Let $A(\mathbb{C};\mathbb{Q})$ be the set of complex numbers algebraic over $\mathbb{Q}$. Show it has no finite field extensions. Hint: Assume all polynomials split over $\mathbb{C}$.
That $A(\mathbb{C};\mathbb{Q})$ is a subfield of $\mathbb{C}$, I know. Since for $a,b\in A(\mathbb{C};\mathbb{Q})$: $\mathbb{Q}(a,b)\subset A(\mathbb{C};\mathbb{Q})$, so it contains $a+b$, $ab$, $a^{-1}$ and $-a$. And it's clearly a non-empty set.
My first thought is to show that for $a$ algebraic over $A(\mathbb{C};\mathbb{Q})$, $A(\mathbb{C};\mathbb{Q})(a)=A(\mathbb{C};\mathbb{Q})$. Since any finite extension is algebraic and can be expressed by adjoining a finite set of elements, showing $A(\mathbb{C};\mathbb{Q})(a)=A(\mathbb{C};\mathbb{Q})$, would solve the exercise.
The only idea I have, but I'm not sure if it is even true, is that the minimal polynomial of $a$ together with the hint of the exercise would imply the degree of the minimial polynomial being $1$ and therefore $a\in A(\mathbb{C};\mathbb{Q})$.
This should be simple, so any hints are appreciated!
Thanks to Arturo Magidin for his comments that shaped this solution.
Continuing from before, if we prove $A(\mathbb{C}:\mathbb{Q})(a)=A(\mathbb{C}:\mathbb{Q})$, we have solved the problem. Assuming the extension is finite, the minimal polynomial of $a$ over $A(\mathbb{C}:\mathbb{Q})[t]$, $m_1(t)$, exists. By the assumption, $a$ must be a complex number. Let $$m_1(t)=t^n+c_{n-1}t^{n-1}+\cdots+c_1t+c_0.$$ Naturally, this polynomial also exists in $\mathbb{Q}(c_{n-1}, \cdots, c_1, c_0)[t]$, so the extension $\mathbb{Q}(c_{n-1}, \cdots, c_1, c_0)(a):\mathbb{Q}(c_{n-1}, \cdots, c_1, c_0)$ is finite (and actually of the same degree). Since every $c_i$ is by assumption algebraic over $\mathbb{Q}$, the extension $\mathbb{Q}(c_{n-1},,\dots,c_0):\mathbb{Q}$ if finite. By the Tower Law, the extension $\mathbb{Q}(c_{n-1}, \cdots, c_1, c_0)(a):\mathbb{Q}$ is finite and therefore algebraic, so $a\in A(\mathbb{C}:\mathbb{Q})$.