On the index of an algebraic integer and beyond (Q2301457)

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On the index of an algebraic integer and beyond
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    On the index of an algebraic integer and beyond (English)
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    24 February 2020
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    The authors prove the following theorem. Theorem. Let \(K=\mathbb{Q}(\theta)\) be an algebraic number field with \(\theta\) in the ring \(A_K\) of algebraic integers of \(K\) having minimal polynomial \(f(x)\) over \(\mathbb{Q}\). Let \(p\) be a prime number and \(i_{p}(f)\) denote the highest power of \(p\) dividing the index \([A_K : \mathbb{Z}[\theta]]\). Suppose that \(\overline{f}(x) = \overline{\phi}_1(x)^{e_1} \dotsb \overline{\phi}_r(x)^{e_r}\) is the factorization of \(\overline{f}(x)\) obtained by replacing each coefficient of \(f(x)\) modulo \(p\) into a product of powers of distinct irreducible polynomials over \(\mathbb{Z}/p\mathbb{Z}\) with \(\phi_i(x) \in \mathbb{Z}[x]\) monic. Let \(t_i \geq 0\) denote the highest power of \(\overline{\phi}_i(x)\) dividing the polynomial \(N(x)\), where \(N(x)\) belonging to \(\mathbb{Z}[x]\) is defined by \(f(x) = \prod_{i=1}^{r} \phi_i(x)^{e_i} + p^lN(x)\), \(l \geq 1\), \(\overline{N}(x) \neq \overline{0}\). Let \(u_i\) stand for the non negative integer given by \[ u_i = \begin{cases} \frac{(e_i-1)l +\operatorname{gcd}(e_i,\, l+1)-1}{2}, & \text{if } t_i>\frac{e_i}{l+1}\\ \max\left\{lt_i, \frac{(e_i-1)(l-1) +\operatorname{gcd}(e_i,l)-1}{2}\right\}, & \text{otherwise}. \end{cases} \] Then the following holds: \begin{itemize} \item[(i)] \(i_p(f) \geq \sum_{i=1}^{r}u_i\,\operatorname{deg}\phi_i(x)\). \item[(ii)] If \(\operatorname{gcd}(e_j,l)=1\) and \(t_j=0\) whenever \(e_j>1\), then equality holds in (i), i.e., \(i_p(f) =\sum_{i=1}^{r} \frac{(e_i-1)(l-1)}{2} \operatorname{deg}\phi_i(x)\). \end{itemize} This result yields a Dedekind criterion which gives necessary and sufficient conditions to be satisfied by \(f(x)\) for \(i_p(f)\) to be zero. Moreover, the authors give a family of irreducible polynomials for which the assumption given in (ii) holds and the \(i_p\) is explicitly given. On the other hand, an example is given showing that the assumption given in (ii) is not necessary for the equality to hold in (i). Finally, a more general version of the above theorem is proved with \(\mathbb{Z}\) replaced by any Dedekind domain.
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    Dedekind domains
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    rings of algebraic integers
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    valued fields
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