I want to show that the ring of integers of the cubic number field $K = \mathbb Q(\alpha)$, where $\alpha$ is a root of $f = X^3 - X - 2$, is equal to $\mathbb Z[\alpha]$.

$(1, \alpha, \alpha^2)$ forms a $\mathbb Q$-basis of $K$ consisting of integers. I know $\mathbb Z[\alpha] \subseteq \mathcal O_K$. To show equality, I'm guessing the following result is useful:

Let $K$ be a number field of degree $n$, $(\alpha_1, \dots, \alpha_n)$ a $\mathbb Q$-basis of $K$ consisting of integers. Suppose there exist $m \in \mathbb N, m > 2$ and $k_1, \dots, k_n \in \mathbb Z$ with $\gcd(m, k_1, \dots, k_n) = 1$ such that $$\frac{k_1 \alpha_1 + \dots + k_n \alpha_n}{m} \in \mathcal O_K$$ then $m^2 \,\big\vert\, \mathrm{disc}(\alpha_1, \dots, \alpha_n)$.

I have $\mathrm{disc}(\alpha_1, \dots, \alpha_n) = -104$, so $m = 2$, hence I want to show that I cannot have $$\frac{a + b \alpha + c \alpha^2}{2} \in \mathcal O_K$$ for integers $a, b, c \in \mathbb Z$ of which at least one is odd. But I'm getting stuck here. How do I show these aren't integral elements? Can anyone give me a hint on how to proceed?

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    $\begingroup$ Brute force checking each of the seven cases might be easiest here, although that depends on how quick you are at computation. $\endgroup$
    – user14972
    Sep 21, 2015 at 17:14
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    $\begingroup$ As pointed out by Hurkyl, you need to show that the seven numbers obtained by setting 0 or 1 for each of $a,b,c$ is not integral. For example, $\alpha/2$ is not integral because it is a root of $4X^3-X-1$, which is irreducible. $\endgroup$
    – Aravind
    Sep 21, 2015 at 17:42
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    $\begingroup$ I'm almost certain that you only need to check the equality $(2) = (2,\alpha)(2,\alpha+1)^2$ as ideals of $\Bbb Z[\alpha]$. Typically, when you lack some elements, the ideals "grow too fast" when you multiply them. $\endgroup$
    – mercio
    Sep 21, 2015 at 17:46
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    $\begingroup$ @mercio Can you elaborate on that? I don't really see where you get the equality from or how it establishes the result I'm trying to prove. $\endgroup$
    – Mark
    Sep 21, 2015 at 19:32
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    $\begingroup$ It can’t make any difference to the computations, but it seems to me that $|\text{disc}|=104$. $\endgroup$
    – Lubin
    Sep 22, 2015 at 21:32

4 Answers 4


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 $\mathcal{O}_K=\mathbb{Z}[\alpha]$.

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    $\begingroup$ Oh that's quite simple indeed. Thanks! $\endgroup$
    – Mark
    Sep 24, 2015 at 18:09
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    $\begingroup$ You are welcome. But note that Stickelberger will not always help you finding the ring of integers. There are cases where there are several options for $[\mathcal{O}_K:\mathbb{Z}[\alpha]]$ and they are all consistent with $\text{disc}(\mathcal{O}_K)\equiv 0,1\bmod 4$. So, no way to rule them out here. But Stickelberger is still a usefull fact to keep in mind since in some cases, it can lead to a quick determination of $\mathcal{O}_K$, as in your example. $\endgroup$
    – russoo
    Nov 9, 2023 at 18:31

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 $\Bbb Q(\alpha)$.

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    $\begingroup$ In any case like this, where you want to see that a candidate ring is or is not the full ring of integers of a number field, Localization Is Your Friend. For instance, you can use this method to see quickly that $\Bbb Z[\sqrt5]$ is not the ring of integers of $\Bbb Q(\sqrt5)$. $\endgroup$
    – Lubin
    Sep 23, 2015 at 3:01
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    $\begingroup$ Do you have a reference for the part about the roots of an Eisenstein polynomial generating the ring of integers in the fraction field? $\endgroup$
    – Steve D
    Aug 10, 2017 at 3:49
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    $\begingroup$ No reference, @SteveD, but here’s an argument. Dealing with a finite extension $k$ of $\Bbb Q_p$ with ring of integers $\mathscr O$ and prime element $\pi$ ; and with an algebraic element $\alpha$ which is root of an $\mathscr O$-Eisenstein polynomial, we see that $v(\alpha)=v(\pi)/n$, where $v$ is the additive valuation and $n$ is the degree. (You see that the ramification index is also $n$.) Now, in a totally ramified extension with prime element $\Pi$, the integers there are $\mathscr O[\Pi]$. Here, you can take $\Pi=\alpha$. $\endgroup$
    – Lubin
    Aug 10, 2017 at 14:52

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.

  • $\begingroup$ the sad thing is that this does not work as is when you start from something not of the form $\Bbb Z[\alpha]$ $\endgroup$
    – mercio
    Sep 22, 2015 at 0:07
  • $\begingroup$ Good stuff! Since my learning curve is what it is nowadays let me attempt to summarize/generalize. Assume that $\Bbb{Z}[\alpha]\subseteq \mathcal{O}_K$, $[K:\Bbb{Q}]<\infty$. Let $p$ be a rational prime, and let $m(x)\in\Bbb{Z}[x]$ be the minimal polynomial of $\alpha$. Assume that in $\Bbb{F}_p[x]$ we have the factorization $$m(x)=\prod_{i=1}^k m_i(x)^{a_i},$$ where $m_i(x)\in\Bbb{F}_p[x]$ are irreducible. Let $I_i$ be the ideal of $\Bbb{Z}[\alpha]$ generated by $p$ and $m_i(\alpha)$. If for all $i$ we have that $p$ is contained in the ideal $I_i^{a_i}$, then $I_i$ is a cancellable ideal $\endgroup$ Sep 22, 2015 at 5:18
  • $\begingroup$ (cont'd) of the ring $\Bbb{Z}[\alpha]$. Because the ideal structure of $\Bbb{F}_p[x]/\langle m(x)\rangle$ is easily understood from $\Bbb{F}_p[x]$ being a PID, all the ideals of $\Bbb{Z}[\alpha]$ containing $p$ are cancellable. Consequently it is impossible to find an algebraic integer $w\in K$ such that $w\notin\Bbb{Z}[\alpha]$ but $pw\in\Bbb{Z}[\alpha]$. This is because such a number $w$ gives rise to a non-cancellable ideal $(p,w)$ of $\Bbb{Z}[\alpha]$. $\endgroup$ Sep 22, 2015 at 5:20
  • $\begingroup$ Thanks! I think this is quite a convenient method for checking whether $\mathbb Z[\alpha]$ is equal to the ring of integers. $\endgroup$
    – Mark
    Sep 24, 2015 at 18:13

Other answers are too advanced. Checking definitions is enough for this question.

We have $\mathrm{tr}\,\frac12(a+b\alpha+c\alpha)=\frac32a+c$, so $a$ is even, leaving $\frac12(b\alpha+c\alpha^2)$ to be considered. We have $\mathrm{N}(\frac12\alpha(b+c\alpha))=-\frac14(b^3-bc^2+2c^3)$, which is an integer iff $b,c$ are both even.


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