Counterexample to the exercise on the homomorphism image of the Jacobson radical

Question

Let $J(R)$ denote the Jacobson radical of the ring $R.$ It is easy to see that if $f:R\to S$ is a subjective ring homomorphism, then $f(J(R))\subseteq J(S).$ We can choose $R=\mathbb{Z}$ and $S=2^2\mathbb{Z},$ the projection $f$ to see that this inclusion need not be an equality. But I don't know what happens if $f$ is not surjective. I also consider $J(R)$ where $R$ is a principal ideal domain, not a field. Please help me.

Answer 1

If $R$ is any local integral domain that isn’t a field, the inclusion map $\phi$ into its field of fractions $Q$ shows $\phi(J(R))\not\subset J(Q)=\{0\}$.