# What is the discriminant of this order $A=\mathbb{Z}[\alpha]\cap\mathbb{Z}[\alpha^{-1}]$?

Let $$f(x) = a_dx^d + a_{d-1}x^{d-1} + \ldots + a_1x + a_0$$ be a non-monic irreducible polynomial in $$\mathbb{Z}[x]$$ and call $$\alpha$$ be one of it's roots. Let $$\mathbb{K}=\mathbb{Q}[\alpha]$$ the field extension induced by $$f(x)$$ and let $$\mathcal{O}$$ be the corresponding ring of integers.

Then the ring $$A=\mathbb{Z}[\alpha]\cap\mathbb{Z}[\alpha^{-1}]$$ is a subring of $$\mathcal{O}$$ (so an order of $$\mathbb{K}$$), and it is generated by the following algebraic integers

\begin{align} \beta _ { 0 } \quad &: = a _ { d } \alpha ^ { d - 1 } + a _ { d - 1 } \alpha ^ { d - 2 } + \cdots + a _ { 3 } \alpha ^ { 2 } + a _ { 2 } \alpha + a _ { 1 }\\ \beta _ { 1 } \quad &: = a _ { d } \alpha ^ { d - 2 } + a _ { d - 1 } \alpha ^ { d - 3 } + \cdots + a _ { 3 } \alpha + a _ { 2 }\\ \beta _ { 2 } \quad &: = a _ { d } \alpha ^ { d - 3 } + a _ { d - 1 } \alpha ^ { d - 4 } + \cdots + a _ { 3 }\\ \vdots \\ \beta _ { d - 3 } \: &: = a _ { d } \alpha ^ { 2 } \:\:\:+ a _ { d - 1 } \alpha \:\:\:\:\:+ a _ { d - 2 }\\ \beta _ { d - 2 }\: &: = a _ { d } \alpha \:\:\:\:\:+ a _ { d - 1 } \end{align}

So we also have $$A=\mathbb{Z} + \sum_{i=0}^{d-2} \beta_i\mathbb{Z}$$. Another representation of the $$\beta_i$$'s is $$\beta_{-1}=0$$ and $$\beta_i = (\beta_{i-1} - a_i)\alpha^{-1}.$$

One can see that $$\omega_d = a_d\alpha$$ and $$\omega_{0}=a_0\alpha^{-1}$$ are algebraic integers, and show that $$\mathbb{Z}[\omega_d]\subseteq A$$ and $$\mathbb{Z}[\omega_0]\subseteq A$$. Furthermore if we call $$\Delta {\omega_d}$$ the discriminant of $$\mathbb{Z}[\omega_d]$$ = discriminant of $${a_d}^{d-1}f(\frac{x}{a_d})$$ the minimal polynomial of $$\omega_d$$, and similarly for $$\omega_0$$ who's minimal polynomial is $${a_0}^{-1}x^df(\frac{a_0}{x})$$, we have \begin{align} \Delta {\omega_d} &= {a_d}^{(d-1)(d-2)}\Delta f\\ \Delta {\omega_0} &= {a_0}^{(d-1)(d-2)}\Delta f. \end{align}

It follows that the discriminant of $$A$$ satisfies $$\Delta\mathcal{O} \mid \Delta A\mid\operatorname{lcm}(a_d,a_0)^{(d-1)(d-2)} \Delta f.$$

Can one deduce more on the properties of $$A$$? Does $$\Delta A$$ always divide $$\Delta f$$? Does $$\Delta \mathcal{O}$$ always divide $$\Delta f$$? Is $$\Delta A$$ equal to $$\Delta f$$? Thanks!

• Yes, but the intersection of $\mathbb{Z}[\alpha^{-1}]$ with $\mathbb{Z}[\alpha]$ is a subring of the ring of integers (at least this is what I read, alas with no proof). If anyone can link me some papers on non-maximal orders that show some properties of this ring I'd be grateful Apr 9, 2019 at 6:27
• Oh yeah! I was overly worried about the problems coming from $\Bbb{Z}[\alpha^{-1}]$ not being an order at all (think $\alpha=\sqrt2$ when $\Bbb{Z}[\alpha^{-1}]$ contains $\Bbb{Z}[1/2]$). Sorry about that. Hmm... Apr 9, 2019 at 6:32
• In case it is of any help: numerical experiment (in low degrees) suggests the answers are Yes, Yes, Yes. Mar 24, 2021 at 22:36
• Thanks, that's one step in the right direction. Mar 25, 2021 at 23:59
• Answering the question in @Kolja's first comment: See math.stackexchange.com/q/791689 for the proof. Mar 27, 2021 at 17:43

The discriminant is $$\Delta(f)$$.

By definition, the discriminant of an order is the determinant of the trace form. In this case, this is the determinant of the $$d\times d$$ matrix $$M$$ whose entries are given by the formula $$M_{ij} = \operatorname{tr} ( \beta_{i-1} \beta_{j-1})$$, taking $$\beta_{d-1} =1$$.

The entries $$M_{ij}$$ are polynomials in $$a_0,\dots, a_d$$. For example $$M_{dd} = \operatorname{tr}(1) =d,$$ $$M_{d,d-1} = \operatorname{tr} (a_d \alpha + a_{d-1} ) = - a_{d-1} + d a_{d-1} =(d-1) a_{d-1},$$ $$M_{d,d-2} = \operatorname{tr} ( a_d \alpha^2 + a_{d-1} \alpha + a_{d-2}) = \frac{a_{d-1}^2}{a_d} -2 a_{d-2} - \frac{ a_{d-1}^2}{a_d} + d a_{d-2} = (d-2) a_{d-2}$$ $$M_{d-1,d-1} =\operatorname{tr} (a_d^2 \alpha^2 + 2 a_d a_{d-1} \alpha + a_{d-1}^2) = a_{d-1}^2 - 2 a_{d-2} a_d - 2 a_{d-1}^2 + d a_{d-1}^2 = (d-1)a_{d-1}^2 - 2 a_d a_{d-2}$$

So the determinant is a polynomial in $$a_0,\dots, a_d$$.

Now your same argument shows that this polynomial divides $$a_0 ^{ (d-1) (d-2)} \Delta(f)$$ and $$a_d ^{ (d-1) (d-2)} \Delta(f)$$ in the ring of polynomials in $$a_0,\dots, a_d$$ over $$\mathbb Z$$. But in that ring, the only common factor of these polynomials is $$\Delta(f)$$ (because the ring has unique factorization into irreducibles), so this polynomial divides $$\Delta(f)$$. Then because $$\Delta(f)$$ is an irreducible polynomial, this polynomial is $$\pm \Delta(f)$$.