Let $\mathbb Q\le K$ finite extension of fields $\alpha\in K$. Then $N(\alpha)=\pm1\Leftrightarrow\alpha$ is a unit Let $\mathbb Q\le K$ (where $K\le\mathbb C$) be a finite extension (say $|K:\mathbb Q|=n$) of fields and let  $\alpha\in K$ be an algebraic integer, i.e. $\alpha$ is a root of a monic polynomial over $\mathbb Z$; let's call $\mathbb A$ the set (which in fact is a ring) of the algebraic integers of $\mathbb C$. Hence we picked $\alpha\in\mathbb A\cap K$.
We know that there exists $\beta\in K$ s.t. $K=\mathbb Q(\beta)$. Call $f\in\mathbb Q[X]$ the minimal polynomial of $\beta$ over $\mathbb Q$, clearly $\partial f=n$ and it has $n$ different roots (in some extension of $K$ or in $K$ itself if we suppose it normal over $\mathbb Q$). Then there are exactly $n$ embeddings of $K$ in $\mathbb C$ (an embedding is an immersion $K=\mathbb Q(\beta)\hookrightarrow\mathbb C$ s.t. $q\mapsto q\;\;\forall q\in\mathbb Q$ and $\beta$ goes to one of the $n$ roots of $f$; clearly if $K$ is normal then the embeddings are automorphisms of $K$); call them $\sigma_i,\;i=1,\dots,n$.
So we can now define a norm on $K$: $N(\alpha):=\sigma_1(\alpha)\cdots\sigma_n(\alpha)\;\;\forall \alpha\in K$.
So we can now face the "problem" (I know it must be almost silly but I can't get find a way out!):
given $\alpha\in\mathbb A\cap K$, prove that
$$
N(\alpha)=\pm1\Longleftrightarrow\alpha\;\; \mbox{is a unit in}\;\; \mathbb A\cap K\;.
$$
I solved $\Leftarrow\;$: by hypotesis there exists $\alpha^{-1}\in\mathbb A\cap K$; then
$$
\alpha\alpha^{-1}=1\Longrightarrow\;1=N(1)=N(\alpha\alpha^{-1})=N(\alpha)\left(N(\alpha)\right)^{-1}.
$$
Observe now that $\alpha\in\mathbb A$ hence in particular it's a root of a polynomial over $\mathbb Z$. Among such polynomials, taking the one of lower degree, we can see it must be irreducible. This is the minimal polinomial of $\alpha$. Hence $N(\alpha)$ will be the constant term of this polynomial, so $N(\alpha)\in\mathbb Z$.
Hence $N(\alpha)=\pm1$.
But how can I prove the other implication?
Thanks all
 A: (Extended) Hint: Assume that $\alpha$ is an algebraic integer with $N(\alpha)=\pm1$. Then the minimal polynomial $f(x)$ of $\alpha$ is something like
$$
f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in\Bbb{Z}[x].
$$
Show that here $a_0=\pm1$. Show that multiplying the equation $f(\alpha)=0$ by $\alpha^{-1}$ allows you to solve $\alpha^{-1}=p(\alpha)$ for a certain polynomial $p(x)\in\Bbb{Z}[x]$. 
Alternatively use the fact that the minimal polynomial of $\alpha^{-1}$ is the reciprocal polynomial
$$
\tilde{f}(x)=x^nf(\frac1x)=a_0x^n+a_1x^{n-1}+\cdots a_{n-1}x+1.
$$
A: You've done fine with one direction, for the other, we use the facts that the product of algebraic integers is an algebraic integer (which follows from the fact that the algebraic integers of a number field form a ring) and the fact that any $\mathbb{Q}$-homomorphic image of an algebraic integer is also an algebraic integer.
So suppose $N(\alpha) = 1$. We can re-write this as $\Pi_{i=1}^n\sigma_i(\alpha)=1$, whence $\sigma_1(\alpha^{-1})=\Pi_{i=2}^n\sigma_i(\alpha)$. In particular, the right hand side of this lies in $\sigma_1(K)$, so we can apply $\sigma_1^{-1}$* to obtain $\alpha^{-1}=\sigma_1^{-1}(\Pi_{i=2}^n\sigma_i(\alpha))$. The facts in the first paragraph show that this expression for $\alpha^{-1}$ is an algebraic integer, so $\alpha$ is a unit in the ring of integers.
A small point: If you assume $K \subset \mathbb{C}$ then one of the $\sigma_i$ will be the identity, which we may as well take to be $\sigma_1$ to simplify the proof a little. However, there is no reason a number field should be a subfield of $\mathbb{C}$. Quite often, to use your notation, we realise $K=\mathbb{Q}(\beta)$ as $\mathbb{Q}[X]/(f)$, since this view gives a good way of explicitly writing the elements of $K$, and of dealing with homomorphisms of $K$. With this viewpoint in mind, there's no "special" $\sigma_i$ so the proof looks a little more complicated.
*Recall that field homomorphisms are injective, so are isomorphisms onto their image. 
A: If $$ \alpha^n+a_{n-1}\alpha^{n-1}+\ldots +a_1\alpha+a_0=0$$
then
$$ \alpha\cdot(\alpha^{n-1}+a_{n-1}\alpha^{n-2}+\ldots +a_1)=-a_0$$
