What are the solid definitions of algebraic number fields $\mathbb{Q}(\theta)$ and $\mathbb{Q}(\theta, \phi)$

Question

I have never seen a solid definition of what these objects are actually defined as or what they contain.

Answer 1

"Never have seen a solid definition". Werd, because this is the most basic notion at the starting point of all algebraic number theory.

1) If the elements added to $k$ are supposed to live in some ambient unspecified field $K$, then $k(\theta),k(\theta,\phi)$, etc. denote the smallest subfields containing $k$ and these elements, independently of any property of algebraicity or transcendance. The notations $k(X), k(X,Y)$, etc. (with capital letters) should be reserved for the fields of rational fractions in $X, (X,Y)$, etc. with coefficients in $k$. One defines analogously the rings $k[X]$ (=ring of polynomials) and $k[\theta]$ (= smallest subring, etc.) Note that the fields $k(X), k(\theta)$ are the fields of fractions of the rings $k[X]$ and $k[\theta]$.

2) The distinction between "algebraic" and "trancendental" appears with the notion of an annihilating (resp. minimal) polynomial $f(X)$ (resp. $m(X)$) of $\theta$. Because of the "universal property" of the polynomials, there exists a surjective homomorphism of rings $\Psi : k[X]\to k[\theta]$, with Ker $\psi = (m(X))$, the principal ideal generated by $m(X)$ (this comes from euclidian division), so that $k[X]/(m(X))\cong k[\theta]$. If $m(X)$ is the zero polynomial, then $k[X]\cong k[\theta]$, hence $k(X)\cong k(\theta)$, and $\theta$ is called transcendental over $k$. If $m(X)$ is not zero, its minimality implies its irreducibility, hence the maximality of the ideal $(m(X))$, so that $k[X]/(m(X))$ is a field, $k[\theta]=k(\theta)$, and $\theta$ is called algebraic over $k$.

3) In a last step, we must prove the existence of an ambient field $K$. If $\theta$ is transcendental, no problem, just take $K=k(X)$. If $\theta$ is algebraic, it is certainly shown in your course (using the axiom of choice ?) that an "algebraic closure" $k^{alg}$ of $k$ can be constructed, and one just take $K=k^{alg}$ ./.