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  "body": "\n## Proof\n\nLet $R$ be a [[unique factorisation domain|06YMXJMXYS]] and let $f = \\sum_i a_i x^i$ and $g = \\sum_j b_j x^j$ be [[primitive polynomials|0SP14DZN0G]] in $R[x]$. We argue by contradiction.\n\n1. The zero polynomial is not primitive: $0$ divides all its coefficients, and $0$ is not a [[unit|019NFGFPR5]] since $1 \\neq 0$. So $f \\neq 0$ and $g \\neq 0$.\n2. Since $R$ is an integral domain, $R[x]$ is an integral domain by [[the theorem on polynomial rings|0B5Z1EC66J]]. So $f g \\neq 0$.\n3. Suppose that $f g$ is not primitive. Then some $r \\in R$ that is not a unit divides every coefficient of $f g$.\n4. If $r = 0$, then every coefficient of $f g$ is $0$, so $f g = 0$, against step 2. So $r \\neq 0$.\n5. By unique factorisation, $r = p_1 \\cdots p_n$ with $n \\ge 1$ and all $p_k$ [[irreducible|052SDEPZH7]]. Put $p = p_1$. Then $p$ divides $r$, so $p$ divides every coefficient of $f g$.\n6. In a unique factorisation domain, [[irreducible elements are prime|0QRCFZKXH8]]. So $p$ is a [[prime element|0MNHXBSX3M]].\n7. Since $p$ is not a unit and $f$ is primitive, $p$ does not divide every $a_i$. Let $s$ be the least index with $p \\nmid a_s$.\n8. In the same way, let $t$ be the least index with $p \\nmid b_t$.\n9. The coefficient of $x^{s+t}$ in $f g$ is $c = \\sum_{i + j = s + t} a_i b_j$. By step 5, $p \\mid c$.\n10. Take a term $a_i b_j$ with $i + j = s + t$ and $i \\neq s$. If $i \u003c s$, then $p \\mid a_i$ by the choice of $s$.\n11. If $i \u003e s$, then $j \u003c t$, so $p \\mid b_j$ by the choice of $t$. In both cases $p \\mid a_i b_j$.\n12. So $a_s b_t = c - \\sum a_i b_j$ is divisible by $p$, where the sum runs over $i + j = s + t$ with $i \\neq s$.\n13. Since $p$ is prime, $p \\mid a_s$ or $p \\mid b_t$. This contradicts steps 7 and 8.\n14. Hence every $r \\in R$ dividing all coefficients of $f g$ is a unit, and $f g$ is primitive.\n\n## Idea\n\nIf $f g$ were not primitive, some prime $p$ would divide all its coefficients. Look at the first coefficient of $f$ and the first coefficient of $g$ that $p$ does not divide. In the matching coefficient of $f g$, every other term is divisible by $p$, so their product would be too, which is impossible for a prime.\n",
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