I'm trying to prove that when $L/K$ finite separable one has $\Omega_{L/k}=L\otimes_K \Omega_{K/k}$. The standard proof for that says that $L=K[X]/(f)=K[\alpha]$ (with $f(X)=\sum_i a_iX^i$) and uses the conormal exact sequence $$ (f)/(f^2)\to L\otimes_{K[X]}\Omega_{K[X]/k} \to \Omega_{L/k}\to 0$$ Then we rewrite \begin{align*} L\otimes_{K[X]}\Omega_{K[X]/k}&=L\otimes_{K[X]}\Omega_{K\otimes_k k[X]/k}\\ &=L\otimes_{K[X]}\left( K[X]\otimes_K\Omega_{K/k} \oplus K[X]\otimes_{k[X]}\Omega_{k[X]/k} \right)\\ &= L\otimes_K\Omega_{K/k}\oplus L\otimes_{k[X]}k[X]dX\\ &= L\otimes_K\Omega_{K/k}\oplus LdX \end{align*} and we see (no?) that under these isomorphisms $$ f\mapsto \sum_i \alpha^i\otimes da_i+f'(\alpha)dX$$ As $L/K$ is separable $f'(\alpha)\neq0$ and here comes my problem: all the author conclude that the term $LdX$ is then cancelled in the coker of $(f)/(f^2)$ and only stays $L\otimes_K\Omega_{K/k}=\Omega_{L/k}$.
But there is also the term $\sum_i \alpha^i\otimes da_i$ and one should get something like $$ L\otimes_K(\Omega_{K/k}/\oplus_i Kda_i)=\Omega_{L/k}$$
Where is my stupid mistake?
What the annulation of $f$ means is that in $\Omega_{L/k}$, $d\alpha$ is in the $L$-vector subspace generated by $\Omega_{K/k}$. In other words, the composition $L \otimes_K \Omega_{K/k} \rightarrow L \otimes_{K[X]} \Omega_{K[X]/k} \rightarrow \Omega_{L/k}$ is surjective.
Now, this composition is injective, because else (here we use the exactness of the sequence) there is some $gf \in (f)/(f^2)$ which is mapped to some nonzero $\sum_i{\alpha^idb_i} + 0dX$.
But, you can check that the coefficient before $dX$ is always $g(\alpha)f’(\alpha)$: therefore $g(\alpha)=0$, hence $f|g$ and the image of $gf$ is actually zero.