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  "body": "\n## Definition\n\nLet $p$ be a prime number and let $n \\ge 1$ and $r \\ge 1$ be integers.\nLet $f \\in (\\Z/p^n\\Z)[x]$ be a [[monic polynomial|0C3QBCR46M]] of degree $r$ whose reduction modulo $p$ is [[irreducible|03FKX7Z5DC]] in $\\F_p[x]$.\nThe *Galois ring* of characteristic $p^n$ and rank $r$ is the [[quotient ring|0076HWHGHQ]]\n$$\\mathrm{GR}(p^n, r) = (\\Z/p^n\\Z)[x] / (f).$$\nHere $\\Z/p^n\\Z$ is the ring of [[integers modulo|03VFZ2WETT]] $p^n$.\nUp to isomorphism, this ring depends only on $p$, $n$ and $r$, and not on the choice of $f$.\n\n## Remarks\n\nTake $p = 2$, $n = 2$, $r = 2$ and $f = x^2 + x + 1$.\nThe polynomial $f$ is monic, and modulo $2$ it has no root in $\\F_2$, so it is irreducible of degree $2$ over $\\F_2$.\nHence $\\mathrm{GR}(4, 2) = (\\Z/4\\Z)[x]/(x^2 + x + 1)$ is a Galois ring with $4^2 = 16$ elements.\nThe ring $(\\Z/4\\Z)[x]/(x^2 + 1)$ is not obtained this way, because $x^2 + 1 \\equiv (x + 1)^2$ modulo $2$ is reducible.\nThe ring $\\mathrm{GR}(p^n, r)$ has $p^{nr}$ elements and characteristic $p^n$.\nIt is a local ring with maximal ideal $p \\, \\mathrm{GR}(p^n, r)$ and residue field the [[finite field|0ZXZ82FTX2]] $\\F_{p^r}$.\nThe special cases are $\\mathrm{GR}(p, r) = \\F_{p^r}$ and $\\mathrm{GR}(p^n, 1) = \\Z/p^n\\Z$.\n",
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