How can I write $\mathbb{Q}(a,b)$ into a set?

Question

I know that

$$\mathbb{Q}(\sqrt{d}):=\{a+b\cdot\sqrt{d}:a,b\in\mathbb{Q}\}$$

while $d\in\mathbb{Z}\backslash\{1\}$ is square free. But how can I describe f.e.

$$\mathbb{Q}(a,b)$$

as a set?

Answer 1

First you describe $\mathbb{Q}(a)$, assuming $a \in \mathbb{C}$ is algebraic over $\mathbb{Q}$. You will need to know the (degree of the) minimal polynomial of $a$ over $\mathbb{Q}$.

Then you repeat to describe $\mathbb{Q}(a, b) = (\mathbb{Q}(a)) (b)$, again with $b \in \mathbb{C}$ algebraic over $\mathbb{Q}$. You will need to know the (degree of the) minimal polynomial of $b$ over $\mathbb{Q}(a)$

Answer 2

I think it depends to a certain extent on just what $a$ and $b$ are, and just what form you want your description to fit.

In all cases, $\Bbb Q(a,b)$ is the set of all $\Bbb Q$-rational expressiona $f(a,b)/g(a,b)$ for which $g(a,b)\ne0$, and where $f$ and $g$ are $\Bbb Q$-polynomials in two variables. For some applications, this all-purpose representation is perfectly satisfactory.

But if both $a$ and $b$ are elements of some big field $\Omega\supset\Bbb Q$, and are both algebraic over $\Bbb Q$, then Andreas Caranti’s description is right on. If, for instance, $a$ is of degree $m$ over $\Bbb Q$ and $b$ is of degree $n$ over $\Bbb Q(a)$, then the set of numbers $$\bigl\lbrace a^ib^j:0\le i<m, 0\le j<n\bigr\rbrace$$ will be a $\Bbb Q$-linear basis for $\Bbb Q(a,b)$.

In case $a$ and $b$ are independent transcendentals over $\Bbb Q$, you’re stuck with the rational-function description, but there’s an intermediate situation, where $a$ and $b$ are transcendental, but with some relation between them, which may be taken to be an irreducible $\Bbb Q$-polynomial $F(X,Y)$ such that $F(a,b)=0$. Then you may describe $\Bbb Q(a,b)$ as the fraction-field of the ring $\Bbb Q[X,Y]\big/\bigl(F(X,Y)\bigr)$, or you may recognize that $\Bbb Q(a)$ is (isomorphic to) the field of rational functions in the indeterminate $a$, and $\Bbb Q(a,b)$ is a finite algebraic extension of $\Bbb Q(a)$, in other words $\Bbb Q(a)[Y]\big/\bigl(F(a,Y)\bigr)$. In this case, if $F(a,Y)$ is a (one-variable) polynomial of degree $n$ over $\Bbb Q(a)$ the set $\{b^j:0\le j<n\}$ is a $\Bbb Q(a)$-linear basis for the big field.