If $A$ is a (shifted) Poisson algebra, what does $A[\varepsilon]$ represent? I have a question which is not really precise, unfortunately.
Let $A$ be a Poisson $n$-algebra, i.e. a graded commutative algebra with a Lie bracket of degree $n-1$ s.t. the bracket is a biderivation w.r.t. the commutative product. Add a new square-zero variable $\varepsilon$ to $A$, possibly of positive degree, to get $A[\varepsilon]$. Clearly this is a graded commutative algebra, and there is also a shifted Lie bracket $[a+\varepsilon b, a'+\varepsilon b'] := [a,a'] + \varepsilon ([a,b'] \pm [a',b])$.
If $A$ has some geometric meaning (and this is the imprecise part), does $A[\varepsilon]$ also has some geometric meaning? I'm open to various notions of "geometric meaning". I found this appearing in an algebro-topological setting, and I'm wondering if there is a bigger picture I don't know about. This may be something very simple. My best guess is that $A[\varepsilon]$ is twice the dimension of $A$, so perhaps this is related to some (co)tangent bundle...?
 A: I can sort of give an answer in the context of derived geometry.
So let $A$,$B$ derived rings (i.e. dg-rings, simplicial rings, ${\mathbb{E}_\infty }$-rings).
For a $B$-module $M$, we can define the a square-zero extension ${B \oplus M}$. In your specific case $M$ would be $B\epsilon$. 
Let $f:Y = SpecB \to X = SpecA$ be a map of affine derived schemes and $Y\left[ M \right] = Spec\left( {B \oplus M} \right)$ the affine derived scheme corresponding to the square-zero extension of $B$ by $M$.
Then we can define a space of derivations of Y over X via a pullback diagram $\begin{array}{*{20}{c}}
  {De{r_X}\left( {Y;M} \right)}& \to &{Map{s_X}\left( {Y\left[ M \right],Y} \right)} \\ 
   \downarrow &{}& \downarrow  \\ 
  {\left\{ {i{d_Y}} \right\}}& \to &{Map{s_X}\left( {Y,Y} \right)} 
\end{array}$
Then there exists a $B$-module ${\mathbb{L}_{Y/X}}$, called the cotangent complex, which acts as a sort of universal derivation. i.e. $De{r_X}\left( {Y;M} \right) \simeq Map{s_B}\left( {{\mathbb{L}_{Y/X}},M} \right)$
The contangent complex plays a very important role in derived geometry, pretty much an analogous role to the cotangent bundle in usual geometry.
