Projective Resolution of Simple Modules over Upper Triangular Matrices Let $R$ be the ring of $n\times n$ upper triangular matrices over a field $F$. I've found that the simple modules $L_i$, $i=1,\ldots, n$ are all $1$-dimensional over $F$, where the module action is given by multiplication by the $i$th diagonal entry. I want to compute $Ext^p(L_i, L_j)$, but I'm having trouble finding a projective resolution for $L_i$. I need that the Ext groups are $0$ when $p>1$, to the projective resolution should only have length $2$, I think. Could I please have a hint?
 A: If you're still looking for a hint, try using the following theorem:

Theorem: (Kaplansky's Theorem) Take a ring $A$ every left ideal of which is projective. Then the following are equivalent for a left $A$-module $P$:
\begin{align}
& P \text{ is projective.}\\
& P \text{ is a submodule of a free } A \text{-module.}\\
& P \text{ is isomorphic to the direct sum of left ideals of } A \text{-modules} \end{align}
Clearly $(3) \implies (1)$ since the direct sum of projective modules is projective. And $(1) \implies (2)$ since each projective is a direct summand of a free module.
$(2) \implies (1)$. Take a left free $A$-module $F = \oplus_{i \in I} A$ with submodule $P$ and let $\{ e_i \}_{i \in I}$ be the standard basis of $F$. We show that $P$ is projective. Put a well ordering $\leq$ on $I$. For each $i \in I$ define $F_{\leq i} = span(\{e_j : j \leq i \})$ and $F_{< i} = span( \{ e_j : j < i \} )$. Write $P_{\leq i} = P \cap F_{\leq i}$ and $P_{ < i} = P \cap F_{ < i}$. Let $\Phi : F_{\leq i} \rightarrow F_{< i} \oplus A$ be the canonical isomorphism, and $\pi : F_{< i } \oplus A \rightarrow A$ the canonical projection. $I_i = \pi \circ \Phi (P \cap F_{\leq i})$ is a left ideal of $A$, and since it is projective the exact sequence $0 \rightarrow P_{< i} \rightarrow P_{\leq i} \rightarrow I_i \rightarrow 0$ splits. Thus $P_{\leq i}$ can be expressed as an internal direct sum of $P_{< i}$ and $\mathfrak{a}_i$, where $\mathfrak{a}_i$ is an isomorphic image of $I_i$ in $P_{\leq i}$.
$\sum_{i \in I} \mathfrak{a}_i \subset P$. If this inclusion is strict then we can take $i = \min \{ j : P_{\leq j} - \sum_{k \leq j} \mathfrak{a}_k \neq \emptyset \}$ and $x \in P_{\leq i} - \sum_{j \leq i} \mathfrak{a}_j$. Writing $x = y + a$ for $y \in F_{< i}$ and $a \in \mathfrak{a}_i$, we see that $y \in F_{\leq j}$ for some $j < i$ in $I$, so that $y \in \sum_{j \in I} \mathfrak{a}_j$ by minimality of $i$. Thus $x \in \sum_{j \in I} \mathfrak{a}_j$, a contradiction. So $P = \sum_{i \in I} \mathfrak{a}_i$.
The sum $\sum_{i \in I} \mathfrak{a}_i$ is direct. Define 
\begin{align*}
 n = \min \left\{ m \in \mathbb{N}_{\geq 0} : \exists F \subset I, |F| = m, \{ a_i \}_{i \in F}, a_i \in \mathfrak{a}_i \text{ such that } \S_{i = 1}^n a_i = 0 \right\} 
\end{align*}
Suppose $n > 0$. We can take $i_1 <  ... < i_n \in I$ and $a_i \in \mathfrak{a}_i$ nonzero such that $\S_{i =0}^n a_i = 0$. $\mathfrak{a}_{i_n} \cap F_{< i_n} = \{ 0 \}$,  $a_{i_n} \in \mathfrak{a}_{i_n}$, and $\S_{j = 1}^{n-1} a_{i_j} \in F_{< i_n}$, so $a_{i_n} = 0$, a contradiction. 
Thus $P \cong \oplus_{i \in I} \mathfrak{a}_i \cong \oplus_{i \in I} I_i$ is the direct sum of left ideals of $A$, as claimed.

Here's one way of continuing:
Let $U$ be the ring of upper triangular matrices.
For each $i, j \in \mathbb{N}_{\geq 1}$ with $1 \leq i \leq j \leq n$, let $e_{ij} \in U$ be the matrix with $1$ in the $(i, i)$th entry and $0$ elsewhere. Let $P_i$ be the ideal $Ue_{ii}$. Elements of $P_i$ are those upper triangular matrices whose nonzero entries lie in the $i$th collumn. $U = \oplus_{i = 1}^n P_i$, and in particular each $P_i$ is projective.
Every left ideal of $U$ is isomorphic to a direct sum of the form 
\begin{align}
\oplus_{k =1}^m P_{a_k}
\end{align}
for integers $\{ a_k \}_{k = 1}^m$ with $1 \leq a_k \leq n$. To see this, take a left ideal $J \subset U$. $J = J \cap \oplus_{i = 1}^n P_i \cong \oplus_{i = 1}^n (J \cap P_i)$. Each $J \cap P_i$ is isomorphic to $P_j$ for some $1 \leq j \leq i$: $J \cap P_i$ is generated by $e_{ij}$ where $j$ is maximal such that there exists an element in $J \cap P_i$ with nonzero $(i, j)$th entry. Hence every left ideal of $U$ has the form posited above. In particular every left ideal is projective.
By Kaplansky's Theorem, shown below, every finite dimensional left module of $U$ can be characterized as a submodule of a free module. Additionally, every finite dimensional left module of $U$ is isomorphic to a finite direct sum of left ideals of $U$, each isomorphic to $P_i$ for some $i$. Note one implication of this: every object $M$ has a projective resolution $0 \rightarrow Q \rightarrow P \rightarrow M \rightarrow 0$, so that $Ext_i ( M, N) = 0$ when $i \geq 2$.
One can also calculate $Ext_0 (S_j, S_k)$ and $Ext_1 (S_j, S_k)$ where $\{ S_l \}_{l = 1}^n$ are the simple modules, rather easily. Just use the projective resolutions of length $2$.
Elaboration upon request. Hope this helps! 
