groups-rings-fields

[vestal snow]

If anyone has any ideas @ me

[next obsidian]

It doesn’t really

Any functor F:C -> D turns into a functor F^op: C^op-> D^op by just declaring the objects and morphisms to be the same thing essentially.

If F and G are an adjoint pair, then once you pass to the opposite categories G^op and F^op form an adjoint pair

If you really don’t want to do a duality argument then you should be able to just take your proof for right adjoints and then just basically do everything backwards

At some point you might need to talk about coimage but in an abelian category image and coimage are naturally isomorphic so it should push through

[unique juniper]

$$\mathbb{Q}[x]/\langle x^2 - 2 \rangle \cong \mathbb{Q}[x]/\langle x^2 +4x+2 \rangle $$

[cloud walrus]

Yes

[unique juniper]

how would i go about showing this?

[hot lake]

by finding an isomorphism

[unique juniper]

yea that would work

but

any idea on an isomorphism ?

[hot lake]

what kind of isomorphism do you need

[unique juniper]

wdym

[hot lake]

isomorphism of groups ? vector spaces ? rings ? fields ? Q-algebras ? sets ?

[unique juniper]

hmm

rings pls

[rustic crown]

sets >.<

[hot lake]

well then first argue that an isomorphism of rings has to come from an isomorphism of Q-algebras

then since they are Q-vector spaces of dimension 2 you just need to find the images of 1 and the class of x; so you need some element x' in the second one that satisfies x'² = 2

[unique juniper]

well we havnt covered Q algebra so im guessing its not necessary ?

[hot lake]

then forget about my rambling and try to find an element of the second ring that squares to 2

because any isomorphism will send the class of x in the ring on the left, to something that squares to 2

[unique juniper]

i was thinking we could somehow apply an isomorphism theorem

[hot lake]

for example if you find a map f from Q[x] to Q[x]/(x²+4x+2) whose kernel is the ideal generated by x²-2 ?

[unique juniper]

yes

[hot lake]

well then you would need to find a suitable candidate for f(x)

[unique juniper]

yes, thats where im kinda stuck

[hot lake]

and guess what, f(x)² would need to be equal to 2

so you are back to trying to find a square root of 2 in Q[x]/(x²+4x+2)

[unique juniper]

hm why

[hot lake]

because (x²-2) is in ker f

so f(x²-2) = 0

but f(x²-2) = f(x)²-f(2) because f is a ring morphism

and f(2)=2 because if f(1) is not 1 you have some serious problems for a ring morphism

so f(x)² = 2

[unique juniper]

what problems lol

[hot lake]

f(1 * 1) = f(1) = f(1) * f(1)

[unique juniper]

oh right

[hot lake]

so uuuh

well both rings are domains in this case

so you get f(1) = 1 or f(1) = 0

and if f(1) = 0 then f(x) = f(1 * x) = 0 * f(x) = 0 forall x

[unique juniper]

ye

[hot lake]

that won't be an isomorphism I'm pretty sure

so f(1) = 1

and you can deduce from this that f(r) = r for any r in Q

[unique juniper]

ok

so we are looking for stuff in Q/x^2+4x+2 that will square to 2

[hot lake]

yes

you could pick some element

say (ax+b)

and square it

and see what happens ?

[unique juniper]

okay

[hot lake]

I'm not sure how useful that would be though

[rustic crown]

(i kinda feel like it would have been more <some word> if we proved both of them are isomorphic to Q(sqrt(2)))

[carmine fossil]

If you want the answer,just try (x+2)^2

So you will be mapping x+<x^2-2> to (x+2)+<x^2+4x+2>

[unique juniper]

i see

so uh

are you saying that

0 + <x^2-2> -> (2) + <x^2+4x+2>

[carmine fossil]

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

[carmine fossil]

How are you concluding that?

[unique juniper]

i dont understand your map

the goal is to find a map with kernal of x^2-2 right?

that is surjective

[carmine fossil]

So, elements of Q[x]/x^2-2 will be elements of form a+<x^2-2>

[unique juniper]

yes

[hot lake]

to define a ring map from Q[x]/stuff into something, you only need to give the image of x

the image of everything else you get to deduce with the properties of ring morphisms

[carmine fossil]

x+<x^2-2> is in that set

And it generates Q[x]/(x^2-2)

[unique juniper]

ok

[carmine fossil]

Similarly (x+2)+<x^2+4x+2> generates Q[x]/x^2+4x+2

You map x+<x^2-2> to (x+2)+<x^2+4x+4>

and deduce the map using the fact that it's a homomorphism

[vestal snow]

How do I prove that if the wedge product is 0 then the vectors are linearly dependent?

[chilly ocean]

iirc the wedge product is alternating and multilinear, so i believe is the same reason as the determinant

[vestal snow]

Which is?

[rustic crown]

did you read Lemma 4.3?

[vestal snow]

Yes, but I don't think we can use that here

[rustic crown]

if v1,... vl are linearly independent then M = <v1, ..., vl> is a vector space of rank l

[vestal snow]

Ah I see

[rustic crown]

and in the proof we give an explicit basis for the module

[vestal snow]

I think that we're sweeping something under the rug

We haven't proved that if M is a submodule of N, then wedge^n M is a submodule of wedge^n N

Wait actually nvm

I think it follows from the universal property

[rustic crown]

no i'm directly applying the lemma on the module M = <v1, ..., vl> which has rank l

the lemma says that the l-fold wedge of M as rank 1

[vestal snow]

Okay, I don't see how it follows from this that they are linearly dependent

[rustic crown]

yea we need this

[vestal snow]

I guess the correct version of that will be that there is map from wedge M to wedge N

[rustic crown]

otherwise i'm only showing that the wedge is non-zero in l-fold wedge of M and not l-fold wedge of V

[vestal snow]

Not injective in general

[rustic crown]

oh lol

[vestal snow]

I think it still works

[rustic crown]

take v1, ..., vl, and complete it to a basis 😛

[vestal snow]

Because the map must take 0 to 0

[rustic crown]

now take the l-fold wedge of V?

[vestal snow]

What if V is infinite dimensional?

[rustic crown]

sad

[vestal snow]

I think it works though

Like consider v_ 1,...,v_l

and the submodule generated by them

It is free of rank l

Then prove that we have 0 = v_1 wedge ... v_l

In the wedge product of this submodule

[rustic crown]

universal properties does show this right

[vestal snow]

Given M a submodule of N, we have a natural map from the M^(x)n to N^(x)n

And then we use the universal property of the kernel

to get a map from the wedge prod of M to wedge prof of N

[rustic crown]

if N --> M is a submodule, then the composition N^l --> M^l --> wedge M is a alternating map

[vestal snow]

which must map zero to zero

[rustic crown]

so we get a natural map wedge N --> wedge M.

oh we're saying the same thing ig

[vestal snow]

Yeah

I went through the tensor product

Okay so in our rank l module we have v_1 wedge ... v_l = 0

[rustic crown]

i'm kinda not sure... is the map we get from wedge M to wedge N not injective?

[vestal snow]

In general I don't think so

[rustic crown]

then this won't work right

[vestal snow]

Oh wait hold on

We're dealing with vectors

So we get flatness

I don't know if the map will be injective or not

Let's assume it is for now

What do we do after that?

[rustic crown]

then there isn't anything left to show

[vestal snow]

one sec I forgot what I wanted to show

[rustic crown]

wedge of v1...vl in the submodule is non-zero as that is the basis for one-dimensional wedge M

if the map is injective then this non-zero can't go to 0

[vestal snow]

Yup that does it

Injectivity follows from the following fact

M -> N is injective implies M (x) M -> N (x) M is injective. We also have M -> N injective implies N (x) M -> N (x) N is injective

you can now repeat the process

To get that the natural map between the tensor products is injective

After that you need to show that the kernel of the map into the wedge product has kernel equal to the submodule of the tensor product of M generated by the alternating elements.

Which I don't think should be too difficult

[oblique river]

Unless there are some hypotheses on M and N i dont think this is true. Consider the M = Z/2Z and N = Z/4Z and the map M -> N sends 1 to 2

[vestal snow]

Vector spaces

[oblique river]

If you tensor with M now, both sides will be Z/2Z but the map will be the 0 map

[vestal snow]

Both are vector spaces

[oblique river]

Oh god why are you calling them M and N

[vestal snow]

Wdym?

[oblique river]

But yes then my bad i didnt read far up enough in the logs

Why not call them V and W haha

[vestal snow]

Oh yeah my bad

[rustic crown]

anyway can we do it without using that the map is injective?

[oblique river]

But yea it is true for vector spaces

[rustic crown]

the problem was the forward direction in

[oblique river]

So if thats all you care about then why arent you in #linear-algebra then you dont have anything to worry about

Oh I see, and V is a vector space over what field?

R or C i suppose

[rustic crown]

i think its arbitrary field

we did define the notion of wedge without "dividing by n!"

[oblique river]

Oh, i feel like things with wedges get weird over char 2

Because the usual proof that x \wedge x = 0 doesnt work

[rustic crown]

yea makes sense

[oblique river]

Sry doing some googling and there are other ways to define it where that’s still ok

So if you want you can take a very computational approach. Pick a basis for the span of the vectors so you can write them in coordinates and then think about row reducing some matrix

[rustic crown]

no so the problem is its easy to see that the wedge over the span W of the vectors is non-zero

but the wedge in W non-zero doesn't imply the wedge in V is non-zero without assuming that the map would be injective

[oblique river]

Oh i see this is the part of the conversaon i joined in on earlier

Well you dont have to assume anything, you can prove it :P

[rustic crown]

if V was finite dimensional, there was an easy argument, just complete v1,...,vl to a basis and now the particular wedge is a basis in the lth wedge of V

was just wondering if we were overcomplicating things by thinking in that direction

[oblique river]

Hmm maybe? Ill ponder it

[fickle brook]

so here's something i'm wondering

is it possible for a field to have roots of unity of all orders, but fail to be algebraically closed?

[sturdy marsh]

probably yeah

take any algebraically closed field

and adjoin an indeterminate or something

yeah that should do it

C(t) has all roots of unity

but is not algebraically closed

[golden pasture]

something nice could be like $\mathbb Q_p\left(\mu_{\infty}\right)$ `union' of all cyclotomic extensions and also happens to be the maximal abelian extension

[cloud walrus]

ariana

[fickle brook]

@oblique river wait can you give an algebraic equation that has no solution in C(t)

[oblique river]

x^2 - t

the roots are \pm \sqrt{t} which don't exist in C(t)

[fickle brook]

ah hm

[oblique river]

another example you could consider over Q is like, take Q and adjoin all the roots of unity

sometimes written Q(mu_\infty)

this isn't algebraically closed because it doesn't contain the cube root of 2

but it does have every root of unity

[unique juniper]

$J/I$ is a prime in $R/I$ iff $J$ is prime in $R$

[cloud walrus]

Yes

[unique juniper]

so if j is prime then (R/I)/(J/I) is an integral domain

not sure how to show its prime tho

[oblique river]

an ideal P is prime if and only if R/P is a domain

prove that R/J is isomorphic to (R/I)/(J/I)

[unique juniper]

yeah

then we need to show j/i is prime

which is where im kinda stuck

[old lava]

just do what buncho said

J/I prime iff (R/I)/(J/I) is an integral domain iff R/J is an integral domain iff J is prime

[unique juniper]

oh right sorry my bad

ty

[wicked stump]

Having trouble with a couple of problems:

Not sure how to tackle part b of this question - I assume the fact that H contains the kernel is important but I don't understand how to use that to set up the isomorphism

Also for this problem, I've seen an argument where m=2 and you can sort of do it exhaustively by cases and checking where g^2 could possibly land. But I'm not sure where to begin when trying to generalize it.

[rustic crown]

consider the composition G --> G' --> G'/H'.

what's the kernel?

[wicked stump]

hmm

the kernel of G to G' is just ker(phi)

the kernel of G' to G'/H' is H' I think

and if H' is the image of H, would the kernel of this composition then be H?

wait

no is it ker(phi)

[rustic crown]

yep, that's what we want it to be.... but notice exactly where we need ker(phi) is contained in H

the kernel is the inverse image of H'

[wicked stump]

Oh... okay I think I sort of see it

If you set up this mapping right, it becomes a homomorphic image of G

and every homomorphic image is isomorphic to a quotient group of G right

[rustic crown]

with the kernel

[wicked stump]

G/K

[rustic crown]

which would be (pi*phi)^-1 ({e}) = phi^-1(pi^-1({e})) = phi^-1(H')

[wicked stump]

Ahh right so the kernel of this mapping is exactly H

[rustic crown]

how do you prove H = phi^-1(H')?

H contained in phi^-1(H') is easy

[wicked stump]

umm

not sure

[inner pine]

just to make sure, primitive polynomials, finite fields, and things along that vein (information theory) count for abstract algebra right?

[wicked stump]

i know that some b in phi^-1(H') has to map to an element in H'

[rustic crown]

so if you have an element g in G such that phi(g) in H', then can we say g actually lies in H?

[wicked stump]

I think no?

because phi isn't one-to-one?

it's only onto

[rustic crown]

(we still haven't used the hypothesis that ker(phi) is contained in H)

so use the fact that its onto!

phi(g) = phi(h) for some h in H

now what can we say about (h^-1)*g?

[wicked stump]

OH

it's in the kernel

so it's in H

[rustic crown]

yea!

h^-1g in H => g in H

so kernel of composition is indeed H

and the composition is clearly an epimorphism (i would use the word surjective, but the problem uses this so okie)

[wicked stump]

Great okay that helps a lot

Thank you very much

[rustic crown]

for this, do you know lagrange's theorem?

[wicked stump]

Yes

[rustic crown]

yea so G/H is a group of order m. and gH in G/H is an element

[wicked stump]

I know Lagrange says m = |G|/|H|

[rustic crown]

yea, it says the order of the subgroup divides the group, if the group is finite. So if G is a finite group, and g is an element, we can consider the subgroup H = <g> generated by g. now g^|H| = e => g^|G| = e

[rustic crown]

using that here, gH is an element and group has order m, so (gH)^m = identity in G/H = eH

so g^mH = eH => g^m in H

[wicked stump]

Ahh okay

That makes sense

But what if H isn't a cyclic subgroup?

It still holds because H has to have an order that divides the order of the entire group anyways, right

Or something like that, I'm just not entirely clear on that point

[rustic crown]

this was just a separate claim... it just says order of an element divides order of the group, and is a special case of lagrange's theorem when the subgroup is cyclic

[wicked stump]

Ok

I think I get it more clearly now, just had to sketch it out myself on paper.

Thank you so much!!

[rustic crown]

[viscid pewter]

someone fact check me, 29b is like 3rd isomorphism theorem just right

[rustic crown]

yep

[viscid pewter]

sick

[rain crescent]

Can someone help me with this question? I was thinking a must be coprime with 48 because if it's not then x^a will get us less elements in the codomain but im not sure if the argument is correct

[unique juniper]

yes looks like it

[vestal snow]

Is it true that if $W$ is a subspace of $V$, then $\wedge^l W$ can be naturally embedded into $\wedge^l V$.?

[cloud walrus]

Have a Banana, Bitch

[unique juniper]

$\mathbb{Z}[x]/\langle x^2 + 1 \rangle \cong \mathbb[Z][i]$

[cloud walrus]

Yes

[unique juniper]

my map was phi : z[x]/x^2+1 to z[i]

phi(m + nx + <x^2+1>) = m + ni

this works right? i was looking at a solution of this and they did it somewhat differently

wdym

oh

is it good now :D

wait did i change the right thing

[unique juniper]

you meant this?

wait i am mapping m to m and ni to nx

1 to 1 and i to x

OHHHHH

i seee lolol

lol thank you

[vestal snow]

Let $V$ be a vector space and $W$ a subspace. Show that $W^{\otimes l}\cap A(V)=A(W)$ where $A(V)$ and $A(W)$ represent the modules generated by alternating pure tensors in $V^{\otimes l}$ and $W^{\otimes l}$ respectively.

[cloud walrus]

Have a Banana, Bitch

[old lava]

can't you just fix a basis of W, extend it to a basis of V

then easily show containment both ways

[vestal snow]

I can show that $W^{otimes l}$ is a subspace of $V^{\otimes l}$ in a natural way

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

And then $A(W)\subseteq W^{\otimes l}\cap A(V)$ follows easily

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

Say we fix a basis B of W and extend it to a basis B' of V

Then we will get bases D and D' of $W^{\otimes l}$ and $V^{\otimes l}$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

respectively

and D' must contain D

[vestal snow]

What should I do now?

Because the bases of A(W) and A(V) might not be subsets of D and D'

[warped bane]

sorry for interrupting, I just wanna make sure of this

is the set that contains continuous functions finite?

[sharp sonnet]

what an odd question

[celest brook]

is the set of polynomials finite @warped bane

[vestal snow]

Also, this probably isn't suitable for this channel

[warped bane]

because I found an ex in a textbook that says: determine the dimension of P and I such that P and I are the vector space of even functions and odd functions respectively, and show that they are supplementary sub-spaces

which got me confused since dim C(R,R) = +inf

so the dimension has to be infinite since card(basis of P)=card(basis of I) = +inf

[upbeat juniper]

yes it is infinite

you can't even construct a basis explicitly

[warped bane]

exactly!!

[upbeat juniper]

but P and I being supplementary (complementary?) is still well defined

[warped bane]

it's just the ex is introduced made me puzzled a little

yea the are supplementary

i don't think that they can be complementary

cuz if so then one of them can't be a sub-space

[upbeat juniper]

I'm pretty sure they are

[warped bane]

since if the first one has 0 then the other one won't

[upbeat juniper]

not necessarily

even if two subspaces are complementary, both must contain 0

ok I gotta leave now sorry

[warped bane]

oh then I just had complementary the wrong way since I'm french

thanks a lot!

[cloud walrus]

slimvesus

[upbeat juniper]

yeah I think he might have confused set theoretic complements and direct sum complements

[fair shard]

well you could also be talking about orthogonal complements

in which case you have three entirely different meanings

but they are only different if you have a fixed inner product

im pretty sure spaces are complementary iff they span V and there exists some inner product with which they are orthogonal complements

idk how that works in infinite dimensional spaces

but its true for finite and it would be sad if it didnt generalise

[upbeat juniper]

it should generalise right

the inner product is completely determined by the inner products of pairs of basis vectors

[fair shard]

yeah

it makes sense

but I'm too tired to think rn

but you have to be careful when the vector space is infinite dimensional

[upbeat juniper]

well if U,W are complementary and u_i, w_j are bases

then let <u_i,u_j> = <w_i,w_j> = 1 and <u_i,w_j> = 0

[latent anvil]

It generalizes but in a shitty way

If you start out with a topology on the vector space, you might not be able to find an inner product which generates the same topology and does what you said

But if you're just looking algebraically then yeah do what wy said, everything is just some huge direct sum of R

[jagged dune]

the kernel of a ring homomorphism is the set of elements that map to the identity, right?

ohhh, additive

so I'm mapping Z6 to Z15

define by 10x mod 15

so nothing maps to 1, multiplicatively

well, nothing maps to 1 additively either in that case, right?

Oh, you answered my question like you knew, sorry lol

I used that term in my question and you still answered it

I'm not saying it's standard

that's just how I worded it

[chilly ocean]

No you didnt

[jagged dune]

"set of elements that map to the identity"

I used mapping

[viscid pewter]

'mapping' is standard

[jagged dune]

actually I get what you mean

[viscid pewter]

'mapping additively' is not

[jagged dune]

I got my words mixed up

I'm sorry

ok here's what I'm saying

0 is the additive identity in Z15

so 0 and 3 would be the kernel of the homomorphism, correct?

Okay, sorry for the confusion before.

Thank you.

Okay another question, what does it mean when referring to the image of the homorphism?

Still referring to the same one.

[viscid pewter]

all the elements that you can get to by taking the homomorphism from one of the elements of the original

[jagged dune]

so it's the set of those elements

[viscid pewter]

yeah

[jagged dune]

in this case it's {0, 5, 10}?

[viscid pewter]

uhhh sounds right

[jagged dune]

sounds good to me , thank you

might have one more question so we'll see how this goes

[fair shard]

:(

[cloud walrus]

det

[rustic crown]

I have seen a different proof of this fact, which doesn't use any heavy machinery, just simple stuff to keep on reducing the degree-tuple of the polynomial. But is there a nice way to see this from galois theory without much computations?

[rustic crown]

any ideas?

[vestal snow]

How do I prove that the induced sequence is exact?

[next obsidian]

Do u know what Tor is

[vestal snow]

Yes

[next obsidian]

It’s then immediate

To the left of M(x) Q you have Tor1(P,Q) = 0 by P being flat

If you can’t do that, I believe you can express P as a quotient of a free module, “tensor the two SESs”, then use the snake lemma

I recall solving this by doing something like that

[vestal snow]

Why does P being flat imply Tor_1(P,Q)=0?

[next obsidian]

P is flat

And that’s a definition of flat

[vestal snow]

I think I'll go with that

[next obsidian]

Tor is a derived functor of the tensor product and flat is defined to be the tensor product acyclic objects.

Also if u want to use the snake lemma

[vestal snow]

The book I'm using is saving the proofs for Tor stuff until the end

[next obsidian]

You express Q as a quotient of a free module

Is this Aluffi

[vestal snow]

Yes

[next obsidian]

I think he mentions that Tor is 0

When F is flat

[vestal snow]

Yeah I can see it

[next obsidian]

Except for Tor 0

[vestal snow]

But he proves it in the last chapter

[next obsidian]

Sure

I mean here’s the proof

[vestal snow]

By this I mean literally. Not that I understand it

[next obsidian]

You make Tor by taking a projective resolution then tensoring with F

and then taking homology

[vestal snow]

What's F?

[next obsidian]

I guess P here

[latent anvil]

There is a torless way to do this

[next obsidian]

So like

I said that as well

[latent anvil]

Ah sorry

Missed that

[next obsidian]

Okay so

You make Tor(M,N)

By either taking a projective resolution of M or N

Then tensoring with the other module

Then taking homology

[vestal snow]

It's okay you don't have to explain it

[next obsidian]

So if one of them is flat

Then tensoring with that is exact

[vestal snow]

I'll take your word for it

[next obsidian]

So when you take homology u get 0

[vestal snow]

For now

[next obsidian]

Doesn’t he like give a not rigorous idea of how Tor is made in the second section?

Like he says how Tor is defined

But not why it’s well-defined or actually has the properties u want

[terse crystal]

using tor is a good idea but... you can also direct prove that statement...

[vestal snow]

Yeah I'm still wrapping my head around this stuff

I had a question

He says that if P is flat, then Tor_1(N,P) is 0

[vestal snow]

Isn't this different

Like we flipped P and Q

[next obsidian]

Oh

Maybe, but it doesn’t matter

It’s symmetric

[terse crystal]

Notice that first you can assume Q to be finitely generated then you can construct an exact sequence 0—>K—>L—>Q—>0 where L is a free module of finite rank.

[next obsidian]

You don’t actually need finite rank

I proved it without it

[terse crystal]

Indeed . Arbitrary free module is flat

[vestal snow]

Okay I see the stuff you're mentioning now

It kinda makes sense

I will definitely need to brush up a bit on this stuff

Also Chmonkey you used Aluffi right?

Did you think that the difficulty of the exercises increased by a lot in Chapter 8?

[next obsidian]

Yeah

Well tbh I didn’t do them before it for the most part haha

[terse crystal]

You will have a 3*3 diagram whose second row and third column are exact then you can directly prove your statement.

[next obsidian]

But I can say I did every exercise in VIII.2

And when my algebra class rolled around to this stuff I had a MASSIVE head up

Having intuition for how to use Tor helped me a shit ton

[vestal snow]

Your class didn't assign homework from the book?

[next obsidian]

We didn’t use Aluffi for my class

Or well...

We did take some exercises from it for the thing I did my freshman year

But that was all like group, ring, field, Galois theory and I don’t know what came from where

A lot of it was just our TA being like “lol here”

[vestal snow]

This stuff shows up a lot in AG right?

[next obsidian]

I mean Tor and Ext will show up in a lot of places, AG included

The general theory of derived functors of which Tor and Ext are special cases is the definition of sheaf cohomology

Which is absolutely indispensable

So getting familiar with it is important

[vestal snow]

I better get to studying these then

[chilly ocean]

Can someone help me using Legendre polynomios to find the minimun of a function?

[vestal snow]

ChChChmonkey, did you do exercise 4.6 from chapter 8?

This one

[chilly ocean]

Whats the meaning of the operator « V reversed »?

Is it an « and » in this context?

Any set containing the zero vector is linearly dependent btw

[paper flint]

I think it stands for the wedge product?

[vestal snow]

wedge product

Somehow I need to show that given a vector space V and a subspace W, there is a way to extend an alternating map on W to an alternating map to V

My strategy was to find a basis of W and extend it to a basis of V

[carmine fossil]

Ye,That should work

[rustic crown]

yea it works

[vestal snow]

I'm an idiot lol

[rustic crown]

if the basis for V is B, then to get a map from V^l to U its sufficient to give the images of the basis elements B^l. If all are in W, then use the alternating map which you have. If even a single basis element is not from W, then map that to 0. We need to check this is indeed alternating, so it's sufficient to check its alternating on B^l. If all elements are in W, then it follows from the fact that the original map was alternating, and if even a single was outside W, then its 0 = -0

[vestal snow]

I switched V and W in my work

[vestal snow]

This is exactly what I was trying to do

[rustic crown]

[rustic crown]

can someone take a look at this? >.<

[vestal snow]

I think I know a proof of this

One sec

[rustic crown]

(I know two nice proofs now, but I wanna use Galois theory to crush it )

[vestal snow]

I think it went something like this

Consider F(x_1,\cdots,x_n)

As a field extension of F(s_1,\cdots,s_n)

Then the galois group of this extension is S_n

[rustic crown]

yea

[vestal snow]

And if you have something fixed by all elements of S_n, then by the galois correspondence, they must be in F(s_1,\cdots,s_n)

[rustic crown]

yep

[vestal snow]

And symmetric polynomials are precisely the elements that are fixed by all elements of S_n

By definition

The main argument here is proving that the galois group is S_n

[rustic crown]

that is okie, what i'm asking is if i start with a polynomial in F(x1,..,xn) then this argument shows that it would be a rational function in F(s1,...,sn). But I want it to be a polynomial.

[vestal snow]

A polynomial in F(x_1,...,x_n) cannot be a rational function in F(x_1,...,x_n) with a non-trivial denominator (in reduced form)

So if your polynomial was a rational function in F(s1,...,sn) and had a non-trivial denominator

It would have a non-trivial denominator in F(x_1,...,x_n)

[rustic crown]

like maybe a bad example would be consider k(t) and the subfield k(x,y) where x = y = t
then x^2/y in k(t) is just t but has a non-trivial denominator

[vestal snow]

I think this a more of a bookkeeping problem than a conceptual problem

I'll look at it later today

[rustic crown]

okie thanks!

[vestal snow]

Maybe try doing something like the integral closure?

The integral closure of F(x_1,...,x_n) over F[x_1,..,x_n]

[rustic crown]

how do you define it?

[vestal snow]

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

I don't think you need it, but it might be worth trying out

[lunar coyote]

can someone help me with this

[rustic crown]

so you need to show that (1) this map is well-defined, (2) its injective, (3) surjective (this will be automatic), (4) its a ring isomorphism.

what all have you tried?

[lunar coyote]

well i tried to show it's injective

but i don't know how to really show that it's injective

[rustic crown]

one way to show injectivity is to check that kernel of the map is {0}

but first we then need to show its actually a homomorphism

[lunar coyote]

tried this but it's 100% wrong

[rustic crown]

minor typo there [a]_m = [b]_m

but what can you say about (a-b)?

what are [a-b]_m and [a-b]_n?

[lunar coyote]

hmm

m and n are coprime

so would they be equal

[rustic crown]

you saying these are equal?

[lunar coyote]

wait

no

[rustic crown]

they live in different places, one lives in Z/m and other in Z/n

[lunar coyote]

that wouldn't make sense

[glossy pulsar]

@lunar coyote when will [x]_n = [0]_n?

[lunar coyote]

when x is a multiple of n

[rustic crown]

recall we have [a]_n = [b]_n => [a-b]_n = ?

[glossy pulsar]

I think it will be easier to first show that the map is a morphism and to then show that its kernel is { [0]_nm }

Or that's how I did it when studying this

[rustic crown]

generally you don't give the map directly, you use old maps to construct new ones, so the map is a hom is always the case

so I have the maps Z --> Z/nZ and Z --> Z/mZ, then (using universal property of products) we get a map Z --> Z/mZ x Z/nZ

then you find the kernel of this map. if something goes to 0, its both a multiple of m and n, hence (if m and n are coprime) a multiple of mn

(now using property about quotients) we get an injective map Z/mnZ --> Z/mZ x Z/nZ

[lunar coyote]

woludn't it be [a]_n - [b]_n

[rustic crown]

and that is [0]_n

[lunar coyote]

yeah

[rustic crown]

so this gives that (a-b) is divisible by n

and similarly by m

[vestal snow]

det did you do this problem?

[rustic crown]

this was just a restatement iirc

page 511

[lunar coyote]

so would that show the kernal?

[rustic crown]

we didn't show that this map is a homomorphism yet right. and we were trying to show that its injective.

what we have is a-b is divisible by mn

hence [a]_mn = [b]_mn which precisely shows the injectivity.

[lunar coyote]

idk why this stuff is so confusing

i've read through the notes

and it seems so abstract

[glossy pulsar]

You'll get there

Math is a wall you'll keep running into until you find a way to climb it

Then it'll seem trivially easy, you'll get there if you keep trying

[lunar coyote]

would my starting point be

trying to complete this question

or try and attempt some easier question or read the notes more

and try understand it

just feels like theres no starting line to this

[rustic crown]

if something feels abstract, try to look for examples

in this case, try to understand the maps for specific values of m and n

say 3 and 4

[glossy pulsar]

After that you might try it for 2 and 4 as well, to see where it goes wrong

Does anyone have some knowledge about Galois Extensions?

[rustic crown]

(depends on how much is some 😶 )

[glossy pulsar]

Basically, I have an extension K < C with K of a positive characteristic and C an algebraically closed field

I want to show that this extensions is seperable (or Galois which is equally hard in this case)

Oh and its a finite extension

[rustic crown]

so first we need that this extension is algebraic, other...

[glossy pulsar]

Yes it is an algebraic extension

[rustic crown]

lol you typed first

[glossy pulsar]

K has a positive non-zero characteristic

To give the complete problem, I am working from the assumption that Char K = p and [C : K] = p. Eventually I'll have to derive a contradiction but the hint told me to show that in this case K is perfect

[rustic crown]

i think i have a solution if [C:K] = p

but i didn't prove that K is perfect...

[glossy pulsar]

shoot, I've been stuck on this for a couple of days so any input is welcome

[rustic crown]

if you pick an element a in C\K, its minimal polynomial over K would have degree p. if this polynomial is separable then C = K(a) is separable! so assume that no such a has a separable minimal poly.
if its inseparable, then the derivative criteria shows that its of the form x^p - b with b in K.
Now what about x^p - a. since C is algebraically closed, there is an element c in C such that c^p = a, but then c doesn't lie in K, as a doesn't.
so minimal polynomial of c over K would be something like x^p - d where d is in K, but this means c^p = a = d in K. a contradiction!

you can use the notion of separable and inseparable degree to simplify the wording of the proof, but the basic idea is you just joined pth root of something, and got an algebraically closed field, but what about pth root of that pth root? how did it automatically get there?

[glossy pulsar]

The thing is, if K < C is seperable then K will be perfect as you can't have any extensions strictly between C and K. The product formula disallows this because p = [C : K] = [C : E] [E : K]

[rustic crown]

oh yes nice

i remember now, any algebraic extension is separable if and only if base field is perfect

[glossy pulsar]

I might be missing something, but why is the minimal polynomial of C - K always of degree p?

[rustic crown]

the degree is [K(a):K]

[glossy pulsar]

Ah yes

Alright this might actually be my answer holy shit. I'll try and get it written down properly thanks for the help

[rustic crown]

okie you an generalize what i did to any finite degree

look at F = set of separable elements in C over K

this will form an intermidiate field and the extension F < C would be purely inseparable!

so if you pick a in C, then by algebraic closedness i can demand c^q = a, but c^q will have to lie in F proving C = F

(here q = [C:F])

[glossy pulsar]

I think you're correct

[paper flint]

Looks good to me.

For future reference, similar questions may be better suited to #proofs-and-logic or #elementary-number-theory . I do think this might have been covered in preliminaries for an intro algebra course.

[uncut girder]

What is the normalizer of the diagonal matricies $diag(t,1/t)$ in $GL(2)$?

[cloud walrus]

PTYamin

[chilly ocean]

You mean $GL_2(\mathbb{K})$ probably

[cloud walrus]

Carla_

[uncut girder]

yeah

I'm figuring it out

Got it

The normalizer consists of the diagonal and anti diagonal matricies

[celest brook]

when $a^{2}$ doesn't mean $a \times a$

[cloud walrus]

Moosey

[celest brook]

[chilly ocean]

I forgot like all my group theory stuff except the basics and important theorems

Saw a problem that went something like

H is a subgroup of G with index 3, can we show that any element^6 is in H

Lagrange gives you that 6 = 2|G|/|H| but eh idk if that is useful

maybe tho? shrug

[hot lake]

you can try to have x act on G/H

[chilly ocean]

idr if it was normal, if so that would probably be useful for sure

[viscid pewter]

iirc you can show g^3 is in H, so g^6 follows

[unique juniper]

How

[viscid pewter]

gimme a sec

you hear that sound, that's my brain wheel squeaking

[unique juniper]

I wanna try this but I’m in bed tryna sleep

[viscid pewter]

oh yeah so

ok this probs isn't gonna be rigorous but

wait i should spoil it in case zd doesn't want the full solution

[chilly ocean]

I do

overheard this from a friend taking abstract and they got their problem sets back last week but he is offline

my prof for abstract went rap god over the little details and I honestly dont remember much of the nitty gritty lmao

abstract ii has been way better for me

[viscid pewter]

||H has three cosets, let them be H, aH and bH, partitioning G, a and b not in H||
||a^3 is in H and b^3 is in H because uhhh considering G/H, it's cyclic of order 3 so every non-identity has order 3, so (aH)^3 = (bH)^3 = H, so a^3 and b^3 are in H||
||for any g in G, assume that g is not in H, as if g is in H it is obvious; assume further WLOG that g = ah for some h in H||
||then g^3 = ahahah||
||= (aha^-1)(a^2ha^-2)(a^3)(h)||
||= (h1)(h2)(1)(h) which is in H||

i assumed H was normal

thaaaat might not be true

[unique juniper]

[viscid pewter]

whoops

[chilly ocean]

if it were normal though ||it being cyclic would be correct since it has to be isomorphic to Z_3 right, just 3 elements yeah?||

[viscid pewter]

yes

i also used it being normal later, in basically the final step

[chilly ocean]

ye ye

[viscid pewter]

so

how do i do this without the easy mode of normality

[unique juniper]

You don’t

[viscid pewter]

wait, is it false?

[chilly ocean]

I forgot how much the very advanced trick of adding and subtracting the same thing is used in group theory lmao

[viscid pewter]

i mean this is sorta a kludge, there's gotta be a more elegant way to do this

[chilly ocean]

I like it

[viscid pewter]

me too but it's a kludge

[unique juniper]

can this be done without normality

[viscid pewter]

assuming zd transcribed the q right, yes

[chilly ocean]

yeah I'm not sure if it is

but nice to explore the question

trying to find counterexample rn

[viscid pewter]

eh, let's have a go

i bet it's right

wait actually i kinda proved it in line 2 bc you can just choose any representative for the coset lmao

all of the algebraic fiddling was unnecessary

[chilly ocean]

before the G/H stuff?

[viscid pewter]

no

[unique juniper]

Yea

I thought so

[viscid pewter]

shorter proof if H is normal:
||for any g in G, assume that g is not in H, as if g is in H it is obvious
let the cosets of H be H, gH and kH
considering G/H, it's cyclic of order 3 so every non-identity has order 3, so (gH)^3 = g^3H = H, so g^3 is in H||

[unique juniper]

Nice

[chilly ocean]

right yeah you ended up using the thing to prove the thing after proving it LOL

[viscid pewter]

i bet i 'did' this proof at some point, as in tried it, gave up and looked the answer up, then forgot it

[chilly ocean]

maybe something like ||
let K be the normalizer of H in G, then 3 = [G:H] = [G:K][K:H], so the only possibilities are 3 and 1 for either. if [K:H] = 3, then [G:K] = 1 so that G = K and your proof would follow
Then if [K:H] = 1, K=H and [G:K] = 3 is a tautology and uhhh idk||

I think that is half of it though

oh wait ||if K=H then xhx^{-1} = a with a not in H for all x in G, so a^6 = xh_2 x^{-1} not sure how to finish this if this would work||

[celest brook]

isn't this given by definition of automorphism? Would that be a better way to say it?

i.e. an automorphism by definition is bijective, so it must map to all of G

[viscid pewter]

lmao what's that blue paragraph

nutty

[celest brook]

oh that's just weird comments my prof makes

before hw problems

[viscid pewter]

your prof is whack ngl

[celest brook]

meaning he's wrong or he's just saying weird stuff unrelated?

[viscid pewter]

both

[celest brook]

mm

[old lava]

well your statement doesn't show surjectivity at all

unless you proved every element of G is of the for g a g^{-1} for some a in G

before hand

all it does it show that the map is in fact a map into G

[viscid pewter]

it seems like it's the prof's notes

[old lava]

oh I thought it was a solution

[celest brook]

red is the profs comment

[viscid pewter]

ohhh

so is this your proof

[celest brook]

yes

it's wrong

i k

:)

[viscid pewter]

so you want to show that it's an automorphism

[celest brook]

yes

doesn't that

come

[old lava]

well surjectivity is that the image of the map is the entire codomain

you do not prove that

[viscid pewter]

so you can't say it's surjective bc it's an automorphism

[old lava]

in any form

[celest brook]

oh wait

fejdiogngm

I read the problem wrong

[viscid pewter]

because you want to show it's

yeah

[celest brook]

ugh

how would i show injectivity and surjectivity for an arbitary group

[viscid pewter]

you want to use the trivial center detail i believe

that's important

[celest brook]

...

wydm

man i wish I wasn't so dumb

[old lava]

I mean you can just show that for any a \in G, g^{-1} a g maps to a under that map

[celest brook]

[old lava]

easy surjectivity

it's just one line

[celest brook]

how

it's not abelian

[viscid pewter]

f(g-1ag) = gg-1agg-1 = a

surjective

[celest brook]

or abelian is not part of it

huh?

[old lava]

I don't see how abelian-ness has anything to do with anything here

?

[celest brook]

nevermind

[viscid pewter]

so you want to show that for all a in G, there exists x in G such that f(x) = a

if x = g-1ag, then it works

[celest brook]

OH

[fair shard]

I could hear this

[viscid pewter]

hear?

[fair shard]

sound it out

it make funny noise

[viscid pewter]

oh i see

[celest brook]

i'm still confused, why does fixing x=g^{-1}ag prove surjectivity, i.e. that it maps all of G

[viscid pewter]

for any given a, let x = g-1ag

then f(x) = a

so for all a there exists x such that f(x) = a

so f is surjective

[celest brook]

okkk so a can be any element

in G

i.e. $\forall$ a $\in$ $G$

[cloud walrus]

Moosey

[viscid pewter]

yes

[celest brook]

okkk

and that also implies injectivity

right?

[viscid pewter]

injectivity is just

if gag-1 = gbg-1, a = b

[celest brook]

ooo

ok

oh wait yeah I already did that

under surjectivity

[old lava]

your injectivity proof is fine

it was just your understanding of surjectivity was off

[celest brook]

so like this?

[old lava]

that works

but I'd word it differently probably

I'd start off with letting a be an arbitrary element of G

then do the rest

[celest brook]

ok.

[eager schooner]

this is a bit of a hard one

Would anyone advise on where to start

okay that is way too small

Let k be a field, and suppose we are given n+1 distinct elements a1,…,an+1∈k and n+1 (not necessarily distinct) elements b1,…,bn+1∈k. Prove that there is a unique polynomial f(x)∈k[x] of degree ≤n that satisfies f(ai)=bi for all 1≤i≤n+1.

I've tried think about it in terms of linear algebra but that seems to be the wrong approach

I've also tried thinking about how perhaps the orders of the a's have something to do with it but no that doesn't seem to be important as the first step at least

[old lava]

I mean a linear algebra approach works

it's just a polynomial regression, which can be written in terms of matrices and such

[eager schooner]

But how would you show that the inverse of the A matrix exists

[old lava]

and you can prove that the matrix when row reduced has exactly one solution or something

each row is linearly independent, you should get an n+1 x n+1 matrix where each row is linearly independent from the rest since a_1, ..., a_n+1 are distinct

[final pasture]

there's an easier approach if you only want existence and unicity

but not to know how the polynomial actually looks like

[old lava]

I was just mostly replying to this

[final pasture]

(still linear algebra though, just not like this)

try looking for the appropriate isomorphism between K_n[X] and K^{n+1}

[eager schooner]

I'm not sure I understand the notation there

[final pasture]

K_n[X] = polynomial with coefs in K of degree at most n
K^{n+1} = cartesian product of K with itself n+1 times

and K the field, ofc

[chilly ocean]

I think uniqueness is not true, unless the field is infinite, or something like that

Or maybe characteristic 0, or something like that

[final pasture]

(it is though)

[old lava]

as long as your field has at least n+1 distinct elements

I think it holds true

[final pasture]

^

condition on the degree guarantees this

[old lava]

all you're really doing is viewing a set of polynomials of degree <= n as like a vector space almost I think

[final pasture]

well K_n[X] is literally a subspace of K[X] viewed as a vector space, so yh

[final pasture]

(is what I said clearer after I explained the notations ?)

[eager schooner]

It is but I'm still wading through the suggestions a bit

[chilly ocean]

Ah hm yeah huh

But anyhow this is a well known problem

I think it is something like lagrange's polynomial or something

[final pasture]

yes

but he shouldn't look it up on google

that'd spoil the problem

[chilly ocean]

True

[eager schooner]

@old lava You said you could row reduce, but how do you row-reduce for an arbitrary field with an arbitrary number of elements in the matrix?

Seems like that gets out-of-hand really quickly

[celest brook]

third post is my edits, but I have a feeling I still have to do x^{-1}Hx subset of H, and then H is a subset of x^{-1}Hx

it doesn't help i have no idea how to do that without being too wordy

[old lava]

not really, gauss-jordan elimination is well developed for any field

[eager schooner]

How would you even know when something is 0?

I guess you would mutliply by the inverses of the elements by jeeze that would get crazy long

So I tried it an yeah this is already getting a bit unwieldy after a single step

[eager schooner]

@final pasture Could you actually expand on what you mean by your suggestion

[final pasture]

yeah. So are you okay with viewing K_n[X] and K^{n+1} as vector spaces over K ?

[eager schooner]

No not really

I understand the isomorphism between them somewhat

I'm just sort of confused how you unroll the input to the function

[final pasture]

Are you familiar with the dimension of a vector space ?

[eager schooner]

yes

[final pasture]

Ok what would be the dimension of K_n[X] (which as a reminder is the space of polynomial of degree at most n)

and same question for K^{n+1}

[eager schooner]

I think for K^{n+1} you would have dimension n+1

[gentle pendant]

Another quite concrete approach that will give the answer quickly:

Find a polynomial pj with pj(aj)=1, and pj(ai)=0 for i=/=j. (It is easy to cook up polynomials that vanish where you want them.)

Then take a linear combination of these. (It should be pretty clear how to choose this.)

[eager schooner]

and if there's an iso. between them K_n[X] would also be n+1

[final pasture]

indeed

(1,X, .., X^n) is a basis of K_n[X]

[eager schooner]

Ah okay so that I get

[final pasture]

if you have an injective linear application between K_n[X] and K^{n+1}, since the dimension are equal, the linear app would be an isomorphism (so bijective), right ?

since, if dim(E) = dim(F), then for a linear app f: E -> F, f is bijective <=> f is injective <=> f is surjective

[eager schooner]

Ah okay so in what sense an isomorphism? Are we still talking about a ring isomorphism? A group isomorphism?

[final pasture]

vector spaces isomorphism

so bijective linear app

[eager schooner]

Okay yeah so that's all clicking a little bit

So then I guess my question is \phi(x) = {......} right?

But how can I change an x value and unroll it

The x value is a function sure, I guess I would be taking the coefficients?

[final pasture]

You're looking for a linear app $f: \mathbb{K}_n[X] \to \mathbb{K}^{n+1}$ that would prove you that there exists $P \in \mathbb{K}_n[X]$ s.t $P(a_i) = b_i$ for all $i \in [![1, n+1!]$

[cloud walrus]

Shika-Blyat

[final pasture]

$P \mapsto (P(a_1), \cdots, P(a_{n+1}))$ may help, do you see why ?

[cloud walrus]

Shika-Blyat

[eager schooner]

Okay a little yes

I'm familiar with Lagrange polynomial from a numerical methods class and that's clearly what this is hinting at

[final pasture]

By proving that this application is bijective, that would prove the existence of Lagrange polynomials for any field yeah

and since it's a linear app between two vector spaces of the same dimension, show it is injective suffices

[eager schooner]

I guess I'm still having trouble seeing what's happening here

[final pasture]

(and to prove that it is indeed injective, you need to use some properties of polynomials )

[eager schooner]

We are taking a polynomial P and then making x_1, x_2, x_3 etc. be multiplied by P(a_1), P(a_2), P(a_3), etc. right?

Or am I missing something

[final pasture]

We're not multiplying anything

I'm probably not really clear because I'm trying to leave some work to you

[eager schooner]

Right okay fair enough

[final pasture]

But I'm failing at doing so properly sry lol

[eager schooner]

How about this then, could you give a concrete example for the isomorphism

Like take p = x+1

what is phi(x+1)

it doesn't need to be the right isomorphism

just any map between the two spaces

so I can see what is happening there

[final pasture]

the isomorphism depends on the a_is

so like

let's say we want to prove that there exists an unique polynomial P with real coefficients of degree <= 2 s.t P(1) = 2, P(2) = 3 and P(3) = 4

so we would have a_1 = 1, a_2 = 2 and a_3 = 3

and b_1 = 2, b_2 = 3, b_3 = 4

We look at the map $\varphi: \bR_3[X] \to \bR^4, P \mapsto (P(1), P(2), P(3))$

[cloud walrus]

Shika-Blyat

[final pasture]

assume that $\varphi$ is bijective

[cloud walrus]

Shika-Blyat

[final pasture]

do you see why we would have proved that there exists such a polynomial and that he is unique ?

(we will prove that phi is bijective after, I just want to explain why we are looking at phi)

[eager schooner]

So no I don't

Okay wait actually

I'm starting to see it

The bijective is really really important here

[final pasture]

yes

If it is bijective, that would precisely mean that there exists an unique polynomial P s.t $\varphi(P) = (2, 3, 4)$

[cloud walrus]

Shika-Blyat

[final pasture]

(existence coming from surjectivity, unicity from injectivity)

[eager schooner]

Yeah yeah okay

Okay then actually I think I can get it from here

Thanks

[final pasture]

right

Feel free to ping me if you still need some help to conclude

Also once you did it, you should probably look at the wikipedia page on lagrange polynomial, 'cause my way of solving the exercise is quite abstract, it doesn't give you the actual polynomial, but you can find it explicitly too (that's what gomez was speaking about a bit earlier here #groups-rings-fields message )

[celest brook]

b=k?

yeah, i mean that makes sense

and then I can twist it around and get what I want

[celest brook]

On the second part. how to prove that $H \subseteq x^{-1}Hx$

[cloud walrus]

Moosey

[bleak abyss]

@sturdy marsh finally did that sl2-triples problem

Actually I realized in class the fact that we already classified irreps of sl2 made life much more pleasant

[sturdy marsh]

[bleak abyss]

We didn't use much of the classification tbh but like

Really it's just writing things as linear maps instead of matrices

And remembering that in representations of sl2, you have raising and lowering operators

And then it's just pattern recognition that a nilpotent Jordan block is a lowering operator

So yeah gg

[sturdy marsh]

I should say gg whenever I end a proof in my next talk

the prof probably wont get it tho

[bleak abyss]

The last proof ends with "gg no re"

[bleak abyss]

Didn't write up full details but good enough

[next obsidian]

de Jong says “we win” on stacks project a lot

[bleak abyss]

Oh Frank Calegari does that too lol

[chilly ocean]

i was wondering, usually when you have a proper subfield K or R, we have [R:K] infinite. So i wondered if theres some proper subfield K of R such that [R:K] is finite or countable. I thought of doing kinda the opposite of a generated extension: Consider the intersection K of all subfields of R that dont contain for example sqrt(2) and hopefully would have [R:K]=2. Not sure if this works, but is this something well known if it does?

[chilly ocean]

yeah, i was thinking something along the lines of zorn's lemma, although i haven't thought/gone through the argument yet

i was worried itd depend on AC lol

i was thinking: consider the set of all subfields of R that do not contain sqrt(2). then zorn's lemma to get a maximal subfield. then maybe this subfield is such that [R:K] is finite?

[final pasture]

I think you just need to show that there exists an uncountable subfield

since if [R: K] was infinite with K uncountable, there would be a cardinality problem

[chilly ocean]

there is a cardinality problem? i don't see it

[final pasture]

well if $\bR$ is of dimension $\mathfrak{c}$ (where $\mathfrak{c}$ is an infinite cardinal ) over $K$, then $\bR$ is isomorphic (as a vector space) to $K^\mathfrak{c}$ right ?

[cloud walrus]

Shika-Blyat

[final pasture]

but if $\mathfrak{c}$ is infinite, then it's greater or equal than $\bN$

[cloud walrus]

Shika-Blyat

[final pasture]

and well, $\bR^\bN$ is already greater in cardinal than $\bR$

[cloud walrus]

Shika-Blyat

[final pasture]

oh wait

i'm assuming the continuum hypothesis here

I need a subfield that is in bijection with R !

[chilly ocean]

i looked it up in google and it says they have the same cardinality

[final pasture]

but you can inject $\mathcal{P}(\bR)$ in $\bR^\mathbb{N}$

[cloud walrus]

Shika-Blyat

[final pasture]

wait

no you can't

god I'm dumb

okay forget everything I just said

[chilly ocean]

if it was N^R you could

[final pasture]

yeah

sorry

[chilly ocean]

why did you delete your basis argument tho, i didnt see what was wrong with it

that was my intuition for it when thinking of this

[final pasture]

I felt like if I take a basis

remove one element

and then consider Q(the basis minus the element)

that would work properly

But I was thinking it'd work properly for cardinality reasons 🐒

[chilly ocean]

i think so too

[final pasture]

maybe it still work, I just don't know how to prove it and I was claiming I knew (before realizing I don't), so I deleted

[chilly ocean]

If B is basis that contains sqrt(2) for R over Q then B-{sqrt(2)} still uncountable

[final pasture]

(that's a nice question though, I like it )

[chilly ocean]

im concerned it might not be a field anymore then tho

[final pasture]

Actually your idea about intersecting every subfield that doesn't contain sqrt(2) is probably a really good idea

[rustic crown]

that's a bad idea

Q doesn't contain sqrt(2)

[final pasture]

oh huh yeah we need to take them big enough, and we're back at the same problem

I should probably go to sleep, that's the third super dumb thing I've said in 10 minutes 🐒

[rustic crown]

i was thinking in these lines take t0 = 2 and t_{n+1} = sqrt(t_n)
then what i want to claim is [K(t_n):K] = 2^n. So was trying some induction argument.
so like the idea is, if it doesn't contain sqrt(2) then it doesn't contain the 4th root as well... and so on... this would intuitively make the extension [R:K] with degree as large as possible

which would contradict that its finite?

[final pasture]

but we want it to be finite, don't we ?

[rustic crown]

maybe such a thing can't happen?

[hot lake]

yeah but to show that K(2^2^-n) : K is 2^n isn't that obvious ?

like, what if I pick sqrt(2) + sqrt(sqrt(2)) right away when building my field

we have to show that sqrt(2) is in that

though in that case it's easy

[rustic crown]

are you saying "isn't that obvious" in the sense that it is obvious or in the sense "that isn't obvious"

context says its latter nvm

another thing i tried was looking at a finite subgroup G of automorphisms of R, that fixes Q, so that [R:R^G] is finite... but it was not that hard to show the only automorphism is identity 😶

[chilly ocean]

[R:R^G]?

[rustic crown]

the elements of R which are fixed by any automorphism in G

its called the fixed point set... if G acts on X then X^G is the set of fixed points.

[chilly ocean]

yea im stupid nvm what i said lol

[final pasture]

^

[rustic crown]

i wanna use this emoji 😶

[hot lake]

I think there is a theorem that says something like any subfield of C of finite index has index 2 but I can't remember the name

[final pasture]

https://math.stackexchange.com/questions/1586050/complex-subfields-of-finite-index

[chilly ocean]

oh

@final pasture what server did you get this emote?

interesting that R is precisely C^G where G acts on C by complex conjugation

[final pasture]

MangaDex's server @chilly ocean

want an invitation ?

@chilly ocean https://www.diva-portal.org/smash/get/diva2:1445629/FULLTEXT01.pdf

[chilly ocean]

whats that

and assuming AC theres weird automorphism groups of C,i wonder what C^G would mean for those

[final pasture]

wait wrong link 🐒

2sec

[chilly ocean]

maybe either of infinite degree or of degree 2 for those

[final pasture]

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.299.5536&rep=rep1&type=pdf

the theorem 3.1 is the theorem the ME answer is speaking about

and there's a proof

but there's some Galois theory

[chilly ocean]

cool thank

[unique juniper]

[rustic crown]

https://cdn.discordapp.com/emojis/676473025994031140.png?v=1

[final pasture]

can't we ask the admins/mods to add it here ?

the server is boosted to level 3, there probably is a few emoji slots left

[molten silo]

Anyone got any ideas?

[paper flint]

Ask them in the discussion channels

[rustic crown]

what does gcd(a,r) = 1 means?

[final pasture]

probably defined in term of ideals det

the only ideal that contains both a and r is R

[rustic crown]

(lol i'm asking him so that he could continue the proof himself)

xD

[final pasture]

Oh sry xD

[molten silo]

1 = as+rt for some r,t

[rustic crown]

try multipying both sides by b

😄

[final pasture]

Now I feel dumb 🐒

[final pasture]

how ? Like am I supposed to ping someone ?

[molten silo]

i think i figured it out

[rustic crown]

[paper flint]

There's a mod in #discussion atm, ask them

[final pasture]

I got ignored

[paper flint]

[rigid cave]

Hello! Can someone explain how $K'm = K'{m-1}(\gamma)$?

[cloud walrus]

older sister