Field extension over $\mathbb{Q}$ I am given a subfield $E$ of $\mathbb{C}$ and asked to show that $[E : \mathbb{Q}] \le 10$ when every element of $E$ is a root of a polynomial in $\mathbb{Q}[x]$ of degree $10$. 
But I don't think this is true. Say that $E$ contains the square roots of the first $5$ primes. Then when I stack these on top of each other, I have a degree $32$ extension inside $E$. Note that each of them satisfies $x^8(x^2 - p)$, $p$ prime.
Should the question instead say minimal polynomial? Thanks.
 A: But $\sqrt{2} + \sqrt{3} + \sqrt{5} + \sqrt{7} + \sqrt{11}$ doesn't satisfy a non-constant degree $10$ polynomial. The element will have minimal polynomial of degree $32$.
A: Your counterexample isn't valid, and here's why:
One implication of the primitive element theorem is that, for any finite extension $K$ over $\mathbb{Q}$, we have $K = \mathbb{Q}[\alpha]$ for some $\alpha \in K$.  Also recall that the degree of an algebraic extension $\mathbb{Q}[\beta]$ over $\mathbb{Q}$ is equal to the degree of the minimal polynomial of $\beta$ in $\mathbb{Q}[x]$.  
Now let $E$ be the field we get by adjoining the square roots of the first five primes to $\mathbb{Q}$.  Applying the above facts, we know that $E = \mathbb{Q}[\alpha]$ for some $\alpha \in E$.  Specifically, $\alpha = \sqrt{2} + \sqrt{3} + \sqrt{5} + \sqrt{7} + \sqrt{11}$ works as Spooky mentioned.  Since $[E: \mathbb{Q}] = 32$, it follows that the minimal polynomial for $\alpha$ must have degree $32$.  
Coincidentally, the primitive element theorem also guarantees the result you are trying to prove!  As mentioned in the comments, however, you will first need to make an argument that $E$ is a finite extension of $\mathbb{Q}$ in order to apply the primitive element theorem.
