groups-rings-fields

[coral spindle]

In fact it is enough to show that g = g'h for some h in H.

[hollow parrot]

Ok thanks 👍

[hollow parrot]

This is handy

[coral spindle]

You should try proving it

[hollow parrot]

I see it immediately now that u mention it

g =g'h
gH = g'hH
gH = g'(hH)
gH = g'H

[coral spindle]

Yup

[hollow parrot]

Thanks

[coral spindle]

No worries

[main needle]

For $n \geq 5$, Want to show that $A_n$ has no normal subgroups of order $p$ odd prime (Without using the fact that $A_n$ is a simple group). My approach is assume it exists and then there is a surjective homomorphism $\phi : A_n \to A_n/N$ and consider the image of 3-cycles since they generate $A_n$ but couldn't derive a contradiction from the case that $p = 3$. Any hints ?

[cloud walrus]

ru0xffian

[chilly ocean]

Is Fraleigh sufficient as a prerequisite for Eisenbud...

?

[rotund aurora]

If R is a ring and k some (infinite) set, what are the necessary and sufficient conditions so that the R-module R^k is free?

[tribal moss]

That sounds like one of those more-complicate-than-it-seems questions.

[rotund aurora]

xD

[main needle]

Any quick look at this ? I must be missing something strightforward

[rotund aurora]

The proof that the An are simple is pretty straightforward, idk why you would stop in a partial result

[main needle]

I am just curious what is the contradiction that we can get from assuming If A_n has a normal subgroup of index 3 for n >= 5

[rotund aurora]

if you have a normal subgroup, then it is closed by conjugation. Therefore, if your normal subgroup N contains a permutation of one "type" it will contain all permutations of that "type". Example: if it contains a 3-cycle, then it will contain all 3-cycles. Then, you just have to show that it will contain at least one 3-cycle

(didn't read the index part)

[tribal moss]

"k is finite" or "R is a field" would each clearly be a sufficient condition, but intuitively I expect there would be plenty of R^k that are free "by accident" without falling into one of those cases.

[main needle]

same cycle shape doesn't imply conjugates in A_n right? it's not the same as in S_n

[rotund aurora]

uhm think you are right. Its been a while since I haven't looked at that, so I cannot really help you.

The 3 cycles do form a single conjugacy class tho

so you just need to show that it contains a 3-cycle

[main needle]

And what is the contradiction that we get from that ?

[rotund aurora]

you just get the whole group

thus showing that if N is a normal subgroup of An, it is An itself, hence there are no proper normal subgroups

[main needle]

Yeah I just noticed but I can't see why the surjective hom. $\phi : A_n \to A_n/N$ must map all the 3-cycles in to the identity if $p = 3$. It can map it to an element of order 3 right ?

[cloud walrus]

ru0xffian

[rotund aurora]

A priori, if you are assuming that N is a proper normal subgroup of An, it should map the 3-cycles to a non-identity element of order 3 (otherwise, you would be saying that N contains a 3-cycle and is hence An). But I'm not sure what result you want to apply to derive a contradiction from that map in the index=3 case. It's been some time since I looked at this, so I cannot really help you more than what I said.

[main needle]

Yeah, thanks a lot I appreciate it !

[thorn delta]

are you asking about "commutative algebra with a view toward algebraic geometry"?

[chilly ocean]

Of course...

[thorn delta]

eiesnbud defines what a ring is, so maybe technically no prereqs. in general i think a first course in algebra should be sufficient to get started in commutative algebra

[ruby sundial]

Is the vanishing ideal of 0 the entire ring?

Yes

lol

thats what it comes down to

[balmy vale]

hi, i've been stuck on this practice problem for a while

the question is

[cloud walrus]

sheeppunk

[balmy vale]

i know the subgroup has to be of order 6

but thats about as far as i can get

[coral shale]

(...)(...431...)(...)

they all have to look like this right?

could just brute force

and see what it looks like maybe

[balmy vale]

i guess

i mean i did that

but im not sure how to tell if its a coset or not

cause there doesnt seem to be an obvious decomp

[coral shale]

how many are there?

if there arent many, you test all of them to be g

supposing your coset is gH

admit its not smart, but at least u get there lol

[balmy vale]

ah yeah i guess that makes sense

[lethal dune]

Given M be a finitely generated injective module over an integral domain R then show R is a field.

Any hint on how to approach

[coral shale]

given M be

[chilly ocean]

R a field or M a field?

[lethal dune]

I said M is a module

[chilly ocean]

No I mean, to show that R is a field or that M is a field

Because for M it's simple, I think

Actually

Yeah, fix some x in R

Define f: M to R via m to xm

Since M is injective there is a map g s.t. fg = id

Now 1 = g(f(1)) = g(x) = xg(1), hence x is invertible

Something like that, I don't have any paper near me

[hidden haven]

Didn't use finitely generated, can't work

[chilly ocean]

Where's the mistake?

[hidden haven]

Idk I didn't read too carefully but ℚ is an injective module over ℤ which is not a field

[chilly ocean]

I guess you could remedy it by defining the above map only on the generators then

[hidden haven]

What do you mean by 1 in M

M has no 1

[hidden haven]

In this step

[chilly ocean]

It's an injective module, it has a left inverse for any map, right?

[hidden haven]

Is that right?

For M → R yes

Because R is projective and this is a surjection

[hidden haven]

Ok first of all this is M → M lol

[chilly ocean]

I am not sure now lol, I should read up on the definitions

[hidden haven]

It should be injective because M is injective

I assume

[chilly ocean]

A map M to M is just a restriction of a map M to R

[lethal dune]

[gritty sparrow]

Hint: injective modules are divisible, and the structure thm for fgen modules over a pid

[next obsidian]

I don’t think you can use the structure thm

[gritty sparrow]

My idea was as follows, if there is an R/rR factor then (0, … , 1,0 ..0) will not be divisible by r. And now take a free factor R then the same sort of element being divided by r shows that r has an inverse as the coefficients in the other factors have to be zero since they are all free

[next obsidian]

No but it’s not a PID

So you literally can’t apply the structure thm

[gritty sparrow]

Oh damn i misread the q lol

[next obsidian]

I think you’re supposed to use Nak

[gritty sparrow]

Ah I see how that works

[lethal dune]

So localize?

[next obsidian]

I don’t remember immediately if localizing preserves injectivity

What I would first do is show mQ = Q

Using injectivity

then localize

[lethal dune]

didn't get it

[solar glacier]

Can someone check this work plz

[ruby sundial]

Nak?

[hollow mica]

so far I have seen three (equivalent) definitions of the tensor product

1. If V, W are vector spaces, and v_1, ..., v_m and w_1, ..., w_n are their respective bases, then V ⊗ W is the vector space with basis v_i ⊗ w_j for all i, j.

2. If V, W are vector spaces, let U = { v ⊗ w | v \in V, w \in W }. Now let X be the subspace generated by the relations < λ v ⊗ w - λ(v ⊗ w), v ⊗ λw - λ(v ⊗ w), (u + v) ⊗w - u ⊗ w - v ⊗ w, u ⊗ (w + x) - u ⊗ w - u ⊗ x >. Then we say V ⊗ W is the vector space U/X.

3. If V, W are vector spaces, their tensor product V ⊗ W is a vector space along with a bilinear map f: V x W -> V ⊗ W, such that for all linear maps g: V x W -> U there exists a linear map h: V ⊗ W -> U such that g = h ○ f.

what is the fundamental bridge of these definitions to multilinear map, i.e., why do people regard elements of these tensor products as multilinear maps?

[ruby sundial]

2 and 3

people regard elements of tensor products as multilinear maps because people sometimes/often consider the elements to be matricies

there is an isomorphism between V tensor V* and End(V) which are square matricies

and when things are finite dimensional V isomorphic to V*

also have you tried to write down multilinear maps in a matrix form?

[hollow mica]

I have not

[hollow mica]

I am trying to construct this isomorphism now

[ruby sundial]

Want any tips?

[pine patio]

what's t?

[chilly ocean]

An indeterminate.

[pine patio]

and alpha ^p = t?

is f an element of F?

[chilly ocean]

f is the element x^p - t of F[x].

[cloud walrus]

rakko

[pine patio]

can i start working with t^(1/p) to factorise the polynomial?

[pine patio]

is (x+a)^p just x^p + a^p mod p?

[chilly ocean]

Using the binomial theorem, yes.

Well, it's a little more delicate than the argument I had just posted. But that's how you do it.

[pine patio]

im still confused by t. i want to say f is the pth power of a degree 1 polynomial

x - t^(1/p)

and since alpha is a root of f then it must be a root of this degree 1 polynomial so theres only 1 root of f in K

[coral spindle]

But t^(1/p) is not meaningful in a ring in general.

In fact, there is nothing that could be called t^(1/p) in F_p(t).

[pine patio]

yeah thats why i dont wanna use it

oh is it x-alpha

[coral spindle]

Alpha is indeed defined to be a root of x^p - t.

[pine patio]

yeah so x-alpha divides f

and then i show that (x-alpha)^p = x^p - alpha^p = x^p - t

[fossil vapor]

Paper of the day
https://www.researchgate.net/publication/266724118_Historical_links_to_a_Lie_theory_of_semigroups

[maiden heath]

I don't understand this question. I'm confused over the wording of the second and third line. can someone please explain what they're saying? thanks

[chilly ocean]

Do you want a mathematical explanation, or are you just having trouble parsing the sentence?

If it's the latter, it's saying: a subalgebra is semisimple if it is generated by the following:
-JM-elements X_i;
-subgroups of S_n;
-centers of subgroup algebras of kS_n.
If it's a mathematical explanation you want, I cannot help.

[noble hedge]

Anyone wanna help out with this?

ik G is congruent to the group of all roots of unity (under multiplication) but I don't see that being helpful here

[maiden heath]

Thanks, that was all, just had trouble with the wording

[chilly ocean]

It is a funny sentence.

[maiden heath]

Yeah it's a problem I get often when reading maths

[coral spindle]

Congruent? Wym by this

I've not heard this term used for groups before

[noble hedge]

isomorphic

sorry

\cong is the latex command for it

\cong gives you the isomorphism symbol and \equiv gives you the congruence symbol, so I tend to confuse them in conversation lol

[coral spindle]

Ah right

[next obsidian]

Congruent modulo isomorphism, duh

[coral spindle]

Trying to think of an appropriate hint for this question

[noble hedge]

I'm not quite sure moving to roots of unity does anything helpful

[coral spindle]

It doesn't.

Right OK here's a good hint

You can of course see elements of G as equivalence classes of fractions

The elements of H are (equivalence classes of) fractions of the form x/p^n

A general fraction will be x/m, for some number m of course.

Hint: think about the factorisation of m.

[noble hedge]

the cannonical map from G to G/H will send x/m to some equivalence class based off its factorization right?

Like if m looks like p^n for some n, then x/m would be in the kernel as it'd be sent to H

but if not

Yeah I got no clue what I'm doing

what I have is that if the subgroup product of two normal disjoint subgroups equals G, then their direct product is isomorphic to G

So I'm thinking there's probably a normal subgroup in G/H whose preimage under the cannonical map from G is what I should be looking for

[formal ermine]

could I perhaps get a hint on how I can show that $$\text{sgn}(\pi) = \prod_{i < j} \frac{\pi(j) - \pi(i)}{j -i}$$ please?

[cloud walrus]

illuminator3 👻(#eric4honorable)

[coral spindle]

show that sgn(sp) = -sgn(p) for any transposition s, and show sgn(id) = 1.

[chilly ocean]

Replace the product sign by a limit and you have something that looks like a derivative. Hmmm...

[coral spindle]

You can hopefully work out why this is sufficient

[chilly ocean]

Sign(pi) is the derivative of pi.

[formal ermine]

transposition is when j > i but pi(j) < pi(i)?

[coral spindle]

Uh no

[formal ermine]

what does transposition mean

[coral spindle]

A transposition is just (i j) for some i =/= j

Have you proved that every permutation is a product of transpositions?

[formal ermine]

no

[formal ermine]

what does the notation (i j) mean?

[coral spindle]

Cycle notation :)

[formal ermine]

ohhhh

I remember

[coral spindle]

Well this is a good time to prove this fact

[formal ermine]

so a transposition is just an element of S_n that only switches two elements?

[coral spindle]

Hold on though

How have you defined sgn(pi)?

[formal ermine]

,, \text{sgn}(\pi) = (-1)^{\abs{\text{Fehlstände von } \pi}}

[cloud walrus]

illuminator3 👻(#eric4honorable)

[formal ermine]

this is our definition

and we showed that that is a homomorphism

[coral spindle]

I do not know what the word fehlstände means.

[formal ermine]

uh

[wooden ember]

Is felhstande pairs of inversion?

[formal ermine]

yes

inversion is the word I was looking for, thanks

amount of inversions in pi

[coral spindle]

Could you remind me how that is defined?

This is not the way I usually think about the sign.

[formal ermine]

an inversion is when j > i but pi(i) > pi(j)

[wooden ember]

It’s how many pairs i,j are inverted under the permutation

[formal ermine]

what does sign(pi) even represented to begin with? we just defined it in the lecture and never said anything about it

[coral spindle]

Great. OK then your product shouldn't be hard to show from this. Notice that this is a big fraction of the following form:

I will finish this first.

[formal ermine]

yes, thanks

[coral spindle]

$\frac{(\pi(2) - \pi(1)) (\pi(3) - \pi(1)) \cdots }{(2 - 1) (3 - 1) \cdots}$

[cloud walrus]

Boytjie

[coral spindle]

Note that every pair where i =/= j has some representative on the top, but either as i - j or j - i

Cancel the top and bottom :)

[formal ermine]

ahhhhhh, I see, thank you very much

[coral spindle]

Great

Now as for what sign(pi) means

You will see soon that every element of S_n can be written as a product of transpositions

[formal ermine]

seems intuitive to me

[coral spindle]

I don't think I agree, but you'll have to prove it anyway.

However, notably, this is not necessarily a unique product of transpositions

In fact it's never unique – this is actually pretty obvious as you can just add copies of a transposition twice, like (1 2)(1 2) to get a different product that produces the same result

However, the number of transpositions is odd or even, and no matter what you choose, for a given permutation it will only be able to be written as either an odd or an even number of transpositions.

This is what the sign counts.

[formal ermine]

-1 means odd and 1 means even?

[coral spindle]

That's right

[formal ermine]

epic

thanks once again

[coral spindle]

No worries

[solar glacier]

think i could bother you for a question

its more of verifcation if i did an exercise correctly

[coral spindle]

Just ask! People who can asnwer will see and do so

[solar glacier]

so i am trying to show a group of order 112 = 7*2^4 is NOT simple.

I know that n2 = 1 or 7 and n7 = 1 or 8

so I assume it IS simple, then neither of the n's can be equal to 1

this implies I have 6*8 elements of order 7

and I have 7 * (2^4-1) elements of order 2? contradicting the order being 112? thus one of the n's need be 1?

[coral spindle]

Yes, since the sylow 7-groups have trivial intersection of course.

I think you should mention that's why, but this is right

[solar glacier]

ok cool

[coral spindle]

As for the other part, I'm not convinced that's true

The 7-subgroups must have trivial intersection otherwise they are equal, since they are all cyclic (being prime order)

[solar glacier]

can I argue that

[coral spindle]

Sure go on

[solar glacier]

using the fact thgat the total order is 112

thus theres 64 elements left

meaning n2=4 which cant be

[coral spindle]

I don't see how that follows

Why would this imply n2=4?

[solar glacier]

because the order of 2sylow subgroups is 2^4=16?

[coral spindle]

Ah right there we go

That is nice, yeah

[solar glacier]

nice ok thanks ! and n2 cant be 4 cause its conreguent to 1 mod 2 and divides 7

[coral spindle]

Yup, indeed you'd already eliminated that case

Hm let me just think for a moment actually

[solar glacier]

sweet. thanks!

okie doke!

[coral spindle]

I just realised there could be a large issue here.

[solar glacier]

whats that

[coral spindle]

concluding n2 = 4 is wrong, in fact

Yeah this is a huge issue. I was wrong to just say yes

All 2-subgroups have a common element, so in the first place this is just incorrect

but more importantly, the conjugate images could have nontrivial intersection too

In fact, if anything we have proven that n2 is not 4, since that would mean that the remaining elements number at most 16*4 - 3 = 61, after we remove the necessary quadruple-counting of the identity.

So in fact this is incorrect, sorry.

[solar glacier]

so how would I go about proving it then? Im all out of arguments lol

[coral spindle]

I'm thinking about it. I'm not very good at Sylow arguments haha

[solar glacier]

whats wrong with just supposing neither n2 nor n7 is equal to 1?

since thats going toward a contradiction

[coral spindle]

There's nothing wrong with that, I'm just thinking about it.

[solar glacier]

because if either equals 1 then were done

ah ok

[coral spindle]

Just checking, you know the part that states that n_p = |G : N_G(P)| right?

[solar glacier]

yes

so if thats 1 then N_G(P)=G

[coral spindle]

Of course.

OK, I think we should proceed by looking at what the intersection between sylow 2-subgroups can be.

Let's say P and Q are distinct Sylow 2-subgroups, both of course of order 16

[solar glacier]

isnt their intersection just e?

if theyre distinct

[coral spindle]

Not necessarily, no

As I stated earlier, they may have nontrivial intersection.

I pointed out even earlier that we can say this for the sylow 7-subgroup since it is cyclic of prime order

[solar glacier]

i thought they had to have trivial intersection lol

but since the order of the 2 subroups is not prime their intersection need not be trivial?

[coral spindle]

Yes, as I stated

Yeah sorry I can't see how to limit this.

I'm going to look this up

[solar glacier]

lol ok

[coral spindle]

Well OK, I had the right idea by trying to bound the intersection, but it seems to be very difficult. https://math.stackexchange.com/a/352005/825884

This person gives a very slick answer at the top though

It's not at all clear to me why it would embed in S_7 though

[solar glacier]

yeah i saw that :/

the last answer i liked the best

[coral spindle]

The one that says the answer is wrong?

[solar glacier]

oh i just saw that part smh

[coral spindle]

Yes and in fact we have been discussing why this answer is wrong

Unfortunately I don't think you're going to be able to prove this in any easy way.

And by "you" I don't mean like, specifically you lol

Just in general.

[solar glacier]

lol

so how about for a group of order 136

which is 2^3 * 17 then the number of 17 sylow subgroups is 1 by Sylows theorem and is thus normal

so certain numbers are harder to prove than others I see

[coral spindle]

Yup

[solar glacier]

im hoping on my midterm I dont get 112 lol

[agile burrow]

The action on the subgroups induces a homomorphism to S7, if it had nontrivial kernel then G wouldn't be simple

[coral spindle]

Oh very nice

Well-spotted

[solar glacier]

why to S7? im lost now lol

how about a group of order 144 which is 3^2*2^4

n3=1 or 16

if =1 done, so suppose it equals 16

then theres 8 times 16 =128 elements of order 3?

and the remaining constitutes a sylow 2 subgroup as theres 16 elements left

and so n2=1 so were done

does this work?

[solar glacier]

would this work? ^^^

[solar glacier]

as Syl_2(G) is non empty

[agile burrow]

The 3 Sylow subgroups could have nontrivial intersection, no?

[solar glacier]

true, so how would you go about proving this one is not simple? so if the order is prime their intersection is trivial but if its non prime then they could have non trivial intersection?

[agile burrow]

Can't n_3 be 4?

[solar glacier]

oh shoot yes! it can be

so do I suppose n_3 isnt 1 and take care of the n3=4 and n3=16 cases indivicually

[agile burrow]

I'm not sure how to rule out n3 = 16

[solar glacier]

so its tru then, certain orders are harder to prove non simplicity than others!

[agile burrow]

But for n3 = 4, the action of G on the 3 Sylow subgroups induces a homomorphism from G to S4

By order considerations, this map is not injective so it has nontrivial kernel. Therefore, G has a nontrivial normal subgroup

[trail stump]

is the first requirement ex=x redundant for the defintion?

because g is group, then g*(ex)= (ge)x=g(x) for all g in G, and x in X, hence ex=x for all g and x?

i guess i am assuming: If g(ex)=g(x), then ex=x

[chilly ocean]

,rotate

[south patrol]

Yeah I mean without axiom 1 there'd be no contradiction to e.g. fixing some y in X and taking g.x = y for all g and x

so the first bit definitely isn't redundant

[trail stump]

you are right, i was goona think of some other example, but this is just brutally simple, thx

[south patrol]

Np

[long nebula]

can you define group rings over infinite groups?

and if so would it still just be finite formal sums?

I think that that would work, but I'm just wondering about the conventions here

[toxic zephyr]

in an arbitrary integral domain, gcds of nonzero nonunits are defined but not necessarily unique. because of this is it bad form to write gcd(a,b)=c even though c is not the only possible gcd? or should i instead just say c is a gcd of and b?

[next obsidian]

Latter

[south patrol]

Ye

[hollow mica]

what's a slick way to show that there is no group of order 6 whose non-identity elements all have order 2

[toxic zephyr]

this is what my professor gave us. it says it's denoted by gcd(a,b). is c=gcd(a,b) just some notational shorthand that tells us it's some GCD of a and b, and not implying that there is a well defined gcd function on R?

didn't realize he wrote it like that on the question before i asked, but now i'm wondering because i can see arguments on both sides for whether or not using it is appropriate/correct

[next obsidian]

I mean for one, I don’t believe any gcd exists in arbitrary integral domains tbh.

There’s things called gcd domains which is when they always exist

But if a gcd exists actually they’ll always differ by units so maybe it is okay to call it gcd(a,b), so long as you’re in a domain

This follows because by definition of c,c’ are two gcds for a and b, then c divides c’ and c’ divides c, which in an integral domain implies they differ by a unit

[toxic zephyr]

right. yeah that's part a, to prove that gcds are unique "up to associates".

[next obsidian]

So idk whether you call it “the” or “a” is up to you, but you want to remember they aren’t exactly unique

They’re basically unique in the same way a prime decomposition is unique in a UFD, but you need to remember the unit

[toxic zephyr]

totally. that makes sense, thank you

[long nebula]

ty! :)

[chilly radish]

Anyone know a tool for generating Cayley Graphs for known groups?

or maybe finding some

[quiet pelican]

Show that a group with only elements of order 2 (except e) has order 2^n for some n (which isn’t too hard, considering a minimal generating set)

Or it’s infinite

[kind temple]

the way i like to do this is to show that its a vector space over Z/2Z

[weary bison]

How do I prove that $Hk_i$ are pairwise disjoint?

[cloud walrus]

Poscat

[weary bison]

oh wait they are cosets 🤦‍♂️

[kind temple]

lul

[lethal dune]

for a finitely generated module M over integral domain R, it is true that M_p ≅ (M/TorM)_p where _p is localisation?

[proud bear]

Let R=k[x] and M=R/(x^2). Then TorM=M because everything is annihilated by x^2. We have M_(x) is nonzero. If x/1=0/1 in M_(x), then p(x)\cdot x=0 has to hold in M for some p in R\(x). But p has nonzero constant term so this can't happen.

I think this works

[lethal dune]

Looks like it works tks

[lethal dune]

How about localizing at R\0

[gritty sparrow]

Localize the exact sequence 0 -> Tor(M)->M->M/Tor(M)->0 at (0). Since localization is exact and Tor(M) localized at (0) is 0, we get the isomorphism you are after

[lethal dune]

nice

[knotty python]

So I've got this question. It's kind of trivial the if statement.

For only if I've made this argument:

1. A previous part stated that if $f(a)=0$ then $x-a$ divides $f(x)$. So $f(x)$ and $f'(x)$ are divisible by x-a. So $f'(x)=(x-a)q(x)$. Then integrating both sides between a and x. $f(x)=\int_a^x q(x)(x-a)dx$ then by parts, differentiating q(x) and integrating (x-a) we get $f(x)=q(x)(x-a)^2/2-\int_a^x q'(x)(x-a)^2/2 dx$. Then this can be repeated and repeated. (integrating the $(x-a)^i$ part and differentiating the $q^i(x)$ part. Then as q(x) is a polynomial in C[x] we know it must eventually differentiate to 0. As all other terms are divisible by $(x-a)^2$ the result is proven.

[cloud walrus]

Max..

[knotty python]

but im not sure if this argument is correct

[past path]

I'd like to prove that if R is a finite ring and a is an element of R, that aR = R implies a is a unit.

Without assuming R has an identity.

Is there a way to do this concisely?

I'm not sure if R has to be finite, but we're given finiteness in the problem statement.

It seems obvious but I don't know if there's an elegant way to prove it.

This explicitly shows that each a in R has an element that acts as its own identity, but I don't know if that's sufficient justification to show that there's a universal identity element.

[paper flint]

I vaguely recall that this is necessary in order to have the multiplying-by-a-on-left map to be surjective as well as injective

I cannot figure out why a universal identity element must exist either

At the very least, a cannot be a zero divisor since the map b->ab is injective

[past path]

I mean, I think it's doable if we do a lot of lines of algebra. I was just wondering if there's a nice way to prove it.

In analysis I try to find some way of proving what's asked.

In algebra I want to prove it in a satisfying and elegant way.

[paper flint]

Fair

[formal ermine]

we proved this in today's lecture now lol, gonna attempt the thing you originally suggested

[south patrol]

Prove it by intuition

[formal ermine]

yeah, in my head it intuitively makes sense kind of, but I wouldn't know how to put it into words

like you can achieve any ordering by swapping two elements N amount of times

just like tower of hanoi kind of?

[south patrol]

Yeah so for me the intuition is: say we have a permutation σ of a set {1,...,n}. We can apply a transposition τ so that τσ sends 1 ->1. This means τσ restricts a permutation of {2,...,n} and by induction on the size of the set this restricted permutation is a product of transpositions.

like we can just "fix" the first element, then the second, etc etc

[formal ermine]

where are we doing induction here?

[rigid cave]

Is this channel still active or can I ask a question?

[lament dawn]

You can ask it's fine

[south patrol]

Maybe I'll add more detail then

[rigid cave]

Okay nice. I'm sorry if this is dumb but I don't see how the last two things here highlighted in red follow

[south patrol]

aha i remember getting stuck on this too

the proof is subtle imo

okay sorry so it turns out that I ended up writing down a different proof in my notes but I'll try to cook up an explanation

[rigid cave]

I would really appreciate that

[south patrol]

Oh yeah so I will also give the like more intuitive proof I've seen as well

Which is the same thing just written differently lol

[formal ermine]

ah wait so it's like for a random given permutation we can get it back to the identity using only transpositions by first fixing 1 -> 1 then 2 -> 2, etc. and the inverses of those transpositions combined is our original permutation?

[south patrol]

Changed notation slightly for psychological reasons, hope that's okay aha

But yes the key point is that by definition of a, the map A -> k is evaluation at a etc

[rigid cave]

ooohh okay nice I see

[south patrol]

another way to write it up:

I suppose the second one is longer but I think it is more intuitive lol

The second one is more or less an alternative way to write up the Rabinowitsch trick (where you append an extra variable to deduce strong from weak) - just I think it's more clear where the extra variable comes from when written up like this lol

[south patrol]

Yup

An illustration I was once told was like lol if gave a child a group of numbered blocks, how would they put them in the right order?

[formal ermine]

epic

[rigid cave]

ye right I see, this is a lot nicer. Thank you so much!

[formal ermine]

thanks

[south patrol]

They'd probably wind up doing smth like what I said

[formal ermine]

another question:

[south patrol]

Sans induction

[south patrol]

Np!

I think the Artin Tate lemma is a bit of a pain to remember though aha

I never remember the exact conditions

Just the application to nullstellensatz

[formal ermine]

we have to show $$\text{sgn}(\sigma \pi) = -\text{sgn}(\pi)$$ and $$\text{sgn}(\text{Id}) = 1$$ for $\pi \in S_n$ and $\sigma$ being a transposition, yes?

[cloud walrus]

illuminator3 👻(#eric4honorable)

[south patrol]

wdym?

or is this some separate problem

[formal ermine]

separate problem

[south patrol]

ah okay

[formal ermine]

I'm given some formula and have to show that it's the sign function

[south patrol]

oh okay

[formal ermine]

is this proof to fermat's little theorem correct?
\
Let $G = \bZ_p^*$ and $x \in G$. Note that $|G| = p - 1$, and by euler's theorem: $$x^p \equiv x^{p - 1 + 1} \equiv x^{p - 1}x \equiv 1x \equiv x \pmod{p}$$

[cloud walrus]

illuminator3 👻(#eric4honorable)

[south patrol]

Eh, I mean sure but often one might be interested in proving flt independently or smth

it is just the special case of euler's theorem in the case of a prime after all

[formal ermine]

alright, thanks

[formal ermine]

bump

[barren sierra]

@formal ermine what is the full context of this

[rotund aurora]

Try proving flt by induction

you can do it without induction, but the induction is fun :p

[formal ermine]

I have to show that $$\text{sgn}(\pi) = \prod_{i < j} \frac{\pi(j) - \pi(i)}{j -i}$$

[cloud walrus]

illuminator3 👻(#eric4honorable)

[formal ermine]

I'll try that once I'm done with my homework

[formal ermine]

(I already did this by expanding the product and cancelling things, but I wanna know if it's also possible by just showing those two things I mentioned earlier)

[formal ermine]

if I have an operation $* : G \times G \to G$, $g_1 * g_2 \coloneqq g_1g_0g_2$ for a fixed $g_0 \in G \setminus \Set{e}$ is it even possible to have an inverse?

[cloud walrus]

illuminator3 👻(#eric4honorable)

[coral spindle]

It is

[chilly ocean]

what does gh mean for example?

[formal ermine]

wdym?

[chilly ocean]

you are defining the "operation" using some unknown operation

huh?

[formal ermine]

oh

G is a group

with some operation defined on it

[chilly ocean]

you have to be more precise about such things

[formal ermine]

yeah, mb

[chilly ocean]

so yeah, clearly this operation you define is a semigroup, and it looks very interesting

[formal ermine]

so we want to solve $g_1g_0g_1\inv = g_0\inv$ for $g_1\inv$

[cloud walrus]

illuminator3 👻(#eric4honorable)

[lavish gull]

if you fix one of g1 or g2, then the function f(g1) = g1*g0*g2 has an inverse

maybe write it as $f_{g_2}(g_1) = {g_1}{g_0}{g_2}$

[cloud walrus]

coderizer

[chilly ocean]

The neutral element f satisfies fg_0f = f, that is f = g_0^-1

it is in fact a monoid

[coral spindle]

They're asking for inverse elements, not an inverse of a function, I think.

[chilly ocean]

so we want to find $h$ such that $gg_0hg_0 = e$, basically

[coral spindle]

It makes no sense to say "fix g1" then define the function in terms of g1...

[chilly ocean]

given some g

[lavish gull]

right, so fix g2

[formal ermine]

would you mind adding tex to that please?

[lavish gull]

it makes sense to say "fix g2" then define the function in terms of g1

[chilly ocean]

pain

[lavish gull]

why would you not pick the interpretation that made sense?

[cloud walrus]

Blitz

[formal ermine]

wait why?

[coral spindle]

It's worth noting actually that e is not going to be the neutral element of this operation. Not sure if illum wants to find an inverse wrt e or that element.

[chilly ocean]

because g_0^-1 is the neutral element

[formal ermine]

yes

[lavish gull]

as far as I see, without being more specific, you can't invert an operator. you can invert an element or a univariate function of a single argument

[formal ermine]

I want to find an inverse element of g_1 in terms of g_0^-1 and g_1

[chilly ocean]

huh? Of course with the neutral element of the monoid (G, *)

[formal ermine]

(G, *) is a group, apparently

[chilly ocean]

yeah, clearly from that equation $h = g_0^{-1}g^{-1}g_0^{-1}$

so it's a group

[formal ermine]

idk how you got the gg_0hg_0 = g_0^-1

[cloud walrus]

Blitz

[lavish gull]

you want to find $f^{-1}$ of $f(g_1) = {g_1}{g_0}{g_2}$?

[cloud walrus]

coderizer

[chilly ocean]

no, e is not g_0^-1

e is the neutral element of G with the original multiplication

[formal ermine]

oh

right

[chilly ocean]

yeah, and here the inverse I write is also wrt that operation

now h is inverse wrt *

[formal ermine]

but don't we need $g_1 * g_1\inv = g_0\inv \implies g_1g_0g_1\inv = g_0$?

[cloud walrus]

illuminator3 👻(#eric4honorable)

[chilly ocean]

no, you're mixing things up

^-1 is used in two different ways in what you wrote

you got to be more careful

[lavish gull]

$f(g_1) = {g_1}{g_0}{g_2} = a \
f^{-1}(a) = a{g_2^{-1}}{g_0^{-1}} = g_1$

[chilly ocean]

anyway I kinda solved it already, so I'll let other people try and alleviate your confusion

[cloud walrus]

coderizer

[formal ermine]

wait, how?

oh

wait I get that

[lavish gull]

@formal ermine are you trying to solve for the g_0 instead?

[formal ermine]

@lavish gull I have no idea what you've been doing this entire time

[lavish gull]

likewise

usually, to solve an equation, you write f(x)=y, then solve for x

what is your equation?

[formal ermine]

let $G$ be a group and $g_0 \in G \setminus \Set{e}$ a fixed element. let $* : G \times G \to G$, $g_1 * g_2 \coloneqq g_1g_0g_2$. show that $(G, *)$ is a group again.

[cloud walrus]

illuminator3 👻(#eric4honorable)

[formal ermine]

I've already shown a neutral element and associativity

I just have to show that an inverse element exists

[lavish gull]

if you are trying to solve f(a,b)=y, you can do that (maybe where f is a group operator) but you get solution pairs of (a,b) for every y

[formal ermine]

what?

[formal ermine]

$g_1 * h = g_0\inv \implies g_1 g_0 h = g_0 \inv$

ah

[cloud walrus]

illuminator3 👻(#eric4honorable)

[formal ermine]

I get what blitz was doing now

$\implies h = g_0\inv g_1 \inv g_0\inv$

oh that's trivial

[cloud walrus]

illuminator3 👻(#eric4honorable)

[coral spindle]

So here's a question: what is this group? We've taken a group G and produced a group (G, *), but what is its relationship as a group to G?

[formal ermine]

interesting question. I'll think about that tomorrow, it's already getting really late here and I still have homework that's due in an hour lol

[coral spindle]

alr

[lavish gull]

looks related to conjugation?

[coral spindle]

Not related to conjugation, no

[lavish gull]

if you let g2 = g1^-1

$g_1 g_0 g_1^{-1}$ is conjugation, no?

[cloud walrus]

coderizer

[chilly ocean]

hmm... I don't know

[coral spindle]

Should I give the answer in a spoiler?

If you're interested

[chilly ocean]

Of course not

[coral spindle]

[chilly ocean]

I think it's some reparametrization

g_0*1 = 1

Observation is that f(g*h) = f(g)f(h) for f(g) = gg_0

ah yeah

[coral spindle]

With that you can kinda see how they came up with this operation

[chilly ocean]

this is like this classic example of shifted non-zero rationals with multiplication

just more abstract

[coral spindle]

uh-huh

[chilly ocean]

You know, where they define multiplication by x*y = (x-1)(y-1) +1 just expand it to hide where it came from

[lavish gull]

how does f(g) = g g_0 relate to the original map he gave as f(g_1,g_2) = g_1 g_0 g_2

it looks like you just get the original group to me \
if you take the group operator $g_1 \times g_2$, then \
$f(g_1, g_2) = (g_1 g_0) \times g_2$ \
now let $u = g_1 g_0$ be any element

[hollow mica]

If V is the two dimensional vector space (F_p)^2 over F_p (the finite field of order p), and G = GL(V) is the group of automorphisms, then V has p+1 one dimensional subspaces, and in particular, there exists a group homomorphism ϕ: G -> S_{p+1} because every automorphism permutes the one dimensional subspaces. How can I describe the kernel and image of ϕ when p = 3?

so far, I know that |V| = (3^2 - 1)(3^2 - 3) = 48, and to find the kernel of ϕ I want all automorphisms that fix each one-dimensional subspace. Trivially the identity is one, but there may also exists automorphisms which move around the elements within each one dimensional subspace.

[cloud walrus]

coderizer

[lavish gull]

p+1 subspaces? why is that

[coral spindle]

I'm going to give V a basis v_1, v_2. Now if g is in the kernel of phi, then g(v_1) = mu_1 v_1, and g(v_2) = mu_2 v_2 for some nonzero mu_1, mu_2. Now additionally, these are assumed to fix every one-dimensional subspace, which is spanned by av_1 + bv_2, so av_1 + bv_2 is proportional to mu_1av_1 + mu_2bv_2, which tells us that mu_1 = mu_2. You can now check that the kernel is now going to be the automorphisms which are just scalar multiplication.

[hollow mica]

there are p^2 - 1 total (nonzero) elements in V, and each one-dimensional subspace contains p - 1 (nonzero) elements, so there are a total of (p^2 - 1) / (p - 1) = p + 1 one-d subspaces

[coral spindle]

I'm going to think about the image now.

Interestingly, this action should be 2-transitive

[hollow mica]

which action?

[tribal moss]

We can also explicitly list them as the subspaces generated by (1,y) for every y in Fp, plus the one generated by (0,1).

[coral spindle]

Your map ϕ: G -> S_{p+1} is an action of G on the set of 1d subspaces

[hollow mica]

oh I see

[coral spindle]

I'm not seeing any obvious characterisation of the image quite yet

The good thing is, for a small enough example, we can just compute it

Tropo already stated how you can list all subspaces, and since I pointed out that the kernel is scalar multiplication, we can focus attention to a much smaller collection of matrices

GL(V) should contain (p^2-1)(p^2-p) elements, so for p=3 that's 8 x 6 = 48, and up to scalar multiples that's only 16 matrices (correction) 24 matrices which generate the image

That's still a considerable amount of course...

[tribal moss]

Well, the size of the image is |G| divided by the order of the kernel.

And then the size of S_{3+1} doesn't leave many options for what the image can be.

[coral spindle]

Ah-hah! True!

Indeed I see I should've divided by 2, not 3

So in fact it is the entirety of S_4

@hollow mica you still around?

[hollow mica]

yeah

I am trying to understand tropo's point

[tribal moss]

The image is a subgroup of S_4 of order 48/2=24.

[hollow mica]

ok so this I see because of first-iso theorem

[tribal moss]

But S_4 itself only has 24 elements.

[hollow mica]

oh I see

that's neat

[coral spindle]

That's very nice

I wouldn't have expected that

[hollow mica]

I have another algebra problem that is mostly unrelated

[tribal moss]

It's only because F_3 is so small.

[coral spindle]

Right, because otherwise S_{p+1} gets really big

[hollow mica]

and for p = 2 I think the same thing is true

[tribal moss]

Yes.

But already for F_4 there will be permutations that can't be made.

[hollow mica]

yeah

ok my next question is

construct a natural map from Sym^2 (Sym^2 (V)) to Sym^4 (V)

where Sym^k is the kth symmetric power

[rotund aurora]

Books for rep theory?

[tribal moss]

If "symmetric power" means what it sounds like, the natural choice would be to map {{ei,ej},{ek,el}} to {ei,ej,ek,el} (where set brackets denote multisets).

[hollow mica]

by symmetric power I mean the vector space of symmetric tensors, i.e., elements of V* ⊗ ... ⊗ V* (d times) that are invariant under the action by S_d

[tribal moss]

Okay yes, then I think my answer matters. (The point is that we can make a basis of Sym^n(V) by taking n-element multisets of a basis for V).

((But if it turns out the scalar field has characteristic 2, I was never here!))

[hollow mica]

hmm I think I just failed a sanity check:

Sym^d is the subset of V* ⊗ ... ⊗ V* whose elements are invariant under switching arguments, but this vector space can be identified (is isomorphic) with the space of d-linear forms (maps V x ... x V -> K), but in order for such a map to be invariant under argument swapping, shouldn't the map be the same in each argument?

[tribal moss]

Hmm, I didn't notice the stars at first. Then I'm less sure of my interpretation. Can we assume that V is finite-dimensional?

[hollow mica]

Yes

[chilly ocean]

kowalski

[coral spindle]

My supervisor swears by Isaacs' character theory

[rotund aurora]

Thanks

[hazy portal]

can someone explain what is Quotient Group

im having hard time understanding this

[rotund aurora]

https://math.stackexchange.com/questions/14282/why-do-we-define-quotient-groups-for-normal-subgroups-only/14315#14315

You can read the answer by Arturo Magidin

If G is a group and H is a normal subgroup, then G/H is the group obtained from G by killing the elements of H, that is, in this new group, if h is in H and g is some element of G, then gh and g will be equal.

[pastel cliff]

R[x,y] is the polynommial ring of two variables - are elements of this ring \Sigma (r_i)(x^i)(y^j)

[coral shale]

yes?

[pastel cliff]

dummy question but i forgor

[coral shale]

R[x,y] is isomorphic to (R[x])[y]

[pastel cliff]

okie

[coral shale]

u can check this is the same as what u described

[coral shale]

or maybe defined to be equal to, cant remember

yh should be definition, pretty sure...

[pastel cliff]

so in R[x,y], (x) (pretend those are ang brackets) is the set of polynomials with no y values right

values isnt the right word but yk what i mean

[chilly ocean]

$$R[x, y]/\langle x \rangle?$$

[cloud walrus]

rakko

[pastel cliff]

im just asking what $\langle x \rangle$ looks like

[cloud walrus]

stμ₂dying

[chilly ocean]

Oops, I read the comma as a quotient / for some reason.

I should get my eyes checked.

I actually don't know what "no y values" means.

[tribal moss]

Polynomials where every term has at least one factor of x.

[pastel cliff]

ah that's what i meant

[delicate bloom]

also no constant term

[pastel cliff]

but there can be terms with a factor of y though

[chilly ocean]

Sure, xy.

[pastel cliff]

[pastel cliff]

too handwavey?

[chilly ocean]

It's a little bit handwavy. If you were submitting this to a strict grader, I'd suggest you change it.

I personally think it's okay. You have a correct idea and you clearly communicated it.

Actually, wait. How come no cancellation occurs? You should pick the terms with only y a bit more carefully.

You have a term with y^i in a and y^j in b, and the product of those terms has y^{i + j}. But what's stopping you from having something with y^{i - 1} in a and y^{j + 1} in b which cancels the first thing out in the sum?

A slightly more careful choice is needed. You had the right idea, but missed out on some technical details because of handwaving.

You can give a really clean proof of this by showing that the quotients are integral domains, by showing that they're isomorphic to polynomial rings over K in one variable.

[little root]

does anyone know why the bottom bullet is true? This is being done in abelian categories, so I don't think it's as obvious as $(\varphi, -\psi) \circ (\psi', \varphi') = (\varphi \circ \psi') - (\psi - \varphi')$. I think I'm missing something really obvious since the author says the assertion is trivial lol

[cloud walrus]

Frank

[pastel cliff]

will try this

[chilly ocean]

If you'd like a hint: ||apply the first isomorphism theorem to the map K[x, y] -> K[y] that sends x to zero.||

[pastel cliff]

showing for x must also suffice for y right

[chilly ocean]

Of course.

No need to do the same proof twice.

[pastel cliff]

somewhat adjacent question, what does $\mathbb{Q}[x] / \langle x(x-1)\rangle$ look like

preamble L

[cloud walrus]

stμ₂dying

[pastel cliff]

$\langle x(x-1)\rangle$ the set of polynomials that can be factored to have x(x-1) somehow right

[cloud walrus]

stμ₂dying

[pastel cliff]

maybe that's a weird way to word things but im getting used to rings still

[chilly ocean]

<x(x - 1)> is by definition the set of polynomials with a factor of x(x - 1).

[pastel cliff]

so quotienting Q[x] by it is just polynomials that don't have that factor

[maiden heath]

I don't understand this step here, $(\mathbb{C}\mathcal{S}n)^{\mathcal{S}{n-1}}$ denotes the subalgebra invariant under conjugation with elements of $\mathcal{S}_{n-1}$

[cloud walrus]

ArtyLeAardvark

[agile burrow]

$\mathrm{End}(V)$ is a representation of $G$ via $(g \cdot f)(v) = gf(g^{-1}v)$. Note that $f$ is $G$-linear if and only if $f(gv) = gf(v)$. Then $(g \cdot f)(v) = gf(g^{-1}v) = f(g g^{-1}v) = f(v)$. In other words, $f$ is fixed under the action of $G$. That is, $\mathrm{End}(V)^G = \mathrm{End}_G(V)$

[cloud walrus]

walter

[maiden heath]

Ohhhhhhhh

That clears up so much thank you!!!

[pastel cliff]

im being asked to find a commutative ring with identity which has a unique non-zero prime ideal - isn't any field an example of this?

also is this ever the case? doesnt this have to be true bc of distributivity

actually is the sum of elements of an ideal necessarily in the ideal

[void cosmos]

wdym a unique non-zero prime ideal?

only one prime ideal?

a field has no nontrivial ideals

only the field itself and the 0

[void cosmos]

try it out

[void cosmos]

try easy examples in Z_n or something

but yea

[pastel cliff]

exactly what i meant, question never said anything about nontrivial examples 😎

[void cosmos]

yea then your right

[agile burrow]

Isn't a prime ideal defined to be a proper ideal

In which case a field wouldn't work

[pastel cliff]

O

[void cosmos]

a prime ideal is an ideal that has the property that whenever ab in is in the ideal either a or b is in the ideal

[agile burrow]

In addition to the ideal not being the whole ring

[void cosmos]

yea

so you cant use the field yea

missed that haha

anyways there is an example

and it is easy to prove that it does infact have only one prime idieal

[pastel cliff]

will be back thn

[void cosmos]

yea

think of polynomial rings

[pastel cliff]

also i feel like this isnt necessarily truee right

[agile burrow]

What's your definition of an ideal

[void cosmos]

no

it depends

for me its true

cuz an ideal is a subgroup in the additive sense

idk if there are other ways of defining so thats why walter asked

[pastel cliff]

yeah probably being a goober rn

[little root]

usually (a + I)(b + I) is just defined to be ab + I. it seems like you're trying to multiply the cosets (a + I) and (b + I) elementwise but i don't think that's a thing people usually do (at least for cosets)

here "coset multiplication" just means (a + I)(b + I) := ab + I, not elementwise multiplication of sets

people do multiply ideals, but a + I isn't an ideal unless a = 0

[pastel cliff]

so that means these conditions never happen right

[little root]

i mean no, but the equation (a + I)(b + I) = {(a + r1)(b + r2) | r1, r2} isn't true

like (a + I)(b + I) is just defined to be ab + I

[pastel cliff]

oh wait i continued being a goober

[pastel cliff]

that's just the defn the problem gives

[little root]

oh what

[little root]

hmm ok that's odd lol

i feel like this is possible

[chilly ocean]

If I is two sided it is not possible

Right I mean just distribute?

(a +I)(b+I) = ab + aI +Ib + I^2 = ab + I

[little root]

oh yeah i think ur right

[chilly ocean]

Ideals are closed under left multiplication right multiplication and squaring

[little root]

they arent closed under squaring right

I^2 could be a proper subset of I (e.g. (2)^2 = (4))

but (ab + I) + I^2 would still just be ab + I since I^2 is a subset of I?

[chilly ocean]

Ohh I understand the question it’s asking if aI + Ib + I^2 can be a propped subset

of I

[pastel cliff]

i just saw that yeah

[little root]

wait and aI isn't equal to I either

[pastel cliff]

fuck discord

Il

[chilly ocean]

Yeah not necessarily

[pastel cliff]

which is i which is L

the world may never know

[little root]

yeah ok i think you can find a counterexample

[pastel cliff]

i think you accidentally gave one lol

[little root]

yeah i did

i just realized

lmao

[pastel cliff]

left is always contained in the right but not vice versa

[little root]

yea

[pastel cliff]

[chilly ocean]

Just try random ideals in Z and integers a,b should be easy to find a counter example for the backwards inclusion

Like I = 7Z and some random a,b

[wooden ember]

Having trouble on this one exercise from my Lie algebra course and the TAs weren’t sure how to proceed either. Given a nilpotent complex Lie algebra with basis {x_1,…,x_n} and a representation p:L -> gl(V) I want to show that the linear map lambda:L->C which takes x_i to an eigenvalue such that there is a non zero vector v belonging to the intersection of the generalized eigenspaces associated to the chosen eigenvalues. Show lambda is a weight

The actual weight part isn’t difficult

What hurts me is just showing that lambda is a representation

Ie that it preserves the lie bracket

I’d somehow want to restrict p to some submodule of V in which I can simultaneously triangularise but I’m struggling to construct this

[toxic zephyr]

we're talking about UFDs in lecture and last week when decomposing elements into irreducibles, my professor did something like
b=b_1*...*b_m
but this week he did
b=p_1^(r_1)*...*p_m^(r_m)
are the powers necessary? it seems like they are. like did my prof just forget to put them, or is it supposed to be understood that b_1 can be some power of an irreducible element?

[proud bear]

i think in the first one, the b_i aren't assumed to be distinct

[hollow mica]

can someone help me understand where my reasoning was wrong here?

I'm thinking of symmetric tensors wrong in some way

[chilly radish]

Subset?

It's a quotient

Not a subset

You quotient by the action of permuting the entires

And I don't see why you conclude what you do

You have functions where f(x,y)=f(y,x) for all x,y without f(x,_) being equal to f(_,y)

[hollow mica]

oh I see

[chilly radish]

For multilinear maps this does have implications on the structure of such maps

Saying "the same function" isn't 100% well-defined

I guess

Ok under this identification it should actually be true I think

But I have to think about it for a second

[chilly radish]

Wait I also didn't notice the star

My bad

It should be a subspace then I think?

I thought you were talking about the symmetric power of the vector space itself, which is a quotient

Ok wait

Mrean

[hollow mica]

Yeah

[chilly radish]

Do you mean those products of functionals who are invariant under premuting their input? i.e. if you have f(x)g then it's in Sym^2 if f(x)g(v(x)w)=f(x)g(w(x)v)

[hollow mica]

these are my definitions

ohhh

I wrote down the definition of B_symm , symmetric bilinear forms

and I thought I could extend that to the definition of Sym^d

but those are two different things

[hollow mica]

definition 18.8 is the standard definition right?

[hollow mica]

so scratch anything I said about functionals

[chilly radish]

Oh ok lol

[chilly radish]

I've seen this definition too

My confusion comes from conflating it with something else

I was confusing the symmetric algebra and the algebra of symmetric tensors

The former is a quotient, the latter a subspace

[hollow mica]

ah I see

[chilly radish]

So yea scratch what I said too

So what is the question now

[hollow mica]

I guess like

[chilly radish]

I think you could expand the definition of B_symm to n-linear forms like you did

And what you said in light of the new context is still false. Not all the factors have to be identical

[hollow mica]

why is the basis at the bottom linearly indep

[chilly radish]

In particular, saying the factors are identical in the tensor product is not well-defined

Because like

[hollow mica]

It’s clear to me why they span

[chilly radish]

(av)(x)v=v(x)(av)

This also shows you can have symmetric tensors who are not just a product of the same linear functional

Ok let's see

[chilly radish]

Is there not a proof on the next page?

[hollow mica]

Nope

these notes are really brief

[chilly radish]

Ok so

[hollow mica]

wait actually

isnt any set of pure tensor linearly independent

[chilly radish]

Not necessarily, eg if you exceed the dimensions

But in particular if you take s linear combination of these, you will get a linear combination of a subset of the basis of the tensor power

Which is of course linearly independent

You just have to make sure there's no like, cancellations under the averaging map (where one component becomes 0), but otherwise it should be pretty evident that this is linearly independent

Because the basis for the tensor power is lin ind

By the basis I mean all the e_i1(x)...(x)e_id

Because you are multiplying each sum by a scalar the computation is gonna be a bit annoying

[hollow mica]

wait hmm

I thought if

v_1, ..., v_n are linearly independent, then so are a_1 v_1, ..., a_n v_n for any nonzero scalars a_i

[chilly radish]

Yea that's true

Actually yea, this should be fine

You're right

[hollow mica]

yay

is this just extra notation

it doesn't mean anything idt

[chilly radish]

Yea it is

I guess what I worried about is there might be cancellations between the sums (when you permute), but this can't happen because the number of elements of each basis elements is always different

So linear independence is evident

[hollow mica]

everythings positive tho?

[chilly radish]

When you take libear combinations I mean

You.might have a as a coefficient of one and -a for another

But that's irrelevant here

[hollow mica]

oh

[hollow mica]

and u dont have to worry about that if you use this right

[tribal moss]

At least in characteristic !=2 they're naturally isomorphic, right?

[chilly radish]

I don't see why they would be?

My defn of symmetric algebra is quotienting the tensor algebra by the action of Sn

And symmetric tensors is the algebra generated by those invariant under the action of Sn

Are you saying the averaging map induces an isomorphism?

[tribal moss]

The map (v otimes w) -> ½(v otimes w) + ½(w otimes v) ought to be compatible with the quotienting and land in the symmetric subspace.

Hmm, for larger products we need to divide by n! instead of 2. Perhaps it's better just to say characteristic 0 for now.

[chilly radish]

Yes

That's the averaging map described in the image mrean sent

This should work

[tribal moss]

To be honest, it confuses me that Mrean's text keeps saying V* in some definitions and V in others, with little notice of which is when.

In definition 18.8 $\mathrm{Sym}^2(V)$ is a subspace of $V^{\otimes 2}$ which I take to mean $V\otimes V$.
But the example (?) at the beginning of section 19.1 then says ${\text{things in } V^* \otimes V^} = \mathrm{Sym}^2(V)$.
Sure $V$ and $V^$ are isomorphic when $V$ is finite-dimensional, but that isomorphism depends on a choice of basis, so it requires at least a bit of care to pass from one to the other.

[cloud walrus]

Troposphere

[hollow mica]

yeah I am trying to understand that too

one thing, though

the definition of a tensor product using the universal property makes most sense to me

but a definition like this:

how is this sufficient?

doesn't one need to define what addition is, what additive inverses are, etc.

somewhere else I read that

(ignore the fact that one is V ⊗ W and the other is U ⊗ V)

[hollow mica]

but I don't understand this, because ⊗ is used to define the vector space, but at the same time, the definition of ⊗ involves a linear map into U ⊗ V... isn't this circular?

[uneven jackal]

not really

the "spanned by e_i o e_j" can be a definition

a syntaxic definition

[tribal moss]

When one says "the vector space with such-and-such basis", it is implicit that the elements of the vector space are (finite) linear combinations of those basis elements. The vector space operations then simply become operations on the coefficients on those basis elements.

In other words "the vector space with basis B" is isomorphic to the space of functions f: B -> R where f(b) is nonzero only for finitely many b. (The function gives the coefficient of each basis element in the vector you're looking at). Addition and multiplication then works pointwise, like in other function spaces.

[hollow mica]

then what is the significance of this

[tribal moss]

There are two different uses on the otimes symbol going on here. There's one whose operands are vector spaces, and it produces a vector space in return. Then there's one whose operands are vectors and it produces an element of the vector space U otimes V. Formally they're two completely different operations; we just typeset them with the same symbol because it helps remember what goes together (once you've become used to it).

And in "the vector space with basis {ei otimes fj}" the otimes is formally just a fancy way to write an ordered pair of ei and fj.

But after the dust has settled and we define the otimes-on-vectors operation, it ends up actually producing the basis element (ei,fj) when applied to ei and fj, so the "ei otimes fj" notation is just anticipating that fact.

[hollow mica]

what exactly is the precise definition of ⊗ (as an operation that takes in a vector from U and a vector from V)

is it

[tribal moss]

That depends on which of the definitions of the tensor product of spaces you're using.

With the universal-property definition, the otimes-on-vectors map is the unique bilinear map from U and V to the mystery space that every other bilinear map from U and V factors through.

[hollow mica]

This is my attempt at defining the tensor product without overloading operators:

Let U and V be two spaces over a field K, where u_1, ..., u_m and v_1, ..., v_n are their respective bases. We will define the tensor product U @ W as the set of all formal linear combinations of the ordered pairs (u_i, v_j) over all i, j, and each ordered pair will be denoted as u_i ⊗ v_j.

[tribal moss]

Exactly.

[hollow mica]

Now how do I formally define this part (with my notation)

[tribal moss]

We can now prove that there is exactly one bilinear function f: U×V -> U @ V such that f(ui,vj) = ui ⊗ vj for all i and j. We then also notate that function with a ⊗ symbol.

$$f(k_1u_1+\cdots+k_mu_m, k'_1v_1+\cdots+k'nv_n) = \sum{i,j} k_ik'_j (u_i\otimes v_j)$$

[hollow mica]

Yes, so f is determined by our choice of f(u_i, v_j)

[cloud walrus]

Troposphere

[hollow mica]

where exactly do we use this new notation

[tribal moss]

It can be a more compact way to write elements of U @ V than to write everything out in basis elements.

And, as noted above, it is an important function because it's the one that appears in the universal property of the tensor product space.

[hollow mica]

okay I see, so for two arbitrary vectors u = a_1 u_1 + ... + a_m u_m and v = b_1 v_1 + ... + b_n v_n,

the element sum_{over all i, j} ((a_i * b_j) u_i ⊗ v_j) is the same as f(u, v), and we abbreviate this as u ⊗ v since we can precisely (and uniquely) define f for some choice of bases of U and V

wait, so big picture:

originally U @ V had no "bilinear" structure right? Like (u_1 ⊗ v_1) + (u_2 ⊗ v_1) is not necessarily the same as (u_1 + u_2) ⊗ v_1 if we consider them as formal linear combinations (here I am using the original definition of ⊗, not the overloaded one that we just defined)

[tribal moss]

In the "original" definition we couldn't even speak of all of u1 ⊗ v1 and u2 ⊗ v1 and (u1 + u2) ⊗ v1, because ⊗ was only defined on basis elements at first, and u1+u2 cannot be a basis element if u1 and u2 are.

[hollow mica]

Oh that is true

I guess my one confusion is, by introducing the map f, didn't we just impose some artificial bilinear structure on V @ W ?

because again, V @ W is just a bunch of formal linear combinations

[tribal moss]

f just maps into V@W -- it doesn't change how the vector space operations inside it works.

It happens to be compatible with the operations that were already there (but that of course deserves some proof).

[hollow mica]

oh, and we defined f / ⊗ in such a way that (u_1 ⊗ v_1) + (u_2 ⊗ v_1) is the same element as (u_1 + u_2) ⊗ v_1 (here I am using our second definition of ⊗)

[tribal moss]

Yes. (This is what allows us to say "f is a bilinear map").

[hollow mica]

what does compatible mean here?

[tribal moss]

Just the bilinearity conditions, such as f(u+v,w)= f(u,w)+f(v,w) and so forth.

[hollow mica]

because the only operation that was already in V @ W is just the "formal linear combination" thing

oh

[tribal moss]

But "formal linear combinations" does tell you how to add such combinations, and how to multiply them by scalars.

[hollow mica]

Sure, but how do you check that f is compatible with those (the formal linear combination)

because I keep writing down a trivial equation lol

[tribal moss]

It does look fairly trivial, yes. There's lots of notation, not a lot of substance. What substance there is would be about being clear when your write + whether it's addition in K, or in U or V, or in U @ V. The various definitions make sure you can pass from one to the other when you need to, which is good -- that means the proof works!

[hollow mica]

We want to check that a ⊗ b := f(a, b) and c ⊗ d := f(c, d), elements of U @ V, when added, equal what exactly?

[tribal moss]

That will not usually have a nice form that we need to check. The conditions on f only speak about what happens when either a=c or b=d.

(Beware that most elements of U @ V are not even in the range of f, so we shouldn't expect to be able to write an arbitrary element of it in a nice form).

[hollow mica]

I thought that we defined this function f so that it already followed the bilinearity conditions (and then we showed it as unique if it sends f(u_i, v_j) to u_i ⊗ v_j). Now, we're trying to show that this f is "compatible" with the addition operation in V @ W ?

or is my last sentence irrelevant

[tribal moss]

"Compatible" and "bilinearity conditions" are the same thing. They're not something to show separtely.

[hollow mica]

how does bilinearity relate to formal linear combinations (the actual operation in V @ W) though

[tribal moss]

If you're getting confused by "that deserves proof" here, I apologize. I didn't mean that as something deep, just what you probably had already convinced yourself of.

Formal linear combinations is not an operation in V @ W -- it's what the elements of V @ W are.

[hollow mica]

If sum a_{i, j} u_i ⊗ v_j is an element of V @ W, and so is sum b_{i, j} u_i ⊗ v_j, then their sum (the vector addition operation in V @ W) is the formal linear combination sum (a_{i, j} + b_{i, j}) u_i ⊗ v_j, right

Ok I think what I am trying to check doesn't make sense

let me try and give a summary of how we defined the tensor product

[tribal moss]

That is right, though.

[hollow mica]

U, V are vector spaces over the same field
U @ V is the set of formal linear combinations of u_i ⊗ v_j (⊗ is just an arbitrary symbol), and this set takes on the structure of a vector space over the same field as U and V

[tribal moss]

Yes.

[hollow mica]

Next we defined a bilinear map f: U x V -> U @ V, and we showed that if f(u_i, v_j) = u_i ⊗ v_j, then the rest of the map is determined by bilinearity, and in particular, f(u, v) = sum (a_i * b_j) u_i ⊗ v_j where u = sum a_i u_i and v = sum b_j v_j

[tribal moss]

Yes.

[hollow mica]

so it makes sense to regard/denote the element f(u, v) as u ⊗ v

[tribal moss]

Yes.

[hollow mica]

and this way, we can show that for any linear map h: U x V -> W, where W is also over the field K, there always exists a unique linear map g: U @ V -> W such that g ∘ f = h

[tribal moss]

Yes.

[hollow mica]

how do we prove this again

[tribal moss]

I don't think we've proved that in the chat yet.

But the basic idea is that g(u_i ⊗ v_j) must be h(u_i, v_j), and everything else then follows by linearity of g (to define it on the other elements of U @ V) and linearity/bilinearity of everything (to show that gof ends up equaling h).

[hollow mica]

Yes, h(u_i, v_j) = g(f(u_i ⊗ v_j) = g(u_i ⊗ v_j)

so define g by
g(u_i ⊗ v_j) = h(u_i, v_j)
then for a general x = a_{i, j} u_i ⊗ v_j, we have by linearity of g:
g(x) = sum g(a_{i, j} u_i ⊗ v_j) = sum a_{i, j} h(u_i, v_j)

we didn't have to use bilinearity anywhere

[tribal moss]

Bilinearity is for making sure that g(f(u,v)) = h(u,v) for all u and v.

[hollow mica]

I only used the fact that h(u_i, v_j) = g(f(u_i ⊗ v_j) = g(u_i ⊗ v_j) though

which is because of the definition of f

[tribal moss]

That shows it when u and v are basis vectors, not when they are arbitrary elements of U and V.

Then we have to decompose each of u and v as combinations of basis vectors and then use bilinearity to get that to be a combination of g(ui,vj) isntead.

[hollow mica]

hmm, where in my proof did I do that

or is my proof just incomplete

[tribal moss]

As far as I can see it's just incomplete so far.

gtg, have a meeting..

[hollow mica]

Ok, thanks for the help

[hollow mica]

It is true that if V, W are vector spaces with respective dimensions m and n, then the maximum rank of an element in V ⊗ W is min(m, n).

If U is an additional vector space of dimension k, isn't it true that the maximum rank of an element in V ⊗ W ⊗ U is min(m, k)? Because V ⊗ W ⊗ U = (V ⊗ W) ⊗ U and dim(V ⊗ W) = mn so we have the bound min(mn, k) = k, and the same argument on V ⊗ W ⊗ U = V ⊗ (W ⊗ U) gives min(m, nk) = m so the maximum rank of an element is min(m, k)

[tribal moss]

Hmm, that doesn't sound right. The tensor product on spaces is commutative (up to isomorphism) so there's no reason why V ⊗ W ⊗ U would know which of the three spaces is the middle one. And if U has dimension 1, then V ⊗ W ⊗ U is isomorphic to V ⊗ W, and then we'd certainly be back to min(m, n).

So e.g. K³ ⊗ K² ⊗ K¹ has a maximal rank of 2, which is neither the mininum nor the maximum of anything meaningful.

(I remember reading somewhere that even though rank decomposition of two-way tensor products is a simple matter of linear algebra, decomposing three-way tensors is deeply mysterious and hard in general).

[tribal moss]

The reason why this won't work is that even though you can always write an element of V ⊗ W ⊗ U as a sum of k things of the form s_i ⊗ u_i where s_i in V ⊗ W, there's no guarantee that the s_i can be a simple element of V ⊗ W.

[hollow mica]

ohh I see

so technically the statement "the maximum rank of an element in V ⊗ W is min(m, n)" is not always true

when V or W are also tensor products

[tribal moss]

Well I'd say technically it's true, but "rank" really ought to be "rank over V and W".

[hollow mica]

ah that would make more sense

problem:
show that ϕ: End(V) x End(V) -> K sending (A, B) -> trace (A ◦ B) is symmetric and bilinear

[brittle pebble]

Must generators of cyclic groups be unique? If not, why is it that the klein four group is not cyclic (a^2, b^2, c^2 can all be generators)?

[next obsidian]

No

The Klein four group doesn’t have a generator

Because no single element generates it

[barren sierra]

tfw I ask a question and realize it's dumb 💀

[tribal moss]

The only cyclic groups with a unique generator are C_1 and C_2.

[barren sierra]

nvm

Yea

Realized that mid typing

[tribal moss]

Oh well, it answers Soda's question too.

[coral shale]

if a generating set contains x, then the same set with x^-1 instead is also generating

[brittle pebble]

yea, I appreciate that. I realized that my definition of cyclic is funky (also a^2, b^2, c^2 generate ^2 elements but not the.... single elements of V4)

[tribal moss]

That's even worse; the only square in V4 is the identity.

[brittle pebble]

f

thank you, i get it now

[toxic zephyr]

so i proved the first 4 things just fine but how am i supposed to do these last two things for an arbitrary integral domain R?

if $F$ is the field of fractions itself, i think it would just be as simple as

$\vfunc{j}{R}{F(R)}{r}{r1_R^{-1}}$

[cloud walrus]

nix

[chilly ocean]

Yes. I misread the question.

Ignore what I wrote, please.

[toxic zephyr]

np

[chilly ocean]

I don't think it's asking you to prove this for a general integral domain R. It's saying that if you have this, then F is isomorphic to the field of fractions of R.

It's basically just restating the theorem before it.

[toxic zephyr]

yeah that's what i thought too, but the prof said we did have to prove those two things. i'm thinking maybe he was just rushed after class and didn't remember his intention/read it closely enough.

[chilly ocean]

Probably. The only way I can interpret "prove these things" is to just refer back to the theorem with the 4 things (which you already proved).

You should ask them for clarification.

[toxic zephyr]

yeah, will do. just wanted to ask here first to see if maybe i was missing something.

thanks @chilly ocean 🙂

[chilly ocean]

No problem. Sorry for the initial confusion.

I thought you were asking how to prove that an integral domain embeds into its field of fractions.

[celest cairn]

How do I do this?

[south patrol]

equivalently you must show thet Q[sqrt(2)] is closed under inverses [as a subring of C i mean]

[gusty thistle]

When do we know that complex conjugation is an element of a Galois group of an extension?

Is there a criterion for this?

[next obsidian]

What does this mean

[gusty thistle]

sorry

[next obsidian]

Are you asking about extensions of Q?

[gusty thistle]

Yes

[next obsidian]

It’s always in there

If your extension lives in R it’s the identity

If it doesn’t, then it’ll satisfy alpha^2 = id

[gusty thistle]

I thought so too but then i found this on stack exchange : (

[next obsidian]

I mean you have to embed the extension into C, but then once you do that this should make sense

[gusty thistle]

but i dont rlly understand it

[gusty thistle]

yeah it seems very logical to me too

[next obsidian]

Well to me a Galois group necessarily is for a Galois extension

[gusty thistle]

but why is it true for Galois extensions and not for non Galois extensions ?

[next obsidian]

So this ain’t an issue

Because it can fail to be normal

[gusty thistle]

yeah for me too usually xD

[next obsidian]

You need that the Galois conjugates stay inside your extension

[gusty thistle]

what does normal mean again : |

[next obsidian]

And failing normality says that you don’t have every root of a polynomial

Sure

Normal means if alpha is in L, every other root of its minimal polynomial is in L

[gusty thistle]

ah yes

[next obsidian]

This is why it could be that the complex conjugate isn’t in the extension

Because you are “missing” elements in your extension

That’s what failing to be normal is

You miss stuff that “should” be there

[gusty thistle]

yes that seems reasonable

let me think abt it for a bit

thanks : ]

[next obsidian]

Anyway the proof is you assume normal is as follows

Let x in L, we want to know alpha(x) is in L whwre alpha is complex conjugation

Let f be x’s minimal polynomial over Q

Then note that alpha(x) is also a root of f, because 0 = alpha(f(x)) = f(alpha(x))

You can do this because alpha fixes all the coefficients of f

Because the coefficients are real numbers and alpha does nothing to them