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A compact object (mathematics) is an object whose represented Hom functor preserves filtered colimits. This preservation condition differs from the extra right-adjoint property defining a tiny object.

Evaluation at $e_i$ identifies a module homomorphism $h:kQ{e}_i\to X$ with $h(e_i)\in e_{i}X$. Given $x\in e_{i}X$, the inverse sends $p$ to $px$. The isomorphism is natural and shows that Hom functor evaluation on a vertex projective is exact.

The Ext functors are the right derived functors of the Hom functor. They can be computed by applying ${\mathrm{Hom}}_R(-,N)$ to a projective resolution of $M$ and taking cohomology.

A graded Hom functor construction consists of maps shifting degree by $j$. One homological convention is $d_{M}f=f{d}_C+(-1{)}^{j-1}{d}_{C^{\prime }}f$. Its degree-zero cycles are chain maps, and its degree-zero boundaries are null-homotopic chain maps, so its zeroth homology is the module of chain homotopy classes. This convention differs by degree-dependent signs from the alternative convention $d^{\prime }f-(-1{)}^{j}fd$.

For $\mathbb{Z}G$-modules $M_1,M_2$,because the Hom functor into a finite direct sum splits degree by degree and taking cohomology preserves that splitting.

Applying the contravariant Hom functor ${\mathrm{Hom}}_R(-,W)$ to $0\to I\to X\to C\to 0$ givesThe connecting map sends $I\to W$ to its pushout of a module extension. Its vanishing means that map extends to $X\to W$. For a hereditary ring, ${\mathrm{Ext}}^2(C,W)=0$, making the indicated restriction on first extension groups surjective. The covariant variable has its corresponding long exact sequence.

Since $M$ is a finitely generated module, choose a surjection from a finite free module, $F_0\twoheadrightarrow M$. A finite free module over a left Noetherian ring is a Noetherian module, so the kernel $K_1$ is finitely generated. Choose a finite free module $F_1$ surjecting onto $K_1$, then compose its map with $K_1\hookrightarrow F_0$. Its kernel is again finitely generated. Repeating constructs an exact sequencewith all $F_i$ finite free modules, hence finitely generated projective modules. Thus the requested projective resolution exists, without asserting that it has finite length.

Assume now that $M\ne 0$ and ${\mathrm{pd}}_{R}M=n<\infty$. We may arrange a length-$n$ projective resolution with all terms finite projective. Here is why the preceding construction gives one. Set $K_0=M$, with short exact sequences $0\to K_{j+1}\to F_j\to K_j\to 0$. Applying the Ext functor, whose positive degrees vanish on a projective first argument, gives by repeated dimension shiftingfor every left module $N$. The final equality follows from the existence of any length-$n$ projective resolution. Vanishing of these degree-one groups means that every short exact sequence ending in $K_n$ splits: its extension class is zero. In particular a free surjection onto $K_n$ splits, so $K_n$ is a projective module. It is finitely generated by the Noetherian kernel construction. Use $P_n=K_n$ and $P_j=F_j$ for $j<n$ to obtainFor a left finite projective module $P$, put $P^{\ast }={\mathrm{Hom}}_R(P,R)$, a right module under $(fr)(p)=f(p)r$. It is a right finite projective module, and its natural evaluation mapis an isomorphism. These claims hold for a finite free module by coordinates, and pass to direct summands, proving them for every finite projective module without any right-Noetherian hypothesis.

Suppose $n\ge 1$ and ${\mathrm{Ext}}_R^n(M,R)=0$. Computing the Ext functor from this resolution shows that its last dual mapis surjective. Since $P_n^{\ast }$ is projective as a right module, this map has a right-module section $h:P_n^{\ast }\to P_{n-1}^{\ast }$, with $d_n^{\ast }h=1$. Dualizing gives $h^{\ast }{d}_n^{\ast \ast }=1$. Under the natural double-dual identifications, $d_n^{\ast \ast }$ is $d_n$, so this gives a left-module retraction of $d_n$.

Therefore $P_{n-1}=d_n(P_n)\oplus Q$ for a finite projective module $Q$. The next differential has kernel $d_n(P_n)$, so restricting it to $Q$ gives an exact resolution with $Q$ replacing $P_{n-1}$ and the initial $P_n$ removed. This has length $n-1$, contradicting the definition of projective dimension. For $n=0$, the nonzero $M$ is a finite projective module and a summand of some $R^d$. A nonzero vector in that summand has a nonzero coordinate, so one coordinate projection restricts to a nonzero map $M\to R$. Thus the degree-zero Hom functor is nonzero as well. We have provedThis is top Ext detects finite projective dimension over a left Noetherian ring. The nonzero-module convention matters when $n=0$: if one assigns projective dimension zero to the zero module, then ${\mathrm{Hom}}_R(0,R)=0$ is an exception. With the convention ${\mathrm{pd}}_{R}0=-\infty$, it is outside the asserted finite nonnegative case.

For an example of infinite projective dimension, take the dual numbers $R=k[\epsilon ]/({\epsilon }^2)$ over a field $k$, and $M=k=R/(\epsilon )$. The periodic resolution over dual numbers isIt is exact because both the kernel and image of multiplication by $\epsilon$ are the ideal $(\epsilon )$, and that is also the kernel of $R\to k$. Applying ${\mathrm{Hom}}_R(-,k)$ gives a copy of $k$ in every degree and zero differentials, since $\epsilon$ acts as zero on $k$. ConsequentlyA finite projective resolution would make these groups zero above its length, so

Take $B=\mathbb{Z}$, $N=\mathbb{Z}/2\mathbb{Z}$, and the short exact sequenceMultiplication by two is injective, and the reduction map has image all of $\mathbb{Z}/2\mathbb{Z}$ and kernel $2\mathbb{Z}$, proving exactness. Any module homomorphism from $\mathbb{Z}/2\mathbb{Z}$ to $\mathbb{Z}$ is zero: its generator must map to an integer killed by two. On the other hand, a homomorphism from $\mathbb{Z}/2\mathbb{Z}$ to itself is determined arbitrarily by the image of its generator, so its Hom group is $\mathbb{Z}/2\mathbb{Z}$. Applying the Hom functor therefore givesThis sequence is not exact at $\mathbb{Z}/2\mathbb{Z}$: the incoming image is zero while the outgoing kernel is the whole nonzero group. In particular, the identity of $N$ cannot lift through the reduction map to a map $N\to \mathbb{Z}$.

Choose a surjection $P_0=A^{r_0}\twoheadrightarrow M$. Since $A$ is a Noetherian ring, a finite free module is a Noetherian module, and its kernel $K_1$ is a finitely generated module. Choose a finite free surjection $P_1\twoheadrightarrow K_1$ and repeat. This constructs a resolution by finitely generated free moduleswhose terms are, in particular, finitely generated projective modules. It may be infinite; the request is finite generation of every term, not finite length.

Now suppose the nonzero finitely generated module $M$ has finite projective dimension $n$. For $n\ge 1$, its $n$th syzygy $K_n$ in this free resolution is projective. Indeed, dimension shifting along the successive kernel sequences givesfor every module $N$. The vanishing follows by computing the latter Ext functor from any length-$n$ projective resolution. Vanishing of ${\mathrm{Ext}}^1(K_n,-)$ makes every short exact sequence ending in $K_n$ split: apply the long exact sequence of Ext groups to a free surjection onto $K_n$ to lift its identity. Thus $K_n$ is a projective module. It is finitely generated by the preceding construction. Truncating giveswith every term finite projective.

Apply the Hom functor ${\mathrm{Hom}}_A(-,A)$ and write $P_i^{\ast }={\mathrm{Hom}}_A(P_i,A)$. At the top degree,If this group were zero, $d_n^{\ast }$ would be surjective. The dual module of a finite projective module is finite projective, so this surjection has a section $s:P_n^{\ast }\to P_{n-1}^{\ast }$. Finite projective modules identify with their double dual modules: the evaluation map is an isomorphism for a finite free module and passes to its direct summands. Dualizing $d_n^{\ast }s=\mathrm{id}$ therefore gives $s^{\ast }{d}_n={\mathrm{id}}_{P_n}$. Hence $d_n$ splits, and $P_{n-1}/d_n(P_n)$ is projective. Replacing the last two terms by this quotient shortens the projective resolution to length $n-1$, contradicting minimality. This proves top Ext detects finite projective dimension:For $n=0$, $M$ is a nonzero finite projective module and embeds as a summand of $A^r$. Some coordinate projection restricts nontrivially to $M$, proving ${\mathrm{Hom}}_A(M,A)\ne 0$ as required.

For infinite projective dimension, take the dual-number algebra $A=k[\epsilon ]/({\epsilon }^2)$ and $M=k=A/(\epsilon )$. Multiplication by $\epsilon$ has image and kernel both equal to $(\epsilon )$, giving the periodic resolution over dual numbersApplying ${\mathrm{Hom}}_A(-,k)$ makes every differential zero, so ${\mathrm{Ext}}_A^i(k,k)\cong k$ for every $i\ge 0$. A finite projective resolution would force all sufficiently high Ext functors to vanish. Thus ${\mathrm{pd}}_{k[\epsilon ]/({\epsilon }^2)}k=\infty$.

The finite-generation hypothesis is essential if the middle sentence's “a module” is interpreted as an unrestricted new module. A concrete vanishing top Ext without finite generation occurs for $A=k[[t]]$ and $K=A[1/t]=k((t))$. This Noetherian ring has the exact free presentationInjectivity of $d$ follows by examining the largest index of a finitely supported input, since $t$ is a non-zero-divisor. For exactness in the middle, use the relations to move a finite sum to its largest index $N$; its remaining coefficient is ${\sum }_{i=0}^N{a}_i{t}^{N-i}$, which is zero precisely when its image in $K$ is zero. The map onto $K$ is surjective because every Laurent series has a lower bound on its exponents.

Applying ${\mathrm{Hom}}_A(-,A)$ gives on products the map $(b_i)\mapsto (b_i-t{b}_{i+1})$. It is surjective: for any $(c_i)$, set $b_i={\sum }_{j\ge 0}{t}^j{c}_{i+j}$. These are well-defined formal power series, because each coefficient receives only finitely many contributions, and $b_i-t{b}_{i+1}=c_i$. Consequently ${\mathrm{Ext}}_A^1(K,A)=0$. But ${\mathrm{Hom}}_A(K,A)=0$: the image of every $t^{-i}$ must lie in ${\bigcap }_{r\ge 0}{t}^{r}A=0$, since it is divisible by every power of $t$. A nonzero projective module is a summand of a free module and therefore has a nonzero coordinate map to $A$. Thus $K$ is not projective, while its displayed resolution has length one. In particular,The nonvanishing conclusion proved above retains the finitely generated $M$ introduced at the start of the question.

The tensor differential. Use homological grading, so each differential lowers degree by one. The tensor product of chain complexes hasfor $x\in C_p$. This Koszul sign rule givesThus the graded tensor product is a chain complex.

The Hom differential. Write $M_j={\prod }_p{\mathrm{Hom}}_R(C_p,C_{p+j}^{\prime })$. For a degree-$j$ element of this graded Hom complex of chain complexes, use the prescribed differentialThe next application uses degree $j-1$, soIn degree zero, $d_{M}f=f{d}_C-d_{C^{\prime }}f$, so its kernel consists exactly of chain maps. A degree-one element $h$ has $d_{M}h=h{d}_C+d_{C^{\prime }}h$, which is exactly the change between two chain maps related by a chain homotopy. ConsequentlyThis is natural: precomposition and postcomposition by chain maps preserve both degree-zero cycles and degree-zero boundaries.

The dual complex and the sign adjustment. Define the reversed dual chain complexHere $d_X\varphi$ is a functional on $C_{1-i}$, so it belongs to $X_{i-1}$, and $d_X^2=0$. The absence of an additional sign in $d_X$ is intentional.

Interpret finite generation of the free chain complexes as finiteness of their total graded free modules. Then only finitely many degrees occur, and the finite-free evaluation isomorphisms assemble into a graded isomorphismwhere $\varphi \in X_i$, $y\in C_j^{\prime }$, and this component is zero outside $C_{-i}$. The dual module construction and evaluation make $\psi$ natural. Finite rank is needed for evaluation to be an isomorphism; finite total support also makes the sums on the tensor side agree with the products on the Hom functor side.

Under $\psi$, the Hom functor differential has the formThe usual tensor product of chain complexes instead has second coefficient $(-1{)}^i$. SetThis is integer-valued for every $j\in \mathbb{Z}$, including negative $j$, and satisfies $\rho (j)-\rho (j-1)=j-1$. Conjugating the usual tensor differential by $A$ leaves the first term unchanged and changes its second coefficient to $(-1{)}^{i+j-1}$. Hence $d_M=\psi AD{A}^{-1}{\psi }^{-1}$, giving the sign conjugation for the tensor-Hom identificationThis is an isomorphism of chain complexes, not merely of their homology. If one instead assumes only degreewise finite rank with unbounded grading, the ordinary tensor need not identify with the product defining $M$; the finite-total convention is essential to this last conclusion.

Let $K=\ker f_{\rho }$ and $C=Y_j/\mathrm{im}{g}_{\rho }$. Consider the map from the arrow term of the extension complex of quiver representations to ${\mathrm{Hom}}_k(K,C)$ that restricts its $\rho$-component to $K$ and then takes the quotient in $Y_j$. This map is surjective: a map $K\to C$ can be lifted to $Y_j$ and extended from $K$ to $X_i$.

Every coboundary is killed by this map, since for $x\in K$,It therefore induces a surjectionBoth vector spaces in the final Hom functor are nonzero, so its dimension is positive. Hence ${\mathrm{Ext}}_Q^1(X,Y)\ne 0$. This is the kernel-cokernel obstruction to splitting a quiver extension and remains valid for loops and repeated arrows elsewhere in the quiver.

Choose a projective presentation $0\to K\stackrel{j}{\to }P\stackrel{\pi }{\to }C\to 0$, with $P$ a projective module. Continuing it to a projective resolution showsIndeed a degree-one cocycle in the Hom functor applied to the projective resolution descends to $K$, while the degree-one coboundaries are exactly restrictions of maps from $P$. This establishes the description of the Ext functor without first assuming the extension classification.

For a module extension with inclusion $i:A\to B$ and quotient map $r:B\to C$, projectivity lifts $\pi$ to $s:P\to B$. Since $rsj=0$, there is a unique map $h:K\to A$ with $ih=sj$. A different lift changes $h$ by $\ell j$ for some $\ell :P\to A$. An equivalence of module extensions, which is the identity on both end modules, also preserves this class. We have therefore defined a map from extension classes to ${\mathrm{Ext}}_R^1(C,A)$.

Conversely, for $h:K\to A$, form the pushout of a module extensionThe map $A\to B_h$ sends $a$ to $[a,0]$, and $B_h\to C$ sends $[a,p]$ to $\pi (p)$. The first is injective because $j$ is injective. If $\pi (p)=0$, write $p=j(k)$; then $[a,p]=[a+h(k),0]$, proving exactness in the middle. The last map is surjective. Thus $B_h$ is a module extension. If $h^{\prime }=h+\ell j$, the map $[a,p]\mapsto [a-\ell (p),p]$ is an equivalence of module extensions from $B_h$ to $B_{h^{\prime }}$. Finally $[a,p]\mapsto i(a)+s(p)$ identifies the pushout of a module extension built from an existing extension with its middle module. These two constructions are inverse, proving the classification:The zero class corresponds to a split short exact sequence. Fixed end modules matter: equivalence does not permit arbitrary automorphisms of $A$ or $C$.

In the second calculation the acting group is the infinite cyclic group. Its modules are modules over the group ring $S=\mathbb{Z}[z,z^{-1}]$, not merely over the underlying ring $\mathbb{Z}$. The trivial representation has $z$ acting as the identity. Its projective resolution isMultiplication by $z-1$ is injective, and the augmentation quotient is $\mathbb{Z}$. Applying the Hom functor into the trivial module makes the differential zero, so ${\mathrm{Ext}}_S^1(\mathbb{Z},\mathbb{Z})\cong \mathbb{Z}$. Taking two copies gives $\boxed{\mathcal{E}({\mathbb{Z}}^2,\mathbb{Z})\cong {\mathbb{Z}}^2}$.

An explicit representative of this extension of trivial modules for an infinite cyclic group is the abelian group $B_{m,n}=\mathbb{Z}u\oplus \mathbb{Z}v\oplus \mathbb{Z}w$, with injection $1\mapsto u$, quotient $v\mapsto (1,0)$, $w\mapsto (0,1)$, and actionThis is an invertible action: the inverse subtracts the same multiples of $u$. Every underlying abelian-group extension splits because ${\mathbb{Z}}^2$ is free, so any group-module extension has this form after choosing lifts of the quotient basis. Replacing those lifts by multiples of $u$ does not alter $m,n$. An equivalence fixing the ends has exactly such changes of lifts, so two representatives are equivalent precisely when their ordered pairs agree. Only $(0,0)$ splits as a group-module extension; overlooking the group action would incorrectly give a single class.

For central $t_1,\dots ,t_n$, define the Koszul complex on central ring elements using formal exterior symbols:with zero terms outside $0\le k\le n$ and differentialThe terms are free bimodules with the formal symbols commuting with coefficients. Centrality makes $d$ a bimodule map, and the terms in $d^2$ cancel in pairs because the $t_i$ commute. This defines the Koszul complex over a possibly noncommutative ring without using an undefined exterior algebra of arbitrary one-sided modules. In particular $K(\mathbf{t}){\otimes }_{R}M$ is a well-defined chain complex of left modules.

A regular sequence on a module means that multiplication by $t_i$ is injective on $M/(t_1,\dots ,t_{i-1})M$ for each $i$, usually with the additional convention that the final quotient is nonzero. The homology vanishing below uses only the injectivity conditions. For one element, the chain complex is $M\stackrel{t_1}{\to }M$, with zero first homology and zeroth homology $M/t_{1}M$.

For the induction let $C=K(t_1,\dots ,t_{n-1}){\otimes }_{R}M$. Adjoining the last generator identifies the new Koszul complex with the mapping cone of multiplication by $t_n$ on $C$. In the convention $\mathrm{Cone}(t_n{)}_k=C_k\oplus C_{k-1}$ its differential is $(a,b)\mapsto (da+t_{n}b,-db)$; the identification is $a+e_n\wedge b$. The long exact sequence in homology of this mapping cone givesThe induction hypothesis is $H_k(C)=0$ for $k>0$ and $H_0(C)=N=M/(t_1,\dots ,t_{n-1})M$. Regularity makes $t_n:N\to N$ injective. Hence the new positive homology is zero and its zeroth homology is $N/t_{n}N$. The augmentation to this quotient induces these homology isomorphisms, giving the quasi-isomorphismwhere the right side is placed in degree zero. The results used are the long exact sequence in homology of a degreewise short exact sequence of complexes and its mapping cone form, together with the stated induction; no unmentioned acyclicity criterion is needed.

For $R=\mathbb{Z}[X]$, $(p,X)$ is a regular sequence: $p$ is a non-zero-divisor in $R$, and multiplication by $X$ is injective in $(\mathbb{Z}/p\mathbb{Z})[X]$, even when $p$ is composite. The Koszul resolution of $R/(p,X)$ isWith $N=\mathbb{Z}/q\mathbb{Z}$ and $X$ acting as zero, applying the Hom functor gives the cochain complexThus, more generally, the Ext groups between polynomial-ring residue modules areand vanish in all degrees above two. For the equal-modulus case,These are $R$-modules through $X=0$ and reduction modulo $p$. If the multiplicative self-Ext functor is desired, the Koszul self-Ext algebra is the exterior algebra on two degree-one generators over $\mathbb{Z}/p\mathbb{Z}$. The two contractions on the Koszul resolution lift these classes, square to zero and anticommute, so this description also holds for composite $p$ and characteristic two. For coprime $p,q$, multiplication by $p$ is invertible on $N$, so both its kernel and cokernel vanish:

Put $A^e=A{\otimes }_k{A}^{\mathrm{op}}$, so an $A$-bimodule is a left $A^e$-module through $(a\otimes b^{\mathrm{op}})m=amb$. The Hochschild cohomology isEquivalently, the Hochschild cochain complex has $C^n(A,M)={\mathrm{Hom}}_k(A^{\otimes n},M)$ and coboundary mapIts cohomology agrees with the displayed Ext functor because the bar resolution of an associative algebra is free over $A^e$ when $k$ is a field. The Hochschild cohomological dimension isThe supremum uses all bimodules, not merely the finitely generated one supplied in the question; an unbounded projective dimension is infinity.

An extension in this classification is a square-zero extension of an algebra: a short exact sequence $0\to M\to E\to A\to 0$, where $E$ and $A$ are unital algebras, $E\to A$ is unital, and $M$ is its two-sided ideal with $M^2=0$ and induced $A$-bimodule structure equal to the prescribed one. An equivalence of extensions is an algebra isomorphism of the middle terms commuting with the maps and inducing the identity on both $A$ and $M$. Arbitrary isomorphisms of middle algebras, or extensions without the square-zero ideal requirement, are not classified by this cohomology group.

Choose a $k$-linear map section $s:A\to E$ with $s(1)=1$. Its multiplication defectis a normalized Hochschild cocycle: $\mu (1,a)=\mu (a,1)=0$, and associativity in $E$ gives $\delta \mu =0$. Changing $s$ to $s+g$, where $g(1)=0$, changes the defect to $\mu +\delta g$, since $M^2=0$.

Conversely, for a normalized Hochschild cocycle $\mu$, put $E_{\mu }=A\oplus M$ as a vector space and defineThe Hochschild cocycle equation is exactly associativity, and $(1,0)$ is the identity. If ${\mu }^{\prime }=\mu +\delta g$, the map $E_{{\mu }^{\prime }}\to E_{\mu }$, $(a,m)\mapsto (a,m+g(a))$, is an equivalence. Conversely, every equivalence has this form after choosing sections. The normalized Hochschild cochain complex computes the same cohomology as the full complex: in the bar resolution of an associative algebra, the degenerate terms containing an inserted identity form a contractible subcomplex. Passing to the normalized bar resolution of an associative algebra, then applying the Hom functor, gives the same cohomology. Thus every class has a normalized representative. We obtain a bijection between $H{H}^2(A,M)$ and equivalence classes of square-zero extensions.

For a formal associative deformation, a completion convention is necessary. The usual star product lives on the formal power series moduleThis is the adic completion of a module applied to the ordinary tensor product, rather than literally the ordinary $A{\otimes }_{k}k[[t]]$ when $A$ is infinite-dimensional. For example, ${\sum }_{n\ge 0}{X}^n{t}^n$ lies in $k[X][[t]]$ but not in the ordinary tensor product, whose coefficient spaces have finite-dimensional span. The two agree when $A$ is finite-dimensional. We interpret the printed notation in this standard completed sense; the infinite iteration below requires that interpretation.

A star product is a $k[[t]]$-bilinear, unital product continuous for the adic topology satisfying associativity of the formIt is a trivial formal deformation if a $k[[t]]$-linear automorphism continuous for the adic topology $T=\mathrm{id}+{\sum }_{r\ge 1}{t}^r{T}_r$, with $T(1)=1$, satisfies $T(a\ast b)=T(a)T(b)$.

If $\mathrm{Dim}(A)\le 1$, then $H{H}^2(A,A)=0$. Suppose changes of coordinates have removed all coefficients below order $r$. The order-$r$ part of associativity then says $\delta {\mu }_r=0$. Hence ${\mu }_r=\delta g_r$ for a $k$-linear map $g_r:A\to A$. Its normalization gives $g_r(1)=0$, since $(\delta g_r)(1,1)=g_r(1)$. Transport the product by $T_r=\mathrm{id}+t^r{g}_r$:The order-$r$ coefficient becomes ${\mu }_r-\delta g_r=0$, and lower coefficients remain zero. Repeating constructs compatible changes of coordinates modulo every $t^N$. They converge in the adic topology to an invertible $T$ fixing $1$, with inverse obtained coefficient by coefficient. The limit product is ordinary multiplication. Therefore every star product is trivial under the completed formal-series convention. In fact, the argument only needs the vanishing of $H{H}^2(A,A)$, not all of Hochschild cohomological dimension at most one.

Put $I={\mathrm{Hom}}_{\mathcal{O}G}(W,W^{\prime })$. Since $W$ is a projective module, its Hom functor is exact. Applied to $0\to \pi W^{\prime }\to W^{\prime }\to W^{\prime }/\pi W^{\prime }\to 0$, it givesThe kernel is $\pi I$ by (ii). Every map to $W^{\prime }/\pi W^{\prime }$ kills $\pi W$ and factors uniquely through $W/\pi W$; this gives the second isomorphism. ThereforeBy (i), $I$ is finite free with rank ${\dim }_K{\mathrm{Hom}}_{KG}(M,M^{\prime })$. Its reduction has dimension equal to that rank. Thus reduction of Hom from a projective group-algebra lattice yieldsThe original PDF has the group-algebra subscripts used here; the supplied TeX drops or corrupts several of them.

Let $W,W^{\prime }$ be choices of integral form of a group representation over a complete p-modular system. Then $I={\mathrm{Hom}}_{\mathcal{O}G}(W,W^{\prime })$ is finite free, and $K{\otimes }_{\mathcal{O}}I={\mathrm{Hom}}_{KG}(K\otimes W,K\otimes W^{\prime })$: clearing denominators proves the spanning assertion. Also ${\mathrm{Hom}}_{\mathcal{O}G}(W,\pi W^{\prime })=\pi I$. If $W$ is a projective module, applying the Hom functor to $0\to \pi W^{\prime }\to W^{\prime }\to W^{\prime }/\pi W^{\prime }\to 0$ yieldsConsequently these ordinary and modular Hom spaces have the same dimension of a vector space. Projectivity is essential: for $G=C_2$ over ${\mathbb{Z}}_2$, the trivial and sign lattices have zero Hom between them, whereas their reductions in characteristic $2$ coincide and have a one-dimensional Hom space.

Put $A=k[[t]]$ and $K=A[1/t]$. A free resolution has countably many generators $e_i$ and relations $e_i-t{e}_{i+1}$. Applying the Hom functor with coefficients in $A$ gives $(b_i)\mapsto (b_i-t{b}_{i+1})$ on sequence products. This is surjective: for prescribed $(c_i)$ use $b_i={\sum }_{j\ge 0}{t}^j{c}_{i+j}$, a well-defined formal power series. Therefore the degree-one Ext functor vanishes. But ${\mathrm{Hom}}_A(K,A)=0$, since every image is infinitely divisible by $t$. A nonzero projective module has a nonzero coordinate map into $A$, so $K$ is not projective. Its displayed resolution proves that its projective dimension is exactly one. The finite-generation hypothesis in top Ext detects finite projective dimension is essential.