Lecture 12
The Primitive Element Theorem
 

Example 10.6  

We know that the polynomial $ x^{p^n}-x$ is the product of all the degree $ d$ monic irreducible polynomials in $ \mathbb{F}_p[x]$ where $ d \mid n.$ This is useful for constructing irreducible polynomials over $ \mathbb{F}_p.$ Let us factorize $ x^{16}-x$ over $ \mathbb{F}_2.$ The irreducible quadratic polynomials are factors of $ x^4-x=x(x+1)(x^2+x+1).$ Hence there is only one quadratic irreducible polynomial over $ \mathbb{F}_2.$ The cubic irreducibles are factors of

$\displaystyle x^8-x=x(x^7+1)=x(x+1)(x^6+x^5+x^4+x^3+x^2+x+1).$

By Gauss' formula $ N_2(3)=2.$ Therefore the irreducible cubics over $ \mathbb{F}_2$ are $ x^3+x^2+1$ and $ x^3+x+1.$ By Gauss' formula, we count irreducible quartics over $ \mathbb{F}_2:$
$\displaystyle 4N_2(4)=\sum_{d\vert 4} \mu(4/d)2^d=\mu(4)2+\mu(2)2^2+\mu(1)2^4=-4+16=12.$

Hence $ N_2(4)=3$. These quartics are factors of $ x^{16}-x.$ The irreducible factors of this polynomial have degrees $ 1, 2$ and $ 4.$ Therefore the irreducible quartics are factors of

$\displaystyle \frac{x^{16}-x}{x(x+1)(x^2+x+1)}=(x^4+x+1)(x^4+x^3+1)(x^4+x^3+x^2+x+1).$
We end this section by an interesting application of finite fields.