Ring of integers of $K=\Bbb Q[u]$ where $u=\sqrt[3]{p^2q}$

Question

Let $p,q$ be distinct prime numbers $\ge 5$ such that $pq^2 \not\equiv 1\mod9$. Let $K=\Bbb Q[u]$ where $u=\sqrt[3]{p^2q}$, and $A$ be the ring of integers of $K$. I have shown that $u,v=pqu^{-1}\in A$, $D(1,u,v)=-27p^2q^2$, $B=\Bbb Z+\Bbb Zu+\Bbb Zv$ is a subring of $A$. Let $Q$ is a prime ideal of $A$ containing $Aq$, then $Aq=Q^3$, $B\cap Q=Bu+Bv+Bq$, $A/Q=B/B\cap Q$, so $A=B+Q$.

Now how can I deduce that $A=B+Aq$? Also how can I show that $3A$ is the cube of a prime ideal of $A$, and $A=B+3A$?

Answer 1

If you're trying to find the ring of integers, then you already have everything you need. I don't know what you're trying to show it is exactly, but $B=\mathbb{Z}+\mathbb{Z}u+\mathbb{Z}v$ is the ring of integers in this case.

Let $f(x)=x^3-p^2q$ be the minimal polynomial of $u=\sqrt[3]{p^2q}$, and let $K=\mathbb{Q}(u)$. A first guess for the discriminant of $K$ is $\Delta(u)=-27(p^2q)^2=-3^3p^4q^2$. Hence if we let $A=\mathcal{O}_K$, the ring of integers, the only rational primes that could divide the index $[A:\mathbb{Z}[u]]$ are $3$, $p$, and $q$. Since $p,q\ge 5$, we know that $p,q\ne 3$.

Modulo $3$. We have a couple of cases. We know that $p^2q\equiv 0\text{ (mod }3)$ is impossible. The squares modulo $9$ are $1$, $4$, and $7$, and the values that $p$ could take modulo $9$ are $1$, $2$, $4$, $5$, $7$, $8$. Hence the possible values of $p^2q$ modulo $9$ are $1$ (but this has been ruled out by assumption), $2$, $4$, $5$, $7$, and $8$. So $\overline{f}(x)=x^3+1=(x+1)^3$ or $\overline{f}(x)=x^3+2=(x+2)^3$. Let's take each of these cases in turn.

• If $p^2q\equiv 4\text{ or }7\text{ (mod }9)$, then $\overline{f}(x)=(x+1)^3$, and $f(-1)=-1-p^2q$. The restrictions of $p^2q$ modulo $9$ imply that $9\nmid f(-1)$, and the prime $r_1=(3,u+1)$ above $(3)$ is invertible.
• If $p^2q\equiv 2,5\text{ or }8\text{ (mod }9)$, then $\overline{f}(x)=(x+2)^3$, and $f(-2)=-8-p^2q$. The restrictions of $p^2q$ modulo $9$ again imply that $9\nmid f(-2)$, and the prime $r_2=(3,u+2)$ above $(3)$ is invertible.

Modulo $p$. We have that $\overline{f}(x)=x^3$, and that $p^2\mid f(0)=-p^2q$, so the prime $\mathfrak{p}=(p,u)$ above $(p)$ is not invertible. This gives rise to the element $v=\frac{1}{p}u^2=\sqrt[3]{pq^2}\in A\setminus\mathbb{Z}[u]$.

Modulo $q$. We have that $\overline{f}(x)=x^3$, and that $q^2\nmid f(0)=-p^2q$, so the prime $\mathfrak{q}=(q,u)$ above $(q)$ is invertible.

Now let $g(x)=x^3-pq^2$ be the minimal polynomial of $v$. We have that $\Delta(v)=-27(pq^2)^2=-3^3p^2q^4$, and so the power of $p$ in the discriminant is reduced.

Modulo $p$. We have that $\overline{g}(x)=x^3$, and that $p^2\nmid g(0)=-pq^2$, so the prime $\mathfrak{p}'=(p,v)$ above $(p)$ is invertible.

At this point, we're done. We find that $A=\mathbb{Z}[u,v]$.

---

Without the assumption $p^2q\not\equiv 1\text{ (mod }9)$, the ring of integers does get a little different. For example $u=\sqrt[3]{11^2\cdot 7}$ has ring of integers $\mathbb{Z}[u,\frac{1}{3}(2+2u+\frac{1}{11}u^2)]$.