Is there some standard way to determine the basis of $K(a_0, a_1, \ldots, a_n)$ considered as a vector space over $K$?

Question

Let $L \supset K$ be a field and consider $K(a_0, a_1, \ldots, a_n)$ a field extension of $K$ with $a_i \in L$.

Is there some standard way to determine a basis of $K(a_0, a_1, \ldots, a_n)$ considered as a vector space over $K$ ?

Answer 1

By induction, we may assume $n=0$. So we want to find a basis of $K(a)$ over $K$. If $a$ is trancendental over $K$, then this vector space is infinite dimensional. You can compute its dimension (I think it is $\max(\aleph_0,|K|)$ ...) but cannot write down a basis, so this is useless anyway.

If $a$ is algebraic, choose its minimal polynomial $f = \lambda_0 + \lambda_1 T + \dotsc + T^n$. Then $K(a) \cong K[T]/(f)$ and we see that $\{1,a,\dotsc,a^{n-1}\}$ is a $K$-basis.

Example. $\mathbb{Q}(i,\sqrt{2})$ over $\mathbb{Q}$. We first look at $\mathbb{Q}(i,\sqrt{2})$ over $\mathbb{Q}(\sqrt{2})$. The minimal polynomial of $i$ over $\mathbb{Q}(\sqrt{2})$ is $T^2+1$ (since $i^2+1=0$ and $i \notin \mathbb{Q}(\sqrt{2})$). Hence, $\mathbb{Q}(i,\sqrt{2})$ has basis $\{1,i\}$ over $\mathbb{Q}(\sqrt{2})$. Next, $\mathbb{Q}(\sqrt{2})$ has basis $\{1,\sqrt{2}\}$ over $\mathbb{Q}$. It follows that $\{1,\sqrt{2},i,\sqrt{2} i \}$ is a basis of $\mathbb{Q}(i,\sqrt{2})$ over $\mathbb{Q}$.