Let $\psi:R[X_1,..,X_d] \rightarrow S$ be a surjective ring homomorphism from polynomial ring $R[X_1,..,X_d]$ to ring $S$ with $\operatorname{Ker}(\psi)=I=I_0R[X_1,..,X_d]$, where $I_0$ is an ideal of $R$. Show that (1) $I_0=\operatorname{Ker}(\psi|_R)$ and (2) $S\cong\psi(R)[X_1,..,X_d]$.
For the first question: From $I_0R[X_1,..,X_d] = \{i_0f:i_0 \in I_0,f \in R[X_1,..,X_d]\} $ follows that $\forall f \in R[X_1,..,X_d]: \psi(f) \neq 0$ we have $\psi(i_0f) = \psi(i_0)\psi(f) = 0$, hence $\psi(i_0) = 0, \forall i_0 \in I_0$. It proves that $I_0 \subseteq \operatorname{Ker}(\psi|_R)$. Also $\forall r \in \operatorname{Ker}(\psi|_R): \psi(r)=0$, so $\operatorname{Ker}(\psi|_R) \subseteq I_0$, hence $\operatorname{Ker}(\psi|_R) = I_0$.
For the second question: isomorphism theorem states that $R[X_1,..,X_d]/I \cong S$, so maybe prove $R[X_1,..,X_d]/I \cong \psi(R)[X_1,..,X_d] $?