$\mathbb{P}^1_{k'}$ is not smooth over $k$ if $k'/k$ finit inseparable Let $k'$ a finite inseparable extension of a field $k$, with $k$ not perfect. I want to prove that $\mathbb{P}^1_{k'}$ is not smooth over $k$.
Because all is local I guess this is equivalent with $\mathbb{A}^1_{k'}$ is not smooth over $k$ that is $\operatorname{Spec}(k'[x]\otimes_k\overline{k})$ is not regular, but I don't see how to simplify this tensor product.
 A: Say $k = \Bbb F_p(t^p)$ and $k' = \Bbb F_p(t)$.
The first thing to do is to find out what exactly is $k' \otimes_k k'$. Since $k'$ is a $k$-vector space of dimension $p$, this is going to be a $k$-vector space of dimension $p^2$, or also a $k'$-vector space of dimension $p$.
If you consider $k' \otimes_k k'$ as a $k'$-vector space with multiplication on the right, a basis is the family $1 \otimes 1, t \otimes 1, t^2 \otimes 1, \ldots, t^{p-1} \otimes 1$.
Let $\varepsilon = 1 \otimes 1 - t \otimes t^{-1}$. It should be easy to check that with respect to this basis the matrix of the family $1 \otimes 1, \varepsilon, \varepsilon^2 \ldots, \varepsilon^{p-1}$ is a triangular matrix with nonzero entries on the diagonal, so that it also is a $k'$-basis for $k' \otimes_k k'$.
Finally, $\varepsilon^p = 1 \otimes 1 - (t^p \otimes t^{-p}) = 1 \otimes 1 - 1 \otimes 1 = 0$. We conclude that $k' \otimes_k k'$ is isomorphic to $k'[\varepsilon]/(\varepsilon^p)$. And yes, everyone has quite a lot of $p$-th roots in this ring.
Similarly, our ring of interest $A = k'[x] \otimes_k \bar k \simeq \bar k[x,\varepsilon]/(\varepsilon^p)$. 
The spectrum of this is the same as the regular $\operatorname{Spec}(\bar{k}[x])$, but where we add $\varepsilon$ to every prime ideal : Since $\varepsilon^p = 0$, $\varepsilon$ is in the radical of $A$, so is included in any prime ideal. Prime ideals of $A$ then correspond to prime ideals of $A/(\varepsilon) = \bar k[x]$.  
The localization at the prime $(x,\varepsilon)$ is a local ring whose maximal ideal $\frak m$ is generated by $x$ and $\varepsilon$. It is not regular because it is generated by $2$ elements while the dimension of the ring is $1$.

if $k \subset k'$ is Galois, then $k' \otimes_k k' = k'^{[k':k]}$ (this is actually a characterisation of Galois-ness), and the spectrum of $k' \otimes_k \bar k$ is $[k':k]$ separated geometric points. From there you can hopefully check the smoothness without problem.
