Extension of nonisomorphic simple objects Let $X$ and $Y$ be two nonisomorphic simple objects in an abelian category. Are all extensions of $X$ by $Y$ trivial? ( $\mathrm{Ext}^1(X,Y)=0$ ?)
 A: No.
Pick a field ${\mathbb k}$ and a finite oriented tree $Q$, considered as a category. Then the simple objects in the abelian functor category ${\mathscr A} := k\text{-Vect}^Q$ (which is called the category of representation of $Q$ and is equivalent to the category of modules over the path algebra ${\mathbb k}Q$ of $Q$ over ${\mathbb k}$) are in bijection with the vertices of $Q$, and given two vertices of $Q$, the dimension of $\text{Ext}^1_{\mathscr A}$ between the corresponding simple objects is the number of edges connecting $v$ and $w$.
For example, taking $Q := \bullet\to\bullet$, ${\mathscr A}={\mathbb k}\text{-Vect}^{\bullet\to\bullet}$ is the category of homomorphisms of ${\mathbb k}$ vector spaces (with commutative squares as diagrams), and ${\mathbb k}\xrightarrow{1}{\mathbb k}$ constitutes a (and, up to scalar, the) nontrivial extension between the simples $0\to{\mathbb k}$ and ${\mathbb k}\to 0$. 
In the commutative world you won't find examples so easily due to the notion of support.
