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  "body": "\n## Proof\n\n1. Let $R$ be an [[integral domain|09S3PFBEPM]] and let $f, g$ be elements of the [[polynomial ring|0Q9EZS90GY]] $R[x]$. We use the convention $-\\infty + d = d + (-\\infty) = -\\infty$ for every $d$, including $d = -\\infty$.\n2. If $f = 0$ or $g = 0$, then $f g = 0$. So both sides are $-\\infty$, by the convention for the [[degree|0DPF6WRZW6]] of $0$.\n3. Now let $f \\neq 0$ and $g \\neq 0$. Write $f = \\sum_{i=0}^{m} a_i x^i$ and $g = \\sum_{j=0}^{n} b_j x^j$ with $m = \\deg f$ and $n = \\deg g$. Then the [[leading coefficients|0P05RJC5XA]] are $a_m \\neq 0$ and $b_n \\neq 0$.\n4. Put $a_i = 0$ for $i \u003e m$ and $b_j = 0$ for $j \u003e n$. By the multiplication of $R[x]$, $f g = \\sum_{k} c_k x^k$ with $c_k = \\sum_{i + j = k} a_i b_j$, over pairs of integers $i, j \\ge 0$.\n5. Let $k \u003e m + n$, and let $i + j = k$. Then $i \u003e m$ or $j \u003e n$, since otherwise $i + j \\le m + n$. So $a_i b_j = 0$, and $c_k = 0$.\n6. Let $k = m + n$, and let $i + j = k$ with $(i, j) \\neq (m, n)$. If $i \u003c m$, then $j = m + n - i \u003e n$. If $i = m$, then $j = n$, which is excluded. So $i \u003e m$ or $j \u003e n$, and $a_i b_j = 0$ for these pairs.\n7. Hence $c_{m+n} = a_m b_n$.\n8. Since $R$ is an integral domain and $a_m \\neq 0$, $b_n \\neq 0$, we get $a_m b_n \\neq 0$.\n9. By steps 5 and 8, $f g \\neq 0$, and the largest $k$ with $c_k \\neq 0$ is $m + n$. So $\\deg(f g) = m + n = \\deg f + \\deg g$.\n\n## Idea\n\nThe highest power that can appear in $f g$ is $x^{\\deg f + \\deg g}$, and its coefficient is the product of the two leading coefficients. In an integral domain this product is nonzero, so no cancellation of the top term can happen.\n",
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