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Let $R \colon= K[[X]]$ be a formal power series ring over a field $K$. We consider a monic polynomial $f(T) \in R[T]$ as follows$\colon$

$$ f(T) = T^e + c_{e-1}T^{e-1} + \ldots + c_1T + c_0. $$

Let $\overline{f(T)} \in K[T]$ be the reduction of $f(T)$ by the ideal $(X)$. Suppose that both $f(T)$ and $\overline{f(T)}$ are irreducible. Let \begin{align*} & S \colon= R[T]/(f(T)) \\ & L \colon= K[T]/(\overline{f(T)}) \end{align*} be finite extensions of $R$ and $K$, respectively.

Choose an element $\alpha \in L$ and consider ${\mathrm{Norm}}(\alpha) \in K$, where ${\mathrm{Norm}}(\alpha)$ is defined to be the constant term of the minimal polynomial of $\alpha$ over $K$. For example, ${\mathrm{Norm}}(\alpha) = \alpha$ for $\alpha \in K$.

Q. Choose an element $\beta \in R$ such that $\beta \mapsto {\mathrm{Norm}}(\alpha)$ via the surjection $R\twoheadrightarrow K$ by ${\mathrm{mod}}\,(X)$. Does there always exist a lift $\gamma \in S$ of $\alpha \in L$ such that the following two conditions are satisfied?$\colon$

\begin{align*} \gamma & \equiv \alpha \quad {\mathrm{mod}}\,(X) \\ {\mathrm{Norm}}(\gamma) & = \beta, \end{align*}

where ${\mathrm{Norm}}(\gamma)$ is the constant term of the minimal polynomial of $\gamma \in S$ over $R$. For example, we have ${\mathrm{Norm}}(X) = X$.

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  • $\begingroup$ If $\alpha=0$, then $\gamma$ must be a multiple of $X$, so its norm is a multiple of $X^e$. So there is no solution if $\beta=X$ and $e>1$. $\endgroup$ – Laurent Moret-Bailly Apr 15 at 20:52
  • $\begingroup$ Thanks Professor Lautrent Moret-Bailly, but I just remedy the definition of Norm, which is defined to be the product of different conjugations over the base ring. $\endgroup$ – Rinmyaku Apr 15 at 21:37
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    $\begingroup$ The true norm is the determinant of the mutiplication by $\gamma$, in which case it commutes with reduction modulo $(x)$ even if $f,\overline{f}$ are not irreducible. For irreducible $f,\overline{f}$ the true norm are $p(0)^{\deg(f)/\deg(p)},q(0)^{\deg(f)/\deg(q)}$ with $p,q$ the minimal polynomials. $\endgroup$ – reuns Apr 16 at 0:10
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Let $R$ be a comlete local ring with a maximal ideal ${\frak m}_R$. Let $I$ be an ideal of $R$ such that $(R, I)$ is a henselian pair. Let we denote by $T \colon= R/I$ the residual ring which is a domain. For an algebraic element $\alpha$ over $T$, we can consider a finite extension $T[\alpha]/T$. By abuse, we denote by $R[\alpha]$ the unqiue extension of $R$ which is induced by the residual extension. We consider a division algebra $D \colon= [T[\alpha],{\mathrm{Norm}}(\alpha)\} \in {\mathrm{Br}}(T)$. $D$ is a trivial element in ${\mathrm{Br}}(T)$.

On the other hand, we choose an arbitrary lift $\beta \in R$ such that $\beta \mapsto {\mathrm{Norm}}(\alpha) \in T$ and consider the division algebra $E \colon= [R[\alpha], \beta\} \in {\mathrm{Br}}(R)$. That is, $E \mapsto D$ by the specialisation modulo $I$. The following result holds$\colon$

Theorem(Gabber). The isomorphism $H_{{\acute{e}}t}^2(R, \mu_n) \cong H_{{\acute{e}}t}^2(R/I, \mu_n)$ holds.

Via this isomorphim, we may send $E$ to $D$ knowing that $E$ is still trivial element in ${\mathrm{Br}}(R)$. This implies that there exists an element $\gamma \in R[\alpha]$ such that $\gamma \mapsto \alpha$ and that ${\mathrm{Norm}}(\gamma) = \beta$. The question is the special case where $R = K[[X]]$ and $I = (X)$.

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