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Is $GR(p^k,m)$ isomorphic to $Z^{m}_{p^k}$ ?

where $GR(p^k,m)$ is the Galois ring with $p^{km}$ elements and characteristic $p^k$; and $Z^{m}_{p^k}=Z_{p^k}\times Z_{p^k}\times \ldots \times Z_{p^k}$ for $Z_{p^k}$ is the set of integers from $0$ to $p^{k-1}$

Thanks a lot.

azimut
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Math_D
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1 Answers1

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Yes and no. As an additive group the Galois ring $GR(p^k,m)$ is isomorphic to the $m$-fold cartesian product $\mathbb{Z}_{p^k}^m$. However, the latter has no really useful ring structure. The product in the Galois ring $GR(p^k,m)$ mimics that of the Galois field $GF(p^m)$ except that the rules about the primitive element are a bit different, and the coefficients of its powers are integers modulo $p^k$ as opposed to modulo $p$. We can construct the Galois ring $GR(p^k,m)$ as a quotient ring of the polynomial ring with coefficients from $\mathbb{Z}_{p^k}$ $$ GR(p^k,m):=\mathbb{Z}_{p^k}[x]/\langle f(x)\rangle, $$ where $f(x)$ is a carefully chosen irreducible monic polynomial of degree $m$ (a Hensel lift of an irreducible polynomial from $\mathbb{Z}_p[x]$).

As a consequence of this we have that we recover the finite field $GF(p^m)$ as a quotient ring of the Galois ring: $$ GR(p^k,m)/p GR(p^k,m)\simeq GF(p^m). $$ This is in sharp contrast with the (mostly useless) `componentwise' product in $R=\mathbb{Z}_{p^k}^m.$ There $R/pR\simeq\mathbb{Z}_p^m$ is not a field.

Jyrki Lahtonen
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    In this answer I give an example of Hensel lifting the polynomial $x^3+x+1\in GF(2)[x]$ to a useful cubic polynomial $f(x)$ such that we get $$GR(8,3)=\mathbb{Z}_8[x]/\langle f(x)\rangle.$$ – Jyrki Lahtonen Mar 25 '13 at 19:18
  • I have found both the answer and the comment useful, thank you. Now, considering the previous question you have replied, I have an idea to define bent functions over Galois rings. Let $f$ be a bent function from $GR(p^2,m)$ to $GR(p^2,1)$. Since we consider the additive part of the domain while definning the walsh transform, can we see the function $f$ from $Z_{p^2}^{m}$ to $Z_{p^2}$ and define: $f$ is a bent function if $|\sum_{x \in Z_{p^2}^{m}}w^{f(x)-z\cdot x}|=p^m$ where $w=e^{2 \pi i/p^2}$ and $z \in Z_{p^2}^{m}$? Is this true? I hope, my question is clear. – Math_D Mar 25 '13 at 20:16
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    @Dilek: Yes, after identifying the additive group of the Galois ring with a power of $\mathbb{Z}_{p^2}$ you can do that! Do observe that the character $x\mapsto w^{Tr(ax)}$ is usually different from the character $x\mapsto w^{a\cdot x}$. But that is of no consequence to the bent-function property, because as the parameter $a$ varies, you get the same set of characters from both recipes :-) – Jyrki Lahtonen Mar 25 '13 at 20:22
  • That is nice:) I will try to prove this. Thanks a lot! – Math_D Mar 25 '13 at 20:37