Draft lesson.

Noetherian generators, filtered injectives and Ext

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

Exact filtered colimits need not preserve injective objects. A noetherian generating family supplies the missing finiteness, including the relations needed to factor maps through a stage. We prove that fact without projective generators, prove a converse under local finite generation, and retain the actual filtered-colimit comparisons for all Ext groups.

Fix universes and choice. Let $\mathcal A$ be a Grothendieck category: it is abelian, has all small colimits and a small generating family, and filtered colimits are exact. Use the full small-subobject and evaluation-presentation arguments of Generators and small quotient families, Sections 2–4. The generator-family extension criterion and functorial injective embeddings are proved in Testing injectivity on generators and constructing enough injectives, Sections 1–4 and 7.

Injective replacements and right derived functors, Sections 2–4, and Derived morphisms through quasi-isomorphism roofs, Section 5, supply the bounded-below injective Hom tests. Hom, tensor and the exactness of module limits, Sections 3–4, proves the filtered group-colimit equality and cohomology comparisons. Finitely presented generators and coherent Ind reconstruction, Sections 1–6, supplies the exact Ind realization once its finite-presentation hypotheses have been proved here.

1. Finite generation and noetherian subobjects

An object $N$ is finitely generated if a directed union of subobjects equal to $N$ has a member equal to $N$. It is noetherian if every ascending sequence of subobjects stabilizes. The category is locally finitely generated, respectively locally noetherian, when it has a small generating family of finitely generated, respectively noetherian, objects. None of these definitions asserts projectivity.

Lemma 1.1. An object is noetherian exactly when all its subobjects are finitely generated. Subobjects, quotients and finite extensions of noetherian objects are noetherian. Quotients and finite sums of finitely generated objects are finitely generated.

Proof. If a subobject $K\subset N$ is a directed union with no member equal to $K$, choose successively a strictly larger member: at each step some member escapes the previous one, and a common upper member contains both. This contradicts the ascending chain condition on $N$. Conversely, the union of an ascending sequence in $N$ is a subobject; if it is finitely generated, some sequence member equals that union, so the sequence stabilizes.

Subobjects inherit the chain condition. Subobjects of a quotient correspond to subobjects containing its kernel, by inverse image and quotient, so quotients inherit it too. For $0\to L\to N\to Q\to0$, an ascending sequence in $N$ has ascending intersections with $L$ and ascending images in $Q$. If both stabilize, its successive inclusions induce isomorphisms on the two ends of the short exact intersection/image rows. The middle inclusion is then invertible: its kernel and cokernel vanish by the exact-row comparison. Thus noetherian $L,Q$ imply noetherian $N$.

For finite generation, inverse images of a directed union under a quotient have union the source. More generally, if $B=\operatorname{colim}B_t$ is a union of subobjects and $f\colon M\to B$, their inverse images have union $M$. Indeed each pullback is the kernel of $(f,-j_t)\colon M\oplus B_t\to B$. Exact filtered colimits identify the colimit kernel with the kernel of $(f,-1_B)$, namely $M$. This also proves that those colimit maps are the specified inclusions. Apply this to a quotient of a finitely generated object to prove the quotient assertion. A finite sum is finitely generated because the inverse images in its finitely many summands reach the whole summands at finitely many stages; one common upper stage contains their images and hence the entire sum. This includes zero and empty sums. $\square$

Lemma 1.2. With a finitely generated generating family, every object is the directed union of finitely generated subobjects. With a noetherian generating family, those subobjects may be chosen noetherian; in this case finitely generated objects are precisely the noetherian objects.

Proof. The canonical evaluation from the coproduct of all generator maps into $X$ is epic. Take the images of its finite partial sums. They form a small directed family of subobjects. Exact filtered colimits of their image factorizations identify their union with the full image, which is $X$. Each finite image is a quotient of a finite sum of generators. Lemma 1.1 gives its asserted finiteness. If $X$ is finitely generated, one of the noetherian partial images is all of $X$. Conversely every noetherian object is finitely generated by Lemma 1.1 applied to itself. $\square$

2. A finite map killed in the colimit dies at a stage

Let $X\colon I\to\mathcal A$ be a small filtered diagram, with colimit $X_\infty$. Fix $i\in I$. The category $i/I$ of arrows out of $i$ is filtered: common targets followed by equalization make common targets commuting with the chosen arrows from $i$, and further equalization handles parallel arrows. Its projection to $I$ is final. For a vertex $j$, common targets give a vertex receiving $j$ and $i$; any two such choices have a common commuting refinement by the finite-graph filteredness test. These facts give the nonempty connected comma categories, or equivalently the unique extension of cocones, required for finality.

Lemma 2.1. As subobjects of $X_i$, $\ker(X_i\to X_\infty) =\bigcup_{a\colon i\to j}\ker X(a). \tag{2.1}$ Consequently, if $N$ is finitely generated and $g\colon N\to X_i$ becomes zero in the colimit, then $X(a)g=0$ for some arrow $a\colon i\to j$. The canonical map $\operatorname{colim}_i\operatorname{Hom}(N,X_i) \longrightarrow\operatorname{Hom}(N,X_\infty) \tag{2.2}$ is injective.

Proof. On $i/I$, the arrows $X_i\to X_j$ are a natural transformation from the constant diagram. Exact colimits preserve its kernel, and finality identifies its target colimit with $X_\infty$. The source constant colimit is $X_i$, since a filtered category is nonempty and connected. Its kernels embed in $X_i$ and are directed by refinement, proving (2.1) with the actual kernel inclusion.

The inverse images of these kernels under $g$ have union $N$, by the pullback-union argument of Lemma 1.1. Finite generation makes one all of $N$, proving stagewise vanishing. For injectivity in (2.2), represent a class by one map, using the filtered group-colimit test. If its image is zero, the preceding vanishing kills that representative at a stage. Differences of representatives may first be put at a common stage. Thus equality is witnessed in the group colimit as required. $\square$

A related consequence used below is that a map from a finitely generated object into an arbitrary coproduct factors through a finite partial sum. Those sums are subobjects: coproducts are exact because they are filtered colimits of exact finite biproducts. They have the coproduct as union by its universal property. Pull back the partial sums and use finite generation. This concerns factorization of the whole map, not just each separate component.

3. Noetherian generators also control the relations

An object $N$ is finitely presented if (2.2) is bijective for every small filtered diagram.

Theorem 3.1. In a locally noetherian Grothendieck category, finitely generated, noetherian and finitely presented objects coincide.

Proof. Lemma 1.2 gives the first equivalence. For noetherian $N$, injectivity of (2.2) is Lemma 2.1. To prove surjectivity, fix $f\colon N\to X_\infty$ and form $P_i=N\times_{X_\infty}X_i. \tag{3.1}$ They constitute a filtered diagram: its transition is identity on the $N$-coordinate and the given transition on $X_i$. Each is the kernel of $(f,-u_i)\colon N\oplus X_i\to X_\infty$. Exact filtered colimits identify $\operatorname{colim}P_i$ with the kernel of $(f,-1)$, which is $N$. In particular the directed images of $P_i\to N$ have union $N$, so some $P_i\to N$ is epic.

By Lemma 1.2 the object $P_i$ is a union of noetherian subobjects. Their images have union $N$; its finite generation supplies a noetherian $P'\subset P_i$ still mapping epically onto $N$. Let $K$ be the kernel of this epimorphism. It is noetherian and hence finitely generated. The other pullback component is $g\colon P'\to X_i$. The map $g|_K$ becomes zero in $X_\infty$, so Lemma 2.1 kills it after some $a\colon i\to j$. Thus $X(a)g$ descends through the cokernel $P'\twoheadrightarrow N$ to a map $\bar g\colon N\to X_j$. Composing to $X_\infty$ gives $f$, since that equality holds after the epimorphism from $P'$. This is the desired stage factorization. It used a finite kernel, not a splitting of that epimorphism.

Finally, if $N$ is finitely presented, apply its Hom comparison to any directed union of subobjects equal to $N$. Identity factors through one inclusion $N_t\hookrightarrow N$; that inclusion is split epic as well as monic, hence invertible. Thus $N$ is finitely generated. $\square$

4. Filtered colimits and sums of injectives

Theorem 4.1. In a locally noetherian Grothendieck category, every small filtered colimit of injective objects is injective. Every small coproduct of injective objects is injective as well.

Proof. Choose noetherian generators $(G_s)$. Let $E=\operatorname{colim}_iE_i$, with every $E_i$ injective. For any subobject $W\subset G_s$, both $W$ and $G_s$ are noetherian. By Theorem 3.1, a map $W\to E$ factors as $W\to E_i\to E$. Injectivity of $E_i$ extends its first component to $G_s\to E_i$. Its composite to $E$ is the required extension. The complete generator-family criterion from the prerequisite now proves injectivity of $E$.

A finite sum of injectives is injective: extend each projected component of a map along the given monomorphism, then combine those finitely many extensions. The coproduct of an arbitrary family is the filtered colimit of its finite partial sums, including the empty sum. Apply the first assertion. No transition monomorphism assumption was needed for that assertion. $\square$

5. The converse and its finite-generation hypothesis

Theorem 5.1. For a locally finitely generated Grothendieck category, the following conditions are equivalent: it is locally noetherian; injectives are closed under all small filtered colimits; injectives are closed under all small coproducts; injectives are closed under countable coproducts.

Proof. Theorem 4.1 gives the first two implications, and arbitrary coproduct closure gives countable closure. It remains to recover noetherian generators from countable closure.

Let $N$ be finitely generated. If it is not noetherian, a nonstabilizing ascending sequence has a strictly increasing subsequence, which we write $N_1\subsetneq N_2\subsetneq N_3\subsetneq\cdots\subset N. \tag{5.1}$ For each $t\ge1$, choose a monomorphism $e_t\colon N/N_t\hookrightarrow E_t$ into an injective, using enough injectives. On $N_n$, the maps $e_tq_t$ are zero for $t\ge n$. Define $h_n\colon N_n\to\bigoplus_{t\ge1}E_t$ by its first $n-1$ components $e_tq_t|_{N_n}$. These are actual finite-biproduct maps. Their extensions by zero to the sum agree under $N_n\subset N_{n+1}$, because the new component kills $N_n$. Consequently they induce $h\colon K=\bigcup_nN_n\longrightarrow\bigoplus_{t\ge1}E_t. \tag{5.2}$ Under the assumed countable closure the target is injective, so $h$ extends along $K\hookrightarrow N$ to $H\colon N\to\bigoplus E_t$. Lemma 2.1's finite-sum factorization consequence places $H$ inside a finite partial sum. Choose $t$ beyond its finite index set. The $t$-th coordinate of $H$ is zero, but on $N_{t+1}$ it is $e_tq_t$. Its image is the nonzero subobject $N_{t+1}/N_t$ followed by a monomorphism. This is a contradiction.

Thus every finitely generated object is noetherian. In particular the given finitely generated generating family is noetherian, proving local noetherianness. This last use of a finitely generated generating family is exactly the local finite-generation hypothesis. The argument has not discarded it for an arbitrary Grothendieck category. $\square$

For any unital, possibly noncommutative, ring $R$, the left regular module is a finitely generated generator. Its being noetherian is equivalent to the ascending chain condition on left ideals. Moreover $R\text{-}\operatorname{Mod}$ is locally noetherian exactly when $R$ is left noetherian: one direction uses this regular generator, and the other applies Lemma 1.2 to the finitely generated object $R$. Theorem 5.1 therefore gives the full left-module sum and filtered-colimit criterion. Passing to $R^{op}$ gives its right-module version with right noetherianness.

6. The canonical Ext comparison in every degree

For objects $N,M$, compute $\operatorname{Ext}^r(N,M)$ as $H^r\operatorname{Hom}(N,I_M^\bullet)$, where $M\to I_M^\bullet$ is an injective resolution starting in degree zero. The K-injective Hom test identifies this naturally with $\operatorname{Hom}_{D^+(\mathcal A)}(N[0],M[r])$: cycles are chain maps into the shift, boundaries are homotopies, and the test removes every incoming quasi-isomorphism roof. In degree zero it is ordinary Hom, and in negative degrees it is zero.

Theorem 6.1. Suppose $\mathcal A$ is locally noetherian and $N$ is noetherian. For every small filtered diagram $(M_i)$, the canonical maps are isomorphisms for all $r\ge0$: $\operatorname{colim}_i\operatorname{Ext}^r(N,M_i) \xrightarrow{\sim} \operatorname{Ext}^r\bigl(N,\operatorname{colim}_iM_i\bigr). \tag{6.1}$ They are natural in $N$, in morphisms of filtered diagrams and in their structural maps. For a filtered diagram of short exact rows they commute with the connecting maps in the second-variable long exact Ext sequence.

Proof. Use the ordinary functorial injective embedding $\eta\colon\operatorname{id}\to J$ constructed in the prerequisite. Define $C^0M=\operatorname{coker}\eta_M$, $C^{n+1}M=\operatorname{coker}\eta_{C^nM}$, and $I^0M=J(M),\qquad I^{n+1}M=J(C^nM). \tag{6.2}$ The differential is the quotient to the next cosyzygy followed by its embedding. These maps give an exact augmented complex $0\to M\to I^0M\to I^1M\to\cdots$. Naturality of $\eta$ induces maps on the cokernels and makes these strict chain maps for every arrow. Identity and composition follow from the functor and cokernel universal properties. This constructs a diagram of resolutions on an arbitrary indexing category. It requires ordinary functoriality of $J$, not additivity of $J$.

Take its degreewise filtered colimit $I_\infty^\bullet$. Exactness of colimits makes the augmented row a resolution of $M_\infty=\operatorname{colim}M_i$. Theorem 4.1 makes each of its terms injective. All resolutions start in degree zero, so this remains a bounded-below injective resolution. Theorem 3.1 gives an isomorphism of complexes $\operatorname{colim}_i\operatorname{Hom}(N,I^\bullet M_i) \xrightarrow{\sim}\operatorname{Hom}(N,I_\infty^\bullet). \tag{6.3}$ Its components are the actual maps induced by the colimit legs. Exact filtered colimits of abelian groups commute with cohomology, with the cycle-class map proved in the prerequisite. Taking cohomology of (6.3) proves (6.1). Since $M_i\to M_\infty$ is lifted by the resolution colimit leg, the K-injective comparison shows that this is precisely the map induced on Ext by each structural arrow, rather than merely an abstract group isomorphism.

Precomposition in $N$ and chain maps of diagrams commute with (6.3), proving the asserted naturality. Different resolutions give the same maps by the unique homotopy-class comparison of injective models. For short exact rows, each structural arrow is a morphism of rows. Naturality of the long exact Ext sequence gives a commuting square between its connecting map and that arrow's Ext maps. Passing to the colimit of those squares gives the desired square for (6.1), since its map is induced by those same arrows. Exactness of colimits also makes the colimit row short exact. This proves connecting compatibility without asserting that the chosen embedding functor preserves short exact rows. $\square$

In particular Ext from $N$ commutes with arbitrary coproducts: view a coproduct as the filtered colimit of its finite partial sums. Finite additivity of the injective-resolution Ext functors follows from finite injective sums and the Hom complex. Equation (6.1) then identifies the comparison with the sum of the specified inclusion maps. The restriction on $N$ is necessary even over a field, as Exercise 8.3 shows.

7. Finite pieces and the full injectivity test

Proposition 7.1. In a locally noetherian Grothendieck category, an object $E$ is injective exactly when $\operatorname{Ext}^1(N,E)=0$ for every finitely presented object $N$. The noetherian objects form an essentially small abelian category $\mathcal B$, and the natural realization and restricted Yoneda give an exact equivalence $\operatorname{Ind}(\mathcal B)\simeq\mathcal A$.

Proof. An injective has the stated vanishing by its degree-zero resolution. Conversely, choose a noetherian generator $G_s$ and $W\subset G_s$. The quotient $Q=G_s/W$ is noetherian and hence finitely presented. For an injective resolution $I_E$, injectivity in every degree makes $0\to\operatorname{Hom}(Q,I_E^\bullet) \to\operatorname{Hom}(G_s,I_E^\bullet) \to\operatorname{Hom}(W,I_E^\bullet)\to0 \tag{7.1}$ an exact row of complexes. Its long cohomology sequence shows that $\operatorname{Hom}(G_s,E)\to\operatorname{Hom}(W,E)$ is onto, since its next group is $\operatorname{Ext}^1(Q,E)=0$. The generator criterion proves injectivity. Thus the test sometimes called fp-injectivity is the full injectivity test here.

Theorem 3.1 identifies $\mathcal B$ with the finitely presented objects. Lemma 1.1 makes it closed under kernels, cokernels and finite sums in $\mathcal A$. It inherits the abelian structure with those actual maps. The complete finite-generator reconstruction theorem in the linked lesson applies: the noetherian generators are finitely presented, its small-skeleton argument gives essential smallness, and its actual filtered comma presentation realizes every object. Its full Hom comparison and natural inverse maps give the exact equivalence asserted here. $\square$

Proposition 7.2. For any essentially small abelian category $\mathcal B$, its Ind-completion is locally noetherian exactly when every object of $\mathcal B$ is noetherian.

Proof. Retain the complete abelian Ind and constant-finite-piece theorem from the prerequisite: the constants are the finitely presented objects, embed exactly, and generate. If the completion is locally noetherian, Theorem 3.1 makes each constant noetherian. A chain of subobjects in $\mathcal B$ remains a chain of subobjects in its completion, so every object of $\mathcal B$ is noetherian.

Conversely suppose every object of $\mathcal B$ is noetherian. Let $K\subset B$ be any subobject of a constant object in the completion. Present $K$ as a filtered colimit of constants $C_i$. The maps $C_i\to B$ have images $D_i$ in $\mathcal B$, because the constant embedding is exact and full. They form a directed family of subobjects of $B$. Exact filtered colimits identify their union with the image of $K\to B$, namely $K$. The noetherian subobject lattice in $\mathcal B$ makes this directed family attain its union: otherwise successive strict increases would contradict its chain condition. Hence $K=D_i$ for some $i$, as an actual subobject of $B$.

Thus every subobject of a constant is constant, and its entire subobject lattice is the one in $\mathcal B$. Every constant is consequently noetherian in the completion. The constant generating family proves local noetherianness. This argument distinguishes local noetherianness from the coherent abelian finite-piece property, which holds for all such $\mathcal B$ by the earlier theorem. $\square$

8. Four graded exercises with complete solutions

Exercise 8.1 — Two different transition maps on Hom and Ext (intermediate)

Fix a prime $p$, set $M_n=\mathbb Z/p^n$, $n\ge1$, and use the injections $M_n\to M_{n+1}$ given by multiplication by $p$. Compute the colimits of $\operatorname{Hom}(\mathbb Z/p,M_n)$ and $\operatorname{Ext}^1(\mathbb Z/p,M_n)$, including the transition maps. Compare them with the groups for the colimit module.

Solution. The map taking $\bar a\in M_n$ to $a/p^n+\mathbb Z$ identifies the diagram union with $P=\mathbb Z[1/p]/\mathbb Z$. It respects the injections, since $pa/p^{n+1}=a/p^n$, and every element of $P$ occurs in this union.

The free resolution $0\to\mathbb Z\xrightarrow p\mathbb Z\to\mathbb Z/p\to0$ gives $\operatorname{Hom}(\mathbb Z/p,M)=M[p],\qquad \operatorname{Ext}^1(\mathbb Z/p,M)=M/pM, \qquad \operatorname{Ext}^r(\mathbb Z/p,M)=0\ (r\ge2). \tag{8.1}$ For completeness, put its two free terms in degrees $-1,0$. Lifting a generator proves projectivity of $\mathbb Z$, and the bounded projective Hom test in the prerequisite computes the derived Hom by the two-term Hom complex $M\xrightarrow{-p}M$ in degrees $0,1$, with the normalized cohomological Hom sign. Its kernel and cokernel are the ones in (8.1), and its induced maps are those of $M$, proving the formula including its natural maps. Integer ideals are principal by the least-positive-element division argument; an ascending chain after its first nonzero member has generators among the finitely many positive divisors of that member's generator and stabilizes. Thus $\mathbb Z$ is noetherian, and the finite object $\mathbb Z/p$ meets the source hypothesis of (6.1).

In $M_n[p]$, write elements as $p^{n-1}a$, with $a$ taken modulo $p$. The injection sends this to $p^na$, so these Hom transitions are identity in the indicated coordinates. Their colimit is $\mathbb Z/p$. On $M_n/pM_n$, the same injection sends $a$ to $pa$, which is zero modulo $pM_{n+1}$. These Ext transitions are zero, so their colimit is zero, even though each stage Ext group is $\mathbb Z/p$.

Multiplication by $p$ is surjective on $P$, and $P[p]$ consists of the classes $a/p$, hence is $\mathbb Z/p$. Formula (8.1) therefore gives Hom equal to $\mathbb Z/p$ and Ext equal to zero at the limit. The Hom comparison sends the coordinate $a$ to $a/p+\mathbb Z$; the Ext comparison is the unique map between zero groups. These are exactly (6.1). $\square$

Exercise 8.2 — A countable sum of one coinduced injective fails (advanced)

Let $V$ have countable basis $(v_n)_{n\ge1}$ over a field $k$, and let $R=k\oplus V$ with $V^2=0$: multiplication is $(a,v)(b,w)=(ab,aw+bv)$. Prove that $E=\operatorname{Hom}_k(R,k)$, with $(r\phi)(s)=\phi(sr)$, is an injective left module. Give an explicit map from the ideal $V$ into $E^{(\mathbb N)}$ which does not extend to $R$.

Solution. For any left module $M$, evaluation at $1$ gives a natural bijection $\operatorname{Hom}_R(M,E)\simeq\operatorname{Hom}_k(M,k). \tag{8.2}$ Its inverse sends $g$ to $m\mapsto(s\mapsto g(sm))$. It is left linear because its value at $rm$ on $s$ is $g(srm)$, the value of $r$ acting on the previous functional. Evaluation and this construction are inverse: for an $R$-linear $f$, $f(sm)(1)=(sf(m))(1)=f(m)(s)$. Every linear functional on a subspace extends after extending a basis. Thus (8.2) gives the full extension property for $E$, proving injectivity without a dimension restriction on $M$.

Let $\epsilon\in E$ be scalar projection $(a,v)\mapsto a$. Then $r\epsilon=a\epsilon$. Define $h(v_n)$ to be $\epsilon$ in the $n$-th coordinate and zero in the others, and extend linearly. Each vector of $V$ has finite basis support, so its image belongs to the coproduct. The action on both that image and $V$ is multiplication by the scalar part; hence $h$ is $R$-linear.

An extension $f\colon R\to E^{(\mathbb N)}$ would be determined by $z=f(1)$, a tuple with finite coordinate support, and would satisfy $f(r)=rz$. Choose $n$ outside that support. The $n$-th coordinate of $f(v_n)$ is zero, whereas that of $h(v_n)$ is the nonzero functional $\epsilon$. No extension exists. The sum is not injective, although every summand is the same injective $E$. The ideals $\langle v_1,\ldots,v_n\rangle$ are strictly increasing, so the cyclic regular generator is not noetherian. This explicitly realizes the failure detected by Theorem 5.1. $\square$

Exercise 8.3 — The source finiteness cannot be dropped (introductory)

Over a field $k$, let $N=k^{(\mathbb N)}$, and let $M_n\subset N$ be the span of the first $n$ basis vectors. Identify the image of the filtered Hom comparison, and explain why local noetherianness alone does not make that map surjective for this source.

Solution. The union of the $M_n$ is $N$. A map $N\to M_n$ assigns an arbitrary vector of $M_n$ to each basis vector of $N$; each input is a finite sum, so this always defines a linear map. The comparison sends it to its composite with the inclusion into $N$. Its image therefore consists precisely of linear maps whose whole image lies in one of the fixed finite coordinate subspaces $M_n$. Two stage maps with equal composites agree after moving to a common stage, because these inclusions are monic. The comparison is injective.

Identity on $N$ has no such finite coordinate bound and is not in its image. The comparison is not onto. The one-dimensional object $k$ is a noetherian generator of all vector spaces, but the ascending coordinate spans show that $N$ is not noetherian. Every positive Ext group here is zero because all vector spaces are injective by basis extension. The failure already occurs in degree zero and has no derived-resolution explanation. $\square$

Exercise 8.4 — Colimit closure does not bound homological dimension (advanced)

Let $A=k[\varepsilon]/(\varepsilon^2)$ and let $S=A/(\varepsilon)$. Prove that sums and filtered colimits of injectives are injective, but $\operatorname{Ext}^r_A(S,S)\ne0$ in every degree $r\ge0$. Also verify directly that $A$ itself is injective.

Solution. Every ascending chain of ideals is a chain of subspaces in the two-dimensional $k$-space $A$, so stabilizes. Thus the regular left module is noetherian. The full module specialization of Theorem 5.1 gives both colimit assertions.

Let $\lambda(a+b\varepsilon)=b$. The map $A\to\operatorname{Hom}_k(A,k)$ sending $x$ to $(y\mapsto\lambda(yx))$ is left linear for the coinduced action. Its pairing matrix in the basis $1,\varepsilon$ is $\begin{pmatrix}0&1\\1&0\end{pmatrix}$, invertible over every field, so the map is an isomorphism. The complete coinduction extension proof (8.2), with $R=A$, makes its target injective. Hence $A$ is injective as well as projective.

The augmented free resolution of $S$ has a copy of $A$ in every nonpositive degree, differential multiplication by $\varepsilon$ in every negative degree, and quotient $A\to S$ in degree zero. It is exact because both the kernel and image of multiplication by $\varepsilon$ are $k\varepsilon$; the augmentation kernel is the same ideal. Free rank-one modules are projective by lifting their chosen generator. The bounded-above projective Hom test applies to this supplied complex. Applying Hom into $S$ gives $k$ in every nonnegative degree and zero differentials, since $\varepsilon$ acts as zero on $S$. Thus $\operatorname{Ext}^r_A(S,S)\simeq k\quad(r\ge0). \tag{8.3}$ Each identification uses the free-generator coordinate of that resolution. In particular the positive groups never vanish, so $S$ has no finite injective dimension: a finite injective resolution would make all sufficiently high groups vanish. The filtered-colimit theorem and its Ext continuity impose no such dimension bound. The computation uses one explicit bounded-above projective resolution and asserts no new unbounded injective construction. $\square$

9. Freely accessible further reading

Henning Krause's freely accessible Homological Theory of Representations, Section 11.2, Lemmas 11.2.2–11.2.5 and Theorem 11.2.12, gives the finite-generation comparisons and the locally finitely generated injective-closure criterion. The argument above uses noetherian subobjects of actual pullbacks, the generator-family extension test, and explicit compatible maps into countable sums; it supplies the natural Ext and Ind interfaces in the stated scope.

The Stacks Project, Injective modules, Lemma 47.3.7, proves the noetherian commutative-ring sum assertion by the ideal extension test. The proofs here retain arbitrary left rings, the categorical generating family and general filtered diagrams.