Solution

Show that the powers of $\alpha = x$ generates the finite field with elements constructed using the irreducible polynomial $f(x) = x^3 + x^2 + 1$ from $\mathbb{Z}_2[x]$.



Note that we are looking at polynomials coming from $\mathbb{Z}_2[x]$, which means the coefficients are being treated modulo $2$.

Further, since we are only considering remainders upon division by $f(x) = x^3 + x^2 + 1$, we may replace $x^3$ with $x^2 + 1$ in order to reduce the degree of any expression we find.

As such,

$$\begin{array}{rcl} \alpha^1 &=& x\\ \alpha^2 &=& x^2\\ \alpha^3 &=& x^3 = x^2 + 1\\ \alpha^4 &=& x^3 + x = x^2 + x + 1\\ \alpha^5 &=& x^3 + x^2 + x = x + 1\\ \alpha^6 &=& x^2 + x\\ \alpha^7 &=& x^3 + x^2 = 1 \end{array}$$

These $7$ elements, in combination with $0$ make up all $8$ elements of the corresponding finite field.