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  "body": "\n## Proof\n\n1. Let $f \\in \\Z[x]$ be [[monic|0BDX4MBKT4]], let $p$ be a prime, and assume that $\\bar{f}$ is [[irreducible|00X2A9G9W3]] in $\\F_p[x]$. Here $\\F_p = \\Z/p\\Z$ is a [[field|0NV6E43P4A]].\n2. Let $\\pi \\colon \\Z[x] \\to \\F_p[x]$ reduce each coefficient modulo $p$. By the [[universal property of the polynomial ring|0DQMDSVSXJ]], applied to $\\Z \\to \\F_p$ and $x \\mapsto x$, the map $\\pi$ is a ring homomorphism. Also $\\pi(f) = \\bar{f}$.\n3. If $g \\in \\Z[x]$ is monic, then $\\pi(g)$ is monic with $\\deg \\pi(g) = \\deg g$, because the leading coefficient $1$ reduces to $1 \\ne 0$.\n4. Over a field $F$, the [[units|019NFGFPR5]] of $F[x]$ are the nonzero constants. Indeed, $u v = 1$ forces $\\deg u = 0$ by the [[degree of a product|0T233AZ1GE]], and a nonzero constant has an inverse in $F$.\n5. By the same degree argument, the units of $\\Z[x]$ are the units of $\\Z$, that is, $\\pm 1$.\n6. Since $\\bar{f}$ is irreducible, it is not a unit, so $\\deg \\bar{f} \\ge 1$ by Step 4. By Step 3, $\\deg f = \\deg \\bar{f} \\ge 1$. So $f$ is nonzero and not a unit of $\\Z[x]$.\n7. Suppose that $f = g h$ with $g, h \\in \\Z[x]$. Their leading coefficients multiply to $1$, so both are $1$ or both are $-1$.\n8. In the second case, replace $g, h$ by $-g, -h$. This does not change the product, and $g$ is a unit if and only if $-g$ is. So we may assume that $g$ and $h$ are monic.\n9. By Step 2, $\\bar{f} = \\pi(g) \\, \\pi(h)$. Since $\\bar{f}$ is irreducible, $\\pi(g)$ or $\\pi(h)$ is a unit, say $\\pi(g)$. By Step 4, $\\deg \\pi(g) = 0$.\n10. By Step 3, $\\deg g = 0$. A monic polynomial of degree $0$ is the constant $1$, so $g$ is a unit.\n11. By Steps 6 to 10, $f$ is irreducible in $\\Z[x]$.\n12. The monic polynomial $f$ is [[primitive|0SP14DZN0G]]: an integer that divides every coefficient divides the leading coefficient $1$, so it is $\\pm 1$.\n13. The ring $\\Z$ is a [[Euclidean domain|000JFQGDXG]], hence a [[principal ideal domain|0QKVE1GD9A]], hence a [[unique factorisation domain|055HT85T3S]], with [[field of fractions|05K3J8J8T2]] $\\Q$.\n14. By [[Gauss's lemma|0QFV5Y1Z5X]], with Steps 11 to 13, $f$ is irreducible in $\\Q[x]$.\n\n## Idea\n\nReduction modulo $p$ is a ring homomorphism, and it keeps the degree of a monic polynomial. So a factorisation of $f$ into monic factors of positive degree would give one of $\\bar{f}$. Gauss's lemma then moves irreducibility from $\\Z[x]$ to $\\Q[x]$.\n",
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