TITLE: Let $\mathbb Q\le K$ finite extension of fields $\alpha\in K$. Then $N(\alpha)=\pm1\Leftrightarrow\alpha$ is a unit
QUESTION [7 upvotes]: 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

REPLY [0 votes]: If $$ \alpha^n+a_{n-1}\alpha^{n-1}+\ldots +a_1\alpha+a_0=0$$
then
$$ \alpha\cdot(\alpha^{n-1}+a_{n-1}\alpha^{n-2}+\ldots +a_1)=-a_0$$
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