Covering derivations of a quotient algebra Let $(\mathcal{A},+,·)$ an algebra and $\mathcal{I}$ an ideal of $\mathcal{A}$.
Is easy to check that if $D\in Der(\mathcal{A})$ with $D(\mathcal{I})\subseteq I$ then $D$ induces a derivation $D_I$ of the algebra $\mathcal{A}/\mathcal{I}$.
The question is given $D'\in Der(\mathcal{A}/\mathcal{I})$ there exist $D\in Der(\mathcal{A})$ with $D(\mathcal{I})\subseteq\mathcal{I}$ sucht that $D_I=D'$.
I suspect that the answer is no but I can't find a counterexemple. And there exist a nice property that guarantee a positive answer?
 A: Let $\mathcal{A}$ be the two-dimensional Lie algebra over a field $k$ with basis $\{x,y\}$ and relation $[x,y] = y$. Let $\mathcal{I}$ be the ideal of $\mathcal{A}$ spanned by $y$. Let $D' : \mathcal{A} / \mathcal{I} \to \mathcal{A}/\mathcal{I}$ be the $k$-linear map that sends the image $\overline{x}$ of $x$ in $\mathcal{A}/\mathcal{I}$ to $\overline{x}$. Since $\mathcal{A} / \mathcal{I}$ is abelian, this is a derivation. Now if $D$ is a derivation of $\mathcal{A}$ that lifts $D'$, then $D(x) = x + ay$ for some $a \in k$, and $D(y) = b y$ for some $b \in k$ because $D(y) = D([x,y]) = [x,D(y)] + [D(x),y]$ is a sum of commutators in $\mathcal{A}$, and as such must be an element of $\mathcal{I}$. Now 
$$by = D(y) = D([x,y]) = [x,D(y)] + [D(x),y] = [x, by] + [x + ay,y] = (b + 1)y$$
forces $y = 0$, a contradiction. So $D$ does not exist.
There is probably also a counterexample if $\mathcal{A}$ is only assumed to be a commutative algebra.
In the positive direction, if $\mathcal{A}$ is a commutative $k$-algebra such that the Kahler differentials $\Omega_{\mathcal{A}/k}$ is a free $\mathcal{A}$-module (for example, if $\mathcal{A}$ is a polynomial algebra in finitely many variables over $k$), then you can always lift $D'$. 
A: Let $A = k[x]/x^4$ where $k$ is a field of characteristic 3, and let $I=(x^3)$ so $A/I \cong k[x+I]/(x+I)^3$. Then there is a derivation $\delta$ of $A/I$ such that $\delta(x+I)=1+I$, whereas in $A$ a derivation $d$ must satisfy $0=d(x^4)=4x^3d(x)=x^3d(x)$ and so $d(x)$ must have degree at least one in $x$.
As Konstantin points out, this can only happen because $\Omega_A$ is not a free $A$-module, indeed $x^3 \Omega_A=0$.
