# Finite extension of $K[[X]]$ and the norm

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 $${\mathrm{Norm}}(X) = X$$.

• 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 $e>1$. – Laurent Moret-Bailly Apr 15 at 20:52
• 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. – Rinmyaku Apr 15 at 21:37
• 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. – reuns Apr 16 at 0:10

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 $$I = (X)$$.