Inseparable, irreducible polynomials The standard examples of irreducible, inseparable polynomials that one encounters in an introductory course on field theory all seem to have only a single root in an algebraic closure. Are there elementary examples of inseparable, irreducible polynomials with multiple different roots (at least one of which is repeated)? Equivalently, can a field extension contain elements which are inseparable, but whose minimal polynomials have more than one distinct root?
 A: EDIT: Discussions in the comments have convinced me to add a bit of introduction to my example, as follows: any separable extension of an inseparable extension ought to provide an example, so here's a very simple separable extension of a very simple inseparable extension: 
Let $F={\bf F}_3(t)$, let $E=F(t^{1/3})$, let $K=E(\sqrt2)$. Note $[K:F]=[K:E][E:F]=2\times3=6$. Show that $K=F(t^{1/3}+\sqrt2)$ by showing that that element is not of degree 2 or 3. Then show that that element is what you're looking for. 
A: Consider
$$
(X-a)^p(X-b)^p\in\mathbb F_p(a^p+b^p,a^pb^p)[X],
$$
where $a,b,X$ are indeterminates.
EDIT. A generalization: Let $q$ be a power of a prime, let $a_1,\dots,a_n$ be indeterminates, put
$$
f:=(X-a_1)^{q^k}\cdots(X-a_n)^{q^k}\in\mathbb F_q[a_1,\dots,a_n,X],
$$
write $K$ for the extension of $\mathbb F_q$ generated by the coefficients of $f$. 
Then $f$ is irreducible in $K[X]$, and any example will be a specialization of this one.
A: Let $p \in \mathbb N$ be prime, $q \in \mathbb N$ coprime to $p$, and let $F = \mathbb F_p(t)$ the field of rational functions of $t$ with coefficients in $\mathbb F_p$. Consider
$$
f(x) = x^{pq} - t.
$$
EDIT : By Eisenstein's criterion, $x^{pq} - t$ is irreducible over $\mathbb F_p[t]$ (because $t$ is a prime in there). By Gauss' Lemma, it is also irreducible over the field of fractions, which is $\mathbb F_p(t)$. Thanks to Sam L. for this part of my argument.
Since the derivative of $f$ is zero in $\mathbb F_p(t)[x]$, the polynomial is inseparable. But the polynomial $x^q - 1$ is separable in $\mathbb F_p(t)[x]$, because its derivative is $qx^{q-1}$, which has no common roots with $x^q - 1$, so that the roots of $x^q - 1$ are distinct. Now letting $\sqrt[pq]t$ be a root of $x^{pq} - t$ and $w$ a $q^{th}$ root of unity. Then the distinct roots of $f$ are $w^i (\sqrt[pq]t)$, with $i$ ranging from $0$ to $q-1$, each with multiplicity $p$.
Hope that helps, 
