= Field norm
{title2=$N_{L/K}(\alpha)$}
{wiki}
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:
$$
N_{L/K}(\alpha)=\det(m_\alpha)\in K.
$$
This does not depend on the chosen <basis>. If $L/K$ is a <separable field extension>, then
$$
N_{L/K}(\alpha)=\prod_{\sigma:L\hookrightarrow\overline K,\ \sigma|_K=\operatorname{id}}\sigma(\alpha),
$$
where 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>”.https://math.mit.edu/classes/18.785/2021fa/LectureNotes4.pdf[Andrew Sutherland, Number Theory I, Lecture 4, norm and trace]{ref}
For $L=\mathbb Q(\sqrt2)$ over $K=\mathbb Q$, write $\alpha=a+b\sqrt2$. In the <basis> $1,\sqrt2$, multiplication by $\alpha$ has columns $(a,b)$ and $(2b,a)$, hence
$$
M_\alpha=\begin{pmatrix}a&2b\\b&a\end{pmatrix},\qquad
N_{L/\mathbb Q}(\alpha)=a^2-2b^2=(a+b\sqrt2)(a-b\sqrt2).
$$
The <determinant> calculation and the product of the two <field embeddings> give the same answer. In particular,
$$
N(\sqrt2)=-2,\qquad N(1+\sqrt2)=-1,\qquad N(3+2\sqrt2)=1,\qquad N(2)=4.
$$
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 $-1$.https://math.mit.edu/classes/18.785/2021fa/LectureNotes4.pdf[Sutherland, Lecture 4, properties of the norm]{ref}https://kconrad.math.uconn.edu/blurbs/ugradnumthy/quadratic-integers.pdf[Conrad, Quadratic Integers, norms and units in quadratic rings]{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, $3+\sqrt2$ has <field norm> $7$, so it cannot divide $2$, whose <field norm> is $4$. It does divide $7$, with quotient $3-\sqrt2$. This necessary condition can reject a proposed factor without solving for its quotient.
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