Yoneda Extensions and Homological Dimension

The two outer vertical maps are isomorphisms, and both rows are exact.

The short five lemma applied to this diagram makes the middle vertical map an isomorphism.

Identity maps give reflexivity, and composing endpoint-identity maps of short exact sequences gives transitivity.

Such a map is an isomorphism on the middle term; its inverse commutes with the endpoint maps, so it witnesses symmetry.

Apply the pullback or pushout universal property from Pullback and pushout of an extension to the middle-object map of the equivalence. It produces a morphism between the induced short exact sequences that is the identity on their new endpoints.

This induced morphism is an isomorphism by A morphism of extensions is an isomorphism, hence it is an equivalence of the induced extensions.

Taking the direct sum of the two equivalence diagrams gives an equivalence between the direct-sum extensions.

Apply Pullback and pushout descend to extension classes first to the diagonal and then to the codiagonal prescribed by The Baer sum of extension classes. The resulting Baer extensions are equivalent.

The Baer sum is independent of representatives makes the operation well-defined. The associativity and symmetry maps of finite biproducts transport the two iterated diagonal-pullback/codiagonal-pushout constructions into equivalent extensions; the required additive identities are those in On a biproduct, the injections and projections satisfy the identity-sum relation.

The split extension is neutral because its diagonal pullback and codiagonal pushout recover the original extension. Pushing out an extension along $-1_N$ gives the inverse: the codiagonal of $1_N$ and $-1_N$ is zero, hence the resulting Baer sum is split.

Pullback along $a$ and pushout along $b$ are well-defined on equivalence classes by Pullback and pushout descend to extension classes. Their universal properties make their composites agree with pullback and pushout along composite maps.

Both constructions commute with the direct-sum, diagonal, and codiagonal steps defining Baer addition. Hence they are homomorphisms for the group structure of Baer sum makes extension classes an abelian group, with the stated variance.

Naturality of the long exact Ext sequence gives a commutative square between the two connecting maps $\mathrm{Hom}(M,M)\to {\mathrm{Ext}}^1(M,N)$.

The endpoint maps are identities, so the square sends $1_M$ to the same element on both sides. These two images are the two extension classes.

The low-degree dimension-shifting sequence from Ext dimension shifting in the first variable represents $\xi$ by a morphism $f:\mathrm{\Omega }M\to N$, modulo morphisms extending across $P$. Push out $0\to \mathrm{\Omega }M\to P\to M\to 0$ along $f$, as defined in Pullback and pushout of an extension.

The resulting bottom row is an extension of $M$ by $N$. Naturality of its connecting homomorphism sends $1_M$ to the class represented by $f$, which is $\xi$ by the definition in The derived Ext-one class of an extension.

Choose a projective presentation $0\to \mathrm{\Omega }M\stackrel{i}{\to }P\twoheadrightarrow M\to 0$. Projectivity lifts $P\to M$ through the middle epimorphism of each extension. Restricting those lifts to $\mathrm{\Omega }M$ gives maps $f,f^{\mathrm{\prime }}:\mathrm{\Omega }M\to N$. In the cokernel description of ${\mathrm{Ext}}^1(M,N)$, equality of the two classes says $f-f^{\mathrm{\prime }}=ti$ for some $t:P\to N$.

Each original extension is equivalent to the pushout of the projective presentation along its map $f$ or $f^{\mathrm{\prime }}$. The relation $f-f^{\mathrm{\prime }}=ti$ induces an isomorphism between these two pushouts, represented on $P\oplus N$ by $(p,n)\mapsto (p,n-t(p))$; it fixes both endpoint maps. Composing the endpoint- identity equivalences proves that $e$ and $e^{\mathrm{\prime }}$ are equivalent.

The assignment is well-defined by Equivalent extensions have the same Ext class, surjective by Every Ext-one class is represented by an extension, and injective by Two extensions with the same Ext class are equivalent.

For $a:M^{\mathrm{\prime }}\to M$, the pullback extension and the original extension form a morphism of short exact sequences; naturality of the connecting morphism sends $1_M$ to the pullback of its class along $a$. For $b:N\to N^{\mathrm{\prime }}$, the analogous pushout square sends the class forward along $b$. These are exactly the two maps in Extension classes are contravariant in the quotient and covariant in the subobject, so the bijection is natural in both variables. The diagonal-pullback and codiagonal-pushout defining Baer sum correspond, by the same two naturality squares, to addition in Ext; hence the bijection is additive.

For an extension $e:0\to N\to E\to M\to 0$, assign its connecting class $c(e)={\delta }_e(1_M)\in S$. By [F1] and [F3], $c(e)=c(e^{\mathrm{\prime }})$ if and only if $e$ and $e^{\mathrm{\prime }}$ are equivalent. These lemmas concern individual extensions and do not assume a set of all extension classes.

The supplied projective resolution gives the presentation $0\to \ker (P_0(M)\to M)\to P_0(M)\to M\to 0$ required by [F2]. Hence every $s\in S$ equals $c(e)$ for some extension $e$. Together with step 1.1 this makes the already given set $S$ a set realization of precisely the extension equivalence classes. No quotient of a proper class or simultaneous choice of representatives is used.

A generating chain map between one pair of representatives can be concatenated with the identity chain map on the other factor; the result is a chain map of the spliced exact sequences that is the identity at both ends.

Such maps generate the equivalence relation of Equivalence of n-fold extensions, so repeated generating replacements do not change the class of the splice defined in The Yoneda splice product.

Endpoint pullback and pushout preserve exact extensions by [F2] and the universal properties of kernels and cokernels. An endpoint-preserving chain map induces a chain map of their pullbacks or pushouts, again fixing the new endpoints. Thus these operations respect every generating map and hence the generated equivalence relation. Together with [F1] and ordinary composition, the product is defined on classes in every pair of degrees.

If all three degrees are positive, both parenthesizations concatenate the same list of middle objects with the same junction maps, so their extensions agree. If all are zero, associativity is the category's associativity of morphism composition.

For two consecutive degree-zero factors, iterated pullback is canonically the pullback along the composite map, and iterated pushout is canonically the pushout along the composite map, by their universal properties. This treats degree patterns $(+,0,0)$ and $(0,0,+)$. For patterns $(0,+,+)$ and $(+,+,0)$, the outer endpoint pushout or pullback affects only the outer endpoint of the concatenation, giving the same extension before or after concatenating.

For the pattern $(0,+,0)$, endpoint pushout and endpoint pullback commute up to canonical equivalence. For extension length at least two, they modify distinct outer middle objects and their universal maps commute. For a short extension $0\to B\to E\to A\to 0$, a quotient map $f:A^{\mathrm{\prime }}\to A$ and a subobject map $g:B\to B^{\mathrm{\prime }}$ give a canonical map $g_{\ast }(f^{\ast }E)\to f^{\ast }(g_{\ast }E)$: the maps to $g_{\ast }E$ and to $A^{\mathrm{\prime }}$ agree over $A$, and the map from $B^{\mathrm{\prime }}$ is supplied by the pushout. It fixes $B^{\mathrm{\prime }}$ and $A^{\mathrm{\prime }}$, so [F3] makes it an isomorphism of short extensions. This proves the remaining pattern with two zero degrees.

For the pattern $(+,0,+)$, let $\xi$ extend $L$ by $N$, let $f:K\to L$, and let $\eta$ extend $M$ by $K$. The two products are $(f^{\ast }\xi )⌣\eta$ and $\xi ⌣(f_{\ast }\eta )$. There is a chain map from the first spliced extension to the second: use the pullback projection on the last middle object of $\xi$, the pushout map on the first middle object of $\eta$, and identities elsewhere and at $N,M$. At the junction its square commutes by the defining equations of that pullback and pushout; all other squares are their endpoint squares or identity squares. Hence these two extensions are equivalent by the generated relation, which needs no isomorphism of all middle objects.

Steps 1.2 and 2.1–2.3 exhaust the eight patterns of zero and positive degrees, proving associativity. Pullback along an identity and pushout along an identity are canonically isomorphic to the original extension by their universal properties. In degree zero the same assertion is the identity law for composition. Thus identity morphisms give both units in every degree.

Fix $P_{\bullet }\to M$ and an extension $E$ with inclusion $i:N\to E_{n-1}$. By [F2], successively lift the augmentation to $u_k:P_k\to E_k$ for $0\le k<n$, respecting differentials. Exactness makes $u_{n-1}{d}_n$ factor uniquely as $ic$ for $c:P_n\to N$. Since $i$ is monic and $d_n{d}_{n+1}=0$, $c{d}_{n+1}=0$, so $c$ is a cocycle.

A cocycle $c:P_n\to N$ factors uniquely as $c=f{\bar{d}}_n$, where ${\bar{d}}_n:P_n\twoheadrightarrow {\mathrm{\Omega }}^{n}M=\mathrm{im}{d}_n$; its kernel is $\mathrm{im}{d}_{n+1}$. Push out $0\to {\mathrm{\Omega }}^{n}M\stackrel{j}{\to }{P}_{n-1}\to \cdots \to P_0\to M\to 0$ along $f$. The new first middle object is $E(f)=\mathrm{coker}((f,-j):{\mathrm{\Omega }}^{n}M\to N\oplus P_{n-1})$. Its inclusion of $N$ is monic by [F2], and its cokernel is the unchanged cokernel of $j$, so this is an exact $n$-extension.

Two lift systems $u,v$ in step 1.1 have the same terminal cocycle class. Indeed, projectivity and exactness construct homotopy components $h_k:P_k\to E_{k+1}$ through $k=n-2$ satisfying $u_k-v_k=d_E{h}_k+h_{k-1}{d}_P$, with $h_{-1}=0$. At $k=n-1$ the remaining difference factors uniquely through $i$ as $it$, with $t:P_{n-1}\to N$. Applying $d_n$ then gives $c-c^{\mathrm{\prime }}=t{d}_n$. For $n=1$ this last factorization is the entire argument. A chain map of extensions fixing endpoints carries one lift system to another with the same cocycle, so the class is constant on the generated equivalence relation.

If $c^{\mathrm{\prime }}=c+t{d}_n$, then $f^{\mathrm{\prime }}=f+tj$. The automorphism $(a,p)\mapsto (a-tp,p)$ of $N\oplus P_{n-1}$ carries the relation $(f,-j)$ to $(f^{\mathrm{\prime }},-j)$, so it induces an isomorphism $E(f)\to E(f^{\mathrm{\prime }})$, identity on both endpoints and on the unchanged tail. This notation denotes a biproduct matrix, and requires no element description of the abelian category. Thus step 1.2 descends from cocycles to cohomology classes.

For an extension and lifts from step 1.1, the map $(i,u_{n-1}):N\oplus P_{n-1}\to E_{n-1}$ kills $(f,-j)$, since $if=u_{n-1}j$ after canceling the epimorphism ${\bar{d}}_n$. It induces a chain map from the pushout extension back to the original, with identity endpoints. Conversely the canonical map $P_{n-1}\to E(f)$, together with the identity maps on the remaining $P_k$, is a lift system for the pushout extension and has terminal cocycle $f{\bar{d}}_n=c$. The two constructions are therefore inverse on classes.

A map $N\to N^{\mathrm{\prime }}$ sends $c$ to its postcomposition and sends $E(f)$ to its endpoint pushout. A map $M^{\mathrm{\prime }}\to M$ has a comparison lift $P(M^{\mathrm{\prime }})\to P(M)$ by [F3]; composing the lift system in step 1.1 with it computes the endpoint pullback, with cocycle obtained by precomposition. Homotopic comparison maps induce cochain-homotopic Hom maps by $s^k(\varphi )=\varphi h_{k-1}$, so these cohomology maps are independent of the comparison. In particular comparisons over identity give resolution independence. This proves both-variable naturality. Direct sums of lift systems give direct sums of cocycles; diagonal pullback followed by codiagonal pushout sends $(c,c^{\mathrm{\prime }})$ to $c+c^{\mathrm{\prime }}$. Hence the bijection respects Baer addition and transports the abelian group laws to extension classes.

By [F4], the target just proved is exactly ${\mathrm{Ext}}_P^n(M,N)$, without using balanced Ext. Apply the same argument in ${\mathcal{A}}^{\mathrm{o}\mathrm{p}}$: injective coresolutions become projective resolutions, extensions reverse endpoints, and pushouts become pullbacks. Translating back yields the asserted natural identification with ${\mathrm{Ext}}_I^n(M,N)$ and its addition.

Use the projective-resolution comparison in Higher Yoneda Ext agrees with derived Ext. A $p$-extension and a $q$-extension are represented by their lifted cocycles on the corresponding projective syzygies. Splicing the extensions concatenates their exact tails; lifting through that concatenated tail is the composite of the two lifted comparison maps.

Therefore the cocycle assigned to the splice is the chain-level composite representing the derived Ext composition. Passing to cohomology identifies the splice with $\alpha \circ \gamma$. Degree zero agrees as well because both products use ordinary morphism composition, with the units fixed in The Yoneda product is associative and unital.

If ${\mathrm{\Omega }}_P^n(M)$ is projective, truncate the fixed resolution after that object. The resulting length-$n$ projective resolution proves $\mathrm{pd}(M)\le n$.

Conversely, compare the fixed resolution with a projective resolution of length at most $n$. Iterating [L3] through their first $n$ projective presentations shows that ${\mathrm{\Omega }}_P^n(M)$ plus a finite direct sum of projectives is isomorphic to the terminal projective of the short resolution plus another finite direct sum of projectives. Hence ${\mathrm{\Omega }}_P^n(M)$ is a direct summand of a projective object and is projective. This also proves independence of the chosen resolution.

Assume $\mathrm{pd}(M)\le n$. If $n=0$, then $M$ is projective and positive Ext vanishes by [L3]. If $n\ge 1$, [L1] makes the $n$th syzygy projective, and repeated [L2] identifies every ${\mathrm{Ext}}^k(M,N)$ with $k>n$ with a positive-degree Ext group out of that syzygy, which vanishes by [L3]. Thus (1) implies (2), and (2) implies (3) by taking $k=n+1$.

Conversely, assume (3). For $n\ge 1$, [L2] gives ${\mathrm{Ext}}^1({\mathrm{\Omega }}^{n}M,N)=0$ for every $N$; for $n=0$, read ${\mathrm{\Omega }}^{0}M$ here as $M$. To see directly that an object $X$ with ${\mathrm{Ext}}^1(X,-)=0$ is projective, take the first stage $0\to K\stackrel{i}{\to }{P}_0\to X\to 0$ of a projective resolution and factor $d_1:P_1\to P_0$ as $P_1\stackrel{\pi }{\to }K\stackrel{i}{\to }{P}_0$. The map $\pi$ is a cocycle representing the identity of $K$ in the usual cokernel description of ${\mathrm{Ext}}^1(X,K)$. By [L5] and the assumed vanishing, $\pi =r{d}_1=ri\pi$ for some $r:P_0\to K$. Since $\pi$ is epic, $ri=1_K$, so the sequence splits and [L4] makes $X$ projective. Apply this to $X={\mathrm{\Omega }}^{n}M$ (or $X=M$ when $n=0$), and use [L1] when $n\ge 1$, to prove $\mathrm{pd}(M)\le n$.

If $\mathrm{id}(N)\le n$, the $n$th cosyzygy is injective. Repeated second-variable shifting from Ext dimension shifting in the second variable reduces every ${\mathrm{Ext}}^k(M,N)$ with $k>n$ to positive Ext into that injective cosyzygy, which vanishes by Positive injective-resolution Ext vanishes on an injective second variable.

Conversely, repeated shifting makes ${\mathrm{Ext}}^1(M,{\mathrm{\Sigma }}^{n}N)$ vanish for every $M$. The defining injective-dimension convention in Injective dimension of an object and the splitting criterion for the final cosyzygy show it is injective, so the chosen resolution has length at most $n$.

Projective dimension at most n iff higher Ext vanishes gives ${\mathrm{Ext}}^k(M,-)=0$ for $k>d$.

If $d=0$, take $X=M$: then ${\mathrm{Ext}}^0(M,M)=\mathrm{Hom}(M,M)$ contains the nonzero identity of $M$. If $d\ge 1$ and every ${\mathrm{Ext}}^d(M,X)$ vanished, the same theorem with $n=d-1$ would give $\mathrm{pd}(M)\le d-1$, contradicting minimality of $d$. Hence in either case some $X$ witnesses nonvanishing in degree $d$.

By Global dimension of an abelian category, global dimension is the supremum of projective dimensions. Projective dimension at most n iff higher Ext vanishes says that each individual projective dimension is exactly the least uniform Ext-vanishing bound.

Taking suprema over objects gives the displayed equality. If there is no nonzero object, both sides are $0$ by the stated convention; otherwise ${\mathrm{Ext}}^0(M,M)$ contains $1_M\mathrm{\ne }0$ for every nonzero $M$, so the degree-zero endpoint is present. When there are also enough injectives, Injective dimension at most n iff higher Ext vanishes gives the same uniform Ext bounds and hence the dual formulation.

If left global dimension is zero, every left module has projective dimension zero and hence is projective. By Equivalent module-theoretic characterizations of semisimple rings, a ring for which every left module is projective is precisely a semisimple ring, equivalently every left module is semisimple.

Conversely, if every left module is semisimple, the same ring characterization makes every left module projective. Its projective dimension is therefore zero, so the supremum in Left and right global dimension of a ring is zero. Replacing $R$ by $R^{\mathrm{o}\mathrm{p}}$ proves the right-module statement.

Choose a surjection $P\to M$ from a free, hence projective, left module $P$. Its kernel $K$ is a submodule of $P$ and is projective by hypothesis.

Thus $0\to K\to P\to M\to 0$ is a projective resolution of length at most one. The definitions Projective dimension of an object and Left and right global dimension of a ring yield the claimed bound; the right proof is identical with sides reversed.

Well-order $B$. For $b\in B$, let $F_{\le b}$ be the span of the $e_c$ with $c\le b$, and put $H_{\le b}=H\cap F_{\le b}$. The $b$-coordinate homomorphism ${\pi }_b:F_{\le b}\to \mathbb{Z}$ has image ${\pi }_b(H_{\le b})=d_b\mathbb{Z}$ for a uniquely determined $d_b\ge 0$. Put $D=\{b\in B:d_b>0\}$.

For each $b\in D$ the set $X_b=\{x\in H_{\le b}:{\pi }_b(x)=d_b\}$ is nonempty. Apply AC to this set-indexed family and fix $x_b\in X_b$ simultaneously for all $b\in D$. These are the only choices of lifts in the argument; they do not depend recursively on previously chosen bases.

We prove by well-ordered induction on $b$ that $H_{\le b}$ is spanned by $x_c$ with $c\in D$ and $c\le b$. Suppose this is proved below $b$, and take $h\in H_{\le b}$. If $d_b>0$, write ${\pi }_b(h)=q{d}_b$ and set $h^{\mathrm{\prime }}=h-q{x}_b$; if $d_b=0$, set $h^{\mathrm{\prime }}=h$. In both cases $h^{\mathrm{\prime }}\in H$ has finite support strictly below $b$. If that support is empty, $h^{\mathrm{\prime }}=0$. Otherwise it has a greatest element $c<b$, and $h^{\mathrm{\prime }}\in H_{\le c}$ is in the required span by the induction hypothesis. Thus so is $h$. Conversely all the indicated $x_c$ belong to $H_{\le b}$. This verifies the inductive implication, including the least-element and limit cases, and [L4] proves the assertion for every $b$.

The family $(x_b{)}_{b\in D}$ is linearly independent. Indeed, in a nonzero finite relation choose the greatest index $b$ with nonzero coefficient $n_b$. Its $b$-coordinate is $n_b{d}_b$, since all smaller-indexed vectors have zero $b$-coordinate. This is nonzero in $\mathbb{Z}$, contradicting the relation. Every nonzero $h\in H$ has a greatest basis-support index $b$ and belongs to $H_{\le b}$, so step 3.1 also proves spanning of all $H$. Hence $(x_b{)}_{b\in D}$ is a basis. If $B$ or $D$ is empty the same reasoning says $H=0$ with empty basis.

Take the canonical surjection $\epsilon :{\mathbb{Z}}^{(A)}\to A$ and put $F={\mathbb{Z}}^{(A)}$, $K=\ker \epsilon$. The sequence $0\to K\to F\to A\to 0$ is exact. Under the stated AC hypothesis, [L2] makes $K$ free and both $K$ and $F$ projective. Hence this is a projective resolution of length at most one, and ${\mathrm{pd}}_{\mathbb{Z}}A\le 1$. No family of resolutions of all groups is selected: this construction applies separately to each $A$.

The group $\mathbb{Z}\mathrm{/}n\mathbb{Z}$ is not projective. Otherwise its identity map would lift through the quotient surjection $\pi :\mathbb{Z}\to \mathbb{Z}\mathrm{/}n\mathbb{Z}$ to a homomorphism $s$ with $\pi s=\mathrm{i}\mathrm{d}$. But $ns([1])=s(n[1])=s(0)=0$, and an integer annihilated by the positive integer $n$ is zero. Thus $s([1])=0$, contradicting $\pi s([1])=[1]\mathrm{\ne }0$ for $n>1$. Consequently [L4] and step 1.1 give ${\mathrm{pd}}_{\mathbb{Z}}(\mathbb{Z}\mathrm{/}n\mathbb{Z})=1$, while step 1.1 bounds every projective dimension by one. Therefore the left global dimension is exactly one.

Left and right $\mathbb{Z}$-modules are both precisely abelian groups with the integer action given by repeated addition and additive inverses. Passing from $ma$ to $am=ma$ preserves the underlying homomorphisms, exact sequences, and lifting diagrams; thus it preserves projectivity and lengths of projective resolutions in both directions. Their two suprema are therefore equal, so the right global dimension is also one.

An extension of $M$ by $N$ has the typed form $0\to N\to E\to M\to 0$; its connecting homomorphism starts at $1_M\in \mathrm{Hom}(M,M)$ and lands in ${\mathrm{Ext}}^1(M,N)$.

Here ${\mathrm{Ext}}^1(\mathbb{Z}\mathrm{/}n\mathbb{Z},\mathbb{Z})\cong \mathbb{Z}\mathrm{/}n\mathbb{Z}$, whereas ${\mathrm{Ext}}^1(\mathbb{Z},\mathbb{Z}\mathrm{/}n\mathbb{Z})=0$ because $\mathbb{Z}$ is projective. Interchanging the endpoints therefore changes the group and disproves the asserted order.

Taking their direct sum gives $0\to N\oplus N\to E_1\oplus E_2\to M\oplus M\to 0$, whose endpoints are $N\oplus N$ and $M\oplus M$, not the required $N$ and $M$.

Pulling this sequence back along $\mathrm{\Delta }:M\to M\oplus M$ restores the quotient $M$, and pushing out along $\mathrm{\nabla }:N\oplus N\to N$ restores the subobject $N$. These two operations, not merely the direct sum of middle terms, define the Baer sum.

The axioms stated here do not bound the possible middle objects $E$ by a set. Thus the definition initially supplies only a class of extensions and an equivalence relation on it. Taking a quotient of a class by an equivalence relation does not, by itself, prove that the quotient is a set.

The cited corollary obtains a set only after a resolution-based classification identifies the classes with a set-sized ${\mathrm{Ext}}^1(M,N)$. That extra smallness discharge is not a consequence of the equivalence-relation definition alone, which refutes the claim as now stated.

Put $V=kx\oplus ky$, so $R=k\oplus V$ and $V^2=0$. A free resolution of $S$ is $P_n=R{\otimes }_k{V}^{\otimes n}$ with $d_n(r\otimes v_1\otimes \cdots \otimes v_n)=r{v}_1\otimes v_2\otimes \cdots \otimes v_n\mathrm{.}$ Its kernel in degree $n$ is $V{\otimes }_k{V}^{\otimes n}$, exactly the image of $d_{n+1}$, so it is exact. Applying ${\mathrm{Hom}}_R(-,S)$ gives zero differentials and identifies ${\mathrm{Ext}}_R^n(S,S)$ with $(V^{\ast }{)}^{\otimes n}$.

Under Yoneda splicing, the two degree-one dual basis classes $u,v\in V^{\ast }$ multiply by tensor concatenation. Thus $uv$ and $vu$ are the two distinct basis tensors of $(V^{\ast }{)}^{\otimes 2}$. In particular the self-Ext algebra is the free associative algebra $T_k(V^{\ast })$, which is not graded commutative. This is a typed counterexample inside one self-Ext algebra.

The two-term resolution $0\to \mathbb{Z}\stackrel{n}{\to }\mathbb{Z}\to M\to 0$ has length one, and ${\mathrm{Ext}}^1(M,\mathbb{Z})\cong \mathbb{Z}\mathrm{/}n\mathbb{Z}\mathrm{\ne }0$, so the projective dimension of $M$ is exactly one.

Yet $0\to \mathbb{Z}\stackrel{c\mapsto (0,c)}{\to }{\mathbb{Z}}^2\stackrel{(a,b)\mapsto na}{\to }\mathbb{Z}\to M\to 0$ is also exact and consists of projectives, but has length two. Hence projective dimension is the least possible length, not the length of every resolution.