[proofplan] We construct $\Phi$ explicitly by the only formula forced by the requirements: a polynomial $\sum_{k=0}^n a_k x^k$ must map to $\sum_{k=0}^n \varphi(a_k) s^k$. We then verify that this map is a well-defined ring homomorphism satisfying both conditions. For uniqueness, we show that any ring homomorphism $\Psi: R[x] \to S$ with $\Psi|_R = \varphi$ and $\Psi(x) = s$ is forced to agree with $\Phi$ on every polynomial, since the ring homomorphism properties determine $\Psi$ on each monomial $a_k x^k$. [/proofplan] [step:Define the candidate homomorphism $\Phi$ by evaluation at $s$] Define the map \begin{align*} \Phi: R[x] &\to S \\ f = \sum_{k=0}^n a_k x^k &\mapsto \sum_{k=0}^n \varphi(a_k)\, s^k, \end{align*} where $a_k \in R$ for $0 \le k \le n$. This map is well-defined: if $f$ has two representations (differing by trailing zero coefficients), the additional terms contribute $\varphi(0_R)\, s^k = 0_S \cdot s^k = 0_S$, so the image is independent of the representation. [guided] The two conditions $\Phi|_R = \varphi$ and $\Phi(x) = s$ completely determine what $\Phi$ must do on every polynomial — if $\Phi$ is a ring homomorphism, then it must preserve sums and products, so \begin{align*} \Phi\!\left(\sum_{k=0}^n a_k x^k\right) = \sum_{k=0}^n \Phi(a_k)\, \Phi(x)^k = \sum_{k=0}^n \varphi(a_k)\, s^k. \end{align*} This calculation tells us both that there is at most one such homomorphism and what formula it must have. We therefore define \begin{align*} \Phi: R[x] &\to S \\ f = \sum_{k=0}^n a_k x^k &\mapsto \sum_{k=0}^n \varphi(a_k)\, s^k. \end{align*} We must check well-definedness: a polynomial can be written with different numbers of terms by appending zero coefficients. If we write $f = \sum_{k=0}^{n} a_k x^k = \sum_{k=0}^{n+m} a_k x^k$ with $a_k = 0_R$ for $k > n$, then the extra terms contribute $\varphi(0_R) s^k = 0_S$ (since $\varphi$ is a ring homomorphism and therefore sends $0_R$ to $0_S$). So the value of $\Phi(f)$ does not depend on the choice of representation. [/guided] [/step] [step:Verify that $\Phi$ restricts to $\varphi$ on $R$ and sends $x$ to $s$] For any constant polynomial $a \in R$ (viewed as $a \cdot x^0 \in R[x]$), we have $\Phi(a) = \varphi(a)\, s^0 = \varphi(a) \cdot 1_S = \varphi(a)$, so $\Phi|_R = \varphi$. For the indeterminate $x = 1_R \cdot x^1$, we have $\Phi(x) = \varphi(1_R)\, s^1 = 1_S \cdot s = s$, using the fact that $\varphi$ is a ring homomorphism and therefore $\varphi(1_R) = 1_S$. [/step] [step:Verify that $\Phi$ preserves addition] Let $f = \sum_{k=0}^n a_k x^k$ and $g = \sum_{k=0}^n b_k x^k$ in $R[x]$, where we extend coefficients by zero so that both sums run to the same index $n$. Then $f + g = \sum_{k=0}^n (a_k + b_k) x^k$, and \begin{align*} \Phi(f + g) &= \sum_{k=0}^n \varphi(a_k + b_k)\, s^k = \sum_{k=0}^n \bigl(\varphi(a_k) + \varphi(b_k)\bigr) s^k \\ &= \sum_{k=0}^n \varphi(a_k)\, s^k + \sum_{k=0}^n \varphi(b_k)\, s^k = \Phi(f) + \Phi(g), \end{align*} where we used the additivity of $\varphi$ and the distributive law in $S$. [/step] [step:Verify that $\Phi$ preserves multiplication] Let $f = \sum_{i=0}^n a_i x^i$ and $g = \sum_{j=0}^m b_j x^j$. The product $fg = \sum_{k=0}^{n+m} c_k x^k$ has coefficients $c_k = \sum_{i+j=k} a_i b_j$. Then \begin{align*} \Phi(fg) &= \sum_{k=0}^{n+m} \varphi(c_k)\, s^k = \sum_{k=0}^{n+m} \varphi\!\left(\sum_{i+j=k} a_i b_j\right) s^k = \sum_{k=0}^{n+m} \left(\sum_{i+j=k} \varphi(a_i)\, \varphi(b_j)\right) s^k, \end{align*} using the additivity and multiplicativity of $\varphi$. On the other hand, \begin{align*} \Phi(f)\, \Phi(g) &= \left(\sum_{i=0}^n \varphi(a_i)\, s^i\right)\!\left(\sum_{j=0}^m \varphi(b_j)\, s^j\right) = \sum_{k=0}^{n+m} \left(\sum_{i+j=k} \varphi(a_i)\, \varphi(b_j)\right) s^k, \end{align*} by the Cauchy product formula for multiplying sums in the commutative ring $S$, using commutativity of $S$ to write $\varphi(a_i)\, s^i \cdot \varphi(b_j)\, s^j = \varphi(a_i)\, \varphi(b_j)\, s^{i+j}$. Comparing the two expressions gives $\Phi(fg) = \Phi(f)\, \Phi(g)$. [guided] This is the most substantial verification. Let $f = \sum_{i=0}^n a_i x^i$ and $g = \sum_{j=0}^m b_j x^j$. By the definition of multiplication in $R[x]$, the product is $fg = \sum_{k=0}^{n+m} c_k x^k$ where $c_k = \sum_{\substack{i+j=k \\ i,j \ge 0}} a_i b_j$. Applying $\Phi$: \begin{align*} \Phi(fg) &= \sum_{k=0}^{n+m} \varphi(c_k)\, s^k = \sum_{k=0}^{n+m} \varphi\!\left(\sum_{i+j=k} a_i b_j\right) s^k. \end{align*} Since $\varphi$ is a ring homomorphism, it preserves finite sums and products: $\varphi\!\left(\sum_{i+j=k} a_i b_j\right) = \sum_{i+j=k} \varphi(a_i b_j) = \sum_{i+j=k} \varphi(a_i)\, \varphi(b_j)$. Therefore \begin{align*} \Phi(fg) &= \sum_{k=0}^{n+m} \left(\sum_{i+j=k} \varphi(a_i)\, \varphi(b_j)\right) s^k. \end{align*} Now we compute $\Phi(f)\, \Phi(g)$ directly. Since $S$ is a commutative ring, we can rearrange products of elements of $S$ freely: \begin{align*} \Phi(f)\, \Phi(g) &= \left(\sum_{i=0}^n \varphi(a_i)\, s^i\right)\!\left(\sum_{j=0}^m \varphi(b_j)\, s^j\right). \end{align*} Expanding by distributivity and using commutativity of $S$ to write $(\varphi(a_i)\, s^i)(\varphi(b_j)\, s^j) = \varphi(a_i)\, \varphi(b_j)\, s^{i+j}$, we collect terms by the total power $k = i + j$: \begin{align*} \Phi(f)\, \Phi(g) &= \sum_{k=0}^{n+m} \left(\sum_{i+j=k} \varphi(a_i)\, \varphi(b_j)\right) s^k. \end{align*} The rearrangement step — collecting $s^i \cdot s^j = s^{i+j}$ and factoring out $s^k$ — is valid because $S$ is commutative and the sums are finite. Comparing the two displayed expressions, we conclude $\Phi(fg) = \Phi(f)\, \Phi(g)$. [/guided] [/step] [step:Verify that $\Phi$ sends $1_{R[x]}$ to $1_S$] The multiplicative identity of $R[x]$ is the constant polynomial $1_R$. By the computation in the second step, $\Phi(1_R) = \varphi(1_R) = 1_S$, since $\varphi$ is a ring homomorphism. Therefore $\Phi$ preserves the multiplicative identity. [/step] [step:Prove uniqueness by showing any such homomorphism must equal $\Phi$] Let $\Psi: R[x] \to S$ be any ring homomorphism satisfying $\Psi|_R = \varphi$ and $\Psi(x) = s$. For any polynomial $f = \sum_{k=0}^n a_k x^k \in R[x]$, the ring homomorphism properties of $\Psi$ give \begin{align*} \Psi(f) = \Psi\!\left(\sum_{k=0}^n a_k x^k\right) = \sum_{k=0}^n \Psi(a_k)\, \Psi(x)^k = \sum_{k=0}^n \varphi(a_k)\, s^k = \Phi(f). \end{align*} Here we used: $\Psi$ preserves finite sums, $\Psi(a_k x^k) = \Psi(a_k)\, \Psi(x^k)$ by multiplicativity, $\Psi(x^k) = \Psi(x)^k$ by induction on $k$ using multiplicativity, $\Psi(a_k) = \varphi(a_k)$ since $\Psi|_R = \varphi$, and $\Psi(x) = s$ by hypothesis. Since $\Psi(f) = \Phi(f)$ for every $f \in R[x]$, we conclude $\Psi = \Phi$. This establishes uniqueness. [guided] Suppose $\Psi: R[x] \to S$ is any ring homomorphism with $\Psi|_R = \varphi$ and $\Psi(x) = s$. We must show $\Psi = \Phi$, i.e., $\Psi(f) = \Phi(f)$ for every $f \in R[x]$. Take any $f = \sum_{k=0}^n a_k x^k$. Since $\Psi$ is a ring homomorphism, it preserves addition: \begin{align*} \Psi\!\left(\sum_{k=0}^n a_k x^k\right) = \sum_{k=0}^n \Psi(a_k x^k). \end{align*} For each monomial $a_k x^k$, multiplicativity of $\Psi$ gives $\Psi(a_k x^k) = \Psi(a_k) \cdot \Psi(x^k)$. Now $\Psi(a_k) = \varphi(a_k)$ because $a_k \in R$ and $\Psi|_R = \varphi$. For $\Psi(x^k)$, we use multiplicativity repeatedly: $\Psi(x^k) = \Psi(x \cdot x \cdots x) = \Psi(x)^k = s^k$. (The base case $k = 0$ gives $\Psi(x^0) = \Psi(1_R) = 1_S = s^0$, using the fact that ring homomorphisms preserve multiplicative identities.) Therefore \begin{align*} \Psi(f) = \sum_{k=0}^n \varphi(a_k)\, s^k = \Phi(f). \end{align*} Since $f$ was arbitrary, $\Psi = \Phi$. The key insight is that the polynomial ring $R[x]$ is generated as a ring by $R$ and $x$. Any ring homomorphism out of $R[x]$ is completely determined by its values on the generators, and the conditions $\Psi|_R = \varphi$ and $\Psi(x) = s$ pin down those values. This is precisely what "universal property" means: $R[x]$ is the free commutative $R$-algebra on one generator $x$, and specifying the image of $x$ determines the entire homomorphism. [/guided] [/step]