groups-rings-fields

[solemn rain]

i mean

it's not that kinda of a big deal

its bad tbh

but i mena

meh

as long as u get the idea its cool

its bad yea but like

i mean who the fuck cares am i right

[vocal depot]

My professor might

[solemn rain]

then thats up to you

to find out

if ur asking wether this is osmething to ask the prof for

id believe it is

1+ is not the same as 1-

[vocal depot]

1+ is not the same as 1-
@solemn rain

Yeah, that's why I thought I'd ask before I proceeded.

I'll just run it by my professor then.

[slate forum]

Anyone know how to show that
$\prod_{i=1}^\infty Hom_\mathbb{Z}(\mathbb{Z},\mathbb{Z})$ is not isomorphic to $\bigoplus_{i=1}^n \mathbb{Z}$.

[cloud walrus]

HelixKirby:

[slate forum]

I was thinking of saying that the former has an element that just has the identity in every slot, whereas the other one has to only have finitely many nonzero terms

[sour plume]

I think that train of thought is too easy, since in principle, an isomorphism from one to the other could map "finite" elements to infinite ones

I think one way to see it is by seeing that, as Z-modules, the infinite product is not free (abelian), but the infinite direct sum is free (abelian). But showing that the infinite product is not free is not trivial

@slate forum blub blub

[supple marsh]

Isn't the product uncountable, while the sum is countable?

[slate forum]

Nah, they are both countable, it's just the sum has only finitely many nonzero terms

[sour plume]

(i think you made a typo in the upper bound of the direct sum)

But yeah, that might actually be a better argument, because I do think the direct product is uncountable, it's basically the space of all integer-valued sequences, so it's baaaaasically the space of decimal representations of real numbers

(stretched out baaasically because you'd need to keep track of 0.99999... = 1 and shit but yeah, it's well known that the space of all sequences like that is uncountable)

[chilly ocean]

@solemn rain yo homeboy Ive got a question

[queen vine]

@slate forum the product is uncountable by Cantor's diagonal argument. It follows that it can't be isomorphic to the direct sum, one can also show that it is not even free.

[slate forum]

Did I construct this map correctly?

[coarse storm]

There is some variable mix-up, making it a but confusing.

[stone fulcrum]

Why do you think it isn't?

Actually, looking at associativity,
(ab)×s = sab
a×(b×s) = a×sb = sba

Does that look right, or am I mixing this up at all?

I'm using × as the operation between the group and the set

And concatenation as the group multiplication

a × b makes no sense, in that case

Well, it does cuz it's all the group, haha yeah that's confusing

But let's pretend s was a member of a set and not a group for the logic

(ab)×s = a×(b×s)

Because b×s is a set member not a group member

Yeah this question is confusing in particular as the action is the group on itself

[golden pasture]

S_n and A_n?

[smoky cypress]

Can someone give me the definition of these two modules

For 4) I’m not quite sure what the operation is. For 5) I don’t even know what a group-algebra is

[carmine fossil]

I am guessing something like x.a being T(a)

For 4)

[smoky cypress]

Oh probably

[supple marsh]

@smoky cypress https://en.wikipedia.org/wiki/Group_ring#Definition its the same thing basically, just remembering the map from k into it makes it a k-algebra

[mint gulch]

Hi

I have a question about Group Rings. i'm reading Dummit and Foote Abstract Algebra and see this

Mi book says the grouo Ring RG is conmutative if and only if G is a conmutative group

Is that true?

I think it don't have to, if R isn't conmutative

I thought in two elements in RG like

a_1e and b1e, the rest is 0, their product is a1b1e, and b1a1e

It doesn't has to be equal

I saw this in Wikipedia too

That's what I think it is

[wind parrot]

It says at the top "fix a commutative ring R"

I.e they already assume the ring is commutative

[mint gulch]

I see, hahaha

I got distracted for "Conmutative of R is not needed" at the end

Thanks

[dawn kiln]

remember hearing about something

take an abelian group (G, +)

and consider End(G)

we can form a ring from End(G), by using + as the (commutative) additive operator, and function composition as the multiplicative operator

with 0: x --> 0, and 1: x --> x as the additive/multiplicative identities

[dawn kiln]

are there any interesting things you can do with this construction?

it seems like End(Z_n) = Z_n, ie. for cyclic groups

End(Z) = Z too i think

and are all rings representable like this?

[next obsidian]

@dawn kiln You can frame a module as being a ring map from a ring R to End(M) for an abelian group M

Let phi be this map, then you define rm = phi(r)(m), and you can see this is the exact same thing as a module structure

This let's you really easily see when an R module is still a module over a quotient of R, indeed you can even apply the first isomorphism theorem

In fact, if you consider M as a module over R/ker phi it's torsion-free, and if R/ker phi is a PID this tells you that M is a free R/ker phi module as long as it is finitely generated via the classification.

ker phi here is just the annihilator of M

[dawn kiln]

ooh that's cool

[next obsidian]

You can admittedly do the same thing showing any R-module is a module over R/Ann(M) but you'd manually have to verify well-definedness but this construction gives it to you for free

granted, it saves you maybe like 15 seconds of work lol but

[dawn kiln]

haven't come across modules much though, so need to google some terms haah

[next obsidian]

I think this is also important for representation theory

Specifically in this case the representation theory of (finite) groups

Specifically you can do stuff relating modules over the endomorphism ring of a group to the group or something

I don't really remember too much tbh 😅

[prime gale]

Doesn't answer question though.

At least not in a simple enough way for me to understand.

[dawn kiln]

it answers the "interesting things to do" part lol

[prime gale]

I see. I was wondering about the other part. Turns out I don't really know anything about infinite abelian groups.

[coarse storm]

You know that they are infinite!

[slate forum]

Can someone help me come up with an example of a zariski continous map that doesn't take closed sets to closed sets?

In particular I need it to be polynomial

[queen vine]

Think about $xy=1$

[cloud walrus]

leoli1:

[slate forum]

That's still polynomial, right?

Is this the map, or is this the closed set?

[full holly]

i'm looking for resources on iterative (non combinatorial) optimization problems in a group context, ie, maximizing some function F(g(x)), (where g is a member of group G acting on elements of set X and F: X -> R) by moving varying which element of G is acting on x. I'm mostly interested in problems where X is R^n and G is some lie subgroup of GL(n, R). are there any optimization methods which naturally restrict the search -- which when unrestricted is through all nxn matrices -- to those in the subgroup of interest, or is it just a matter of encoding the group structure as constraints?

[slate forum]

@queen vine

[queen vine]

Take $V$ to be the zero set of $xy-1$ and look at the projection $V\to\mathbb{A}_k^1$ onto the first component

[cloud walrus]

leoli1:

[slate forum]

It needs to be a map from $\mathbb{A}^n \rightarrow \mathbb{A}^m$

[cloud walrus]

HelixKirby:

[queen vine]

the same works, take $\mathbb{A}^2\to\mathbb{A}^1$ to be the projection onto the first component and look at the image of $V$.

[cloud walrus]

leoli1:

[slate forum]

Ok.... So If I take $f:\mathbb{A}^2 \rightarrow \mathbb{A}^2$ given by $f(x,y)=x$, then If I look at $f(Z(xy-1))$, this essentially projects the hyperbola onto the $\mathbb{A}^1$ axis, but it's missing $0$, so its image is $\mathbb{A}^1-0$.... How do I show this isn't closed, I mean, I guess I could say that there's no polynomial that has infinitely many roots.... I don't know if that's good enough though

[cloud walrus]

HelixKirby:

[slate forum]

@queen vine

[queen vine]

yes, exactly, if a polynomial has infinitely many roots then it has to be the zero-polynomial but this can't be the case here since 0 is not in the set.

of course we would need to assume that the ground field has infinitely many elements

[slate forum]

Ok, thanks! I'm really bad at this stuff, I wish I had a better understanding

[elder valley]

Is F(x) always a proper extension of F? Equivalently, is it possible for a field F to be isomorphic to F(x)?

[prime gale]

Those are two different questions. Not sure what you mean by proper extension, but the map Spec F(x)->Spec F is not proper. If you mean that F->F(x) is always a non trivial extension, then yes--it is always non trivial. Yes. I think it is possible for F to be isomorphic to F(x). Take F to be Q(x_i for i in N).

[elder valley]

i mean F being a proper subfield of F(x)

that example looks good though

actually i don't know what i mean. too tired right now 🤣

[prime gale]

Let C be complex numbers. Is algebraic closure of C(x) isomorphic to C as fields?

[quick kernel]

yes

algebraically closed fields are characterized by absolute transcendence degree and characteristic

[prime gale]

Thanks!

Is it easy to prove?

[quick kernel]

there is a sketch here: https://ncatlab.org/nlab/show/algebraically+closed+field

[prime gale]

Thanks. I found something also. It looks pretty easy.

[kindred mist]

Ok I am pretty sure they are using the division algorithm here but for some reason it is giving me pause. I get that f' can't properly divide f and be a nonunit, as f is irreducible and f' is a polynomial over the same field, but what is the rest of the inference going on here?, like I am trying to see why f' not equal to zero would mean that gcd(f,f') is 1 or something else up there doesnt hold. Also we are assuming f is nonconstant, so I believe that is important here somehow

[carmine fossil]

Let's say f and f' are not coprime and f' is nonzero,then there is some non constant polynomial g which divides f

[kindred mist]

yeah

[carmine fossil]

But then f is irreducible

[kindred mist]

hold up

yeah but why should that g be over the same field

as opposed to an extension

[carmine fossil]

I think you can use the Euclid's divison lemma to show such a g exists in this field

[hollow comet]

There is no need to use Euclid, the gcd is by definition in the same field

[kindred mist]

ah ok there we go

ty

[cinder bone]

any tips to go about this?

@open torrent so i have to show a^11 results in contradiction?

do i do that by showing gcd(10,11)=1?

I showed that the order for each a^k is less than 10 does that work to show they are distinct?

yes!

right

?

[next obsidian]

Here's what Viburnum had in mind, suppose that |a| > 10, then we want to show a^1,..., a^11 are distinct. If a^k = a^j for j < k, with k,j <= 11 then e = a^k(a^j)^{-1} = a^{k - j} this implies that the order of a is <= k - j, but k - j <= 10 so that |a| <= 10

[mint gulch]

@cinder bone, do you know Lagrange Theorem?

If not, by contradiction, suppose that |x| > 10

Imagine that |x| = 11

What would happen?

[thorn delta]

the best way to approach this is just to induct on the length of words in the group, and show that they all reduce to something of the form a^kb^n with 0 <= k < 8 and 0 <= n < 2, right?

Also, if those are the only relations, then how could the group ever have order less than 16? I ask because the problem seems to suggest that.

[carmine fossil]

They might be further simplified through some manipulation

For example,This seems to have 12 elements, but there is actually just 1

[thorn delta]

ah okay interesting

[carmine fossil]

With presentations,you can only give an upper bound as to how many elements are present in the group

[cerulean wraith]

Hi, anybody undertand how to solve this? Thanks!

What I undertand from this is that the operation is * where a an element of G and b an element of G is multiplied together and powered, it becomes a^n b^n

[scarlet estuary]

for part (a), you might want to note that $(ab)^n = (ab)^{n-1}(ab)$

[cloud walrus]

Namington:

[scarlet estuary]

this is because $(ab)^n = (ab)(ab)(ab)\dots(ab)$, $n$ times

[cloud walrus]

Namington:

[cerulean wraith]

Ok, but why does $(ab)^n = (ab)^{n-1}(ab)$ happen?

[cloud walrus]

faTe:

[scarlet estuary]

uh

i just stated

$(ab)^n = (ab)(ab)(ab)\dots(ab)(ab)$, where there are $n$ $(ab)$s on the right hand side

[cloud walrus]

Namington:

[cerulean wraith]

Ooooh okok

[scarlet estuary]

$(ab)^{n-1} = (ab)(ab)(ab)\dots(ab)$, where there are $n-1$ $(ab)$s on the right hand side

[cloud walrus]

Namington:

[scarlet estuary]

so if we introduce another $(ab)$

[cloud walrus]

Namington:

[cerulean wraith]

Yes, when multiplied, they are the same

[scarlet estuary]

anyway, this in particular tells us that $(ab)^n = (ab)^{n-1}(ab)$

[cloud walrus]

Namington:

[cerulean wraith]

Ok

And then... I don't understand why they switch places

[scarlet estuary]

well we know $(ab)^n = a^nb^n$

[cloud walrus]

Namington:

[cerulean wraith]

yES

[scarlet estuary]

so $(ab)^{n-1}(ab) = a^nb^n$

[cloud walrus]

Namington:

[cerulean wraith]

Ok...

And then just divide?

Or $(ab)^{n-1} = a^nb^n (ab)^{n-1}$

[cloud walrus]

faTe:

[scarlet estuary]

hold on

where does that come from

[cerulean wraith]

From left side to right side

[scarlet estuary]

??

what did you multiply by

the only valid operations in a group are multiplication and taking inverses

[cerulean wraith]

From here $(ab)^{n-1}(ab) = a^nb^n$?

[cloud walrus]

faTe:

[scarlet estuary]

how do you get from that to what you said

[cerulean wraith]

I just switch the (ab) to the right side?

[scarlet estuary]

then you would have $(ab)^{n-1} = a^nb^n(ab)^{-1}$

[cloud walrus]

Namington:

[scarlet estuary]

not $a^nb^n(ab)^{n-1}$

[cloud walrus]

Namington:

[cerulean wraith]

Oh ok, my bad

[scarlet estuary]

what you can do is observe that $(ab)^{n-1}ab = a^nb^n$ implies $(ab)^{n-1}a = a^nb^{n-1}$ through right-multiplication by $b^{-1}$

[cloud walrus]

Namington:

[cerulean wraith]

And then after that, it becomes $a^{n-1}b^{n-1}$?

[cloud walrus]

faTe:

[scarlet estuary]

how?

[cerulean wraith]

since $(ab)^{-1} = a^{-1}b^{-1}$

[cloud walrus]

faTe:

[cerulean wraith]

So when $a^{n}b^{n} a^{-1}b^{-1} = a^{n-1}b^{n-1}$

[cloud walrus]

faTe:

[cerulean wraith]

?

Does that apply?

Or I'm missing some steps in the process

[scarlet estuary]

how do you know $(ab)^{-1} = a^{-1}b^{-1}$?

[cloud walrus]

Namington:

[scarlet estuary]

this is not true in many groups

$(ab)^{-1} = b^{-1}a^{-1}$ is true in all groups

[cerulean wraith]

In the beggining it said that it is the operation, no?

[cloud walrus]

Namington:

[scarlet estuary]

it said that $(ab)^{n} = a^nb^n$ is true for some $n > 1$

[cloud walrus]

Namington:

[scarlet estuary]

not necessarily for -1

ok, hold on

let me try an alternate explanation approach

lets look at what each side of the equation $(ab)^n = a^nb^n$ means

[cloud walrus]

Namington:

[scarlet estuary]

$(ab)^n = (ab)(ab)(ab)\dots(ab)$, where there are $n$ $(ab)$s

[cloud walrus]

Namington:

[scarlet estuary]

whereas $a^nb^n = aaa\dots abbb\dots b$, where there are $n$ $a$s and $n$ $b$s

[cloud walrus]

Namington:

[scarlet estuary]

so we have $(ab)(ab)(ab)\dots(ab) = aaa\dots abbb\dots b$

[cloud walrus]

Namington:

[scarlet estuary]

where again, there are n of "everything"

since $n > 1$ we can safely rewrite $(ab)^{n}$ as $(ab)(ab)^{n-2}(ab)$, taking the convention that $(ab)^0 = e$

[cloud walrus]

Namington:

[scarlet estuary]

and similarly we can rewrite $a^nb^n$ as $aa^{n-1}b^{n-1}b$

[cloud walrus]

Namington:

[scarlet estuary]

so $ab(ab)^{n-2}ab = aa^{n-1}b^{n-1}b$

[cloud walrus]

Namington:

[scarlet estuary]

let's multiply by $a^{-1}$ on the left, and $b^{-1}$ on thje right

[cloud walrus]

Namington:

[scarlet estuary]

this gets us $b(ab)^{n-2}a = a^{n-1}b^{n-1}$

[cloud walrus]

Namington:

[scarlet estuary]

and we can apply associativity to the left hand side to get $(ba)^{n-1} = a^{n-1}b^{n-1}$

[cloud walrus]

Namington:

[cerulean wraith]

Wait you can multiply $a^{-1} and b^{-1}$?

[cloud walrus]

faTe:

[scarlet estuary]

you can multiply both sides of an equation by the same thing

that's middle school algebra

so what i did was, i took $ab(ab)^{n-2}ab = aa^{n-1}b^{n-1}b$

[cloud walrus]

Namington:

[scarlet estuary]

and multiplied by $a^{-1}$ on the left

[cloud walrus]

Namington:

[scarlet estuary]

which gets us $a^{-1}ab(ab)^{n-2}ab = a^{-1}aa^{n-1}b^{n-1}b$

[cloud walrus]

Namington:

[scarlet estuary]

and of course, $a^{-1}a = e$

[cloud walrus]

Namington:

[cerulean wraith]

Ok, sure, that means that cancels out the a and b in the right side

[scarlet estuary]

so this simplifies to $b(ab)^{n-2}ab = a^{n-1}b^{n-1}b$

[cloud walrus]

Namington:

[scarlet estuary]

and you can get rid of the bs on the right hand side through a similar process

right-multiplying by b^-1

we generally call this "the cancellation law"

but what we're actually doing is multiplying

(this distinction is important since in, say, arbitrary rings, you cant necessarily cancel since inverses dont necessarily exist)

(though in groups they always do)

[cerulean wraith]

Ok, so it doesn't matter to the group

[scarlet estuary]

well, this is literally how you prove cancellation works in a group

[cerulean wraith]

Oh, you mean the inverses?

Ok, so cancellation works in all groups?

Is that what you're trying to say?

[scarlet estuary]

as long as you "cancel on the correct side", yes

[cerulean wraith]

But not on rings?

[scarlet estuary]

but we cant take, say, $xy = yx$ and ``cancel the $x$ out" to get $y = y$

[cerulean wraith]

I'm not on rings yet btw

[cloud walrus]

Namington:

[scarlet estuary]

since the x are on "different sides"

[cerulean wraith]

Ok, so only can cancel on one of the sides

By multiplying the inverse of the element we are trying to cancel

Which works because the identity element axiom supports it

[scarlet estuary]

and yeah, you cant necessarily cancel in rings; for example, in the ring of integers modulo 8, 2 * 3 = 2 * 7, but this does not mean 3 = 7

[cerulean wraith]

What how do you read that and what is modulo?

Is it the one where we count the remainder?

[scarlet estuary]

don't worry about it if you havent covered it yet

i'm just saying for future reference

[cerulean wraith]

Ok

So, I think we got side tracked lul

[scarlet estuary]

well, i gave a (very rough) proof outline for part (a)

you can clean it up a bit and write it up more formally

[cerulean wraith]

So, I can't use $(ab)^{-1}$ but I can use $(ab)^{n-1}$

[cloud walrus]

faTe:

[cerulean wraith]

Switching side doesnt matter also right

[scarlet estuary]

i don't know what you mean

[cerulean wraith]

$a^{n-1}b^{n-1} = b^{n-1}a^{n-1}$

[cloud walrus]

faTe:

[scarlet estuary]

that is not true

(in general)

for example, the quaternion group satisfies (ab)^n = a^nb^n, but it does not satisfy that.

[also sorry, but it's way too late for me to be awake, so i have to go to sleep]

[hopefully someone else can help you connect the dots here]

[cerulean wraith]

Ok, it's alright, thank you for the help @scarlet estuary . I had a feeling it wasn't true in general, I just don't know how to prove it lol

[cloud walrus]

slimvesus:

[solemn rain]

@cerulean wraith do you still need help

[cerulean wraith]

I submitted this

Lul

[solemn rain]

what do you wnt to prove

[obsidian path]

I'm confused with the definition of the order of a group and the order of the elements in the group. If you were to look at the cyclic group Z8, why is the order of the elements 2 and 6 equal to 4? if order of element is defined in the formula I have below, it should be order 3 for 2 is what it looks like to me.

[next obsidian]

You've messed up the operation

the operation there is addition

[obsidian path]

Oh.. is it because of the lagrange's theorem..?

[next obsidian]

so a^n actually is na

as in, a^n = a + a + a +... + a where you add it n times

[obsidian path]

right

I don't get what you mean

[next obsidian]

why do you think the order of 2 is 3?

You're thinking 2^3 = 8 = 0 mod 8

correct?

[obsidian path]

yes

[next obsidian]

but 2^3 = 6

since the operation here is not multiplication

[obsidian path]

oh my god

right

[next obsidian]

yeh

common mistake

[obsidian path]

tyvm

im struggling with groups

[next obsidian]

it's fairly common for abelian groups to write the opeartion as addition

so like you'll say 2 + 2

when you're working in Z/8Z

but it's not very common to write na to mean like, the operation done on a n-times

they still generally do a^n

so you need to remember that a^n actually means adding

it's just some notational problem, but you'll get used to it

[obsidian path]

so always consider the operation given?

[next obsidian]

yup

[obsidian path]

I haven't seen any groups with division, is that a thing?

just curious

[next obsidian]

I mean they all have division

if you consider the opeartion as multiplication

you can think of a/b as being a*b^{-1}

[obsidian path]

oh right.. division is not considered a real operation.. or something of the sort?

since its just the inverse

[next obsidian]

yeah it's just multiplication by inverse

[obsidian path]

right right

[next obsidian]

this ends up mattering more when you go to rings

where you have an additive structure that's abelian

and a multiplication which distributes

the prototypical example being Z

you can add numbers

multiply them

and it distributes over addition

BUT you don't have multiplicative inverses

since 2^{-1} doesn't exist

so this is where you can sort of say "division doesn't exist"

but really you just mean multiplicative inverses don't exist

[obsidian path]

honestly didn't catch that :'))

[next obsidian]

My point is just that division = inverses existing

so for groups since you require inverses

you won't ever have a group without division, if it even makes sense to talk about taht

since if your group is considering multiplication, you need inverses to even be a group

but you can have stuff where you can add and subtract, and multiply, but you can't divide

Like Z

[obsidian path]

I don't get why 2^{-1} doesn't exist

in that scenario

but I haven't touched rings yet so

I'm assuming thats why

we've only started groups this week

[next obsidian]

I mean 2^{-1}

under multiplication

should be 1/2

but that isn't in Z

that's all I'm saying

[obsidian path]

omg

right

sorry I'm not thinking right

[next obsidian]

don't worry about it haha

[chilly ocean]

Let k be an n bit string denoting a natural number. Let 1 be the n bit string consisting of all 0’s except a 1 at the end, and let 2 be the n bit string consisting of all 0’s except for 1 at the second to last place. Let XOR denote bitwise XOR, and let + denote natural number addition. Then under what circumstances does ((k XOR 1) + 1) XOR 1 = (k+1) XOR 2?

[supple marsh]

Interesting: https://math.stackexchange.com/questions/441314/if-g-is-a-finite-group-of-order-n-why-is-it-isomorphic-to-its-centralizer-i ... can anyone help me interpret Qiaochu's comment on this?

I thought Yoneda for groups was Cayley's theorem, but maybe there is more structure I'm not aware of?

[supple marsh]

The proof Mikko gives is not that bad.

Is the same thing true for a group representation -- e.g take the regular representation of a group, then if you are C[G] endomorphism you're basically given by where you send 1... acting on the right. I guess this is the thing about the endomorphism ring of a module being the opposite ring that Jack mentions?

[supple marsh]

resolved with Ultraproduct in math-discussion

[tacit saffron]

I have a homework question that involves a homomorphism and i'm somewhat confused at what a homomorphism is

[next obsidian]

depends on context fam

in general it's a map which "preserves structure" but that changes depending on what you're working with

for a group it just means f(gh) = f(g)f(h)

for rings it's that plus f(g + h) = f(g) + f(h)

etc

[tacit saffron]

im being asked to consider the group of polynomials Rx with real coefficients undert addition, with map phi from Rx to Rx where phi(f(x)) = f'(x)

so if its closed under addition i just do that f(g+d) = f(g) + f(h) thing?

[next obsidian]

yup

since the group operation on both sides is just +

you'll see that it isn't multiplicative, as in f(gh) =\= f(g)f(h)

because of the product rule

but that's okay, since you want to consider the stuff under +

not multiplication

[tacit saffron]

i just got confused cuz in my notes i just wrote phi(ab) = phi(a)phi(b) and thought i had to multiply, which doesn't work cuz of product rule, but i was told to prove that it was homomorphic

[next obsidian]

yeah that's just because when you say

ab

you mean a*_Gb as in like, how the opeartion is definedcx

[tacit saffron]

you're assumign some computation yeah

[next obsidian]

and here *_G means addition

the same thing happens with something earlier in here

consider Z/8Z under addition

then the order of 2 is 4

because 2^4 = 8 = 0 mod 8

[tacit saffron]

hmmmm what about proving phi to be onto or one to one

[next obsidian]

becasue 2^4 actually means 2 + 2 + 2 + 2

in that context

are you asking how to do that?

[tacit saffron]

ooo

[next obsidian]

Its onto but not one to one

to show it's onto just consider any polynomials antiderivative

this exists and is still in R[x]

not one-to-one because...

for groups one-to-one <==> kernel is trivial

if you don't yet know what that means

consider what maps to 0 under differentiation

[tacit saffron]

just an integer?

[next obsidian]

nah

what about 1/2

or pi

[tacit saffron]

*number

i bad with word

[next obsidian]

right

a constant

[tacit saffron]

yeah

C

[next obsidian]

but then it clearly isn't one-to-one

I wouldn't say C

just say any constant

[tacit saffron]

ok

[next obsidian]

because in algebra land C = complex numbers

[tacit saffron]

oh right

im stuck in calc land

[next obsidian]

no worries

but yeah that's the right idea

so clearly f(a) = 0 for any constant a

so all you'd have to do is "prove" there's more than one constant and f isn't one-to-one

big surprise, R[x] has more than one constant lol

[tacit saffron]

oh really wouldn't have thought

wow that makes much more sense now

[next obsidian]

NP

anyway, this is actually the way to show one-to-one for group homomorphisms

in any group you have an idea of what 0 is right?

(or e)

The identity

you can show that if f:G -> H is any groupo homomorphism that f being one-to-one is equivalent with the condition that if f(x) = 0, then x = 0

i.e. the only thing mapping to 0 (in H) is 0 itself (in G)

[tacit saffron]

would the kernel be all constants

because its mapped to identity 0

[next obsidian]

not a constant

but the set of constants

yup

[tacit saffron]

idk how to find what the image would be. in linear it would be the span of the vectors of the linear transformation

my book writes it as "im phi = {x in G' I x = phi(a) for some a in G},"

[solemn rain]

the image is all the values 'x' such that there exist an element a such that x=phi(a)

[tacit saffron]

im being asked to consider the group of polynomials Rx with real coefficients undert addition, with map phi from Rx to Rx where phi(f(x)) = f'(x)
so in my case, the image would be all polynomials with real coefficents because any polynomials have a polynomial antiderivative?

[tacit saffron]

i swear the longer i stare at my textbook the more i get annoyed at math but the longer i spend in here the more i understand and like math

the duality

[tacit saffron]

how can i know when subgroups of Sn are cyclic

[thorn delta]

the same way you know any subgroup is cyclic i guess

[tacit saffron]

i'm being asked to formulate an informal conjecture about when the subgroup ⟨(a b), (c d)⟩ is cyclic, i just dunno where to start

like ik i should start with random examples of a b c d and n

but idk how to know if its cyclic

[thorn delta]

the way i see it, there are two cases: (a b) and (c d) are disjoint cycles or b = c
EDIT: also there is (a b) = (c d) which is kind of the trivial case

[chrome hinge]

Im kinda struggling with Sn rn too. I need to show that Sn is a semi-direct product of An and Z2

I found a bijection between $S_{n}$ and $Z_{2} \times A_{n}$ but its not an homomorphism :(

[cloud walrus]

Maikel:

[thorn delta]

keep in mind that An is a subgroup of index 2 in Sn. So $S_n = gA_n \cup A_n$ for some $g \in S_n$. $g$ has to be a transposition (iso to $Z_2$). You can prove this, but it is also kind of intuitively clear since $gA_n$ gives you all the odd permutations of the group.

[cloud walrus]

kxrider:

[chrome hinge]

Mi idea was similar. I fixed a transposition

The thing i dont get is how should i define the semi direct product

[thorn delta]

you don't have to. there are sufficient conditions that allow you to conclude that Sn is a semi direct product

[chrome hinge]

Like, $Aut(Z_{2}) = {id}$

[cloud walrus]

Maikel:

[chrome hinge]

you don't have to. there are sufficient conditions that allow you to conclude that Sn is a semi direct product
@thorn delta Really???

[tacit saffron]

i'm being asked to formulate an informal conjecture about when the subgroup ⟨(a b), (c d)⟩ is cyclic, i just dunno where to start
so would it only be cyclic if one number in each cycle equaled each other so it could be rewritten as a normal single cycle

[thorn delta]

For example, if $H, N$ are subgroups of $G$ with $N$ normal in $G$, then if $H \cap N = 0$ and $HN = G$ then $G$ is the semidirect product of $H$ and $N$.

[cloud walrus]

kxrider:

[slate forum]

When y'all got the chance, I don't really understand this proof

[chrome hinge]

I know that result

For example, if $H, N$ are subgroups of $G$ with $N$ normal in $G$, then if $H \cap N = 0$ and $HN = G$ then $G$ is the semidirect product of $H$ and $N$.
@thorn delta

[cloud walrus]

Maikel:

[chrome hinge]

But i need a normal subgroup...

[thorn delta]

A_n is normal

[slate forum]

A_n is

[chrome hinge]

Well, i feel so stupid now lmao

[slate forum]

Actually, any index two subgroup is normal

[chrome hinge]

Yeah your absolutely right

Thank you guys!

I just wasnt seeing that

[slate forum]

External semi direct products are weird

[chrome hinge]

Im kinda tired you know

Oh, yeah ive heard there are outer and inner semi direct products

Whats the exact difference between them?

[slate forum]

They're isomorphic, but one happens within a bigger group

The other just combines two that a priori have nothing to do with each other

Well... the external one you need like a hom from one group into the automorphisms of the second

So I guess they're kind of related

[chrome hinge]

Oh, so in this case $Z_{2} \times A_{n}$ would be an inner semi direct product in Sn?

[slate forum]

It's either or

[cloud walrus]

Maikel:

[slate forum]

Though I'm guessing you're realizing them as subgroups of S_4

Err S_n

It's kind of like in linear algebra where you can do an internal direct sum and an external one

[chrome hinge]

Ok i think i get it

[slate forum]

One happens inside a bigger vector space, the other kind of just creates a new one from old ones

But uhh, can I have some help?

I can repost my screenshot

[chrome hinge]

Probably wont be able to help you, sorry

Thanks for your help man, much appreciated

Both

[slate forum]

I meant it towards @thorn delta

[thorn delta]

i think @tacit saffron still needs help?

[slate forum]

Oh yeah, crap

[thorn delta]

if b = c then you can write (a b) (b d) as a 3 cycle. so <(a b), (b d)> contains <(a b d)>, but the inclusion does not go the other way

@slate forum i think u can go ahead and post ur question. if Zempro still needs help I can move to another channel

[slate forum]

Aight

Prop 2.1.10

I understand the plan

But I don't know where this zero locus of I(X)+J comes from

[tacit saffron]

the homework assignment i was working on was due a few minutes ago so i don't really need help anymore

i kinda BSed an answer but its fine bad time management on my part

ima go to my profs office hours to go over it so i don't really need help

but thank you @thorn delta

[slate forum]

Sorry

How do I conclude the zero locus of I(X)+J is a subset of Z(f)?

It seems like, maybe we show something not in Z(f) must not be in I(X)+J?

@thorn delta

[thorn delta]

bro i have nooooo idea that is chinese to me

[mint gulch]

Why the relation must have at the quotient ring?

[slate forum]

Oh... crap, anyone know AG?

[thorn delta]

I think @latent anvil and @next obsidian ask AG questions here every now and then. (sry if y'all don't like being pinged)

[next obsidian]

I know some AG

I'm kind of trash at varieties stuff but I can give it a shot

what's the qwuestion?

[slate forum]

Ugh

Err

Uhhh

[next obsidian]

what part of it?

[slate forum]

1.1.10

2.1.10

[next obsidian]

what part of that lol

[slate forum]

Oh

Showing the zero locus of I(X)+J is a subset of the zero locus of f

Technically speaking

f isn't even in k[x]

[upbeat juniper]

Why the relation must have at the quotient ring?
@mint gulch if f is reduction mod n then f(x)^2+f(y)^2 = f(x^2+y^2) = f(3z^2) = 3f(z)^2

[slate forum]

It's in the coordinate ring

[next obsidian]

I mean where specifically does your understanding break down

What they did was show that for any p such that f(p) =\= 0 there exists a g in J such that g(p) =\= 0

so if q is in the zero locus of I(X) + J, then in particular g(p) = 0, so that necessarily f(q) = 0

so q is in Z(f)

Also f is basically in k[x_1,...,x_n]

Ai priori it's in k[x_1,...,x_n]/I, but you can take a representative of f in k[x_1,...,x_n]

but then you'd have to show that f(p) is well-defined, but this is true because for any g in I, g(p) = 0 for all p in X

Since X is just the zero-locus of I

[robust veldt]

I'm also having difficulties with algebraic geometry right now lol

[next obsidian]

As am I 😔

Nope

nopenopenopenope

soz

2 hard 4 me

[robust veldt]

I don't feel like I understand local rings well enough to understand why mod F and G it could be considered a vector space here

[next obsidian]

what is F and G here lol

It just says plane curves?

[robust veldt]

Two plane curves, yeah.

[next obsidian]

how do you mod out by that

okay, so is there a natural map from k into there?

[robust veldt]

$\mathbb{O}_P(\mathbb{A}^2)$ is a subring of the rational functions defined at the point P, and $(F,G)$ is the ideal generated by $F$ and $G$ as polynomials in this ring. The curves are the zero-loci of $F$ and $G$

[cloud walrus]

megaman:

[next obsidian]

okay I get this I think

so O_P(A^2) has a way to be considered a vector space?

Like if you're a rational function of the form f/g can't you just define a(f/g) to be (af)/g

is that still well-defined once you quotient by F and G?

[robust veldt]

A vector space over k, which I think means that every rational function in this quotient ring is equivalent to some polynomial in the quotient ring

[next obsidian]

I mean doesn't the above construction work?

I mean f is actually a polynomial in k[x,y] no?

To be clear here

my variety knowledge is garbage

I only know schemes

so I'm tryna apply what I know to the classical setting haha

[robust veldt]

I don't think I understand what you mean. Are f and g general polynomials?

[next obsidian]

Sorry, so O_P(A^2)

is this not just rational functions of polynomials in k[x,y] where the denominator doesn't vanish at P?

or is it something else

you said it's a subring of this?

[robust veldt]

Yes, that is what that means (I think lol)

[next obsidian]

Okay then you can define a vector space structure on just O_P(A^2) the following way

This also gives you the one in the quotient for free

So you actually define an algebra structure

This is a map phi: K-> O_P(A^2)

then the way you interpret like af where a in K and f in O_P(A^2) is by saying that af = phi(a)f

So you can define a map from k -> O_P(A^2) by sending a to the constant function a

like, consider a as a polynomial which is literally just the constant a

this is like the inclusion of K into K[x,y]

then this still makes sense mapping into O_P(A^2), if you want it rational

just do like a/1

[robust veldt]

Can you give a concrete example?

[next obsidian]

of?

I'm saying define phi:K -> O_P(A^2) by phi(a) = a/1

where a/1 is considered as a rational function

[mint gulch]

@mint gulch if f is reduction mod n then f(x)^2+f(y)^2 = f(x^2+y^2) = f(3z^2) = 3f(z)^2
@upbeat juniper f is the canonical projection?

[upbeat juniper]

yep

[mint gulch]

I see, thanks

[robust veldt]

So the natural homomorphism phi:K->O_P(A^2)/<F,G> where phi(a)=a/1 is certainly injection from a 1-d vector space to the ring, right? But if the intersection number of F and G at P is bigger than 1 then the theorem says that the ring should be isomorphic to a larger vector space.

[next obsidian]

Isn't that fine?

isn't that just saying you have a copy of K as a 1-dimensional subspace?

[robust veldt]

The theorem says that the dimension of the ring over k equals the intersection number of F and G at P

[next obsidian]

that's still fine

I mean there's two problems here

How do we know phi is still injective after quotienting by <F,G>?

The map O_P(A^2) -> O_P(A^2)/<F,G> might not be injective

secondly, even if it is injective, if it isn't surjective this isn't an issue

I have an injective map R -> R^2 given by f(a) = (a,a)

but R^2 is still a 2-dimensional vector space over R

[robust veldt]

If phi(a)=phi(b) then a-b is in <F, G>. If a does not equal b then a-b is a unit, which would mean that <F, G> is the entire ring

So I think this could only be the case if F and G didn't intersect at P

But this is interesting because I think this is the dimension 0 case then for the theorem

[next obsidian]

Oh right, it's a map out of a field

of course this is injective

[slate forum]

Sorry I'm back

[next obsidian]

Right, so it's injective, that's good

but even if it is injective this isn't a problem

Since there should be copies of K in any K-vector space

[slate forum]

Why is it that for any f(p)!= 0 there's a g in J such that g(p)!=0

[next obsidian]

by definition of J

you can represent phi as h/g, and g cannot vanish at p (since this representation is at the O_P thing)

then it's obvious that g is in J by definition of J since gphi = g(h/g) = h

[slate forum]

Sure

Why does g(p)=0 imply f(p)=0?

[next obsidian]

because its the contrapositive of the statement I did right before it

you showed f(p) isn't 0 implies g(p) isn't 0

[slate forum]

Hmm

So let p in Z(I(X)+J), then z(j)=0 for all j in J

[robust veldt]

I don't want to interrupt so I will take a break for a few minutes, but I'll leave with this:
By concrete example I meant that the theorem they gave me should be true for when P=(0,0), F=Y-X^2, G=Y. The intersection number at P=(0,0) between F and G is 2, so the theorem says there is an isomorphism between a 2-vector space over k and the ring of rational functions defined at (0,0) mod the ideal generated by <F,G>. I am looking for an explicit construction of an isomorphism, and I would think it would look something like phi(b0*k0 + b1*k1)=T*k_0 + S*k_1 where b0 and b1 are basis elements and T and S are forms in terms of X and Y. But I'm not sure what T and S would be or why such a function might end up being surjective over all rational functions mod F and G

[slate forum]

So then I can say if f(p)!=0, then p is in X_f, so then phi=h(x)/g(x), with g(p)!=0, so g(x) is in J, but then we have to have g(p)=0

[next obsidian]

No, g is not in J because g(p) != 0

g is in J because phi = h/g

you get g(p)!= 0 as an additional consequence of having phi = h/g in O_P

[slate forum]

I'm confused, g isn't in J?

[next obsidian]

g is in J

but that's because you said phi is in O_p, i.e. it has the form h/g for some h,g

[slate forum]

So if p is in Z(I+J)

[next obsidian]

then necessarily g(p) = 0

since g is in J

[slate forum]

That's not good

[next obsidian]

right

[slate forum]

I wonder if we can phrase it more directly

[next obsidian]

I mean you could but why bother

it would just be changing orders on stuff

this is just 1 application of a contrapositive, doesn't seem worth worrying over

[slate forum]

If f(p)!=0, then p in X_f, so phi = h/g where g(p)!=0, and so g is in J, p is not in Z(J)

[next obsidian]

I don't want to interrupt so I will take a break for a few minutes, but I'll leave with this:
By concrete example I meant that the theorem they gave me should be true for when P=(0,0), F=Y-X^2, G=Y. The intersection number at P=(0,0) between F and G is 2, so the theorem says there is an isomorphism between a 2-vector space over k and the ring of rational functions defined at (0,0) mod the ideal generated by <F,G>. I am looking for an explicit construction of an isomorphism, and I would think it would look something like phi(b0*k0 + b1*k1)=T*k_0 + S*k_1 where b0 and b1 are basis elements and T and S are forms in terms of X and Y. But I'm not sure what T and S would be or why such a function might end up being surjective over all rational functions mod F and G
@robust veldt I wouldn't hold out on getting an explicit isomorphism. I gave it a k-vect structure, and by some general linear algebra it is isomorphic to some k^n, but the way it's embedded and stuff is probably all sorts of screwy

Idk, I gotta do other stuff

Hopefully you at least got one proof of it

[slate forum]

I guess, it's tricky

We're showing that if i+j vanishes at p for every i+j in I+J, then we have to have f vanishing at p, as otherwise we can find a j where j(p)!=0

This j in particular is the denominator of phi

Is that right? @next obsidian

[next obsidian]

I have no clue dude

I worked through the proof given in the book and it made sense to me

[slate forum]

And I can do this because p is in X_f

Oh alright

[obsidian path]

its difficult for me to think of the properties required for the subgroups? like how do construct a subgroup properly?

[thorn delta]

If ur familiar with Lagrange's theorem, you might recall that any non trivial subgroup should have order 2, which is cyclic, or 4, which is either cyclic or the klein 4 group.

[next obsidian]

I think if they’re phrasing it in terms of symmetry group of a square and not D_4, I don’t imagine they know Lagrange’s theorem

Problems like this usually appear at the start of a group theory course

[scarlet estuary]

flashback to my group theory course which covered cosets and lagrange's theorem week 2

[slate forum]

Jesus, that's fast

@scarlet estuary

I mean, that's what my graduate level algebra class did

[next obsidian]

My algebra class had us prove |g| | |G| on the first hw lol

[scarlet estuary]

the proof of lagrange's is straight up given as an exercise

[slate forum]

What the fuck

[scarlet estuary]

(although admittedly lagrange's is just applying a lemma or two which we covered in class, so you just had to cite those lemmas)

(not very conceptually tricky if you actually get the definitions)

specifically, it was using this lemma to realize how coset decomposition works

once you have coset decomposition lagrange's follows "obviously"

[slate forum]

If I'm looking in the coordinate ring $A(X)=k[x]/I(X)$ and I define an ideal $J={g\in A(X)\mid g\phi\in A(X)}$, is it really ok to talk about $I(X)+J$? Like, $J$ is subset of $A(X)$, so how can I do this?

[cloud walrus]

HelixKirby:

[slate forum]

@next obsidian

[next obsidian]

Everything in J has a representation in k[x]

And when evaluated at points of X these give well-defined functions

I’m pretty sure

Also I actually don’t really know any of this stuff lol

[slate forum]

Oh, ok

But does this ideal $I(X)+J$ like correspond to the ideal in $A(X)=k[x]/I(X)$?

[cloud walrus]

HelixKirby:

[slate forum]

I mean, I think this is just a commutative algebra question at this point

[next obsidian]

Yes it does haha

I mean, wait

[slate forum]

I thought it would be more like $J/I(X)$ or something

[cloud walrus]

HelixKirby:

[next obsidian]

Yeah, it should be

This is just the correspondence theorem

[slate forum]

mk

[next obsidian]

Ideals containing I in bijection with those in R/I

[slate forum]

I usually see it has $J\subset R$ corresponds to $J/I$ in $R/I$.

[cloud walrus]

HelixKirby:

[next obsidian]

Isn’t that exactly what you wrote tho?

[slate forum]

Not quite

I think it kind of goes backwards here

[next obsidian]

Yeah but that’s how it goes

You send a J in R/I to its preimage under mod I

Aka to J + I

[slate forum]

You start with an ideal $J\subset R/I$, and then it's corresponding ideal is $J+I$?

[cloud walrus]

HelixKirby:

[slate forum]

Oh ok

Talking about ideals in quotients without explicitly writing them as $J\subset R$ with $J/I$ confuses me I guess?

[cloud walrus]

HelixKirby:

[slate forum]

So, $J/I \leftrightarrow J$ but also $J\subset R/I \leftrightarrow J+I$

[cloud walrus]

HelixKirby:

[slate forum]

no wait, that's not right, I should technically write $\bar{J}\subset R/I$ or something

[cloud walrus]

HelixKirby:

[slate forum]

How can I show that $Z(y-x^2)\cap Z(xy-1)$ intersect in at most four points, I mean, I just did the math and it seems like they always intersect in 3 points

[cloud walrus]

HelixKirby:

[slate forum]

I don't think I know that theorem yet

It looks like this corresponds to the fact that the ideal $(y-x^,xy-1)$ is a product of at most four maximal ideals, is that right?

[cloud walrus]

HelixKirby:

[slate forum]

@open torrent

[leaden finch]

can someone help me with this one

i got stuuck

[chilly ocean]

i think the whole point is proving that $ab$ is not 0 (zero divisors cant be 0 right?)

[cloud walrus]

88ddda:

[leaden finch]

yes

but i got stuck and idk what to do next

ab is zero divisor *

[chilly ocean]

what is $abc$ equal to?

[cloud walrus]

88ddda:

[leaden finch]

hmm not sure

[stone fulcrum]

@leaden finch
bc = 0

What's abc?

[leaden finch]

thats the deifntion

[scarlet estuary]

you know bc = 0

what does this tell you about abc

[chilly ocean]

When a question says "operation of multiplication", are they referring to the normal multiplication or the modular one?

[scarlet estuary]

normal multiplication

modular multiplication woudlnt even make much sense as theres no modulus involved here

[chilly ocean]

Ok

I was confused because another question asked me to make the multiplication table of a set and for that we had to use modular multiplication

[scarlet estuary]

U(8) and R^* are different groups.

and the term "multiplication" is a bit overloaded in group theory

sometimes the abstract operation in a group is called "multiplication"

with no further qualifiers

[chilly ocean]

yes its very confusing lol

How am I supposed to know I need to use modular multiplication for U(8) and not R*

Like is it when a set is finite?

[scarlet estuary]

just by definitions of the groups involved really

U(n) is always modular multiplication

whereas when it says R* is a group "under the operation of multiplication", this is referring to "standard" multiplication, since theres no further qualifiers

the operation is fundamentally "part of" the group definition

U(8) and the set {1, 3, 5, 7} under the operation "ab = a + b - 1 mod 8" would have the same elements

but theyre fundamentally different structures

[chilly ocean]

Ok thank you very much

Is there a reason U(n) is assumed to use modular multiplication?

Or I guess it would be fundamentally "part" of its definition too?

[scarlet estuary]

yes

thats how we define U(n)

[vestal snow]

I had a question about function fields

There's a theorem which says that any element has as many poles as zeroes when counted properly

What does counting properly mean in this context?

[supple marsh]

@vestal snow Maybe someone else can give a more in depth answer, but it is the analogue of counting the function (say, on the complex plane) x^2 + x^3 as having a zero of order 2 at 0.

Or 1/x^3 as having a pole of order 3 at 0.

[vestal snow]

I think we're talking of different things here

Maybe

[supple marsh]

No, it's the same. In the above case the field is C(x)

and there are certain discrete valuation rings associated with it, which are the points on the projective line.

[vestal snow]

I was interpreting it in the following way: Say the place P (w/ valuation ring O) is a zero of x. Then x = t^n u where P = tO. Then the zero P must be counted n times

Is this what you meant?

[supple marsh]

yeah that's right, and thats the same as above because x^2 + x^3 = x^2 ( 1 + x) , and (1 + x) is a unit in C[x]_{(0)}

[vestal snow]

Oh okay

Gotcha

Thanks

[supple marsh]

no problem. 🙂

but maybe someone else can give a better answer, I don't know the number theory side of this story as well as I want to.

The point is that if you function field came about as the field of birational functions on a projective curve, then each (closed) point of the curve gives you a discrete valuation on your ring, which corresponds to "order of vanishing / pole" at this particular point. (This also lets you reconstruct the curve, at least over algebraically closed fields IIRC.) If you stare it for a while one can see that the definition of a valuation is basically supposed to mimic the behavior of 'order of vanishing at a point'

"theorem which says that any element has as many poles as zeroes when counted properly" this is geometrically intuitive, sort of. You can imagine the rational function from your curve to P^1. Then it asserts that 'counted properly' f^{-1}(0) and f^{-1}(infinity) have the same size. Away from branch points of the cover of P^1 that you get from your rational function, counted properly just means the size. At branch points, say of order k, k sheets come together and that's why it is counted as order k. So, if you from 0 to infinity and watch the number of points in the fiber (counted correctly if they come together at branch points), then the number doesn't change.

[chilly ocean]

Since they only give the set of the groups, does that mean the groups would stay different no matter the operation?

wut?

they are giving the 5 groups of order 8, up to isomorphism, if thats what u mean

So the Z8 is a group?

I thought it was a set

well, it is a set, but it also has a group structure

in fact, i don't know any setting where you talk about Z_8 without a group structure

yeah ok I get it

sorry im asking dumb questions but my prof doesnt explain shit

[quaint ivy]

Since they only give the set of the groups, does that mean the groups would stay different no matter the operation?
@chilly ocean the usual operation is assumed on $\b Z_n$, $D_4$ and $Q_8$. Algebra courses and textbooks often do this since it's redundant to specify the operation everytime.

[cloud walrus]

bastian.uwu:

[quaint ivy]

also (and this should be clear if you've seen isomorphisms) there's many groups that could have the same structure as, say $\b Z_8$

[cloud walrus]

bastian.uwu:

[chilly ocean]

i have never seen that word in lectures but I will try to find a textbook thank you

[quaint ivy]

np

you'll probably see it later, no worries

usually one tries to see examples of groups or whichever structure you're definining (apparently your course is here), then cosets and quotients and only then homomorphisms, the isomorphism theorems and stuff

[chilly ocean]

ok yeah

Is there a quick way to calculate the highest element order for Z2 X Z4?

Or do we have to calculate all of them manually

there is obviously an element of order 4, there is no element of order 8 because it is not Z8

and order of an element must divide the order of the group, so highest must be 4

ohh

ok thank you didnt know

i was calculating manually lmao

well, even, there is no element of order 8 because (x, y)^4 = (x^4, y^4) = (1^2, 1) = (1, 1)

how do you know the 1^2?

cuz x is in Z2, so x^2=1

and y^4 = 1?

(tbh it wasn't really necessary to write x^4=1^2, rather than more direct x^4=1)

yah

[sour plume]

Given two vector spaces $V, W$, one can fairly easy see:
$\Lambda^2 (V \otimes W) \cong \left(\Lambda^2 V \otimes S^2 W \right) \oplus \left( S^2 V \otimes \Lambda^2 W \right)$. I'm guessing I could interpret this as a decomposition into irreducible submodules under the $S_2$-action?

Is there an easy way to see how this generalizes to something like $\Lambda^3( V \otimes W)$ using Young symmetrizers or something like that?

[cloud walrus]

Lartomato:

[hot lake]

good question

I think I knew at one point how to decompose Sn and Ln of A+B, but not of A * B

I forgot all about it meh

[sour plume]

Ye, that's just a distributivity formula iirc, so $S^\bullet (A \oplus B) = S^\bullet A \otimes S^\bullet B$ in terms of chain complexes (so e.g. $S^2(A \oplus B) = S^2A \oplus (A \otimes B) \oplus S^2 B$)

[cloud walrus]

Lartomato:

[sour plume]

Tensor products seem more spoopy, pretty sure this will involve some insights about this whole young symmetrizer business

[cloud walrus]

Lartomato:

[sour plume]

did an oopsy, am actually just interested in having two vector spaces, not three, so the above version is what i want

[chilly ocean]

Quick question: Im currently doing dihedral groups. This is greek to me
"Represent the elements both in form r^i*s^j with i in {0,...,n-1} and j in {0,1} and also in cycle notation. "

Firstly, does anyone know what r^i and s^j those letters stand for? Im trying to look for them, but I can't find it in my textbook or online. Maybe my professor uses a different notation?

[golden pasture]

r^i=rrrrrrr i times

similarly for s^j

[dusk summit]

r, s are symbols for generators in the group, and i and j are exponenets

Geometrically, they stand for repeated rotations or reflections

[chilly ocean]

oh so r & s are "rotation & reflection" and i & j are just "n"?

[dusk summit]

i and j stand for the number of times you rotate or reflect the shape, respectively

[chilly ocean]

yeah, ok. is there a reason why r&s have to have different named exponents?

[dusk summit]

so since j is in {0,1}, it follows that s^2=s^0=e. Geometrically this means that 2 repeated reflections lead to the same shape you started with

rotations and reflections are 2 different actions, so while on any shape 2 consecutive reflections about the same line lead to the shape you started with, on an n-gon you need n rotations to get back to the original shape

And then you can also combine rotations and reflections, which is why r and s can be "multiplied" together

[chilly ocean]

2 repeated reflections lead to the same shape you started with
Isnt this true for any j?

[dusk summit]

So for dihedral groups describing the symmetry of a polygon we number each vertice and then track those vertices as we perform reflections and rotations

so while the shape "looks the same," the corners have actually moved around

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

[chilly ocean]

oh yeah you're right

[dusk summit]

This might have a better explanation I'm still earning group theory so idk if I'm explaining it well xD

[chilly ocean]

no no i get it

thanks, really appreciate it

[dusk summit]

np!!

[cinder bone]

the order of each unit is itself right?

[olive mirage]

no, the order is 2

[cinder bone]

oh since its multiplied by itself

[olive mirage]

bingo!

[cinder bone]

thx :3

[olive mirage]

"order" is a very confusing unintuitive word

[cinder bone]

yeah i get confused by the order of a group versus the order of the elements in that group

[next obsidian]

If $A$ is an integral domain (you can assume Noetherian as well, but I don’t think it matters), and $P$ is a prime ideal of $A$, denote by $P[x]$ the prime ideal $A[x]P$, the extension of $P$ in $A[x]$. What then, is $A[x]{P[x]}$? I have reason to believe this is $A_P[x]{PA_P}$, but I don’t think you can do some neato tensor product rewrite trick to get this.

[cloud walrus]

Chmonkey:

[cinder bone]

so an element of any group must have an order that divides the order of the group?

by lagrange's?

[smoky cypress]

If your group is finite

[cinder bone]

yeah

okay cool

[next obsidian]

Ummm

Even works in infinit case

[smoky cypress]

elaborate

[next obsidian]

There is a sense in which it makes sense to say anything divides infinity

Namely in classifying maps from a cyclic group to any group

Since it’s in bijection with elements of the target with order dividing the order of the cyclic group

And to extend this result to Z just say anything divides infinity

(The map is given by send generator to the element with order dividing the order of the cyclic group)

[smoky cypress]

hmmm ok

actually I think I've seen somewhere that

|G|=|H|[G:H] when you interpret these as cardinal numbers

[next obsidian]

This is also true

You can do the same proof

[smoky cypress]

Yeah

[next obsidian]

This just boils down to showing a bijection from G and H x G/H

Where G/H might not be a group, but you do set theoretic quotient

[scarlet estuary]

this is literally the general form of lagrange's theorem

note though that it only works for finite groups

[next obsidian]

Yeh

I mean the proof works in the infinite case to show the bijection I showed

[scarlet estuary]

sure

[next obsidian]

which gives you a cardinal version of that equality of sizes

[cinder bone]

okay another question lol

so if a group of order 4 contains no element of order 4, then 3 elements have order 2 and the identity element has order 1

which means every element operating on itself results in the identity element right

[next obsidian]

Right

[cinder bone]

what about the elements operating on one another?

obviously the identity would result in that element

but say we have a,b,c for the others

[next obsidian]

You get either itself when you multiply by identity

Or for the other 3, you get the third

Because else the two would be inverses

Consider like ab, this can’t be a or b

[cinder bone]

ahhhh

[next obsidian]

Since then a or b = e

Can’t be e since then a = b^-1

So ab = c

[cinder bone]

that makes sense

[next obsidian]

This is actually why the automorphism group is S_3

Basically a,b,c are symmetric

You could literally just rewrite them as b,a,c and nothing changes

So the automorphism group is given by shuffling those 3 elements around

aka S_3

If that doesn’t make sense yet, maybe write this down and come back to it in a few weeks

[cinder bone]

no i think it does make sense

there has to be 4 distinct elements

so if ab equals anything but c we have a contradiction

because then that means either a or b are not really distinct from the other elements

i appreciate the help, its hard to learn this stuff on zoom at 9 am

[cinder bone]

so for this id show something similar right

but also that any nonidentity element is commutative with any other nonidentity element

which i find a little less intuitive

obviously a*a=e but what about a * b etc

[urban acorn]

@cinder bone try to write the multiplication table of this group

label the elements 1,a,b,c

i think that'll be helpful in this case

[cinder bone]

dunno the order of the group though

[urban acorn]

ohhh

[cinder bone]

is the only different with this question

[urban acorn]

Take two elements a and b in G. Consider the properties of their product ab.

ab is another element in the group, what do we know about it?

[cinder bone]

mmm it also has order 2 or is the identity element

[urban acorn]

how can you express that as an equation?

[cinder bone]

(ab)^2=e

or (ab)^2=1 i use e

[urban acorn]

is there any manipulation you can do on (ab)^2 = abab = e?

let me pose another direction: What do we know about the inverse of a product of two elements?

[cinder bone]

ab=(ab)^-1

[urban acorn]

alright

is there an expression for (ab)^-1?

[cinder bone]

(a)^-1 * (b)^-1 ?

[urban acorn]

it's not generally true that $(ab)^{-1} = a^{-1}b^{-1}$.

[cloud walrus]

Intel:

[cinder bone]

but in this case it is right

[urban acorn]

indeed

in general, $(ab)^{-1}ab = 1$, so we can multiply both sides by $b^{-1}a^{-1}$ from the right to get $(ab)^{-1} = (ab)^{-1}aa^{-1} = (ab)^{-1}abb^{-1}a^{-1} = b^{-1}a^{-1}$

[cloud walrus]

Intel:

[cinder bone]

;o

[urban acorn]

do you follow?

[cinder bone]

yeah which proves that its commutative

[urban acorn]

yeah

[cinder bone]

where are you multiplying $b^{-1}a^{-1}$

[cloud walrus]

therealjoshua:

[scarlet estuary]

$(ab)^{-1}ab = 1$, therefore $(ab)^{-1}abb^{-1}a^{-1} = b^{-1}a^{-1}$

[cloud walrus]

Namington:

[scarlet estuary]

just multiply on the right

[urban acorn]

because in general $(ab)^{-1} = b^{-1}a^{-1}$, but in this case we also have $b^{-1}a^{-1} = ba$ and $ab = (ab)^{-1}$

[cloud walrus]

Intel:

[cinder bone]

ohhhhh

okay got it thanjks

[urban acorn]

@cinder bone in general, you can manipulate equations in groups by "dividing from the right or left" like this

Think of it like ab is doing b then a, and so to undo it, you gotta undo a then undo b, so you gotta do a^-1 then b^-1 so you gotta do (b^-1)(a^-1)

[mild valley]

I need to show that with $n \ge 3$ and $0\le k \le 1$ then $\mu\rho^k\in D_n$ is a reflection about the line with angle $\frac{-\pi k}{n}$ measure from the x-axis

[cloud walrus]

iVeRn:

[mild valley]

Any advice on how to do this?

I've thought about trying to represent the points as points on the unit circle with cos and sin, but not sure how to go from there

[urban acorn]

it's probably already given to you with k = 0 by merit of the way the dihedral group is defined with respect to the symmetries of the polygon

[mild valley]

oh, and I meant k les than or equal to n -1, not 1, sorry

[urban acorn]

oh

[mild valley]

If it was two cases, I don't think that would be too hard

[urban acorn]

you can look at the way these permutations act on the corners of the polygon

if you label them 1 through n, where it sends 1 and 2 should already specify the symmetry

imagine that line, and imagine reflection about it acts on the corners

btw mu and rho instead of s and r is weird

[mild valley]

Ok, I'm gonna try thinking about where the points will end up generally

Also, that's just how my book taught it

[cinder bone]

how do i manage to show this?

i did something similar to show a & its inverse have the same order

rough work

[solemn rain]

induct on n naturalo

[leaden finch]

can someone explain too me how they got this whole thing in yellow?

[next obsidian]

What?

0 = n mod n

n = pq

Then break it up

This is just saying like, mod 6, you know 0 = 6 =2•3 so 2 and 3 are zero divisors in Z/6Z as an example

[leaden finch]

how did they come up with this one ? [o]= [n]

where did the 0 coem from

[golden pasture]

0 = n mod n

[leaden finch]

is that wat zero divisor in Zn means?

[golden pasture]

0=n in that ring

0=n=pq and p,q are non zero

so p,q are 0 divisors

[sharp oyster]

how do i manage to show this?
@cinder bone
$$(b a b^{-1})^n = b a b^{-1} b a b^{-1} b a b^{-1} ... b a b^{-1}$$
All the $b^{-1} b$ eliminate, so we end up with just
$$b a ... a b^{-1} = b a^n b^{-1}$$

[cloud walrus]

ConfusedReptile:

[solemn rain]

@sharp oyster induct

on n

[sharp oyster]

more formally - by induction

[cinder bone]

Thanks 🙏🏻

[leaden finch]

how do write 0_R in latex?

addtive identity

[smoky cypress]

$0_R$

[cloud walrus]

Whoever:

[smoky cypress]

xD