Field norm ($N_{L/K}(\alpha )$)

For a finite field extension $L/K$, multiplication by $\alpha \in L$ is a $K$-linear map $m_{\alpha }:L\to L$. Its determinant is the field norm:This does not depend on the chosen basis. If $L/K$ is a separable field extension, thenwhere the product runs over all $[L:K]$ field embeddings into an algebraic closure of $K$. Embeddings are counted even when several give the same value on $\alpha$: for $r\in K$, the result is $r^{[L:K]}$. This avoids ambiguity in the phrase “product of algebraic conjugates”.^[ref]

For $L=\mathbb{Q}(\sqrt{2})$ over $K=\mathbb{Q}$, write $\alpha =a+b\sqrt{2}$. In the basis $1,\sqrt{2}$, multiplication by $\alpha$ has columns $(a,b)$ and $(2b,a)$, henceThe determinant calculation and the product of the two field embeddings give the same answer. In particular,A field norm is an arithmetic invariant; it is not a nonnegative norm measuring length, and need not be positive on nonzero elements.

The field norm is multiplicative, $N(\alpha \beta )=N(\alpha )N(\beta )$, because $m_{\alpha \beta }=m_{\alpha }{m}_{\beta }$ and determinants are multiplicative. It vanishes exactly at zero. For a number field over $\mathbb{Q}$, it takes algebraic integers to integers, and an algebraic integer is a unit exactly when its field norm is $1$ or $-1$.^[ref][ref]

This makes the field norm a useful test for divisibility. If nonzero $\alpha ,\beta \in {\mathcal{O}}_L$ and $\alpha$ divides $\beta$, then the integer $N(\alpha )$ divides $N(\beta )$. For example, $3+\sqrt{2}$ has field norm $7$, so it cannot divide $2$, whose field norm is $4$. It does divide $7$, with quotient $3-\sqrt{2}$. This necessary condition can reject a proposed factor without solving for its quotient.

Let $A$ be an integrally closed domain with fraction field $K$, and let $L/K$ be finite. If $b\in L$ is an integral element over $A$, its monic minimal polynomial lies in $A[T]$: its roots are integral, so its coefficients are integral elements of $K$, hence belong to $A$. If that polynomial has degree $r$ and constant coefficient $a_0$, then $N_{L/K}(b)=((-1{)}^r{a}_0{)}^{[L:K(b)]}\in A$. This remains true for inseparable extensions. An infinite algebraic extension has no usual determinant field norm.

For a finite extension of local fields with normalized discrete valuations, the field norm satisfies the displayed equality. The extended absolute values on a field of all conjugates agree, so $|N(x){|}_K=|x{|}_L^{[L:K]}$. Using $[L:K]=e(L/K)f(L/K)$ and $v_L{|}_K=e(L/K)v_K$ gives the equality. Thus the field norm is a unit precisely when the original element is a unit.

For a finite number field $L/\mathbb{Q}$ with a rational basis ${\omega }_1,\dots ,{\omega }_n$, the norm form is the homogeneous polynomial

For a rational prime $p\ne 229$, representation by this norm form is equivalent to complete splitting in the Hilbert class field of Q of square root 229. Its Frobenius is a permutation of the three roots of $X^3-4X-1$. The quadratic residue condition $(p/229)=1$ makes this permutation even; a root modulo $p$ gives a fixed point, so the even permutation is the identity. Complete splitting makes the prime above $p$ principal. Its generator has norm $\pm p$, and the unit of norm minus one changes the sign to positive, giving the representation.

If $x=y^2$ in a number field, then $N(x)=N(y{)}^2$; a norm that is not a square in the base field therefore obstructs square roots.

For a number field $K$ and an algebraic integer $\alpha \in {\mathcal{O}}_K$, the criterion isHere the absolute field norm means the field norm down to $\mathbb{Q}$; its absolute value is one. Necessity follows from $N(\alpha )N({\alpha }^{-1})=1$ with both factors integers. Conversely, multiplication by $\alpha$ on an integral basis is a matrix with integer entries and determinant $\pm 1$. Its inverse also has integer entries, and applying that inverse to $1$ gives ${\alpha }^{-1}\in {\mathcal{O}}_K$. The integrality hypothesis matters: $\alpha =(3+\sqrt{2})/(3-\sqrt{2})=(11+6\sqrt{2})/7$ has field norm one, but is not in $\mathbb{Z}[\sqrt{2}]$.

In $\mathbb{Z}[\sqrt{2}]$, the criterion becomes $a^2-2b^2=\pm 1$. Thus $(1+\sqrt{2}{)}^{-1}=\sqrt{2}-1$ and $(3+2\sqrt{2}{)}^{-1}=3-2\sqrt{2}$ are again algebraic integers. These units produce infinite families of solutions to the Pell equation.