Motivation/Setup: Suppose $R$ is a domain. I'd want to "adjoin $p$th roots of an element $a \in R$." where $p$ is a prime (which i am mostly interested in $p$-adic) This seems to be a cmomon operationo in $p$-adic goemetry. See the rings of interst below.
Goal: I would like to express these objects as a quotient of polynomial.
- Let $K$ be the algebraic closure of fraction field of $R$, and make a choice of $a^{1/p}$ for $a \ inR$
- Suppose $a^{1/p} \notin R$.
- There is a map
$$ R[x]/(x^{p}-a) \rightarrow R[a^{1/p}] \hookrightarrow K$$
$$x \mapsto a^{1/p}$$
Question: Are there simple criteria to test is if first map is iso? For instance, i'd like to adress the three cases below. I could explain 1. in example.
Examples of interest:
-
$R=\Bbb F_p((t)), a=t$. This is adressed below combining the two observations below in my thoughts. In otherwords, I bellieve I have proven :
$$F_p((t))[t^{1/p}] \simeq F_p((t))[y]/(y^p-t)$$ -
$R=\Bbb Z_p, a=p$. I would like to adjoint $p$th roots of unit as
$$\Bbb Z_p[\xi_p] \simeq \Bbb Z_p[y]/(y^{p-1}+\cdots +1)$$
- $R=S[x]$, $a=x$, where $S$ is some domain like $\Bbb Z, \Bbb Z_p, \Bbb Q_p$. I would like to state something as
$$\Bbb Z_p[x][x^{1/p}] \simeq \Bbb Z_p[x][y]/(y^p-x)$$
Thoughts:
The map is clearly surjectiev. So we are down to the case of innjectivitiy.
Let $I:= \ker(R[t] \rightarrow R[a^{1/p}])$. This is a prime ideal.
Condition 1: When all prmies of $R[t]$ are principal/$R$ is a field These are equivalent condition.
Condition 2: $R$ is of charateristic $p$ The question is equivalent to finding minimal polynomial of $a^{1/p}$ over $R$. Over $K$, $x^p-a=(x- a^{1/p})^p$. Thus, if in $R[t]$, it is reducible, we must have $a^{k/p} \in R$ for some $1 \le k <p$, and $(k,p)=1$ would imply $$(a^{lk/p}a^b)^p=a$$ for some integers $b,l\in \Bbb Z$, this shows we have found a $p$th root.
Best Answer
For $p$ prime $x^p-a\in K[x]$ is irreducible iff it has no root in $K$.
Let $b$ be a root in the algebraic closure and $f$ its minimal polynomial, of degree $m$.
In characteristic $p$: $x^p-a=(x-b)^p$ so $f=(x-b)^m, f(0)=(-b)^m\in K$, if $m<p$ then $b\in K$.
In characteristic not $p$:
$f=\prod_{j=1}^m (x-\zeta_p^{c_j} b)$ so $f(0)=(-1)^m \zeta_p^r b^m\in K$. If $m<p$ then $\zeta_p^{rs} b\in K$ where $sm=1\bmod p$.
And hence $x^p-a$ has a root $\zeta_p^{rs} b\in K$.
Conversely if $x^p-a$ has no root in $K$ then $m=p$ ie. $x^p-a$ is irreducible and for any subring $R\subset K$ containing $a$, $R[b]\cong R[x]/(x^p-a)$.