Product formula for a number field

Write $x=a\mathrm{/}b$ with $a,b\in {\mathcal{O}}_K\setminus \{0\}$. Indeed, $K\mathrm{/}\mathbb{Q}$ is finite so $x$ is algebraic over $\mathbb{Q}$ and has a monic minimal polynomial $m(X)=X^d+c_{d-1}{X}^{d-1}+\cdots +c_0\in \mathbb{Q}[X]$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element); choose $M\ge 1$ with all $M^j{c}_{d-j}\in \mathbb{Z}$, and set $a=Mx$. Then $a^d+M{c}_{d-1}{a}^{d-1}+\cdots +M^d{c}_0=M^{d}m(x)=0$, a monic integer polynomial relation, so $a\in {\mathcal{O}}_K$ by the minimal-polynomial criterion (Minimal-polynomial criterion for algebraic integers); with $b=M\in \mathbb{Z}\setminus \{0\}\subseteq {\mathcal{O}}_K$ this gives $x=a\mathrm{/}b$.

For the archimedean factors, [F3] gives $N_{K\mathrm{/}\mathbb{Q}}(x)={\prod }_{\psi }\psi (x)$ over all $[K:\mathbb{Q}]$ embeddings, and the embeddings consist of the $r_1$ real embeddings together with the $r_2$ conjugate pairs $\{{\tau }_j,\overline{{\tau }_j}\}$; taking absolute values and using $\mathrm{\mid }zw\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }$ and $\mathrm{\mid }\bar{z}\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }$ from [F4], $\mathrm{\mid }{N}_{K\mathrm{/}\mathbb{Q}}(x)\mathrm{\mid }={\prod }_{\psi }\mathrm{\mid }\psi (x)\mathrm{\mid }=\left({\prod }_{i=1}^{r_1}\mathrm{\mid }{\sigma }_{i}x\mathrm{\mid }\right)({\prod }_{j=1}^{r_2}\mathrm{\mid }{\tau }_{j}x\mathrm{\mid }\mid \overline{{\tau }_{j}x}\mid )=\left({\prod }_{i=1}^{r_1}\mathrm{\mid }{\sigma }_{i}x\mathrm{\mid }\right)\left({\prod }_{j=1}^{r_2}\mathrm{\mid }{\tau }_{j}x{\mathrm{\mid }}^2\right)$, which is exactly the product of the archimedean normalized absolute values.

In the language of fractional ideals (Fractional ideals, The field of fractions $\mathrm{Frac}(D)=(D\setminus \{0\}{)}^{-1}D$ of an integral domain) one has $(x)=(a)(b{)}^{-1}$, hence $v_{\mathfrak{p}}(x)=v_{\mathfrak{p}}((a))-v_{\mathfrak{p}}((b))$ for every prime $\mathfrak{p}$; by [F1] write $(a)={\prod }_{\mathfrak{p}}{\mathfrak{p}}^{e_{\mathfrak{p}}}$ and $(b)={\prod }_{\mathfrak{p}}{\mathfrak{p}}^{f_{\mathfrak{p}}}$ with finite supports, so $v_{\mathfrak{p}}(x)=e_{\mathfrak{p}}-f_{\mathfrak{p}}$ vanishes outside the finite union of those supports and ${\prod }_{\mathfrak{p}}N{\mathfrak{p}}^{-v_{\mathfrak{p}}(x)}=\left({\prod }_{\mathfrak{p}}N{\mathfrak{p}}^{f_{\mathfrak{p}}}\right)({\prod }_{\mathfrak{p}}N{\mathfrak{p}}^{e_{\mathfrak{p}}}{)}^{-1}=N((b))\mathrm{/}N((a))=\mathrm{\mid }{N}_{K\mathrm{/}\mathbb{Q}}(b)\mathrm{\mid }\mathrm{/}\mathrm{\mid }{N}_{K\mathrm{/}\mathbb{Q}}(a)\mathrm{\mid }=1\mathrm{/}\mathrm{\mid }{N}_{K\mathrm{/}\mathbb{Q}}(x)\mathrm{\mid }$, the third equality by [F2] applied to the two finite factorisations and the last by [F3], since $N_{K\mathrm{/}\mathbb{Q}}(a)=N_{K\mathrm{/}\mathbb{Q}}(x)N_{K\mathrm{/}\mathbb{Q}}(b)$.

Multiplying the finite product of step 2.1 and the archimedean product of step 1.2 gives ${\prod }_v\mathrm{\mid }x{\mathrm{\mid }}_v=\left({\prod }_{i=1}^{r_1}\mathrm{\mid }{\sigma }_{i}x\mathrm{\mid }\right)\left({\prod }_{j=1}^{r_2}\mathrm{\mid }{\tau }_{j}x{\mathrm{\mid }}^2\right)\left({\prod }_{\mathfrak{p}}N{\mathfrak{p}}^{-v_{\mathfrak{p}}(x)}\right)=\mathrm{\mid }{N}_{K\mathrm{/}\mathbb{Q}}(x)\mathrm{\mid }\cdot \mathrm{\mid }{N}_{K\mathrm{/}\mathbb{Q}}(x){\mathrm{\mid }}^{-1}=1$, and only the finitely many primes in the supports of $(a)$ and $(b)$ contribute a finite factor different from $1$, so the product is over a finite set of places; the only Choice in the argument is [A1], the norms, moduli and logarithms being computed without further selection.