Let $p$ be a prime and set $g = X^2 + X - 1\in \mathbb{F}_p[X]$. Prove the following: $$\text{$g$ is irreducible in $\mathbb{F}_p[X]$ iff $g$ is irreducible in $\mathbb{F}_{p^3}[X]$}$$ One side is really easy to prove, namely
$\Leftarrow$ We know that $\mathbb{F}_p \subset \mathbb{F}_{p^3}$. As $g$ is irreducible in $\mathbb{F}_{p^3}[X]$ and $\deg(g) = 2$, then $g$ has no zeros in $\mathbb{F}_{p^3}$. Therefore, it has no zeros in $\mathbb{F}_p$, thus it is irreducible in $\mathbb{F}_p$.
Any suggestions about the other implication are welcome.
We will assume that $g$ is irreducible in $\mathbb{F}_p[X]$. On the contrary, suppose that $g$ is not irreducible in $\mathbb{F}_{p^3}[X],$ i.e., suppose there exists a root $\alpha$ of $g$ in $\mathbb{F}_{p^3}.$ By definition, $[\mathbb{F}_p(\alpha):\mathbb{F}_p]$ is the degree of the minimal polynomial of $\alpha$ over $\mathbb{F}_p$. As $g$ is irreducible, we conclude that the minimal polynomial is $g$ and $[\mathbb{F}_p(\alpha):\mathbb{F}_p] = \deg(g) = 2$. Also, recall that $[\mathbb{F}_{p^3}:\mathbb{F}_p] = 3$. As $\mathbb{F}_p \subset \mathbb{F}_p(\alpha) \subset \mathbb{F}_{p^3},$ we have that $$\begin{align*} [\mathbb{F}_{p^3}:\mathbb{F}_p] &= [\mathbb{F}_{p^3}:\mathbb{F}_p(\alpha)]\cdot[\mathbb{F}_p(\alpha): \mathbb{F}_p] \text{ so that} \\ 3 &= [\mathbb{F}_{p^3}:\mathbb{F}_p(\alpha)]\cdot 2. \end{align*}$$ But this is a contradiction: $3$ is not divisible by $2$. Hence, $g$ is irreducible in $\mathbb{F}_{p^3}$.