Is $\mathbb{Q}[2^{1/3}]$ a field? Is $\mathbb{Q}[2^{1/3}]=\{a+b2^{1/3}+c2^{2/3};a,b,c \in \mathbb{Q}\}$ a field?
I have checked that $b2^{1/3}$ and $c2^{2/3}$ both have inverses, $\frac{2^{2/3}}{2b}$ and $\frac{2^{1/3}}{2c}$, respectively. 
There are some elements with $a,b,c \neq 0$ that have inverses, as $1+1*2^{1/3}+1*2^{2/3}$, whose inverse is $2^{1/3}-1$.
My problem is that is that I can't seem to find a formula for the inverse, but I also can't seem to find someone who doesn't have an inverse.
Thanks for your time.
 A: A neat way to confirm that it is a field:
Since $\mathbb{Q}$ is a field, $\mathbb{Q}[x]$ is a PID.  $\mathbb{Q}[2^{1/3}] \cong \mathbb{Q}[x] / (x^3 - 2)$.  Now, $x^3 - 2$ is irreducible over $\mathbb{Q}$, since if it weren't, there would be a rational root to $x^3 - 2$. Because $\mathbb{Q}[x]$ is a PID and the polynomial is irreducible over $\mathbb{Q}$,  $(x^3 - 2)$ is a maximal ideal in $\mathbb{Q}[x]$.  
By the Correspondence Theorem of Ideals, we see that as $(x^3 - 2)$ is maximal, $\mathbb{Q}[x] / (x^3 - 2)$ must be a field.  
A: It is a field, and you don't need to find an inverse for each element to prove that each element has an inverse. You can prove that if $\alpha$ is in the set then $\lbrace\,1,\alpha,\alpha^2,\alpha^3\,\rbrace$ is a linearly dependent set over the rationals, then deduce that $\alpha$ satisfies an equation of degree (at most) 3 over the rationals, then if $A\alpha^3+B\alpha^2+C\alpha+D=0$ you have $\alpha(A\alpha^2+B\alpha+C)=-D$, from which you can see an inverse to $\alpha$. 
A: Actually, the inverse of $x=2^{1/3}$ can be computed in a very elementary way as follows. Since $x\neq0$, the equality 
$$
x^3-2=x\left(x^2-\frac2x\right)=0
$$
yields
$$
\frac1x=\frac12x^2\in{\Bbb Q}[x].
$$
This "trick" extends immediately to any algebraic number.
A: HINT $\ $ Since $\rm\:f(x) = x^3 - 2\:$ is irreducible in the Euclidean domain $\rm\:\mathbb Q[x]\:,\:$ any lower degree polynomial $\rm\:g(x) = a + b\:x + c\:x^2\:$ is coprime to it in $\rm\:\mathbb Q[x]\:,\:$ therefore, by the Euclidean algorithm, there is a Bezout equation $\rm\ a\ f + b\ g = 1\ $ which yields $\rm\ b(2^{1/3})\ g(2^{1/3}) = 1\ $ for $\rm\:x = 2^{1/3}\:.$ 
