The Stacks project

Proof. Consider the ring $A' = A[x]/(x^ n - \pi )$ and denote $\pi '$ the image of $x$ in $A_1$. As $A'$ is finite free of rank $n$ as an $A$-module, the element $\pi $ is a nonzerodivisor in $A'$, and hence so is $\pi '$. Moreover, $A'/\pi 'A'$ is isomorphic to $A/\pi A$ which is a field. Hence $A'$ is a discrete valuation ring with uniformizer $\pi '$. It follows that the fraction field $K'$ of $A'$ is an extension of degree $n$. Clearly, $K'$ is obtained from $K$ by adjoint an $n$th root $\pi ' = \pi ^{1/n}$ of $\pi $, i.e., $K' \cong K_1$. Since $A'$ is normal, we see that $A'$ is the integral closure of $A$ in $K'$, i.e., $A' = A_1$. This proves (1) – (5).

The assertion that $K_1/K$ is tamely ramified if $n$ is invertible in the residue field is immediate from Definition 15.112.7 and (3) and (4). To prove the very last statement, it suffices to show that every $n$th root of unity $\zeta \in K'$ is contained in $K$, see Fields, Lemma 9.24.3. It is clear that $\zeta \in (A')^*$. We can write $\zeta = a + b$ with $a \in A$ and $b \in A \pi ' \oplus \ldots \oplus A (\pi ')^{n - 1}$. If $b$ is nonzero, then we can write $b = u(\pi ')^ t$ where $t \geq 1$ and $b$ a unit of $A'$. Then

Since $n$ and $a$ are units of $A$, this would imply that $b = u(\pi ')^ t$ is congruent to $(1 - a^ n)/(n a^{n - 1}) \in A$ modulo $(\pi ')^{t + 1}$. The direct sum decomposition $A' = A \oplus \ldots \oplus A(\pi ')^{n - 1}$ shows that this is impossible because the image of multiplication by $(\pi ')^{t + 1}$ is a corresponding direct sum whose intersection with $A(\pi ')^ t$ is $(\pi A)(\pi ')^ t$. Thus $b = 0$ and $\zeta \in A$ as desired. $\square$