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  "body": "\n## Definition\n\nLet $(R, \\mathfrak m)$ be a [[local ring|MZ02BKPHJA]] with [[residue field|NJ3APQ8BHH]] $k = R/\\mathfrak m$.\nFor $f \\in R[x]$, write $\\bar f \\in k[x]$ for the polynomial obtained by reducing the coefficients modulo $\\mathfrak m$.\nThe ring $R$ is *Henselian* if the following holds for every [[monic polynomial|0C3QBCR46M]] $f \\in R[x]$.\nIf $\\bar f = g_0 h_0$ with $g_0, h_0 \\in k[x]$ monic and coprime, then there are monic $g, h \\in R[x]$ with $f = gh$, $\\bar g = g_0$ and $\\bar h = h_0$.\n\n## Remarks\n\nAn equivalent condition: for every monic $f \\in R[x]$ and every $a_0 \\in k$ with $\\bar f(a_0) = 0$ and $\\bar f'(a_0) \\ne 0$, there is $a \\in R$ with $f(a) = 0$ and $a \\equiv a_0 \\pmod{\\mathfrak m}$.\nThat is, simple roots modulo $\\mathfrak m$ lift to roots.\n\nExample: every complete Noetherian local ring is Henselian, by the general form of [[Hensel's lemma|010QVB71T2]]; this includes the $p$-adic integers $\\Z_p$ and $k[\\![t]\\!]$ for a field $k$.\nEvery field is Henselian, since $\\mathfrak m = 0$ and $\\bar f = f$.\n\nNon-example: $R = \\Z_{(5)}$ is not Henselian.\nFor $f = x^2 + 1$, one has $\\bar f = (x - 2)(x - 3)$ in $\\Z/5\\Z[x]$, and the factors $x - 2$ and $x - 3$ are coprime because $2 \\ne 3$.\nA lift would give a root $a \\in \\Z_{(5)} \\subseteq \\Q$ of $x^2 + 1$, but $-1$ has no rational square root.\n",
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