Lemma 15.11.11. Let $J$ be a set and let $\{ (A_ j, I_ j)\} _{j \in J}$ be a collection of pairs. Then $(\prod _{j \in J} A_ j, \prod _{j\in J} I_ j)$ is Henselian if and only if so is each $(A_ j, I_ j)$.

Proof. For every $j \in J$, the projection $\prod _{j \in J} A_ j \rightarrow A_ j$ is an integral ring map, so Lemma 15.11.8 proves that each $(A_ j, I_ j)$ is Henselian if $(\prod _{j \in J} A_ j, \prod _{j\in J} I_ j)$ is Henselian.

Conversely, suppose that each $(A_ j, I_ j)$ is a Henselian pair. Then every $1 + x$ with $x \in \prod _{j \in J} I_ j$ is a unit in $\prod _{j \in J} A_ j$ because it is so componentwise by Algebra, Lemma 10.19.1 and Definition 15.11.1. Thus, by Algebra, Lemma 10.19.1 again, $\prod _{j \in J} I_ j$ is contained in the Jacobson radical of $\prod _{j \in J} A_ j$. Continuing to work componentwise, it likewise follows that for every monic $f \in (\prod _{j \in J} A_ j)[T]$ and every factorization $\overline{f} = g_0h_0$ with monic $g_0, h_0 \in (\prod _{j \in J} A_ j / \prod _{j \in J} I_ j)[T] = (\prod _{j \in J} A_ j/I_ j)[T]$ that generate the unit ideal in $(\prod _{j \in J} A_ j / \prod _{j \in J} I_ j)[T]$, there exists a factorization $f = gh$ in $(\prod _{j \in J} A_ j)[T]$ with $g$, $h$ monic and reducing to $g_0$, $h_0$. In conclusion, according to Definition 15.11.1 $(\prod _{j \in J} A_ j, \prod _{j\in J} I_ j)$ is a Henselian pair. $\square$

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