I know that since $a$ is algebraic over $Q$, this means that $Q(a)$ is a finite extension of $Q$ so $[Q(a) : Q] \leq n$
so we can definite a basis $\{v_1, ..........., v_n\}$ for $Q(a)$
Im stuck on how to proceed. Any help would be much appreciated
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Forget about bases and consider minimal polynomials. – Georges Elencwajg Apr 21 '14 at 17:55
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Hint: Is $\Bbb Q$ a subset of $F$?
More hints: Note that $1$ is in $F$. So $1+1$ is in $F$, $1+1+\ldots+1$ is in $F$. Also, $\frac{1+1}{1+1+1+1+1}$ must be in $F$...
ajotatxe
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that is what it says in my notes but i don't really understand why I can assume it to be the case to answer the question? if i take that as a given, I think I can then argue that Q(a) is a subset of F(a) and so my basis {v1....vn} of Q(a) is contained in F(a) and then I would only need to show that my basis spans F(a) but I am having trouble understanding why Q is a subset of F? – user144712 Apr 21 '14 at 17:23