Given the rational numbers $\mathbb{Q}$, I am wondering what is the algebraic closure of $\mathbb{Q}$, i.e., I need to find a field extension $\mathbb{K}/ \mathbb{Q}, $ such that for an arbitrary polynomial $f$ over $\mathbb{Q},\, \exists a\in \mathbb{K}, $ such that $f(a)=0.$
Given the definition, it must be the smallest algebraically closed field containing $\mathbb{Q}.$ One needs to adjoin roots of elements of $ \mathbb{Q}$ including $i$. How does the algebraic closure of $\mathbb{Q}$ look like?
$\endgroup$ 21 Answer
$\begingroup$The algebraic closure $\mathbb A$ of $\mathbb Q$ is the field of algebraic numbers, which consists of those complex numbers which are roots of some non-zero polynomial in one variable with rational coefficients. It is a countable set and therefore $\mathbb{A}\varsubsetneq\mathbb C$.
$\endgroup$
