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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 $\phi(ab)=\phi(a) b +a \phi (b)$)

  1. Is $\Der_R(A,-)$ an exact functor? I am able to prove this a left exact functor. Is this a right exact functor?
  2. 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 $\Der_R(A,-)$?
  3. (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 $\mathbb{A}^n$.)

KReiser
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1 Answers1

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  1. By Stacks 00RO, $\operatorname{Der}_R(S,M)\cong \operatorname{Hom}_S(\Omega_{S/R},M)$, which is left exact but not right exact in general.
  2. Please don't name your fields $R$. What you're finding here is that $\Omega_{A/R}$ is isomorphic to the free $A$-module on the symbols $dx_i$, so by the material from (1), the functor in question is exact.
  3. As explained here, a tangent vector is something you can take a derivative along.
KReiser
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