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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!

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  • $\begingroup$ 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 $\endgroup$
    – Kolja
    Apr 9, 2019 at 6:27
  • $\begingroup$ 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... $\endgroup$ Apr 9, 2019 at 6:32
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    $\begingroup$ In case it is of any help: numerical experiment (in low degrees) suggests the answers are Yes, Yes, Yes. $\endgroup$ Mar 24, 2021 at 22:36
  • $\begingroup$ Thanks, that's one step in the right direction. $\endgroup$
    – Kolja
    Mar 25, 2021 at 23:59
  • $\begingroup$ Answering the question in @Kolja's first comment: See math.stackexchange.com/q/791689 for the proof. $\endgroup$ Mar 27, 2021 at 17:43

1 Answer 1

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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 $\pm \Delta(f)$.

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