Having trouble with something that Dummit and Foote is saying. On page 591 it says "Note that over $\mathbb{Q}$ or over a finite field (or, more generally, over any perfect field) the splitting field of an arbitrary polynomial $f(x)$ is the same as the splitting field for the product of the irreducible factors of $f(x)$ taken precisely once, which is a separable polynomial."
It is the last sentence that doesn't make sense because...
take $F_p$ to be our finite field. Take some element $\alpha$ that is not in $F_p$ so that $\alpha^p$ is. Then $f(x):=x^p-\alpha^p$ is irreducible over $F_p$ because $f(x)=(x-\alpha)^p$ so if this is a product of two factors, $f(x)=(x-\alpha)^i(x-\alpha)^{p-i}$ and the $x^{i-1}$ coefficient of the first factor is $i\alpha$ which is not in $F_p$. We also see that this irreducible polynomial has multiple roots in the extension field containing $\alpha$ which is evidently a splitting field for this polynomial. So $f(x)$ is not separable.
Thus $f(x)$ is the polynomial we get by taking the irreducible factor $x^p-\alpha^p$ once. But $f(x)$ is certainly not separable as dummmit and foote says it should be.
Remember Fermat's little theorem: $a^p=a$ for all $a\in\mathbb F_p$.
In particular, for every $a\in\mathbb F_p$ we have $x^p-a=x^p-a^p=(x-a)^p$, so $a$ is a $p$-fold root of $x^p-a$, and thus the only possible root of $x^p-a$ in any field extension of $\mathbb F_p$.
So there cannot be any $\alpha$ outside $\mathbb F_p$ whose $p$th power is in $\mathbb F_p$, and your construction is impossible from the outset.
If you try to create such an $\alpha$ by force, say by adjoining a new $p$th root of $a$ to $\mathbb F_p$, what you get is a ring with nontrivial zero divisors in it (namely, $(a-\alpha)^p=0$ but $a-\alpha\ne 0$), which cannot be extended to a field.