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  "body": "\n## Statement\n\nLet $R$ and $S$ be [[commutative rings|0GYGHKX834]], let $\\varphi \\colon R \\to S$ be a [[ring homomorphism|0GGE0Z0JQ1]] and let $s \\in S$.\nThen there is exactly one ring homomorphism $\\Phi \\colon R[x] \\to S$ such that $\\Phi(r) = \\varphi(r)$ for all $r \\in R$ and $\\Phi(x) = s$.\nIt is given by $\\Phi(a_0 + a_1 x + \\dots + a_n x^n) = \\varphi(a_0) + \\varphi(a_1) s + \\dots + \\varphi(a_n) s^n$.\n\n## Remarks\n\nWith $S = R$, $\\varphi$ the identity and $s = c$, the map $\\Phi$ is the [[evaluation homomorphism|0125QVNCFQ]] at $c$.\nIn the language of algebras: for each commutative $R$-algebra $A$ and each $a \\in A$, there is exactly one $R$-algebra homomorphism $R[x] \\to A$ that sends $x$ to $a$.\nThis property characterizes $R[x]$ up to a unique isomorphism.\n\n## Explanations\n\n### L1 human\n\nA ring homomorphism from $R[x]$ is fixed by two choices: what it does on the coefficients, and where it sends $x$.\nEvery pair of choices works, and it gives exactly one homomorphism.\nEvaluation at a number is the main example.\n",
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