Derivations vs automorphisms of deformations Let $A$ be a ring, and $p : R'\rightarrow R$ a surjection of $A$-algebras with kernel $I$ a square zero ideal - ie $I^2 = 0$.
Let $f,g : R'\rightarrow R'$ be automorphisms of $R'$ over $R$ (ie, $pf = pg = p$), then it is easy to check that $f-g : R'\rightarrow R'$ has image in $I$, and actually defines an $A$-linear derivation, ie an element of $Der_A(R',I)$. Here we probably want to assume that $f,g$ are also $A$-linear.
Now let $f\in\text{Aut}_A(R')$ respecting the map $p : R'\rightarrow R$, and $D\in Der_A(R',I)$. Must $f+D$ also be an automorphism of $R'$?
If not, what are the "minimal reasonable assumptions" we need to impose for this to be true?
I can show that $f+D : R'\rightarrow R'$ is a homomorphism of $R$-algebras, but am having trouble proving bijectivity.
Variations of this are given in Hartshorne's "Deformation theory", where it's given as exercise 5.2, and in Sernesi's book "Deformations of Algebraic Schemes", where it's "Lemma 1.2.6", but he doesn't actually prove that $f+D$ is an automorphism.
 A: In the generality you state, the answer is no, the map $f+D$ need not be a bijection. 
Here is an ad-hoc example. Let $A=R$ be a field and let $R'$ be the polynomial ring in infinitely many variables $e_i, i\geq 1$ with the relations $e_ie_j=0$ for all $i,j$. Then we have a natural surjection $p:R'\to R$ and the kernel $I$ is the $A$-vector space generated by the $e_i$s and $I^2=0$. Any element in $R'$ can be written uniquely as $\alpha+\sum a_ie_i$ with $\alpha, a_i\in A$, and the sum is finite. Consider the derivation $D:R'\to I$ given by $D(\alpha+\sum a_ie_i)=\sum a_ie_{i+1}$. Then the map $I+D:R'\to R'$ where $I$ is the identity map is not onto. One easily checks that the element $e_1$ is not in the image of this map.
In deformation theory, one of the natural conditions that can be imposed to make everything work is not to take an arbitrary derivation $D:R'\to I$, but take a derivation $D:R\to I$ (note that $I$ is an $R$-module) and then consider it as a derivation from $R'$ using $p$. Then, $D\circ D=0$ and you will have $f+D$ a bijection as you desire. There may be other conditions which might yield the same result.
