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If global dimension is zero, every module is a projective module. Each short exact sequence splits because its quotient is projective. Conversely, splitting every short exact sequence makes each module a direct summand of a free module and therefore a projective module. Splitting also makes every module an injective module: for $A\hookrightarrow B$ choose a retraction $B\to A$ and compose it with any prescribed map out of $A$.

For a finite-dimensional algebra over an algebraically closed field, basic algebra means $A/J(A)\cong k^n$, so its simple modules are one-dimensional and occur once in the semisimple quotient of the regular module. A hereditary ring has every submodule of a projective module projective; equivalently its global dimension is at most one.

Choose representatives $S_1,\dots ,S_n$ of the simple modules. Our Ext quiver convention gives one vertex $i$ for $S_i$ andWith primitive orthogonal idempotents $e_i$, put $P_i=A{e}_i$ and $J=J(A)$. Applying ${\mathrm{Hom}}_A(-,S_j)$ to $0\to J{P}_i\to P_i\to S_i\to 0$ givesThe restriction map from $\mathrm{Hom}(P_i,S_j)$ is zero since every map to a simple kills $J{P}_i$, and $P_i$ is projective. Thus the arrow multiplicity is $\dim e_j(J/J^2)e_i$. This specifies the orientation rather than silently replacing the quiver by its opposite.

Because $A$ is hereditary, $J{P}_i$ is projective. Its semisimple top has $a_{ij}$ copies of $S_j$, where $a_{ij}$ is the arrow multiplicity. Its projective cover therefore supplies an epimorphismThis splits by projectivity of $J{P}_i$. Its kernel is a summand contained in the radical of the source; projection onto that summand and Nakayama lemma force the kernel to be zero. HenceEvery arrow $i\to j$ forces $\dim P_i>\dim P_j$, so $Q$ has no directed cycles.

Choose lifts in $e_{j}J{e}_i$ of bases of the arrow spaces. Send the vertices and arrows of the path algebra to these idempotents and lifts, multiplying paths in the order $\beta \alpha$ for consecutive arrows $\alpha :i\to j$, $\beta :j\to l$. This defines an algebra homomorphism $kQ\to A$. It is surjective: vertices span $A/J$, arrows span $J/J^2$, and products show that paths of length $r$ span $J^r/J^{r+1}$. Nilpotence of $J$ then gives all of $A$.

Let $N_i$ be the number of directed paths starting at $i$, including the length-zero path. Acyclicity gives the same recursion $N_i=1+{\sum }_j{a}_{ij}{N}_j$, with value one at a sink. Induction from sinks gives $N_i=\dim P_i$. ThereforeThe surjective homomorphism is an isomorphism. This proves the basic hereditary algebra path-algebra theorem:

For each connected component $C$ of the underlying unoriented graph, $e_C={\sum }_{i\in C}{e}_i$ is a central idempotent. Paths never connect distinct components, so $kQ={⨁}_C{e}_{C}kQ$ as a direct sum of two-sided ideals. To prove each is a block, let $z$ be any central idempotent. Modulo the arrow ideal it has coordinates $c_i\in \{0,1\}$. Commuting $z$ with an arrow $i\to j$ and reducing modulo the square of that ideal gives $c_i=c_j$. Thus these coordinates are constant on each component. If they are zero on a component, the restriction of $z$ there is a nilpotent idempotent, hence zero; if they are one, apply that argument to $1-z$. Consequently

Set $S=R\setminus P$, a multiplicative subset. The localization at a prime ideal is $R_P=S^{-1}R$, whose elements are fractions $r/s$. Equality of two fractions means that $u(r{s}^{\prime }-r^{\prime }s)=0$ for some $u\in S$. Likewise the localization of a module is $M_P=S^{-1}M$, withAddition uses a common denominator, and the module action isThe equivalence relations make these operations well-defined, so $M_P$ is an $R_P$-module.

If $M\ne 0$, choose $0\ne m\in M$. Its annihilator is proper, so it lies in a maximal ideal $P$. The element $m/1$ cannot vanish in $M_P$: vanishing would mean $sm=0$ for some $s\notin P$, contrary to ${\mathrm{Ann}}_R(m)\subseteq P$. The converse is immediate by localizing the zero module. ThusThis proof of localization detects zero elements actually applies to arbitrary modules, without finite-generation or Noetherian assumptions. We will use that extra generality for an Ext functor module below.

An injective module $E$ has the extension property: for each inclusion $A\hookrightarrow B$, every map $A\to E$ extends to a map $B\to E$. Equivalently, ${\mathrm{Hom}}_R(-,E)$ is exact. A projective module $Q$ has the lifting property: for each surjection $B\twoheadrightarrow C$, every map $Q\to C$ lifts to $Q\to B$. Equivalently, ${\mathrm{Hom}}_R(Q,-)$ is exact, or $Q$ is a direct summand of a free module.

For the local criterion for injectivity over a Noetherian ring, recall the Baer criterion: $E$ is injective precisely when every map from an ideal $J\subseteq R$ into $E$ extends to $R$. Through the short exact sequence $0\to J\to R\to R/J\to 0$, this is equivalent toWe also need localization of Ext over a Noetherian ring. Because $R$ is Noetherian, the module $R/J$ has a free resolution with every term finitely generated: all successive kernels are finitely generated, so this can be built recursively. For a finite free term $F$, the natural mapis an isomorphism, as is clear from a finite basis. Exactness of localization lets us pass to cohomology of the Hom complex. ThereforeThis explains the finiteness hypothesis needed for the localization argument, rather than assuming that localization preserves injectivity automatically.

If $M$ is injective, the left-hand side vanishes for every $J$ and $P$. Every ideal $L$ of $R_P$ is $J{R}_P$, where $J$ is its contraction to $R$: if $r/s\in L$, then $r/1=s(r/s)\in L$, and conversely localization of a member of the contraction stays in $L$. Hence all ideal tests for $M_P$ vanish, and the Baer criterion over $R_P$ makes $M_P$ injective.

Conversely, suppose every $M_P$ is injective. For each ideal $J$, the displayed Ext functor localization is zero at every prime. The zero-detection argument above gives ${\mathrm{Ext}}_R^1(R/J,M)=0$, without needing this Ext module to be finitely generated. Applying the Baer criterion over $R$ provesIn fact, under the Noetherian ring hypothesis this equivalence holds for arbitrary $M$; the printed finite-generation assumption is more than is needed.

The global dimension iswhere projective dimension is the smallest length of a projective resolution, or infinity if there is no finite one. If $\mathrm{gldim}R=0$, every module is projective. Given an inclusion $A\hookrightarrow B$, the quotient $B/A$ is then projective, so the exact sequence $0\to A\to B\to B/A\to 0$ splits. A retraction $\rho :B\to A$ exists. For any module $E$ and any map $f:A\to E$, the composition $f\rho :B\to E$ extends $f$. Thus every module is also injective. Global dimension zero makes all modules both projective and injective, as recorded by global dimension zero and split exact sequences.