I have the following question,
Let $\theta $ be the root of qubic equation $\theta^3 + \theta + 1 = 0$. Let $ D = \mathbb{Z} + \mathbb{Z} \theta + \mathbb{Z}\theta^2 = \{a + b\theta + c\theta^2 \, : a,b,c \in \mathbb{Z} \}$. Show that $D$ is an integral domain.
It is easy to see that $D$ is a commutative ring with unity, so it is enough to show that there are no zero divisors. Let $a + b\theta + c\theta^2 $ and $x+ y\theta + z\theta^2 \, $ be in $D$. We assume that $a + b\theta + c\theta^2 \ne 0$ and we need to show that $x = y = z = 0$ when $(a + b\theta + c\theta^2 ) \cdot (x + y\theta + z\theta^2) = 0$.
Expanding this we get the following system of equation
$$ \left\{ \begin{aligned} ax- cy -bz &= 0 \\ bx+ (a-c)y + (-c-b)z &= 0\\ c x+b y+ (a-c)z &= 0 \end{aligned} \right. $$ To show $x = y = z = 0$ we need to show that the determinant of the matrix \begin{pmatrix} a & -c & -b \\ b & a -c & -c-b \\ c & b & a -c \end{pmatrix} is non zero. Does there exist integers $a,b$ and $c$ such that at least one of them in non zero and the determinant is $0$? I suspect that the determinant is always non zero but I am not able to prove it. If any two of $a,b,c$ are zero then we can easily see that the determinant is non zero but how to do in general?
Let $f(x)=x^3+x+1$. By the rational root test, the only possible rational roots of $f(x)$ are $\pm 1$, which are not roots. Therefore $f(x)$ is irreducible in $\mathbb{Q}[X]$. Since $f(x)$ is also primitive in $\mathbb{Z}[X]$ it follows that $f(x)$ is irreducible in $\mathbb{Z}[X]$. Since $\mathbb{Z}[X]$ is a UFD, $f(x)$ is a prime element so $(f)$ (the ideal generated by $f(x)$ in $\mathbb{Z}[X]$) is a prime ideal, so the quotient $\mathbb{Z}[X]/(f)$ is an integral domain. Elementary enough?