# What is the algebraic closure of $\mathbb{Q}$?

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?

• This is known as the field of algebraic numbers. It is not clear what you mean by "How does it look like?" Feb 7 '18 at 11:42
• en.wikipedia.org/wiki/Algebraic_number#Properties Feb 7 '18 at 11:46

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 $$\mathbb{A}\varsubsetneq\mathbb C$$.