Let $R$ be a Dedekind domain with fraction field $K$, and let $\mathfrak p$ be a maximal ideal of $R$. Let $f\in R[x]$ be a monic, separable polynomial and let $N/K$ be a splitting field of $f$.

A well known theorem of Dedekind in the case $R=\mathbb Z$ states that if $\bar f(x)$ factors as $\bar g_1(x)\cdots \bar g_m(x)$, then the decomposition group of any prime of $N$ lying over $\mathfrak p$ can be generated by an element whose permutation action on the roots of $f(x)$ is a product of cycles of length $(\deg g_i)$ for every $i$.

My first question is whether there are any generalizations of this result to other Dedekind domains. I'm particularly interested in the case where $R$ is the polynomial ring $\mathbb Q[T]$. Of course Dedekind's result uses the fact that $\mathbb Z$ has finite quotients, which won't hold more generally, but is there anything that can be said in the general case, or in the case $R=\mathbb Q[T]$?

My second question concerns a complement to Dedekind's theorem due to Beckmann (Journal of Algebra 164, pp 415-429), also in the case $R=\mathbb Z$. Assuming that $f(x)=g_1(x)^{e_1}\cdots g_m(x)^{e_m}+p\cdot h(x)$, where $\bar h(x)$ is not divisible by any $\bar g_i(x)$ and $p$ does not divide any $e_i$, Beckmann shows that the inertia group of any prime of $N$ above $p$ is generated by an element whose permutation action on the roots is a product of $(\deg g_i)$ $e_i$-cycles for each $i$.

Can Beckmann's result be generalized to other Dedekind domains $R$? And is there in the literature a proof of this result based on standard theory of decomposition and inertia groups?

  • $\begingroup$ Dedekind's theorem relies on the residue fields of the Dedekind domain being finite, and thus having cyclic Galois groups. It extends to a base ring equal to the algebraic integers of any number field, whose residue fields are also finite fields. But with $\mathbf Q[T]$ as the base ring, its residue fields are number fields and decomposition groups need not be cyclic. What kind of substitute would you want? $\endgroup$ – KConrad Feb 22 '17 at 4:05
  • $\begingroup$ I am suspicious about the way you describe Beckmann's result. Consider the polynomial $f(x) = x^6 - 35x^4 + 3x^2 - 225$. It is irreducible over $\mathbf Q$ and $f(x) \equiv x^4(x^2+1) \bmod 3$, so in your notation $g_1(x) = x$, $e_1 = 4$, $g_2(x) = x^2+1$, and $e_2 = 1$. You are saying that in the splitting field $N$ of $f(x)$ over $\mathbf Q$ that any prime lying over 3 has an inertia group generated by an element whose permutation action on the roots of $f(x)$ "includes" (do you mean equals?) a product of one 4-cycle and two 1-cycles. That means ${\rm Gal}(N/\mathbf Q)$ contains a [contd] $\endgroup$ – KConrad Feb 22 '17 at 4:10
  • $\begingroup$ permutation of the roots that is a 4-cycle on four roots and fixes the other two roots, which is an odd permutation of the roots. But the discriminant of the polynomial is a perfect square, so the action of ${\rm Gal}(N/\mathbf Q)$ on the roots is by even permutations only: the Galois group contains no 4-cycle on the roots. So something is wrong. I got this example from an answer I posted on MO a number of years ago: mathoverflow.net/questions/21247/… $\endgroup$ – KConrad Feb 22 '17 at 4:14
  • $\begingroup$ @KConrad: I omitted an additional condition in Beckmann's result: writing $f(x)=g_1(x)^{e_1}\cdots g_m(x)^{e_m}+p\cdot h(x)$ in $\mathbb Z[x]$, the polynomial $\bar h(x)$ should not be divisible by any of the $\bar g_i(x)$. This condition is not met in your example because $h(x)=-12x^4 + x^2 - 75$, so $\bar h(x)=x^2$ is divisible by $\bar g_1(x)=x$. $\endgroup$ – 352506 Feb 22 '17 at 12:28
  • $\begingroup$ Unlike Dedekind's theorem, Beckmann's result has a chance of being generalized almost verbatim to the case $R=\mathbb Q[T]$, since the inertia groups would still be cyclic. I was wondering if this has been worked out somewhere. $\endgroup$ – 352506 Feb 22 '17 at 12:31

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