= Number field
{title2=$[K:\mathbb Q]<\infty$}
{wiki}
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,
$$
\mathbb Q(\sqrt2)=\{a+b\sqrt2:a,b\in\mathbb Q\},\qquad [\mathbb Q(\sqrt2):\mathbb Q]=2.
$$
The <basis> $1,\sqrt2$ makes arithmetic explicit: use $(\sqrt2)^2=2$ to reduce products. Every nonzero element has its inverse in the same <field>:
$$
(a+b\sqrt2)^{-1}=\frac{a-b\sqrt2}{a^2-2b^2}.
$$
The denominator cannot vanish unless $a=b=0$, since $\sqrt2$ is not a <rational number>. For instance, $(3+\sqrt2)^{-1}=(3-\sqrt2)/7$ 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 $\mathbb Q(\sqrt2)$ these are the two real maps
$$
\sigma_+(a+b\sqrt2)=a+b\sqrt2,\qquad \sigma_-(a+b\sqrt2)=a-b\sqrt2.
$$
Both preserve addition and multiplication. Keeping both values, rather than just the positive choice of $\sqrt2$, is what makes the <field norm> and the <Minkowski embedding of a number field> useful.
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