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?
Why every polynomial over the algebraic numbers $F$ splits over $F$?
5
$\begingroup$
field-theory
splitting-field
-
0Isn't that more or less the definition of [algebraic numbers](http://en.wikipedia.org/wiki/Algebraic_number)? – 2011-03-08
-
1More or less, but not quite. This is saying that every polynomial in $F[X]$ factors over $F$. – 2011-03-08
-
2The question is why the field of algebraic numbers is algebraically closed. – 2011-03-08
-
0Thank you. I was reading that every polynomial in $\mathbb{Q}[x]$ splits over $F$ ;-) – 2011-03-08
-
0In my opinion, you should add that it is the field of all algebraic numbers, in case of misread, thanks. – 2011-03-08
-
0This 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