What are the elements of GF(9)? What is the 'addition' and 'multiplication' operations on this field?

Question

I have read methods to construct GF(p^m). I have understood the primitive polynomials and other concepts but I have not understood how the p and m are entering into the discussion. And finally the proof the pudding. What are the elements of it, what are the two operations on this set? How do we show that they satisfy the requirements of a field? For simplicity let us take GF(3^2 =9). What are the elements of this field and what are the operations on this?

Answer 1

To give $\def\F{\mathbf F}\F_9$ (or $\F_{p^m}$ in general), we start with $\F_3$ ($\F_p$) an an irreducible polynomial of degree $2$ over $\F_3$. As $p(X) = X^2 + 1$ has no roots in $\F_3$ ($p(0) = 1$, $p(1) = 2$, $p(2) = 2$), $p$ is irreducible. Now we define $\F_9 := \F_3[X]/(p)$. That is the elements of $\F_9$ are (in this representation), the classes $\pmod{(p)}$ represented by $$ 0, 1, 2, X, X+1, X+2, 2X, 2X+1, 2X+2 $$
Addition and multiplication is Addition and multiplication of polynomials $\pmod p$. For example, $$ (X+2) \cdot (X+1) = X^2 + 3X + 2 = X^2 + 2 = (X^2 + 1) + 1 = 1 \pmod {X^2 +1} $$ That $\F_9$ is a field follows from the irreducibility of $p$.

Answer 2

$GF(9)$ is by definition the splitting field of $X^9-X$. So the elements are the roots of $X^9-X$ (except the root $0$ it's not easy at all to find them. We usually need a computer). In general, $GF(p^m)$ is the splitting field of $X^{p^m}-X$. Now, it can be interesting for you to prove that $$\{x\mid x^{p^m}-x=0\}$$ is really a field.

Answer 3

$$GF(9)=:\Bbb F_9=\Bbb F_3[x]/\langle\,x^2+1\,\rangle$$

and the elements of the quotient ring can be expressed in the form $\;aw+b\;,\;\;a,b\in\Bbb F_3\;,\;\;w^2=-1\;$ , so we actually get nine elements.

The fact $\;\Bbb F_9\;$ is a field is because $\;x^2+1\in\Bbb F_3[x]\;$ is irreducible , so the ideal generated by it is maximal in this polynomial ring.

Thus, we have that

$$\Bbb F_9=\left\{\;0,\,1,\,2,\,w,\,2w,\,w+1,\,2w+1,\,w+2,\,2w+2\;\right\}$$

and with the addition and multiplication rule determined by addition and multiplication modulo $\;3\;$ and by $\;w^2=-1\;$ , for example:

$$(w+1)(2w+1)=2w^2+\overbrace{w+2w}^{=3w=0\cdot w=0}+1=2(-1)+0+1=-2+1=-1=2$$

and since $\;2^{-1}=2\pmod3\;$ , we get that

$$1=2\cdot2=(w+1)(2(2w+1))=(w+1)(w+2)\implies (w+1)^{-1}=w+2$$

in this field. Now you can play around with this.