Lecture 12
The Primitive Element Theorem
 


Proposition 10.7  

The polynomial $ x^4+1$ is irreducible in $ \mathbb{Z}[x]$ but it is reducible over $ \mathbb{F}_p$ for every $ p.$


Proof.
Let $ f(x)=x^4+1.$ Then $ f(x+1)$ is irreducible over $ \mathbb{Z}$ by Eisenstein's criterion. For $ p=2,$ we have $ x^4+1=(x+1)^4.$ Now let $ p$ be odd. Then $ 8 \mid p^2-1.$ Hence

$\displaystyle x^4 +1 \mid x^8-1 \mid x^{p^2-1}-1 \mid x^{p^2}-x.$

The splitting field of $ x^{p^2}-x$ over $ \mathbb{F}_p$ is the finite field $ F=\mathbb{F}_{p^2}.$ Hence $ [F:\mathbb{F}_p]=2.$ Therefore the roots of $ x^4+1$ in $ F$ have degree $ 1$ or $ 2.$ Therefore $ x^4+1$ cannot have a cubic or quartic irreducible factor over $ \mathbb{F}_p.$ Hence it is reducible over $ \mathbb{F}_p$ for each prime $ p.$

11. The Primitive Element Theorem

Since $ \mathbb{F}_{{q}^{n}}^{\times}$ is a cyclic group, $ \mathbb{F}_{{q}^{n}} = \mathbb{F}_{q} (\alpha)$ where $ \alpha$ is a generator of $ \mathbb{F}_{{q}^{n}}^{\times}$We say that $ \alpha$ is primitive element of the field extension $ \mathbb{F}_{q} \subset \mathbb{F}_{{q}^{n}}$. In this section we discuss existence of primitive elements in finite algebraic field extensions. We will show that in a finite separable extension, primitive elements always exist.


Definition 11.1  

Let $ E/F$ be a field extension. An element $ \alpha \in E$ is called a primitive element of $ E$ over $ F$ if $ E = F(\alpha)$.