Ytukyg

Let $ngeq 3$ a natural number and let $omega$ be a primitive $n$-th root of unit.

1. Let $a, b in mathbb{Q}, aneq 0$ show that $N_{mathbb{Q}(omega)/mathbb{Q}}(aomega +b)=a^{phi (n)}Phi_n (-b/a)$.




2. If $ngeq 3$ is odd, show that $N_{mathbb{Q}(omega)/mathbb{Q}}(1+omega)=1$.




3. If $n=p$ is an odd prime show that $$N_{mathbb{Q}(omega)/mathbb{Q}}(aomega +b)=begin{cases}(a^p+b^p)/(a+b) & text{ if } bneq -a \ pa^{p-1} & text{ if } b=-aend{cases}$$

$$$$

Could you tell me how $N_{mathbb{Q}(omega)/mathbb{Q}}(aomega +b)$ is defined in general?

Let $K$ be a number field (that is, a finite field extension of $mathbb{Q}$).

It turns out that there are exactly $n=[K:mathbb{Q}]$ embeddings $sigma_i:K to mathbb{C}.$

$Big[$For example, if $K=mathbb{Q}(sqrt[3]{2}),$ then $n=3$ and the embeddings are defined by
$$sigma_1(sqrt[3]{2})=sqrt[3]{2},;; sigma_2(sqrt[3]{2})=omega sqrt[3]{2},;; sigma_3(sqrt[3]{2})= omega^2 sqrt[3]{2}$$
where $omega$ denotes a primitive cube root of unity.$Big]$

For any $x in K,$ one defines its norm $N_{K/mathbb{Q}}(x)$ to be the product $sigma_1(x)sigma_2(x)ldotssigma_n(x).$

$Big[$For example, if $K=mathbb{Q}(zeta_p)$ where $p$ is an odd prime and $zeta_p$ is a primitive $p$th root of unity, then
$$N_{K/mathbb{Q}}(1-zeta_p)=(1-zeta_p)(1-zeta_p^2)ldots(1-zeta_p^{p-1}).$$
Noting that
$$Phi_p(X)=X^{p-1}+ldots+X+1=(X-zeta_p)(X-zeta_p^2)ldots(X-zeta_p^{p-1})$$
we get
$$N_{K/mathbb{Q}}(1-zeta_p)=Phi_p(1)=p.Big]$$
Hope this helps :)

EDIT: In response to the OP's comments below.

As you say in your comment, we have $[mathbb{Q}(zeta_n):mathbb{Q}]=phi(n).$

In fact, we have that
$$Phi_n(X)=prod (X-zeta_n^j)=(-1)^{phi(n)}prod (zeta_n^j-X)$$
where $j$ varies over the integers between $0$ and $n$ which are coprime to $n.$

As you also say in your comment, we have
$$N_{mathbb{Q}(zeta_n)/mathbb{Q}}(azeta_n+b)=N_{mathbb{Q}(zeta_n)/mathbb{Q}}(a)cdot N_{mathbb{Q}(zeta_n)/mathbb{Q}}left(zeta_n+dfrac{b}{a}right)=a^{phi(n)}N_{mathbb{Q}(zeta_n)/mathbb{Q}}left(zeta_n+dfrac{b}{a}right).$$
Combining these two things, it is easy to show that
$$N_{mathbb{Q}(zeta_n)/mathbb{Q}}left(azeta_n+bright)=(-a)^{phi(n)}Phi_n(-b/a).$$
I've got an extra minus sign here...

Let's try a theoretical introduction.

1. 
   

   Determinants of endomorphisms

   
   
   

   Consider first the case of a commutative field $K$, a finite dimensional $K$-vector space $V$ of dimension $n in mathbb{N}$ and an endomorphism $f in mathrm{End}_{K}(V)$ (in other words a map from $V$ to $V$ that preserves the internal addition of vectors and the external scalar multiplication by which $K$ acts on $V$). You might have heard of the notion of representation matrix: upon fixing an ordered basis, say $a in V^{n}$ (by which we mean a family of $n$ vectors which are at the same time linearly independent as well as a generating system for $V$), you can write down the coordinates of each $f(a_k)$ with respect to the basis $a$, and gather this numeric data (in the sense that it is a collection of scalars from the field $K$) into what is known as the representation matrix of $f$ with respect to the chosen basis $a$. Each new choice of an ordered basis will in principle yield a different representation matrix w.r.t that basis, however there is one key invariant: all these representation matrices have the same determinant, regardless of the ordered basis they are associated to. This invariant object will be henceforth called the determinant of the endomorphism $f$, denoted as $mathrm{det}_{K}(f)$ (it makes sense to reference the field of scalars, as the determinant of a given $K$-endomorphism $f$ can change when you are considering the same map $f$ as the endomorphism of the $L$-vector space $V$ obtained by restricting scalars to a subfield $L$ of $K$ such that the degree $[K:L]$ is finite and hence $mathrm{dim}_{L}V=[K:L]mathrm{dim}_{K}V$ is also finite. Consider the very simple example of $mathbb{C}$ viewed as a space over itself, with endomorphism given by multiplication with the imaginary unit $i$, endomorphism whose determinant will no longer be the same if you restrict scalars to $mathbb{R}$).

   
   


2. 
   

   Introducing algebras

   
   
   

   Now, let us consider the additional structure of a finite dimensional unitary associative $K$-algebra $A$. Specifically, I am referring to an algebraic structure rigorously given as a quadruplet $(A, +, cdot,cdot)$, where I have committed the abuse of sacrificing rigour for simplicity, and denoted two distinct operations with the same symbol, that of abstract multiplication. This kind of a structure satisfies the following axioms:

   
   
   

   i) The additive law + is an internal binary operation on $A$ such that the structure $(A, +)$ be an abelian group.

   
   
   

   ii) The second of the multiplicative symbols (identical in syntax, but not in semantics!) is an external left multiplication by which $K$ operates on $A$, in other words a map $cdot: Ktimes A rightarrow A$ such that the abelian group structure mentioned above become a $K$-vector space when equipped with this additional external law (operation). This vector space structure is said to be subjacent to (i.e. part of the larger) algebra structure, and it is the dimension of this vector space over $K$ that we are going to refer to as the dimension of the algebra.

   
   
   

   iii) The first of the multiplicative symbols denotes an internal binary multiplication on $A$, which is at the same time $K$-bilinear and also such that the structure $(A, cdot)$ be a monoid (hence the terminology ''associative and unitary algebra''). As $K$-bilinearity implies biadditivity, this internal operation is distributive with respect to the addition mentioned above, so that it yields a ring structure when adjoined to the group structure at 1. This ring is also said to be subjacent to the unitary associative algebra.

   
   

In such a general context, for any $x in A$ one considers the map $gamma_{x}: A rightarrow A, gamma_{x}(t)=xt$, the map of left multiplication by $x$ (the dual map $delta_{x}$ can also be considered, and in the absence of any hypothesis on the commutativity of the algebra $A$ the left and right multiplications given by a certain element will in principle be different maps). One notices right away that both $gamma_{x}$ and $delta_{x}$ are $K$-linear, in other words endomorphisms of the subjacent $K$-vector space structure of the given algebra. As the algebra is of finite dimension, one can consider their determinants as endomorphisms and thus formulate the following definition:

for any $x in A$ let $mathrm{N}_{A/K}(x)=mathrm{det}_{K}(gamma_{x})$ define the norm of element $x$ over $K$.

In the particular setting of your problem, $mathbb{Q}$ is a subfield of $mathbb{Q}(omega)$, leading to an obvious $mathbb{Q}$-algebra structure on $mathbb{Q}(omega)$. The dimension of this algebra is of course finite, as it is the degree of the extension obtained by adjoining an algebraic element to the ground field. It helps immensely to notice that, thanks to the hypothesis $a neq 0$ one has
$$mathbb{Q}(a omega+b)=mathbb{Q}(omega)$$
(try to justify this, it's easy). Next, notice that $f in mathbb{Q}[X]$ is an annihilating polynomial for $a omega+b$ if and only if $f(aX+b)$ is an annihilating polynomial for $omega$ ; in other words the $mathbb{Q}$-algebra automorphism $sigma$ of $mathbb{Q}[X]$ given by substitution of the indeterminate $X$ with $aX+b$ maps the annihilating ideal of $a omega+b$ to the corresponding ideal for $omega$, ideal which is generated by $varPhi_{n}$, the minimal polynomial of $omega$. The inverse of $sigma$ is given by $sigma^{-1}(X)=frac{X-b}{a}$ and thus the annihilating ideal of $aomega+b$ is generated by $sigma^{-1}(varPhi_{n})=varPhi_{n}left(frac{X-b}{a}right)$; this latter polynomial has dominant (leading) coefficient equal to $frac{1}{a^{varphi(n)}}$, hence the minimal polynomial of $aomega+b$ will be equal to the unique monic polynomial associated to this polynomial, namely to $$a^{varphi(n)}varPhi_{n}left(frac{X-b}{a}right)$$

You can continue from here by resorting to the following theorem, which can be easily proven straight from the general definition of the norm:

let $K$ be a commutative field, $A$ a unitary associative $K$-algebra and $a in A$ an element algebraic over $K$, of minimal polynomial $f in K[X]$. Then the subalgebra $K[a]$ is finite dimensional over $K$, $mathrm{dim}_{K}K[a]=mathrm{deg} f$ and $ mathrm{N}_{K[a]/K}(a)=(-1)^{mathrm{deg} f} f(0_K)$.

This already takes care of the first requirement in your exercise.

P.S. (as I realise that in the course of writing my lengthy answer someone beat me to it, so to speak). The mathematical claims (relations and equalities) made in the previous answer are correct, however I would like to make a comment on method of expound: one does not start by defining norms of elements in number fields in the way mentioned by the previous user (to whom I mean no ill-will), but defines them in the general way formulated in the paragraphs above. Later on, upon developing the theory of number fields one realises that the norm can always be expressed as the product of all conjugates as a subsequent characterisation. It is one aspect of the beauty of field theory that such a characterisation is possible on the basis of what would otherwise appear as an arid definition.

$mathbf{mathrm{II^{nd}}}$ P.S. The following result is known:

If $n in 2mathbb{N}+1$ and $ngeqslant 3$ then $varPhi_n(-1)=1$.

Sketch of proof : we evaluate the Dedekind decomposition relation $X^n-1=prodlimits_{k|n} varPhi_{k}$ at $-1$ in order to obtain, taking into account the parity of $n$ :
$$-2=prodlimits_{k|n}varPhi_{k}(-1)=varPhi_{1}(-1) prodlimits_{substack{k|n \ k neq 1}} varPhi_{k}(-1)$$
As $varPhi_{1}=X-1$ we can cancell out $-2$ on both sides in the above relation leaving:
$$prodlimits_{substack{k|n \ k neq 1}} varPhi_{k}(-1)=1$$
Since $varPhi_{3}=X^2+X+1$ and thus $varPhi_{3}(-1)=1$ you can argue by induction on $n$ to conclude the argument. $Box$