0

Let $A$ be a commutative integral domain over a field $k$ of characteristic zero. Assume that the invertible elements of $A$ consists of $k^*$ only. Let $a \in A-\{0\}$ (it is not known whether $a$ belongs to $k^*$ or not), and let $b$ be an algebraic element over $A$ (taken from an algebraic closure of $A$. $b$ is not known to be integral over $a$, just algebraic). Let $f$ be the minimal polynomial of $b$ over $A$, $f(t)=a_nt^n+a_{n-1}t^{n-1}+\cdots+a_1t+a_0$ ($f$ is irreducible).

Define $\hat{B}:=A[b][a^{-1}]$.

I am curious to know if the following question has an easy answer (which I have missed):

When the invertible elements of $\hat{B}$ are contained in $Q(A)$, the field of fractions of $A$?

Remarks: (1) I wish to further assume that $B:=A[b]$ is an integral domain (of course, $Q(A) \subseteq Q(B)$. Further assume that $Q(A) \subsetneq Q(B)$.

(2) If it happens that $a_0=a$, then $0=f(b)=a_nb^n+a_{n-1}b^{n-1}+\cdots+a_1b+a$, so $b(a_nb^{n-1}+a_{n-1}b^{n-2}+\cdots+a_1)a^{-1} =(a_nb^n+a_{n-1}b^{n-1}+\cdots+a_1b)a^{-1}=-aa^{-1}=-1$, which means that $b$ is invertible in $\hat{B}$. In this case the invertible elements of $\hat{B}$ are not contained in $Q(A)$, since if they were contained in $Q(A)$, then $b \in Q(A)$, and then $Q(A)=Q(B)$, which contradicts our assumption that $Q(A) \subsetneq Q(B)$.

(3) If $\tilde{B} \ni u=\sum a_{ij}b^ia^{-j}$, $a_{ij} \in A$, is invertible in $\tilde{B}$, then there exists $\tilde{B} \ni v=\sum a'_{ij}b^ia^{-j}$, $a'_{ij} \in A$, such that $uv=1$. Then $(\sum a_{ij}b^ia^{-j})(\sum a'_{ij}b^ia^{-j})=1$, and then we get some polynomial $g$ over $A$, having coefficients in $k[\{a_{ij}\},a]$, such that $g(b)=0$, so $f$ divides $g$. But I am not sure how to continue.

user237522
  • 7,263
  • Apparently there is no relation between $a$ and $b$; you are just adjoining an algebraic element over (the field of fractions of) $A$, and then independently the inverse of another element of $A$? – Marc van Leeuwen May 07 '17 at 12:27
  • Yes, this is exactly what I meant. Please, do you see any easy answer to my question? I am afraid that there is some connection between $a$ and $b$, for example, if $a$ is the last coefficient of the minimal polynomial of $b$ over $A$, then $b$ is invertibe in $\hat{B}$, as I mentioned in remark (2) above. But otherwise, is it true that the invertible elements of $\hat{B}$ are contained in $Q(A)$? – user237522 May 07 '17 at 12:58
  • More precisely, is this (=$a$ is the last coefficient of the minimal polynomial of $b$ over $A$) the only case in which $b$ becomes invertible in $\hat{B}$? If $b$ does not have a minimal polynomial of such form, then, is it true that $b$ is necessarily not invertible in $\hat{B}$, and then this guarantees that the invertible elements of $\hat{B}$ are contained in $Q(A)$? (and probably they are of the form $a^j$, $j \in \mathbb{Z}$?). – user237522 May 07 '17 at 13:13

0 Answers0