# Struggling to understand a field theorm's corollary

1. Feb 27, 2015

### PsychonautQQ

Theorem: Let F be any field. If G is a finite subgroup of the multiplicative group F* of F, then G is cyclic. In particular, is F is finite, then F* is cyclic.

Corolarry 1: GF(p^n) = Z_p(u), where u is any primitive element for GF(p^n).

So <u> = GF(p^n)*, so |u| = GF(p^n) - 1.

I'm now trying to imagine what Z_p(u) would look like, maybe:

{a_0 + a_1*u + a_2*u^2 + ...... a_(n-1)*u^n-1 | a_i are elements of Z_p, u^n = (0?)}

This would make sense because this field would have order n... but it would also mean that u^n = (0?) or something, when u^n should just equal u^n because |u| = GF(p^n) - 1 > n.

Anyone understand my dilemma? If anyone could drop some knowledge on this topic in general it'd be appreciated.

2. Feb 27, 2015

### micromass

Staff Emeritus
Well, let's look at an example. Take $\mathbb{F}_4 = \{0,1,a,a+1\}$, where $a^2 = a+1$. The primitive element is clearly $a$ and this has order $3$ because
$$a^2 = a+1~\text{and}~a^3 = a(a+1) = a^2 + a = a+1+a =1$$
We indeed have $\mathbb{F}_4 = \{\alpha + \beta a~\vert~\alpha,\beta\in \mathbb{Z}_p\}$. But you see that this does not imply that $a^2 = 1$. The issue is that $\alpha + \beta a$ and $a^n$ are very different notations which might coincide. In this situation, we have $a^2 = a+1$. So if you look at the cyclic element $a$, then we have a representation $\{0,a,a^2,a^3\}$ and when you look at it your way then we have $\{0,1,a,1+a\}$. These are two very different notations.

I encourage you to try other examples to see this more clearly.

3. Feb 27, 2015

### PsychonautQQ

why is a^2 = a + 1? As a field, all these elements need additive inverses do they not? what is the additive inverse for 1?

4. Feb 27, 2015

### micromass

Staff Emeritus
1 is its own additive inverse

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