---
tag: 1YS91YH4V6
kind: proof
name: Proof that the algebraic numbers form a field
msc: [11R04]
proves: 0QFTETCHKB
uses: [0NPW1DWECM, 07QY5RPFBR, 0859D58WPD, 0N5QW9K7MH, 0A7KJ68XHT, 0DFYTNZ6AM, 0MA5B744K3]
sources:
  - url: https://en.wikipedia.org/wiki/Algebraic_number
    locator: Field
---

## Proof

1. Let $\overline{\Q}$ be the set of complex numbers that are [[algebraic|0NPW1DWECM]] over $\Q$.
2. Every $c \in \Q$ is a root of $x - c$, so $\Q \subseteq \overline{\Q}$. In particular $0, 1 \in \overline{\Q}$.
3. Let $a, b \in \overline{\Q}$. Let $E = \Q(a)$ and $F = E(b)$, [[subfields of $\C$ generated|0MA5B744K3]] by these elements.
4. By the [[degree of a simple extension|0N5QW9K7MH]], $[E : \Q]$ is the degree of the minimal polynomial of $a$, so it is finite.
5. 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.
6. By the [[tower law|0A7KJ68XHT]], $[F : \Q] = [F : E] \cdot [E : \Q]$ is finite.
7. The field $F$ contains $a$ and $b$, so it contains $a - b$ and $a b$, and also $b^{-1}$ if $b \ne 0$.
8. 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 $\overline{\Q}$.
9. By Steps 2 and 8, $\overline{\Q}$ contains $1$ and is closed under subtraction and multiplication. So it is a [[subring|0859D58WPD]] of $\C$.
10. By Step 8, it also contains the inverse of each of its nonzero elements. So $\overline{\Q}$ is a [[subfield|07QY5RPFBR]] of $\C$.

## Idea

Two 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.