Let $A$ be any ring and $B$ be its subring.
Is it true that $J(A)\cap B\subseteq J(B)$?
If not, is it true that for a finitely generated algebra $A$ over a local commutative noetherian ring $\mathcal{O}$ and a subalgebra $B$ of $A$,
$J(A)\cap B\subseteq J(B)$?
I first tried to note that $J(A)\cap B$ is a nilpotent ideal of $B$, so it must be included in $J(B)$, but then I realized $J(A)$ is nilpotent for left artinian rings, but not in general.
Best Answer
No this is not true. Take any polynomial ring $K[T]$ over a field. Then $J(K[T]) = \{0\}$. Now you want the left hand side to be nonzero, so you take $A$ as the localization of $K[T]$ at the maximal ideal $(T)$ and we have $J(K[T]_{(T)}) \cap K[T] = (T)$.