About Galois rings 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.
 A: 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.
