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The diagonal multiplicative group of a number field is discrete in its idele group. Require local units at every finite place and an archimedean neighbourhood of one. A diagonal element $x$ meeting these restrictions has integral $x-1$. If $x\ne 1$, its nonzero integer field norm cannot have absolute value less than one, while sufficiently small archimedean differences would force exactly that.

Let $L$ be a fixed number field and let $V\subset L$ be a fixed rational linear subspace for which the associated norm form has no unit-family degeneracy. There are effective constants $A,C>0$ such that every $z\in V\cap {\mathcal{O}}_L$ with nonzero field norm $m$ satisfiesTo prove this, factor the principal ideal $(z)$, write its generators as a bounded factor times powers of fundamental units, use the linear relations defining $V$, and apply the Baker lower bound for a homogeneous linear form in logarithms to bound the unit exponents by $O(\log |m|)$. For $L=\mathbb{Q}(\alpha )K$ with $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$, $[K:\mathbb{Q}]=2$, and $V={\mathrm{span}}_{\mathbb{Q}}\{1,\alpha ,\omega \}$ for a basis $1,\omega$ of $K$, the coprime degrees rule out the degeneracy.

For a degree-$m$ extension $L/K$, substitute a $K$-basis of $L$ into a homogeneous degree-$d$ polynomial in $N>d$ variables over $L$, then take its field norm. The resulting degree-$md$ polynomial has $mN>md$ variables over $K$. A nonzero zero over a C1 field gives a nonzero zero of the original form because the norm of a field element vanishes only at zero. Separability is unnecessary.

In $K=\mathbb{Q}(\sqrt{-14})$, the Minkowski bound for ideal classes is less than five. The ideals $\mathfrak{p}=(2,\sqrt{-14})$ and $\mathfrak{q}=(3,1+\sqrt{-14})$ generate all classes, with ${\mathfrak{p}}^2=(2)$ and $(2-\sqrt{-14})=\mathfrak{p}{\mathfrak{q}}^2$. No element has field norm two, so $[\mathfrak{p}]$ has order two and $[\mathfrak{q}]$ has order four.

The product formula uses finite-place factors given by ideal norms, the ordinary modulus at real embeddings, and the modulus squared at complex places. The complex square accounts for the pair of conjugate embeddings. Equivalently ordinary complex modulus can be used with weight two in the product. These normalizations cancel the absolute global field norm.

The ring $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain for the absolute field norm. Approximate both rational coefficients of a quotient by nearest integers. If the errors are $u,w\in [-1/2,1/2]$, then $|u^2-2w^2|\le 1/2<1$, giving a remainder of strictly smaller absolute norm. In particular this ring is a principal ideal domain and has trivial ideal class group.

For a finite extension $L/K$, an unramified or general cyclic character $\chi$ over $K$, and $a\in L^{\times }$, transfer satisfies ${\mathrm{Cor}}_{L/K}(\mathrm{Res}\chi ,a)=(\chi ,N_{L/K}a)$. It is the cup-product projection formula combined with the field norm on multiplicative degree-zero coefficients. With the norm definition of Brauer transfer it also covers purely inseparable steps.

For a finite extension of local fields $L/K$, its norm subgroup is the image of the field norm $N_{L/K}:L^{\times }\to K^{\times }$. For an abelian extension, it is the kernel of the induced local Artin map to $\mathrm{Gal}(L/K)$.