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  "body": "\n## Proof\n\n1. Let $\\overline{\\Q}$ be the set of complex numbers that are [[algebraic|0NPW1DWECM]] over $\\Q$.\n2. Every $c \\in \\Q$ is a root of $x - c$, so $\\Q \\subseteq \\overline{\\Q}$. In particular $0, 1 \\in \\overline{\\Q}$.\n3. Let $a, b \\in \\overline{\\Q}$. Let $E = \\Q(a)$ and $F = E(b)$, [[subfields of $\\C$ generated|0MA5B744K3]] by these elements.\n4. By the [[degree of a simple extension|0N5QW9K7MH]], $[E : \\Q]$ is the degree of the minimal polynomial of $a$, so it is finite.\n5. A nonzero polynomial in $\\Q[x]$ with root $b$ also lies in $E[x]$. So $b$ is algebraic over $E$, and $[F : E]$ is finite by the same result.\n6. By the [[tower law|0A7KJ68XHT]], $[F : \\Q] = [F : E] \\cdot [E : \\Q]$ is finite.\n7. The field $F$ contains $a$ and $b$, so it contains $a - b$ and $a b$, and also $b^{-1}$ if $b \\ne 0$.\n8. Since [[finite extensions are algebraic|0DFYTNZ6AM]], every element of $F$ is algebraic over $\\Q$. So $a - b$, $a b$ and, for $b \\ne 0$, $b^{-1}$ lie in $\\overline{\\Q}$.\n9. By Steps 2 and 8, $\\overline{\\Q}$ contains $1$ and is closed under subtraction and multiplication. So it is a [[subring|0859D58WPD]] of $\\C$.\n10. By Step 8, it also contains the inverse of each of its nonzero elements. So $\\overline{\\Q}$ is a [[subfield|07QY5RPFBR]] of $\\C$.\n\n## Idea\n\nTwo algebraic numbers $a$ and $b$ generate a field $\\Q(a, b)$ of finite degree over $\\Q$, by the tower law. Every element of a finite extension is algebraic. So sums, products and inverses built from $a$ and $b$ stay algebraic.\n",
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