Is $\mathbb{Q}(\sqrt[3]3)$ = $\{a+b\sqrt[3]3+c\sqrt[3]9 | a,b,c \in \mathbb{Q}\}$ a field?
My first attempt was to prove that $\{a+b\sqrt[3]3\}$ was a field. But, I ran across a problem where I am unable to find an inverse for every element.
I know that the inverse would have to be of the form $\frac{1}{a+b\sqrt[3]3}$. I then multiplied top and bottom by the conjugate and got $\frac{a-b\sqrt[3]3}{a^2-b^2\sqrt[3]9}$. But, this isn’t of the form $\{a+b\sqrt[3]3\}$. Had I found an inverse of the proper form, I would’ve used induction to prove that my initial question was a field.
You may have seen $$ x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - yz - zx - xy) $$
We are going to introduce $ d^3 = 3$ then make changes to the identity using $ x \mapsto x, y \mapsto d y, z \mapsto d^2 z$ to arrive at $$ x^3 + d^3y^3 + d^6z^3 - 3d^3xyz = (x+dy+d^2z)(x^2 + d^2y^2 + d^4z^2 - d^3yz - d^2zx - dxy) $$ for $$ \frac{1}{x+dy+d^2z \; } \; = \; \frac{ \; x^2 + d^2y^2 + d^4z^2 - d^3yz - d^2zx - dxy \; }{x^3 + d^3y^3 + d^6z^3 - 3d^3xyz} $$
You need $d^3 = 3$