galois theory - Find the elements of the extension field using primitive polynomial over $GF(4)$

Let $p(z) = z^2 + z + 2$ be a primitive polynomial. I want to construct the elements of the extensional field $GF(4^2)= GF(16).$

Since $p(z)$ is primitive polynomial , it should generate the elements of the extension field.

Let $\alpha$ be zero point then :

• $\alpha^2 + \alpha + 2 \rightarrow \alpha^2 = -\alpha-2 \rightarrow \alpha^2 = 3\alpha + 2$.

Using $\alpha$ :

• $\alpha^1 = \alpha$


• $\alpha^2 = 3\alpha+2$ and according to the book, $\alpha^2 = \alpha+2$ but $-1 \mod 4 = 3$.

Since the base is $4$ how can it be that the correct value of $\alpha^2 = \alpha + 2$?

I have mostly solved problems where the base is a prime number and the same procedure i have used here.

Answer

You seem confused about what $GF(4)$ is: it's not $\mathbb{Z}/(4)$ (which is not even a field). You can represent $GF(4)$ as an extension of $GF(2)=\mathbb{Z}/(2)$ by an element $\beta$ such that $\beta^2=\beta+1$. You can identify $GF(4)$ with the set $\{0,1,2,3\}$ by mapping $\beta$ to $2$ and $\beta+1$ to $3$, which appears to be what the book you refer to has in mind, but this does not mean you are working mod $4$ (the addition and multiplication operations are not ordinary mod $4$ addition and multiplication on $\{0,1,2,3\}$).

In particular, $1+1=0$ in $GF(4)$ and so $\alpha+2$ and $-\alpha-2$ are the same thing, and so you have $\alpha^2=\alpha+2$.