How to prove that $Nil_*M_n(R) = M_n(Nil_*R)$?,
$Nil_*R$ is the Baer's lower nilradical (It is the smallest semiprime ideal in R, and is equal to the intersection of all the prime ideals in R)
$M_n(R)$ is the ring of matrices over R.
2026-03-25 07:48:21.1774424901
How to prove that $Nil_*M_n(R) = M_n(Nil_*R)$?
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This follows immediately from ideal correspondence. That is, the ideals of $M_n(R)$ are exactly of the form $M_n(I)$ where $I\lhd R$.
It is not hard to see that prime ideals correspond and intersections correspond in the obvious way.