Smallest subfield of $\mathbb{C}$ containing $\sqrt[5]{2}$ I was thinking that the minimal subfield of $\mathbb{C}$ containing $\sqrt[5]{2}$ would include all elements of the form $\{x + y\sqrt[5]{2} \}$ where $x$ and $y$ are in $\mathbb{Q}$. However, why is it required to include additional powers of $\sqrt[5]{2}$ in the minimal subfield? In other words, I believe the right answer would be that there would be elements of the form $\{a + b\sqrt[5]{2} + c(\sqrt[5]{2})^2 + d(\sqrt[5]{2})^3 + e(\sqrt[5]{2})^4 + f(\sqrt[5]{2})^5 + g(\sqrt[5]{2})^6\}$ where $a,b,c,d,e,f,g$ are in $\mathbb{Q}$. Why is this the case, though? 
Thank you!
 A: Because the field contains $\sqrt[5]2$, and is a ring, it also needs to contain $\sqrt[5]2^2$ by closure of multiplication. But $\sqrt[5]2^2$ is not possible to write as $x+y\sqrt[5]2$ with $x, y$ rational, so the set $\{x + y\sqrt[5]2\}$ is not a ring. The same goes for $\sqrt[5]2^3$ and $\sqrt[5]2^4$.
A: You might like to approach this problem by providing an explicit construction of the field. The smallest field containing an algebraic (over $\mathbb{Q}$) number $\alpha$ will be isomorphic to $\mathbb{Q}[x]/\langle f(x) \rangle$, where $f(x)$ is the minimal polynomial of $\alpha$.
The minimal polynomial of an element $\alpha$ is an irreducible polynomial which has $\alpha$ as a root. The irreducibility allows you to say this field is the smallest such field. 
The isomorphism from $\mathbb{Q}[x]/\langle f(x) \rangle$ to the form embedded in $\mathbb{C}$ is just the morphism which sends $x \mapsto \alpha$. 
So find the minimal polynomial of $\sqrt[5]{2}$, then construct the quotient field and apply the isomorphism. You should arrive at precisely the representation you have without the 5th and 6th powers.
