Let $p(x)$ be an irreducible element in $F[x]$. Then by Theorem 1206, we know that $K = F[x] / \left( p(x) \right)$ is a field extension of $F$, and $K = F + F\theta + \cdots + F\theta^{n-1}$ with $\theta = \overline{x} \in F[x] / \left( p(x) \right)$. Hence $K = F[\theta] = F(\theta)$, and $K / F$ is a simple extension, and $\theta$ is a primitive element for $K / F$.
$\mathbb{F}_{p}(t) / \mathbb{F}_{p}$ is simple and $t$ is a primitive element for $\mathbb{F}_{p}(t) / \mathbb{F}_{p}$. In this case, $\mathbb{F}_{p}(t) \neq \mathbb{F}_{p}[t]$.
Let $F$ be a field. Then $$F(x) = \left\{ \left. \frac{a(x)}{b(x)} \right\vert a(x), b(x) \in F[x], b(x) \neq 0 \right\}$$ is a field with $$\begin{align*}+ &: \frac{a(x)}{b(x)} + \frac{c(x)}{d(x)} = \frac{a(x) d(x) + b(x) c(x)}{b(x) d(x)} \\ \times &: \frac{a(x)}{b(x)} \cdot \frac{c(x)}{d(x)} = \frac{a(x) c(x)}{b(x) d(x)}\end{align*}.$$$F(x) / F$ is simple and $x$ is a primitive element for $F(x) / F$. $F[x] \neq F(x)$.