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let $A$ be an associative ring, let $\mathfrak{a} \subseteq A$ be a two-sided ideal and let $\mathfrak{J}(A) \subseteq A$ be the Jacobson radical, defined as the intersection of the maximal left ideals.

Question: When does the equality $\mathfrak{J}(A/\mathfrak{a}) = [\mathfrak{J}(A) + \mathfrak{a}]/\mathfrak{a}$ hold?

The inclusion from right to left does always hold, is this also true for the other inclusion?

Context: I'm reading the paper "A proof of the strong no loop conjecture"* and at some point the author claims that the statement is true in the following special case:

Let $A$ be an artin algebra, $e$ an idempotent of $A$ such that $S_e$ has finite injective dimension. Then the statement is true for $\mathfrak{a} = A(1-e)A$. Here $S_e$ refers to the semisimple module supported by $e$, so $S_e = Ae/\mathfrak{J}(A)e$

Why does this hold?

*Authors: Kiyoshi Igusa, Shiping Liu, Charles Paquette; appeared in Advances in Mathematics Volume 228, Issue 5, 1 December 2011, Pages 2731-2742

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If your algebra $A$ is Artinian (left or right), then the equality holds. Indeed, in that case, the Jacobson radical is the only ideal $I$ of $A$ such that $I$ is nilpotent and $A/I$ is semisimple. Since $(\mathfrak{J}(A)+\mathfrak{a})/\mathfrak{a}$ is nilpotent in $A/\mathfrak{a}$, and since $$ \frac{A/\mathfrak{a}}{(\mathfrak{J}(A)+\mathfrak{a})/\mathfrak{a}} \cong \frac{A}{(\mathfrak{J}(A)+\mathfrak{a})} $$ is semisimple (it is a quotient of the semisimple algebra $A/\mathfrak{J}(A)$), the Jacobson radical of $A/\mathfrak{a}$ is $(\mathfrak{J}(A)+\mathfrak{a})/\mathfrak{a}$.

If you don't assume that $A$ is Artinian, then the result is not necessarily true anymore. Let $K$ be any field, and let $A = K[x]$. Then the Jacobson radical of $A$ is zero. Now, let $\mathfrak{a} = (x^2)$. Then $A/\mathfrak{a} = K[x]/(x^2)$ has non-zero radical (it is equal to $(x)/(x^2)$).