Rupture field of $X^p+T$ equals its splitting field

Question

Let $K$ be a field of prime characteristic $p$. Let $P(X)=X^p+T$ be a polynomial from $K(T)[X]$. $P$ is irreducible over $K(T)$ by Eisenstein criterion. Show that a rupture field of $P$ is also a splitting field.

Answer 1

Just to be sure about the starting point: a splitting field is the smallest field extension, such that the polynomial splits and a rupture field is a field extension generated by a single root.

In this case, there is one observation that seems to solve the question, namely $$(X+\sqrt[p]{T})^p = X^p+T+p\cdot(\ldots). $$ Since $K$ is a field of characteristic $p$, the latter terms vanish and $(X+\sqrt[p]{T})^p = X^p+T$ remains. By adjoining $\sqrt[p]{T}$ we thus get a splitting field of $P$ over $K(T)$ and this is also a rupture field.