groups-rings-fields

[chilly ocean]

Lol

[coral shale]

yes 2 and 3 aren't inverses of each other

2.3 = 0

which is not the identity

[gilded gull]

yes 0 is not the identity lol

[coral shale]

Its impossible for 5 to have an inverse?

what?

how is that remotely related (or even true)

[gilded gull]

wait

nvm

so 5k/6=1

[coral shale]

Maybe do your sheet Q first, but after that, I would strongly recommend constructing a cayley table

[coral shale]

no. thats not how its going to happen ..........

[gilded gull]

?

[coral shale]

that makes no sense.

[gilded gull]

wouldn't that have to be true for some k between 1 and 5 for

ok not 6 necessarily

but

mod 6

[coral shale]

Where is that equation even from

[gilded gull]

Ok hold up

so the identity is 1

[coral shale]

Well yes.

[gilded gull]

Then 5 times some element = 1

[coral shale]

mod 6

[gilded gull]

for it to be an inverse

[coral shale]

but alright

[gilded gull]

mod 6

[coral shale]

5x === 1 mod 6

[gilded gull]

right

x can't even be higher than 5 though ?

because then its not an element of the group

[coral shale]

We're working with equivalence classes of integers

not actual integers

0, 1, 2, 3, 4, 5 are merely representatives

of their class

[gilded gull]

OHH

[coral shale]

The classes themselves are integers which have the same remainder upon divison by 6

[gilded gull]

I forgot it was the equiv. classes

[coral shale]

So it doesn't make sense to say something like this

[gilded gull]

I get it now

and then for this example x = 5 works

[coral shale]

yes, 5 is self inverse.

[gilded gull]

but then there's two self -inverses

and they aren't equal

[coral shale]

?

[gilded gull]

1*1 =1

[coral shale]

ok

is something the problem with that though?

[gilded gull]

does a group need a unique identity?

[coral shale]

thats something to prove

[gilded gull]

i mean

not a unique idenittyu

[coral shale]

State what an identity is...

[gilded gull]

Not a unique identity

mistype

[coral shale]

.

[gilded gull]

Ik what an identity is lol

[coral shale]

I disagree.

State what it is.

[gilded gull]

Any element times the identity is the element

[coral shale]

And so how is this the case

[gilded gull]

It isn't

i mistyped

[coral shale]

Now I'm confused

[gilded gull]

as I said right after

[coral shale]

not a unique identity

[gilded gull]

there is still a unique identity

[coral shale]

This is what you meant yes?

[gilded gull]

No

[coral shale]

ok... sure

[gilded gull]

Ok so im still looking for some contradiction here

[coral shale]

also note we need ex = xe = x (both directions)

forall x

Anyways, yes we are.

[gilded gull]

ok so am i possibly looking for an order that doesn't divide the order of the group?

[coral shale]

🤔 🤔 🤔

[gilded gull]

Or

[coral shale]

Just stick to the group axioms.

[gilded gull]

Ok

So closure is fine

There's still an identity

So then inverses

5 is a self-inverse

4x === 1 mod 6

[chilly ocean]

In Z mod n, think which element does not have an inverse such that Z × x = 1

[coral shale]

keep going yes

[gilded gull]

ok does 4 not have one?

[coral shale]

because

[gilded gull]

oh because they're both even

right?

[coral shale]

4x = 1 + 6k

yes, parity error

But as pointed out there is a much easier example

[gilded gull]

a much easier example?

[coral shale]

well at least an example that generalises

[gilded gull]

Ok so for Z_{n}?

[coral shale]

Just keep going for Z/6Z

[gilded gull]

oh ok

[coral shale]

Might as well check out the rest

[gilded gull]

I mean 2 mod 6 wouldn't work either then

for the same reason

3 mod 6 wouldn't

it would just flip between 0 and 3

is that the generalized part

n/2 just flips between 0 and n

[coral shale]

you havent even checked everything

[gilded gull]

i mean 0 and n/2

[coral shale]

what if n is odd

[gilded gull]

wait 0

0 can never be the inverse

and that is much easier

[coral shale]

Yes.

[gilded gull]

wow

I can't believe im this obvious

oblivious*

[coral shale]

Now can you figure out for general n, which elements don't have multiplicative inverse

this is not so easy

But focus on an algebraic approach

[gilded gull]

well two are clearly 0 and n/2

[coral shale]

I gave a hint above on how to write such an argument

[gilded gull]

this one?

[gilded gull]

and here

[coral shale]

.

[gilded gull]

o

[coral shale]

Have you done number theory?

diophantine equations?

[gilded gull]

Not much tbh

I know what they are

[coral shale]

euclidean algorithm?

[gilded gull]

Uh is that one for gcd?

[coral shale]

yes

[gilded gull]

then yes

[coral shale]

but theres also a reverse euclidean algorithm

[gilded gull]

O.o

[coral shale]

Well maybe quickly look it up again

because it's totally relevant here

and somewhat relevant to some proof methods you will see at the start of group theory

[gilded gull]

ok so you subtract off a multiple of the smaller one

and repeat

[coral shale]

if you want to answer this question, you will basically need to rely on euclid ig

'Which elements in Z/nZ do not have multiplicative inverse'

You don't actually need to know the algorithm itself here

I guess

More like - what are the conditions for using the reverse algorithm

'Bezout's lemma'

🤔

[gilded gull]

tbh it sounds interesting

[coral shale]

Well I don't want to give it away - but try it out first

[gilded gull]

but i have so many problems left

[coral shale]

Maybe you have enough tools to try proving

Well yeah do your problems first ig

[gilded gull]

I will try it out at some point but rn i have to do homework lol

yeah

tysm

[zealous flame]

how do i use texit?

[spice whale]

for a group epimorphism $\varphi: G\to H$, $$G/\ker\varphi \cong H$$

[cloud walrus]

all functions are alison

[winter solstice]

In #resources, there is a texit cheatsheet.

[fallow plume]

I'm really confused here. are we making vectors in a vector space polynomials?

i'm also confused why there's a free rank of 0. is that 'cause all of the basis elements are nonzero and we're multiplying by t^i?

if more context is needed lmk but I'm lost as to the motivation here

[fallow plume]

okay yea I think this is why, 'cause he says they're all monic and this is from V as a K module (?) to a K[t] module (?)

[zealous flame]

Thankyou

I was gonna write up a FAT proof from when shuri (thanks for helping me with that question) was helping me earlier but I'm too tired now

[hidden haven]

k[t] is infinite dimensional as a k vector space, so it can't be a summand of V which is finite dimensional

[hidden haven]

We're finding an isomorphic k[t] module which is built out of polynomials

[fallow plume]

i think i kinda get it

hopefully I'll understand it after I see why

[chilly ocean]

Any good resources on permutation representations/how to find permutation representations?

[hybrid island]

can someone walk me thru this? no idea how to start

[kind temple]

try to start with the definition of an orthogonal matrix

[little root]

Is it true that an injective map between free modules of the same rank (over commutative rings) is bijective?

[next obsidian]

No

Consider Z -> Z given by multiply by 2

It’s true that a surjective map is bijective tho

One can use a variation of Nakayama’s lemma to prove it for example

[little root]

oh sweet that's a simple example, thanks!

that's so strange, so a strict submodule of a free module can have the same rank (like 2Z and Z in this case)

another unrelated question, does anyone know where I might be able to find a proof of this fact (that follows the sketch written here)? My prof did something like this but the lecture was incomprehensible to me

[next obsidian]

I wish I knew normal forms, I really should lol

Maybe try Lang?

[little root]

lemme check

don't think lang covers normal forms 😦

[next obsidian]

Surely he does????

Maybe it’s buried as statements about modules

[little root]

wait there's multiple langs

i found some stuff on jordan canonical form but not smith/rational

[terse crystal]

Jacobson covers it

First volume Third chapter

[fallow plume]

I know Aluffi covers Jordon/Smith

I've never heard of rational tbh

[fallow plume]

^Yea try Aluffi

it also has rational it seems

VI 7.2 I think

pg 373

[little root]

Will look at these in the morning tysm yall

[fallow plume]

np

[next obsidian]

Which Aluffi tho?

[lethal dune]

[fallow plume]

Aluffi deez n--

i feel like he would be very disappointed in me rn

[next obsidian]

I’m serious!

There’s two textbooks

[chilly ocean]

What's the difference between a projection and a morphism

[vestal snow]

Let $R$ be a PID and $k$ be its fraction field. Suppose we have a free finitely generated $R$-module $N$ and a submodule $M$ where the rank of $M$ is the same as the rank of $N$. Let $v_1,v_2,\ldots, v_m$ form a $R$-basis of $M$ and $1\otimes_R v_1,\ldots,1\otimes_R v_m$ form a $k$-basis of $k\otimes_R M$. Given an element $v\in N$, find the basis of the $R$-module generated by $v_1,\ldots,v_n,v$.

[cloud walrus]

Finitely Many Bananas

[vestal snow]

What is the algorithm for this?

I'm pretty sure this algorithm shows up in FTFAG over PID, but I suck with matrices so I can't remember

If it makes things simpler, let $R$ be $\mathbb{Z}$ and $k$ be $\mathbb{Q}$.

[cloud walrus]

Finitely Many Bananas

[vestal snow]

It would be even better if someone had the sage code for doing this

[vestal snow]

Actually, here's a cleaner formulation: Let $R$ be a PID and $N$ be a finitely generated free module over $R$. Let $M$ be a submodule of $N$ with the same rank as that of $N$. Given an $R$-basis $v_1,\ldots,v_m$ of $M$ and $v\in N$, find a basis of the module generated by $v_1,\ldots,v_m,v$

[cloud walrus]

Finitely Many Bananas

[dull root]

I have the following question about determining when a polynomial is irreducible over a poly ring.

Let p = x^4 - 2. It is clear p is irreducible over Q by Eisenstein' s criterion with prime p = 2.

1. p has 4 roots, and we see the splitting field of p over Q is Q(2^(1/4),\omega) where \omega is a primitive 4-th root of unity.
2. From this, it is clear that p is NOT irreducible over the simple extensions Q(2^(1/4)) and Q(\omega). It also shows p is not irreducible over fields extensions which contain either of the two listed above.

However, my question is what if we had another simple extension where it is not quite as clear whether p is irreducible or not for example Q(\sqrt 3)? This is an extension that is not contained or contains those described in 2. What techniques could I use to determine the irreducibility of p over such a field. My hunch is that p is still irreducible over Q(\sqrt 3), and I think it might have something to do with the splitting fields, but I am not sure

[coral shale]

'Normal extension' comes to mind

[dull root]

So I know normal extensions are the one where if I take any element in the field, then its minimal poly splits as root factors over the extension. I also know normal extensions L of K iff L is a splitting field of some p in K[x] over K

[coral shale]

For your specific example, you can show Q(rt3) is normal. And proceed

in particular show one of the fields does not contain the other (maybe both ways needed, not thinking about it much)

[dull root]

If Q(\sqrt(3)) were to be a normal extension, and p was reducible, would that mean that it splits p to its 4 root factors over Q(rt 3)

[coral shale]

yes

check the defn of normal

[dull root]

Yes, that is clearly a contradiciton since a primitive 4th root of unity can not be in there

Wait, the def on normality is that if Q(rt 3) is normal, p is irreducible AND has a root in Q(rt 3),, then it splits as root factors. However, why does that say if p is reducible, then it splits into roots factors?

[coral shale]

my defn is

L : K extension

This is normal iff

• Algebraic extension
  and
• All (non-constant) irreducible polynomials in K[t] either split into linear factors in L[t] or have no roots in L

[dull root]

I see, then that means if p was reducible in Q(rt 3), it must split into its linear factors and we get that its split field is contained in Q(rt 3)

[coral shale]

yes

[dull root]

If p is an irreducible poly in K[x] with some root in L, then p splits into linear factors in L[x]. However, it is also possible p to not have a root in L but still be factorized in L[x] where the factor is not linear though right?

If so, then we don't have the statement I made above, since p being reducible in L[x] could still mean that it factorizes into factors with deg > 1

[iron vessel]

It must also be an algebraic extension to be normal

[dull root]

yes, but the same issue still remains if Im not wrong

[iron vessel]

Maybe i'm misunderstanding your question, but since p is in K[x] and L/K is an algebraic extension, then $K \subset L$ and thus L contains all roots of any polynomial in K[x], so also for p, and p must split entirely in L[x]

[cloud walrus]

Évariste Galois

[dull root]

I'm having trouble showing this defn implies what you wrote: L is a normal extension of K if

1. [L:K] is finite. This part I see in what you wrote.
2. If x \in L, then the minimal polynomial of x over K splits into linear factors in L. Equivalently, if p is irreducible in K[x] AND p has a root in L, then p splits into linear factors in L[x].

My main question is with 2. Why is it that the condition implies that if I take any irreducible poly p in K[x], then it must have a root in L

[iron vessel]

Because any finite extension is algebraic

[dull root]

L being an algebraic extension means that for every element in L, it has a minimum poly in K[x] right?

[iron vessel]

Yes

[dull root]

And any irreducible poly in K[x] is the minimal poly of some element in L?

[iron vessel]

I was mistaken

so yeah every element in L is algebraic over K

[dull root]

But still that does not mean that any irreduiclbe poly in K[x] is the minimal poly of some element in L right? I know that for any irreduible poly in K[x], there exists a simple extension K(a) such that a has that minimal poly

[iron vessel]

maybe youre right

if it is an algebraic extension doesnt mean it is algebraically closed

[coral shale]

you probably want to specify 'monic' but sure (what you said isnt wrong, but we dont rlly care about non-monics, they are never minimal polynomials in a polynomial ring over a field)

[dull root]

@coral shale Then I don't see how the idea you said goes:

We have p = x^4 - 2 an irreducible poly in Q[x]. We want to determine if p is reducible over Q(\rt 3).

1. I see Q(\rt 3) is a normal extension of Q.
2. IF a root of p exists in Q(\rt 3), then we would be done and p is reducible, in fact it splits as linear factors by normality.

However, it is not true that if p is reducible over Q(\rt 3) that it must split into linear factors. For example, p can split into factors of degree 2 each

[coral shale]

it is true.

if L:K normal

Then all irreducible polynomials in K[x] must split into linears when considered in L[x]

or not at all

3) is an impossibility given a normal extension (since p is irreducible in Q[x])

[dull root]

So I am using this definition:

so what you claim is that every irreducible poly in K[x] must all its root in L if L is a normal extension of K, or it has no roots in L at all -- this I see. However, I really don't understand from this how to get that if p is irreducible in K[x], say of degree n > 3, but reducible in L[x], that it must split as linear factors just because of the irreduciblity of p in K[x]

[coral shale]

Think over it carefully with examples

I don't see the misunderstanding issue...

I suggested above to show one field cannot contain the other

But there may be better approaches

[dull root]

so p = ab where a,b \in L[x] and deg(a), deg(b) < n is all that reduciblity of p in L[x] gives. It is true that the roots of p must appear in a or b, but it can happen that even if a,b are irreducible in L[x], non of their roots are in L

[coral shale]

Let L:K extension. Let f in K[x] irreducible of degree n

If f splits in L, I believe that [L:K] >= n

I think this was the result you were trying to use but couldn't see why?

It would be nice to show this - I'm pretty sure it is true, and don't think it should be too hard

[dull root]

you mean L:K \leq n?

[coral shale]

No I don't...

[dull root]

nvm i see what you mean, yes that part is clear to me

[coral shale]

Yh it is true btw

Then using this fact, you can show what you needed originally

[coral shale]

a stronger result is that n must divide the degree of the extension... if I'm not mistaken

[hidden haven]

Yes, n divides the degree of the extension

You don't need splitting, having any root is enough

[dull root]

Yes, that is true. But I'm still having trouble understanding the general framework.

1. p = x^4 -2. I want to show p is not irreducible over Q(\sqrt 3) (this is clearly a degree 2 extension).
2. The splitting field of p is generated by Q(2^(1/4), omega) where omega is a primitive 4-th root of unity. 4 divides the degree of this extension.

[coral shale]

Let L:K extension. Let f in K[x] irreducible of degree n

If f has a root in L, n | [L:K]

(for myself)

(Oh I knew this smh)

[hidden haven]

Why do you think p reduces over that extension?

[dull root]

If p reduces, then that means it it can be factored into irreducibles in Q(\sqrt 3), if any of these factors were linear, we are done since p must split completely in Q(\sqrt 3) which is a contradiciton as then Q(\sqrt 3) extends the splitting field of Q. However, the only issue is if p, which has degree 4, reduces into factors of degree 2 each.

I can't wrap my head around why this case can not happen

[hidden haven]

Yeah but p doesn't reduce

[dull root]

yes, I was trying to reach a contradiction if it reduces

[hidden haven]

Also, I don't think reducibility implies splitting, having a root implies splitting

Having a root is stronger

[dull root]

if it reduces into a linear factor, then it must have a root in Q(\rt 3), which is normal extension

thus it must split completely into roots

[hidden haven]

Yes

[dull root]

so reducing where one of the factors is linear gives that Q(\sqrt 3) contains the splitting field of p

so I understand why that case is a contradiciton

[coral shale]

I havent completely followed but ?

L:K normal
if irreducible f in K[x] splits in any kind of way in L[x], it must split completely into linears
?

[hidden haven]

No, x^4 - 2 factors in the splitting field of x^2 - 2, but doesn't split

[coral shale]

huhhh

[dull root]

we require that the polynomial has a root in that field for it to split

[coral shale]

uh oh, big oops

time to correct this thought

mb

[dull root]

So I think the problem reduces to showing (x^4 - 2) = (x^2 + rt 2)(x^2 - rt 2), and that the factors (x^2 +- rt 2) are not irrreducible in Q(\rt 3)

[chilly ocean]

does anyone have any good sources for understanding group actions? I am having a hard time adjusting to them

[coral shale]

this sounds annoying to do more generally, I feel there surely must be a way to avoid...

[dull root]

This problem would be so much easier if it was x^3 - 2 instead, since it forces one of the factors to be linear,

[coral shale]

im just thinking here, we have 2 normal extensions. Does considering their intersection or something not help

maybe this is just equivalent to the original problem - not an approach, but a way to rephrase

=========
If an irreducible polynomial f in K[x] is reducible in some normal extension L over K,

the splitting field of f and L are both normal over K.

Their intersection must be a non-trivial normal extension of K??? Is this true I think?

Well even if it is true I suppose it isn't very helpful...

[dull root]

Ok I'm trying to take a step back at the problem.

WTS: p = x^4 - 2 is irreducible over Q(\sqrt 3).

I know for sure:

1. The splitting field of p is Q(2^(1/4), \omega) where omega is a primitive 4-th root of unity. This is a degree 8 extension over Q.

2. Q(\sqrt 3) is a degree 2 NORMAL extension of Q.

3. If p reduces into any linear factor over Q(\sqrt 3), we have a contradiction, since it must split into linear factors which means Q(2^(1/4), \omega) \subset Q(\sqrt 3). This contradicts the degree of extensions (2 > 8).

4. If p does not reduce over Q(\sqrt 3), we are done. By 3, p not reducing is equivalent to showing that p (which has degree 4) CAN NOT be reduced as factors of degree 2 polynomials which are irreducible in Q(\sqrt 3)[x]. The only such factorization of p is

x^4 - 2 = (x^2 + \sqrt 2)(x^2 - \sqrt 2). If we want a contradiction, it should be that (x^2 +- \sqrt 2) \notin Q(\sqrt 3)[x]

Wait...isn't 4 true because sqrt 2 \notin Q(\sqrt 3)

So it def can not happen

[coral shale]

yh i agree it boils down to showing Q(rt 2) not in Q(rt 3)

If it does reduce, Q(rt2) should be the intersection of the 2 fields

(split of p) intersect Q(rt 3)

[sharp sonnet]

start here https://gowers.wordpress.com/2011/11/06/group-actions-i/

[coral shale]

10 yrs later and still good

[dull root]

But I don't think we need any of that. If we know that \sqrt 2 \notin Q(rt 3), then such factors already do not exists in that polynomial ring, which means a factorization of x^4 - 2 into two degree two factors can not happen over Q(\sqrt 3).

[coral shale]

well no we don't, but when talking about a more general example this is what I'd think about

[woeful flint]

What is $\text{Hom}(A,C_2)$ where $C_2$ is the cyclic group of order 2 and $A$ is any $\mathbb{Z}$-module. Am I right in saying that it's isomorphic to $A$? I can't make my mind up

[cloud walrus]

Kraft Macaroni

[chilly radish]

It is, in general, not going to be isomorphic to A

For example, take A=Z

[woeful flint]

How would I go about computing the quotient $\mathbb{F}{16}^\times/\mathbb{F}{4}^\times$?

[cloud walrus]

Kraft Macaroni

[trim grove]

By using an example how can i say that Q under addition is not a finitely generated group?

[sharp sonnet]

those groups are cyclic, right?

[sharp sonnet]

using an example?

[trim grove]

some contradiction

[woeful flint]

I'm not entirely sure if the first one is

i.e. F_16

[sharp sonnet]

probably use the fundamental theorem of finitely generated abelian groups then

[sharp sonnet]

oh, it is

so i would show that, then this is simple

alternatively you construct F_16 as some F_2[x]/(f(x)) and then how F_4 lies inside that

and then you can work it out explicitly

[trim grove]

but how i can use this here , can you please elaborate a little bit?

[delicate orchid]

I think there might be a simpler way

[sharp sonnet]

there is

[delicate orchid]

By considering the fractions 1/p with p prime

You should be able to arrive at a contradiction from the fact that there are infinite primes, unless there’s a way to generate 1/p’’ from 1/p’ and 1/p that I’m missing

[sharp sonnet]

you can also use the fact that it is divisible

just not sure if this exactly qualifies as "by contradiction"

[delicate orchid]

Assume it’s finitely generated => there exists a 1/p that is unreachable by the generating set, contradiction

Is what I had in my head

[woeful flint]

tried doing that but its easier said than done ahaha

[sharp sonnet]

what exactly is the issue?

[trim grove]

hmmm that looks something intresting , Thnks

[woeful flint]

finding the elements which precisely coincide with F_4 within F_16 is not so clear

[sharp sonnet]

you need an irreducible polynomial of degree 4 in F_2[x] right?

[woeful flint]

as in I dont know how to do it

yeah

x^4 + x + 1 works

[sharp sonnet]

then you can just write down all elements now

[woeful flint]

yeah

[sharp sonnet]

in F_16

[woeful flint]

thats what I did

[sharp sonnet]

look at the subgroups each element generates

[woeful flint]

Maybe its easier if I do it write F_16 down

one sec

[woeful flint]

thats a good idea tbf

[sharp sonnet]

one of them must generate the subgroup

this uses the fact that its cyclic but you dont need to know that to confirm that

(this is just super tedious)

[woeful flint]

Ok I've done it

[cloud walrus]

Kraft Macaroni

[woeful flint]

I have to determine for which values of $n$ this sequence is split

[cloud walrus]

Kraft Macaroni

[woeful flint]

I was thinking of computing the baer sum for the sequence above and either proving that the baer sum is split or that it isnt split but it seems a bit hard. Would anyone recommend/discourage this idea?

[cloud walrus]

Kraft Macaroni

[woeful flint]

there was a typo there

[hot lake]

it's split if and only if (4^n-1)/3 is coprime with 3 I think ?

[woeful flint]

Hmmm

[cloud walrus]

Kraft Macaroni

[woeful flint]

Here the C's are cyclic groups and the square brackets are the p(n)-torsion elements

If I'm able to show that the baer sum of the sequence above with itself is split then I'm basically done because that would imply that the order of the sequence above divides 2 and so it must correspond to zero in the first cohomology group

[woeful flint]

Why do you think this is the case?

[cloud walrus]

Shadow08

[hot lake]

because if it's coprime then it's a direct product and if it's not then since the groups are all cyclic there is only one possible group morphism for the split and the elements of order 3 are sent to the image of F4* which are sent to 0 on the way back.

[mystic bolt]

Everything single module I've had on abstract algebra is missing a ton of the motivation behind the concepts. Am I right in thinking ideals essentially group together numbers with some kind of property?

[spice whale]

specifically, they're closed under addition
and if you multiply any element of the ring by an element of the ideal, the ideal slurps it up

for example, 2Z is an ideal in the ring Z

because the even numbers are closed under addition

and if you multiply an integer by an even number it becomes an even number

[delicate orchid]

ideals are a generalisation of the concept of a "multiple" or a "times table" if you wanna go back to middle school

[hot lake]

start by looking for an identity element ?

[woeful flint]

ah thanks I got it

[wraith obsidian]

Do we know anything about normal forms of tensor products? Like, is there a way to turn the tensor product relations into a confluent+terminating rewriting system?

(tensor product of R-modules, R commutative unital)

Is there even a decidable procedure to determine whether two representations are equal?

Or can one find the word problem for groups in there by considering some group ring

[dull root]

I have a basic fact that I can not see. Let L be a field, and G:= Aut(L). If I take a subgroup H of G, then the fixed pt set L^H is a subfield of L.

It is also true that Aut(L: L^H) = H. It is clear from the definition that H is a subgroup of Aut(L: L^H), but I don't see the other inclusion

[subtle ivy]

so what are you trying to prove? that L^H is a subfield?

[dull root]

no, the 2nd part

that Aut(L:L^H) = H

1 inclusion is trivial

[subtle ivy]

oh ok lol

i think this is false. if we take H to be trivial, then L^H=L, and so aut(L:L)=aut(L) which is not H in general, no?

unless im not understanding what aut(L:H) is.

@dull root

[dull root]

No, L^H are the elements x in the field L such that hx = x for every h \in H. Then if H=1, L^H = L, and Aut(L: L^H) = Aut(L:L) = 1 = H, since the only automorph in L that fixes everything in L is the identity automorphism

[hybrid island]

can someone help me w this

[vast quiver]

I found this paper related to the noncommutative case, haven’t read it yet. I would guess that the answer for the commutative case is also no, but I’m not an expert.

https://www.sciencedirect.com/science/article/pii/S0747717184710674/pdf?md5=e6b3a4c8b628b832ae9bacfaed591baf&pid=1-s2.0-S0747717184710674-main.pdf

[vestal snow]

Let $M$ be a free finitely generated abelian group of rank $n$ and let $M_0$ be the $Q$-vector space obtained by localizing at 0. Let $v_1,\ldots,v_n$ be a set of $Q$-linearly independent elements and let $v$ be an arbitrary element of $M$. Then we have that $v=a_1v_1+\cdots+a_nv_n$ for some $a_i\in Q$. Find a Z-basis for the abelian group generated by $v,v_1,\ldots, v_n$ in terms of $v,v_1,\ldots v_n,a_1,\ldots,a_n$.

[cloud walrus]

Finitely Many Bananas

[next obsidian]

Try the 1-dim case first

Any element of M_0 forms a basis

Then maybe try 2-dim then see if you can figure it out

[vestal snow]

Is it going to be some form of SNT/row reduction/other matrix wizardry?

[next obsidian]

Idk

Just my suggestion

[vestal snow]

1 dimension is pretty straightforward I think

[next obsidian]

Yeah but it probably illustrates the general idea

Or rather, once you go to 2-dim

[vestal snow]

Once you do it for 2 dim I think you should be able to induce it up to n

Or maybe not

[wraith obsidian]

Uhh, that is seriously interesting (and a very accessible paper), thank you!

[chilly ocean]

explain this plz..

[upper pivot]

the relations is a normal subgroup right

as you wanna mod out by it and that doesnt make sense unless its normal

[chilly ocean]

i think the words should be g(abab)g^(-1) is in R for any g in D_n

[upper pivot]

sure thats true, but they are using in particular g=b to get what they want

[chilly ocean]

why..@upper pivot

[upper pivot]

well they want to get baba

and they recognize that b(abab)b^-1 is equal to this

[next obsidian]

baba

[vestal snow]

Do you know how to do the two dimensional case? Suppose $\frac{a_1}{b_1}v_1+\frac{a_2}{b_2}v_2=v$ where $a_i$ and $b_i$ are integers.

[cloud walrus]

Finitely Many Bananas

[vestal snow]

Somehow we want to write each $\frac{1}{b_i}v_i$ as a $Z$-linear combination of things in the module generated by $v,v_1,v_2$

Or at least if we can do this we're done

[cloud walrus]

Finitely Many Bananas

[vestal snow]

I worked on this problem a bit more and essentially I need to do the following:
Let $v=\frac{a_1}{b_1}v_1+\cdots+\frac{a_n}{b_n}v_n$ where the $a_i$ and $b_i$ are integers. Let $M$ be $n \times (n+1)$ matrix which is the $n \times n$ identity matrix and the $(n+1)^\text{th}$ column is has entries $\frac{a_i}{b_i}$ for the $i^\text{th}$ slot. Using elementary column operations (switching columns, adding to one column an integer multiple of another column, multiplying all entries of a column by a unit (in this case, $\pm 1$)), get a new matrix where one of the columns has zeroes in every slot.

[cloud walrus]

Finitely Many Bananas

[delicate orchid]

Gaussian elimination ohh yeahhh

[lapis trail]

I'm sure it's easy but I'm stuck

[lethal dune]

try contrapositive argument

[chilly ocean]

,rotate

[chilly ocean]

Is this consistent?

S_p(0) is kernel of A ->S^-1A where S=A-p

[fallow plume]

is prime ideals a sharp condition?
I could totally see an argument that it could be maximal ideals instead, but I think I might be missing something since it doesn't say maximal

[fallow plume]

looks good to me

[fallow plume]

Do you mind spoilering/deleting this? I'm not supposed to receive outside help and I think your proof is for the problem--not what my question is. (The problem was just related so I posted it with)

Thank

[dull root]

I have the following question: For this question everything is working as subfields of C.

If I have a subfield K, and a poly p in K[x]. If L is the splitting field of p over K, then is L automatically a Galois extension. I know Gal extension iff normal and separable, so it amounts to saying that L is separable extension. I am wondering what the role of C is playing here, that is, why is it important we work over C

[lethal dune]

C is acting here as an algebraic closure. Like we can construct fields by adjoining roots to it, for adding a root, we must show it exists somewhere

like R[i], we know i exists in C(complex) so R[i] makes sense

[dull root]

I see. But it is not true that if If L is the splitting field of p over K, then is L automatically a Galois extension right? Are there Splitting fields which are not separable

[shell brook]

does eisensteins work if i'm working in a splitting field over Q

I dojn't think so

[sharp sonnet]

you need generalizations of gauss's lemma and of eisenstein

[shell brook]

I dont think I know of any that would work

[terse crystal]

C is of characteristic 0

Any extension field of a characteristic 0 field is separable

[sharp sonnet]

you want to show some polynomial in your splitting field is irreducible?

[shell brook]

Yeah

I was thinking I could do the derivative thing

but it's not easy to find the gcd

[sharp sonnet]

i think you need that the ring of integers is ufd and then eisenstein becomes a statement about prime ideals

to use it

so thats probably not the intended solution if you dont have that

[shell brook]

ah yeah I've never seen this

does the polynomial that it splits over even matter

I don't think it does

it splits over x^11 - 1 and the poly is x^7 - 3x^2 + 3x + 3

[sharp sonnet]

i think you just have to calculate the gcd

sage can do it

[shell brook]

the gcd calculation is disgusting

mathematics

[sharp sonnet]

funnily enough

[shell brook]

i typed it in wrong but same vibes

[sharp sonnet]

eisenstein works as well

but you dont have the tools to show that

[shell brook]

[sharp sonnet]

i dont know if you are allowed to use sage, but its 3 lines in there

[shell brook]

we are not

[sharp sonnet]

i never compute polynomial gcds by hand

or anything really

[shell brook]

we havent either

i think we just mentioned it

[sharp sonnet]

i mean its just long division

good luck, have fun

(there is possibly just some other way to solve it)

[shell brook]

I will simply not do it

probably

[dull root]

Just do it, a mindless calculation is less taxing on the brain than a complex idea

[coral shale]

Is the subdirect product a simple definition or quite involved (is it rep theory???)
(scanned wiki, didnt get it, but dont want to put too much time into it if it's not worth)
If there's a simple intuition would be appreciated 🙏

[dull root]

You mean the definition of the semidirect product?

[coral shale]

Sorry, this is the Rubik's Cube group for context

I know what the semidirect product is

But here, it talks about the subdirect product

You see the ^0.5

I am writing a project on this topic --- I would include this if I could understand it

[sharp sonnet]

do you the link to what Griess links to

[coral shale]

just the person but one sec

[sharp sonnet]

oh well, then no idea

the definition of subdirect product doesnt seem hard to get

but it doesnt explain this notation

[coral shale]

https://en.wikipedia.org/wiki/Rubik's_Cube_group#Group_structure

uhh

I haven't done representation theory

Am I missing knowledge to understand?

Or in fact, this definition is within the context of cat theory?

Oh no - maybe not. Universal algebra

[sharp sonnet]

you dont need this

you need to know what an algebra is

[coral shale]

right, I don't

so nvm ig

[sharp sonnet]

well, that you could look up

tbf i have never seen this before

[coral shale]

I mean I could have a good guess

[sharp sonnet]

but i get the definition and the two examples i looked at

but i cant intuit this notation

so im afraid you would have to look at the work of Griess

[coral shale]

Alright, thanks

I think I'll just miss out on this

and leave it as a semidirect product

[sharp sonnet]

thats probably best

[shell brook]

does this become easier if I only care about if it has a root

[sharp sonnet]

you know the roots of x^11-1 and can check

[shell brook]

Right because the splitting field is just Q(roots of x^11 - 1)

fuck im so dumb lmao

[obtuse turtle]

@spice whale what are monoids and semigroups?

lol

[spice whale]

so basically

if you have a set

of anything

you can define a way to combine two elements in the set and get another element in the set

[mellow fern]

yeahh I'm now advanced in math

[obtuse turtle]

cool! like the union?

[spice whale]

if you have a set of sets then yes

[candid kiln]

more like an operation really

[obtuse turtle]

oh cool

[spice whale]

you take two sets in your set of sets
and take the union

[obtuse turtle]

wow, i get it now.

[spice whale]

you just need to make sure that you'll always get a result inside your set of sets

[obtuse turtle]

what about semigroups?

cool

[spice whale]

so

[obtuse turtle]

and where can I learn more about it?

[spice whale]

if you have this set, and a way to combine two things, you get a magma

[obtuse turtle]

magma? I'm sorry, I'm dumb haha

[spice whale]

that's just what it's called

[obtuse turtle]

oh wow

oh yesss, i get it now

[spice whale]

if you package up a set with a "binary operation" (way to combine two things in the set to get another thing in the set) you get a "magma"

[obtuse turtle]

cool, I'm just reading about it in wiki now

[spice whale]

yeah

[obtuse turtle]

thanks for this wonderful topic

[spice whale]

so

[obtuse turtle]

where can I learn more bout this?

[spice whale]

a semigroup

[spice whale]

uh

[obtuse turtle]

lol, and from the basics?

[spice whale]

a good book is "algebra" by michael artin

[obtuse turtle]

this explains it!!

[spice whale]

it doesn't introduce monoids and semigroups though afaik

[obtuse turtle]

what are the prerequisites?

[spice whale]

familiarity with proofs

[obtuse turtle]

cool, I just need the basics. I can read the monoid and semigroup stuff later

[obtuse turtle]

yay!!! thank you

[spice whale]

oh

[obtuse turtle]

wow, it's legally available for free!

just downloaded it

[spice whale]

and if you're not particularly advanced in maths, some of the examples might not make much sense lol

[candid kiln]

you kinda need to be comfortable with proofs and the language of proofs

[spice whale]

yeah

it's quite dense without

[obtuse turtle]

I can only understand 1st chapter lol

[candid kiln]

tho u can definitely become comfortable over the course of doing the book

[obtuse turtle]

oh, where can I learn that?

[spice whale]

oh you're completely fine then lol

[candid kiln]

this chapter is very different from the matrices in the class 12th chapter

[obtuse turtle]

ohh

[candid kiln]

a lot of people used Vellemans book on proofs

i never did any proof book, i just yolo'ed it and it kinda worked

[spice whale]

same

[obtuse turtle]

yay, thanks a lot! gonna start from today!

[spice whale]

yep

hope you enjoy

[candid kiln]

[spice whale]

we caught another one algebra army

[obtuse turtle]

thanks y'all!

[spice whale]

yw

[candid kiln]

wew would be proud

[obtuse turtle]

vallemans book on "how to prove it"?

[candid kiln]

yes

try reading artin first

[spice whale]

yeah

[candid kiln]

if it feels a bit weird or hard

then go to vellemans book

[obtuse turtle]

I already have it, never read it haha

cool

[iron vessel]

Found this answer under: Why is x^4 + 2 irreducible in F_125[x]?

For some reason i cannot understand the part right after "As gdc(4,3) = 1,"

How does he obtain that expression for the degree of $[F_{5^3}(\alpha):F_5]$?

[cloud walrus]

Évariste Galois

[coral shale]

I was informed this is a scam

[terse crystal]

E and K are both extension of F, [E:F]=m,[K:F]=n. If gcd(m,n)=1 them. [EK:F]=mn

[candid kiln]

wdym

[coral shale]

the implications do not hold in all cases or something

[candid kiln]

[coral shale]

What was it. Apparently there are associative quasigroups that aren't inverse semigroups?

Or the other way round

[iron vessel]

Ok i am familiar with this, but does $\mathbb{F}_{5^3}\mathbb{F}(\alpha) = \mathbb{F}__{5}(\alpha)?

For some reason the latex cant compile with two _

[spice whale]

yep

i believe

[candid kiln]

😵‍💫

i dont know enough algebra to verify

[terse crystal]

Do you mean why F_5^3(α)=(F_5^3)(F_5(α))?

[iron vessel]

yes

[spice whale]

associative quasigroups don't need inverses for everything

[iron vessel]

Also, how does he conclude that [F_5^3(\alpha):F_5^3] = 4 just from the fact that [F_5(\alpha):F_5] = 4?

Is it done by applying the tower law to [F_5^3(\alpha):F_5]?

[terse crystal]

$F_{5^3}F_{5}(α)\cong F_{5^3}\otimes F_{5}[x]/I \cong F_{5^3}[x]/IF_{5^3}[x] \cong F_{5^3}(α)$

[cloud walrus]

Cogwheels of the mind

[iron vessel]

Ahh ok

[terse crystal]

It’s just a commutative diagram. [EK:E]=[K:F] and [EK:K]=[E:F]

[iron vessel]

ok that makes sense. thanks!

[shell brook]

hi everyone i need to show that x^3 + 5x + 5 is not irreducible in E[x] where E is splitting field of x^11 - 1

so i think cause deg3 it must have linear factor

so the only way I can think to show it is to show that it's roots aren't in E

but it's roots are awful

,w roots x^3 + 5x + 5

[cloud walrus]

Results provided by WolframAlpha

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

Requested by jesse#3659

[shell brook]

So idk if there is a better wya to do this

[small bison]

eisenstein maybe?

[shell brook]

how can I eisenstein when I'm not in Q

I'm in a splitting field of Q

oh I don't have generalized eisenstein

[small bison]

i think you just need to show that 5 is still prime in E

[shell brook]

Why does that do anything

[small bison]

i think you need 5 to be prime if you want to use the generalized eisenstein

idk though

[shell brook]

oh okay

I don't have gen eisenstein

[small bison]

prove it in your homework and use it

[shell brook]

[small bison]

or maybe you can do some argument by looking at towers

[shell brook]

hows that

okay well maybe

I guess the splitting field is pretty nice

[small bison]

like assume it has a root alpha in E

[shell brook]

so maybe I can just look at its degree?

[oblique river]

Do yku know the gslois group?

[small bison]

and consider something like Q -> Q(alpha) -> E maybe

[oblique river]

Is it S_3 or A_3?

[shell brook]

No I don't know the galois group of this

[oblique river]

Then calculate it. It’s easy for cubics, just find the discriminant

[shell brook]

what does that do for me

[oblique river]

If it’s S_3, youre done

[shell brook]

we haven't really studied the galois group much

enough that I could use this probably

[oblique river]

The galois group of E/Q is abelian

And so all wuotients must be abelian

(In fact the galois group must be S_3 since this cubic has nonreal roots)

[shell brook]

why is S_3 good

[prisma ibex]

it's as big as possible

[oblique river]

It’s not abelian

[prisma ibex]

that too

[shell brook]

I have no idea why that implies the poly is irreducible though

[oblique river]

And therefore it’s not a quotient of an abelian group

Im not proving (directly) that this poly is irreducible over E

Im proving that it doesnt have a root in E. If it did then all of its roots must be in E (corrollary of E/Q being galois) and therefore the solitting field of this poly is a subfield of E

[shell brook]

I think maybe there is something being unmentioned that i have not learned

[oblique river]

Wait. There is a much easier argument sorry lmso

Just look at degrees

[shell brook]

yeah okay

I figured

[oblique river]

The degree of E/Q is 10

Which isnt divisible by 3

[shell brook]

okay

[oblique river]

So you cant have a root of an irreducible cubic

[shell brook]

yeah that makes sense

[oblique river]

Sorry i should have seen that first

[shell brook]

smh

thank you bunhco

i appreciate it

[oblique river]

Np sorry for not listening lol

[prisma ibex]

in general if you have f(x) of degree n over K with splitting field F irreducible then Gal(F/K) is transitive on S_n.
In the case n=3 the only transitive subgroups of S_3 are S_3 itself and A_3

so if you find some Galois group of a cubic that isn't these you immediately have a reducible cubic and get a linear factor that splits off

[shell brook]

I see

I think the lecture that I haven't watched yet starts in on this stuff

[shell brook]

wait isn't it 11

x^11 - 1

[south coral]

To prove that a subgroup $H$ is in the center of a group $G$ for which I have a presentation with generators $a_1\dots a_n$, I would only need to check that the elements of $H$ commute with the generators of $G$, right? Since $ha_i=a_ih$ implies $a_i^{-1}h=ha_i^{-1}$ and $ha=ah,hb=bh$ imply $hab=ahb=abh$

[cloud walrus]

Astianthus

[small bison]

it's not irreducible cause 1 is in Q

[shell brook]

what do you mean

I'm saying isnt the degree of E/Q 11 because the degree of the thing it splits over is 11

[small bison]

i think the polynomial needs to be irreducible for that to be true right?

[shell brook]

oh it needs to be minimal I guesss

[small bison]

yeah you'd need to factor that bad boy

[shell brook]

damn

how do I easily see the degree of something

[small bison]

look up the cyclotomic polynomials

[shell brook]

oh yeah I guess it factors really nicely

do cyclotomics have a nice degree?

[small bison]

no not really xd

[shell brook]

[small bison]

actually for primes i think it's easy

[prisma ibex]

uhh they have degree Euler totient of n

[shell brook]

ah it is the euler totient

yeah

[prisma ibex]

yeah

since the Galois group is just (Z/nZ)*

[shell brook]

right so thats how its 10

I see

[vestal snow]

Do you know how to do this?

[delicate orchid]

not immediately by looking at it

[shell brook]

Okay last question I have \alpha transcendental over F and \beta = \alpha + \alpha^3 obviously also transcendental. how do I know \alpha algebraic over F(\beta)

I guess maybe a better question is what does an element in F(\beta) even look like lol

cause I figure it's not so hard to write down a polynomial

[vestal snow]

I don't even think its possible

For example, consider the matrix [1 3/7]

There is no way to get 0 for either entry using Z-invertible column operations

[delicate orchid]

that is quite convincing

hmm

[vestal snow]

Essentially, the problem is the following. Suppose we have v = 3w/7 and this is the only relation on v and w. Find a basis (of only one element) of the group generated by v and w

We are in an abelian group here

[delicate orchid]

yeah we're in Z or Q depending

[vestal snow]

I guess the formal way to state this would be this

[delicate orchid]

<1/7> is my guess

[terse crystal]

The minimal polynomial of α in F(β) is f(x)=x^3+x-β I think

[vestal snow]

How would you show that 1/7 is indeed in that subgroup though?

[shell brook]

How do you know this

[terse crystal]

Because it’s irreducible

[hidden haven]

You don't need minimality to get algebraic

[shell brook]

I just need a polyniomial

[delicate orchid]

(7w-2v)/7 is my guess, cause we're in a subgroup of Q right?

[hidden haven]

As long as there's a polynomial, it's algebraic

So you're done

[shell brook]

I just need to know its in the field

oh okay and it obivously is

thank you

i overcomplicated

😎

[terse crystal]

If it has a root , g(β), then g(β)^3+g(β)-β=0

Then this contradicts to the fact that α is transcendental

[vestal snow]

Does (7w-2v)/7 generate both w and v as a Z module?

Wait

I think I got it

[terse crystal]

Because h(α)=0 and h(x)=(g(x+x^3))^3+g(x+x^3)-(x+x^3)

[delicate orchid]

please, tell me

[vestal snow]

Suppose w = av/b for a and b integers

Then we can assume that a,b are coprime

[delicate orchid]

makes sense

oh bezout's memes? I smell a bezout meme
haha I was right

[vestal snow]

So there are integers k,c such that ak+bc=1

This implies that (ak+bc-1)/b * v = 0

I.E. we get kw + cv = v/b

[delicate orchid]

Following so far

[vestal snow]

So v/b is in the subgroup gen'd by v and w

Clearly v/b generates this subgroup as well

So v/b is the required basis element

[delicate orchid]

nice

so the original space is 1 dimensional, meaning that there will be some kernel space => some basis that you can write the matrix in such that one column is zero? I think? Wait nevermind maschke's might not apply on Z it's not a field

[vestal snow]

And now I need to generalize this to n-gens

Is rationalization of an abelian group a term that's used? I've heard complexification of a real vector space

[delicate orchid]

it's just an extension of scalars

[vestal snow]

Yeah I was just wondering if it had a special name

[delicate orchid]

if you mean going from a Z-module to a Q-module, that is

[vestal snow]

like complexification of a real VS

[delicate orchid]

I like rationalisation so that is now what it is called

I decree it

[vestal snow]

I spelled it with a z :p

Thanks for the help

[delicate orchid]

you did everything ngl

[vestal snow]

Emotional help

[delicate orchid]

but you're welcome anyway

[trail stump]

can anyone show a proof of Theorem 23.11?

or at least some intuitions behind why it works? (if possible, not too complicated lol)

or any links to already done proof

[next obsidian]

It’s a special case of something called Gauss’s lemma

You can find a proof of Gauss’s lemma in nearly any textbook on abstract algebra

And I’m sure a quick google search will turn it up

[trail stump]

oh ok i will look into it, thank you!

[next obsidian]

I mean that works but

Normal also means (as in this is equivalent) for all g that gHg^-1 = H

And that’s just baked into the definition of N_G(H) so you don’t need the argument to even be this involved

[winter hound]

I thought that as well

so if for all g in gHg'=H, then that means gHg' is in H ?

[next obsidian]

Uhh

What does “in H” mean

Idk what your definition of normal is

But it is very easily going to be seen to be equivalent to the statement that gHg^-1 = H for all g in G

[winter hound]

same thing: for all g in G and h in H, ghg' in H

[next obsidian]

Well no this is a different statement

But it’s equivalent

[winter hound]

for it to be =, it's equivalent

[next obsidian]

gHg^-1 is a set

You can’t talk about gHg^-1 being in H

[winter hound]

sorry, I didn't know to word that part

got it, thank you

[vestal snow]

Okay so the general problem has been reduced to the following:

Given integers $a_i,b_i$ for $i=1,\cdots,n$ such that $(a_i,b_i)=1$ for all $i$, fix $k\in{1,\ldots, n}$. Compute an integer solution to the following system of equations in $x_1,\cdots,x_n,x$:

$b_ix_i+a_ix=0$ if $i\neq k$\
$b_kx_k+a_kx=1$

[cloud walrus]

Finitely Many Bananas

[vestal snow]

Which would just be solving the following matrix
[b_1 0 ... 0 a_1 | 0]
[0 b_2 ... 0 a_2 | 0]
.
.
.
[0...0 b_k 0...a_k| 1]
.
.
.
[0... b_n a_n | 0]

In integers

Anyone know how to do this?

I feel like the idea is to show that the set of solutions to [0...0 b_k 0...a_k| 1] is not empty in the integers using Euclidean algorithm and then inductively showing that its intersection with the solution sets of [0...0 b_i 0...a_i| 0] is non-empty for all i less than some m

Wait I think I got it

Since $a_k$ and $b_k$ are coprime, we can find integers p and q such that $pb_k+qb_k=1$. So the problem reduces to finding an integer c such that $\frac{(q-cb_k)a_i}{b_i}$ is an integer for all $i\neq k$

[cloud walrus]

Finitely Many Bananas

[vestal snow]

Yeah this isn't true in general

[upper cape]

Is a map of short exact sequences just something like this?

where together, the maps a,b,c define a map of short exact sequences?

[hidden haven]

The squares should commute

[robust pollen]

Let $E$ be a field with an action of a finite group $G = {\sigma_1, ..., \sigma_n}$ (by automorphisms), and set $F = E^G$. Let ${b_1, ..., b_n}$ be a basis of $E/F$.
It is claimed that by faithfulness of the $G$-action, we can apply Dedekind's lemma to conclude that the matrix $(\sigma_i b_j)_{ij}$ is invertible.

But where does faithfulness actually enter? I don't get what's wrong with this: define characters $\chi_j (g) = g b_j$, these are distinct without any appeal to faithfulness. Thus $\det (\chi_1 \ldots \chi_n) \neq 0$ by Dedekind's lemma?

[cloud walrus]

expectTheUnexpected

[oblique leaf]

How do I extend this proof so that it is also valid for K1 and K2 being algebraic extensions (right now they are finite)?

I know that algebraic extensions are unions of their finite subextensions, but I am not sure what the basis for K1K2 would be if K1 had basis U{a_i}_j and K2 basis U{b_i}_j

[terse crystal]

Those {αβ} generates K_1K_2

So the dimension is smaller than or equal to the product of dimensions.

Because you have that condition, the dimension equals the product of dimensions, those generators become a basis

For a linear space V of dimension n, and you find n elements who span V, then those elements form a basis

[oblique leaf]

I know about this but this is true for the finite case

at least dummit proves it for the finite case

with K1 and K2 being finite extensions

i wonder whether the basis is still {ab} for algebraic extensions that aren't necessarily finite

[terse crystal]

Then probably no general result…

Even the finite case, if the condition doesn’t hold, we don’t know generally what the basis is

[oblique leaf]

are you sure it isn't the other way around? I think the condition you are talking about here is actually a consequence of {ab} being a spanning set

[terse crystal]

We know it’s a spanning set

But it might not be a basis if that condition doesn’t hold

I don’t know what basis is if the condition doesn’t hold

Except it’s a subset of them

[coral shale]

If G = Aut(L)

I reckon Fix will be the prime subfield of L

Is this obvious 🤔

I think it is fairly obvious Fix must contain the prime subfield (homomorphism defns)

but i dont see why the other way is obvious (if it is) for now...

[terse crystal]

I am thinking about construct a counterexample of the form L=F_p[x]/(x^(p^r)-a). Then maybe x+(x^(p^r)-a) is also in the Fix(Aut(L))

Because x^(p^r)-a=(x-A)^(p^r) in the splitting field where A^(p^r)=a, if I can construct such a field then any automorphism f of L, f(x bar) has to be x bar because x bar is the only root

[maiden ocean]

Theres also just R

Every automorphism of R is the identity

So Fix_R(Aut(R)) = R rather than Q

[terse crystal]

I see. Oh, my L also, Aut(L) has only identity

[coral shale]

oh is it

[maiden ocean]

Yes

The process is basically

Every automorphism is continuous -> it fixes Q -> it fixes R

[coral shale]

wait wait why do they need to be continuous

[maiden ocean]

you can show that any automorphism is increasing

[coral shale]

I swear

isnt R : Q an infinite extension

[maiden ocean]

hence its only discontinuities can be jump discontinuities, and that would make it not surjective

[terse crystal]

f(y)-f(x)=f(y-x)

[maiden ocean]

Ergo its continuous

[coral shale]

what is going on with my thinking here

Gal(R : Q) = {e}
You're saying?

as a consequence

I always thought it was infinite

[maiden ocean]

No

[coral shale]

oh wait

Gal isnt automorphisms

it is K-homomorphisms? edit: nvm it has to be auto - gal is a group

Gal(L : K)

[maiden ocean]

Gal is homomorphisms of L that fix K yes

But again every automorphism of R is the identity so its pretty trivial regardless

I think this isnt that weird when you realize that R contains Qbar and hence isnt algebraic over Q

So you wouldnt expect it to behave very well

[coral shale]

ima just have to think this over again . . .

[maiden ocean]

I think you're probably confusing transcendental and algebraic degree

[maiden ocean]

actually nah not really to be clear this is true

Like R is not algebraic over Q so many of the usual theorems about galois groups do not apply

[coral shale]

So just to be clear

Gal(R : Q) = ???

[maiden ocean]

Its trivial

[coral shale]

ok ima have a think

[maiden ocean]

More strongly Aut(R) is itself trivial and Gal(R/Q) is the subgroup of Aut(R) that fixes Q

But R/Q is not even algebraic so the Galois correspondence doesnt hold

And its degree as an extension doesnt have to have anything to do with the size of the Galois group

[coral shale]

Gal(Q bar : Q) is infinite ?

oh wait

ok ok so
Gal(C : Q) is infinite ?

[maiden ocean]

This too

[coral shale]

I think I confused some examples

[maiden ocean]

Okay like

You have to recall that the size of a galois group and the degree of an extension do not always correspond when the extension is not galois

[coral shale]

Yes I know that

I'm not basing it on that, though

[maiden ocean]

Right to say anything about the size of Gal(C/Q) you need to either show its Galois or think about it explicitly

And C/Q is not Galois

Its not even algebraic

[coral shale]

I am struggling to get what Aut(F) looks like I guess

I can consider it to be the P-automorphisms of F where P is the prime subfield of F?

In which case I can think of it in terms of polynomials in P and permuting roots of those

[maiden ocean]

Uh I literally just meant automorphisms of R

Lol

[coral shale]

Yes I understand, but I'm thinking in general

[coral shale]

to approach this definition

[maiden ocean]

In general Aut(F) is just the automorphisms of F. It will fix the prime field necessarily but thats not a stronger condition or anything

[coral shale]

Right. But then at least I can see it is necessary for roots to map to roots

of polynomials in P

I just wanted some intuition for why my thinking earlier was wrong 🤔

[maiden ocean]

I mean I think you were just not thinking about the less pleasant examples where L is not Galois over its prime field

[coral shale]

Right now I can't intuit if Gal(C/Q) or Gal(Q bar/Q) is 'bigger'

I think Gal(C/Q) must be smaller now, but previously I thought it was bigger...

[maiden ocean]

Its not smaller

Its very big

unlike R there are many non trivial automorphisms of C

Its bigger than Gal(Qbar/Q)

But also I dont think you necessarily need intuition for these kinds of pathologic cases right now?

[coral shale]

No I probably don't with regards to my course

[maiden ocean]

This is less a matter of understanding the Galois theory and more a matter of just knowing things about Q, R, and C

[coral shale]

Yes, for one thing I kept thinking Q bar in R for whatever reason earlier

Today will be a thinking day for me ig

[maiden ocean]

Like Gal(C/Q) is very big whereas Gal(R/Q) is not mostly because Aut(R) ends up being trivial because its totally ordered and the field automorphisms respect that order which isnt true for C

A lot of these infinite degree examples can be very complicated or pathological (even nicer ones like Gal(Qbar/Q) are like, very much not explicitly known to us and areas of a lot of active research)

[coral shale]

'every automorphism is continuous'

In general or just for R? I haven't thought about continuity in this topic

[maiden ocean]

Just for R

I mean there might not be a notion of continuity for an arbitrary field right?

But even if there is

This happens because R is totally ordered and its notion of continuity is tied to that, and the automorphisms respect that order

[coral shale]

well yeah that was what i was confused about there when u said it - wasnt sure

ok I see

Thanks, talking this thru rlly helped 🙇‍♂️

[maiden ocean]

np

[coral shale]

Ah more reading and I see other than identity and complex conjugatiom, you need AoC to construct Automorphisms of C

[coral shale]

So I should probably just think about this definition in the context of fields which are algebraic over their prime field...

[maiden ocean]

Um not necessarily?

Im not sure why you're contextualizing this as a statement about like, extensions

You should just think of this as a collection of subfields of L defined by subgroups of its automorphism group

[coral shale]

I mean within my course (like for examples, I don't think we will consider Aut(C))

uhh well I will see

[rapid bramble]

In subrings, the definition we were given is closed under + and *, and "shares the same 1"

Does this mean 1 has to be in the subring?

[hot lake]

yes

[south patrol]

Ye, it's also natural since we want the inclusion to be a homomorphism

[rapid bramble]

So then is it false that every ideal is a subring?

[hot lake]

that's very false

[rapid bramble]

Ok cause Google gave conflicting answers

[coral shale]

I can think of 1 counterexample

[rapid bramble]

Yeah 0 generator

[coral shale]

?

The entire ring is an ideal

[rapid bramble]

<0>

[coral shale]

and subring

[coral shale]

not sure what that is

the smallest subring containing 0?

[rapid bramble]

It's the ideal with {0} as the generating set

[coral shale]

that might not be a ring depending on course definition

The answers to this stuff varies on definition

[rapid bramble]

Right but it's an ideal

[coral shale]

yes its an ideal.

[rapid bramble]

And not a subring since 1 notin it

Course assumes 1 and 0 distinct*

[coral shale]

Some might define a ring to not necessitate unity

in which case all ideals are subrings

[rapid bramble]

Rings don't need to have units yeah

And the converse is false too then, since Z is a subring of Q

Which isn't an ideal in Q

[coral shale]

Yes fields only contain 2 ideals

The trivial and improper ideals

[south patrol]

The common thing I know (ig it's what the category Ring is based on for example) is the convention where rings have a 1 a(not necessarily distinct from 0 as there is the 0 ring) and homomorphisms are to preserve this 1, but if you don't enforce rings to be unital then it seems more the convention that subrings are just 'rings under the restriction of the operations' and thus needn't have the 1

[rapid bramble]

Yeah we proved that R and <0> are the only ideals iff R is a field on the same assignment

[south patrol]

nice

[rapid bramble]

Well... I tried to lol

I think I did it

[coral shale]

The key is to consider the set of units R^x := {a in R : a has multiplicative inverse}

[mystic tiger]

Not sure how to approach this

[terse crystal]

That thing in the bracket to the power of 3 is -1

[chilly ocean]

think about what the vectors in vectorspace look like

[mystic tiger]

Just wanna know if I'm on the right track or not?

[coral shale]

Looking for intuition for the discriminant in algebraic number theory

https://i.imgur.com/RXEQuEn.png

[lethal dune]

yes

[alpine island]

The natural numbers and exponentiation are what? A loop?

[delicate orchid]

just a unital magma I think

[hidden haven]

Not unital

[delicate orchid]

oh yeah lol 1 is just a one sided identity

so it's a magma

[alpine island]

Can't it be unitial if it has an identity for each side?

Wait no, that wouldn't work

[delicate orchid]

yeah what would be the left identitiy

[tender mist]

In the proof, that is true because K1/F (or K2/F) is a finite extension, and hence algebraic over F, otherwise we couldn't deduce it, right?

[alpine island]

I was going to say 1 because 1^n=1 but that is not how identities work lol

[terse crystal]

It’s still span K_1K_2, we just don’t have basis either

[simple mulch]

I don't understand why there's the nz

[tender mist]

Mm, I am not sure it would span it because how can you assure that the inverse of a linear combination of a_ib_j is still a linear combination of a_ib_j?

[lapis trail]

In general is there a method to prove a polynomial over a field is minimal with respect to a root?

[next obsidian]

It’s equivalent to show it’s irreducible. Equivalently, any polynomial has a root in some algebraic closure, so your question is equivalent to asking if there’s a general method to find if a polynomial is irreducible, which is a very hard problem. I think it’s like undecidable or whatever computability term you’re supposed to put there, even over Q

[lapis trail]

Oh rip, the only know a couple irreducibility tests

Thanks for the response