groups-rings-fields

[pastel cliff]

first of all finite extensions are algebraic?

[barren sierra]

I think I fucked something up here

I need to give an explicit isomorphism between the two group algebras at the top

I do this by chaining a bunch of isomorphisms

and the end result when I checked by hand seems wrong

X^1 has order 4 in the complex group algebra but the element I got at the end is not of order 4

I think my f(x) and g(y) are fine?

what did I do wrong man

anyone?

I have checked a bunch that f(1) = 1, f(-1) = i, g(1) = -1, and g(-1) = -i

so that's definitely right

unless I've lost all basic algebra abilities, very possible

I feel maybe the very last part where I map the chosen (f(x), g(y)) is wrong?

specifically how I map the constants

but whatever I map those to, they should cancel right?

nvm maybe I found it

yea I found it I'm a moron

[solar shore]

$(h_1h_2^{-1})(h_1h_2^{-1})^{-1}g(h_1h_2^{-1})(h_1h_2^{-1})^{-1} = g$

[cloud walrus]

blanket

[solar shore]

they just created identities on either side

[solar totem]

these really helped me do well in a class i expected to fail
god bless you

[wraith cargo]

Yes

[formal ermine]

ignore me lol

[dusty verge]

Don't paradox yourself on a math discord

[dim widget]

What does that mean?

[novel parrot]

How to calculate Z/mZ (X) Q

I can see that all elements have order <= m

Its just 0?

Since a x p/q = a x p/q * (m/m) = ma x p/q * 1/m = 0

am i correct

[delicate orchid]

it’s 0 yeah

Tensoring any finite abelian group with Q is 0

[dim widget]

If you have a module M and a module N, and r \in R acts invertibley on N and not on M, then automatically M \otimes N = 0.

[formal ermine]

only over Z?

[dim widget]

????

[novel parrot]

groups are Z modules

[dim widget]

What else are you planning on tensoring a finite abelian group with Q over?

[delicate orchid]

Something something extension of scalars

Irregardless of the fact that we’ve just established trying to extend the scalars of a finite abelian group to Q gives u 0

[dim widget]

You always have a surjective map M \otimes_Z N \to M \otimes_R N anyway. So if the tensor over Z is zero then it is the same over any other ring.

[novel parrot]

What kind of example works for the last part?

f,g both injective but f x g is not

[delicate orchid]

$M \otimes_{\bZ} N \rightarrow M \otimes_{\bR} N$ for the people in the back!!

[cloud walrus]

Wew

[dim widget]

This is a good exercise! Maybe you should give it a little thought. Iirc either you or sebb asked a related question recently.

[delicate orchid]

Something something initial object yeah

[novel parrot]

Should try looking at Z/mZ ?

[delicate orchid]

Yeah these are good exercises, the 2nd one is basically showing it’s right-exact too

[novel parrot]

Hm so i want to go from a large group to a small group

f: Z/12 -> Z/13Z and g: Z/12Z -> Z/15Z

then we get f x g : Z/12Z -> Z/Z = 0

[delicate orchid]

You need f and g to be injective

[novel parrot]

f and g are just identity

i think they are homomorphism?

[delicate orchid]

f has to be the zero map there

[dim widget]

Z/13 has only 13-torsion and Z/12 has only 4, 3, 2, 6, 12 torsion.

[delicate orchid]

Alternatively look at the additive subgroups, you’d need to map the generator (which is order 12) of Z/12Z to an element of order 13, which only leaves the trivial map as 13 is prime

[novel parrot]

oh

the identity map is not injective

[dim widget]

It is not well-defined even

[novel parrot]

nope

[delicate orchid]

Do you recall what Z/mZ (x) Z/nZ is isomorphic to?

[novel parrot]

yes

Z/(m,n)Z

the gcd

[delicate orchid]

The gcd is a good way of getting a small number from two big numbers

[novel parrot]

yeah

thats what i was trying

just i need to make f and g injective

[delicate orchid]

Yeah so you’ll need maps that look like Z/mZ -> Z/kmZ for some k

[dim widget]

So one of f or g can be the identity

At least one of the abelian groups has to be torsion.

[novel parrot]

ah

so we take m and n to be coprime

then we get a map from Z -> Z/kZ

understood

[formal ermine]

itself

[frigid lark]

Love when Lang disguises a theorem as an excercise

That was 6-8 hours down the drain

[dim widget]

Q won't be a module over that!

[novel parrot]

How does S (x) M become an S module?

Since s * (t (x) m) = st (x) m, should it also be t (x) sm ?

[dim widget]

No

because the tensor product is over R

Which means only elements of R can pass from left to right and vis versa

The structure is that s*(1 \otimes m) = (s \otimes m)

[novel parrot]

i see

[dim widget]

In general the structure of an S-module is just the structure of a map $m: S \otimes_Z M \to M$ such that both of the maps $S \otimes_Z S \otimes_Z M \to S \otimes M \to M$ (one multiplying $s_1$ by $s_2$ first then multiplying the result by $M$, and one multiplying $s_2$ by $M$ first then multiplying by $s_1$) are the same map.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

So somehow for any abelian group $M$ $S \otimes M$ can be made "the trivial $S$-module" by defining the map $m: S \otimes_Z (S \otimes_Z M) \to S \otimes_Z M$ to just be multiplication in the first two factors.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

Then $S \otimes_R M$ is just a quotient of this.

[cloud walrus]

Topos_Theory_E-Girl

[novel parrot]

@dim widget did you finish university already?

you have more knowledge on algebra than my professor

[dim widget]

yeah I am not in university.

[novel parrot]

I (x) M is not a submodule of R ?

[dim widget]

Yes it is not

[novel parrot]

So how does this quotient work?

[dim widget]

More interestingly it is not even a submodule of M

The quotient M/IM? Which quotient do you mean?

[novel parrot]

R/ I (x) M

[dim widget]

That is $(R/I) \otimes M$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

The parentheses matter.

[novel parrot]

ah ...

mis read it

[dim widget]

It's I guess ambiguous. But that's what they mean.

[rancid totem]

Would this be false since 0.999... = 1?

[frigid lark]

Isn't f just floor(10x) - 10floor(x)?

And floor(x) can be defined as the supremum of the set {y in Z | y < x}

[dim widget]

You're giving an interpretation of the map which is well-defined. But the whole point of questions like "is this map well-defined" is to learn that sometimes intuitive-seeming mathematical ideas can be deceptive. The way that sentence is written the map is not well defined, but it's easy to make a version of it that is by making more choices (like you did).

[frigid lark]

What additional choice did I make?

[dim widget]

You gave a specific way to interpret "first element to the right of the decimal point" which is unambiguous.

Like for the example of .99999999999... your definition does not agree with the naive definition, so you have disambiguated the meaning of "first digit to the right of ." in a way which gives an unambiguous answer.

You could've chosen to disambiguate differently, so that all integers are mapped to 9 for instance.

[frigid lark]

Is there a definition of a decimal expansion?

[formal ermine]

@dim widget nice color

[dim widget]

There is the usual map from subspace of sequences ${0, \dots, 9}^{\mathbb{Z}}$ which are all zero for entries indexed by $N << 0$ to $\mathbb{R}$ given by taking $(n_i) \to \sum_i n_i/10^{i}$.

There are other definitions as well.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

But I would call any fiber of this map over a point x a decimal expansion of x.

[frigid lark]

Mmmm

What is N << 0?

[dim widget]

N is an index of an element of the sequence $n_i$, and $N<<0$ just means "for $N$ sufficiently small and negative"

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

So in other words this map is only defined on sequences $n_i$ such that $n_i = 0$ for any $i$ which are less than $N$ for some $N$.

[cloud walrus]

Topos_Theory_E-Girl

[frigid lark]

Ok

Is this map even well defined? Cause of the infinite sum?

[pastel cliff]

Topos_Theory_E-Girl

[dim widget]

Yes, because of the condition about the sequence vanishing it is well-defined, I have ensured that all such sequences converge

[frigid lark]

Sorry for all of these questions

[dim widget]

That's okay! But these questions about decimal expansions would probably go better in a different # (perhaps #real-complex-analysis ), arguably even the first question that was asked about this was off topic. However I do think it's very much worth asking about

A lot of people took decimals to be the definition of real numbers for most of their childhood without really considering what that entailed.

[dim widget]

[pastel cliff]

you got that real fasrt

[frigid lark]

One last question, in $\mathbb{R}$ does $\sum_{i\in \mathbb{N}} x_i$ mean $\lim_{n\to \infty}\sum_{i=1}^n x_i$?

[cloud walrus]

Parrot Tea

[rotund aurora]

[dim widget]

It depends on what you mean by $\mathbb{N}$ i.e. are you French or English speaking.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

But yes the idea that it's defined by a limit is correct.

[formal ermine]

my prof is doing an intro to classical ag

[frigid lark]

Whatever I want it to be at a certain point

[formal ermine]

schemes >

[rotund aurora]

Tbh I never write $\sum_{i\in \mathbb N} x_i$

[cloud walrus]

Croqueta

[rotund aurora]

because there is no order specified

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

and then things are dangerous

unless you prove absolute convergence and things

[frigid lark]

I do it when convergence isn't an issue

[pastel cliff]

https://math.stackexchange.com/questions/2419674/abi-is-algebraic-over-mathbbq-iff-a-and-b-are-algebraic-over-mathb
repost from yesterday but im confused about the backwards implication here - are all finite extensions algebraic

[frigid lark]

Yes

[rotund aurora]

guess its fine then

[dusty verge]

If we ignore him, then we'd ignore him saying to ignore him, meaning we don't have to ignore him

[frigid lark]

Say [K:k] = n, then 1,a,a^2,...,a^n are linearly dependant over k, for any a in K

[formal ermine]

finite => algebraic
algebraic and fg => finite

[dusty verge]

Are there stronger anologies between direct limits and "actual" limits than them just kinda feeling the same?

[dim widget]

But if I don't paradox myself, then I don't not not paradox myself, which is the same as (not)^i paradoxing myself for all i odd, which in the limit is the same as .....(not)(not)(not) paradoxing myself, but this is the same as ...(not not)(not not) paradoxing myself which is the same as paradoxing myself. But this is itself a paradox?

[dusty verge]

Nah

[frigid lark]

Is this what the category theorists' call abstract nonsense?

[dim widget]

Haha god knows what category theorists call abstract nonsense. This is what people call a "swindle." Sometimes it's based on doing something silly and invalid but sometimes it's genuinely correct reasoning.

[dusty verge]

Why are you assuming \sum^{\infty} log(not) is absolutely convergent?

[dusty verge]

Sorry meant to say log not

[dim widget]

An actual limit seems more naturally like an inverse limit not a direct limit to me. Also this should be in #category-theory I guess.

[pastel cliff]

so the idea here is that adjoining i to Q is a finite extension?

[oblique river]

Yes

And a and b apparently

(Whatever they represent)

[pastel cliff]

well a and b are just elements of Q

[oblique river]

Oh

[dusty verge]

I mean, that depends on how you look at it, since you could argue that a direct limit's a lot like the limit of a sequence, yeah?

[pastel cliff]

it's this

but dont the () mean "smallest subfield containing"

[oblique river]

But yes adjoining i gives a finite extension

[dusty verge]

Also, I hate how in math, a lot of the time, ideas make a ton of sense, but then people go and make them more general, which makes it more confusing

OH yeah, a direct limit is the limit of a direct system. You know, instead of a damn sequence

[oblique river]

?

[dusty verge]

It annoys me

Even though I understand why

[oblique river]

People dont just make up definitions for the hell of it

[dusty verge]

Yeah, it just annoys me when stuff is made more confusing because it makes the definition more general

[oblique river]

The reason it’s defined for a directed system is because that construction shows up in that form in other problems

So if you only defined it for a literal sequence then you wouldnt be able to apply it when necessary

[dim widget]

I don't really understand why a direct limit is more like a limit than an inverse limit except that the notation looks the same. This is not an argument imo

[dusty verge]

Uh, because you could take a direct limit by taking a sequence of groups with homomorphisms between them, and construct the direct limit from that

You can definitely make a direct limit out of a sequence, it's just that direct systems allow for more generality.

[oblique river]

And what’s wrong with that?

That generality is necessary

[dusty verge]

Nothing, it just annoys me because it's more effort to understand

[oblique river]

Welcome to math? Lol

[dusty verge]

Yeah, fuck math :(

[dim widget]

This is again just repeating the definition, it is not an argument for why it's analogous. In what category are you thinking that the direct limit of a sequence is the limit of the sequence in say R?

[oblique river]

Tteg i dont think they are on that topic anymore

I think they are now just complai ing about the definition of direct limit

[dusty verge]

Uh, because if you draw an anology between the sequence "looking" more and more like the direct limit itself the "farther" down into the direct system you get, that's analogous to the sequence getting closer and closer to its limit

[oblique river]

Nvm they still want to argur this

Halliday i was trying to give you an out

[oblique river]

You could say the same thing about inverse limits

Which means this isnt a good argument

[dim widget]

I think all of these various limits are just generalizations of all procedures in math which possibly involve doing an infinite process in math, and possibly not. They're not meant to just generalize limits of sequences in any way, they were far more directly motivated by certain operations in algebra.

[dusty verge]

Well yeah, it's an argument for a subjective opinion about one thing looking like another thing

[narrow brook]

I don’t really think of limits and colimits as limits (like in analysis). I just think of the universal property.

[dusty verge]

It's not gonna be rock solid. It's not necessarily even an argument, just an explanation of my pov

[oblique river]

In some sense i guess they are the same since direct limits and inverse limits are categorical duals

[dim widget]

Yes in any category which you construct in which the objects are elements of R and the morphisms are such that direct limits agree with limits of sequences, I'll show you another one where the same is true but with direct replaced with inverse.

[oblique river]

Yes i know that lol thats why i said it

[dim widget]

Yeah I was just unpacking what you said.

Since this is a 3-person conversation where one person is just learning category theory.

[dusty verge]

What makes you think I'm just learning category theory?

[dim widget]

This conversation?

[dusty verge]

Again, as I've said multiple times, I'm explaining my subjective perspective on the analogies I have in my head that help my understanding

How exactly is those analogies not being rigorous making you think I don't know what I'm talking about?

[oblique river]

For one, you started with this question

[dusty verge]

Also why the fuck does conversation get shortened to convo, when it realistically should be conver or conve or something?

[dim widget]

I didn't say you didn't know what you're talking about. Just that you might still be learning which is fine.

[oblique river]

Which comes across as you not knowing the answer and asking for feedback from someone who might know the answer

If someone asks “can someone help me understand the definition of a group” then it certainly comes across like they are just learning abstract algebra for the first time

[dusty verge]

Also, I'm pretty sure there's a pretty natural way of describing the limits of sequences as the direct limits

Which is more natural than taking the dual of it would be

[oblique river]

What way is that?

[dusty verge]

Attach each object to an element of the sequence, morphisms are "distance to the limit is smaller".

okay leq

[oblique river]

That presumes you know what the limit is to begin with

[dim widget]

This is a terrible definition, a.) it presuposes the existence of the limit in order to be well-defined, b.) then there are not necessarily morphisms $a_n \to a_{n+1}$

[dusty verge]

Fair

[cloud walrus]

Topos_Theory_E-Girl

[oblique river]

Ok i have to go to work. Tteg you can take it from here lmao

[dusty verge]

Yeah, the idea was more to take a sequence we can already describe and attach a category to it such that the limit of the sequence makes sense as the direct limit in the category

[dim widget]

Yeah I think there are likely infinitely many ways to do this since R is not naturally a category in any way shape or form. For each one of them there will be a dual notion which looks just as natural.

[dusty verge]

Sure

Although now I want to see if I can work cauchyness into this idea, because that fixes issue a. that you pointed out

Don't take this as me arguing my point or anything btw, I'm just curious about a math idea I had at this point

[dim widget]

I think maybe that is the way to salvage some educational mathematics from this discussion, as that's a very good basic exercise in understanding the process of "categorifying" certain classical constructions in mathematics and using the actual definitions of these concepts.

[dusty verge]

I mean maybe, but honestly it just seems like a fun exercise to me

[dim widget]

No one said (correctly) that educational things can't be fun!

[dusty verge]

Tons of people have said that

Including me a couple times at least

but sure I agree with you

[pastel cliff]

sorry to repeat but - my question here was that i was under the impression that Q(a,b,i) would mean "smallest subfield of Q containing a,b,i", but a,b,i aren't necessarily elements of Q so how can there be a subfield containing them

[dim widget]

It isn't really a well-defined notion unless a, b, i are considered as elements of an algebraic closure or some other over-field

[dusty verge]

i definitely isn't an element of Q

As a side note, writing i in lowercase is the hardest part about math

[pastel cliff]

so the () notation is just weird?

[dim widget]

The actual definition is that $Q((a_i))$ is usually just the algebra of elements $\sum_i b_{i, j} a_i^{n_j}$ where $b_{i, j} = 0$ for all but finitely many $i, j$ and $n_i \in \mathbb{Z}$, modulo some algebraic relations that the $a_i$ are assumed to satisfy.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

Also the $b_{i, j} \in \mathbb{Q}$

[cloud walrus]

Topos_Theory_E-Girl

[pastel cliff]

[dusty verge]

Polynomials in a_i, with possibly negative coefficients

[dim widget]

But like what are a, b in your example?

[dusty verge]

Algebras are just polynomials but meaner

[pastel cliff]

elements of R

made a mistake before

[dim widget]

Okay then since all the elements lie naturally in C the definition of Q(a, b, i) is the smallest subfield of C containing {a, b, i}

[pastel cliff]

ohhh

how do we show that it's a finite extension

[dusty verge]

It's gonna be a subfield of $f[a, b, i]$, right?

[cloud walrus]

Halliday

[dim widget]

What are a and b? Depending on what a, b are it may not be finite.

[dusty verge]

Idk if the degree of an extension is nice enough for that to help though

[dim widget]

Do you know that Q(a), Q(b) are finite for instance?

[dusty verge]

I think that's an assumption of the problem, based on the phrasing of the picture sent

[pastel cliff]

they're just arbitrary reals

[dim widget]

That can't be true

There must be more context not in your screenshot

[pastel cliff]

https://math.stackexchange.com/questions/2419674/abi-is-algebraic-over-mathbbq-iff-a-and-b-are-algebraic-over-mathb

[dim widget]

For instance if a = \pi Q(\pi) is not finite.

[dim widget]

Okay this is easier.

[pastel cliff]

a,b \in R

[dusty verge]

Woah, this guy managed to say something wrong on MSE without getting murdered

that's a first

[dim widget]

But a, b are actually in $\bar{\mathbb{Q}} \cap \mathbb{R}$ in your question which is significant.

[cloud walrus]

Topos_Theory_E-Girl

[dusty verge]

Oh, wait yup

Since they're algebraic over Q, they're the root of a polynomial in Q, so Q[a] and Q[b] are finite

[dim widget]

Anyway the actual exercise should be generalized. Suppose that a, b are algebraic over a field K, show that K(a, b) is algebraic over K.

[dusty verge]

I think the point of the exercise is to show that the algebraic closure of Q has to include complex numbers

Which is obvious if you know it already, but not if you don't

[narrow brook]

This isn’t true like C is not a subfield of \bar{Q}

[pastel cliff]

ok wait simpler question about the () notation again

F(x) \subseteq F[x]?

[lethal dune]

[pastel cliff]

my notes say that it's the fraction field of F[x]

ryu

[lethal dune]

x not algebraic?

[dim widget]

It is true when x is algebraic. It is an iff

[dusty verge]

You've done a classic blunder

[dusty verge]

The arrow goes the other way, it's that \bar{Q} is a subfield of C

[dim widget]

In that case x is transcendental, because it does not satisfy any (nonzero) polynomial by definition.

That is what it means for x to be an "indeterminate"

[dusty verge]

It satisfies the polynomial x - x

[dim widget]

That is just 0

[dusty verge]

:)

Everything's a root of zero

[dim widget]

I don't think that is helpful.

[lethal dune]

[dusty verge]

It tells us that F[x] / (0) is the quotient we get when we associate zero with zero

That's not abstract algebra

[vale raptor]

Whatd be the best channel for this then?

[narrow brook]

It’s probably musical isomorphism

[dim widget]

Not here, I would put this in #modeling or #advanced-analysis or #real-complex-analysis

[vale raptor]

Ty

Put it in #modeling

[dim widget]

That seems like the best fit.

[dusty verge]

So you're familiar with quotient rings and ideals, yeah? Well one way to think of the quotient ring over an ideal generated by an element is as the quotient ring you get when you set that element to zero. For example, if you take Q[x]/(x^2 - 2), you're basically taking polynomials in Q, except you're also stating that $x^2 - 2 = 0$, right?

[cloud walrus]

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

[dusty verge]

So basically what you're doing is taking the ring of polynomials over some field, and declaring that some polynomial is equal to zero

[dim widget]

Here are answers to all of your questions: first Q(a + bi) is algebraic iff a, b are algebraic. Suppose a, b are algebraic, then you can show that if the degree of the minimal polynomial of a is n and the degree of the minimal polynomial of b is m then Q(a, b, i) has Q-rank less than 2mn because you can explicitly make a spanning set of this size. Because Q(a, b, i) is finite dimensional over Q, you know that any element satisfies a polynomial, since the first (2mn +1) powers of that element are linearly dependent. In the other direction, if Q(a + bi) is finite dimensional, then so is Q(a + bi, i) by the same argument as the above (construct an explicit basis). But Q(a + bi, i) = Q(a, b, i) so that means that a, b are algebraic.

Now about your question about Q(x) for x some element of some field L such that Q \subset L. If Q[x] = Q(x) then there exists p = \sum_i a_ix^i such that px = 0, whence x satisfies a polynomial over Q. On the other hand this logic is invertible, so this shows that this is iff.

[formal ermine]

akckchkchually x^2 + 2 \equiv 0 🤓

[dusty verge]

you can't use this to add x to the field, because x is the thing you're defining the value of

[dusty verge]

akchkckckckckckckckhualy it's x^2 - 2

[formal ermine]

touché

[dusty verge]

But yeah, basically when you do a field extension, you add in an element x to your field, and then define its value by quotienting out the ideal of the polynomial you want it to be a root of. You can't define x to equal x, because then you just get the polynomial ring back, i.e. you get Q[x]/(x - x).

[dusty verge]

There, I made it helpful just to spite you

[dim widget]

In the statement a, b are real numbers which are algebraic. So all they're claiming is that (for a choice of complex place of $\bar{Q}$) $\bar{Q} = (\bar{Q} \cap \mathbb{R})[i]$ which is true.

[cloud walrus]

Topos_Theory_E-Girl

[narrow brook]

I was replying to this. But I guess this has two interpretations. I interpreted them as saying $\mathbb{C} \subset \bar{\mathbb{Q}}$ which is false.

[dusty verge]

Oh gotcha, I meant it includes /some/ complex numbers

[delicate orchid]

If they said “has to include THE complex numbers” then sure

But they didn’t

[dim widget]

I mean that's not the point of the exercise anyway even if it is interpreted in the way in which it was intended, I would just ignore it.

[dusty verge]

Is catking a good thing or a bad thing? Wew keeps doing it to me

[delicate orchid]

I approve of this message ^

[limber vale]

Why would b or c have to be in P

[delicate orchid]

It’s a prime ideal

[limber vale]

yes

it is

that is the definition indeed

thanks

https://tenor.com/view/facepalm-really-stressed-mad-angry-gif-16109475

[delicate orchid]

Np np

[dusty verge]

What's a group?

[long nebula]

felt this

[delicate orchid]

Who gon tell ‘im

[formal ermine]

an object in Grp

[delicate orchid]

A groupoid with one object

[dusty verge]

It's an integer module but more general

[long nebula]

A model of the theory of groups

[delicate orchid]

Absolutely no model theory allowed in these hallowed grounds

[dusty verge]

Wassat?

[long nebula]

A group

[rancid totem]

Does this work for showing a^2 = pb^2 is false for any nonzero integers a and b and primes p?

This should probably be in number theory lol, but I put it here because it's in the abstract algebra book im reading (dummit)

[dusty verge]

I deserved that, but also what is it?

[long nebula]

How much logic do you know

[dusty verge]

Divide by b^2

[long nebula]

Are you allowed to assume that each integer has a unique prime factorization?

If so then this seems fine

[dusty verge]

Not enough to know how to answer that question

[rancid totem]

yup

[long nebula]

Okay it basically just means a set that satisfies the group axioms lol

[dusty verge]

I can take the hard definition I promise

[long nebula]

Well

It would take a while

[dusty verge]

I'm using silence to tempt you into explaining a thing you find interesting

[long nebula]

This is literally what it means in this case though

[dusty verge]

Wait shit

[long nebula]

The theory of groups is just all first-order sentences which are true about groups

For example "for every element x, there is an element y such that x * y = e"

[dusty verge]

What's a first order sentence?

[long nebula]

Oh boy

Basically one that only uses "or", "and", "not", "for all", and "there exists"

:)

Gonna handwave the details

You can find them in any intro mathematical logic textbook

[dim widget]

We need a "CatChad" react on this server.

[dusty verge]

What's the point of that then?

[long nebula]

A model of the theory of groups is just a set (and corresponding definition of the group operation and the identity element) such that all of the sentences in the theory of groups are satisfied

[long nebula]

The point of what

[delicate orchid]

that's what catking is for

[dusty verge]

Actually, is it that plus undefined things?

The second question is a better one I think

[dim widget]

It's not the same though. There is a subtle difference

[dusty verge]

Like is it those "operations" or whatever they're called plus variables?

[long nebula]

Being able to use variables is included in being able to use "for all" and "there exists"

But also yes

There can't be any free variables if it's a sentence though

[dusty verge]

How strict are the rules here?

I'm assuming sentence has a particular meaning

[long nebula]

They are mathematically precise

Yes

It is a well-formed formula with no free variables

Maybe let's move to #foundations

[dim widget]

The era of abstract-algechill is at an end 🙌

[solar shore]

just wanted to say thank you to this section for helping build my confidence with abstract algebra

i was doing some more exercises and ive been feeling hella good about it because i compare my answers and i keep gettin em right

🫡

[long nebula]

(same ^)

det and chmonkey are the best

🥺

and tterra

and everyone else who answers my questions lol

[solar shore]

literally just having a community that answers is crazy to me

its so wild and so helpful and its not being spoonfed either

[pastel cliff]

will come back to this late but ty

[pastel cliff]

where did TTerra go anyways

[formal ermine]

wasn't he close to graduating

[long nebula]

They're too good for this place

Especially #discussion

[pastel cliff]

no access moment

i sent them a friend request but no response

sadge

[novel parrot]

How can we assume that B is generated be those elements?

[dim widget]

They are probably proving the reverse direction first.

[novel parrot]

oh

[dim widget]

wait no

I misread

[dim widget]

Finite is not the same as "finitely generated as an A-algebra"

[novel parrot]

in this context, they mean finitely generated as an A-algebra

[dim widget]

The A-subalgebra generated by b_1, \dots, b_n is by definition finitely generated by b_1, \dots, b_n

[novel parrot]

The A-subalgebra generated by $b_1, \dots, b_n$ is by definition finitely generated by $b_1, \dots, b_n$

[cloud walrus]

ActiveChapter

[novel parrot]

yeah i understand that

but not too sure how B is generated?

Heres the full thing

[south patrol]

if you're still interested in why we can assume those elements gen B as an algebra, the point is that the conclusions are unchanged if we consider B' = A[b1,...,bn] instead of B

[novel parrot]

i dont understand this

[south patrol]

Okay so like

[novel parrot]

B is just a ring with multiplication defined from A also?

[south patrol]

B' is the subalgebra generated by b1,...,bn

[novel parrot]

A-module but with multiplication with in B also

[south patrol]

the point is like

b_i is integral over A as an element of B' iff it is integral over A as an element of B

So we may as well make our life easier by passing to B'

[novel parrot]

oh is ee

finitely generated only means that the list of generating elements is finite right

[dim widget]

@novel parrot is becoming too powerful, we need to start giving them incorrect advice

[south patrol]

lol

[novel parrot]

hhaa

you can give me wrong advice after my finals 😄

[dim widget]

Yes, the list of generators "as an algebra" meaning allowing positive integer powers of those elements as well.

So to generate something as an algebra using an element a you can also use linear combinations of a, a^2, a^3, etc. over the base ring.

[south patrol]

Another way to phrase it is that the elements b1,...,bn generate B as an A algebra iff the map A[x1,...,xn] -> A sending x_i -> b_i is a surjection, lol

Which is often useful

[novel parrot]

oh

it says that B is finitely generated as A module, this doesnt include powers of stuff in B does it?

[dim widget]

Yes so this means that there is a finite list of elements b_1, \dots, b_n such that every element in B is a linear combination of the elements b_1, \dots, b_n with coefficients in A.

It is much stronger of course because A[x] is finitely generated as an algebra but there is no finite list of polynomials which span all polynomials (since any finite list will be missing polynomials of a high degree).

[novel parrot]

so we have two different definitions of finite algebras?

[dim widget]

One is "finite" and one is "finitely generated"

Some people accidentally mix them up but it is really good practice to keep them straight in your head.

[novel parrot]

i see

[novel parrot]

i like this definition

[formal ermine]

"two complexes f, g : A^* -> B^* are homotopic if there is a collection of morphisms k^i : A^i -> B^i - 1 for each i such that f - g = dk + kd"

what does the f - g = dk + kd mean?

[dim widget]

It is just equality of morphisms? Are you asking for intuition?

[formal ermine]

I don't understand the notation

[dim widget]

Ah okay so maybe here is what is confusing:

[formal ermine]

what does the + mean? union?

[dim widget]

call a morphism of cochain complexes a collection of maps f^i: A_i \to B_i such that df^i = f^{i+1}d'

[delicate orchid]

How do you take the union of two functions lol

[formal ermine]

yes that's my question

f and g are sets

[dim widget]

No the + is just addition.

[delicate orchid]

Are they?

[formal ermine]

[dim widget]

Yes it is a common abuse of notation. We think of them as tuples of morphisms.

[formal ermine]

the - is component wise?

[dim widget]

Yes exactly

[formal ermine]

ah I see

dk is d circ k?

[dim widget]

And composition is done in a funky way so that formulas make sense

[dim widget]

Yes but what that means is a little unclear

Since d is degree-shifting

and so is k

[formal ermine]

yeah I can guess what it means sort of

[dim widget]

Yeah you just compose in the only logical way.

[formal ermine]

yip

hi det

[elder wave]

what are you reading

[rustic crown]

homowogicaw awgebwa

[formal ermine]

I am trying to procrastinate my german homework so I'm reading the intro to derived functors in hartshorne

[rotund aurora]

functor up your german teacher

[formal ermine]

we are doing nature lyric

idk if that makes sense in english

like poetry about nature

from 400 years ago

[rotund aurora]

yes

locus amoenus

[formal ermine]

goethe n schiller

[rustic crown]

thinking about indices is annoying, so it's always nicer to write something like f-g = dk+kd, and this is uwu because you can think of d as a map d : A --> A[1] or d : B[-1] --> B and k : A --> B[-1] etc

(ofc these are not (co)chain maps)

[pastel cliff]

[rustic crown]

amoenus is all i see >.<

[dim widget]

(For the record the notation $A[j]$ means the complex $A[j]^i = A^{i+j}$)

[cloud walrus]

Topos_Theory_E-Girl

[delicate orchid]

Just draw the squares….

[dim widget]

Idr if that's in hartshorne

[rustic crown]

(and some people might like to swap the sign of the differentials by (-1)^j)

[formal ermine]

perhaps a stupid question but sqrt(f) = sqrt(g) iff f = g right?

[dim widget]

no

[lethal dune]

Just ignore signs, they don’t affect the homology and annoying to keep track of

[rustic crown]

but they affect +

.<

[formal ermine]

huhh

[rustic crown]

(non-negative reals? )

[formal ermine]

ring elements

sqrt is radical

lol

[lethal dune]

Lol

[rustic crown]

(yee that was the joke >.<)

[lethal dune]

[dim widget]

?

But that's not true either

[agile burrow]

are you asking when the radical of the ideal generated by f is equal to the radical of the ideal generated by g?

[formal ermine]

wait

ok so in my case

I want to show that Z(f) = Z(g) iff f = g

[lethal dune]

#real-complex-analysis

[rustic crown]

Z(f) = Z(f^n)

[dim widget]

What about g = f^n

That's the whole point of the radical.

[lethal dune]

Society is it was true

[formal ermine]

istg this is what my prof wrote

oh I forgot to mention

[rustic crown]

are f and g like irred?

[formal ermine]

f and g have no multiple factors

like every factor has multiplicity 1

[dim widget]

Are you sure they didnt write Z(f) = Z(g) iff \sqrt(f) = \sqrt(g)?

[novel parrot]

So if B is a finite A-algebra, then all elements of B are integral over A (A is noetherian)?

[formal ermine]

there was this proposition

which apparently immediately implies this

but I cannot see how

[formal ermine]

it's:

"let f(x, y) in C[x, y] \ C without duplicate factors. then I(Z(f)) = <f>"

[dim widget]

Catking-ing yourself is v catking energy

[novel parrot]

This result feels powerful

https://tenor.com/view/ok-okay-gokuok-goku-danielrds-gif-20976699

[rustic crown]

https://tenor.com/view/yui-hirasawa-yui-hair-dryer-kon-gif-22748047

super saiyan yui

[novel parrot]

lol

[dim widget]

So you can prove this by induction on the number of factors. Can you see why it's true for irreducible elements? Alternatively you can use that Z(f) = Z(g) iff rad(f) = rad(g)

[elder wave]

show me

[dim widget]

They wrote it above.

[rustic crown]

hauptideal = principal ideal

det can read (that was a lie)

[elder wave]

Notation verwirrt mich grade, sekunde

[formal ermine]

irreducible components are right after this, how would I do it without that

[formal ermine]

mich auch

[formal ermine]

like with those it's obvious

[dim widget]

You don't need that, just that C[x, y] is a ufd.

[elder wave]

ahh this is what I_f is

[dim widget]

You can get around actually having a definition of irreducible components that way

[formal ermine]

<f>

[elder wave]

ye

[formal ermine]

yeah for irreducible elements it's obvious innit

they're just x - something

and if they share the same root then they must be equal

[dim widget]

Be careful! That's not true.

We are in the world of two dimensions

[formal ermine]

I forgor

[dim widget]

Although yes I suppose they are all x - (x - P(x, y)) where P is irreducible

so you were not wrong

[formal ermine]

what are the irreducibles? like I would've thought they are just x - a and y - b

or wait

[dim widget]

That's a very good question! Can you think of why that is not true?

[formal ermine]

I see the issue

[rustic crown]

https://tenor.com/view/anya-spy-x-family-anya-forger-anime-spy-family-gif-25843273

[formal ermine]

my last guess

x^a y^b + k x^c + l y^d + m

with like

if k = l = 0 then m must not be 0

[rustic crown]

[formal ermine]

how wrong is it

frick

[delicate orchid]

What’s the ring :pack:

[formal ermine]

I forgot about stuff like x^2 + y^2

[formal ermine]

C[x, y]

[dim widget]

I am not even really following your thought process here, some of these are definitely reducible.

Here is another good example of something irreducible y^2 = x^3 + x

[rustic crown]

that be like (x+iy)(x-iy)

[formal ermine]

yea

[rustic crown]

and in general homoegeneous poly would factor because you can like do stuff with y/x

[dim widget]

Anyway basically the point is that every irreducible algebraic shape in the plane gives you a new irreducible polynomial

So there are lots

Fortunately to solve this problem we don't need to classify all of them

[delicate orchid]

Spec(C[x, y]) 😵‍💫… wacky set

[dim widget]

But any argument you make assuming they're all lines is doomed to fail

[delicate orchid]

Consider the irreducible affine varieties in C^2 or something along those lines

[formal ermine]

returning to this

how do we do it

with the irreducibles

[rustic crown]

is there a way to do it, or like you do it in R^2 and pretend that R is C

oh you wrote consider

.<

det sweepy >.<

[dim widget]

,ti det

[cloud walrus]

68 members found matching det!

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[formal ermine]

gmt+1

[pastel cliff]

,ti topos_theory

[cloud walrus]

The current time for Topos_Theory_E-Girl is 07:19 PM (BST) on Mon, 24/04/2023.
stμ₂dying is 5 hours behind, at 02:19 PM (EDT) on Mon, 24/04/2023.

[dim widget]

That's actually incorrect, i'm in cali bb

[cloud walrus]

Member selection timed out.

Apr 24, 2023, 6:19 PM

[rustic crown]

,ti

[cloud walrus]

The current time for det is 08:19 PM (CEST) on Mon, 24/04/2023.

[rustic crown]

det slept 3 hours yesterday night >.<

[pastel cliff]

det :(

[dim widget]

Okay so can you show that two distinct irreducible factors have distinct zero sets?

I guess I'm saying the first thing is to prove it when f, g are irreducibles

[formal ermine]

isn't that true for every polynomial?

[dim widget]

Poor thing 😦

[dim widget]

No we just went over this. Like if f = g^n etc.

[formal ermine]

nvm lol

yeah I see

[dim widget]

Like I mean if P, Q are irreds and P \neq \lambda*Q you have to show that Z(P) \neq Z(Q) and vis versa

[formal ermine]

is it like

f = g <=> Z(f) = Z(g)

=> is trivial

for <= we know that f and g are separable (because char 0) so they are uniquely determined by their roots?

[dim widget]

I think this is begging the question personally. What do you mean that they are uniquely determined by their roots?

Can you phrase that in such a way that it does not mean just "Z(f) = Z(g) => f = \lambda g"

[formal ermine]

can't we wlog make f and g monic

[dim widget]

We can yes

Might as well

Although what you mean by monic has to be carefully defined

[formal ermine]

nvm the nvm the nvm

wait nvm that again

so f and g are both irreducible

so Z(f) and Z(g) are both irreducible?

then f and g are prime so I(Z(f)) = sqrt(f) = f

and thus f = g?

[dim widget]

I mean basically that I(Z(f)) = sqrt(f) is tantamount to what we're trying to prove. Do you have the nullstellensatz already?

[formal ermine]

oh lmao

[dim widget]

Sorry I guess if you're german you probably write it as zero star theorem

[formal ermine]

it immediately follows from that 🤦

why did my prof write that it follows from that one weird proposition

[formal ermine]

cuz like

we have that g | f^n and f | g^m but since they do not have any repeated factors it must be g | f and f | g => f = g

[dim widget]

yep

[formal ermine]

why did I not realize this 40 minutes ago

thanks for the help tho topo

wait

what do I call you

topo is already reserved for toposphere

[next obsidian]

E-Girl

[delicate orchid]

That’s not his name lol

[dim widget]

People have been going with tteg

[formal ermine]

lol

mandela effect

[delicate orchid]

lol

[next obsidian]

Litwrally@only you did that

Kekw

[formal ermine]

I swear to god his name has always been toposphere

[formal ermine]

tteg

epic

[dim widget]

I like Topos though!

fwiw

[formal ermine]

time for curves in projective space

[next obsidian]

Ummmm the plural of topos is topoi (or also toposes???)

[dim widget]

No I mean as a name lol

[next obsidian]

It was a joke

that’s why I had 4 m’s

Kekw

Walter

[dim widget]

Haha there is too much unironic mansplaining it's hard to sort out

[formal ermine]

hmm

[agile burrow]

I'm projective

[formal ermine]

how does my prof's definition of irreducible algebraic set match up with the one I'm used to

[next obsidian]

He has a very ample line bundle

[formal ermine]

his is that Z(f) is irreducible if f is irreducible

but how does f generate a prime ideal

[next obsidian]

Aren’t you working over a field?

That’s a UFD

[formal ermine]

I thought it was only true in pids

[dim widget]

Again f should be squarefree!

[formal ermine]

the definition here doesn't say that 😭

[next obsidian]

People just don’t care very much about non-reduced stuff when dealing with varieties

[dim widget]

Ah wait there's only one f

[next obsidian]

And then just say sfuff without thinking very hard kekw

[dim widget]

I thought it was iff

[next obsidian]

It feels like that’s suppoEd to be a definition tho

So I would agree you should worry about reducedness

[formal ermine]

"Z(f) is called irreducible if f is irreducible"

[dim widget]

It would be better to say

"Z is called irreducible if Z = Z(f) for some f irreducible"

[delicate orchid]

Ermmmm excuse me please take this conversation to #algebraic-geometry thank you

[formal ermine]

scary place

[dim widget]

Tbf last time @formal ermine posted there they got yelled at.

[formal ermine]

...by you

[next obsidian]

My professor put a HW which in scheme language was that given varieties Spec A, Spec B, Spec C and dominant maps, the fibered product as varieties is Spec A (x)_C B

Which is just wrong because you can pickup nilpotence

[patent ocean]

I've seen answers using rings and ideals but isn't it easier to show this by using the fact F is a field.

[formal ermine]

yeah

k[x] pid iff k field

[next obsidian]

He just figured it’s probably true and thought you can emulate the proof to show product of varieties is reduced

Illu, that doesn’t prove anything

Unless you show that irreducible elements generate maximal ideals in a PID

[delicate orchid]

….

[dim widget]

To know that you want to use that F[x] is euclidean

[patent ocean]

oh wait I confused F with F[x] oops

[dim widget]

Or at least that is the easiest way

[dim widget]

F[x] is not a field.

[patent ocean]

yeah i realised

[next obsidian]

Topos who is your pfp of

[patent ocean]

i'm revising and forgot definitions

[dim widget]

Sophie Scholl

[next obsidian]

Who’s that

Oog

[formal ermine]

oh I see where my misunderstanding lies

at the beginning he once again said "f has no duplicate roots" lol

[next obsidian]

[patent ocean]

sorry just going though definitions again, why is F[x] not a field?

[next obsidian]

Varieties language really is unwieldy sometimes because of nilpotence

[next obsidian]

x has no inverse

[patent ocean]

oh yes polynomials

thanks

[formal ermine]

so yeah sqrt(f) = { g in A | g^r = f^n for some r,n > 0 } but because f has no duplicate roots g must be f so sqrt(f) = { f } which is a prime ideal

I should start using punctuation

[formal ermine]

wait this feels really really wrong

[next obsidian]

There’s no way you can show that g is f

Write g and f in terms of irreducible components

Also that’s not the definition of the radical of an ideal

sqrt(f) = {g in A| some n so that g^n in (f)}

Write down f and g in terms of irreducible components

[formal ermine]

nvm

bro what is wrong with me today

[next obsidian]

Ideals have to be closed under multiplication by A

[formal ermine]

yeah

[next obsidian]

And write down g^n = r•f

See that all irreducible components of f show up in g, but we can only say they show up with multiplicity 1

If f is square free, then that’s enough to see f divides g

[novel parrot]

how is V = that direct sum without the free part?

V is torsion?

[dim widget]

V has to be torsion because it's finite dimensional over k

So End(V) is a finite (dim(V)^2) dimensional vector space

If there were a torsion free part then k[X] would inject into End(V)

Alternatively you could just say that X satisfies its characteristic polynomial in End(V)

So you know the map cannot be injective.

[dim widget]

.

[novel parrot]

im not so well versed in linear algebra

but if V if finite dimensional over k

then doesnt that mean that V has to be torsion free

[dim widget]

It is torsion free over k

But not over k[X]

The only finite dimensional (over k) modules for k[X] are the torsion ones.

Because torsion free ones contain k[X] as a sub-k-vector-space

Which is infinite dimensional

[novel parrot]

i didnt understand why V over K[X] is torsion 😄

is it because of the cayley hamilton theorem

[novel parrot]

i think this is the cayley hamilton theorem

[dim widget]

yes

[novel parrot]

thats 0 on endomorphims

[dim widget]

Yes \chi_A(A) = 0

[novel parrot]

and $\chi_A(X) \in K[X]$

[cloud walrus]

ActiveChapter

[novel parrot]

the f_i are all non constant because other it would be a unit and then R/(f_i) = 0 and we just remove them right

[dim widget]

Yes constants are units unless they are zero

In all rings

#all-rings-are-fields

[novel parrot]

yep

not quite sure on how we get that sum that dimV = sum of the degree of the f_i

[dim widget]

V = \oplus_i k[X]/f_i, now count dimensions

[novel parrot]

$V = \oplus_i k[X]/f_i$

[cloud walrus]

ActiveChapter

[novel parrot]

ah

stupid question

[patent ocean]

isn't a subring automatically an ideal?

[novel parrot]

no

[patent ocean]

actually i see a counter example I can't read definitions

[hollow mica]

an ideal is not a special type of subring

[fleet pelican]

ideals don't have identities, subrings do

[novel parrot]

not all rings have identity right

[lavish nexus]

rng

pronounce it however you will

[novel parrot]

some text say that identity is optional in rings

[fleet pelican]

Ew

[hollow mica]

unital ring

[fleet pelican]

🤮

[rustic crown]

all spaces are compact

[novel parrot]

I dont understand the last part, how is f_i are power of an irreducible?

we can factor f_i into irreducibles

[rustic crown]

by CRT you can break each f_i into prime powers

[novel parrot]

whats CRT

[rustic crown]

chinese remainder theorem

[novel parrot]

oh

[rustic crown]

[novel parrot]

so is $R/(f) \oplus R/(f) = R/(f^2)$ ?

[cloud walrus]

ActiveChapter

[rustic crown]

nu

can only break if they coprime

if I + J = 1, then R/IJ = R/I x R/J

[novel parrot]

oh

understood

[novel parrot]

it says that either one of the two is true

[hot lake]

not at the same time

[novel parrot]

when we break into powers of f_i, we possibly lose the first condition right

[hot lake]

but either condition makes the decomposition unique

otherwise you can have plenty of ways to combine the factors

[novel parrot]

this is how to lose the first condition by turning into powers right

[dim widget]

https://www.youtube.com/watch?v=ST42VjR0LYk

[patent ocean]

multiplicative identity?

[patent ocean]

does the hint even help for the proof? My problem is that what is the two groups are not even closed under multiplication.

in this case we are fine but for a general set of abelian groups how does this make any sense

[dim widget]

Eh? Why not?

[patent ocean]

like what if the abelian group is not even closed under multipication

[next obsidian]

This is impossible

Definitionally

[patent ocean]

so an abelian group must be closed under + and x ?

[next obsidian]

Asking a group to be closed under x doesn’t make sense

But it’s a ring

[dim widget]

No but if that abelian group is a ring R or an ideal in that ring, or the quotient of R by that ideal, then it is closed under the multiplication coming from the ring.

[next obsidian]

So x is an operation on it so it’s closed

[patent ocean]

oh i thought the hint was referring to a random abelian group and not just a ring

[delicate orchid]

the hint is basically asking you to show that the standard isomorphism in the first isomorphism theorem extends to a ring homomorphism

so, if f is that isomorphism

not only do we have f(x+y) = f(x)+f(y), but we also have f(xy) = f(x)f(y)

[patent ocean]

but doesn't that just follow from phi being a ring homomorphism

😕

[next obsidian]

I mean basically

The point is f is constructed such that this suffices

But there’s something you need to check

[delicate orchid]

this is what you have to show

[patent ocean]

oh so i just need to construct phi(r1+kerphi) -> phi(r1) ?

then it just follows?

[next obsidian]

Do you know the first isomorphism theorem

For groups

[patent ocean]

yh it's similar just with G instead of R i think.

Have not done groups in a while

[next obsidian]

Do you remember what the specific isomorphism is

How the map is defined

[patent ocean]

T: g+H -> T(H) ?

T(g) *

[next obsidian]

What’s T

I’m very confused

[delicate orchid]

he named the isomorphism after himself

[patent ocean]

phi : g+H -> theta(g)

where theta is the homorphsim defined initially

is that right

[next obsidian]

Right, yes

So the point now is

If you have a map of rings R -> S

This is in particular a map of abelian groups, just forgetting R and S are rings right now

So we get a map f:R/ker phi -> im(phi)

This is an isomorphism of groups which means f(x + y) = f(x) + f(y)

And it’s bijective

So now all you need to do is verify f(xy) = f(x)f(y)

So then you unwind the definition of f, if = g + ker phi

[patent ocean]

which follows from the intial homomorphism phi right?

[next obsidian]

Then f(g + ker phi) = phi(g)

Yeah

Because f is defined as it is

[patent ocean]

yeah

that's what i was trying to get at but ig I didn't explain it well

[next obsidian]

I see

[patent ocean]

so whenever we have an isomorphism between two abelian groups this is true?

[next obsidian]

No

[patent ocean]

^that's what I got from the hint so I'm asking did I not understand the hint or is it just not clear

[next obsidian]

It isn’t true in general

But it’s true here because f is defined in terms of phi basically

But you can have non-multiplicative maps between rings which are isos of their underlying abelian group

[patent ocean]

oh ok

so they just wouldn't be a ring homomorphism since it doesn't have the "multiplication-homomorphism" property?

[next obsidian]

Sure that’s one way to say it

[patent ocean]

🤣 appreciate the help

very helpful and good explanations

sorry guys watching the chat that was probably hard to watch 😅

[rapid junco]

How to prove that every linear operator over an odd dimensional subspace has a one dimensional invariant subspace?

[dim widget]

Over the real numbers you mean

[rapid junco]

yes

[dim widget]

So what is a one-dimensional invariant subspace?

[rapid junco]

span of an eigenvector

[delicate orchid]

eigenvector, surely

[rapid junco]

lmfao

yes

[pastel cliff]

that has to be the quickest honorable ive ever seen

[delicate orchid]

believe it or not it's taken days for the "internal moderator debate " for this honourable

[pastel cliff]

honorable*

sorry typo

[dim widget]

My weekend what has happened to my poor weekend

I was supposed to be working way harder on my revisions

[delicate orchid]

welcome to the maths discord blackhole

[dim widget]

Okay so when does a matrix have an eigenvalue?

[rapid junco]

when its shifted kernel by the eigenvalue is non trivial

[dim widget]

Okay and can you think of a more computable way to check that?

Like something that can be computed efficiently that tells you when Ker(T - a) is nontrivial?

[rapid junco]

when its characteristic polynomala has a root

[dim widget]

Yes, so when do polynomials over R definitely have a root?

[rapid junco]

oh fuck i see

odd degree

[dim widget]

I also ask that when I ask these questions you respond with "yes socrates it is evident"

[rapid junco]

wtf

[elder wave]

Lmao

[rustic crown]

the notations definitely do :p

[dim widget]

Unfortunately.

[fallen trout]

hey, i'm confused about an aspect of a proof. why does b not a unit imply that there is a prime elt p st p|b?

this is in a unique factorization domain, so i assume that causes it somehow?

OH

ok yeah nvm it's trivial

[next obsidian]

Swag

[mild valley]

Do y'all have any intuition for the antipode of a Hopf algebra, I understand the definition and can see a few examples, but I fail to get it's place and necessity in the definition of a general Hopf algebra

[dim widget]

In what context are you seeing them? If it's in topology you should maybe ask in #point-set-topology to get better answers.

The antipode is just the "coalgebra" version of the map which sends an element of a group to its inverse.

[mild valley]

I would personally say category, but I'm reading through Kassels Quantum Groups so the book is more about representations from what I can tell

So my best attempt at intuition would be similar, that it's generalizing retaining the inverse maps of groups after pushing them to algebras

[dim widget]

Okay maybe I can say something a little more precise. Hopf algebra structures show up on objects which are "dual" to group objects in certain categories (in particular contravariant functors from objects to algebras). The two key classes of examples to start with are the ring of functions on an affine algebraic group scheme over a field and the cohomology of a topological space which is a group (or even just an H-space).

In both cases the comultiplication is induced by multiplication on the group space/group scheme, the antipode is dual to the inverse map, and the counit or whatever its called is induced by the identity of the group.

So hopf algebras are algebras which are somehow associated to some "group-like" object.

[white oxide]

Hi all, I'm trying to prove that if E is an extension field of F and F(a1, a2, ..., an) denotes the smallest subfield of E that contains F and the set {a1, a2, ..., an}, then it is equal to the intersection of all the subfields of E that contain F and the set {a1, a2, ..., an}. so i'm doing this by showing that they're subsets of each other; that is, i'm letting x in X1 \cap X2 \cap ... \cap Xn where each Xj is a subfield of E that contains F and the set {a1, a2, ... am}. My issue is that if I let x in this intersection, it need not be a1 or some element of F, because what if it's some element e in E that is not in the subfield F(a1, a2, ..., an)? what appreciate a hint plsss

right now i'm going for contradiction and assuming x is not in F(a1, a2, ... an), and since i'm letting x be arbitrary maybe i can show that that statement is not true for all x (ie i can let x equal some aj) and that will prove one side of the inclusion idk

the other inclusion is obvious

[mild valley]

Ah okay, thank you for the detailed answer! This makes some more sense, but I'll have to think about it and go through those examples you suggested lol

[west valley]

are adjacent transpositions in $S_n$ reflections?

[cloud walrus]

Adam

[dim widget]

What do you mean adjacent transpositions and what do you mean reflections?

[west valley]

when you swapped two adjacent numbers in a permutation

i think reflections are generators for the group

[dim widget]

I mean there are many sets of generators for the group

But the (i i+1) transpositions form a very natural set of generators from the point of view of the root system of Gln. Maybe that's your definition of a (simple) reflection?

[west valley]

only reason i ask this cause i saw this in my textbook

yea simple reflections sorry i forgot thats the common terminology

[mild valley]

You can view Sn as acting on an n-dim vector space by permuting basis elements, then transposition give you so called permutation matrices which are in fact reflections

If that's sort of what you're after

[west valley]

that kinda helps, do you know where i could find some visuals to aid that?

my professor went over this and i didn't understand it well

[mild valley]

I'm not so sure about visual aids, but here's a write up by Keith Conrad:
https://kconrad.math.uconn.edu/blurbs/grouptheory/gpaction.pdf
He has a lot of writing on a variety of math topics, and in my opinion he's very good at explaining them all

[west valley]

appreciate it, thanks!

[empty rose]

visuals might be difficult for spaces that have more than 3 dimensions which is most of them

[south patrol]

Lol

[white oxide]

how do we know that F is a two-dimensional vector space over Z3? is it because we can regard {a, <x^2 + x + 2>} as a basis?

for a in Z3

[south patrol]

It is a standard fact that if f is a degree n irreducible polynomial over k then k[x]/(f) is a field and an n dimensional k vector space with basis 1,x,..,x^(n-1)

[white oxide]

it's clear to me that it's a vector space, just not sure of why it's 2-dimensional

[white oxide]

oh that theorem is on the next page lmao

[south patrol]

Oh lol

Also, (x^2 + x +2) is the zero element

So like it can't be in any basis

[white oxide]

oh right oops

[south patrol]

Here a basis is given by (the image of) 1 and x

[white oxide]

how would i figure it out without that standard fact

oh i see

[south patrol]

Uhh I mean yeah you can use 1 and x as a basis, using the polynomial to reduce higher powers of x to smaller ones

Or use euclidean division

[white oxide]

oh ok

would {1, 1 + <x^2 + x + 2>} work?

[wraith cargo]

These are the same element in the quotient

[white oxide]

also how does it being a vector space over Z3 imply that it's a splitting field

[white oxide]

huh

i thought that was like

1 in Z3

vs 1 in Z3[x]/<x^2 + x + 2>

i guess isomorphically they're the same

$1 \mapsto 1 + \langle x^2 + x + 2 \rangle$ right

[cloud walrus]

okeyokay

[wraith cargo]

take the basis to be {1 +<x^2 + x + 2>, x + <x^2 + x + 2> }

then since x^2 + <x^2 + x + 2>= -x-2 + <x^2 + x + 2>

[south patrol]

An element of a basis of a vector space has to be an element of the vector space...

[wraith cargo]

you can reduce all higher powers to the linear term

[white oxide]

are you talking about me wondering if 1 would be part of the basis

wait now i'm confused because i thought 1 in Z3 was included in the extension field F

so it would be a part of the vector space right

[white oxide]

ahh ok that makes sense

[white oxide]

ah i got it

thank uu

[wraith cargo]

it contains both zeroes of the polynomial

the fact that it's a vector space over Z3 is just there to say like yeah this field is an extension over Z3

[white oxide]

yea but wouldn't it have to be the smallest field that contains the zeroes of the polynomial and Z3

so like how does it imply that it's the smallest

like i know it contains Z3 and the zeroes

[white oxide]

ah ok

idk why they did that then lol since it's clear that it's an extension by the construction featured earlier in the chapter

[wraith cargo]

sorry fuck I got in some weird headspace cuz my prof defined splitting fields to be any field that contains all the zeroes over a polynomial so I have to get used to the usual definition

[white oxide]

oh yea no worries

maybe i can try to show it on my own

is it because like

[wraith cargo]

OK yeah I don't think it's hard
firstly note that it contains all zeroes
And also take note of the fact that it's two dimensional

[white oxide]

the vector space is 2-dimensional, so the basis is of length 2

and so by definition the field would have to contain all linear combinations/multiples of the zeroes

which theare are two of them

[wraith cargo]

Think of it more like

is there a smaller field than this that contains both zeroes

[white oxide]

hmmmmmmm ok

thanks for the help! ppreciate it

[wraith cargo]

npp

[jaunty glacier]

Why is a necklace a group action on the the Cyclic group?

[rapid junco]

Suppose X and Y are two 5 dimensional subspaces of R^8, show that X int Y != {0}