• Oct 18th 2011, 06:38 AM
mathdigger
I was wondering if someone could explain to me how the Galois Field GF(p^n) compares to the ring Z/(p^n)Z i.e. the ring of integers modulo p^n, where p is a prime and n is any integer.

I know that the first one is a field while the second one isn't. However, they seem to have equal number of elements. I just want an intuitive comparison of these two.

I wish to know this because GF(p) and Z/pZ seem to be exactly the same structures!
• Oct 18th 2011, 03:34 PM
Deveno
normally, $\text{GF}(p^n)$ is constructed as a quotient ring $\mathbb{Z}_p[x]/(f(x))$ where f(x) is an irreducible polynomial of degree n in $\mathbb{Z}_p[x]$. this lets us think of $\text{GF}(p^n)$ elements in 3 different ways:

1)polynomial expressions in a root of f(x)
2)a vector space over $\mathbb{Z}_p$, of dimension n
3)an extension field of $\mathbb{Z}_p$

now $\mathbb{Z}/p^n\mathbb{Z}$ isn't n copies of $\mathbb{Z}_p$ interacting....it's one long loop of length $p^n$. and although $p$ is prime, $p^n$ is not, which means it has divisors. and these divisors "get in the way" when we try to divide:

$p(p^{n-1}) = 0\ (mod\ p^n)$, which is bad behavior for a field, it means p has no inverse!

geometrically, you can think of it this way: $\text{GF}(p^n)$ is like n loops of length p, all tied together at 0. $\mathbb{Z}/p^n\mathbb{Z}$ is what you get when you untie the loops, tie the end of one to the beginning of the next one, until you have one big long loop. even though we have the same number of elements, it seems unreasonable to expect these will behave "the same".
• Oct 20th 2011, 06:14 AM
mathdigger
Right. Got it. Another question: I read somewhere that GF(p^n) has characteristic p. What exactly does that mean, in simple words, and what significance does that have?
• Oct 20th 2011, 02:28 PM
Drexel28
Quote:

Originally Posted by mathdigger
Right. Got it. Another question: I read somewhere that GF(p^n) has characteristic p. What exactly does that mean, in simple words, and what significance does that have?

Not to step on Deveno's toes, but the characteristic of a unital ring $R$ is the ADDITIVE order of its multiplicative identity (where we define contradictory to this definition, for a reason which becomes clear from experience, that the characteristic of a ring whose multiplicative identity has infinite order to be zero). So, for example $\mathbb{Z},\mathbb{C},\mathbb{C}[x],C[a,b]$ all have characteristic , $\mathbb{Z}/m\mathbb{Z}$ has characteristic $m$, and $\mathbb{Z}_2^{\infty}=\left\{f:\mathbb{N}\to\mathb b{Z}_2\right\}$ has characteristic $2$. It's a common fact that an integral domain has characteristic either zero or a prime (if $R$ is an integral domain and $\text{char}(R)=c=ab$ then $0=(c)1=(a1)(b1)$ and so one of $a1,b1$ has to be zero, and so if $a,b>0$ this implies that $|1|\leqslant \max\{a,b\} which is a contradiction). So, what is the significance? Intuitively every unital ring $R$ contains a unique smallest unital subring, call this $R_0$. It's pretty easy when you write it out to prove that $R_0=\langle 1\rangle$ and so $|R_0|=\text{char}(R)$. Thus, the characteristic of a ring measures how large the "core" of the ring is. For finite fields this is particular nice because one can prove that if $\mathbb{F}$ is a finite field then $\mathbb{F}$ is a vector space over $\mathbb{F}_0$. We then have from what I said previously that $|\mathbb{F}_0|$ is zero or prime. It can't be zero because that woud imply that $\mathbb{F}$ is infinite and so $|\mathbb{F}_0|=p$ for some prime $p$. From this and the fact that $\mathbb{F}$ is a vector space over $\mathbb{F}_0$ we may conclude that, as vector spaces, $\mathbb{F}\cong \mathbb{F}_0^{k}$ where $k=\dim_{\mathbb{F}_0}\mathbb{F}$. In particular, $|\mathbb{F}|=p^k$. This is the reason why finite fields must have order equal to a prime to a power.
• Oct 20th 2011, 06:41 PM
Deveno
we basically get two choices for this field: $\mathbb{Q}$, if char(F) = 0, or $\mathbb{Z}_p$ if char(F) = p.