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  "body": "\n## Proof\n\nLet $R$ be a [[unique factorisation domain|06YMXJMXYS]] with [[field of fractions|0VF11ZTTMD]] $K$. We identify $a \\in R$ with $a/1 \\in K$, so $R[x] \\subseteq K[x]$. Let $f \\in R[x]$ be [[primitive|0SP14DZN0G]].\n\n1. The rings $R$ and $K$ are [[integral domains|09S3PFBEPM]]. So $R[x]$ and $K[x]$ are integral domains by [[the theorem on polynomial rings|0B5Z1EC66J]], and [[degrees add|0T233AZ1GE]] in products.\n2. If $u v = 1$ in $R[x]$, then $\\deg u + \\deg v = 0$, so $u$ and $v$ are constants of $R$. So the [[units|019NFGFPR5]] of $R[x]$ are the units of $R$.\n3. In the same way, the units of $K[x]$ are constants. Every nonzero constant is a unit, since $K$ is a [[field|0QA8CHH8MG]]. So the units of $K[x]$ are the polynomials of [[degree|0DPF6WRZW6]] $0$.\n4. Let $p \\in R$ be [[prime|0MNHXBSX3M]] and let $G = \\sum_i a_i x^i$ and $H = \\sum_j b_j x^j$ in $R[x]$. Suppose $p$ divides every coefficient of $G H$, but not every $a_i$ and not every $b_j$.\n5. Let $s$ and $t$ be the least indices with $p \\nmid a_s$ and $p \\nmid b_t$. In the coefficient $\\sum_{i + j = s + t} a_i b_j$, each term with $i \u003c s$ or $j \u003c t$ is divisible by $p$.\n6. Every term other than $a_s b_t$ has $i \u003c s$ or $j \u003c t$. So $p \\mid a_s b_t$, and $p \\mid a_s$ or $p \\mid b_t$. This is a contradiction.\n7. So if a prime $p$ divides every coefficient of $G H$, then $p$ divides every coefficient of $G$ or every coefficient of $H$.\n8. Claim: if $r \\in R$, $r \\neq 0$, and $r f = G H$ with $G, H \\in R[x]$, then $f = G' H'$ with $G', H' \\in R[x]$, $\\deg G' = \\deg G$ and $\\deg H' = \\deg H$.\n9. Every nonzero $r$ is $r = u p_1 \\cdots p_n$ with $u$ a unit, $n \\ge 0$ and all $p_k$ [[irreducible|052SDEPZH7]]. We prove the claim by induction on $n$.\n10. If $n = 0$, then $r = u$ is a unit. Take $G' = u^{-1} G$ and $H' = H$.\n11. If $n \\ge 1$, put $p = p_1$ and $r' = u p_2 \\cdots p_n$. Since [[irreducible elements are prime|0QRCFZKXH8]] in $R$, $p$ is prime.\n12. The prime $p$ divides every coefficient of $r f = G H$. By step 7 we may assume that $p$ divides every coefficient of $G$, so $G = p G_1$ with $G_1 \\in R[x]$ and $\\deg G_1 = \\deg G$.\n13. Then $p (r' f) = p (G_1 H)$. [[Cancelling|0GJM24EXP9]] $p \\neq 0$ in $R[x]$ gives $r' f = G_1 H$. The induction hypothesis for $r'$ proves the claim.\n14. The polynomial $f$ is not $0$, since the nonunit $0$ divides every coefficient of $0$. If $\\deg f = 0$, then $f = a \\in R$ divides itself, so $a$ is a unit of $R$.\n15. Then $f$ is a unit of $R[x]$ and of $K[x]$. So $f$ is [[irreducible|00X2A9G9W3]] in neither ring, and the statement holds. From now on, let $\\deg f \\ge 1$.\n16. By steps 2 and 3, $f$ is a nonzero nonunit both in $R[x]$ and in $K[x]$.\n17. Suppose that $f$ is irreducible in $K[x]$, and let $f = g h$ with $g, h \\in R[x]$. Then $g$ or $h$ is a unit of $K[x]$, say $g$.\n18. By step 3, $g$ is a nonzero constant $a \\in R$. Then $a$ divides every coefficient of $f = a h$. Since $f$ is primitive, $a$ is a unit of $R$, hence of $R[x]$.\n19. So $f$ is irreducible in $R[x]$.\n20. Conversely, suppose that $f$ is irreducible in $R[x]$. Suppose that $f = g h$ in $K[x]$ with $g$ and $h$ not units of $K[x]$.\n21. Since $f \\neq 0$, also $g, h \\neq 0$. By step 3, $\\deg g \\ge 1$ and $\\deg h \\ge 1$.\n22. Write the coefficients of $g$ as $u_k / v_k$ with $u_k, v_k \\in R$ and $v_k \\neq 0$. Let $d$ be the product of the $v_k$. Then $d \\neq 0$ and $G = d g$ lies in $R[x]$.\n23. In the same way, there is $e \\in R$ with $e \\neq 0$ and $H = e h \\in R[x]$. Then $(d e) f = G H$ in $R[x]$, with $\\deg G = \\deg g$ and $\\deg H = \\deg h$.\n24. By the claim of step 8, $f = G' H'$ with $G', H' \\in R[x]$ of degrees at least $1$. By step 2, neither is a unit of $R[x]$.\n25. This contradicts the irreducibility of $f$ in $R[x]$. So $f$ is irreducible in $K[x]$, which proves the equivalence.\n\n## Idea\n\nA factorisation over $K$ can be moved to a factorisation over $R$: clear the denominators, then remove the extra constant one prime at a time. Each prime of the constant divides all coefficients of one factor, because it divides all coefficients of the product. Primitivity rules out constant factors over $R$, so the two notions of irreducibility agree.\n",
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