Let $K$ a field of characteristic $p$ and oneconsider a purely inseparable field extension $L = K(t)$. Therefore there exist a minimal $n \in \mathbb{N}$ with $t^{p^n} := k \in K$ (equivalently: $P(X)= X^{p^n} -k$ is minimal polynomial of $t$.
I want to know how to calculate the Kaehler Differentials $\Omega_{L/K}$ of this field extension.
My problem is that in contrast to separable case I guess that $\Omega_{L/K} \neq 0$ since I cant use here the fact
$$0 = d_{L/K}(P(t))= P'(t)d_{L/K}(t)$$
to conclude that $d_{L/K}(t)$ since $P'(t)=0$ because purely inseparable.
I only know that the $t^j$ ($j \in \mathbb{N}$) form a $K$ base of $L$ and therefore the $d_{L/K}(t^j)= jt^{j-1}d_{L/K}(t)$ generate $\Omega_{L/K} $ but from here I'm stuck.
If $d_{L/K}(t)=0$ then $$\Omega_{L/K} = 0$$
but what about the case $d_{L/K}(t) \neq 0$?
I suppose that then $\Omega_{L/K} = L$ but how to show it?
Best Answer
In this case, you indeed have $\Omega^1_{L/K} = L.$ You can use the standard exact sequence $$ I\otimes_S S/I\to\Omega^1_{S/R}\otimes_S S/I\to\Omega^1_{(S/I)/R}\to 0 $$ associated to a sequence of morphisms of rings $R\to S\to S/I$ to compute this.
First, recall that the map $I\otimes_S S/I\to\Omega^1_{S/R}\otimes_S S/I$ is induced by the universal differential \begin{align*} d : I&\to\Omega^1_{S/R}\\ x&\mapsto dx. \end{align*}
Now, say $L = K(t),$ where $t$ has minimal polynomial $X^q - \alpha\in K[X].$ Then we have a sequence of morphisms $K\to K[X]\to K[X]/(P)\cong L,$ and we may see what the exact sequence has to say about it. We find $$ (P)\otimes_{K[X]} L\to\Omega^1_{K[X]/K}\otimes_{K[X]}L\to\Omega^1_{L/K}\to 0, $$ and since $\Omega^1_{K[X]/K}\cong K[X]\,dX,$ it follows that $$ \Omega^1_{K[X]/K}\otimes_{K[X]}L\cong K[X]\,dX\otimes_{K[X]}L\cong L\,dt. $$ Thus, \begin{align*} \Omega^1_{L/K}&\cong \frac{\Omega^1_{K[X]/K}\otimes_{K[X]}L}{(\textrm{im}((P)\otimes_{K[X]} L\to\Omega^1_{K[X]/K}\otimes_{K[X]}L))}\\ &\cong \frac{K[X]\,dX\otimes_{K[X]}L}{\langle d(PF)\otimes\alpha\mid F\in K[X]\rangle}\\ &\cong \frac{K[X]\,dX\otimes_{K[X]}L}{\langle (P(X)F'(X) + P'(X)F(X))dX\otimes\alpha\mid F\in K[X]\rangle}\\ &\cong \frac{L\,dt}{\langle \alpha P(t)F'(t)dt\mid F\in K[X]\rangle}\\ &\cong L\,dt\cong L, \end{align*} as $P(t) = 0.$