I'm trying to show (although I don't know if the statement is even correct) that the ideal $\mathfrak{p}_2$ is not principal, where
$$\mathfrak{p}_2:=(2,x+1) \text{ in the ring of integers } \mathfrak{o}_K=\Bbb{Z}[x]/(x^3-x^2+x+1),$$
where $K$ is the field generated over $\Bbb{Q}$ by a root of the polynomial $X^3-X^2+X+1$.
I supposed the contrary: if $(2,x+1) = (\alpha)$ for some $\alpha \in \mathfrak{o}_K$, then $\alpha \mid 2$ and $\alpha \mid x+1$, so there exists $\beta, \gamma \in \mathfrak{o}_K$ such that $$2 = \alpha \cdot \beta;$$ $$x+1 = \alpha \cdot \gamma.$$
Taking the norms on both side we get that $N(\alpha)\mid 2$, since $N(2)=2^3=8$. I hoped i would have got a contradiction, but when I evaluated the norm of $x+1$ i got $N(x+1)=2$ since this value is the determinant of the matrix $$\begin{pmatrix}1&0&-1\\1&1&-1\\0&1&2 \end{pmatrix},$$ which is the matrix of the linear map which multiplies all the elements of the basis of $\mathfrak{o}_K$ by $x+1$. So no contradiction as desired (or better, not yet).
Other ideas I had did not lead to any results, unfortunately. Could someone kindly give me a hint?
Hint $\ $ You have that $\,(2,x\!+\!1)\supseteq (x\!+\!1)\,$ both have norm $= 2\,$ hence $\,\ldots$
Or, directly $\ f(-1) = -2\,$ so $\,f = (x\!+\!1)g - 2,\,$ so $\,{\rm mod}\ f\!:\ (x\!+\!1)g\equiv 2,\,$ i.e. $\,x\!+\!1\mid 2\ \ $