Why is $Z(y^2 - x^p - t)$ a regular scheme? Let $k_0$ be a field of characteristic $p > 0$ and $k = k_0(t)$. Now we let $C \subset \mathbb{A}^2_k$ be the curve defined by $y^2 = x^p - t$. We wish to show that although $C$ is regular, it is not smooth over $k$. Denote $A = k[x,y]/(y^2 - x^p - t)$ the coordinate ring of $C$.
For smoothness, we note that the module of differentials over $k$ is given by $$\Omega = (Adx \oplus Ady)/(d(y^2 - x^p - t)) = (Adx \oplus Ady)/(2ydy) = Adx \oplus A/(y) dy$$
Now let $P \in C$ be a closed point, associated to a maximal ideal $\mathfrak{m}$ of $A.$ If $y \notin \mathfrak{m}$, then $y$ is a unit in $A_\mathfrak{m}$ so that $A_{\mathfrak{m}}/(y) = 0$ and $$\Omega_P = \Omega_{A_\mathfrak{m}/k} = A_\mathfrak{m}dx$$ which implies that $\Omega_P \otimes k(P) = k(P)dx$ has rank $1$. On the other hand, if $y \in \mathfrak{m}$ then $\mathfrak{m} = (y)$ since $A/(y) = k(t^{1/p})$ is a field, so that $$\Omega_P = k(P) dx \oplus k(P) dy.$$ Hence, $C$ is not smooth over $k$ since the dimension of the fibers of $\Omega_P \otimes k(P)$ change as the closed points vary.
Now I want to show that $C$ is a regular scheme. It suffices to show that $A_\mathfrak{m}$ is a regular local ring for all maximal ideals $\mathfrak{m}$ of $A$. One case is easy: if $(y) = \mathfrak{m}$ then $A_{\mathfrak{m}}$ is a DVR and hence regular. The other case is not so clear. Perhaps you can use some fundamental exact sequence for differentials to get the right inequality? I'm quite stuck on this part.
Does anyone have any ideas to show that $\dim_{k(x)} \mathfrak{m}_x/\mathfrak{m}_x^2 \leq 1$ for the other closed points? Also, is my reasoning fo concluding $C$ isn't smooth over $k$ sound?
Thanks!
For context, this is the first exercise in Hartshorne III.10.
EDIT: Perhaps the jacobian criterion works in the positive characterstic non-algebraically closed case?
 A: Let's show that $X$ is regular.
We deal with the characteristic two case on it's own: since $y^2-x^2=(x+y)^2$, we have that $k[x,y]/(y^2-x^2+t)\cong k[x,y]/(y^2+t) \cong k(t^{1/2})[x]$ which is clearly a regular ring.
When $p\neq 2$, we have slightly more work to do.
As $y^2-x^p+t$ is irreducible, the local ring of the generic point of $X$ is a field and therefore a regular local ring.
Next, for a closed point, let $\mathfrak{m}\subset k[x,y]$ be a maximal ideal containing $(y^2-x^p+t)$.
Then the assertion that $X$ is regular at $\mathfrak{m}$ is that $$\frac{\mathfrak{m}/(y^2-x^p+t)}{\mathfrak{m}^2/(y^2-x^p+t)\cap\mathfrak{m}^2}=\frac{\mathfrak{m}}{\mathfrak{m}^2+(y^2-x^p+t)}$$ is one-dimensional, or that $y^2-x^p+t$ is not in $\mathfrak{m}^2$.
When $y\notin\mathfrak{m}$, we can evaluate $\frac{\partial}{\partial y}(y^2-x^p+t)$ at $\mathfrak{m}$: if $y^2-x^p+t\in\mathfrak{m}^2$, this would vanish, but it gives $2y\notin\mathfrak{m}$, so $y^2-x^p+t\notin\mathfrak{m}^2$.
When $y\in\mathfrak{m}$ and therefore $\mathfrak{m}=(x^p-t,y)$, we can see that $y^2-x^p+t\notin\mathfrak{m}^2$: if it were, $x^p-t$ would be in $\mathfrak{m}^2=(y^2,y(x^p-t),x^{2p}-2x^pt+t^2)$ too, but by degree considerations this is impossible.
Thus $X$ is regular.
Your work showing that $X$ is not smooth via the module of differentials is correct. The suggestion of the Jacobian actually provides a slightly quicker way to show that $X$ is not smooth: let $\overline{k}$ be an algebraic closure of $k$ and consider $X_{\overline{k}}$: if $X$ was smooth over $k$, this would be regular by example III.10.0.3, but $X_{\overline{k}}$ is not regular at $(x-t^{1/p},y)$ because the Jacobian $\begin{pmatrix} px^{p-1} & 2y\end{pmatrix}$ vanishes there. (Jacobians detect geometric regularity aka smoothness, not regularity.)
