Let $f(x) = x^8+a \in \mathbb{F}_{49}[x]$ with $a \in \mathbb{F}_{49}\setminus \{0\}$. Find all $a$ such that $f$ is reducible over $\mathbb{F}_{49}$!
What I know is that $\mathbb{F}_{49} \cong \mathbb{F}_7/(t^2+2)$, hence $a = b_1t+b_2$ with $b_1$, $b_2 \in \mathbb{F}_7$. The brute force method would be to try out every single $a$...
But can I just use Eisenstein and say that as long as $a \neq 1$ and $a$ irreducible over $\mathbb{F}_7$, $f$ is irreducible? Hence I only need to consider the case that $a = b_2$. Is this correct?
Eisenstein has nothing to do with the case. Rather, this story intimately involves the facts that the multiplicative group of a finite field is cyclic, and that the degree of an irreducible polynomial is the same as the field extension degree $[k(z):k]$ if $z$ is any one of the roots of the polynomial.
To make typing easier for myself, I’m going to ask about the reducibility of $X^8-b$, and leave it to you to make the substitution $b=-a$. And I’ll simplify things by writing $k=\Bbb F_{49}$.
What’s the meaning of adjoining the eighth root of an element $b\in k$? If the order of $b$ is $m$ (i.e. $b^m$ is the lowest power of $b$ that’s equal to $1$), then the order of $\sqrt[8]b$ will be $8m$. Adjoining $\sqrt[8]b$ to $k$ will involve an extension of degree $\le8$. This degree will be only $1$ exactly when the order of $b$ is $\le6$, i.e. when $b\in\Bbb F_7$. When this degree is less than $8$, $X^8-b$ will be reducible; when equal, $X^8-b$ will be irreducible.
Thus we can say that $X^8-b$ is reducible if and only if $\left[k(\sqrt[8]b\,):k\right]<8$. Let $k_d$ be the (unique, up to isomorphism) field of degree $d$ over $k$, so that $k=k_1$. We have $|k_d|=49^d$ and $|k_d^\times|=49^d-1$. Now, $k_d$ has an element of order $n$ if and only if $n|(49^d-1)$; still calling $m$ the order of $b$, $k_d$ has an eighth root of $b$ if and only if $8m|(49^d-1)$. Putting all this together, $X^8-b$ is reducible if and only if there is $d<8$ with $8m|49^d$. Direct computation shows that this is the case for $m=1,2,3,4,6,8,12,24$, but not $16$ nor $48$. Since the numbers on the first list are just the divisors of $24$, we can say that $X^8-b$ is reducible if and only if $b^{24}=1$ in $k$.
All that labor, and the elephant delivered a mouse. I could have said, simply, that if $b^{24}=1$ and $z^8=b$, then $z^{192}=1$ and since $192|(49^4-1)$, you get $z\in\Bbb F_{49^4}$, so that $X^8-b$ is reducible. But would you have understood my motivation?