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  "body": "\n## Proof\n\nLet $R$ be a [[commutative ring|0GYGHKX834]]. We identify $r \\in R$ with the constant polynomial $(r, 0, 0, \\dots)$ of the [[polynomial ring|0Q9EZS90GY]] $R[x]$.\n\n1. The ring $R[x]$ is commutative, because $R$ is commutative and the coefficient $\\sum_{i + j = n} a_i b_j$ of a product is symmetric in the two factors.\n2. For $r, s \\in R$, the product of the constant polynomials $r$ and $s$ is the constant polynomial $r s$. This follows from the product formula, since only $c_0 = r s$ can be nonzero.\n3. Two constant polynomials $r$ and $s$ are equal exactly when $r = s$, since polynomials are equal when all their coefficients are equal.\n\nFirst suppose that $R$ is an [[integral domain|09S3PFBEPM]].\n\n4. The identity of $R[x]$ is the constant $1$ and its zero is the constant $0$. By step 3 and $1 \\neq 0$ in $R$, we get $1 \\neq 0$ in $R[x]$.\n5. Let $f, g \\in R[x]$ with $f \\neq 0$ and $g \\neq 0$. Their [[degrees|0DPF6WRZW6]] are integers $\\deg f \\ge 0$ and $\\deg g \\ge 0$.\n6. By the [[degree of a product|0T233AZ1GE]], $\\deg(f g) = \\deg f + \\deg g \\ge 0$. So $\\deg(f g) \\ne -\\infty$, and $f g \\neq 0$.\n7. By steps 1, 4 and 6, $R[x]$ is an integral domain.\n\nNow suppose that $R[x]$ is an integral domain.\n\n8. The ring $R$ is commutative by hypothesis.\n9. In $R[x]$ the constants $1$ and $0$ differ. By step 3, $1 \\neq 0$ in $R$.\n10. Let $a, b \\in R$ with $a b = 0$. By step 2, the product of the constants $a$ and $b$ is the constant $0$ in $R[x]$.\n11. Since $R[x]$ is an integral domain, the constant $a$ or the constant $b$ is $0$. By step 3, $a = 0$ or $b = 0$.\n12. By steps 8, 9 and 11, $R$ is an integral domain. This proves both directions.\n\n## Idea\n\nThe leading coefficient of a product of two polynomials is the product of their leading coefficients. If $R$ has no zero divisors, this coefficient is not zero, so the product is not zero. Conversely, $R$ sits inside $R[x]$ as the constants, so zero divisors in $R$ would be zero divisors in $R[x]$.\n",
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  "name": "Proof that R[x] is an integral domain iff R is",
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