Inseparable elements over $\mathbb F_p(t)$ Let $\mathbb F_p(t)$ denote the field of rational polynomials with coefficients in $\mathbb F_p$
I am asked to show that if some element $x$ is inseparable over $K = \mathbb F_p(t)$ then $K(x)$ contains a $p$th root of $t$.
To my understanding, this is equivalent to saying that the polynomial $X^p - t$ is reducible over $K(x)$.
Now, $x $ inseparable means that if $f(X) \in K\left[X\right]$ is the minimal polynomial for $x$ over $K$, then $gcd(f(X), f'(X)) \neq 1$
As $char(K) = p > 0$ this tells us that since minimal polynomials are irreducible, we must have $f(X) \in K\left[X^p\right] = \mathbb F_p (t)\left[X^p\right]$
Hence $p \mid \left[K(x) : K\right] = deg(f)$.
I am stuck here and unsure how I might be able to show that $X^p - t$ is reducible over $K(x)$.
Is there perhaps another approach to this question that might turn out to be simpler?
 A: My mental Model T spun its wheels for quite a while, but here’s my analysis. I need to talk about purely inseparable extensions, but I prefer to call them “radicial”, a word that’s never been too fashionable. Note the spelling, it rhymes with official, and has very little to do with radical extensions.
When $k$ is a perfect field, the pure transcendental extension $K=k(t_1,t_2,\cdots,t_n)$, transcendence degree $n$, has $K^{1/p}$ radicial of degree $p^n$, since $k(t_1,\cdots,t_n)^{1/p}=k(t_1^{1/p},\cdots,t_n^{1/p})$, using the fact that $k$ is perfect. In particular, of course, for $n=1$. I’m going to make use of a little-observed but easy Lemma, true in rather greater generality, but we need it only in this form:
Lemma. Let $k$ be a perfect field of characteristic $p>0$, and $F$ an extension of $k$ that is finite and separable over the pure transcendental field $k(t)$. If $E\supset F$ is a radicial extension of degree $p^m$, then $E=F^{1/p^m}$. In other words, $F$ has precisely one radicial extension of each possible degree $p^m$.
Now to the proof of your proposition: We have $K=k(t)$ and $x$ inseparable over $K$. Thus if $f(X)=\text{Irr}(x,K[X])$, the minimal polynomial for $x$ over $K$, we have $f(X)=g(X^{p^m})$, with $g(X)$ irreducible and separable in $K[X]$. And the hypothesis that $x$ is not separable over $K$ means just that $m\ge1$. I’ll deal with the simplest case $m=1$, everything boils down to that. Looking at $f(X)=g(X^p)$, we can set $\xi=x^p$, a root of $g$.
Now, in our situation, and subject to the above simplification, let $L=K(x)$, an extension that might well have some separable part (if $g$ is not linear), but certainly has inseparable degree $[L:K]_i=p$. We let $L_0=K(\xi)$, it’s the maximal separable extension of $K$ in $L$, but we don’t make use of that. Then $K\subset L_0\subset L$, the lower layer being separable, the upper being radicial of degree $p$. The hypotheses of the Lemma apply to the extension $L_0\supset k(t)$, so that we get:
$$
K(x)=L=L_0^{1/p}=\bigl(K(\xi)\bigr)^{1/p}=\bigl(k(t,\xi)\bigr)^{1/p}=k(t^{1/p},x)\,,
$$
and thus $t^{1/p}\in K(x)$.
