Ring of integers of cubic number field

Solution 1:

Here a totally different (but quite simple) approach to the problem:

The submodule of $\mathcal{O}_K$ generated by $(1,\alpha,\alpha^2)$ is clearly $\mathbb{Z}[\alpha]$. You have already shown that $$\text{disc}(1,\alpha,\alpha^2)=-104$$ Now, note that the following formula is true (it can be found in most introductory textbook on algebraic number theory): $$\text{disc}(1,\alpha,\alpha^2)=[\mathcal{O}_K:\mathbb{Z}[\alpha]]^2\text{disc}(\mathcal{O}_K)$$ It follows that $[\mathcal{O}_K:\mathbb{Z}[\alpha]]$ equals $1$ or $2$. In the latter case, it follows that $\text{disc}(\mathcal{O}_K)=-26$. But this cannot be true, since $\text{disc}(\mathcal{O}_K) \equiv 0$ or $1 \text{ mod } 4$ (this fact is known as "Stickelberger's theorem on discriminants" which is also contained in a lot of textbooks on the topic). Hence $[\mathcal{O}_K:\mathbb{Z}[\alpha]]=1$ and therefore $\mathcal{O}_K=\mathbb{Z}[\alpha]$.

Solution 2:

Here’s another argument, much, much more advanced. You’re worrying about two rings, $\Bbb Z[\alpha]$ and $\mathscr O$, the integers of $\Bbb Q(\alpha)$. They’re equal if and only if the smaller ring is integrally closed. Now I’ll argue that $\Bbb Z[\alpha]$ is indeed integrally closed. And this is the case if and only if it’s true locally at every prime $p$ of $\Bbb Z$. The only difficult primes are those where there’s ramification, and these are $2$ and $13$, ’cause the discriminant ideal is $(8\cdot13)$. The latter is not a problem, as you have observed, because the discriminant is divisible only by $13$ to the first power. So let’s localize at $2$ and see what’s going on.

We might as well go all the way to the completion, which means looking at the ring $\Bbb Z[\alpha]\otimes_{\Bbb Z}\Bbb Z_2$, which we want to see to be integrally closed (in a sense that will soon be clear). Now, I’ll write the ring $\Bbb Z[\alpha]=\Bbb Z[\alpha-1]$, where the defining polynomial of $\alpha-1$ is $g(x)=f(x+1)=x^3+3x^2+2x-2$. To look at the tensored-with-$\Bbb Z_p$ ring is to look at $\Bbb Z_2[x]/(x^3+3x^2+2x-2)$. But just look at the Newton Polygon of this polynomial: it has vertices at $(0,1)$, $(2,0)$, and $(3,0)$. That means that the polynomial factors into a linear and an Eisenstein quadratic, and this in turn means that $\Bbb Z_2[x]/(x^3+3x^2+2x-2)\cong\Bbb Z_2\oplus\Bbb Z_2[x]/(q(x))$, where $q(x)$ is that Eisenstein quadratic. Are you with me? Our ring is no longer a domain, but the two pieces are certainly integrally closed, since over a local ring, a root of an Eisenstein polynomial generates the whole ring of integers in the fraction field. So $\Bbb Z[\alpha]$ is integrally closed, and thus equal to the full ring of integers of $\Bbb Q(\alpha)$.

Solution 3:

Suppose $w \in \Bbb Z[\alpha]$ is not a multiple of $2$ and $w/2$ is an algebraic integer.

Consider the ideal $I = (2,w)$. We have $I^2 = (4,2w,w^2)$ and $I^3 = (8,4w,2w^2,w^3)$. However, $w/2$ being an algebraic integer of degree $\le 3$, we get that $w^3$ is a combination of $8,4w,2w^2$ with integer coefficients, and so $I^3 = (8,4w,2w^2) = 2I^2$, and so on, which shows that unique factorisation of ideals fails : $(2)I^2 = I.I^2$, but $(2) \neq I$, so $I^2$ is not "cancellable", and neither is $I$.

To show that an ideal $I$ is cancellable it is enough to show it is a factor of cancellable ideal, for example any principal ideal, say, $(2)$ : If $(2)= IJ$ and $AI = BI$ then $AIJ = BIJ$, so $2A = 2B$ and thus $A=B$.

Let $I$ be any ideal containing $(2)$. Since $\Bbb F_2[X]$ is a PID, $\Bbb Z[\alpha]/I$ has to be isomorphic (via the evaluation map) to $\Bbb F_2[X]/Q(X)$ where $Q(X)$ is a divisor of $X^3+X$ (the reduction modulo $2$ of $X^3-X-2$), and then $I = I_Q = (2,Q(\alpha))$.

The connection between the divisor lattice of $X^3-X-2$ mod $2$ and the ideals containing $(2)$ goes further :

If $2 \in I_Q I_R$, then $I_QI_R = (2,4,2Q(\alpha),2R(\alpha),Q(\alpha)R(\alpha)) = (2,Q(\alpha)R(\alpha)) = I_{QR}$.
In particular, if $Q$ and $R$ are coprime, then $1$ is a linear combination of $Q,R$ and $2$, and so $2 \in I_Q I_R$.

So if we show that $(2)$ factors properly as $I_{X}I_{X+1}^2$ then we are done, because this would show that $I_X$ and $I_{X+1}$ are cancellable and their products give all the ideals containing $(2)$.

The only way this can possibly go wrong is if $2 \notin I_{X+1}^2$, so let us do the computation :

$I_{X+1}^2 = (2,\alpha+1)^2 = (4,2\alpha+2,\alpha^2+2\alpha+1)$.
It contains $\alpha.(\alpha^2+2\alpha+1) = \alpha^3+2\alpha^2+\alpha = 2\alpha^2+2\alpha+2$, so it also contains $2(\alpha^2+2\alpha+1)-(2\alpha^2+2\alpha+2) = 2\alpha$, and then $(2\alpha+2) - 2\alpha = 2$.

Hence $I_{X+1}^2 = (2,\alpha^2+1) = I_{(X+1)^2}$, which is all we needed.