 Solution

The ideal ${\mathfrak{p}}_2=(2,\alpha )$ has ideal norm two. Its square is contained in $(2)$ and has norm four, so $(2)={\mathfrak{p}}_2^2$. A generator would have field norm $u^2+26v^2=2$, impossible for integers $u,v$. Thus its ideal class has order two.

Modulo three, $X^2+26=(X-1)(X+1)$, giving $(3)={\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$ with ${\mathfrak{p}}_3=(3,1-\alpha )$. A generator of ${\mathfrak{p}}_3$ would have norm three, again impossible. If ${\mathfrak{p}}_3^2$ were principal, its generator would have norm nine. The only possibilities are $\pm 3$, whose ideal is ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$, not ${\mathfrak{p}}_3^2$.

Finally $N(1-\alpha )=27$. This element lies in ${\mathfrak{p}}_3$ but not ${\overline{\mathfrak{p}}}_3$, because $\alpha \equiv -1$ in the latter quotient. Unique unique factorization of ideals in a number field and norms therefore give