[proofplan]
We prove $i_{L/K}(\sigma) = v_L(\sigma(\alpha) - \alpha)$ by establishing both inequalities. The upper bound $i_{L/K}(\sigma) \leq v_L(\sigma(\alpha) - \alpha)$ is immediate from the definition of $i_{L/K}(\sigma)$ as a minimum. The lower bound uses the hypothesis $\mathcal{O}_L = \mathcal{O}_K[\alpha]$: every $x \in \mathcal{O}_L$ is a polynomial in $\alpha$ with coefficients in $\mathcal{O}_K$, and $\sigma(\alpha) - \alpha$ divides $\sigma(\alpha)^i - \alpha^i$ in $\mathcal{O}_L$, which forces $v_L(\sigma(x) - x) \geq v_L(\sigma(\alpha) - \alpha)$.
[/proofplan]
[step:Establish the upper bound $i_{L/K}(\sigma) \leq v_L(\sigma(\alpha) - \alpha)$]
By definition, $i_{L/K}(\sigma) = \min_{x \in \mathcal{O}_L} v_L(\sigma(x) - x)$. Since $\alpha \in \mathcal{O}_L$, taking $x = \alpha$ in the minimum gives
\begin{align*}
i_{L/K}(\sigma) \leq v_L(\sigma(\alpha) - \alpha).
\end{align*}
[/step]
[step:Express any $x \in \mathcal{O}_L$ as a polynomial in $\alpha$ over $\mathcal{O}_K$]
By hypothesis, $\mathcal{O}_L = \mathcal{O}_K[\alpha]$, so every $x \in \mathcal{O}_L$ can be written as
\begin{align*}
x = g(\alpha) = \sum_{i=0}^{d} b_i \alpha^i
\end{align*}
for some polynomial $g(t) = \sum_{i=0}^d b_i t^i \in \mathcal{O}_K[t]$.
[/step]
[step:Show $\sigma(\alpha) - \alpha$ divides $\sigma(\alpha)^i - \alpha^i$ for each $i \geq 1$]
For $i \geq 1$, the algebraic identity
\begin{align*}
\sigma(\alpha)^i - \alpha^i = (\sigma(\alpha) - \alpha)\sum_{j=0}^{i-1} \sigma(\alpha)^j \alpha^{i-1-j}
\end{align*}
shows that $\sigma(\alpha) - \alpha$ divides $\sigma(\alpha)^i - \alpha^i$ in $\mathcal{O}_L$ (since $\sigma(\alpha), \alpha \in \mathcal{O}_L$ and the sum on the right is an element of $\mathcal{O}_L$). In particular,
\begin{align*}
v_L(\sigma(\alpha)^i - \alpha^i) \geq v_L(\sigma(\alpha) - \alpha) + (i-1) \cdot 0 = v_L(\sigma(\alpha) - \alpha),
\end{align*}
where we used $v_L(\sigma(\alpha)^j \alpha^{i-1-j}) \geq 0$ for each $j$ (since $\sigma(\alpha), \alpha \in \mathcal{O}_L$).
More precisely, $v_L(\sigma(\alpha)^i - \alpha^i) \geq v_L(\sigma(\alpha) - \alpha)$ for all $i \geq 1$.
[/step]
[step:Establish the lower bound $v_L(\sigma(x) - x) \geq v_L(\sigma(\alpha) - \alpha)$ for all $x \in \mathcal{O}_L$]
Let $x = g(\alpha) = \sum_{i=0}^d b_i \alpha^i$ with $b_i \in \mathcal{O}_K$. Since $\sigma$ fixes $K$, $\sigma(b_i) = b_i$ for each $i$. Therefore
\begin{align*}
\sigma(x) - x = g(\sigma(\alpha)) - g(\alpha) = \sum_{i=0}^{d} b_i \bigl(\sigma(\alpha)^i - \alpha^i\bigr).
\end{align*}
The $i = 0$ term vanishes ($\sigma(\alpha)^0 - \alpha^0 = 1 - 1 = 0$). For $i \geq 1$, each term satisfies
\begin{align*}
v_L\bigl(b_i(\sigma(\alpha)^i - \alpha^i)\bigr) = v_L(b_i) + v_L(\sigma(\alpha)^i - \alpha^i) \geq 0 + v_L(\sigma(\alpha) - \alpha),
\end{align*}
using $v_L(b_i) \geq 0$ (since $b_i \in \mathcal{O}_K \subseteq \mathcal{O}_L$) and the divisibility from the previous step. By the ultrametric inequality,
\begin{align*}
v_L(\sigma(x) - x) \geq \min_{1 \leq i \leq d}\, v_L\bigl(b_i(\sigma(\alpha)^i - \alpha^i)\bigr) \geq v_L(\sigma(\alpha) - \alpha).
\end{align*}
Taking the minimum over all $x \in \mathcal{O}_L$ gives $i_{L/K}(\sigma) \geq v_L(\sigma(\alpha) - \alpha)$.
[guided]
The strategy is to reduce the computation of $i_{L/K}(\sigma) = \min_{x \in \mathcal{O}_L} v_L(\sigma(x) - x)$ to a single element $\alpha$ that generates $\mathcal{O}_L$ as an $\mathcal{O}_K$-algebra.
Why does a single generator suffice? Because $\sigma$ fixes $\mathcal{O}_K$ pointwise, the action of $\sigma$ on $\mathcal{O}_L = \mathcal{O}_K[\alpha]$ is entirely determined by where it sends $\alpha$. Concretely, for any $x = g(\alpha)$ with $g \in \mathcal{O}_K[t]$:
\begin{align*}
\sigma(x) - x = g(\sigma(\alpha)) - g(\alpha) = \sum_{i=1}^{d} b_i(\sigma(\alpha)^i - \alpha^i).
\end{align*}
Each difference $\sigma(\alpha)^i - \alpha^i$ factors as $(\sigma(\alpha) - \alpha) \cdot (\text{sum of terms in } \mathcal{O}_L)$, so $v_L(\sigma(\alpha)^i - \alpha^i) \geq v_L(\sigma(\alpha) - \alpha)$.
This means every element of $\mathcal{O}_L$ is moved by $\sigma$ at least as much (in terms of $v_L$-valuation) as $\alpha$ is. Therefore $\alpha$ achieves the minimum in the definition of $i_{L/K}(\sigma)$.
[/guided]
[/step]
[step:Combine both bounds to conclude]
From the upper bound $i_{L/K}(\sigma) \leq v_L(\sigma(\alpha) - \alpha)$ and the lower bound $i_{L/K}(\sigma) \geq v_L(\sigma(\alpha) - \alpha)$, we conclude
\begin{align*}
i_{L/K}(\sigma) = v_L(\sigma(\alpha) - \alpha).
\end{align*}
[/step]
