The Stacks project

Proof. If $u \in 1 + I$ then for some $j \in J$ we see that $u$ is the image of some $u_ j \in 1 + I_ j$. Then $u_ j$ is invertible in $A_ j$ by Algebra, Lemma 10.19.1 and the assumption that $I_ j$ is contained in the Jacobson radical of $A_ j$. Hence $u$ is invertible in $A$. Thus $I$ is contained in the Jacobson radical of $A$ (by the lemma).

Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_0 h_0$ be a factorization with $g_0, h_0 \in A/I[T]$ monic generating the unit ideal in $A/I[T]$. Write $1 = g_0 g'_0 + h_0 h'_0$ for some $g'_0, h'_0 \in A/I[T]$. Since $A = \mathop{\mathrm{colim}}\nolimits A_ j$ and $A/I = \mathop{\mathrm{colim}}\nolimits A_ j/I_ j$ are filtered colimits we can find a $j \in J$ and $f_ j \in A_ j$ and a factorization $\overline{f}_ j = g_{j, 0} h_{j, 0}$ with $g_{j, 0}, h_{j, 0} \in A_ j/I_ j[T]$ monic and $1 = g_{j, 0} g'_{j, 0} + h_{j, 0} h'_{j, 0}$ for some $g'_{j, 0}, h'_{j, 0} \in A_ j/I_ j[T]$ with $f_ j, g_{j, 0}, h_{j, 0}, g'_{j, 0}, h'_{j, 0}$ mapping to $f, g_0, h_0, g'_0, h'_0$. Since $(A_ j, I_ j)$ is a henselian pair, we can lift $\overline{f}_ j = g_{j, 0} h_{j, 0}$ to a factorization over $A_ j$ and taking the image in $A$ we obtain a corresponding factorization in $A$. Hence $(A, I)$ is henselian. $\square$