The algebraic numbers form a field

Statement

Let $\overline{\mathbb{Q}}$ be the set of the complex numbers that are algebraic over $\mathbb{Q}$. Then $\overline{\mathbb{Q}}$ is a subfield of $\mathbb{C}$.

Remarks

The proof: if $a, b$ are algebraic over $\mathbb{Q}$, then $\mathbb{Q}(a, b)$ is a finite extension of $\mathbb{Q}$, by the tower law. So $a + b$, $a - b$, $a b$ and $a/b$ for $b \neq 0$ are algebraic, because finite extensions are algebraic. More generally, for every extension $L/K$, the elements of $L$ that are algebraic over $K$ form a subfield of $L$. The field $\overline{\mathbb{Q}}$ is countable and algebraically closed. It is an algebraic closure of $\mathbb{Q}$.

Explanations

L1 human

The sum and the product of two algebraic numbers are algebraic. For example, $\sqrt{2} + \sqrt{3}$ is a root of $x^4 - 10x^2 + 1$. The proof does not need to find such a polynomial: it uses only dimensions of vector spaces.