groups-rings-fields

[woven delta]

Lol

I know

[oblique river]

what you wrote wouldn't even be true though

since 1 = 0 in the trivial ring

[woven delta]

Yeah I figured something like that would screw it up

[oblique river]

x = 1*x = 0*x = 0 for every x in the module

[latent anvil]

abelian groups are really shitty modules over the trivial ring

[sly nexus]

true

[lean blaze]

someone help me wtih my homework

https://cdn.discordapp.com/attachments/661238560980205569/734935913624240199/unknown.png

[smoky cypress]

Not funny

[shy bluff]

Oh update on that question about the CRT from yesterday.... turns out that the prof hadn't finsihed making his lecture videos and that we covered that in the lecture video that he posted today

[latent anvil]

@lean blaze it looks like your homework would be more appropriate for the physics server. See #old-network for a link

[elder valley]

@shy bluff haha nice

[lean blaze]

no it's all goods i solved it

[elder valley]

it did seem odd that it was assigned as a homework problem

what about the localization one though

[shy bluff]

no we covered localization earlier on in the week

that was part of it

[elder valley]

oh i just meant that the question wasn't true as written

so we all speculated at what was actually meant

[shy bluff]

oh

Yea idk

proffy hasn't responded to my eamils

😦

[elder valley]

ah

[thorn delta]

Let $\sigma$ be an $m$-cycle. Show that $\sigma^i$ is an $m$-cycle iff $i$ and $m$ are relatively prime. \ \
I proved one direction like this: if $\sigma^i$ is an $m$-cycle, then $k = m$ is the smallest number such that $m | ik$ which implies that $\gcd(i, m) = 1$.

[smoky cypress]

Sorry what’s k?

[thorn delta]

ehh, just a dummy variable. It was that or say "m is the smallest number such that m | im"

[smoky cypress]

Oh I meant the k in the statement

[thorn delta]

oh oops, that should be an m

[cloud walrus]

kxrider:

[thorn delta]

anyway, I don't really know of a good way to characterize m cycles for the other direction of the proof. In the direction I wrote above, I could just use the fact that the order of an m cycle is m, but ofc that is not a sufficient condition to prove a permutation is an m cycle

[elder valley]

i think you can just use the isomorphism of <sigma> with Z/mZ

m-cycles in <sigma> correspond to generators of Z/mZ

maybe that implicitly uses what you're trying to prove

[thorn delta]

I understand what you are saying, but I am interested in the sort of lower-level interpretation. This question comes from D&F who doesn't even introduce homomorphisms for a few more sections I think

Hmm I think I got it, and the argument is kind of similar. If $\sigma = (a_1 \dots a_m)$ then for any $k$ we have $\sigma^{ik}(a_j) = a_{j+ik}$ where $k + ij$ is reduced modulo $m$ whenever it exceeds $m$. Since $\gcd(i, m) = 1$, the smallest value of $k$ for which $j + ik \equiv j \pmod m$ is $m$.

[cloud walrus]

kxrider:

[next obsidian]

TFW you’re gonna have to learn what the separable closure and perfect closure of a field is to do a Hartshorne problem 😔

On that note, are those widely used notions in like, ANT or something???

[timid hull]

II.3?

[next obsidian]

Yeah lol

I heard that problem is very very hard though

[timid hull]

Idk abt ANT but I also encountered those stuffs while doing AG and yeah II.3 is really wild lol

[next obsidian]

Yeah

I was told like half a year ago by an upperclassman in the dept that II.3 was really hard compared to II.2 and he showed me the problem which involved the perfect and separable closure and was like “yeah I’ve been stuck on this forever”

[olive mirage]

I use separable closures a good bit.

like if you're working with k(t) as your ground field, you really don't want the algebraic closure because it isn't Galois.

[next obsidian]

What's the separable closure?

Like,

can you view it as a subfield of the algebraic closure

with everything that's separable over the base field?

[olive mirage]

Yes

[next obsidian]

Okay, nice

Is there like a nice theory around it?

Like with integral extensions / integral closure

since I'm trying to judge how much I need to know to do the problem

[olive mirage]

everything about integrality over fucntion fields is messed up

if you're thinking about integrality, you should probably be using algebraic geometry instead

[next obsidian]

Like maybe it can fall out jsut by the definitions, but maybe it'll involve theory I don't know

Sure, but we have theory around integral extensions

is there theory around separable extensions too?

And I can't even guess what the perfect closure is, I feel like maybe I heard a definition of an element being perfect like 8 months ago but I don't remember lmfao

[olive mirage]

yeah I have not heard that term. Basically, I just think of hte separable closure as being the thing I mean when I intuitively want the algebraic closure

[next obsidian]

LOL

nice

[olive mirage]

I have a guess as to what the perfect closure is, but if I'm right I don't want to think about it haha

but all of this is in Dummit and Foote I think, about how to deconstruct an arbitrary field extension into a transcendental part and algebraic part, and then the algebraic part into separable and inserpable

[next obsidian]

Hmm, okay I'll look into it

I sort of figured that separable closure

probably appears in ANT

[olive mirage]

it's just "so you use Galois theory"

[next obsidian]

because to me weird parts of comm. alg that don't seem like they show up in AG I just classify as "probably show up in ANT"

[olive mirage]

I mean, it depends on what ANT means. If ANT is the study of number fields, then it literally never comes up

[next obsidian]

Idk ¯_(ツ)_/¯

ANT to me might as well be Commutative Algebra \ stuff used in AG

LMAO

[olive mirage]

but as soon as you're looking at function fields over finite fields then you can't not look at it

but to me, function fields, though you can give them an ANT interpretation, are fundamentally geometric in nature

[next obsidian]

Interesting

I still need to figure out what "geometry" even is

To me morally it's like the Euclidean geometry stuff

and then I've forcibly put in what people say

so like manifolds, or "consider different sheaves"

but I would be completely lying if I said any of it felt geometric

So AG to me is totally like

Pretend it's algebra

[olive mirage]

something like "the study of function sheaves"

[next obsidian]

and then I use topology to glue togehter

I just

don't see how that relates to what I feel geometry is

like the euclidean stuff

[olive mirage]

but that study is fundamentally motivated by Euclidean geometry, and all the intuition comes from that

[next obsidian]

I realized topology kind of fucked it for me

like obviously

a curve isn't a line

curve meaning, curvy

in like a euclidean geometry sense

but topologically who cares

so for AG stuff

I really struggle to find the middle ground where it isn't as rigid as euclidean geometry, but not as fluid as topology

[olive mirage]

to me the distinction between topology and geometry is hazy. I can draw the line in specific contexts, but in general the two are very closely connected

[next obsidian]

My teacher said geometry = topology plus structure

so like the sheaf provides the structure

which like, i guess restricts it

but he posed this

if you take a variety

[olive mirage]

I mean, you can define the sheaf purely in terms of continuous functions

[next obsidian]

that's a line, with three intersecting lines

So like

[olive mirage]

I'd be kind of inclined to say it's geometry once you can take derivatives

or talk about smooth functions

or whatever

[next obsidian]

Are these all isomorphic

So like my geometry brain would say

that if you mess up with the angles they can't be

but topologially all of these seem the same

I think in an AG context

you have like 1 degree of freedom

intuitively you can sort of like shift i

and like, rotate some stuff

to make the first two lines line up

by scaling you can get the middle lines also lined up

then the last one is just sort of free

so in some sense you get like a 1-dimensional isomorphism class

which ends up being right, I think

but like what's kosher and what isn't isn't at all clear to me

so it's hard to get an intuitive understanding of "what should be true"

The way I did this was all algebraic

There's also some like weird action of the automorphism group or something I remember him mentioning

I just feel so clueless on it overall

[olive mirage]

yeah, algebraic geometry, differential geometry, and topological manifolds are all quiet different

[next obsidian]

I've heard some people say after they did like, manifolds

that the idea of what geometry is made more sense

and a lot of examples in AG are motivated from that

like 1-dimensional locally free stuff is a "vector bundle"

or whatever

[olive mirage]

yeah, I think "manifold" is the fundamental object and everything is motivated by taht

[next obsidian]

I just feel like

I don't want to go into the like needlessly abstract world if I don't have to before it's the time to

And I feel like being so focused on algebra and not the geometry makes it easy to fall into that trap

I guess

[olive mirage]

haha well then Hartshorne is the wrong book haha

[next obsidian]

It feels less rooted

No I mean like

the Lurie vein of stuff

the derived AG stuff

It feels like if you don't see any "geometry" you just redefine geometry to mean "contravariant functor"

and go to infinity category land

and sites and stuff

Is the impression I've gotten haha

[olive mirage]

I am definitely very concrete in my thinking, and I only care about abstraction if it helps me answer a concrete question. A lot of AG people clearly just want everything to be as abstract as possible.

[next obsidian]

I think I slowly am starting to get it though

Maybe

I do draw pictures of schemes sometimes, except like it isn't a picture of a scheme, it's just two circles

I look at the sets

and draw a covering to try and guess at what might happen

but getting more familiar with the localization and how that plays in an affine scheme

sort of gives me some reallly fuzzy idea

[olive mirage]

I think it really helps to understand 19th century geometry stuff, and Weil conjectures stuff, before getting into this stuff super deeply.

[next obsidian]

I bet it honestly does

Sadly it isn't the current track

[olive mirage]

I really recommend Miranda... like you can read it in like a day

[next obsidian]

for how AG is taught

[olive mirage]

and it does a great job tying all this together.

[next obsidian]

I think what also bites is that I hated varieties with a passion

I really did not enjoy them whatsoever

and that's supposed to give intuition for the geometry in what's going on with schemes

[olive mirage]

I think of curves as the fundamental object

and all other things are just different ways of generalizing them

[next obsidian]

hmm

sadly hartshorne puts the stuff on curves

after the two hard chapters on a general theory of schemes and cohomology 😔

From the stuff in the class it really seemed like that's where things opened up

and got more geometric in flavor and like... cool in the applications

But it can be such a slog to get there

[olive mirage]

well youre not going to learn anything about curves from hartshorne. I more mean like, classical/arithmetic stuff about curves

and using that as your foundation, and then you can appreciate the abstraction

[next obsidian]

Can you expand on what you mean by arithmetic stuff about curves?

My knowledge on curves is like

if you do it right you have bezout's theorem for intersections

lmao

[olive mirage]

yeah, that's like, a neat parlor trick to me. It's oddly complicated to get right.

But all the divisor/line bundle stuff, Riemann Roch, the notion of genus, the various interpretations of (co)homology, all of that is super helpful to understand.

and by arithmetic stuff, I mean understanding the Weil conjectures. What do they say about point counting on curves in terms of genus? Why does a good cohomology theory imply them? etc.

[next obsidian]

You can talk about that with schemes though and the stuff at the end of our course in spring talked about that. It was just hard to understand like, what the genus really is, because it isn't the same as like, genus of a torus right?

[olive mirage]

well, I would argue, it is the same, but that definition is veyr narrow.

[next obsidian]

hmmm

I asked and told like, it's a numerical invariant

[olive mirage]

and so one has to generalize it so you can apply it in more contexts. E.g. for curves over finite fields.

[next obsidian]

I mean intuitively genus = holes

[olive mirage]

or perhaps for singular curves. There are a few different ways to generalize it (annoyingly)

[next obsidian]

But what does it mean for a curve to have holes?

[olive mirage]

well by "hole" you should be thinking "non-trivial element of H1"

[next obsidian]

haha, well at that point I lose the geometry

[olive mirage]

not if you're picturing, e.g., a torus and you literally picture two circles

[next obsidian]

two circles?

Like as S^1 x S^1?

[olive mirage]

These two.

[next obsidian]

Okay yup

[olive mirage]

they are non-trivial because they cannot be contracted to a point

[next obsidian]

sure

I guess that makes sense I think'

if I go to the complex plane

hole to me means like

[olive mirage]

so generally dimension of H1 = 2g

and this is your touchstone to connect to that.

[next obsidian]

not a simply connected domain

[olive mirage]

and I'm intentionally writing H1 insteand of H_1 or H^1 or whatever, because it shouldn't matter.

[next obsidian]

and the barrier to simply connected is holes

but also you can state it in terms of it not begin contractible to a point

or something about homotopy and homology which is the same thing right?

[olive mirage]

Yeah. And in the context of Riemann surfaces, that 2 there in the 2g has a very specific interpretation

namely it's coming from the fact that a curve over C is a surface over R.

[next obsidian]

okay

[olive mirage]

but that 2 is always there in algebraic geometry, although you lose track of that nice vision.

[next obsidian]

That makes sense how a 2 could appear there

[olive mirage]

e.g. the statement Dimension of H1 = 2g is still just as true for curves over F7

but of course you have no picture like this in that case.

[next obsidian]

yeah

😔 this stuff can be so hard

[knotty mason]

wait Miranda i'm finding is 400 pagse

[olive mirage]

it's a long day.

[next obsidian]

LOL

[olive mirage]

For me, the story of algebraic geometry is that you start with this vast differential geometry theory for manifolds and riemann surfaces over R/C. You identify that a lot of the best examples of these are algebraic in nature. Indeed, in the case of Riemann surfaces, all of them are. You view that differential geometry has told you the "right" answer for all of these homology theories and genera and so on, but you try bit by bit to avoid using analysis/topology/etc, and see if you can strip away all references to them, and still get the same answers.

And once you can get all the same answers without any reference to the geometric/topological structure, then you can replace C with an arbitrary field.

so you can ask Diophantine geometry questions by replacing C with Q, or arithmetic questions by replacing C with F_7, etc.

[next obsidian]

For the differential geometry bit of that story, is a class on smooth manifolds you figure enough to talk about it?

[olive mirage]

If you're especially batty, you can try to replace all number theory questions by replacing C with Z

Yeah smooth manifolds and algebraic topology is what you really need.

[next obsidian]

There's hope then

[olive mirage]

But yeah, Miranda is one of the few math books taht I think makes good bedtime reading. It is not a text where every sentence feels like a war fought over days.

[next obsidian]

Unlike some books...

[olive mirage]

haha yes, and those books have their place too

but that is what I think makes Miranda such a good supplementary book.

[next obsidian]

I'm bummed I didn't pay much attention to my riemann surfaces class

It was during spring quarter when everything moved online

and I was focused on staying on top of my algebra stuff so it just sort of entered the background

I believe I would've gotten a lot more if classes were in-person

[olive mirage]

I think a lot of professors were not good at that pivot

[next obsidian]

Which is fair

They’re people too and just as much as us had to suddenly just switch gears

[olive mirage]

yeah, they were not hired for their skills in this area.

[vestal snow]

For the uniqueness part of the proof, I feel like you need that $\cap a_i$ is a minimal decomposition to get that the $p_i$ are all the isolated primes belonging to the $0$ ideal

[cloud walrus]

Have a Banana:

[vestal snow]

You can prove that it indeed is minimal, but I think that if you needed minimal, the book would've mentioned it

[smoky cypress]

For what positive integer $m$ does the ring $\bZ/m\bZ[x]$ satisfy the unique factorization theorem

[cloud walrus]

Whoever:

[woven delta]

What's the unique factorization theorem?

Do you mean property?

Also like isn't a Z/m an integral domain iff m is prime (for obvious reasons) so Z/m [x] is a ufd iff it's a PID iff m is prime

@smoky cypress

Unless there is some other unique factorization thing you are talking about

Which makes this question nontrivial

[smoky cypress]

Unique factorization for non zero divisors

So for Z/6Z[x], the unique factorization theorem holds for non zero divisors

[woven delta]

Can you give me an example of a non unit non zero divisor?

[smoky cypress]

Like x?

[woven delta]

Oh ok

I was asking about in Z/m

Are the only 2 options units or 0 divisors?

I guess that follows from coprime stuff

[smoky cypress]

Yeah

[woven delta]

Can you show unique factorization in Z/6[x]?

[smoky cypress]

x = (3x+4)(4x+3)

[woven delta]

Lol

Yeah that makes sense

I'd imagine you can always do something like that

[smoky cypress]

I think you can always do that for square free m

[woven delta]

Sure

[next obsidian]

Liquid, you can strengthen that so that Z/m[x] is a ufd iff pid iff euclidean domain right?

[oblique river]

whoever and I talked about this before for squarefree things

[smoky cypress]

Yeah

[oblique river]

if two rings R and S satisfy a unique factorization property for nonzerodivisors, then so does RxS

the irriducibles in RxS are of the form (r,1) or (1,s) for irreducibles r in R or s in S

[smoky cypress]

I guess now we just need to look at when m is a prime power

[oblique river]

I feel like once you have nilpotents that could just kind of ruin everything

[woven delta]

I don't think that's true whoever since the statement buncho made is not an if and only if

[oblique river]

I mean, then let's make it into an iff

[woven delta]

Although the counterexample I'm thinking of is for Z/pq for distinct primes p and q

So maybe it's nicer for other cases

[oblique river]

what is that a counterexample for?

[woven delta]

Wait one sec

I'm a bit confused

Was x=(3x+4)(4x+3) not a counterexample to Z/6[x] having unique factorization?

Or was that supposed to be something else

[oblique river]

no

that is the factorization of x

in that ring

[woven delta]

Oh okay

I was being silly

[oblique river]

x is not irreducible

but those other polys are

[woven delta]

Okay so I guess the statement that we want to prove is that if m and n are coprime, then Z/mn[x] has the property iff Z/m[x] and Z/n[x] do

[oblique river]

that follows from the statement about products of rings

[woven delta]

Oh ok

[oblique river]

Z/mn[x] = Z/m[x] x Z/n[x]

[woven delta]

Yes, but I thought only 1 direction did

Namely the n, m implies mn direction

I guess the other direction is obvious

From the embedding of Z/m[x] and Z/n[x]

Although actually maybe that's not so clear

One sec

Oh that's what you wrote lol

😅

Okay then as whoever said it's down to the prime powers

Do you guys have an example of a non unique factorization?

[oblique river]

at this point, that is equivalent

if there is a nonunique factorization, it must come from some kind of nilpotence situation

if whoever had an example, that would answer his question

[woven delta]

Oh I see

I forgot what his question was lol

[delicate bloom]

I guess something like $(1+px)^n = 1+pnx$ would be a problem

[cloud walrus]

Merosity:

[delicate bloom]

in Z/p^2Z[x]

[smoky cypress]

I found a paper that says $$x^2+7\equiv(x+1)(x+7)\equiv(x+3)(x+5)\pmod{8}$$ where all the four factors are irreducible

[cloud walrus]

Whoever:

[delicate bloom]

I don't see how those are irreducible when you can factor $1 = (1+2x)^4$

[cloud walrus]

Merosity:

[oblique river]

that's not a "factorization"

both 1 and 1+2x are units

[delicate bloom]

oh right lol

[smoky cypress]

That is the definition of a unit

[delicate bloom]

I was telling myself they were nilpotent for some reason

[oblique river]

unless the factors in the two factorizations of x^2 + 7 happen to differ by units

it seems like that might settle this

at lesat for things like Z/p^3

[smoky cypress]

Btw here's the link: https://apps.dtic.mil/dtic/tr/fulltext/u2/a327984.pdf

[plucky flicker]

Hello guys! It might be a stupid question, but if i have a polynomial ideal I in K[x_1,...,x_n], and G is its Gröbner basis, if i take two different polynomials from K[x_1,...,x_n] (which are not in I), can the normal forms of these two polynomials be equal to each other?

[knotty mason]

sorry!

[plucky flicker]

they will if they are in the ideal I

but i'm interested in if they are not in the ideal

[vestal snow]

Can you have p^n = p^{n+1} for a non zero prime ideal?

We're in an integral domain

[solemn rain]

by p^n do you mean (p^n)

[vestal snow]

No

p is not necessarily generated by a single element

I think I got it

[solemn rain]

i still dont understand shit but good ;D

[elder valley]

any artinian ring with nontrivial prime ideals i think will have that

like Z/nZ with n nonprime

[vestal snow]

The ring isn't artinian

But it is noetherian of dimension 1

[vestal snow]

This is what prompted that question

How do we know that a primary ideal q can only be written as a power of its prime p in just one way?

[next obsidian]

I think that’s not possible right @vestal snow

[vestal snow]

You mean it's not possible for p^k = p^{k+1} in such a ring?

[next obsidian]

So if p^n = p^{n+1} it’s easy to show that p^n = the intersection of p^n over all n to infinity. Localize at p, so that you’re in the local Noetherian case, then p is maximal, and then Krull’s Intersection Theorem says the intersection of the p^n is 0 which I think implies that the intersection of p^n in the non-localized version is 0

I think so

I haven’t worked out the details

But this is my basic idea

[vestal snow]

I think there might be a simpler proof

[next obsidian]

So this implies p^n = 0 which is fake because it’s a domain

[vestal snow]

because krull hasn't been introduced yet

[next obsidian]

Sure, what is it?

[vestal snow]

Oh idk lol

[next obsidian]

Oh hahaha

[vestal snow]

I just said that there probably is one given that the book used this result

[next obsidian]

Maybe there’s some other way to show that that intersection is 0

Or maybe there’s a like

More obvious way to see it

[vestal snow]

One of things with this book is that it sweeps a lot of the stuff under the rug

[next obsidian]

I think that becomes common around this point

[vestal snow]

Maybe, this is my first graduate textbook

[next obsidian]

What does 9.1 say?

It could be hidden in that proof

Even if not stated explicitly

And yeah graduate texts will do that

[next obsidian]

I remember vividly Hartshorne says “since f_1,...,f_n generate the unit ideal so does f_1^N,...,f_n^N” or something to that effect

You can prove this because if (f_1^N,...,f_n^N) were proper it’s contained in a prime which then contains all f_i, so it contains (1) so it isn’t prime

But he doesn’t say anything about it

And I was like !!!!

[vestal snow]

Haha

[next obsidian]

Oh so uniqueness follows from that

The p_i are all pairwise coprime

So they’re different

[vestal snow]

Yes

[next obsidian]

I think

Or no

They’re all distinct maximal

Which implies coprime I think

[vestal snow]

isolated

Yes exactly

since they're maximal, none contains the other

therefore, each is isolated

[next obsidian]

Idk I’m pretty tired so I might just be like mega off rn idk haha

[vestal snow]

Np

[next obsidian]

Anyway, this is around where my knowledge of stuff in A-M fades

[vestal snow]

O no

[next obsidian]

And I enter the Matsumuraverse which does things in a different order haha

[vestal snow]

I'm thinking of stopping here

doing the exercises of the previous chapters

and then continuing

[next obsidian]

That seems good

There isn’t too much left right?

[vestal snow]

3 more chapters

Though I think I'll only do 1.5

because the rest is mostly about algebraic geometry which I won't be doing at least until next year

[next obsidian]

Fair enough

Definitely cover the one which has Krull

Since it builds up like graded module stuff

And then does like I-adic completion or whatever the hell I forget

The like topology stuff

Seems cool

I only just looked at it to get Krull’s intersection theorem

[vestal snow]

Also, which anime is your pfp from?

[next obsidian]

Little busters

[vestal snow]

@next obsidian Found a solution

For every $p$, define $p' = {y \in Frac(A): yp \subseteq A}$

[cloud walrus]

Have a Banana:

[vestal snow]

then p' is a A-module

and pp' = A

so you can "cancel" terms in the equality p^n=p^{n+1} to get p = (1)

which is bs

http://virtualmath1.stanford.edu/~conrad/154Page/handouts/dedekind.pdf

[next obsidian]

Weeeeeird

This doesn’t use the Noetherian or Dim 1 hypotheses does it?

[vestal snow]

I don't think so

I'll prove the lemma once I wake up from my nap

[smoky cypress]

(2993x+14966)(2993x+17959)(2993x+26938)(2993x+29931)(18615x+33508)(18615x+43436)(18615x+21098)(18615x+31026)(29274x+16729)(29274x+42518)(29274x+698)(29274x+26487) = x^4 - 4 (mod 50881)

@woven delta

[woven delta]

Lol

Good to know

[glad juniper]

,w PolynomialMod[(2993x+14966)(2993x+17959)(2993x+26938)(2993x+29931)(18615x+33508)(18615x+43436)(18615x+21098)(18615x+31026)(29274x+16729)(29274x+42518)(29274x+698)(29274x+26487),50881]-50881

[cloud walrus]

Results provided by WolframAlpha

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

Requested by dirib#6336

[smoky cypress]

@glad juniper you’re welcome

[final dove]

Can someone help me proof verify the relationship between zeros of a system of equations and K-algebra morphisms? https://math.stackexchange.com/questions/3766418/proof-verification-bijection-between-solutions-to-a-system-of-equations-and-k-a

[plucky flicker]

@final dove it might help you out

i have another question regarding to this topic: here says that, not every bijective morphism is an isomorphism, and gives a brief example. My question is that, this assumption is based on the fact that there is always a bijective connection between the Mor(X, Y) and Hom_K(K[Y], K[X])? And if so, then in the given example that function is not an isomorphism, because the induced function between the coordinate rings are not bijective?

[plucky flicker]

i guess i've found the answer

[olive mirage]

That is correct. If you like thinking about varieties over finite fields, Frobenius is always a bijection on points, but is never an isomorphism because the map on rings is plainly never surjective.

[plucky flicker]

Thanks @olive mirage

[olive mirage]

This has a bunch of confusing consequences because it is so unlike other things I'm used to working with. You might even have a morphism that is the identity on points and is not the identity.

[plucky flicker]

that sounds strange

[plucky flicker]

can someone explain to me, why is the red framed expression true?

[frozen oasis]

@plucky flicker where are you reading this from just curious

[plucky flicker]

Computational Invariant Theory, by Harm Derksen and Gregor Kemper

It's a decent book, but a bit high tech

[next obsidian]

@plucky flicker

So basically

Consider that f = a_nx^n + ... + a_1x + a_0

Or ugh

There are multiple variables

But notice that J has g_i - x_i for all I in there

When you do f(g_1,...,g_n) - f

The first one puts a g_i in place of all x_i

So by like matching those up

This becomes a polynomial in the x_i - g_i

Which is then in the ideal

[plucky flicker]

I got it, thank you @next obsidian

[elder valley]

it's not very clear to me when products are involved. like f = x_1 x_2 leads to g_1 g_2 - x_1 x_2

[plucky flicker]

f = x_1 x_2 leads to g_1 g_1. You just change the x_1 to g_1 and x_2 to g_2

[elder valley]

you subtract f though

[plucky flicker]

Yep. So take a simple example

Suppose f = x_1 x_2

Then f(g_1, g_2) = g_1 g_2

So when you subtract you get f(g_1, g_2) - f(x_1, x_2) = g_1 g_1 - x_1 x_2

[elder valley]

sure, and i see that you can add and subtract g_1 x_2 to get f(g_1,g_2) - f = g_1(g_2-x_2) + x_2(g_1 - x_1) which is in J. but for a general monomial it's harder to think about for me

[plucky flicker]

g_1 and g_2 are polynomials in x_1 and x_2

[elder valley]

they are polynomials in y

[plucky flicker]

Ugh sorry, yes

[next obsidian]

Hmmm

[elder valley]

i think i see a proof

since f is a sum of monomials, if we show the statement works for monomials then we're done right

[next obsidian]

Yeah

I was gonna say I think it suffices to show for something like that, I thought linear ones but that’s not true I think

[elder valley]

so assume $f = x_1 ^{e_1} \cdots x_n ^ {e_n}$, so you get $g_1^{e_1} \cdots g_n ^ {e_n} - x_1^{e_1} \dots x_n ^{e_n}$

[cloud walrus]

Auvera:

[next obsidian]

Yeah

[elder valley]

add and subtract to get $g_1^{e_1} \cdots g_n ^ {e_n} - x_1^{e_1} \dots x_n ^{e_n} = g_1^{e_1} \cdots g_n^{e_n} - g_1^{e_1} \cdots g_{n-1}^{e_{n-1}} x_n^{e_n} + g_1^{e_1} \cdots g_{n-1}^{e_{n-1}} x_n^{e_n} - x_1^{e_1} \cdots x_n^{e_n} = g_n^{e_n}( g_1^{e_1} \cdots g_{n-1}^{e_{n-1}} - x_1^{e_1} \cdots x_{n-1}^{e_{n-1}}) + x_1^{e_1} \cdots x_{n-1}^{e_{n-1}}( g_n ^{e_n} - x_n^{e_n})$

[cloud walrus]

Auvera:

[elder valley]

use induction on the first term in parenthesis, and the second term in parenthesis is divisible by g_n - x_n

[next obsidian]

Ahhhhh I see

Yeah this kind of smelled like induction

But I didn’t see how to reduce the number of variables, that’s good

[elder valley]

i messed up that equation, but the idea is there

[shy bluff]

why is x^2 + 1 and x^2 comaximal?

[elder valley]

because you can get 1 by just subtracting them

[shy bluff]

oh

[elder valley]

which means (x^2+1) + (x^2) = Z[x]

[vestal snow]

@next obsidian about the prime problem earlier, turns out you can do it using nakayama's lemma

[next obsidian]

Hmm

How?

Do you have to localize first?

[vestal snow]

If p^n = p^{n+1} then pM = M

Therefore, there exists an x=1 mod p such that xM=0

[next obsidian]

Wait wait

[vestal snow]

Since x is not 0 and we're in a domain

[next obsidian]

How do you get there?

[vestal snow]

Let M be p^n

[next obsidian]

Ohhh

Sure sure haha

Lol

Ohhh gotcha

Lol

[vestal snow]

So by properties of a domain p^n = 0

Therefore p=0

[next obsidian]

Yeah gotcha I see

That makes sense

Technically you need to know that x isn’t 0 but

If it were then 1 in p

[vestal snow]

Which is bs

[next obsidian]

Yup

Yeah that makes sense

Nakayama man lol

Pops up outta nowhere

[chilly canyon]

Hi ! Define G=]0,+oo[-{1}, and define • on G as : a•b=a^log(b).
Now this makes G a group whose neutral element is 1=exp(1), and where the inverse of a is e^1/log(a).
What could G be isomorphic to ?

I first thought of exp:R*→G

But it turns out it's not a group homomorphism, sadly

Oh wait

It is !

Sorry 'bout that, it's working like a charm !

[next obsidian]

I mean, you could ike

say it's isomorphic to G lol

but it's interesting that it is isomorphic to R*

[oblique river]

I think i'm being a little dense here but I don't see how that's a group

1*a = 1^log(a) = 1

but that can't ever happen in a group unless a is the identity

but that equation is true for all a

no?

[chilly canyon]

1 is not in G !

[elder valley]

i think his notation excludes 1

[chilly canyon]

The neutral is exp(1)

[oblique river]

ohh I see, thanks

when you wrote 1 = exp(1) I just parsed the 1

[chilly canyon]

Discord didn't like the \

[oblique river]

you can do \\

two slashes produces one slash

[chilly canyon]

When I typed it in, it didn't render it, I think markdown screwed up trying to escape the curly bracket

Oh okay thanks !

[eager frost]

T shirts are on sale for 20% off. Tasha paid $8.73 for a shirt.  What is the regular price of the shirt? There is no tax on clothing purchases under $175.
Let p represent the regular price of these t-shirt. Which of the following equations is correct?
A
0.8p=$8.73
B
$8.73+0.2∗$8.73=p
C
1.2∗$8.73=p
D
p−0.2∗$8.73=p```

[oblique river]

#prealg-and-algebra

[hazy leaf]

Consider the vector space $\mathbb{Q}$ of all quadratic polynomials with coefficients in $\mathbb{C}$. Consider the linear operator $T:\mathbb{Q}\to\mathbb{Q}$ defined by $$T(f) =f+f′+f′′$$ Find a Jordan basis of $\mathbb{Q}$ for $T$

[cloud walrus]

Konoha:

[delicate bloom]

should be #linear-algebra but whatever, seems occupied right now

what have you tried @hazy leaf

[hazy leaf]

honestly, I don't know how to start

I would like a hint

[elder valley]

find the eigenvalues

[hazy leaf]

sorry, I still don't get it

[elder valley]

@hazy leaf do you know how to find eigenvalues

[hazy leaf]

lambda(f+f'+f'')=\lambda f

[elder valley]

it's better to use the characteristic equation

find a matrix representing T first, then compute eigenvalues

[hazy leaf]

oh, now I got it

I keep thinking some un-related theorem

[steady axle]

does a solvalbe group always have a proper normal subgroup?

[woven delta]

If you mean nontrivial then Z/p doesn't

[steady axle]

ya but can we say that for non cyclic group?

[left kite]

@round lagoon

am downloading something is 5%/5 min how much more time if im 15 minutes in

sir

[round lagoon]

[river fern]

(the following is from a fall 2019 hw assignment:)

a) and b) seem to contradict each other

By a), shouldn't b) be false?

Since H is a proper subgroup of G

So the union of the conjugates of H must be a proper subgroup of G

So the union of the conjugates cannot equal G

[wind parrot]

a) says finite group, which GL_2(C) isnt

[next obsidian]

You can strengthen the statement of a) actually. No group is the union of all conjugates of a subgroup of finite index

In a finite group all subgroups are of finite index

[next obsidian]

@latent anvil did you know you can make some estimates on the number of generators of a finite A-module when A is suitably nice? As long as m-Spec A is a Noetherian topological space you can make an estimate, but I don’t think computing it in practice is very practical, and the estimate seems hard to use even just theoretically

It’s interesting though that you can say it’s generated by <= r elements for some r though which has a formula sort of. It’s a supremum across some set, but it’s cool

[chilly canyon]

Hi ! Does Boolean Algebra fit in this channel ?

In case it does, is there a faster method to write all 2^4=16 binary operations on Z/2 as the binary operations given by the union, intersection and complement of sets, as mentinoned in https://math.stackexchange.com/a/3754822/259363, than verifying it by hand ?

[...] it is a simple exercise in Boolean algebra to build every binary operation on {0,1} out of the basic Boolean operations.

Does it suffice to construct all of the following four tables ?

Because summing them allows to make any table ? Do I also need the table whose last column is all zeroes ? Or am I missing something ? (I've never done any Boolean algebra before.)

[shy bluff]

Hrm

When given an arbitarry quotient ring, how does one tell if it's an integral domain?

Similarly, how does one tell if it's a PID?

Say for example, here

Clearly b is not a pid or an integral domain because we have 3 * 2 = 0 in Z/6Z

And a is clearly a PID because 5 is prime?

[next obsidian]

For integral domain it’s equivalent to showing the ideal is prime

[chilly canyon]

I prime ideal <=> A/I integral

[latent anvil]

that's kind of neat magician

[next obsidian]

For a PID determining the ideal is maximal means it’s a field, so it’s also a PID

Besides that I think it’s kind of hard to show it’s a PID without something clever

[chilly canyon]

Yup

Usually, the first iso theorem

Or the third

[next obsidian]

You can try and show it isn’t a PID though by constructing a non-principal ideal (kind of hard) or by showing unique factorization doesn’t hold so it isn’t even a UFD

[latent anvil]

Or by finding a finitely generated module which doesn't decompose as a direct sum of cyclic modules :P

[next obsidian]

Lmfao okay

One thing to consider is that any quotient of a PID that’s an integral domain is a field because prime <=> maximal in a PID and that a single variable polynomial over a field is a Euclidean Domain, and this a POD

This isn’t true for (0) though I guess haha

[shy bluff]

hrm

soh ow would you show taht an ideal is prime?

[next obsidian]

¯_(ツ)_/¯

Usually the best way

Is to find a way to show that R/I is an integral domain via the 1st isomorphism theorem

But that’s kind of getting circular

There’s some other ways, like if it’s principal that its generator is prime and such

Also I realized for my example if you’re in a PID that R/(0) is still a PID lol

So every quotient of a PID that’s an integral domain is a PID

[shy bluff]

somewhat circular reasoning

[woven delta]

Lol

[chilly canyon]

Which one do you have trouble with ?

[woven delta]

Obviously deciding if an ideal is prime is hard

And depends on the situation you're working with

[chilly canyon]

In general, if x is irreducible, then (x) is maximal, thus A/(x) is a field, thus integral

(or is it (x) is prime ? I'm never sure)

In any case, A/(x) is integral

[next obsidian]

If x is prime then (x) is prime

[woven delta]

In a PID x irreducible => (x) prime => (x) maximal

[chilly canyon]

Isn't that the definition too ?

(for x being prime)

[woven delta]

Well those implications are iffs

[shy bluff]

uh

[next obsidian]

Yeah if you want to put it that way

[chilly canyon]

No, maximal need not be prime

[next obsidian]

What? A maximal ideal is prime

[chilly canyon]

Oh in a PID sorry

[shy bluff]

My question is, given an arbitarry quotient ring, how would I know if it's an integral domain, and possibly a pid?

[next obsidian]

You can’t

You need more info

[woven delta]

Yeah

This sort of question is what computablity is for btw

[next obsidian]

And to tackle it on a case by case basis, I mean for ID it’s equivalent to the ideal being prime

[woven delta]

Once you actually phrase it in a nicer way

[chilly canyon]

You need inspection of the quotient, there's no general method

[shy bluff]

Computability?

[next obsidian]

For a PID I don’t know any condition that’s equivalent

[shy bluff]

hrm

So you just sorta have to compute each by hand?

[next obsidian]

Yeah

[shy bluff]

like I was given this

[next obsidian]

Those have context

Z/5Z is a field

[shy bluff]

ANd I can clearly see that a is a PID, but b isnt'

[woven delta]

You just have to think about them individually

Yeah

[shy bluff]

Yea

but I'm not sure about the rest

[next obsidian]

Z:6Z isn’t even an integral domain

[woven delta]

Finite rings are integral domains iff fields

[shy bluff]

Exactly the same wording

[woven delta]

Is a good fact to know

[next obsidian]

Q[x]/(x+7) is just Q

For d)

[shy bluff]

Wait why is Q[x]/(x + 7) just Q?

[next obsidian]

That says x = -7

[chilly canyon]

But in the example you linked, some things are natural, such as :
a) is a field, b) has zero divisors, c) is iso to Q with the 1st iso
d) Use the third iso to have it equal (Z[X]/(X²+1))/2, e) is the same method
f) x irreducible => (x) maximal => A/(x) is a field (in this case it is Z[i] the Gauss integers)

[next obsidian]

Evaluate any polynomial at x = -7

[shy bluff]

Oh I see

[next obsidian]

For e)

Show that poly is irreducible

[chilly canyon]

Generally, the method to use is natural with some experience

[next obsidian]

So that the ideal is prime <=> maximal

Cuz you’re in a PID

So the quotient is a field

[shy bluff]

wait question

[next obsidian]

Matplot your f) is wrong

[shy bluff]

Is the "show that the polynomial is irreducible" thing valid for any uh polynomial ring?

[next obsidian]

Z[i] is only a Euclidean domain

Yeah

Irreducible is prime

In an UFD

[shy bluff]

Oh ok

[chilly canyon]

Oh yeah what did I write x)

[next obsidian]

The two are equivalent

[woven delta]

Yeah

[shy bluff]

We have'nt gotten to UFDs yet

[next obsidian]

For f)

You just manually show Z[i] has a Euclidean algorithm

[woven delta]

If you know pids then you're fine

[shy bluff]

what do youm ean by manually show that?

[next obsidian]

Also irreducible always implies prime right?

You show it exists

I forget if prime => irreducible or if it’s the other way around

Like

That you can do division with remainder in Z[i]

That shows it’s a Euclidean domain

[chilly canyon]

The absolute value is a stathm on Z[i]

Btw, what was I missing to say A/(max) is a field ?

What's the additionnal thingy ?

[next obsidian]

No that’s correct

[shy bluff]

stathm?

[chilly canyon]

Oh so it's x irreducible => (x) prime then

[next obsidian]

Yeah

[chilly canyon]

Okay !

[next obsidian]

So the absolute value

The norm on Z[i]

Actually gives a norm in the sense of the requirement for a Euclidean domain

So you can do division with remainder using it

[chilly canyon]

A stathm (not sure about the terminology in eng) is some application on the ring to allow for Euclidean division (e.g. deg on R[X])

[next obsidian]

We call it a norm

[chilly canyon]

Oh okay !

[next obsidian]

What language is stathm from?

[chilly canyon]

"Stathme" is in French

[next obsidian]

Ah I see

[shy bluff]

oh yes a norm

ok

[chilly canyon]

I assumed it had latin origins and translated the same

[next obsidian]

What you want to think about when trying to prove it

Basically given two Gaussian integers

Like umm

I just wanna say z and w

You can form z/w as an element of C

Then z[i] forms a grid in the complex plane

You want to basically estimate z/w by the closest point on the grid

Then the remainder is whatever you need to shift by to get there

Is the general idea

[shy bluff]

hrm

I see

[chilly canyon]

^
This also works similarly on like Z[j] where j=exp(2iπ/3)

[next obsidian]

I think showing that the remainder is smaller in norm

Is basically just Pythagorean theorem

[chilly canyon]

But not in all the Z[sqrt(-d)] though !

[next obsidian]

Since you move less than 1/2 in both the x and y direction

Or I guess at most if you’re dead center

Yeah actually so given z/w let u be the like “estimate” for it

Then the remainder is like r = z/w - u

So we get z/w = u + r

You want to show that |r| < 1

Then when you multiply by w

You get z = wu + wr

And then you show wr is a Gaussian integer

Then |wr| < |w|

So this is the Euclidean algorithm

[shy bluff]

I see

[next obsidian]

Matsumura is so crazy. They don’t prove Noether Normalization except in some thing right at the end lol

They somehow prove stuff just without using it lol

Does anyone want to think about why for a polynomial ring over a field, A = k[x1,...,xn] that ht p + dim A/p = n?

basically this means @latent anvil lmao

Altho I think maybe you’re taking a math break which is totally fair

I don’t really see how to use induction for it

Actually maybe you can do induction?

If p is ht 0 then it’s 0 so you’re done

If p isn’t ht 0, quotient out by a ht 1 prime q contained inside of p

Then you have A/q which has dim < n??

Is that true hmmm

I mean, yes it is, by this problem but

Oh yeah it has to lol we know easily that ht p + dim A/p <= dim A

So you can extend this to saying that for a fg ID over a field, that ht p + dim A/p = dim A

By Noether Normalization

So A/q is still a finitely generated k-algebra and has dim < n

So doing induction I think you get the result

Hmmm I guess really we don’t even need q

Wait hmmm...

Actually this doesn’t work I think since we’d need to know that dim A/q = n - 1 I think

Ekdoekoddkoww

I guess you can get this by showing that dim A[x] = dim A + 1 for Noetherian A...

[elder valley]

Does anyone want to think about why for a polynomial ring over a field, A = k[x1,...,xn] that ht p + dim A/p = n?
@next obsidian i thought this was immediate from the correspondence theorem

[next obsidian]

Like going up?

[elder valley]

like take a maximal length chain of prime ideals that passes through p

[next obsidian]

This fails for other types of rings though

How do you know that that maximal length is n?

All you know is that there’s some chain with length n

This gives you that ht p + dim A/p <= n

But equality appears to be a lot harder

[elder valley]

the krull dimension of that ring is n though

[next obsidian]

Yes but

There’s no reason that for every prime ideal that maximal chains through p are length n

In an arbitrary ring

Krull dimension n means that there exists some chain of primes of length n, but it doesn’t mean every one is such

[elder valley]

right

hmm

[next obsidian]

It turns out if you’re an integral domain fg over a field this is true

By this theorem

So finding a counterexample will be annoying

I checked my class’s notes since we covered this and it just had a reference to eisenbud lol

[shy bluff]

Hrm

Is finding a Euclidean norm just one of those intuition things

[next obsidian]

Honestly I’ve never done it besides for Z[i]

It seems like it’d be incredibly tedious

And I guess a polynomial ring

[elder valley]

like defining the norm? or computing the norm of an element?

[shy bluff]

uh

Like here they go and come up with a norm

[next obsidian]

If you know something is a Euclidean domain and you can embed it into something with a natural norm, like C or R or R^2 or something.

[shy bluff]

looks like they just pulled it outta thin air to me

[next obsidian]

I feel like it might be possible to use like a a^2 + b^2 norm there

Maybe??

It looks like that was constructed so that the norm of 1 + sqrt(-5) = 6 so it’s a nice number

Or maybe you do need that that’s norm is larger than that of 2

[shy bluff]

Pardon?

[next obsidian]

Oh actually so

That’s the normal norm on C

If you write a + bsqrt(-5) = a + i(bsqrt(5))

Then the norm of that inside of C is a^2 + 5b^2

So that’s where it came from

[elder valley]

my vague recollection is that the norm is just the product of all the conjugates, and in cases like Z[sqrt(-d)] with d positive square free the conjugates are just the complex conjugates, so it's just the regular C norm

[shy bluff]

oh

[elder valley]

i suppose there could be other norms. not sure. that's the usual one though

[pure crest]

hey guys

how can i approach this question

Let a be a root of $f(x) \in F(\sqrt{\alpha})[x]$, where $a \in F$, then a is a root of some $g(x) \in F[x]$

[stone fulcrum]

F(√a) is a field
f(x) is a polynomial
I'm not able to parse f(x) ∈ F(√a)

[pure crest]

sorry i meant

[stone fulcrum]

I'm maybe missing something?

[cloud walrus]

boat:

[stone fulcrum]

Oh yeah haha okay

[pure crest]

haha sorry

so i took $f(x) = a_0 + a_1 x + ... + a_n x^n$, then substituted $0 = a_0 + a_1 (a) + ... + a_n a^n$

[cloud walrus]

boat:

[pure crest]

at least one of $a_i = \sqrt{\alpha} $

[cloud walrus]

boat:

[stone fulcrum]

I'm assuming F is a field

[pure crest]

yup

F is a field

[stone fulcrum]

Then any of the ai could be of the form a + b√[α] ya

[pure crest]

wait

why?

[stone fulcrum]

Where a and b are in F

Basically all of the elements in F(√α) look like a + b√[α]

[pure crest]

the elements of $F(\sqrt{\alpha})$ are of the form $\frac{b_0 + b_1 (\sqrt{\alpha}) + b_2 \sqrt{\alpha}^2 + ... + b_n\sqrt{\alpha}^n}{c_0 + c_1 \sqrt{\alpha} + ... + c_n (\sqrt{\alpha})^k}$

because it is a simple extension

[cloud walrus]

boat:

[stone fulcrum]

It so happens that the inverse of √α lies in F[√α], and so that can be written in the form a + b√α

But I think I'm tangenting pretty hard. Hmm

[pure crest]

hmm

but we learned that that simplifies to just the denominator part when $\sqrt{\alpha} $ is algebraic over F

[cloud walrus]

boat:

[pure crest]

i mean numerator

sorry

i am not sure why it will be reduced just to a + b√α

actually i can see why it is true in this case

[stone fulcrum]

(√α)² = α ∈ F, so it fits into the a
(√α)³ = α√α ∈ F[√α], so it fits into the b

[pure crest]

because alpha is in F

[stone fulcrum]

And that pattern continues

[pure crest]

right

so all a_i is of the form a + b√α

[stone fulcrum]

So we're easily down to
(a + b√α) / (c + d√α)

The next trick is realizing that there's some numerator version of 1/(c + d√α)

[pure crest]

but since x^2 - alpha is in F[x], then its algebraic so we only need to consider the numberator

[stone fulcrum]

Which I'll ignore the proof if one can accept that F[√α] is a field

[pure crest]

isnt a simple field extension always a field?

[stone fulcrum]

Yus haha

Which isn't a trivial thing, but it's a pretty base fact to field theory

[pure crest]

haha 😂 its so trivial that the prof forgot to mention it

i was quite shocked when i found out actually

[stone fulcrum]

It's actually pretty cool to show that such an extension always has inverses

Let me see if I can find it on this discord I know I was in that convo

Anyway I've talked a lot but we're not closer to solving this mb

[pure crest]

haha 😂

i have seen the proof in the book ( but only for Krockner's extentions tho), the book however says that you prove for general field extensions in similar way.

so i think i can seperate the terms a + b√α into two polynomials

one in terms of a and the other in terms of b√α

idk if that will help

[timid hull]

If you know some basic field theory then its easy to see that adjoining $a$ to $F(\sqrt{\alpha})$ is a finite extension of $F$ since finite extensions are algebraic hence $a$ is a root of some $g(x)\in F[X]$

[cloud walrus]

WhyamIsohot?:

[pure crest]

i dont understand how tho

do you mean that $F(\sqrt{\alpha})(a)$ is finite over F

[cloud walrus]

boat:

[timid hull]

I mean if you have field extensions $L\subset K\subset M$ then $[M:L]=[M:K].[K:L]$

Yes

texit dead rip

[pure crest]

right

then M is algebraic right

ohh right

so we have $F \subset F(\sqrt{\alpha}) \subset F(\sqrt{\alpha}) (a)$

[cloud walrus]

boat:

[timid hull]

Yes and by given hypothesis $F(\sqrt{\alpha})(a)$ is a finite extension of $F(\sqrt{\alpha})$

[cloud walrus]

WhyamIsohot?:

[pure crest]

$[F(\sqrt{\alpha})(a) : F] = [F(\sqrt{\alpha}) (a): F(\sqrt{\alpha}) ] [F(\sqrt{\alpha}) : F]$

[cloud walrus]

boat:

[pure crest]

since $a$ is a root, there exists a polynomial of degree n. and so on so its finite

i see

[cloud walrus]

boat:

[pure crest]

but

[stone fulcrum]

Oh yeah that makes sense haha

[pure crest]

i dont see why $F(\sqrt{\alpha}) (a) $ algebraic over F says there is a polynomial with root a in F[x]

[cloud walrus]

boat:

[timid hull]

that's the definition of being algebraic

[pure crest]

im still kinda confused about that

ohh

a is in $F(\sqrt{\alpha})(a) $ so it is a root of some polynomial in F[x]

ohh i see

[cloud walrus]

boat:

[pure crest]

since its algebraic

thank you @timid hull

thank you @stone fulcrum

you guys were a lot of help!

[charred pewter]

This is a T/F question

I just came here to make sure this isn't a typo or something.

Like shouldn't it be (h'^-1)h' in H?

Oh wait, nevermind

This is for subgroups

I was thinking about groups lmao

[smoky cypress]

(h'^-1)h' is always the identity

[charred pewter]

Yeah, that's what I was thinking about

But I was thinking about groups and not subgroups

[pure crest]

what does it mean for a automorphism to have order 6? I understand the order with elements and groups, but i am not so sure how to conceptualize order with respect to homomorphism/ automorphisms

[next obsidian]

automorphisms form a group

it just means that if your automrophism f that f^6 = e

which means that for all x, f(f(f(f(f(f(x)))))) = x for all x

identity for automorphisms is the identity function

and the group operation is composition

[pure crest]

for any element?

[next obsidian]

Yeah

since the identity morphism

just sends x to x for all x

[upbeat burrow]

Do rings contains non unique elements? We’re told the zero of a ring is unique. Aren’t all elements in a ring unique?

[next obsidian]

It means up to isomorphism

any way to make a ring with 1 element is isomorphic

[upbeat burrow]

Because if I think of the elements as a set, {0,1,2,3} is the same as {0,1,2,2,3}

[next obsidian]

oh lol

you meant the element

[upbeat burrow]

Yeah

Sorry

[next obsidian]

it just means that if two elements satisfy

okay pretend you had two 0s

call them um

0 and o

[upbeat burrow]

No I get that proof

[next obsidian]

oh lol

Then what do you mean?

[upbeat burrow]

But aren’t all elements in a ring unique?

[next obsidian]

no so

Err

by "0"

it doesn't mean an element labelled 0

it means an element that satisfies the property 0 does

like in an endomorphism ring

the 0 map is the 0

[upbeat burrow]

It means the additive identity

[next obsidian]

yeah

It's like even stronger than that though, since actually if x + y = x for anything then you know that y = 0

so you don't even need that y is the additive identity on everything

[upbeat burrow]

But if I look at polynomial rings, matrices, etc, aren’t all elements unique?

[next obsidian]

yes but

it means you don't have two elements that both are the additive identity

if x + y and x' + y = y for all y

then x = x'

is all that means

0 in the way it's used here isn't like some specific element its just anything which when added to anything returns what was originally given

but the fact that it's unique

implies that it actually is a specifric element

but ai priori there's no reason to assume only 1 thing has that property

[pure crest]

thats an interesting question. I always thought all elements were unique but all we every proved was the the identity element is unique

[next obsidian]

Like obviously if your set is like

{0,1,2,3,4}

cuz you're representing Z/5Z

then like obviously each of 0,1,2,3,4 are all unique

since they're literally different things

but this uniqueness is more about for variables

things that are like existential quantifiers or universal ones

this just say that since 0 is the "zero" in this ring that if you had x + 3 = 3 then x = 0

But this is sort of contrived since in that ring the thing that "acts as zero" is represented by the symbol 0

but you can have rings where 0 is jsut

$\begin{bmatrix}
0 && 0\
0 && 0
\end{bmatrix}$

[cloud walrus]

Mathemagician:

[upbeat burrow]

So in that ring aren’t all other elements also unique?

[next obsidian]

Bruh you're missing the point tho

0 in this sense isn't referring to that element

it's any 2x2 matrix A such that for all other matrices B, A + B = B

[upbeat burrow]

So it’s the property

[next obsidian]

Yes

but via proof

you show any 2 things that satisfy it are the same

so that 0 is unique

so that that matrix I showed above

is "the" 0

but in the proof you need inverses

[upbeat burrow]

Ya

[next obsidian]

so like

for algebraic structures without it

theoretically you could have two different elements

that "act as 0"

Or at least I think it's inverses you need

I forget lol

[upbeat burrow]

Do you?

Because monoids still have a zero

Don’t they?

[next obsidian]

¯_(ツ)_/¯

really I'm thinking about the stronger property that

if x + y = y for any y

Just any single y

then x = 0

[upbeat burrow]

Yeah

[next obsidian]

but this isn't true for rings under multiplication

if it holds for all

then you're the 1

[upbeat burrow]

Right

[next obsidian]

but if it holds for some

then you might not be

[upbeat burrow]

But then what’s your other operation?

[next obsidian]

I just mean that

under multiplication

a ring is a semigroup

[upbeat burrow]

The identity is unique

[next obsidian]

since you don't have inverses

Or I guess a monoid