Let $Z$ be the set of all multiindices indexed by $J$. Let $k_j$ be the $j$th component of a multiindex $k$. Let $\displaystyle f = \sum_{k \mathop \in Z} a_k \prod_{j \mathop \in J} X_j^{k_j}$ be a polynomial over $R$. Define: :$\displaystyle \phi \left({f}\right) = \sum_{k \mathop \in Z} \psi \left({a_k}\right) \prod_{j \mathop \in J}s_j^{k_j}$ It is clear that $\phi$ extends $\psi$. If $\displaystyle g = \sum_{k \mathop \in Z} b_k \prod_{j \mathop \in J} X_j^{k_j}$, then: {{begin-eqn}} {{eqn | l = \phi \left({f + g}\right)       | r = \phi \left({\sum_{k \mathop \in Z} \left({a_k + b_k}\right) \prod_{j \mathop \in J} X_j^{k_j} }\right)       | c = {{Defof|Addition of Polynomial Forms}} }} {{eqn | r = \sum_{k \mathop \in Z} \left({\psi \left({a_k + b_k}\right)}\right) \prod_{j \mathop \in J} s_j^{k_j}       | c = Definition of $\phi$ }} {{eqn | r = \sum_{k \mathop \in Z} \left({\psi \left({a_k}\right) + \psi \left({b_k}\right)}\right) \prod_{j \mathop \in J} s_j^{k_j}       | c = {{Defof|Ring Homomorphism}} }} {{eqn | r =\sum_{k \mathop \in Z} \psi \left({a_k}\right) \prod_{j \mathop \in J} s_j^{k_j} + \sum_{k \mathop \in Z} \psi \left({b_k}\right) \prod_{j \mathop \in J} s_j^{k_j}       | c = Ring Axioms of $S$ }}  {{eqn | r =\phi \left({f}\right) + \phi \left({g}\right)       | c = Definition of $\phi$ }} {{end-eqn}} Therefore $\phi$ preserves addition. Also: {{begin-eqn}} {{eqn | l = \phi \left({f g}\right)       | r = \phi \left({\sum_{k \mathop \in Z} \left({\sum_{p \mathop + q \mathop = k} a_p b_q}\right) \prod_{j \mathop \in J} X_j^{k_j} }\right)       | c = {{Defof|Multiplication of Polynomial Forms}} }} {{eqn | r = \sum_{k \mathop \in Z} \psi \left({\sum_{p \mathop + q \mathop = k} a_p b_q}\right) \prod_{j \mathop \in J} s_j^{k_j}       | c = Definition of $\phi$ }} {{eqn | r = \sum_{k \mathop \in Z} \left({\sum_{p \mathop + q \mathop = k} \psi \left({a_p}\right) \psi \left({b_q}\right)}\right) \prod_{j \mathop \in J} s_j^{k_j}       | c = {{Defof|Ring Homomorphism}} }} {{eqn | r = \left({\sum_{k \mathop \in Z} \left({\psi \left({a_k}\right)}\right) \prod_{j \mathop \in J} s_j^{k_j} }\right) \left({\sum_{k \mathop \in Z} \left({\psi \left({b_k}\right)}\right) \prod_{j \mathop \in J} s_j^{k_j} }\right)       | c = Ring Axioms of $S$ }} {{eqn | r =\phi \left({f}\right) \phi \left({g}\right)       | c = Definition of $\phi$ }} {{end-eqn}} This shows that $\phi$ is a homomorphism. Now suppose that $\phi'$ is another such homomorphism. For each $j \in J$, $\phi'$ must satisfy $\phi'(X_j)=s_j$ and $\phi' \left({r}\right) = \psi \left({r}\right)$ for all $r \in R$. In addition $\phi'$ must be a homomorphism, so we compute: {{begin-eqn}} {{eqn | l = \phi' \left({\sum_{k \mathop \in Z} a_k \prod_{j \mathop \in J} X_j^{k_j} }\right)       | r = \sum_{k \mathop \in Z} \phi' \left({a_k \prod_{j \mathop \in J} X_j^{k_j} }\right)       | c = $\phi'$ preserves Ring Addition }} {{eqn | r = \sum_{k \mathop \in Z} \phi' \left({a_k}\right) \prod_{j \mathop \in J} \phi' \left({X_j}\right)^{k_j}        | c = $\phi'$ preserves Ring Product }} {{eqn | r = \sum_{k \mathop \in Z} \psi \left({a_k}\right) \prod_{j \mathop \in J} s_j^{k_j}       | c = as $\phi' \left({X_j}\right) = s_j$ and $\phi' \left({r}\right) = \psi \left({r}\right)$ for all $r \in R$ }} {{end-eqn}} and therefore $\phi' = \phi$. This concludes the proof. {{qed}}