Basic algebra: why are direct summands projective?

Question

I was reading my lecture notes, where they define an algebra $A$ to be basic if $A$ as a right $A$-module has a decomposition $$A=I_1\oplus\cdots\oplus I_n$$ in indecomposable right $A$-modules $I_i$ with $I_i\not \cong I_j$ if $i\neq j$. After the definition, it says that the $I_i$'s are projective, although I don't see the reason for this to happen. Can someone help me see this?

Answer 1

Not sure which definition of "projective module" you have, but here are the standard three equivalent ones:

Theorem. Let $P$ be a (left) $R$-module. The following are equivalent:

1. For every (left) $R$-modules $M$ and $N$ and surjective morphism $f\colon M\to N$, if $g\colon P\to N$ is a morphism, then there exists a morphism $h\colon P\to M$ such that $f\circ h = g$: $$\begin{array}{rcccl} &&P\\ &&\downarrow g\\ M&\stackrel{f}{\longrightarrow}&N&\longrightarrow & 0 \end{array} \qquad\implies \qquad \begin{array}{rcccl} &&P\\ &h\swarrow&\downarrow g\\ &M\stackrel{f}{\longrightarrow}&N&\longrightarrow & 0 \end{array}$$
2. For every (left) $R$-module $M$, if $f\colon M\to P$ is a surjective morphism, then there exists a morphism $h\colon P\to M$ such that $fg=\mathrm{id}_P$.
3. $P$ is a direct summand of a free (left) $R$-module; that is, there exists a module $Q$ and a free module $F$ such that $F\cong P\oplus Q$.

Proof. $1\implies 2$: Take $g=\mathrm{id}_P$ and $1$ guarantees the existence of an $h\colon P\to M$ such that $fh=\mathrm{id}_P$.

$2\implies 3$: Every (left) $R$-module is a quotient of a free module, so let $f\colon F\to P$ be a surjection, and let $Q=\mathrm{ker}(f)$. Let $h\colon P\to F$ be such that $fh=\mathrm{id}_P$. Then it is easy to verify that $F=Q\oplus h(P)$, and since $h$ is one-to-one, we have $h(P)\cong P$.

$3\implies 1$: Let $F$ be free on $X$, with $F\cong P\oplus Q$. Identify $P$ with its image in $F$. Given $f$ and $g$, let $\gamma\colon F\to N$ be given by $\gamma(p,q) = g(p)$ for all $p\in P$, $q\in Q$. For each $x\in X$ let $m_x\in M$ be such that $f(m_x) = \gamma(x)$; $m_x$ exists because $f$ is surjective. Since $F$ is free, there exists a module homomorphism $\eta\colon F\to M$ such that $\eta(x)=m_x$ for each $x\in X$. Note that $f\eta(x) = \gamma(x)$, so $f\eta=\gamma$. Now let $h=\eta|_{P\oplus\{0\}}$ be the restriction of $\eta$ to $P\oplus\{0\}$. For every $p\in P$, we have $$fh(p) = f(\eta(p,0)) = (f\eta)(p,0) = \gamma(p,0) = g(p),$$ so $fh=g$, as required. $\Box$

In the situation at hand, each $I_j$ is a direct summand of $A$, which is free as an $A$-module. So $I_j$ satisfies item 3 of the theorem, hence satisfies 1 and 2; that is, $I_j$ is projective.