Let $K$ be a number field (therefore a finite algebraic extension $K/ \mathbb{Q}$). Since this extension is separable the primitive element theorem says $K= \mathbb{Q}(\alpha)$. Denote by $A$ its ring of integers.
Remark: $A:= \{a \in K \vert \exists f \in \mathbb{Z}[X]: f(a)=0 \}$
Denote by $N:K \to \mathbb{Q}$ the norm map.
My question is how to deduce that is for $a \in A$ if we have $N(a)= \pm 1$ then $a \in A^*$ (=invertible elements in $A$)
My attempts: I know that $N$ is a multiplicative map and $A$ a local ring.
I suppose that assuming that $a= pq$ with $p \text{ or} q \in A \backslash A^*$ lead to a contradiction. But why?
Let $L$ be the Galois closure of $K$, and let $B$ be the ring of integers of $L$. Note that $A\subseteq B$. To avoid confusion, let $N_K$ be the norm map of $K$.
If $N_K(a)=\pm 1$, then the product of all Galois conjugates of $a$ (possibly raised to some power, if $\mathbb{Q}(a)$ is properly contained in $K$), is $1$ or $-1$; all of these conjugates lie in $B$, and so $a$ is a unit in $B$: for if the conjugates of $a$ are $a,\sigma_2(a),\ldots,\sigma_m(a)$, then $a^{-1}=\pm\sigma_2(a)\cdots\sigma_m(a)$ (or, if $\mathbb{Q}(a)\subsetneq K$, something of the form $\pm a^{r-1}(\sigma_2(a)\cdots\sigma_m(a))^r$ for some $r$).
But that means that $a^{-1}\in B$. Thus, $a^{-1}$ is an algebraic integer; since $a\in K$, $a^{-1}\in K$ and is an algebraic integer, it follows that in fact $a^{-1}\in A$; that is, $a$ is a unit in $A$.