equation

to make it real

makes sense

we can move on

[thorn delta]

okay, so isomorphism here just means bijective homomorphism. To show f is a homomorphism, you need to show that f((z+R)+(w+R)) = f(z+R)+f(w+R).

[brittle hemlock]

oh ok and that's pretty easy right

(z+R) + (w+R) would be a real number plus the complex part of z + w, so after the function, we have the complex part of z + w

and f(z+R) would be the complex part of z, and f(w+R) would be the complex part of w, so we also have the complex part of z + w

[thorn delta]

yea, pretty much. Now for bijectivity, (by definition) you just have to show that f is injective and surjective. Surjectivity is probably the easier one. i.e. why does everything in R get mapped to by f from something in C/R?

[brittle hemlock]

since we can set any real number to the complex part of z and w

[thorn delta]

not sure what the point of mentioning w is, but sure. Now, for injectivity, what do you have to show?

[brittle hemlock]

I'm mentioning w since we can offset the complex part of z with that of w but yeah just z is enough

does injectivity mean 1 to 1?

idk what it refers to

[thorn delta]

yea. one to one = injective and onto = surjective

[brittle hemlock]

is it 1 to 1?

I feel like it isn't for some reason

I'm trippin

oh wait LOL

it's just bc b's are unique

I'm stupid as hell

every b is mapped to b in R

[thorn delta]

yea, i think your feeling is warranted. for example a + bi \neq c + bi as complex numbers for a \neq c while their image under f is the same, but a+bi + R = c+bi + R, and this is all we need to show for a map from C/R to R to be injective

[brittle hemlock]

great that's why we didn't do f(a + bi + R) = a

thanks man

are we done?

[thorn delta]

welp, we've shown that f is a well defined bijective homomorphism. Thats all there is to it.

[brittle hemlock]

alright, sorry for taking up so much of your time I'm just very new to this stuff

[thorn delta]

np. you'll probably find out later on that you can do a problem like this in two seconds with the first isomorphism theorem tho

[brittle hemlock]

lmao my bad

I was just pretty confused

appreciate it tho

[thorn delta]

nah its np. first iso can be something to look forward to.

[brittle hemlock]

yup for sure

[next obsidian]

Okay, coomutative algebraists, I need help with something.
If you take a proper ideal I of a Noetherian ring A, consider the multiplicative set S = 1 + I. Then I want to show that the IA_S-adic completion of A_S (the completion of A at S) is isomorphic to the I-adic completion of A.

I thought I had it, because the I-adic and IA_S-adic topologies on A_S are the same, then I used a theorem which says that the I-adic completion of a finite module M is just M (x)_A A^, so when M = A_S you look at
A_S (x)_A A^. If you view the latter as a localization of A^ it doesn't do anything so you should get that (A_S)^ = A^... the only issue is that A_S doesn't have to be a finite module
SAD!

If no one can help me I will be sad :(

[wraith obsidian]

refresh my memory here. The valuation at an ideal I was something like a dimension measure (in what I^k is x contained for the first time?)

I never do anything with valuations, so bear with me

[next obsidian]

If you’re a DVR yeah

Since then k makes sense as an integer

Although you embed the DVR into the fraction field, and the valuation is defined on all of the field

So if v(x) is negative it means that x^-1 is in I^k

For a valuation ring it’s a bit less simple but it’s something like that

[wraith obsidian]

okay, so we kinda subtract how much closer we needed to extend our ring towards the fration field to be able to talk about the valuation

[next obsidian]

Idk what that means

[wraith obsidian]

it's the kind of sentence that makes sense if you have the right picture in mind, just that my picture may be very wrong here

[next obsidian]

I think of it like

[wraith obsidian]

I am the opposite of an authority figure when it comes to this stuff lol

[next obsidian]

Imagine your fraction field is like a big circle

Then inside that a good bit you have a circle for your DVR

Then you have more and more concentric circles

Each one representing m,m^2,...

[wraith obsidian]

Ah, yeah, and the growing circles are the fractional ideals

[next obsidian]

Right basically

The valuation tells you which annulus you’re in

Higher is further inside

[wraith obsidian]

huh, that's a good way of wording things

in a noetherian ring, we can have infinitely long descending chains of ideals, so powers of m can get smaller and smaller, right?

[next obsidian]

Yes

The intersection of them all will be empty if it’s an integral domain tho but for completely different reasons

[wraith obsidian]

Well so will the intersection of all (-1/n,1/n), that doesn't mean that sequence stagnates

Stupid question. Is 1+I the same as the complement of I?

they need to be disjoint if I is proper, because otherwise 1 in I, right?

ah, but it need not be all of the complement, counterexample I := (3) sub ℤ , 1 + I does not contain e.g. 2

Do we have that A/I is the same as A_S/IA_S?

My visualization for the localization is the decreasing sequence AS, IAS, I²AS, I³AS, etc. , while we previously had A, IA, I²A, … (written verbosely)

So elements in the first completion should be (compatible) sequences in (AS/IAS, AS/I²AS, …) whereas elements in the second completion are (compatible) sequences in (A/I, A/I², …). Did I understand that correctly?

Hold on, that was assuming that (I AS)² = I² AS

Okay, atiyah tells me that extension of ideals (wrt to a ring extension) commutes with multiplication of ideals, and since IAS ist just extension of I along the ring extension A≤ AS we have that (IAS )² = I²AS

sorry for uncoordinated spitballing, this will have helped myself more than you

[median tree]

I'm not sure this is where this belongs, but I'm not clear on how this

is used to conclude

these are quaternions, obviously

but I'm assuming that the ij=k and ji=-k rules were derived from the above equation

So that clearly indicates that right and left multiplication are different.

[scarlet estuary]

take ijk = -1

right-multiply by k

ijk^2 = -k, hence ij = k

ji = -k is a couple more steps, but the same idea

might be good to try it on your own.

[median tree]

Okay, thanks for clarifying!

[vestal snow]

Can someone explain the notation for me? It seems like the 0th cohomology group of something related to the curve X and the first order holomorphic differentials over it

[latent anvil]

could it be sheaf cohomology?

I don't know this stuff really, but that's what it looks like to me

[vestal snow]

Might be

[latent anvil]

Although that would just be Ω^1(X) wouldn't it...

[vestal snow]

My advisor warned me that this has some AG in it

Wait

Whats the difference between Ω^1(X) and Ω^1_{X/k}?

[latent anvil]

no idea

oh wait

Is Ω^1_{X/k} a group or a sheaf?

if it's a group this could just be singular cohomology with coefficients in that group

would be weird though

might be easier to tell if you post context

[latent anvil]

(although I'm the blind leading the blind rn)

@prisma ibex seems to be online and can say things with more confidence than me

[vestal snow]

Shamrock what does Ω^1(X) mean to you?

[rich ravine]

@next obsidian The usual approach would be to look at what happens to the exact sequence $0 \to A/I^n[S] \to A/I^n \to A_S/I^nA_S \to 0$. Now an element of $S$ is invertible in $A/I^n$ (take the formal power series inverse) so there is no $S$-torsion in $A/I^n$. Therefore, passing to the projective limit in the compatible isomorphisms $A/I^n\simeq A_S/I^n A_S$ gives the result

[cloud walrus]

Othenor

[rich ravine]

I'm not sure you even have to have any Noetherian hypothesis

[next obsidian]

What is A/I^n[S] here?

Clearly it’s got to be the kernel of the latter map, but I’m not familiar with the notation

[chilly ocean]

Wait shamrock why did you go to UW if you are from california? Why not UC?

[next obsidian]

He’s not. He’s from Washington, but his family moved to CA so he went with

[chilly ocean]

Oh

[rich ravine]

The kernel of the projection, and it's also the S-torsion: we have $A_S/I^n A_S=S^{-1}A/I^n$ and the kernel of $M\to S^{-1}M$ is always the S-torsion

[next obsidian]

He was planning to come back to Washington for spring quarter I think but then covid so why leave

[cloud walrus]

Othenor

[next obsidian]

Ah right

I’ve done this before. Is [S] a common notation for S-torsion?

[rich ravine]

I don't think so lol

[next obsidian]

Oh okay haha

But this makes sense

[rich ravine]

Wait it's not surjective

[next obsidian]

This is pretty similar to what I ended up doing when I forgot that M^ = M (x)_A A_S was only for finite M.

Oh no

This might be a good start at least. I believe there’s a pretty natural map A^ -> A_S^ (this comes from the maps A/I^n -> A_S/I^nA_S)

I wasn’t sure how to argue that it was surjective however

Err, I suppose unless it’s surjevtive on the lowest level of A/I -> A_S/IA_S there isn’t a hope of it being surjevtive

But maybe after you quotient it ends up being surjevtive magically

[latent anvil]

to me it means the R-vector space of covector fields on a smooth manifold

because I am not an algebraic geometer

[vestal snow]

Ah right

[rich ravine]

Yes it seems like it's surjective actually

[latent anvil]

sorry I can't help more 😛

[next obsidian]

Do not ask me what it means... because I don’t know differentials

[vestal snow]

In the thing I'm doing my research in, I think it is used for differentials

[next obsidian]

😭

[rich ravine]

An inverse image of x/s will be x(1-y+y^2-...) where s=1+y

[next obsidian]

It is

[vestal snow]

That makes two of us

[latent anvil]

so i am lying a little

covector field = differential form is analogous to a differential in the algebraic sense

[rich ravine]

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

[latent anvil]

so I am thinking of the same thing

but in a different model of geometry

[next obsidian]

Are you able to then cut this off at some point because you’re modding out by I^n?

[rich ravine]

Yep

[prisma ibex]

Wait has someone answered the question already

[rich ravine]

If you guys are doing general AG then I'm pretty sure it's Zariski cohomology of the cotangent sheaf

[prisma ibex]

Sorry I was getting ready for bed

[next obsidian]

Are projective limits actually exact though?

[vestal snow]

Not really

[rich ravine]

We are taking projective limit of isomorphisms

[next obsidian]

Ah right

[prisma ibex]

So if X is over K this Omega^1_X/k is the cotangent sheaf of X over k

[vestal snow]

nGroupoid, we could move to the ANT channel sn=ince Chmonkey is asking his question here

[prisma ibex]

Sure

[rich ravine]

So you just use that projective limit is a functor

[next obsidian]

Right

In general does it have any sort of exactness property?

One-sided exactness?

[rich ravine]

Uh I think it's right exact, and you can derive on the left

[next obsidian]

Oh right

That’s what lim1 is for

[rich ravine]

In abelian groups this sort of gives what we call "Mittag Leffler condition"

[next obsidian]

Right

This seems familiar

[rich ravine]

There will be more lim^i if you're not in Ab though

[next obsidian]

I see

Thanks!

This problem was giving me a real headache

😅

[vestal snow]

Chmonkey what schools did you apply to?

[next obsidian]

Just Columbia

Since I wasn’t thinking of graduating this year anyway

And it was just “oh they reopened apps? Why not”

[vestal snow]

Oh right, you're a junior

[next obsidian]

Indeed

[latent anvil]

hnng solved my analysis problem but the solution feels sus

amogus moment

[next obsidian]

xD

Solution claimed to be in electrical but I saw them in cams by medbay

xDDDDDDDD

[latent anvil]

xDDD

[next obsidian]

xDDDDDD

Congrats on not dropping analysis sham

[latent anvil]

yes

[next obsidian]

You will have slayed the beast and can do fun fun comm alg with Sándor next quarter

[latent anvil]

i have 1 more problem

and then the final

and them im out

[next obsidian]

Forever

[latent anvil]

well

for a year and a half?

[next obsidian]

At UW

;)

[latent anvil]

unless I do my thesis on analysis stuff

in which case I learn it on my own

[next obsidian]

Lol

Somehow I find it hard to believe that’ll happen

considering the 16 other things you’re interested in

[latent anvil]

atiyah singer would require me to learn a whole lot of analysis

and ive been saying it's my top choice for a while

[next obsidian]

Hmmm

[delicate hawk]

is what i did decent?

it seems too ez for an abs alg class

[cyan marten]

I didn't check the calculation, but that's how I'd do it

[cyan marten]

Because it's purely computational.

Boolean rings are Bézout, i.e., finitely generated ideals are principal. What's the significance of this fact?

(a, b) = (a + b + ab)

[shut halo]

Can someone explain why the ideal generated by 5 is not prime is the ring of Gaussian integers? I can see that 5 = (2+i)(2-i)

[carmine fossil]

5=(2+i)(2-i) . Now note that neither (2+i) nor (2-i) is in <5>,but (2+i)(2-i) is, implying <5> is not a prime ideal

If I is prime, ab is in I implies a is in I or b is in I

Take the contrapositive of that

[shut halo]

I see thanks. Also for Gaussian integers the definitions is {a + ib | a,b integers}, here we allow a or b = 0 right?

[carmine fossil]

Yes

[delicate hawk]

I and J are comaximal

i alrdy proved that I^{n} + J^{m} is in R (the ring)

but how do you go the other way

proving R is in I^{n} + J^{m}, thus proving theyre co maximal?

[carmine fossil]

Show 1 is in I^n+J^m

[delicate hawk]

yeah, and that should complete the proof since we know that I^n + J^m is an ideal of R, so it only needs 1 to be R itself

lemme

we have I^n \in I and J^m \in J, I^n + J^m \in I + J

1 \in I + J

[next obsidian]

For things like this

There’s usually an easier way than to directly do it via a computation like this

Suppose that I^n + J^m were a proper ideal

It must be contained in some maximal / prime ideal then right?

Can you think about why this leads to a contradiction?

Hint: if x^n is in a prime ideal p, then x is in p

[delicate hawk]

ehh but does x^n + y^m \in p imply that x + y \in p tho?

[next obsidian]

Try taking a special value of y

And then a special value of x

Do it in 2 steps

[carmine fossil]

||x=(a)(a)(a)... a in I, y=0||

[delicate hawk]

drake can you explain what you mean?

[next obsidian]

p is an ideal

So it suffices to show that x is in p, and y is in p

Take y = 0

You can get x in p like that

Likewise for swapping x and y

I think you can probably finish the proof after this?

[carmine fossil]

Elements of I^n are r_1(a_11 a_12 ... a_1n)+r_2(a_21 a_22 a_23... a_2n) +... r_k(a_k1 a_k2... a_kn) a_ij in I,r_i in R.
Now take a_11=a_12=a_13=a_1i=a(some element of I) and r_1=1 for all i and set k=1

Doing that,you get a^n is in P for all a in I

[hot lake]

is R commutative ?

[next obsidian]

Oh right, that’s a thing

I’m assuming yes since I’ve only seen comaximal in that context but...

¯_(ツ)_/¯

[carmine fossil]

What changes if R is not commutative?

[next obsidian]

Prime ideal doesn't make sense

so you'd have to manually show that you have some way to write 1 as a sum of elements in I^n + J^m

[delicate hawk]

yeah the question is specifically general

commutative vs non

[next obsidian]

D:

[delicate hawk]

but chomsky, your idea is I^n + J^m are in p becuz I^n in p (from setting j = 0) and vice versa?

doesnt that only proof the niche case where I and J are (0)?

[next obsidian]

No

I^n < I^n + J^n < p

Since 0 is always in J

Likewise J^n < p

[delicate hawk]

oh

[steady axle]

https://math.stackexchange.com/questions/3081760/p-is-a-p-sylow-subgroup-of-g-and-h-is-a-subgroup-of-g-show-that-p-ca

[steady axle]

is there an easy example where the containment is strict

[next obsidian]

Take any subgroup with two distinct p sylow subgroups P, Q

then P\cap Q is a strict subset of both P and Q

but p-sylow subgroups of P and Q are just P or Q themselves

@steady axle

[steady axle]

oh nice, thanks!

[paper flint]

chomsky

[cyan marten]

Another way to see this is
rad(I^n + J^m) = rad(I^n) + rad(J^m) = rad(I) + rad(J) = R, and thus I^n + J^m = R. Here rad is the radical.

[pallid wing]

help with 2.14

i already did 2.13 but no idea what 2.14 is even asking

[viscid pewter]

so delta is an operation on the power set of X

and call that long string of all those subsets deltaed together Y

then you want to prove that the elements in Y are the elements that are in an odd number of the subsets deltaed together to make Y

[pallid wing]

ohhh okay ty dude 👍

this author uses delta to denote symmetric difference btw :p

[viscid pewter]

oh. well that makes it easier

[next obsidian]

Delta is pretty common for symmetric difference

[celest brook]

i'm confused by this question. Would it just be as simple as removing the stuff after '11 divides 99 indeed' Aaaaaa

[paper flint]

#proofs-and-logic or #elementary-number-theory would be more suitable.

[celest brook]

ah, this is for my abstract algebra class, so was the reason I was confused

[next obsidian]

You're done for the case p = 11

[celest brook]

yes

[next obsidian]

Also... what the heck is the rest of the proof

[celest brook]

no clue

¯_(ツ)_/¯

[next obsidian]

You only needed one n

but the bigger issue is

you're supposed to let p be an arbitrary prime > 5

you can't just prove the case p = 11

you need to prove p = 7, 13, 17, 19,...

and you won't do this by doing cases on each prime

since there's infinitely many

[celest brook]

so it would be easier if I let n=1

[next obsidian]

no

[celest brook]

wait

no

no primes <5 divide 9

I mean >5

[next obsidian]

rivht

[celest brook]

...proving all the primes greater than 5 that are less than 100 seems like a headache

[next obsidian]

you don't

you cannot do it by cases

unless you do like all of them < 100 and > 100 or something

[celest brook]

I can't use modulus either right??

since it's abritary?

[next obsidian]

you won't ever have a proof if you try to go 1 prime at a time and find a specific n

[celest brook]

or would I do like

mod p

[next obsidian]

no, you probably do use modulus

[celest brook]

so like n=2

[next obsidian]

alternatively

[celest brook]

I'm allowed to do that

[next obsidian]

sure

but it won't work

[celest brook]

??

[next obsidian]

if you set n = 2

just take p = 13 or something

13 does not divide 99

n is going to probably depend on p

you don't need to find one n which works for all p

[celest brook]

but ti's just saying that p is a prime number greater than 5

[next obsidian]

this is impossible

[celest brook]

i'm so confused

[next obsidian]

There's no way an arbitrary prime > 5 will divide 10^n - 1 for n = 2

[celest brook]

11

[next obsidian]

but it needs to work for all primes

11 is not all primes

[celest brook]

it doesn't say ALL PRIMES

Uduihujwe

[next obsidian]

Yes it does

[celest brook]

it says a prime

[next obsidian]

when it says "let p > 5"

you are letting p be an arbitrary prime greater than 5

if p happened to be 11 then n = 2 works

[next obsidian]

Look

[celest brook]

a prime

not all primes

[next obsidian]

if you don't understand what that statement means

then that's on you

[celest brook]

ok..

[next obsidian]

this means "for all p in P, where P is the set of prime numbers, and p > 5, show that there is an n..."

This is how it works

you start with p a prime, and p > 5

now we need to find an n which satisfies p divides 10^n - 1

[celest brook]

so I would simply need to find a case where it doesn't work

like 7

[next obsidian]

no, this statement is treu

so n will always exist

but it'll change depending on what p is

if p = 11, then n = 2 works

if p = 7, a different n will work

but this isn't too bad

reduce both sides mod 5

I think

or rather, I guess reduce mod p, then use some stuff about that yeah

Okay, so p dividing 10^n - 1 is the same as saying that 10^n - 1 = 0 mod p right?

[celest brook]

ye

so 10^n=1 mod p

[next obsidian]

sure

so

10 mod p is not 0 right?

since p > 5

[celest brook]

yes

[next obsidian]

Now what's the best way to explain that this n exists

if you know that (Z/pZ)* is cyclic then you'd be done but

[celest brook]

haven't learned that yet sadly

[next obsidian]

Actually no, this is okay

All you need to know is that (Z/pZ)* is a group

do you know what that symbol means?

(Z/pZ)*

[celest brook]

no

put it in latex?

[next obsidian]

maybe you've seen it as (Z/pZ)^x

you've seen Z/nZ right?

have you seen the multiplication version of that?

[celest brook]

I've seen $\mathbb{Z}_{n}$

[cloud walrus]

Moosey

[next obsidian]

okahy

integers mod n right?

[celest brook]

yes

[next obsidian]

have you seen the multiplication version of that

[celest brook]

...I...don't think so

[next obsidian]

sadge

[celest brook]

we

we've covered cycles, groups and rings

the basics of them

and also some modulus properties and whatnot

[next obsidian]

Oh then have you seen Z_n as a ring?

[celest brook]

yes

[next obsidian]

take the group of units of Z_n

all the units of Z_n are the element coprime to n

[celest brook]

is there a simpler way to prove this

[next obsidian]

I mean you can use Bezout's theorem

that's ultimately what it is

[celest brook]

ahhh

[next obsidian]

I guess we can go back to the drawing board

ugh, this is still kind of annoying tho tbh, the easiest way to do this is to note that this is a group

Bezout's theorem gives you an n such that

n10 = 1 mod p

but now we need to turn this into a statement like 10^m = 1 mod p

If you can figure out a number theoretic way to turn the statement
n10 = 1 mod p into 10^m = 1 mod p then you could do it that way

otherwise I would just say that [10] is a unit in Z_p, and thus it has a finite order since the units of Z_p is a finite group, letting you conclude that [10]^order([10]) = 1 mod p

Sorry, I think if you just started group theory you wouldn't know what Lagrange's theorem is

[viscid pewter]

... is this just fermat's little theorem

[celest brook]

OH WAIT

YES

I WAS LITERALLY LOOKING THAT UP

hahaha

this is literally what I want

[viscid pewter]

like. with FLT it's a two-line proof

idk, have you learned that yet?

[prisma ibex]

FLT also originally had a two line proof (that fit in the margins)

[celest brook]

yes

we skimmed over it, so that's why I forgot

haha

[prisma ibex]

(the joke is that FLT usually stands for Fermat's last theorem)

[viscid pewter]

yeah

idk, i've usually seen FLT used for the little one lol

gods i'm tired