I have been brushing up on cubic number fields. Specifically, let $s$ be a root of the polynomial $x^3 + x^2 + 3x + 17$, and consider $K = \mathbb{Q}(s)$; we have $\mathcal{O}_K = \mathbb{Z}[s]$, and

$$2\mathcal{O}_K = (2,s+1)^3,$$

this by the theorem of Dedekind. My favourite computer algebra system tells me that $(2,s+1)$ is not principal, but I would like to justify this myself.

How may I show that this ideal is not principal?

I can see how my question is equivalent to asking why the ring $\mathbb{Z}[s]$ has no element of norm $\pm 2$, or why 2 is irreducible in this ring. In principle I could answer this by writing down the norm of a general element $a + bs + cs^2$ for $a,b,c \in \mathbb{Z}$; but this is something that I really do not want to do. Rather I'm looking for a way to answer the question whilst keeping my hands clean.


In general, principal ideal tests require knowledge of the unit group, which means that you cannot expect to keep your hands clean. Quite likely even writing down the norm form is not going to help you (in the sense that you should expect a contradiction modulo some integer).

There are exceptions when genus theory is strong enough to tell you something; this requires completely ramified prime ideals. Assume e.g. that $(p) = P^3$ and write $(2,s+1) = (\alpha)$. Then $\alpha \equiv a \bmod P$ for some integer $a$, hence $a^3 \equiv \pm 2 \bmod p$ in the integers. If you're lucky, $2$ is not a cube modulo $p$, and then the ideal is not principal.

  • $\begingroup$ But how does it follow that $a^3 \equiv \pm2$ mod $p$? We could have that $\alpha^3 = 2u$, with $u$ a unit right? Thanks! $\endgroup$ – Willem Beek Dec 11 '15 at 20:27

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