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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](http://en.wikipedia.org/wiki/Algebraic_number)?2011-03-08
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    More or less, but not quite. This is saying that every polynomial in $F[X]$ factors over $F$.2011-03-08
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    The question is why the field of algebraic numbers is algebraically closed.2011-03-08
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    Thank you. I was reading that every polynomial in $\mathbb{Q}[x]$ splits over $F$ ;-)2011-03-08
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    In my opinion, you should add that it is the field of all algebraic numbers, in case of misread, thanks.2011-03-08
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    This is the transitivity of algebraic extension. If $L$ is an algebraic extension of $K$ and $K$ is an algebraic extension of $F$ then $L$ is also an algebraic extension of $F$. Here we use $F = \mathbb{Q}$.2016-03-31

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