# Interesting property of a polynomial ring with coefficients algebraic over $\mathbb{Q}$

I read that if $F$ is the field of algebraic numbers over $\mathbb{Q}$, then every polynomial in $F[x]$ splits over $F$. That's awesome! Nevertheless, I don't fully understand why it is true. Can you throw some ideas about why this is true?

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Isn't that more or less the definition of algebraic numbers? – Fabian Mar 8 '11 at 8:34
More or less, but not quite. This is saying that every polynomial in $F[X]$ factors over $F$. – 4math Mar 8 '11 at 8:48
The question is why the field of algebraic numbers is algebraically closed. – Yuval Filmus Mar 8 '11 at 8:57
Thank you. I was reading that every polynomial in $\mathbb{Q}[x]$ splits over $F$ ;-) – Fabian Mar 8 '11 at 9:11
In my opinion, you should add that it is the field of all algebraic numbers, in case of misread, thanks. – awllower Mar 8 '11 at 11:29

The fact you mention is that the field $\overline{\mathbb{Q}}$ of algebraic numbers is algebraically closed. This is true because it is the algebraic closure of the field $\mathbb{Q}$ inside the larger algebraically closed field $\mathbb{C}$, i.e., the set of all complex numbers which satisfy some nonzero polynomial equation with $\mathbb{Q}$-coefficients.

Notwithstanding the terminology, that the algebraic closure of a field in an algebraically closed field is algebraically closed requires (nontrivial) proof! See for instance $\S 4$ of my notes on field theory, especially Proposition 17 and Corollary 18.

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Consider some polynomial $$x^n = \sum_{i=0}^{n-1} c_i x^i,$$ where the $c_i$ are algebraic numbers. Thus for each $i$ we have a similar identity $$c_i^{n_i} = \sum_{j=0}^{n_i-1} d_{i,j} c_i^j,$$ where this time the $d_{i,j}$ are rationals.

Suppose that $\alpha$ is a root of the original polynomial. By using the above identities, every power of $\alpha$ can be expressed as a linear combination with rational coefficients of terms of the form $$\alpha^m c_0^{m_0} \cdots c_{n-1}^{m_{n-1}},$$ where $$0 \leq m < n \text{ and } 0 \leq m_i < n_i.$$ Putting all these $N = nn_0\cdots n_{n-1}$ elements into a vector $v$, we get that there are rational vectors $u_k$ such that $$\alpha^k = \langle u_k,v \rangle.$$ Among the first $N+1$ vectors $u_k$ there must be some non-trivial rational linear combination that vanishes: $$\sum_{k=0}^N t_k u_k = 0.$$ Therefore $$\sum_{k=0}^N t_k \alpha^k = 0,$$ and so $\alpha$ is also algebraic.

This proof is taken from these lecture notes, but it's pretty standard.

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