Proving $\frac{-\theta + \theta^2}{2}$ is an algebraic integer in $K = \mathbb{Q}(\theta)$, given that $\theta^3 + 11\theta - 4 = 0$ As the title says, given that $\theta^3 + 11\theta - 4 = 0$, I'm trying to prove that  $\frac{-\theta + \theta^2}{2}$ is an algebraic integer in $K = \mathbb{Q}(\theta)$.
I know that $x^3 + 11x -4$ is irreducible in $\mathbb{Q}[x]$ since it's irreducible in $\mathbb{F}_3[x]$.  I also know that the set of algebraic integers forms an integral domain and thus I know that $-\theta + \theta^2$ is an algebraic integer, unfortunately that's the best I can do with that method since $\frac{1}{2}$ is specifically not an algebraic integer.  
Clearly I need to somehow use the polynomial to solve this, but I can't see how, can anyone point me in the right direction?  Thanks.
 A: The minimal polynomial of $(\theta^2-\theta)/2$ is ${z}^{3}+11\,{z}^{2}+36\,z+4$.
One way to get this is: if $t = (\theta^2-\theta)/2$, express  $t^3 + b t^2 + c t + d$ as a rational linear combination of $1$, $\theta$ and $\theta^2$, and solve the system of equations that say that the coefficients of $1$, $\theta$ and $\theta^2$ are all $0$.
A: This demonstrates what Robert Israel suggests.
Suppose that
$$
\theta^3+11\theta-4=0
$$
and
$$
\alpha=\frac{\theta^2-\theta}{2}
$$
Then
$$
\begin{align}
\alpha^0&=\frac22\\
\alpha^1&=\frac{\theta^2-\theta}{2}\\
\alpha^2&=\frac{-5\theta^2+13\theta-4}{2}\\
\alpha^3&=\frac{19\theta^2-107\theta+36}{2}
\end{align}
$$
and
$$
\begin{bmatrix}36&-107&19\end{bmatrix}
\begin{bmatrix}
2&0&0\\
0&-1&1\\
-4&13&-5
\end{bmatrix}^{-1}
=\begin{bmatrix}-4&-36&-11\end{bmatrix}
$$
Therefore,
$$
\alpha^3+11\alpha^2+36\alpha+4=0
$$
and $\alpha$ is an algebraic integer.
A: Also, since $\theta$ is an algebraic integer, to show $\theta(\theta-1)/2$ is an algebraic integer, it suffices to show that it is $2$-adically integral. 
Thus, the plan is to prove that either $2|\theta$ or $2|(\theta-1)$, in $\mathbb Z_2$.
Hensel's lemma shows that $x^3+11x-4=0$ has solutions $1,3$ mod $4$, and that both these give solutions in $\mathbb Z_2$. Thus, since the thing is a cubic, the third root is also in $\mathbb Z_2$, and (by looking at the constant term) is divisible by $4$, in fact. The solutions $\theta_1,\theta_2$ in $\mathbb Z_2$ congruent to $1,3$ mod $4$ both have the property that $2|(\theta_j-1)$, as desired.
True, this did not determine the minimal polynomial of $\theta(\theta-1)/2$.
