$@fretty$ if you don't mind, could you explain how $Frob_{p}(a+ib)\equiv a+(-1)^{\frac{p-1}{2}}bi\pmod{\mathfrak{p}}$ , is it by definition? and does $\frak{p}$ lies above $p$. if so what happens for $Q(\sqrt{-3})/Q$

Yes, $\mathfrak{p}$ lies above $p$ and so when working mod $\mathfrak{p}$ (i.e. in the residue field) you will be working in characteristic $p$. By definition Frob$_p (a+ib) \equiv (a+ib)^{p} \bmod \mathfrak{p}$. You get what I write from this...

If you do it for $\mathbb{Q}(\sqrt{-3})/\mathbb{Q}$ you will find the Legendre symbol $\left(\frac{-3}{p}\right)$ hidden in there for unramified $p$. I gather the workings will not be as nice. You should read D.Cox's book "Primes of the form $x^2 + ny^2$", there is a nice section on Frobenius elements, class field theory, Artin reciprocity and how the weak/strong reciprocity laws follow from it (it is these ones that explicitly give you quadratic, cubic, quartic, Eisenstein reciprocity laws plus many others).

Thanks Fretty, Thank you very much

You should try the calculations for this extension by yourself...it is not too hard. The hard bit is in understanding the general proof of quadratic reciprocity above.

The real thing that is going on here is that for any abelian extension if you use any representation $Gal(K/\mathbb{Q}) \longrightarrow \mathbb{C}^{\times}$ then there will be a corresponding mod $N$ Dirichlet character for some $N$ that agrees with it on the level of primes and Frobenius elements. In the case of quadratic extensions if we define the representation given by $\text{id}\mapsto 1$ and $\text{conj} \mapsto -1$ then the corresponding Dirichlet character agreeing with Frob$_p$ is the Legendre symbol $\left(\frac{\Delta}{p}\right)$ where $\Delta$ is the discriminant of the extension.

Thanks fretty, got it

Let $x\in K=Q(\sqrt{d})$ and $Frob_{p}(x)\equiv x\pmod{\mathfrak{p}}$ such that $p$ splits completely in $Q(\sqrt{d})$, Then the Artin symbol of $p$ is trivial right?

The artin symbol is defined on elements of the ring of integers of the number field in question. It is a general fact that if prime ideal $\mathfrak{p}$ of $\mathcal{O}_K$ is unramified in an abelian extension $L/K$ then the Frobenius element is trivial if and only if $\mathfrak{p}$ splits completely in $L$.

thanks @fretty, also, if $p=\mathfrak{p1}\mathfrak{p2}$ in $K$, such that, $x\equiv 1\pmod{p}$, and $Frob_p(x)>0$ then is it true that, $x\equiv 1\pmod{\mathfrak{p1}}$ and $x\equiv 1\pmod{\mathfrak{p2}}$, i.e, the corresponding Artin symbols over $\mathfrak{p1}$ and $\mathfrak{p1}$ are trivial

What does Frob$_p(x)>0$ mean? It makes no sense in general fields.

I'm sorry, the corresponding automorphism, $\sigma_p(x)$

No my point is there is not necessarily an ordering on a number field...would you say $i>0$?

I think you misunderstand the Frobenius element...it is a unique automorphism that satisfies Frob$_p(x) \equiv x^{N(p)} \bmod \mathfrak{p}$ for EVERY $x\in\mathcal{O}_K$.

For eg: here in $\frac{\sqrt{2}^{p}-1}{\sqrt{2}-1}\equiv 1\bmod{7}$, for $p\equiv 1\bmod{3}$, and is totally positive, so is it true that,$\frac{\sqrt{2}^{p}-1}{\sqrt{2}-1}\equiv 1\bmod{-1+2\sqrt{2}}$

$-7=(-1+2\sqrt{2})(-1-2\sqrt{2})$

Totally positive is a different thing to what you write. It is a condition under each real embedding.

Where ordering makes sense!

It makes no sense mod p

But tbh I don't understand your question...

thanks for the answer, my question is that if for eg in $Q(\sqrt{2})$, $\frac{\sqrt{2}^{7}-1}{\sqrt{2}-1}\equiv 1\pmod{7}$, so what about this

$\frac{\sqrt{2}^{7}-1}{\sqrt{2}-1}\pmod{-1+2\sqrt{2}} and$\frac{\sqrt{2}^{7}-1}{\sqrt{2}-1}\pmod{-1-2\sqrt{2}}

What has this got to do with Frobenius elements, or quadratic reciprocity?

nothing with both of them

The Artin symbol is 1 when evaluated at 7. is it true at other 2 equations too

Then why did you invite me to chat on a question about Frobenius elements and quadratic reciprocity?

I'm sorry. Kindly excuse

You haven't asked a question yet...

Are you simply asking what this quantity is mod p?

sorry mod (-1 +- 2rt(2))

if so then surely it is still 1...

it is regarding Artin symbol of an element $x\in K$ when evaluated at mod $p$ in $Q$ and when same $x$ is evaluated at mod $\mathfrak{p1}$ and $\mathfrak{p2}$, where $p=\mathfrak{p1}\mathfrak{p2}$

if a = b mod r and s|r then a = b mod s

surely it depends on what K is...

yeah, thank you very much

if K/Q is abelian then the frob element depends only on p not p1 or p2

all frob elements are equal

this was the point that was confusing me

thank you really

if it isn't abelian then they are conjugate by an element of the Galois group such that p1 -> p2

ok, good

so in your case you have an abelian extension...hence frob_p1 = frob_p2

yes right

you should find that both are the identity

is it obvious from a=b mod r and s|r, so a=b mod r?

if s|r then <r> is contained in <s>

so if a=b mod r then a-b is in <r> which is in <s>

so a=b mod s

or should we evaluate in some case

I have just proved it...

say (2+\sqrt{2})^{7}-1)/(1+\sqrt{2}) mod 7 case, can I substitute sqrt{2}=4 and proceed?

I don't understand why you are trying to work out this quantity...where has it come from?

tbh, in the Numerator (\sqrt{2}) it is an associate of $2+\sqrt{2}$ in the earlier example

I really have no idea what you are wishing to find...

Are you trying to find the frob of 7 in Q(rt(2))/Q?

sorry to disturb you, this was the point at which I was doubting, whether to substitute 4 or 3 for \sqrt{2}, to simplify the calculations, and reprsent the numbers in the form $x^2+7y^2$

in fact was working out

thank you very much for all the help you gave thanks for your time too

That's fine...I just don't know what I am giving help towards

I had many doubts cleared, thanks

I still don't know what the question is lol

are you merely trying to work out something mod 7?

yes

if so that has nothing to do with frobenius elements

it is basic algebraic number theory

ok

can I substiute for $\sqrt{2}$ as 4 or 3, while evaluating mod 7

u is a unit with inverse u^(-1) = (-1+rt(2))

so

so ((2+rt(2))^7 - 1)/u

= ((rt(2))^7 u^7 - 1)/u = (rt(2))^7 u^6 - u^(-1) = 8rt(2) u^6 + (-1+rt(2))

lets work out u^6

(1+rt(2))^2 = 3+2rt(2)

(1+rt(2))^4 = 17 + 10rt(2)

(1+rt(2))^6 = 91 + 67rt(2)

so we are trying to work out

8rt(2)(91+67rt(2)) + (-1+rt(2)) mod 7

= rt(2)(4rt(2)) + (-1+rt(2)) mod 7

= 8 + (-1+rt(2)) mod 7

= rt(2) mod 7

so, we need to evaluate not substitute for sqrt{2} is it so

Oh hang on u^6 = 99 + 70rt(2)

so it should be

8rt(2)(99+70rt(2)) + (-1+rt(2))

= rt(2) + (-1+rt(2)) mod 7

= -1 + 2rt(2) mod 7

hence the Artin symbol of ((2+\sqrt{2})^{7}-1)/1+sqrt{2}is trivial right

I really don't understand how you conclude that from this calculation

multiply u^{6} to 2^3, subtract -1

That isn't the definition...

then

L/K Galois extension number fields. Let P be a prime ideal of ring of integers of L

lying above prime ideal Q of ring of integers of K

then Frob_P is the unique element of decomposition group of P such that Frob_P(x) = x^{N(Q)} mod P for all x in ring of integers of L

thank you very much,

surely you know this though?

not really all, but some from discussions

You find it must exist by the transitive action of the Galois group on the factorisation

Gal(Q(rt(2))/Q) has two elements

id and complex conjugation

given ANY prime p, it will have a factorisation in Q(rt(2))

the elements of the Galois group will permute these prime ideals

the action is transitive

so you can take one of the prime ideal factors P of <p> and look at the stabilizer

the so called Decomposition group

it turns out that if p is unramified in L then the decomposition group is cyclic, isomorphic to Gal(F_P / F_p), the Galois group of the residue fields...

but these are finite fields, and so the Galois group is cyclic generated by a Frobenius element

So the Decomposition group has a canonical generator that induces this map...

i.e. there is a unique automorphism satisfying sigma(x) = x^{N(p)} mod P

this is the Frobenius element/Artin symbol

ok @fretty nice of you to give some exposure, on Artin map, Frobenius elements, would you mind if I ask some doubts in future

If you haven't seen this stuff then you really shouldn't be trying to calculate them

ok

You should learn about it first...why it is important