# Why the kernel of $\phi_0: Der_R(B, N) \to Der_R(A, N)$ is $Der_{A}(B, N)$?

I have some questions of the book Linear algebraic groups by T.A.Springer. On page 57, it is said that the kernel of $\phi_0: \operatorname{Der}_R(B, N) \to \operatorname{Der}_R(A, N)$ is $\operatorname{Der}_{A}(B, N)$.

Let $\operatorname{Der}_R(A, M)$ be the set of all $R$-derivation of $A$ in $M$, where $R$ is a commutative ring, $A$ is an $R$-algebra and $M$ is a left $A$-module. Let $\phi: A \to B$ be a homomorphism of $R$-modules. Define $\phi_0: \operatorname{Der}_R(B, N) \to \operatorname{Der}_R(A, N)$ by $\phi_0(D) = D\circ \phi$. Why the kernel of $\phi_0: \operatorname{Der}_R(B, N) \to \operatorname{Der}_R(A, N)$ is $\operatorname{Der}_{A}(B, N)$? Take $a \in A$ and $D \in \operatorname{Der}_{A}(B, N)$. We have $D(a)=0$. But $\phi(a) \in B$. It seems that it is possible that $D \circ \phi (a) \neq 0$.

$\boldsymbol{4.1.1.}$ We recall the definition of a derivation. Let $R$ be a commutative ring, $A$ an $R$-algebra and $M$ a left $A$-module. An $R$*-derivation* of $A$ in $M$ is an $R$-linear map $D:A\to M$ such that for $a,b\in A$ $$D(ab)=a.Db+b.Da.$$ It is immediate from the definitions that $D1=0$, whence $D(r.1)=0$ for all $r\in R$.
$\quad$ The set $\operatorname{Der}_R(A,M)$ of these derivations is a left $A$-module, the module structure being defined by $(D+D')a=Da+D'a$ and $(b.D)(a)=b.Da$, if $D,D'\in\operatorname{Der}_R(A,M)$, $a,b\in A$.
$\quad$ The elements of $\operatorname{Der}_R(A,A)$ are the derivations of the $R$-algebra $A$. If $\phi:A\to B$ is a homomorphism of $R$-algebras and $N$ is a $B$-moudle, then $N$ is an $A$-module in an obvious way. If $D\in\operatorname{Der}_R(B,N)$ then $D\circ\phi$ is a derivation of $A$ in $N$ and the map $D\mapsto D\circ\phi$ defines a homomorphism of $A$-modules $$\phi_0:\operatorname{Der}_R(B,N)\to\operatorname{Der}_R(A,N),$$ whose kernel is $\operatorname{Der}_A(B,N)$. Thus we obtain an exact sequence of $A$-modules $$0\to\operatorname{Der}_A(B,N)\to\operatorname{Der}_R(B,N)\to\operatorname{Der}_R(A,N).\tag{12}$$

Keep in mind that the $A$-algebra structure of $B$ is given by $\phi$. That is, for $a \in A, b \in B$ we have $\phi(a)b = a\cdot b$. $\def\Der{\mathop{\rm Der}}$Now let $D \in \Der_R(B, N)$, for $a \in A$ we have $$\phi_0(D)(a) = D\bigl(\phi(a)\bigr) = D(a \cdot 1) = a\cdot D(1) = 0.$$