I asked [this question](https://math.stackexchange.com/q/2783558/660) on Mathematics Stackexchange, but got no answer.


Is the product 
$$
\prod\_{i\in I}A\_i
$$ 
of a family $(A\_i)\_{i\in I}$ of [Jacobson rings](https://en.wikipedia.org/wiki/Jacobson_ring) 
a Jacobson ring?


(Here "ring" means "commutative ring with one".)



Is the product of Jacobson rings a Jacobson ring?
The answer is **no** in general.


Take $R = \prod\_{n \in \mathbb{N}\_{> 0}} \mathbb{Z}/2^n\mathbb{Z}$. Then the Jacobson radical of $R$ is $\prod\_{n \in \mathbb{N}\_{> 0}} 2\mathbb{Z}/2^n\mathbb{Z}$, and it contains a non-nilpotent element, namely $(2 + 2^n \mathbb{Z})\_n$. Therefore the Jacobson radical of $R$ doesn't coincide with the nilradical of $R$.


All credits go to Georges Elencwajg, see this [MO post](https://mathoverflow.net/questions/90980/what-is-the-dimension-of-the-product-ring-prod-mathbb-z-2n-mathbb-z?rq=1).


**Side note.** Let $J(R)$ denote the Jacobson radical of a commutative unital ring $R$. Knowing that $x \in J(R)$ holds if and only if $1 + rx \in R^{\times}$ holds for every $r \in R$, the identity $J(\prod\_{i \in I} A\_i) = \prod\_{i \in I} J(A\_i)$ is immediate.


