If $K$ is a finite separable extension of $F$, show that there is a $K$-vector space isomorphism $Der_k(F)\otimes_F K \simeq Der_k(K)$ I'm studying about derivation algebra in Morandi's book Field and Galois Theory.
There are one of problem as following:
If $K$ is a finite separable extension of $F$, show that there is a $K$-vector
space isomorphism
$Der_k(F)\otimes_F K \simeq Der_k(K)$
where $Der_k(F)$ is the $k$-derivation on $F$.
 A: $$\newcommand{\Der}{\mathrm{Der}}$$
$$\newcommand{\tensor}{\otimes}$$
Although one could prove this via modules of Kähler-differentials quite easily, using standard propositions, I will try to give a direct proof:
First let for rings $k \to R$ and an $R$-module $M$ be $\Der_k(R,M)$ the $R$-Module of derivations from $R$ to $M$ which vanish at $k$. Further let, following your notation, $\Der_k(R) = \Der_k(R,R)$.
About the field $K$, we note $K=F(\alpha) = F[x]/f(x)$ with $f \in F[x]$ irreducible of degree, say, $n$. 
Now we have first
$$\Der_k(F) \tensor_F K = \Der_k(F,K)$$
The maps here are as follows: Let $\delta = \sum_i \delta_i \tensor_F \gamma_i$ be an element of the left side. It gives a derivation, element of the right side by
$$\hat{\delta}(f) = \sum_i \delta_i(f) \gamma_i$$
where one easily checks $\hat{\delta}(f_1 f_2) = f_1 \hat{\delta}(f_2) + f_2 \hat{\delta}(f_1)$.
Starting from the right side, one notices 
$$\Der_k(F,K) = \Der_k(F,F^n) = \Der_k(F,F) \tensor_F F^n = \Der_k(F) \tensor_F K$$
as $F$-modules.
To put it more concrete: Calling 
$$p_i(\sum_{j=0}^{n-1} a_j \alpha^j) = a_i$$
we associate to $\delta \in \Der_k(F,K)$ the $n$-tuple $(\delta_i = p_i \circ \delta \in \Der_k(F))_{i=0,\ldots,n-1}$. Then we set 
$$\widetilde{\delta} = \sum_{i=0}^{n-1} \delta_i \tensor_F \alpha^i \in \Der_k(F) \tensor_F K$$
The associations $\delta \mapsto \hat{\delta}$ resp. $\delta \mapsto \widetilde{\delta}$ from left to right resp. from right to left are inverse to each other, as one checks by a simple calculation. They are both $F$-linear, and the first is $K$-linear. So the second, as the inverse of the first is $K$-linear too.
Now we have to prove
$$\Der_k(F,K) = \Der_k(K,K)$$
For this we introduce for $k \to R \to S$ and $x \in S$ and $g \in R[x]$ and $\delta \in \Der_k(R,S)$ the map
$$g(x) = \sum_j a_j x^j \mapsto g^\delta(x) = \sum_j \delta(a_j) x^j$$
Starting from $\delta \in \Der_k(F,K)$ we define $\hat{\delta} \in \Der_k(K,K)$ by ($g(x) \in F[x]$):
$$\hat{\delta}(g(\alpha)) = g^\delta(\alpha) + g'(\alpha) \hat{\delta}(\alpha)$$
Here $\hat{\delta}(\alpha)$ is fully determined by $\delta$ as
$$0 = \hat{\delta}(f(\alpha)) = f^\delta(\alpha) + f'(\alpha) \hat{\delta}(\alpha)$$
so
$$ \hat{\delta}(\alpha) = -f^\delta(\alpha)/f'(\alpha)$$
It is easy to calculate that $\hat{\delta}(h(\alpha)f(\alpha)) = 0$ (one uses here $(g h)^\delta = g^\delta h + g h^\delta$), so $\hat{\delta}$ is well-defined. Also $\hat{\delta}(g(\alpha) h(\alpha)) = g(\alpha) \hat{\delta}(h(\alpha)) + h(\alpha) \hat{\delta}(g(\alpha))$ is just a simple calculation. Finally one has to verify that the map is $K$-linear, that is $\widehat{\gamma \delta} = \gamma \hat{\delta}$ for all $\gamma \in K$.
The inverse map $\delta \in \Der_k(K,K) \mapsto \widetilde{\delta} \in \Der_k(F,K)$ is simply $\delta \mapsto \delta \circ i$, where $i:F \to K$ is the canonical inclusion.
