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For a nonzero element $x$ of a number field $L$, normalized absolute values satisfyFor an algebraic integer $x$, factor the principal ideal $(x)$ into prime ideals. Its ideal norm is both the product of the finite-place contributions with inverse exponent and the absolute value of the product of its Archimedean conjugates. Equating the two expressions proves the formula for algebraic integers, and writing an arbitrary $x\in L^{\times }$ as a quotient proves the general case.

For an algebraic number $\alpha$ in a number field $L$, the absolute multiplicative Weil height isIt is independent of the field $L$ containing $\alpha$, satisfies $H({\alpha }^n)=H(\alpha {)}^{|n|}$, and obeys $H(\alpha +\beta )\le 2H(\alpha )H(\beta )$.

A place of a number field $K$ is an equivalence class of absolute values on a field. Its Archimedean places come from real embeddings and conjugate pairs of complex embeddings. Its non-Archimedean places correspond to nonzero prime ideals of the ring of integers of a number field.

Equivalence is taken among absolute values on a field, so changing a representative within one place does not change the associated topology.

If $f(X)=a_d{\prod }_{j=1}^d(X-{\alpha }_j)$, its Mahler measure is

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.

The polynomial $X^3-5X+8$ is irreducible over $\mathbb{Q}$. Forone hasThe field has ring of integers $\mathbb{Z}[\beta ]$ and field discriminant $-307$.

An integral basis is a $\mathbb{Z}$-basis of the ring of integers of a number field.

An element $x$ is integral over $\mathbb{Z}$ exactly when it acts stably on some faithful nonzero finitely generated $\mathbb{Z}$-module containing one.

The discriminant of a biquadratic field is the product of the fundamental discriminants of its three quadratic subfields.

Let $d\ne 0,1$ be a square-free integer, and let $K=\mathbb{Q}(\sqrt{d})$ be the associated quadratic field. ThenThe pair $1,{\omega }_d$ is an integral basis. The field discriminant is $d$ in the first case and $4d$ in the second. This statement requires $d$ to be a square-free integer: for example, $\mathbb{Q}(\sqrt{8})=\mathbb{Q}(\sqrt{2})$ but $\mathbb{Z}[\sqrt{8}]$ omits $\sqrt{2}$.^[ref]

For $d=2$, the integral basis is $1,\sqrt{2}$, and the field discriminant is $8$. For $d=5$, it is $1,\varphi$ with $\varphi =(1+\sqrt{5})/2$, and the field discriminant is $5$. The identity ${\varphi }^2=\varphi +1$ makes $\mathbb{Z}[\varphi ]$ closed under multiplication. Since $\sqrt{5}=2\varphi -1$, the smaller order in a number field $\mathbb{Z}[\sqrt{5}]$ consists precisely of $a+b\varphi$ with even $b$. This computes its index two and shows concretely why adjoining a square root does not always supply every algebraic integer.

In this second integral basis, the field norm isIn particular, $N(\varphi )=-1$ and ${\varphi }^{-1}=\varphi -1$: the extra algebraic integer is also a unit. The units of the quadratic field Q square root of five describe all its powers.

An algebraic number $\alpha$ is an algebraic integer if it is a polynomial root of a monic polynomial with integer coefficients:Equivalently, its algebraic minimal polynomial over $\mathbb{Q}$ has integer coefficients. The requirement that the leading coefficient be one matters: $1/2$ satisfies $2X-1=0$ and is an algebraic number, but is not an algebraic integer. In fact the rational numbers that are algebraic integers are exactly the integers.^[ref]

Some useful computed examples are:

Algebraic integers are closed under addition, subtraction, and multiplication. Thus they provide an arithmetic ring, while division is allowed only when the result is again an algebraic integer. The algebraic integer module criterion explains this closure without having to find a new polynomial for every sum and product.

A number field is a finite field extension $K/\mathbb{Q}$ of the rational numbers. Its degree of a field extension $[K:\mathbb{Q}]$ is its dimension as a vector space over $\mathbb{Q}$. Every element is an algebraic number; the primitive element theorem also says that $K=\mathbb{Q}(\theta )$ for some single algebraic number $\theta$.

For example,The basis $1,\sqrt{2}$ makes arithmetic explicit: use $(\sqrt{2}{)}^2=2$ to reduce products. Every nonzero element has its inverse in the same field:The denominator cannot vanish unless $a=b=0$, since $\sqrt{2}$ is not a rational number. For instance, $(3+\sqrt{2}{)}^{-1}=(3-\sqrt{2})/7$ belongs to the number field, although it is not an algebraic integer.

A degree-$n$ number field has exactly $n$ distinct field embeddings into the complex numbers fixing $\mathbb{Q}$, because every finite field extension in characteristic zero is a separable field extension. For $\mathbb{Q}(\sqrt{2})$ these are the two real mapsBoth preserve addition and multiplication. Keeping both values, rather than just the positive choice of $\sqrt{2}$, is what makes the field norm and the Minkowski embedding of a number field useful.

Integer polynomials form a countable set and each has finitely many roots, so the algebraic numbers are countable.

The golden ratio is the positive numberwhich satisfies ${\phi }^2=\phi +1$ and $1/\phi =\phi -1$.

Intersection theory studies algebraic cycles modulo rational equivalence and their products.

Toric geometry studies algebraic varieties containing an algebraic torus as a dense open subset, with the torus action extended to the whole variety. Rational polyhedral fans translate their geometry into lattice combinatorics.

After coordinates place $p=[0:0:1]$, projection away from $p$ is $[x:y:z]\mapsto [x:y]$. Its restriction to a plane curve avoiding $p$ is a morphism to ${\mathbb{P}}^1$ whose degree equals the degree of the curve.

If $C$ is a smooth projective curve of genus at least three, then the smooth projective surface $C\times C$ contains no curve of geometric genus below three. On the normalization of any curve in the product, at least one coordinate projection to $C$ is nonconstant, and the Riemann-Hurwitz formula cannot decrease genus.

For a finite morphism $f:X\to Y$ of smooth curves, the ramification divisor isThe Riemann-Hurwitz formula states $2g_X-2=(\mathrm{deg}f)(2g_Y-2)+\mathrm{deg}{R}_f$.

The coefficients assemble the ramification indices of a morphism of curves.

A plane curve is the zero set of a polynomial in two variables, or a one-dimensional curve embedded in a plane.