When is $\Omega^1_{E/K} \otimes_E L \to \Omega^1_{L/K}$ injective? My question refers to an exercise from Bosch's "Commutative Algebra and Algebraic Geometry (page 355, Ex. 8.1.8).
Let $L/K$ be a field extension and $E$ an intermediate field. As we can interpret $E,L$ as $K$-algebras we obtain a short exact sequence
$$\Omega^1_{E/K} \otimes_E L \to \Omega^1_{L/K} \to \Omega^1_{L/E} \to 0$$
where we denote for a morphism $f: A \to B$ of $R$-algebras the $A$-module $\Omega^1_{B/A} $ of differential forms.
The aim is to verify that the left map 
$$\Omega^1_{E/K} \otimes_E L \to \Omega^1_{L/K}$$
from the sequence is injective iff every $K$-derivation $E \to L$ admits an extension as a $K$-derivation $L \to L$.
I don't know how can I solve the problem. 
My attempts: "$\Leftarrow$" is seems to work as following: Assume $d_E:E \to L$ is $K$-derivation. By assumption it extends to $d_L:L \to L$. Then $E \to L \to \Omega^1_{L/K}$ is also a $K$-derivation so factorizes over $\Omega^1_{E/K}$ by universal property of $\Omega^1_{E/K}$. But why does it imply that $\Omega^1_{E/K} \otimes_E L \to \Omega^1_{L/K}$ is already injective?
Other direction : no idea
 A: You can use the fact that $L$ is an injective $L$-module (because $L$ is a field). This means that the functor $\operatorname{Hom}_L(.,L)$ is exact.
Let $N$ be the kernel of $\Omega^1_{E/K}\otimes_E L\to \Omega^1_{L/K}$. Apply the exact functor $\operatorname{Hom}_L(.,L)$ to get the exact sequence :
$$ 0\leftarrow\operatorname{Hom}_L(N,L)\leftarrow\operatorname{Hom}_L(\Omega^1_{E/K}\otimes_E L,L)\leftarrow\operatorname{Hom}_L(\Omega^1_{L/K},L)$$
Look at the map on the right, using extension of scalar adjunctions and universal properties, it is :
$$\operatorname{Der}_K(E,L)=\operatorname{Hom}_E(\Omega^1_{E/K},L)=\operatorname{Hom}_L(\Omega^1_{E/K}\otimes_E L,L)\leftarrow\operatorname{Hom}_L(\Omega^1_{L/K},L)=\operatorname{Der}_K(L,L)$$
And this map is simply the correspondence $D\mapsto D|_E$. Putting, all this together, we get an exact sequence :
$$0\leftarrow\operatorname{Hom}_L(N,L)\leftarrow\operatorname{Der}_K(E,L)\leftarrow\operatorname{Der}_L(L,L)$$
It follows that a $K$-derivation $E\to L$ extends to a $K$-derivation $L\to L$ iff its image in $\operatorname{Hom}_L(N,L)$ is zero. Hence, every $K$-derivation $E\to L$ extends iff $\operatorname{Der}_L(E,L)\to\operatorname{Hom}_L(N,L)$ is the zero-map, but this is possible iff $N=0$, in other words iff $\Omega^1_{E/K}\otimes_E L\to \Omega^1_{L/K}$ is injective.
