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Sep 2, 2018 at 16:59 history edited abx CC BY-SA 4.0
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Sep 2, 2018 at 5:50 comment added Keith Kearnes The argument is not yet complete: you show that $D(J(A))\subseteq J(A)$, but not that $D(A)\subseteq J(A)$. Continue (in the local case): if $x\in A-J(A)$ does not satisfy $D(x)\in J(A)$, differentiate the minimal polynomial of $x$ over $K$ modulo $J(A)$ to get a contradiction.
Sep 2, 2018 at 3:29 comment added Yemon Choi This doesn't seem so stupid to me coming from the background of Banach algebras: it's an old result of Singer and Wermer that if A is a unital commutative Banach algebra over the complex field and $D:A\to A$ is a continuous complex-linear derivation, then $D(A)\subseteq J(A)$.
Sep 2, 2018 at 3:21 history answered abx CC BY-SA 4.0