Is $x+y\alpha + z\alpha^2$ a subfield? Let $\alpha = \sqrt[3]{2}$. I want to Prove that the set of all numbers $x+y\alpha+z\alpha^2$, for $x,y,z \in \mathbb{Q}$, is a subfield of $\mathbb{C}$. 
I have showed that this set is a subring, but I'm not sure how to show for any $a\neq 0$ in our set has a multiplicative inverse. 
I have tried picking two arbitrary elements, $m,n$ in our set. Let $m= x+y\alpha+z\alpha^2$ and let $n=a+b\alpha+c\alpha^2$. Then setting our product $m\cdot n$ equal to 1 gets us:
$$(ax+2zb+2yc) + (bx+ay+2zc)\alpha + (cx + yb +za)\alpha^2 = 1.$$
I was wondering if this could lead me to my proof? 
 A: Recall the algebraic identity
$$x^3 + y^3 + z^3 -3xyz = (x+y+z)(x^2 + y^2+z^2 - xy -yz - zx)\\
\implies
\frac{1}{x+y+z} =  \frac{x^2+y^2+z^2 - xy-yz-zx}{x^3+y^3+z^3 - 3xyz}
$$
Substitute $(x,y,z)$ by $(x,y\alpha,z\alpha^2)$, one get
$$\frac{1}{x + y\alpha + z\alpha^2}
= \frac{(x^2-2yz) + (2z^2-xy)\alpha + (y^2-zx)\alpha^2}{x^3+2y^3 + 4z^3 -6xyz}$$
Aside from justifying why $x^3+2y^3+4z^3-6xyz \ne 0$, this gives you 
a formula for the multiplicative inverse of $x + y\alpha + z\alpha^2$.
A: You don't need to exhibit inverses, just to prove that they exist. Here is a roadmap:


*

*$\mathbb Q(\alpha)$ is a finite-dimensional vector space over $\mathbb Q$.

*Let $\beta \ne 0$ be an element of $\mathbb Q(\alpha)$. Consider the map $T: x \mapsto \beta x$.

*$T$ is an injective $\mathbb Q$-linear operator on $\mathbb Q(\alpha)$. 

*Therefore, $T$ is surjective. 

*Thus, there is $\gamma \in \mathbb Q(\alpha)$ such that $1=T(\gamma)=\gamma\beta$.
A: You have to solve
$$
\begin{cases}
ax+2bz+2yc=1\\
ay+bx+2cz=0\\
az+by+cx=0
\end{cases}
$$
in the variables $a,b,c$. The notation here is somewhat of counterintuitive with respect to the usual one.
The matrix associate to the non-homogeneous linear system is
$$
A(x,y,z)=\begin{pmatrix}
x & 2z & 2y\\
y & x & 2z\\
z & y & x
\end{pmatrix}
$$
and the determinant is $x^3+2y^3+4z^3-6xyz$. Assuming $A(x,y,z)$ invertible, you can determine $(a,b,c)$ as $\displaystyle A(x,y,z)^{-1}\cdot \begin{pmatrix} 1\\0\\0 \end{pmatrix}$.
It remains to show that $\det A(x,y,z)=x^3+2y^3+4z^3-6xyz=0$ iff $(x,y,z)=(0,0,0)$.
Set $X=x$, $Y=y\sqrt[3]{2}$ and $Z=z\sqrt[3]{4}$. 
Hence the equality above becomes: $x^3+2y^3+4z^3-6xyz=X^3+Y^3+Z^3-3XYZ=0$
It is known that 
$$X^3+Y^3+Z^3-3XYZ=\frac12(X+Y+Z)\big((X-Y)^2+(Y-Z)^2+(Z-X)^2\big)$$
hence we get: 
$X^3+Y^3+Z^3-3XYZ=\frac12(X+Y+Z)\big((X-Y)^2+(Y-Z)^2+(Z-X)^2\big)=0$.
There are two possible cases:


*

*$X+Y+Z=x+y\sqrt[3]{2}+z\sqrt[3]{4}=0$. Hence $x=y=z=0$ because $x,y,z\in \Bbb Q$. You can easily check that $1,\sqrt[3]{2},\sqrt[3]{4}$ are linearly independent over $\Bbb Q$.

*$\big((X-Y)^2+(Y-Z)^2+(Z-X)^2\big)=0$. Since this is a sum of three squares, the total amount is zero iff each summand in zero, that is $(X-Y)=(Y-Z)=(Z-X)=0$ iff $X=Y=Z$ iff $x=y\sqrt[3]{2}=z\sqrt[3]{4}=0$ iff $x=y=z=0$ because $\sqrt[3]{2},\sqrt[3]{4}\notin\Bbb Q$.
Therefore: $\det A(x,y,z)=0$ iff $x=y=z=0$, iff $m=x+y\alpha+z\alpha^2=0$. Hence any element different to zero has a multiplicative inverse.
A: A demonstration which is relies less on explicit calculation and more on abstract properties of fields, vector spaces, and linear maps:
With
$\alpha = 2^{1/3}, \tag 0$
let
$S = \{1, \alpha, \alpha^2 \}; \tag 1$
$S$ is a linearly independent set over $\Bbb Q$.
For 
$\sigma = x + y\alpha + z\alpha^2, \; x^2 + y^2 + z^2 \ne 0, \tag 3$
i.e., for
$\sigma \ne 0, \tag 4$
multiplication by $\sigma$ acts as a $\Bbb Q$-linear map
$\sigma:\text{span}(S) \to \text{span}(S) \tag 5$
via
$v \mapsto \sigma v \tag 6$
for
$v \in \text{span}(S); \tag 7$
as such, $\sigma$ is injective:
$\sigma v_1 = \sigma v_2 \Longrightarrow \sigma(v_1 - v_2) = 0 \Longrightarrow v_1 - v_2 = 0 \Longrightarrow v_1 = v_2; \tag 8$
note that here we have only relied upon the fact that the ring
$\text{span}(S) \subset \Bbb C \tag{9}$
is an integral domain, being a sub-ring of the field $\Bbb C$; we have not yet explicitly invoked invertability of $\sigma$, which is fortunate since that is what we are (still engaged in) trying to prove; now since
$\dim_{\Bbb Q} \text{span}(S) = 3 < \infty, \tag{10}$
it follows that any injective linear mapping $\text{span}(S) \to \text{span}(S)$ is also surjective; in the present case this means that there is some
$\tau \in \text{span}(S) \tag{11}$
with
$\sigma \tau = 1_{\Bbb Q}; \tag{12}$
since the binary operation on the left of this equation is ordinary multiplication in $\Bbb C$, we also have
$\tau \sigma = 1_{\Bbb Q}; \tag{13}$
(12) and (13) together by definition mean
$\tau = \sigma^{-1}, \tag{14}$
and we have shown that $\text{span}(S)$ is a field as per request.
