Number fields and Integral Closure I am trying to solve the following problem: 

Consider the fields $K=\mathbb{Q}[\omega]$, where $\omega^2+\omega+1=0$, and $L=K(x)[y]$ where $y^3=1+x^2$, and the rings $R=K[x]$, and $\mathcal{O}_L=\{R$-integral elements in $L\}$. Show that $R[y]=\mathcal{O}_L$.

My thoughts: $R[y]\subseteq\mathcal{O}_L$ is obvious since $y$ is $R$-integral. The challenging part is to show that $\mathcal{O}_L\subseteq R[y]$. 
We have 
$$R[y]\simeq\mathbb{Q}[\omega,x,y]/I, \qquad I=(\omega^2+\omega+1, y^3-x^2-1)$$
Let $\alpha\in \mathcal{O}_L$. Since $R$ (being a PID) is integrally closed we know that the coefficients of the minimum polynomial $f$ of $\alpha$ are in $R$. I wonder how to use this information to show that $\alpha\in R[y]$. Some ideas?
 A: I think that this follows from the following bits. The field extension $L/K(x)$ is actually cubic Galois (by Kummer), the generating automorphism $\sigma$ is defined by $\sigma(y)=\omega y$. So if $z=a_0+a_1y+a_2y^2, a_0,a_1,a_2\in K(x)$ is integral over $R$, then the minimal polynomial
$$
m_z(T)=(T-z)(T-\sigma(z))(T-\sigma^2(z))\in K(x)[T]
$$
must have coefficients in $R$. But here the coefficient of the quadratic term is the negative of the trace
$$
tr(z)=z+\sigma(z)+\sigma(z^2)=3a_0.
$$
Therefore we already know that $a_0$ must be in $R$.
But we also know that the integral elements form a ring, and obviously $y$ is integral. Therefore so are the elements $zy$ and $zy^2$. We easily see that
$$
tr(yz)=3a_2(x^2+1)\qquad\text{and}\qquad tr(y^2z)=3a_1(x^2+1).
$$
Therefore we know that both $(x^2+1)a_1,(x^2+1)a_2\in R$. In other words, $(x^2+1)z\in R[y]$ and we're done up to that extra possible factor $x^2+1$ in the denominator.
Consider an element of the form
$$
z'=\frac{b_1(x)}{x^2+1} y+\frac{b_2(x)}{x^2+1}y^2
$$
with $b_1(x), b_2(x)\in K[x]$.
Then the norm of $z'$, $N(z')=z'\sigma(z')\sigma^2(z')$ is
$$
\begin{aligned}
N(z)&=\frac{1}{(x^2+1)^3}(b_1 y+ b_2 y^2)(b_1\omega y + b_2\omega^2y^2)(b_1\omega^2y+b_2\omega y^2)\\
&=\frac{\omega^3y^3}{(x^2+1)^3}
(b_1+b_2 y)(b_1+b_2\omega y)(b_1+b_2\omega^2y)\\
&=\frac1{(x^2+1)^2}(b_1^3+b_2^3(x^2+1)).
\end{aligned}
$$
Here it is impossible for the polynomial $q=b_1^3+b_2^3(x^2+1)$ to be divisible
by $x^2+1$ unless $x^2+1\mid b_1$. But in that case $(x^2+1)^3\mid b_1^3$. Repeating the dose we see that for $q$ to be divisible by $(x^2+1)^2$ it is necessary for both $b_1$ and $b_2$ to be divisible by $x^2+1$. Therefore $z'\in R[y]$ proving your claim.
Some details were left out, but I'm sure you will manage!
A: Inspired by the answer of Jyrki above I want to try a (possible) simplification of that reasoning (For convenience, I repeat here from beginning): 
$\Bbb{K}(x)\hookrightarrow \Bbb{L}$ is a Galois cubic with generating automorphism  $\sigma(y)=\omega y$. Let $z\in \Bbb{L}$,
$$z=a_0+a_1y+a_2y^2, \qquad a_0,a_1,a_2\in \Bbb{K}(x)$$
Assume $z\in\mathcal{O}_\Bbb{L}$. We claim that each $a_j$ is in $R$. Since $z\in\mathcal{O}_\Bbb{L}$ we know its minimal polynomial
$$\begin{aligned}
m_z(T)&=(T-z)(T-\sigma(z))(T-\sigma^2(z)) \\
&= T^3-tr(z)T^2+k(z)T-N(z) .
\end{aligned}$$
must have coefficients in $R$, so we have
$$\begin{aligned}
tr(z) &=z+\sigma(z)+\sigma(z^2)=3a_0\in R \\
k(z) &=3a_0^2-3(x^2+1)a_1a_2\in R \\
N(z) &= a_0^3-3(x^2+1)a_0a_1a_2+(x^2+1)a_1^3+a_2^3(x^2+1)^2 \in R
\end{aligned}$$
The first one gives $a_0\in R$. From the second one we get $(x^2+1)a_1a_2\in R$. 
Now, the third condition gives
\begin{equation}
(x^2+1)[a_1^3+a_2^3(x^2+1)]\in R=\Bbb{K}[x] \tag{$\heartsuit$}
\end{equation}
The second condition $(x^2+1)a_1a_2\in R$ allows two possible cases:


*

*$a_1\in R$ and $a_2(x^2+1)\in R$. Then $\heartsuit$ gives $a_2^3(x^2+1)^2\in R$ which is possible only if $a_2\in R$.

*$a_2\in R$ and $a_1(x^2+1)\in R$. Then $\heartsuit$ gives $a_1^3(x^2+1)\in R$ which is possible only if $a_1\in R$.

