Let $A$ be an $R$-algebra (for $R$ a commutative ring). Let $\def\Der{\operatorname{Der}}\Der_R(A,-): A-\mathrm{mod}\to A-\mathrm{mod}$ be the covariant functor, where $\Der_R(A,M)$ is the set of all $R$-derivations of $A$ in $M$. (I mean the set of all $R$-linear maps $\phi: A\to M$ with $\phi(ab)=\phi(a) b +a \phi (b)$)
- Is $\Der_R(A,-)$ an exact functor? I am able to prove this a left exact functor. Is this a right exact functor?
- Let consider $\mathbb{A}^n$, affine $n$-space over field $R$. $A:= R[x_1,\cdots,x_n]$. Let $p\in X$, and let $ev_p:A \to R$ be the evaluation map which is homomorphism of $A$-modules with $\ker ev_p=M_p\subset A$ of all polynomials vanishes at $p$. Now, $\tilde{ev_p}:\Der_R(A,A)\to \Der_R(A,R)$ be the induced map with kernel $\Der_R(A,M_p)$. In this particular case we see $\Der(A,-)$ respects exactness of A-modules $0\to M_p\to A\overset{ev_p}\to R \to 0$. Will this be always the case for $\Der_R(A,-)$?
- (Keeping notations as in 2) The Zariski tangent space $T_p(\mathbb{A}^n)$ is isomorphic to $\Der_R(A,R)$ as A-module. But how to think this isomorphism intuitively/geometrically. The natural recipe from differential geometry is considering $\Der_R(\mathcal{O}_p,R)$. Instead, how does one land at $\Der_R(A,R)$ at first.
(PS: I am not thinking $p$ as a point in subset $X\subsetneq \mathbb{A}^n$, but as a point in $\mathbb{A}^n$.)