groups-rings-fields

[formal ermine]

isn't that the same as the polynomial ring in a

[pastel cliff]

fuck it i'll brb

[formal ermine]

it would intuitively make sense if the smallest intermediate ring that contains a would just be the polynomial ring in a

[elder wave]

but i'm not sure what your problem is here

[pastel cliff]

it's me im the problem

[elder wave]

for all s in S phi(s) = s but in S^-1R and therefore invertible

[rustic crown]

when you say K[x] is the polynomial ring in x, then it should be the case that two polynomials are equal if and only they're coeffcients exactly match up

but for example in Q[sqrt2] you have sqrt2^2 - 1 = 1

[formal ermine]

right

yeah ok I understand this then

[rustic crown]

you can also see that with your definition

if R is any ring containing both K and a, then it will contain g(a) for any polynomial g in K[x]

[formal ermine]

yeh

[rustic crown]

but the set {g(a) : g in K[x]} is actually a ring

and clearly contains both K and a

so this must be the smallest one

yee. so a^-1 in K[a] directly gives you that

[formal ermine]

wait if we evaluate a polynomial at a, how can it give us a^-1 though?

doesn't a polynomial only have non negative powers?

[chilly ocean]

I didn't say that

[formal ermine]

mb

I misunderstood you then

[rustic crown]

oh oopsie, i didn't read this

[formal ermine]

np

[rustic crown]

K[a] is the set of all polynomials evaluated at a

and a^-1 lies there

so there must be a polynomial whihc evaluates to a^-1

[formal ermine]

yeah

but how

if a is not in K

[rustic crown]

cause you're assuming K[a] is a field

[formal ermine]

we could only possibly have a^0, a^1, a^2, ... in our evaluation

[rustic crown]

if a is there and a is non-zero then a^-1 is also in K[a]

[formal ermine]

yes I understand how we get there logically

but it in itself makes no sense to me

[rustic crown]

oh because a satisfies some polynomial equation

say f(a) = 0

this looks like f(x) = c + x*g(x)

you could assume c is non-zero

as if it was zero then just look at g(x) and g(a) would still be 0

[formal ermine]

lol I'm stupid

[rustic crown]

so a * g(a) = -c

a^-1 = -1/c * g(a)

something

[formal ermine]

we have a polynomial b_na^n + b_n-1a^n-1 + ... + b_0, which can of course evaluate to a^-1

that's the part that I was confused with

[rustic crown]

its basically the argumetn for (2) => (1) lol

you confuse me lol >.<

you're confused with the part which is "of course"?

[elder wave]

this will get you question banned from stackexchange

[rustic crown]

oh you were confused with that?

[formal ermine]

I already deleted my post quickly enough

[rustic crown]

and not are anymore?

[formal ermine]

yes

I thought e.g. b*a could never be a^-1 lol

which is obviously not true

[solar glacier]

leet R be Euclidean domain and let m be the minimum integer in the set of norms of non zero elements of R. prove every nonzero element of R of norm m is a unit.

[chilly ocean]

divide 1 by ur element

see what happens

[cloud walrus]

MyMathYourMath

[solar glacier]

where r has minimum norm

[chilly ocean]

by "divide 1 by ur element" i mean use the euclidean domain assumption

assume r is of norm m and write 1 = qr + remainder

i've already spoiled half the exercise

[solar glacier]

i think i see it now!

and use minimalirt of the norm as N(remainder)<N(r)

[chilly ocean]

something like that

||remainder is either zero or has norm strictly less than r. by minimality the remainder must vanish, so 1 = qr||

[pastel cliff]

dum dum moment

primes in Z are prime in Z[sqrt{d}] too right?

also are units irreducible by defn?

intuitively i say obviously but 1 and -1 in Z?

[south patrol]

not necessarily

[pastel cliff]

thinking about 2 and 3 in particular

[south patrol]

oh well for example in Z[sqrt(3)], 3 is no longer prime

[pastel cliff]

oh ofc

but something like Z[sqrt(-5)] it still is right

just depends on what's being adjoined ig

[south patrol]

well not thought about Z[sqrt(-5)]

but in Z[sqrt(-5)], e.g. 29 = (3 + 2sqrt(-5))(3-2sqrt(-5)) i guess

[south patrol]

like this is false in general and in general it is very non-trivial to decide what stays prime

[pastel cliff]

ok adjacent question

how do i prove that something is prime in Z[sqrt(-5)]

defn is obviously that x is prime if it can only be expressed as a product of a prime and itself

but simply throwing this defn into a proof doesnt feel very valid

[elder wave]

you use a trick with the norm of the elements

[pastel cliff]

OH

fucking uhhhhh

euclidean domains

wait no im thinking of the fact that euclidean domains detect units

and also it's not a euclidean domain lol

for sqrt(-5) at least

[elder wave]

Z[sqrt(-5)] is not an euclidean domain

oh you said that

[pastel cliff]

possibly weird question but then why do we get to use the norm

[shell agate]

If you have a non prime x, then you can write x=(a+b*sqrt(-5))(c+d*sqrt(-5)). If x is an integer, then the only possibility is to have cb+ad=0. So you have x=k(a+b*sqrt(-5))(a-b*sqrt(-5)), for some integer k, and then you can write x=k(a²+5b²).

(a²+5b²) is the norm

[pastel cliff]

i know but i thought that was only "defined" for euclidean domains

or should i really be saying that the norm only has nice properties in euclidean domains

[shell agate]

Some euclidean domains have norm, but not all of them

Here we're just dealing with subgroups of C

[pastel cliff]

when do domains have norm then

ig my questions have now changed to be: when should i think about the norm and what's it's connection with irreducibility in Z[sqrt{d}]

[shell agate]

This is the connection with irreducibility.

And the norm arises from calculation, in fact it's not really relevant.

[pastel cliff]

im asking about all of this bc i want to see if (1 + sqrt(-5)) is irreducible in Z[sqrt(-5)]

[shell agate]

If x=yz, then |x|=|y||z|. In your case we have |x|²=6. So |y|²=2 and |z|²=3. But that's not possible since |y|²=a²+5b², it can't be 2 or 3.

[pastel cliff]

yeah that's actually it lol, i wanted to show distinct factorizations of 6

but x^2 isnt 6 is it

x = (1 + sqrt(-5))

[south patrol]

Norm squared is 6

but not x

[pastel cliff]

does this become true for a prime ideal?

[south patrol]

No

e.g. Z[sqrt(-5)] = Z[x]/(x^2 +5)

[pastel cliff]

:O

[rustic crown]

uwu

[pastel cliff]

[restive birch]

the book is telling me that every group of order 4 is isomorphic to either V_4 or Z_4 (the cyclic group of order 4). how is a group G := {1, x, x^2, y | x^3 = y^2 = 1} isomorphic to either?

[coral spindle]

Was this meant to be a group presentation? The notation you're using isn't clear.

[barren sierra]

?

[restive birch]

yea, sorry

[barren sierra]

How is that not clear notation

Isn't that standard notation?

[coral spindle]

No, I'm not at all familiar with this.

[restive birch]

i just meant a group with two elements of order 3 and one of order 2 (and the identity)

[coral spindle]

OK, but in a group of order 4 there cannot be an element of order 3

whoops, typo

Again!

[restive birch]

tpyo lmao

[coral spindle]

So I think you are mistaken

[restive birch]

oh, duh

im a dumb

because xy, ofc

[barren sierra]

It's a valid question lol

[restive birch]

my bad

[barren sierra]

but yea, order 3 is a no-go

[restive birch]

alright, so that only leaves three elements of order 2 (V_4) or one element of order 4 (Z_4). got it

[chilly ocean]

I am doing this proof: 6. Prove: if |x| = |y| = |xy| = 2 in a group, then xy = yx.

x* any a = either x or x^2. because y is not identity, xy cannot equal x, so xy=x^2. Similarly, yx cannot equal y, so yx=y^2. I'm not really sure what to do now

[void cosmos]

we know that

(xy)^2 = e

and (xy)^2 = xyxy

[south patrol]

showing xy = yx is equivalent to showing that xy(yx)^-1 = 1

[void cosmos]

--> xy=y^-1x^-1

how do weee use |x| = 2 tho

and |y| = 2

to show xy=yx

@chilly ocean

[coral spindle]

If |x| = 2, then as you have seen we have x^2 = e. What does this tell us about x^-1? Try using x^2 = e to get some information here.

[void cosmos]

are u asking me

or him

[pastel cliff]

super late but i forgot to say, this was in a PID!

claim to prove is that in a PID $R$, there are only a finite number of distinct ideals containing a given ideal $I$.

[cloud walrus]

sebbb

[pastel cliff]

wait but then

a PID is noetherian

the number of ideals containing (0) must be finite bc of ACC no?

[next obsidian]

Kekw

No

Consider (0) in k[x] and the ideals (x-a)

[pastel cliff]

non moment

[next obsidian]

Consider what it means for (x) < (y)

And then show that up to unite there’s only finitely many y which work

[pastel cliff]

hm wait i was just thinking that was containment

oh well having a PID means we have a UFD

[pastel cliff]

guessing but is there some divisibility consequence here

or is this a consequence of correspondence theorem using R/I

[thorn delta]

(x) < (y) means x = ry for some r in R

[next obsidian]

It’s a trivial consequence of the definitions as kxrider pointed out

[chilly ocean]

I don't know what it tells us about x^-1

Like legit do not

[pastel cliff]

[chilly ocean]

x. Inverse x = e

So x *x^-1 = x^2

So x = x inverse?

[pastel cliff]

yes x is it's own inverse when it has order 2

[chilly ocean]

Yeah stillnnit sure where this leads us

[pastel cliff]

repost your question

[pastel cliff]

though im not sure how this leads to finiteness?

ohhhh wait

it's going down not up

(x) ⊇ (y) (for example (3) ⊇ (6)) means that x | y so there exists an r such that xr = y, and this is just the process of finding and irreducible factorization

like this process must end because we arrive at such a factorization of irreducibles and thus the ideals containing it must also end

such a factorization exists bc a PID is a UFD

i think i see it

[chilly ocean]

I am doing this proof: 6. Prove: if |x| = |y| = |xy| = 2 in a group, then xy = yx.

x* any a = either x or x^2. because y is not identity, xy cannot equal x, so xy=x^2. Similarly, yx cannot equal y, so yx=y^2. I'm not really sure what to do now

[pastel cliff]

since xy has order 2 xy = (xy)^-1

(xy)^-1 = y^-1 x^-1

so xy = y^-1 x^-1

what can you do from there

[pastel cliff]

imma hope this was right tho

[next obsidian]

you don't ever have to deal with chains

[chilly ocean]

Oh

[next obsidian]

Noetherianness doesn't tell you anything about how many ideals lie over any given ideal because this is a question which specificially isn't about chains

you have to say something about how many ideals lie over I, but they don't have to be contained in each other in any fashion

[chilly ocean]

XY=y-1 x-1=yx

[next obsidian]

Also, the problem is false when I = (0)

[chilly ocean]

Thank you

[next obsidian]

so you ahve to assume non-zero

[pastel cliff]

yeah i am

wdym "lie over" though

[next obsidian]

J> I

[pastel cliff]

what i was saying is basically that an element "contains" all its factors, and this kinda translates to the ideal that element generates containing all its ideals generated by its factors

no that's backwards

*is contained in

this definitely isn't enough tho

[restive birch]

how often are lattices used?

just wondering because i think theyre fun to draw and hope theyre not just never used

[pastel cliff]

you mean like maps of subgroups/rings?

[restive birch]

yea

[pastel cliff]

correspondence theorem comes up often and is importnat

also called the lattice isomorphism theorem

[restive birch]

havent come across it yet, but glad to know they have a utility

they just look so cool

does this seem accurate? group definition is at the top

oh, and remind me- if one group is abelian, and another isnt, they cant be isomorphic, right?

[pastel cliff]

what do you think

[chilly ocean]

yes

[ruby sundial]

So there are no surjections between abelian to nonabelian groups. Is there a stricter conditioning on homomorphisms we can get?

[restive birch]

right, because if G is abelian and H isnt, then phi(ab) = phi(a)phi(b) and phi(ba) should equal phi(ab) but phi(a)phi(b) doesnt equal phi(b)phi(a) for all a,b

[chilly ocean]

not "for all a, b", but rather some

"H is not abelian" does not mean no pairs of elements commute, but rather that there is one pair of elements which doesn't

[restive birch]

yea i meant that it doesnt hold for all a,b

i phrased it wrong

[chilly ocean]

if G and H are groups and f: G -> H is a surjective homomorphism, then H is abelian if G is: any two elements of H are of the form f(g), f(g'), and f(g)f(g') = f(gg') = f(g'g) = f(g')f(g)

high-brow version: H is a quotient of G and G is abelian so H is

[restive birch]

havent learned quotients yet, but the first proof makes sense

[restive birch]

im struggling to classify what the subgroup <sr, r^2> of D_16 is

[ruby sundial]

cool!

[restive birch]

like, what its "isomorphism type" is

[ruby sundial]

well lets think about it?

[wooden ember]

It’s true for a PID but not a noetherian ring

[restive birch]

oh is it just D_8 again

hang on let me check

[ruby sundial]

🙂

[restive birch]

yea, i think it is, with sr -> s and r^2 -> r

[ruby sundial]

!

[formal ermine]

prove that that is well defined and an isomorphism

[coral shale]

galois my beloved

[crimson ibex]

Hello, at an algebra class the professor said that given <A, +> a semigroup, the following is equivalent:
1)<A,+> is a group
2)A admits right neutral element and right inverse.

Question is: from the theory I know that a group requires the existence of both the neutral element and the inverse (with respect to an operation), why is requiring the existence of only the right neutral element and right inverse sufficient for <A, +> to be a group?

[autumn kite]

Semigroups can have left identities/inverses but not right identites/inverses

[crimson ibex]

yes i know but why is 1) and 2) equivalent then?

thats the question

i think somehow the fact that it has both right neutral element and right inverse (taken together) implies that it also have left neutral element and left inverse (I do not know why that is the case tho)

[autumn kite]

Actually i see what ur saying

[rotund aurora]

Is order theory a part of algebra?

[crimson ibex]

i would say so

[rotund aurora]

its weird, the book Im reading started a brief digression in order theory before the main Galois theory. I already know what the main theorems say, so I know why he is doing it. But I was wondering if things like Galois connections show up in algebra (excluding order theory) somewhere else

I suppose they do, because Galois connections seem natural, but idk if that abstraction is worth it just to talk about Galois theory, specially since I assume that it is in field theory the first time students usually see this kind of order properties (being relevant)

[untold basin]

Hello, I have a zeta in the algebraic closure of Q which is a root of X⁶+X³+1. I am asked to find all the embeddings from Q(zeta) to the algebraic closure of Q.
Must I find explicitly the minimal polynomial of zeta on Q or not ? It seems that I have to calculate a lot so I wonder if I can do something else

[rotund aurora]

zeta needs to be sent to the roots of x^6+x^3+1

So I think you just have to find the roots of the polynomial

Notice that this is a quadratic in disguise

and then you just have to extract a cube root

[formal ermine]

I want to solve $$\begin{cases} x \equiv 1 \pmod{2} \ x \equiv 2 \pmod{3} \ x \equiv 3 \pmod{5} \end{cases}$$ via the chinese remainder theorem

so I have $\bZ/2\bZ \times \bZ/3\bZ \times \bZ/5\bZ \cong \bZ/30\bZ$

but I have no actually clue of how to do it

[untold basin]

I did all of that but must I do it ? It seems to be a rough way of solving the exercice

[woven obsidian]

The polynomial is irreducible so it must be the minimal polynomial

So you have six embeddings, sending zeta to any of the six roots of x^6+x^3+1

[untold basin]

How do you see that ?

I have that intuition for P of degree 2 and 3

[rotund aurora]

uhm I thought you wanted the explicit embeddings

like its not tedious anyway, its high school algebra

[woven obsidian]

Two ways to show it here: https://math.stackexchange.com/questions/3095081/the-polynomial-x6x31-is-irreducible-over-mathbbqx

[untold basin]

In my course we say like the embedding
2 |-> 2

[woven obsidian]

Still you need to show this is the minimal polynomial

[rotund aurora]

no

uhm yeah

[untold basin]

Because we say that our morphism determines all the element

[rotund aurora]

but well

if you find the roots I think you would see it? Maybe not... you are right

[woven obsidian]

Zeta must be sent to a root of its minimal polynomial, not just roots of any polynomial

[untold basin]

Mmh can you say why ?

[woven obsidian]

Embeddings respect multiplication and addition

[rotund aurora]

try figure it out

[untold basin]

Ok I will see my course for that

[chilly ocean]

Is there a simple example of why the tensor-hom adjunction equation fails when you reverse the arguments?

[woven obsidian]

And the fact that sending an element to a root of its minimal polynomial gives an embedding is a basic result from field theory

In my book it is referred to as Kronecker's theorem

It's essentially just the statement that if $f$ is the minimal polynomial of $\alpha$, then $K(\alpha) = K[X]/(f)$

[lost drum]

hi, does anyone understand herman-mauguin notation for crystallographic point groups?

sorry if this is the wrong channel

im specifically confused about why the achiral tetrahedral group is 4-bar 3m, but includes 180-degree rotations

[untold basin]

I understand that now but I still need to find the minimal polynomial

[woven obsidian]

The link shows why x^6+x^3+1 is irreducible

And an irreducible polynomial is the minimal polynomial of all its roots

Here's an alternative solution: First zeta is a root of x^6+x^3+1. Consider then Q(zeta). In this field zeta^2 is an element of degree 3 over Q, while zeta^3 is an element of degree 2 over Q. As 2 and 3 are distinct primes it follows that $[Q(zeta^2,zeta^3):Q] \geq 6$ and thus as we already know $[Q(zeta):Q]\leq 6$ we must have equality. Thus the minimal polynomial of zeta has degree 6, and it follows that x^6+x^3+1 is the minimal polynomial (and is also irreducible).

[untold basin]

I don't want to spoil me so I won't read it fully now
Thanks a lot though

[woven obsidian]

Np, it all amounts to finding your favourite arguments for showing irreducibility

[cloud walrus]

i believe in mathemagic
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

i believe in mathemagic

i believe in mathemagic

[chilly ocean]

I saw this statement

Does this mean Z4 is isomorphic to Z2 + Z2?

That case confuses me

[woven obsidian]

No

Z4 is cyclic, Z2 x Z2 isn't

2 is not coprime to 2

[chilly ocean]

Oh yeh

[formal ermine]

let K be a field and K(X) the field of rational functions with coefficients in K. let f = X^3/(X^2+1) in K(X). show that X is algebraic over K(f).
the field extension we're looking at is K(X)/K?

and the intermediate field is K(f)?

[rotund aurora]

You are looking at K(f)<K(x)

and you want to show that there is a polynomial p with coefficients in K(f) such that p(x)=0

[formal ermine]

<?

[rotund aurora]

K(x)/K(f)

[solar glacier]

question

if f:Q^x -> Q^x is the function that swaps powers of 2 and 3

in showing its a homomorphism does it suffice to let x,y \in Q^x

then if they are both coprime to 2 and 3 then its the identity map which is trivially a homomorphism

if not let $x=2^{a_1}3^{b_2}m$ and $y=2^{a_2}3^{b_2}n$

where $n,m$ are coprime to 2 and 3

then just compute f(xy)

[lethal dune]

say I have a representation with character χ. Ker χ= {g ∈ G: χ (g) = χ (1)} then is it true that it ρ(g) i.e. the matrix is identity for all g ∈ kerχ?

[coral spindle]

I assume we're working over an algebraically closed field of characteristic zero. Our representation is V. Suppose g is in Ker chi, and consider rho(g). We know this matrix is diagonalisable so we can assume that it is diagonal by change of basis. This diagonal matrix now consists of roots of unity, which are in particular complex numbers z with |z| = 1. We assume that tr(rho(g)) = chi(g) = chi(1) = dim V, so if the entries of rho(g) are lambda_1, ..., lambda_n, we can ignore the complex parts since the sum is real. Hence dim V = sum Re(lambda_i). However Re(lambda_i) <= 1, and there are exactly dim V entries in the sum, so the only possibility is that lambda_i = 1 for each one.

In short: yes.

[lethal dune]

Cool

That’s the proof i had in mind but wasn’t sure if it’s even true

[coral spindle]

It's true.

[lethal dune]

Thanks for confirming

Also since we are here, can you explain why central elements act on irreducible representations by a scalar?

[coral spindle]

Schur's Lemma.

[lethal dune]

It just states it’s Schur’s lemma but that’s not the one im familiar with

[lethal dune]

Lol

[coral spindle]

The action of a central element is, in particular, an equivariant map. Hence it acts by a scalar.

[lethal dune]

Which Schur’s lemma is this

[coral spindle]

An equivariant map between two irreducible representations is zero if they are non-isomorphic, and scalar multiplication otherwise.

[lethal dune]

So b is just schur’s lemma then?

[coral spindle]

I do not see that myself.

[pastel cliff]

how can i show that least common multiple always exist for a pair of elements in a UFD.

[lethal dune]

Ok G is not commutative, doesn’t work

[lethal dune]

Factorize

[coral spindle]

If I gave you a prime factorisation of two integers, would you be able to give me their lcm?

[pastel cliff]

ok then formally

[lethal dune]

Ok then how do I go about b

[pastel cliff]

take some element we know that there is a factorization into a finite number of irreducibles

sorry ryu

[coral spindle]

I think (b) is simply false. If you take the usual faithful irreducible 2-dimensional representation of Dih(8), for example, not every matrix acts by scalar multiplications.

The inclusion is an isomorphism, of course, if and only if Z(rho) = G.

[pastel cliff]

boytjie are you a teacher?

just curious^

[coral spindle]

No, but I do some TA work nowadays

I'm just a postgrad

[lethal dune]

Lol ok my instructor really needs to check the questions before putting them in the assignments

[coral spindle]

Frustrating!

[lethal dune]

Not the first time

[pastel cliff]

ok a bit of a silly question but

[lethal dune]

Ok I shouldn’t be criticising him publically but even he’s not able to solve most of the questions

[coral spindle]

😬 that's not great

[lethal dune]

He puts them in the assignment because he’s just found them “interesting”

[pastel cliff]

how do you know that for r = uk1...kn, where ki is irreducible, some ki is greater than kj

[coral spindle]

What do you mean by greater than in this case?

[pastel cliff]

mandelbrot conjecture is interesting he should put that on

[pastel cliff]

that's exactly my point lol

for any two elements, they either have an irreducible in common or not

if they dont, lcm is a unit. if they do it's the smallest k found in both factorizations

[lethal dune]

He gave the entire structure theorem of Dedekind domains in the hw

And never actually defined it in the class

Dogshit assign

[pastel cliff]

my concern is how we define "smallest" among things that cant be factored further if we dont necessarily have an ordering

that's a poopy butthole prof tho

[coral spindle]

Firstly, I think you're thinking of the gcd rather than the lcm. Secondly, you're not quite right there. Thirdly, the ordering is by containment of principal ideals: We will write a <= b if (a) is contained in (b). This is not a partial order but it is a preorder.

[pastel cliff]

.

[coral spindle]

Think about divisibility, therefore, rather than size.

[pastel cliff]

[coral spindle]

OK

[pastel cliff]

ye i'll brb, ignore the fact that i messed up lcm and gcd

[coral spindle]

The lcm of a and b is the unique (up to multiplication by a unit) element L such that if K is such that a | K and b | K, then L | K too.

So the "greatest" is by divisibility. Hope that's clear.

[coral spindle]

It's easy to do.

The three-letter acronyms don't help.

[formal ermine]

does the chinese remainder theorem only tell us when stuff is solvable or also how it's solved? I only think the former but I'm not sure

[pastel cliff]

PID

UFD

GCD

CRT

[elder wave]

[coral spindle]

It doesn't give us a way of calulating the isomorphism in all forms that I've seen it in.

[formal ermine]

okk

[coral spindle]

It guarantees an isomorphism though

[formal ermine]

so if I have x = 1 mod 2, x = 2 mod 3, x = 3 mod 5 (= should be equiv), we know that there must exist a solution due to 2, 3, and 5 all being coprime to each other?

[coral spindle]

Yes. In practice it's not hard to find of course. Try coming up with an algorithm to find a solution to the two-equation case.

[formal ermine]

yeah yeah I can do it by hand easily

but I didn't understand the crt properly, now I (hopefully) do. thanks!

[coral spindle]

No worries

As a tidbit

Are you familiar with Euler's totient function, and its multiplicative property?

[formal ermine]

yes

it's used in the proof to euler's theorem in group theory or w/e it's called, no?

the totient function

[coral spindle]

Well recall that phi(n) = |(Z/nZ)*|, the order of the group of units of Z/nZ.

[pastel cliff]

totient looks like a fake word

[coral spindle]

Now the chinese remainder theorem in one form states that if n and m are coprime, then Z/nZ x Z/mZ is isomorphic to Z/nmZ

The group of units of Z/nZ x Z/mZ is just the product of the group of units of its parts.

Therefore phi(nm) = phi(n) x phi(m).

[formal ermine]

only if n and m are coprime

that's big brain

[coral spindle]

I like this proof a lot. It is one of my favourite proofs.

[formal ermine]

yeah

it's smooth

can we "combine" the equivalent relation things?

like x = 1 mod 2, x = 2 mod 3, x = 3 mod 5

into x = ? mod 30

or is that already the solution lmao

[coral spindle]

That would be finding a representative, yes

[rotund aurora]

Where does one usually learn about inverse limits of groups?

I mean, would you just learn that while studying category theory?

[coral spindle]

You see them when studying profinite groups

[formal ermine]

if $K/\bR$ is a finite field extension then $K$ can only be $\bC$, no? or am I having a wrong idea

[cloud walrus]

i believe in mathemagic

[rotund aurora]

By the fundamental theorem of algebra, C is algebraically closed

[formal ermine]

yea

so it has no non trivial finite field extension

[coral spindle]

I think you are right, as the inclusion of any additional (necessarily complex) element would mean the whole of C is in the extension.

Furthermore any finite extension of C is C, so that finishes things.

[rotund aurora]

What do you mean? Because I have seen the complete opposite. In elementary NT, it is in fact given as an algorithm. Maybe Im missing some context tho

[coral spindle]

I have seen it non-constructively.

[rotund aurora]

thats not the common thing I believe

[coral spindle]

I only reported on my experience

[rotund aurora]

I just asked

[formal ermine]

this is because [C : R] = 2 which is prime, right?

so it has no trivial intermediate fields

[coral spindle]

That's a nice way of showing that, yeah

[formal ermine]

do you think this is formal enough?

"Due to [C : R] = 2 (which is prime and therefore C/R does not have non trivial intermediate field) and C being algebraically closed (fundamental theorem of algebra) K must be C."

as in formal enough for my homework

[coral spindle]

Why could there not be a different field extension of degree 2? You have argued that there is no field between C and R, but why is there not simply a field that is not between the two?

I think you need to explain why algebraic closure shows this in more detail.

[formal ermine]

hmm lemme think

[coral spindle]

Hint for how to begin: suppose K/R is a finite extension, and let a in K. Considering the minimal polynomial of a...

[formal ermine]

K/R is a fine extension by assumption

therefore it's algebraic

[coral spindle]

And what does this tell you about its relation to C

[formal ermine]

this sounds ultra mega wrong but a thought I just had...
2 = [C : R] = [C : K] * [K : R] = [C : K] * [R(a) : R] = [C : K] * 2
=> [C : K] = 1 => C = K

because any a in K will be algebraic over R so R[a] will be a field(?)

[coral spindle]

2 = [C : R] = [C : K]
You hav assumed that K is in C. Why?

[formal ermine]

because if R subseteq K and R subseteq C and because C is algebraically closed it must follow that K subseteq C

[coral spindle]

That's only true for algebraic extensions K of R, yes. This is all there is to it.

You just need to argue that any finite extension of R lies between R and C.

[formal ermine]

yeah and the only ones for that are R and C

[coral spindle]

Yes

[formal ermine]

okay wait I lost track myself

let me reread

K/R is a finite extension (by assumption) therefore it's algebraic.

assuming (for contradiction) C is a real subset of K (i.e. K lies outside of C).
because R subseteq C, K/C must be algebraic. this contradicts the algebraic closure of C.
therefore K must lie between R and C.

because [C : R] = 2, which is a prime, it has no non-trivial intermediate fields, meaning that K must either be C or R. but because K/R is finitely generated, K cannot be R, therefore it must be C.

does that look gucci?

[rustic crown]

nope >.<

[formal ermine]

what's wrong

[rustic crown]

the claim that either C lies in K or K lies in C

what if both are false?

also R/R is finitely generated as well

[rotund aurora]

[formal ermine]

oh

I thought our generating system couldn't be the empty set

[rustic crown]

it is {1}

[formal ermine]

oh....

[formal ermine]

wait what

I have to show that if K/R is a finite field extension then [K : R] is even

but R/R contradicts that...?

[rotund aurora]

Idk what you were asked to do, but just suppose K is not R, otherwise everything is trivial lol

[formal ermine]

oh

I confused finitely generated and finite

[rotund aurora]

mmh

how

R is both finitely generated and finite over R

[formal ermine]

yeah but I confused it nonetheless

but either way

my exercise says:

"Let K/R be a finite field extension. Show that [K : R] is even."

but [R : R] is a counter example, no?

[rotund aurora]

yes

Like I would consider R/R a finite field extension, it has degree 1

[formal ermine]

yeah

[rotund aurora]

but whatever, no one cares

cuz its trivial

tbh that question is weird

[formal ermine]

I'll just email my ta what he actually means in that exercise

like if he wants us to show that counter example or...

[rotund aurora]

no like the exercise is straightforward, but it seems weird to ask that when you can just ask "show that K/R finite implies K=R or K=C"

[formal ermine]

yeah...

[crimson ibex]

Hello sorry for the repost but it didnt get solved earlier:

At an algebra class the professor said that given <A, +> a semigroup, the following is equivalent:
1)<A,+> is a group
2)A admits right neutral element and right inverse.

Question is: from the theory I know that a group requires the existence of both the neutral element and the inverse (with respect to an operation), why is requiring the existence of only the right neutral element and right inverse sufficient for <A, +> to be a group?

[glossy crag]

This is a standard exercise in group theory.

Show that 1 is the only idempotent in A

i.e. x^2=x => x=1

Then if y is the right inverse of x, i.e. xy=1, show yx is an idempotent.

[crimson ibex]

i dont get it tho

my course is moslty logic and i've barely done any group theory

[glossy crag]

You know what a group, semigroup, right/left identity, right/left inverses are I assume?

[crimson ibex]

ye

[glossy crag]

OK, assume you have a semigroup with right identity and right inverses, to show it's a group you need to show the right identity is also a left identity, same for the inverses.

[crimson ibex]

ye im ok up to here

[glossy crag]

One considers the idempotents of A, elements x of A with the property x^2=x.

Let e be the right identity of A, we show the only idempotent is e.

[crimson ibex]

ok up to now i got it

[glossy crag]

Suppose x^2=x. x is right invertible, so there's a y s.t. xy=e. If you multiply both sides of x^2=x with y from the right you get xxy=xy => xe=e => x=e, since xe=e.

Now I want to show that not only is y a right inverse, it's also a left inverse (i.e. that yx=e). So far all I know is that all idempotents are equal to e, so I show yx is an idempotent.

[crimson ibex]

am i multiplying my x^2 or by x?

[glossy crag]

(yx)^2=(yx)(yx)=y(xy)x=(ye)x=yx => yx is an idempotent => yx=e.

[glossy crag]

you have an equation x^2=x

you multiply both sides by y from the right

[crimson ibex]

ok

[glossy crag]

As a trivial example from arithmetic, if x+2=3, then you can substract 2 from both sides to get x=1, same idea here.

[glossy crag]

I showed that xy=yx=e, now it remains to show ex=x.

ex=(xy)x=x(yx)=xe=x (the last equality because e is a right identity by assumption)

I showed e is an identity and all elements have inverses, we're done.

[crimson ibex]

im medidating on it give me a bit

xd

[crimson ibex]

i think theres an issue here

because we are considering x^2 = x so x=e

issue is that the idempotent does not have an inverse in many cases

[glossy crag]

Yes, there are no idempotents (except the identity) in groups precisely because of that.

We use that fact to show that xy=1 implies yx=1.

There's no issue.

[crimson ibex]

but x here is e

[glossy crag]

IDK what you mean.

you seem to be confused by the two uses of x

[crimson ibex]

you are multiplying both sides by y

[glossy crag]

in x^2=x and xy=e

[crimson ibex]

but in this equation clearly x = e

so x is the identity

and you are trying to show that x * y = e

[glossy crag]

You have it backwards

xy=e

I want to show that if x^2=x, then x=e

I can use different letters if that'd confuse you less, but you need to get used to symbol reusal in algebra.

[crimson ibex]

haha i understand that part now thanks, give me one more moment to think about the rest

[formal ermine]

@rotund aurora my ta replied

K is not R

i.e. a real extension field

[crimson ibex]

@glossy crag ok i fully got what you did now, thanks a lot for the help

[glossy crag]

You're welcome.

[solar glacier]

Can I ask a question

[chilly ocean]

you're trying to ask a question? Unbelievable

here out of all places

channel designed for it!

[solar glacier]

Ok

[chilly ocean]

of course you can ask a question

[solar glacier]

if f:Q^x -> Q^x is the function that interchanges powers of 2 and 3

and the identity if theres no factors of 2 or 3

so showing this is homomorphism

cani say if the numbers x,y \in Q^x have no powers of 2 or 3 its trivial, if not then

let x=2^a_13^b_1m, y = 2^a_23^b_2n

and compute f(xy)

[chilly ocean]

You mean that if q = 2^n 3^m y where y is rational co-prime to 2, 3 then f(q) = 3^n 2^m ?

[solar glacier]

yes

[chilly ocean]

I'm not sure if you can say coprime in this context but whatever, unique factorization still applies

[solar glacier]

cause i showed assuming what I said up top

[chilly ocean]

I guess, q can be negative

[solar glacier]

that f(xy)=f(x)f(y)

[chilly ocean]

so you also demanding f(-...) = -f(...) right

[solar glacier]

why that

[chilly ocean]

whats f(-1)

[solar glacier]

-1

[chilly ocean]

eh

[solar glacier]

its the identity if the number contains no factors of 2 or 3

[chilly ocean]

-1 contains no factors of 2 or 3

[solar glacier]

so its just -1

[chilly ocean]

that's why I asked, what do you do with minus signs

from the way you're saying it, it could have been just 1

calculate f(xy), f(x)f(y), see that they're equal, qed

[solar glacier]

if

i let x=2^a_13^b_1m, y = 2^a_23^b_2n

i have another question

[delicate bloom]

you don't need to ask for permission to ask

[solar glacier]

how do I show such a function is 1-1 and onto

since i have

x=2^a_13^b_1m, y = 2^a_23^b_2n

[barren sierra]

what are m and n

[solar glacier]

rational numbers coprime to 2 and 3

[ripe basalt]

if im computing the degree of the splitting field of x^3 - 2 over Q, i think the answer is 6. Is this just because x^3 - 2 has a root exp(2pi * i / 3) * \sqrt3 and Q(exp(2pi * i / 3) * \sqrt3 ) must be degree 6 over Q I think? I thiink by the tower theorem this means that Q(exp(2pi * i / 3)) is a degree 2 extension of Q, is that true?

nvm i got it

\sqrt{3}i has min polynomial x^2 + 3

so it is a degree 2 extesion

[lunar ledge]

The way you usually would? What's the closed form expression

[solar glacier]

but I have

x,y \in Q^x

if they have a 2 or 3 then

x=2^a_13^b_1m, y = 2^a_23^b_2n, n and m rationals coprime to 2 and 3

how do I suppose x\neq y

[pastel cliff]

how might i approach this

not even sure what vanish formally refers to

[chilly ocean]

"vanishes at a point" means evaluates to zero at that point

"does not vanish at any point" therefore means is never zero

[pastel cliff]

then any hints on starting this?

[chilly ocean]

prove that an element of R which does not vanish anywhere is a unit

and prove that a unit of R does not vanish anywhere

probably unhelpful hint but i really don't know what else to say without literally spoiling it. just work through the definitions

[pastel cliff]

will try, ty for not spoiling

[foggy merlin]

The 0 ideal over a ring R as an R module is considered to be free right?

[chilly ocean]

Yes, the empty set is a basis

[foggy merlin]

Thank you. Another chapter in the adventures of empty sets

[chilly ocean]

empty sets, trivial yet confusing

[chilly ocean]

trivially confusing

[median pawn]

What's GF(2)?

probably the finite field with two elements, because that's what makes sense

but I haven't seen the notation GF(n) before

[formal ermine]

yes

GF stands for general field I believe

just like GL

[chilly ocean]

Galois Field

[median pawn]

GL*?

yeah

[chilly ocean]

here's a meme so you never forget what GF means

[south patrol]

Aw

In fact that implies is an iff innit

[chilly ocean]

funny

[median pawn]

unique upto isomorphism lmao

[glossy crag]

This is probably the only proper "math" meme that I've seen that's funny.

I've got another good one, but idk if we're allowed to post analytic memes here.

[chilly ocean]

no memes

[glossy crag]

But he got to post one

[chilly ocean]

memes for sake of memery... there's a channel for that

and respect someone's pronouns

that's what they are for

[glossy crag]

Didn't know she's a she, lol

[south patrol]

lol tfw pronoun roles

[glossy crag]

Not used to that, didn't think to check

[formal ermine]

just... use they then?

[shell agate]

There's no need to specify n=1, (x+y)^p always equal x^p+y^p in characteristic p.

when n=1 you get x^p=x though

[chilly ocean]

ah, yeah, true. I didn't even notice the asterisk

[south patrol]

me neither lol

[shell agate]

I'm a professional meme proofreader

[chilly ocean]

hmm i didn't notice it either lol wth

[crimson ibex]

@glossy crag sorry if I I bring this up again, I was thinking about the proof you gave me yesterday about right hand inverse etc. Why di we need to check that the identity is single? Isn’t that always the case?

[glossy crag]

Wdym "check the identity is single (and ready to mingle)"?

[chilly ocean]

right identity doesn't imply uniqueness

in a semigroup you can have multiple right identities

[crimson ibex]

Ah I see

[chilly ocean]

it's only when you have right and left identites it turns out that there's a single "identity"

[crimson ibex]

Thanks

[glossy crag]

We at no point checked that the identity is unique, we showed that the right identity is also a left identity (this implies uniqueness automatically, but I never mentioned this explicitly).

[chilly ocean]

this follows from if e is a right identity and f is a left identity, then f = fe = e

[crimson ibex]

Why are we able to set the equation x^2 = x @glossy crag ?

When we know that x has a right identity only

[chilly ocean]

is?

you mean x is a right identity?

all right/left identities are idempotent, it follows from their definition

[crimson ibex]

@chilly ocean I was Talking about a previous question, you can see it here

[chilly ocean]

this is a famous exercise

same holds for left identity and left inverses (with respect to that left identity)

[crimson ibex]

But why do we begin by saying x^2 = x?

I don’t get why we can do that

[chilly ocean]

this doesn't hold for left identity and right inverses with respect to that identity though

and neither does it hold for right identity and left inverses with respect to it

If x is any element, there is y such that xy = 1, so if x^2 = x then x1 = 1 that is x1 = x = 1

not sure what's the argument here

[crimson ibex]

Why x^2 =x

[chilly ocean]

we say x is idempotent if x^2 = x

[crimson ibex]

But why are we saying that

[chilly ocean]

if you suspect something is a group then it's often useful to check for elements which are idempotents first to see if this set is unique, and this will be your candidate for the identity element

[crimson ibex]

The question never said anything about idempotence

[chilly ocean]

¯_(ツ)_/¯

this exercise should just follow from some algebraic manipulations

[crimson ibex]

It wasn’t an exercise

I just took a theorem and derived this from it

[chilly ocean]

[crimson ibex]

It’s just pure coincidence that this is also a famous exercise

[chilly ocean]

if you derived it from it, then that means you already proven it

[crimson ibex]

So I’m not sure why I’m starting the proof by using that x is an idempotence

[crimson ibex]

Yes but I was trying to see why the theorem was true

[chilly ocean]

idempotent*

not to be confused with impotence

[crimson ibex]

Anyways why are we saying that x is idempotent?

[chilly ocean]

ask Ocean Man

[crimson ibex]

@glossy crag

[glossy crag]

What do you mean "why are we saying"? Do you mean why we settle on this particular proof strategy? As Blitz said above, the identity is the only idempotent in a group, so if we suspect something might be a group, we should probably take a look at the idempotents to ensure everything is in order.

[crimson ibex]

No I mean is

What if x is not an idempotent

How are we sure that there is an idempotent

Is there always an idempotent?

[crimson ibex]

also why are we trying to show that x= e?

[glossy crag]

But why would we care about that?

Our goal is to show that there are NO idempotents besides e.

So we assume x is an idempotent, i.e. x^2=x, and we derive x=e from that.

[glossy crag]

e itself is an idempotent and in the beginning, we don't know if there are any others, so we check.

Turns out there aren't any (besides e).

This implies that if xy=e, then yx is also equal to e, because yx is an idempotent (if xy=e).

[crimson ibex]

Ok thanks

Got it

[formal ermine]

I'm still having trouble showing that $\bC$ is the only real finite extension of $\bR$. i.e. if $K/\bR$ is a real finite extension then $K \cong \bC$.

we know that $[\bC : \bR] = 2$, so $K$ cannot lie between $\bC$ and $\bR$. $K$ also cannot lie over $\bC$ because $\bC$ is algebraically closed. but what about the case when $K \cap (\bC \setminus \bR) = \nullset$?

[cloud walrus]

i believe in mathemagic

[rustic crown]

by "real" finite extension, do you mean non-trivial?

[formal ermine]

$\bR \subsetneq K$

[cloud walrus]

i believe in mathemagic

[rustic crown]

right, so the hint is to look at K(i)

[formal ermine]

but C/K isn't a field extension in the last case

[rustic crown]

ah oopsie, i meant look at K(sqrt(-1))

i.e. the splitting field of x^2+1 over K

basically force it to contain C

then argue like you're doing

[formal ermine]

we haven't done splitting fields yet

[rustic crown]

okie so just avoid saying that

consider the polynomial f = x^2+1 in K[x]

i think you've already done the part where f has a root

because in that case K is also an extension of C

so assume f is irreducible, and in this case you can consider the extension L = K[x]/(f) which is finite over R and so reduces back to the first case

in the second case L = C, so K = R

as [L:K] = 2

uwu?

ig i can also say it another way

which might be a little more direct

if K/R is a non-trivial finite extension, then consider an element a in K which is not in R

what can you say about the minimal polynomial of a?

[formal ermine]

[R[a] : R] = degree of the minimal polynomial?

[rustic crown]

right

[formal ermine]

or do you mean R[x]/(minimal polynomial) iso R[a]

[rustic crown]

can you show that degree of the minimal polynomial has to be 2?

[formal ermine]

I'd only know a way if a is algebraic

[rustic crown]

right, for this you would need to compare R to C

[formal ermine]

wdym

[rustic crown]

i'm trying to somehow consider the extension C(a)/C and conclude that a lies in C

[formal ermine]

isn't that because C is algebraically closed

[rustic crown]

right, the only problem here lies in that fact that you don't really have a god given big field which contains both a and C

which is not really a problem if you know that splitting fields are a thing :p

normally you define C(a) as the smallest subfield of ?? containing both a and C

but right now i wanted to try to somehow get around this fact >.< that you can find such a ??

anyway, did you like the first argument i wrote above?

lemme try to rephrase the second idea

okie i'll just say it this way

[rustic crown]

continuing from here

the minimal polynomial is a polynomial in R[x]

and any real polynomial can be factored over C

now since this was a polynomial over R, you can pair up the factors (x - alpha) with (x - alpha conjugate)

so this will show that only irreducible polynomials over R are linear polynomials and irreducible quadratics

as the product (x - alpha) * (x - alpha conjugate) lies in R[x]

since we chose the element outside of R, its minimal polynomial can't be linear, and hence is an irreducible quadratic

but then R[a] is isomorphic to C

so you're done

[pastel cliff]

gonna interject but im a little stuck finishing that - i dont think it's enough to say that just bc a function doesnt vanish it must be invertible - how does an arbitrary function in this ring get inverted then?

like is it enough to say that just bc f is never equal to 0 it has an inverse? i feel like not

[formal ermine]

pair up?

what's alpha?

[rustic crown]

like factorize the minimal polynomial say p(x) over C, then for each non-real factor that looks like (x - alpha) you must have another factor (x - alpha conjugate) because conjugating p(alpha) = 0 gives p(alpha conjugate) = 0

now (x - alpha) * (x - alpha conjugate) is a polynomial in R[x]

as both (alpha + alpha conjugate) and alpha * (alpha conjugate) are real

so this gives you a non-trivial quadratic factor of p

which must be all of p

as p was irreducible

in general if you have any polynomial over R[x], you can factorize it as a product of linear and quadratic polynomials

just factor over C and combine conjugate pairs

[rustic crown]

are these continuous "real-valued" functions?

[pastel cliff]

this is all i get

so i'll assume so

[rustic crown]

no i mean the target of these functions is important

at least they should be valued in some field

[pastel cliff]

i know but this is all the question states

[rustic crown]

so that non-zero is equivalent to invertible

at each point

[pastel cliff]

so im assuming the necessary things are assumed

[rustic crown]

okie then assume real valued, they ideally shouldn't mean anything else :p

basically you wanna define the inverse at each point, and check that this inverse is still continuous

and that's standard analysis

define g(x) = 1/f(x)

since f never vanishes, this is continuous

[pastel cliff]

oh ig it is simple then

kek

i hate vague questions

does that have anything to do w the second part too

showing that f(x) = x doesnt have an irreducible factorization

that's such a random question

like

[pastel cliff]

for this same reason\

[rustic crown]

ah so do write f(x) =x

which means the target must be reals :p

[formal ermine]

what's a trivial factor

[rustic crown]

1

[formal ermine]

why

[rustic crown]

because R[x]/(irred quadratic) = C

if alpha is a root of the quadratic, then define R[x] --> C by sending x to alpha

the kernel is exactly that ideal generated by the quadratic

[formal ermine]

how does this help us show the original thing

[rustic crown]

because this shows that R(a) = C so the original K/R contains a copy of C

[formal ermine]

"contains"?

[rustic crown]

yea, it contains R(a)

[pastel cliff]

is eisenstein's criterion also able to show that something is not irreducible

[formal ermine]

no like what does it mean for K/R to contain a copy of C

[woven obsidian]

no

[pastel cliff]

damn

[rustic crown]

ah just K contains a copy of C

[formal ermine]

isn't K = C then

[rustic crown]

yep and that's the argument >.<

[formal ermine]

ok so:

[rustic crown]

i think just go with the first argument, i kinda made it super weird with the second one

it's basically the same

just more work

[formal ermine]

I barely understood anything in the first argument

[rustic crown]

okie i'll try to explain that again

[formal ermine]

thanks

[rustic crown]

so say K is a finite extension of R

we want to show that K is either R, or isomorphic to C

[formal ermine]

K can't be R

by assumption

[rustic crown]

why

just saying finite doesn't remove K = R

you have to say something like "proper extension" otherwise

[formal ermine]

yeah

that's what they want me to show

[rustic crown]

anyway

[formal ermine]

.

[rustic crown]

the idea is to consider the extension K(sqrt(-1)) and show that this is isomorphic to C

[formal ermine]

K(sqrt(-1))/K?

[rustic crown]

wait so are you asking if that's an extension of K?

[formal ermine]

idk what you mean with extension

[rustic crown]

a bigger field which has a map from the smaller field

[formal ermine]

a map from the smaller field?

[rustic crown]

okie it's weird since we're working with slightly different definitions >.<

[pastel cliff]

another vague question here but is R referring to anything specific?

[rustic crown]

it's usually inconvenient to always what the field extension to set theoretically contain the base field

for instance if you have Q and then look at Q[x]/(x^2+1)

then if you're very precise then Q isn't a subset of Q[x]/(x^2+1)

Q isn't even a subset of Q[x]

only thing you have is that there is a subset isomorphic to Q

[glossy crag]

I think it's a typo and R is meant to be F.

[pastel cliff]

ok thank god

imma hope it's that

[rustic crown]

so instead of requiring it being a subset, a slightly different way to define a field extension is by just defining it as a map k --> F, all field maps are automatically injective so we don't even need to say that

[formal ermine]

k is the smaller field?

[rustic crown]

yee

[formal ermine]

ah okk

[rustic crown]

i was just being imprecise there lol

ig one would call it the base field or something

"smaller" doesn't make sense directly :p

anyway back to the problem

lemme try to read the question carefully

wait so can you first tell me how would you show that there are no finite proper extensions of C?

[formal ermine]

because C is algebraically closed

[rustic crown]

no i mean how would you write it down, i'm just asking so i know what type of definitions you're using :p

or is that your definition of algebraic closure?

[formal ermine]

this is my definition

[pastel cliff]

[rustic crown]

does "endliche" mean finite?

[formal ermine]

yes

[rustic crown]

okie nice

okie so lets first show that if K is a finite extension of R such that x^2+1 has a root in K, then K is isomorphic to C

[formal ermine]

is what I already had here wrong

[rustic crown]

do you see how to do this?

[formal ermine]

no

[rustic crown]

here you have the additional assumption that x^2+1 has a root

if you call the root j, then you get that R[j] is isomorphic to C, right?

[formal ermine]

isn't R[x]/(x^2 + 1) iso R[sqrt(-1)] iso C a definition of C

[rustic crown]

yup

but now K is a finite extension of R[j]

so K = R[j] which is iso to C

this was the first case

now assume that x^2+1 doesn't have a root in K

so it's irreducible in K[x]

in this case you can consider the field L = K[x]/(x^2+1) and identifying K with the canonical copy inside L, you see that L is also a finite extension of R

but now L is a finite extension of R where x^2+1 does have a root, so by what we said above L must be isomorphic to C

but since [L:K] = 2, this force K = R (which you assumed is not the case)

[glossy crag]

Hoerst du Koerpertheorie oder einfach Algebra?

[rustic crown]

yea, R[j] iso R[x]/(x^2+1) iso C

[formal ermine]

yeah I completely misunderstood what you said here at first lol

I thought you meant K[x]/(x^2+1)

[rustic crown]

ah >.<

[rustic crown]

ig only delicate part for you is considering K[x]/(x^2+1) as an extension of K

[formal ermine]

algebra

[formal ermine]

canonical copy?

[rustic crown]

yea, idk how else to say it

[rustic crown]

i said that here as well

how would you usually do that?

i mean considering Q[x]/(x^2+1) as an extension of Q

[formal ermine]

I'm totally lost right now - do what exactly?

[rustic crown]

cause i'm guessing your definition requires Q to be an actual subset of Q[x]/(x^2+1)

elements of Q[x]/(x^2+1) are cosets

so like a general element would look like the coset [a + b*x] containing a + b*x

[formal ermine]

yes

[rustic crown]

even though you want to think of this as an extension of Q, it really isn't

there are two ways to fix that

in one case you look at the subset Q' = {[a] : a in Q} of Q[x]/(x^2+1) and look at the field F = Q[x]/(x^2+1) \ Q' u Q

physically cutting Q' out and replacing with Q

the boring part then is to define the operations again and verify it's a field

another way is to treat Q[x]/(x^2+1) as an extension of Q' instead of Q

does that make some sense? >.<

[formal ermine]

yeah I get what you mean here

but how does all of this help us solve the original problem

[rustic crown]

it doesn't, this was just the technicality of using different definitions

[rustic crown]

the proof ended here

[formal ermine]

why are we looking at the root of x^2 + 1

[rustic crown]

cause you know how to show it for extensions of C

looking at a root of that will let you put C inside the field and use what you know about C

the first case was when you can directly put C inside K

and the second what when you have to get a bigger field to do this

[formal ermine]

what was wrong with my attempted proof

[lethal dune]

what's the 🟦🔶?

ok nvm

[rustic crown]

pronoun roles

[restive birch]

Quotient groups are obliterating my brain

this section has like 9 definitions and 7 theorems in about 6 pages

[rustic crown]

you only did it when either C is contained in K or K is contained in C

[formal ermine]

yes

[lethal dune]

get used to it

[formal ermine]

I wasn't sure how to do it when neither of those apply

so when K intersect (C \ R) = empty

[rustic crown]

yea, that's why i asked you to reduce it to the case where one can assume it contains a copy of C

[rustic crown]

this also won't finish it right

[formal ermine]

ahhhhh that's what we were trying to show

[rustic crown]

there is also the case some but not all the elements of K\R and C\R are common

[formal ermine]

oh

[rustic crown]

see nobody likes to work with fields where you just change the name of one element and call it a different field

i can look at C

and just define a different set C \ {i} u {j}

[lethal dune]

wait that's illegal

[rustic crown]

just change 1 element and define the field structure again

[rustic crown]

yes >.<

let this be your K

then K intersection C = C \ {i, j}

[formal ermine]

yes ok

so we know that when K subseteq C then K = C

so we wanna show that K subseteq C, yes?

[rustic crown]

yea, but that's false in the above example

K neither contains C, nor is contained in C

it differs by exactly 1 element

[formal ermine]

ok I'm really really confused right now

can we start all over

[rustic crown]

(btw just to say this, K still contains things like 2+i and -i etc, only thing it doesn't contain is i. instead of i you have j)

(and you define the operations pretending that j was i, so 2 + j := 2+i, 2i + (-i) = j, etc)

[rustic crown]

what is the part that confuses you?

[formal ermine]

everything

[rustic crown]

[formal ermine]

so when R \subsetneq K with [K : R] finite, to show that K must be C, what do we have to do

[rustic crown]

first K may not be C

you wanna show it is isomorphic to C

[formal ermine]

yes that's what I meant

mb

[rustic crown]

so the argument should be to get hold of an isomorphic copy of C inside K

[formal ermine]

what does that mean

[rustic crown]

find a C' subseteq K such that C and C' are isomorphic

[formal ermine]

ah ok

[rustic crown]

since C' is algebraically closed, this would prove that K = C' so we're done

[formal ermine]

yes

[rustic crown]

so how do you find this C'?

i gave two ideas

one was, somehow force that K contains solution j to x^2+1, in which case we can let C' = R[j]

and other was, pick any arbitrary a in K which is not in R, and then show that even this C' = R[a] works. but here we had to understand the structure of irreducible polynomials in R[x], that they're either linear or quadratic and irreducible.

[pastel cliff]

gonna interject super quick - was in the middle of this proof but i think my reasoning might be wrong

i still need to define what the lcm is for the general case of arbitrary x,y but i eventually need to be specific about it actually being least right

[formal ermine]

ah ok I get this now

[glossy crag]

If x_p is the multiplicity with which every prime shows up in the factorisation of x (and y_p for y), define the lcm as \prod_p p^{max(x_p,y_p)}. Manually check that this is indeed a least common multiple (i.e. x&y divide it and if z is a common multiple, it divides z).

Or you could prove that if R is a GCD-domain (gcd's of all pairs exist), then xy/gcd(x,y) is an lcm of x,y (I haven't checked if this is always true, I think it should be).

[hard flame]

Gotta wired question, I have a subgroup $H \leq G$ and its prime factorization is such that $|H| = \prod_i p_i^{k_i}$ and $p_i > [G:H]$. Can I claim this is normal by counting?

[coral spindle]

I don't see how it would follow that H is normal in G.

[hard flame]

I believe thats true

i cant find a counterexample

any ideas on proving it tho

[coral spindle]

I have been thinking about it. I can't see any particular approach, and I doubt it's true.

[solar glacier]

silly question

but to show something is an ideal you just need to show for all x,y \in I that x-y \in I and for all r \in R (the ring) rx \in I for every x \in I?

[hard flame]

no its true i found as a fact

just cant figure out what direction to go

[rotund aurora]

It should be closed under addition and scalar multiplication by elements of the ring, yes

[solar glacier]

and we do x-y to take care of inverses, correct?

[rotund aurora]

you dont need that (as an extra condition)

I mean, thats not something extra

-x=-1*x

by simple proeprties of scalar multipication

where -1 is the additive inverse of 1 in the ring

[solar glacier]

cause for showing radI is an ideal i showed if a,b \in radI then a+b \in radI but he took off points saying i needed to show a-b \in radI

[rotund aurora]

Uhmm

radl is your name of your ideal? I dont know what it is if you refer to something specific

[solar glacier]

and he wrote "for subring you need to show a-b \in radI"

[rotund aurora]

Wait

[solar glacier]

radical I

[rotund aurora]

Ideal and subring is not the same

[solar glacier]

no not at all ideals are special subrings

[next obsidian]

subring

Mf’s that don’t require rings to have a 1 shaking my head

[solar glacier]

yeah in our class all of our rings have a 1

[next obsidian]

Then an ideal isn’t a subring

Since they aren’t a ring

[solar glacier]

theyre additive subgroups closed under mult by r \in R?

[rotund aurora]

The definition of an ideal I is that its closed under addition and scalar multiplication by elements of the ring R. If I satisfies this properties, then for every a,b in I, -b in I and so a+(-b)=a-b in I

Also idk why Im saying "scalar multiplication" as if we were doing modules, lmao

[next obsidian]

My point is that if an ideal contains 1 then it is equal to R because then r•1 = r is in I for all r

[rotund aurora]

^

[solar glacier]

true ^^

[next obsidian]

So ideals aren’t subrings since they aren’t a ring

[rotund aurora]

But like ideals are submodules of R as a module over itself. I guess thats the right way to look at it

[solar glacier]

so i looked it up, ideals ARE subrings

[chilly ocean]

unity btfo

[next obsidian]

No they aren’t

Ughhhhhhhhhhhhhhhhh

[rotund aurora]

[next obsidian]

You are not using your head to think, whatever you read doesn’t require rings to have unity

If you do they aren’t subrings because THEY DONT HAVE A 1

[solar glacier]

got it

so for a ring with 1 ideals are not subrings but rather additive subgroups?

[next obsidian]

Yes

And they satisfy a very strong closure property for multiplication

[solar glacier]

for every r \in R and every a \in. I ra \in I

right