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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?

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

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