Consider $f(x)=x^5-21$, $z\in \mathbb{C}$ such that $f(z)=0$, and $K=\mathbb{Q}(z)$. Consider as well that $w:=z-2$. I want to find a $\mathbb{Z}$-basis $(v_1,v_2,v_3,v_4,v_5)$ of the ring of integers of $K$, $O_K$, such that $(v_1,v_2,v_3,v_4,N_{K/\mathbb{Q}}(w)v_5)$ is a $\mathbb{Z}$-basis of the ideal $(w)$.
My thoughts: I’ve found that the absolute discriminant of the field $K$, $\Delta_K$, is in this case equal to the discriminant of $f$:
$$\Delta_K=\operatorname{Disc}(f(x))=3^4 7^4 5^5.$$
I’ve also found that $\mathbb{Z}[z]=\mathbb{Z}[w]$, that $N_{K/\mathbb{Q}}(w)=11$ (not sure if that’s useful), and that $$\operatorname{Disc}(f(x))=\operatorname{Disc}(f(x+2)).$$
I’m not really sure how to continue from there, any help would be appreciated.
Since $\Delta_K=disc(f)$, it follows that $(1,z,\ldots,z^4)$ is a $\mathbb{Z}$-basis of $O_K$.
Now, $(w,wz,\ldots,wz^4)$ is a $\mathbb{Z}$-basis of $(w)=wO_K$. The matrix of this basis of $(w)$ in the basis of $O_K$ above is $C=\pmatrix{-2& 0&0&0&21\cr 1& -2& 0 & 0& 0\cr 0& 1&-2&0&0\cr 0&0&1&-2&0\cr 0&0&0&1&-2}$ Applying the usual algorithm to get the Smith normal form of $C$, we get $UCV=diag(1,1,1,1,11)$, where $V\in GL_4(\mathbb{Z})$ is some matrix we don't care about and $U=\pmatrix{0&1&0&0&0\cr 0&0&1&0&0\cr 0&0&0&1&0\cr 0&0&0&0&1\cr -1 & -2 & -4 & -8 & -16}$.
Since $U^{-1}=\pmatrix{-2& -4&-8&-16&-1\cr 1 & 0 & 0 & 0 & 0\cr 0& 1&0&0&0\cr 0&0&1&0&0\cr 0&0&0&1&0}$,
a basis fulfilling the desired property is given by $(-2+z,-4+z^2,-8+z^3,-16+z^4,-1)$.