If Q(a,b) is a field extension, can we always choose an equivalent extension Q(c) such that c=a+b?

Question

If we have two complex numbers $a,b$ that are algebraic over $\mathbb {Q} $, we can make an extension $\mathbb {Q}(a,b)$ that is equal to an extension $\mathbb {Q}(c)$ for some $c\in \mathbb {C} $. That is, we can always choose an extension by just one element that's equal to an extension by more than one element.

As an example, we have $\mathbb {Q}(\sqrt 2,\sqrt 3)= \mathbb {Q}(\sqrt 2 + \sqrt 3)$. Here, $c $ is a linear combination of $a,b$, with each term multiplied by $1$ ($c=1a+1b$).

Can this always be the case? I can't think of any counterexample. Is the answer different in other fields?

Answer 1

See the primitive element theorem,

https://en.wikipedia.org/wiki/Primitive_element_theorem

Constructive results is the exact section you want, and the general idea is: sometimes. There are some situations that you have to be careful of, but the answer is yes: you can always find a $c \in \mathbb{Q}$ such that $\mathbb{Q}(\alpha, \beta) = \mathbb{Q}(c)$.

It won't always be $c = \alpha + \beta$, though.

Answer 2

The primitive element theorem answers your question. It states:

Let $E/F$ be a finite field extension. Then $E=F(\alpha)$ if and only if there are finitely many fields $K$ such that $F\subseteq K\subseteq E$.

Notably in the case of fields of characteristic $p$, if $[E:F]=p^2$ then there are actually infinitely many intermediary fields and so it's not the case that $E$ is simple.

The fact that $c=a+b$ is working is a fluke of low degree. That always works on degree $2$, but even in degree $3$ cases it stops working.