---
tag: 8Y0M6X8Q2A
kind: definition
name: Galois ring
msc: [13M05, 11T30]
defines: H9S7RMBMFD
uses: [03VFZ2WETT, 0C3QBCR46M, 03FKX7Z5DC, 0076HWHGHQ, 0ZXZ82FTX2]
sources:
  - url: https://en.wikipedia.org/wiki/Finite_ring
---

## Definition

Let $p$ be a prime number and let $n \ge 1$ and $r \ge 1$ be integers.
Let $f \in (\Z/p^n\Z)[x]$ be a [[monic polynomial|0C3QBCR46M]] of degree $r$ whose reduction modulo $p$ is [[irreducible|03FKX7Z5DC]] in $\F_p[x]$.
The *Galois ring* of characteristic $p^n$ and rank $r$ is the [[quotient ring|0076HWHGHQ]]
$$\mathrm{GR}(p^n, r) = (\Z/p^n\Z)[x] / (f).$$
Here $\Z/p^n\Z$ is the ring of [[integers modulo|03VFZ2WETT]] $p^n$.
Up to isomorphism, this ring depends only on $p$, $n$ and $r$, and not on the choice of $f$.

## Remarks

Take $p = 2$, $n = 2$, $r = 2$ and $f = x^2 + x + 1$.
The polynomial $f$ is monic, and modulo $2$ it has no root in $\F_2$, so it is irreducible of degree $2$ over $\F_2$.
Hence $\mathrm{GR}(4, 2) = (\Z/4\Z)[x]/(x^2 + x + 1)$ is a Galois ring with $4^2 = 16$ elements.
The 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.
The ring $\mathrm{GR}(p^n, r)$ has $p^{nr}$ elements and characteristic $p^n$.
It is a local ring with maximal ideal $p \, \mathrm{GR}(p^n, r)$ and residue field the [[finite field|0ZXZ82FTX2]] $\F_{p^r}$.
The special cases are $\mathrm{GR}(p, r) = \F_{p^r}$ and $\mathrm{GR}(p^n, 1) = \Z/p^n\Z$.