How do I show that $\overline{\mathbb{Q}} := \{\alpha \in \mathbb{C}\mid \alpha\text{ is algebraic over }\mathbb{Q} \}$ is algebraically closed?

Question

How do I show that $\overline{\mathbb{Q}} := \{\alpha \in \mathbb{C}\mid \alpha\text{ is algebraic over }\mathbb{Q} \}$ is algebraically closed? I am thinking about solving this using that $\mathbb{C}$ is algebraically closed but I don't know hot to proceed from here.

Answer 1

Let $f=a_0+a_1x+\dots+a_nx^n$ be a polynomial with coefficients in $\bar{\mathbb{Q}}$; then $f$ has coefficients in $K=\mathbb{Q}[a_0,\dots,a_n]$, which is finite dimensional over $\mathbb{Q}$. A splitting field for $f$ over $K$ is finite dimensional over $K$, hence over $\mathbb{Q}$. So the roots of $f$ are in a finite dimensional extension of $\mathbb{Q}$, which means they're algebraic over $\mathbb{Q}$.

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Notes

1. If $b$ is algebraic over $\mathbb{Q}$, then $\mathbb{Q}[b]=\mathbb{Q}(b)$ is finite dimensional over $\mathbb{Q}$

2. By induction, if $b_1,\dots,b_k$ are algebraic over $\mathbb{Q}$, then also $\mathbb{Q}[b_1,\dots,b_k]$ is finite dimensional over $\mathbb{Q}$.

3. The splitting field $K$ can be assumed to be a subfield of $\mathbb{C}$ because $\mathbb{C}$ is algebraically closed.

4. The splitting field of a polynomial in $K[x]$ is finite dimensional, because it is generated by the roots of $K$, which are algebraic over $K$ and the same argument as in 1 and 2 can be used.

Answer 2

It is a proposition more general: Let L be a extension of K.If L is algebraically closed then the set of elementes from L algebraic over K is also algebraically closed. In your case L is C and K is Q!