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  "body": "\n## Proof\n\nLet $R$ and $S$ be [[commutative rings|0GYGHKX834]], let $\\varphi \\colon R \\to S$ be a [[ring homomorphism|0GGE0Z0JQ1]] and let $s \\in S$. We identify $r \\in R$ with the constant polynomial $(r, 0, 0, \\dots)$ in the [[polynomial ring|0Q9EZS90GY]] $R[x]$, and $x = (0, 1, 0, \\dots)$.\n\n1. Each $f \\in R[x]$ is a sequence $(a_0, a_1, \\dots)$ with $a_n = 0$ for all but finitely many $n$. Define $\\Phi(f) = \\sum_{n \\ge 0} \\varphi(a_n) s^n$.\n2. This sum has only finitely many nonzero terms, because $\\varphi(0) = 0$. So $\\Phi$ is a well defined map $R[x] \\to S$. For $f = a_0 + a_1 x + \\dots + a_n x^n$, it gives the formula of the statement.\n3. Let $f = (a_n)$ and $g = (b_n)$. Then $f + g = (a_n + b_n)$, and $\\varphi(a_n + b_n) = \\varphi(a_n) + \\varphi(b_n)$. Summing over $n$ gives $\\Phi(f + g) = \\Phi(f) + \\Phi(g)$.\n4. The product is $f g = (c_n)$ with $c_n = \\sum_{i + j = n} a_i b_j$. Since $\\varphi$ is additive and multiplicative, $\\varphi(c_n) = \\sum_{i + j = n} \\varphi(a_i) \\varphi(b_j)$.\n5. So $\\Phi(f g) = \\sum_n \\sum_{i + j = n} \\varphi(a_i) \\varphi(b_j) s^{i + j}$, where we used $s^n = s^i s^j$ for $i + j = n$.\n6. Expand $\\Phi(f) \\Phi(g) = \\big(\\sum_i \\varphi(a_i) s^i\\big) \\big(\\sum_j \\varphi(b_j) s^j\\big)$ by distributivity. Since $S$ is commutative, the term for $(i, j)$ is $\\varphi(a_i) \\varphi(b_j) s^{i + j}$.\n7. Grouping the terms of step 6 by $n = i + j$ gives the sum of step 5. So $\\Phi(f g) = \\Phi(f) \\Phi(g)$.\n8. The constant $r \\in R$ has $a_0 = r$ and $a_n = 0$ for $n \\ge 1$. So $\\Phi(r) = \\varphi(r)$. In particular $\\Phi(1) = \\varphi(1) = 1$.\n9. By steps 3, 7 and 8, $\\Phi$ is a ring homomorphism with $\\Phi(r) = \\varphi(r)$ for all $r \\in R$.\n10. The polynomial $x$ has $a_1 = 1$ and all other coefficients $0$. So $\\Phi(x) = \\varphi(1) s = s$. This proves existence.\n11. For uniqueness, let $\\Psi \\colon R[x] \\to S$ be a ring homomorphism with $\\Psi(r) = \\varphi(r)$ for $r \\in R$ and $\\Psi(x) = s$.\n12. Let $f = a_0 + a_1 x + \\dots + a_n x^n$. Since $\\Psi$ preserves sums and products, $\\Psi(f) = \\sum_{k=0}^n \\Psi(a_k) \\Psi(x)^k$. Here $\\Psi(x)^0 = \\Psi(1) = 1$.\n13. By the hypotheses on $\\Psi$, this is $\\sum_{k=0}^n \\varphi(a_k) s^k = \\Phi(f)$. So $\\Psi = \\Phi$, and $\\Phi$ is the only such homomorphism.\n\n## Idea\n\nA ring homomorphism respects sums and products, and every polynomial is built from constants and $x$ by sums and products. So its values on the constants and on $x$ decide it, which gives uniqueness. For existence, one checks that \"replace $x$ by $s$ and each coefficient by its image\" respects products; this uses that $S$ is commutative.\n",
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