Lemma 59.81.4. Let $R$ be an $A$-algebra which is absolutely integrally closed. Let $I, J \subset R$ be ideals with $I + J = R$. There exists a $g \in G(R)$ such that $g \bmod I = \sigma _0$ and $g \bmod J = \sigma _1$.
Proof. Choose $x \in I$ such that $x \equiv 1 \bmod J$. We may and do replace $I$ by $xR$ and $J$ by $(x - 1)R$. Then we are looking for an $s \in R$ such that
$1 + \pi s$ is a unit,
$s \equiv 0 \bmod xR$, and
$s \equiv 1 \bmod (x - 1)R$.
The last two conditions say that $s = x + x(x - 1)y$ for some $y \in R$. The first condition says that $1 + \pi s = 1 + \pi x + \pi x (x - 1) y$ needs to be a unit of $R$. However, note that $1 + \pi x$ and $\pi x (x - 1)$ generate the unit ideal of $R$ and that $1 + \pi x$ is an $\ell $th root of $1$ modulo $\pi x (x - 1)$1. Thus we win by Lemma 59.81.1 and the fact that $R$ is absolutely integrally closed. $\square$
[1] Because $1 + \pi x$ is congruent to $1$ modulo $\pi $, congruent to $1$ modulo $x$, and congruent to $1 + \pi = \zeta $ modulo $x - 1$ and because we have $(\pi ) \cap (x) \cap (x - 1) = (\pi x (x - 1))$ in $A[x]$.
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