2

How can I determine which rings are isomorphic to the Galois ring $GR(p^r,m)$ for specific values of $p$, $r$ and $m$? For instance, $GR(2,2)$ is isomorphic to $\mathbb{F_4}$, while $GR(2,3)=\mathbb{F_8}$ and $GR(3,2)=\mathbb{F_{27}}$ but I don't know which rings are isomorphic to $GR(4,3)$ and $GR(8,2)$. Please help me. Are Galois rings always isomorphic to rings of the form $\mathbb{F_q}$? If so, are there formulas to find such $\mathbb{F_q}$? Sorry, I am new at this.

Haded1027
  • 63

1 Answers1

2

The Galois ring $GR(p^r, m)$ is a quotient of the polynomial ring $(\mathbb{Z}/p^r\mathbb{Z})[X]$ by a monic irreducible polynomial of degree $m$. When $r = 1$, $GR(p, m)$ is indeed isomorphic to the finite field $\mathbb{F}_{p^m}$ (it is a field extension of $\mathbb{F}_p$ of degree $m$). When $r > 1$, $GR(p^r, m)$ cannot be isomorphic to any $\mathbb{F}_q$ because it is not a field (for instance, $p$ has no inverse since $p \cdot p^{r-1} = 0$ in $GR(p^r, m)$). The 'about' page https://math.stackexchange.com/tags/galois-rings/info might also be helpful.

Community
  • 1
user2150659
  • 101
  • Thanks but I tried $GR(2,2)$ so it is $(\mathbb{Z}/p^r\mathbb{Z})[X]$ over a monic irreducible polynomial right? In this case, I used the polynomial $X^2+X+1$ but I found that $(\mathbb{Z}/p^r\mathbb{Z})[X]/(X^2+X+1)$ has elements ${0,1,X,X+1}$ and I noticed that the order of the non-zero elements are all $2$. However, for $\mathbb{F}_4$ the elements $1$ and $3$ have order $4$ but only the element $2$ has order $2$. Where did I go wrong? Is the polynomial $X^2+X+1$ not irreducible? Are the elements ${0,1,X,X+1}$ incorrect? Am I wrong about the order of elements? – Haded1027 Nov 26 '16 at 04:04
  • Oh sorry. My bad. The elements $1$, $2$ and $3$ with orders $4$, $2$ and $4$ respectively mentioned in my comment above are elements of $\mathbb{Z}_4$ and not $\mathbb{F}_4$. I get it now. Thanks! – Haded1027 Nov 26 '16 at 04:34