Subsequent full independent model review: reviewer model not specified, Ultra; correction closed through exact-version replay. This review covers the recorded version of this lesson and its specified prerequisite interfaces.

Ideal-generated modules and nonstrict quotients

A full subcategory of modules can keep all colimits and exact filtered colimits while changing individual kernels. The change comes from its generator: only the part of an ambient kernel generated by that object remains visible inside the subcategory. This gives a concrete setting where a detecting family is closed under finite coproducts but is not dense, because an epimorphism fails to be strict.

Fix any field $k$, put $A=k[x,y]$, and let $\mathfrak a=(x,y)$. All modules and indexing sets belong to a fixed universe. Let $\mathsf C_0$ be the full subcategory of $A$-modules $X$ admitting a surjection $\mathfrak a^{(I)}\to X$ for some small set $I$. Such modules are called $\mathfrak a$-generated. Let $\mathsf F$ consist of the finite sums $\mathfrak a^{\oplus n}$, including $n=0$, and let $\mathsf G$ consist of the finite free modules $A^n$.

Use Formal colimits and compact presentations for finite presentation, Coimages, images and composition of quotients for intrinsic strictness and ordinary module constructions, and Dense probes and reconstruction from colimits for density and representability. All constructions here concern ordinary modules.

1. One presentation and its matrix category

The ideal has the finite presentation

\begin{gathered} A\xrightarrow{\,d\,}A^2 \xrightarrow{\,\pi\,}\mathfrak a\longrightarrow0,\\ d(t)=(-yt,xt),\\ \pi(u,v)=xu+yv. \end{gathered} \tag{1.1}

Indeed, if $xu+yv=0$, reduction modulo $x$ in the integral domain $k[y]$ forces $v=xt$ for some $t\in A$. Cancellation of $x$ then gives $u=-yt$. The reverse inclusion in the kernel is immediate. This proves exactness and finite presentation without a Noetherian argument.

An $A$-linear map $f:\mathfrak a\to A$ is determined by $p=f(x)$ and $q=f(y)$, satisfying $yp=xq$. Reduction modulo $x$ gives $p=xt$; cancellation then gives $q=yt$. Thus $f$ is multiplication by the unique polynomial $t$. Multiplication by every polynomial also preserves $\mathfrak a$, so

\begin{gathered} \operatorname{Hom}_A(\mathfrak a,A)=A,\\ \operatorname{End}_A(\mathfrak a)=A. \end{gathered} \tag{1.2}

For finite sums, component inclusions and projections identify morphisms $\mathfrak a^{\oplus m}\to\mathfrak a^{\oplus n}$ with $n$-by-$m$ matrices over $A$. Composition is matrix multiplication with the same order of factors as for $A^m\to A^n$. Sending $\mathfrak a^{\oplus n}$ to $A^n$ and retaining the matrix is therefore an equivalence

\mathsf F\simeq\mathsf G. \tag{1.3}

Zero rows and columns describe the zero object and all maps involving it.

2. The nerve factors through a fully faithful module test

Give $\xi(X)=\operatorname{Hom}_A(\mathfrak a,X)$ its $A$-module structure by scalar multiplication on the target. Commutativity of $A$ makes this the same as precomposition by the endomorphisms (1.2). For an $A$-module $M$, define the presheaf

\begin{gathered} \eta(M)(Y)\\ =\operatorname{Hom}_A(Y,\mathfrak a)\otimes_A M \quad(Y\in\mathsf F). \end{gathered} \tag{2.1}

An arrow of $\mathsf F$ acts by precomposition in the first tensor factor. At $Y=\mathfrak a^{\oplus n}$, this is $M^n$, with the usual matrix action by precomposition.

There is a natural map

\begin{gathered} \eta\xi(X)(Y)\longrightarrow\operatorname{Hom}_A(Y,X),\\ p\otimes h\longmapsto h p. \end{gathered} \tag{2.2}

It is balanced because $h(ap)=a(hp)=(ah)p$. If $Y=\mathfrak a^{\oplus n}$ and $\rho_j:Y\to\mathfrak a$ are its projections, the inverse sends $g:Y\to X$ to $\sum_j\rho_j\otimes g\iota_j$, where $\iota_j$ are the inclusions. The identities $g=\sum_jg\iota_j\rho_j$ and $p=\sum_j(p\iota_j)\rho_j$ verify both inverse compositions. For $n=0$, both modules are zero. Composition in either variable proves naturality, so the nerve $\Phi$ has the factorization

\Phi(X)=\operatorname{Hom}_A(-,X)|_{\mathsf F} \simeq\eta\xi(X). \tag{2.3}

Under (1.3), $\eta(M)$ identifies with $\operatorname{Hom}_A(-,M)|_{\mathsf G}$: both have value $M^n$, and the same matrices act on those values. The single rank-two free-probe theorem from the density prerequisite, followed by enlargement to $\mathsf G$, proves that this functor is fully faithful. Hence $\eta$ is fully faithful on all $A$-modules. This uses the already proved stronger probe result with the finite-sum identification, including the zero object.

The composite $\Phi$ need not be full. Write $\overline k=A/\mathfrak a$, with its residue-field module structure. The assignment $x\mapsto1,\ y\mapsto0$ defines a surjection $\mathfrak a\to\overline k$, because the relation in (1.1) becomes zero. Thus $\overline k\in\mathsf C_0$, and

\xi(\overline k)=\overline k^2. \tag{2.4}

Indeed, the two generator images are arbitrary because $x$ and $y$ act by zero. Endomorphisms of $\overline k$ are scalar multiplications, and $\xi$ sends the scalar $c$ to $cI_2$. Projection onto the first summand of $\overline k^2$ is $A$-linear and is not scalar. Through the fully faithful $\eta$, it gives an endomorphism of $\Phi(\overline k)$ that is not $\Phi(f)$ for any $f:\overline k\to\overline k$. This proves that the finite ideal-sum nerve is not full.

3. The trace coreflection and filtered exactness

For any $A$-module $X$, define

T(X)=\sum_{h:\mathfrak a\to X}\operatorname{im}h \ \subset X. \tag{3.1}

The Hom set is small, so the sum is the image of a small direct sum of copies of $\mathfrak a$. Thus $T(X)$ is $\mathfrak a$-generated. A module is in $\mathsf C_0$ exactly when $T(X)=X$. Every map from an $\mathfrak a$-generated module $Z$ to $X$ lands in $T(X)$, by composing with its generating maps. Therefore inclusion $I:\mathsf C_0\to\mathsf{Mod}(A)$ has right adjoint $T$, with the inclusion $T(X)\hookrightarrow X$ as counit. This property determines its action on arrows and its idempotence.

Small coproducts and quotient modules of $\mathfrak a$-generated modules remain $\mathfrak a$-generated. Ambient colimits are quotients of coproducts, so $I$ creates all small colimits, including the empty one. For a diagram $X_j$ in $\mathsf C_0$, its limit is

\lim_{\mathsf C_0}X_j =T\!\left(\lim_{\mathsf{Mod}(A)}I(X_j)\right). \tag{3.2}

To verify the maps as well as the object, map any $Z\in\mathsf C_0$ into the right side. Coreflection and the ambient limit property identify these maps with compatible maps $Z\to X_j$, uniquely. In particular, finite products are ordinary direct sums, already $\mathfrak a$-generated, whereas a kernel is the trace of its ambient kernel.

Proposition 3.1. The trace functor $T$ commutes with every small filtered colimit of ambient $A$-modules.

Proof. Finite presentation (1.1) and the finite-presentation/Hom criterion retained in the prerequisite give

\begin{gathered} \mathop{\rm colim}_i\operatorname{Hom}_A(\mathfrak a,X_i)\\ \simeq\operatorname{Hom}_A \!\left(\mathfrak a,\mathop{\rm colim}_iX_i\right). \end{gathered} \tag{3.3}

Thus every map from $\mathfrak a$ to the colimit lifts to one stage. An element of its trace is a finite sum of values of such maps. Move the finitely many lifted values to a common stage; their sum belongs to that stage's trace. The comparison $\operatorname{colim}_iT(X_i)\to T(\operatorname{colim}_iX_i)$ is consequently surjective.

The compatible inclusions $T(X_i)\hookrightarrow X_i$ remain injective after filtered colimit. This is the module exactness interface of Stacks, Lemma 10.8.8. Its directed-set formulation applies to any small filtered category by the retained cofinal directed-poset replacement Stacks, Lemma 4.21.5. Hence the trace comparison is also injective. Its map into the ambient colimit is precisely the canonical inclusion, proving the asserted natural isomorphism. $\square$

Ambient filtered colimits commute with finite products by simultaneous representatives at one common stage, and with kernels by the same exactness interface. They therefore commute with finite limits. Equations (3.2)–(3.3), together with creation of colimits, show that filtered colimits in $\mathsf C_0$ commute with finite limits as well. The empty finite diagram causes no exception: its terminal object is zero, and a filtered category is nonempty.

In particular filtered colimits are stable under base change in $\mathsf C_0$. Apply finite-limit commutation to the pullback diagrams over a fixed base. The colimit of each constant base object is that object, since the filtered index is nonempty and connected.

4. Detection survives, but a quotient loses strictness

Proposition 4.1. The object $\mathfrak a$ detects isomorphisms in $\mathsf C_0$.

Proof. Suppose $f:X\to Y$ induces a bijection on $\operatorname{Hom}_A(\mathfrak a,-)$. Surjectivity on this Hom means that every generating map $\mathfrak a\to Y$ lifts to $X$. Their images generate $Y$, so $f$ is surjective as a module map.

Let $K$ be its ambient kernel. Injectivity on Hom makes $\operatorname{Hom}_A(\mathfrak a,K)=0$. If $0\ne m\in K$ and $\mathfrak a m\ne0$, multiplication $a\mapsto am$ is a nonzero map $\mathfrak a\to K$. If instead $\mathfrak a m=0$, the map $\overline k\to K$ taking $1$ to $m$ is injective: a nonzero scalar of $k$ cannot annihilate $m$. Composing it with the surjection $\mathfrak a\to\overline k$ from Section 2 again gives a nonzero map. Both cases contradict vanishing of Hom. Thus $K=0$, and $f$ is invertible. $\square$

Epimorphisms in $\mathsf C_0$ are exactly ambient surjections. Surjections are epimorphisms in every full subcategory. Conversely, if $f:X\to Y$ is not surjective, its nonzero ambient cokernel $Q$ is a quotient of the generated module $Y$, hence belongs to $\mathsf C_0$. The nonzero quotient $Y\to Q$ and the zero map agree after $f$, disproving epimorphy.

Consider the surjection

\begin{gathered} q:\mathfrak a\oplus\mathfrak a\longrightarrow\mathfrak a^2,\\ q(u,v)=xu+yv. \end{gathered} \tag{4.1}

Here the target $\mathfrak a^2$ denotes the square ideal, rather than a twofold direct sum. It is generated by $x^2,xy,y^2$ and is a quotient of the source, hence an object of $\mathsf C_0$. The ambient kernel is

K=\{(yt,-xt):t\in A\}\simeq A. \tag{4.2}

The kernel calculation is the divisibility argument of (1.1), with the signs reversed. Since (1.2) says that every map $\mathfrak a\to A$ is multiplication, its trace is $T(A)=\mathfrak a$. Accordingly, the intrinsic kernel in $\mathsf C_0$ is

T(K)=\{(yt,-xt):t\in\mathfrak a\} \ \subsetneq K. \tag{4.3}

Write $X=\mathfrak a\oplus\mathfrak a$. The ambient kernel pair of $q$ is $X\oplus K$, via $(u,k)\mapsto(u,u+k)$. Trace commutes with finite direct sums, because maps into a direct sum are precisely pairs of maps. Its intrinsic kernel pair is therefore $X\oplus T(K)$, with projections $(u,k)\mapsto u$ and $u+k$.

Since colimits are created in modules, the coequalizer of these projections is $X/T(K)$. Hence

\begin{gathered} \operatorname{Coim}_{\mathsf C_0}(q)=X/T(K),\\ \ker_{\mathsf{Mod}(A)} \bigl(X/T(K)\to\mathfrak a^2\bigr)\\ =K/T(K)\simeq\overline k\ne0. \end{gathered} \tag{4.4}

The canonical coimage-to-target map is not an isomorphism, so this epimorphism is not strict. An explicit failure of descent is the quotient $X\to X/T(K)$: it equalizes the intrinsic kernel pair, but it cannot factor through $q$, because it does not kill the larger ambient kernel $K$.

Thus $(\mathsf C_0,\mathsf F)$ has small colimits, finite limits, filtered base-change stability and a finite-coproduct-closed detecting family, yet its nerve is not full. The failed hypothesis of the strict-quotient density criterion is exactly the requirement that every epimorphism be strict.

5. Representability still holds

Corollary 5.1. Every functor $H:\mathsf C_0^{\rm op}\to\mathsf{Set}$ preserving all small limits in its domain is representable.

Proof. In terms of $\mathsf C_0$, the condition means that $H$ sends small colimits to limits of sets. The category is locally small, has all small colimits and finite limits, and has the detecting object $\mathfrak a$. Its filtered colimits are stable under base change by Section 3. The detecting-object representability criterion in the density prerequisite therefore applies directly. That criterion constructs a small finite-colimit-closed family of probes before reconstructing $H$; it does not require that the particular family $\mathsf F$ already be dense. $\square$

There is also a checked solution-set route. Quotients by epimorphisms in $\mathsf C_0$ are ambient module quotients, controlled up to isomorphism by submodules of the fixed source. Since submodules form a small set, $\mathsf C_0$ is co-wellpowered. The cocomplete, co-wellpowered, detecting-object criterion in Solution sets and universal representations gives the same result. Nonstrictness of (4.1) prevents neither route.

6. Graded exercises with complete solutions

Exercise 1 — Foundation: the detecting object depends on the category. Prove that $\mathfrak a\hookrightarrow A$ induces an isomorphism on $\operatorname{Hom}_A(\mathfrak a,-)$. Show that $A$ is not $\mathfrak a$-generated. Explain why this does not contradict Proposition 4.1.

Solution. Both Hom modules identify with $A$ by (1.2). Inclusion sends multiplication by $t$ on $\mathfrak a$ to that same multiplication as a map into $A$, so the induced map is the identity in these identifications. But the trace of $A$ is the sum of the ideals $t\mathfrak a$, equal to $\mathfrak a$, not $A$. Thus $A\notin\mathsf C_0$. The inclusion is not surjective, so $\mathfrak a$ fails to detect isomorphisms in the entire ambient module category. Proposition 4.1 requires both source and target to lie in $\mathsf C_0$.

Exercise 2 — Intermediate: all modules killed by the ideal. Let $M$ be an arbitrary $k$-vector space, regarded as an $A$-module annihilated by $\mathfrak a$. Prove $M\in\mathsf C_0$, and identify $\xi(M)$ and the map $\operatorname{Hom}_A(M,N) \longrightarrow\operatorname{Hom}_A(\xi(M),\xi(N))$ for two such modules. Give a non-induced natural transformation when $M=N\ne0$.

Solution. For each $m\in M$, the map $\mathfrak a\to M$ sending $x$ to $m$ and $y$ to zero respects the single relation. Their images generate $M$, so the evaluation from their small direct sum is surjective. Relation (1.1) vanishes on $M$, giving $\xi(M)=M\oplus M$ naturally. Ambient $A$-linear maps between these annihilated modules are exactly $k$-linear maps. A map $f:M\to N$ is carried to $\operatorname{diag}(f,f)$. All four blocks of a map $M\oplus M\to N\oplus N$ can otherwise be arbitrary $k$-linear maps. For nonzero $M=N$, projection onto the first factor is not of this diagonal form. Full faithfulness of $\eta$ turns it into a natural transformation of the finite ideal-sum nerve that is not induced by a map of the modules. If $M=0$, all relevant maps are zero and this particular obstruction disappears.

Exercise 3 — Advanced: a family of strict and nonstrict ideal quotients. Write $\mathfrak a^0=A$. For $m\ge1$, let $\begin{gathered} q_m:\mathfrak a^m\oplus\mathfrak a^m \longrightarrow\mathfrak a^{m+1}, \\ (u,v)\longmapsto xu+yv. \end{gathered}$ Prove these objects belong to $\mathsf C_0$, calculate the ambient kernel, and prove that $q_m$ is strict exactly when $m\ge2$.

Solution. For $j\ge1$, the degree-$(j-1)$ monomials give maps $\mathfrak a\to\mathfrak a^j$ by multiplication. Their images generate every monomial of degree at least $j$; these are precisely the polynomials in $\mathfrak a^j$. Thus $\mathfrak a^j\in\mathsf C_0$. The equality $x\mathfrak a^m+y\mathfrak a^m=\mathfrak a^{m+1}$ proves surjectivity.

As in (1.1), its ambient kernel has pairs $(yt,-xt)$. Both entries lie in $\mathfrak a^m$ exactly when $t\in\mathfrak a^{m-1}$: multiplication by a single variable raises the degree of each nonzero homogeneous term by one, so a term of degree below $m-1$ cannot meet the condition. Hence the kernel is isomorphic to $\mathfrak a^{m-1}$. For $m=1$, this is $A$, whose trace is the proper ideal $\mathfrak a$, and Section 4 proves nonstrictness. For $m\ge2$, the kernel is itself $\mathfrak a$-generated. Its trace equals the full kernel, so the intrinsic kernel pair agrees with the ambient one and its coequalizer is the ordinary quotient by that kernel, namely $\mathfrak a^{m+1}$. The canonical coimage map is an isomorphism, proving strictness.

Exercise 4 — Expert: trace and strictness for an arbitrary finitely presented probe. Let $R$ be any unital ring, possibly noncommutative, and $P$ a finitely presented left $R$-module. Let $\mathsf C_P$ consist of modules generated by small sums of copies of $P$, and define $T_P$ by the sum of all images of $P$. Prove that inclusion has right adjoint $T_P$, creates all colimits, and that filtered colimits commute with finite limits in $\mathsf C_P$. For any ambient surjection $f:X\to Y$ between its objects, prove $\begin{gathered} f\text{ is strict in }\mathsf C_P\\ \Longleftrightarrow\quad \ker_{\mathsf{Mod}(R)}f\in\mathsf C_P. \end{gathered} \tag{6.1}$ Does this alone say that $P$ detects isomorphisms?

Solution. The Hom set from $P$ is small, so $T_P(X)$ is the image of a small sum of $P$'s. Every map from a $P$-generated module into $X$ lands there. This proves the adjunction, its idempotence and the description of limits as $T_P$ of ambient limits. Coproducts and quotients remain generated, proving creation of all small colimits.

The arbitrary-ring finite-presentation theorem in the prerequisite makes every map $P\to\operatorname{colim}X_i$ lift to a filtered stage. Every element of the trace is a finite sum of such values, which has one common stage. This proves surjectivity of the trace comparison. For injectivity, an element represented in $T_P(X_i)$ that becomes zero in the ambient colimit becomes zero at a later stage, by the filtered module element rule; its image there is still in the trace. Thus the comparison is an isomorphism, without commuting ring coefficients. Ambient filtered colimits commute with kernels by the same representative-and-vanishing rule, and with finite products by common-stage representatives. Combining these facts with the trace limit description proves finite-limit commutation in $\mathsf C_P$, including the terminal zero object.

Put $K=\ker f$. The ambient kernel pair is $X\oplus K$. Trace commutes with finite direct sums by component inclusions and projections, so the intrinsic kernel pair is $X\oplus T_P(K)$, and its coequalizer is $X/T_P(K)$. The canonical map to $Y=X/K$ is invertible exactly when $T_P(K)=K$, which is precisely (6.1). Surjectivity makes $f$ an epimorphism, so this is the strict-epimorphism criterion as well.

Detection need not hold. Let $R$ be the ring of lower triangular two-by-two matrices over $k$, with diagonal idempotents $e_1,e_2$, and put $P=Re_1$. The decomposition $R=Re_1\oplus Re_2$ makes $P$ finitely presented; explicitly, the kernel of $r\mapsto re_1$ is the left ideal generated by $e_2$, giving a presentation with one generator and one relation.

Let $Q=k$, on which a matrix acts by its upper-left entry. The map $f:P\to Q$ sending a first-column matrix to its upper-left entry is a surjective module map. Thus $P,Q\in\mathsf C_P$, but $f$ is not an isomorphism: its kernel contains the nonzero lower-left matrix unit. For every left module $X$, evaluation at $e_1$ identifies $\operatorname{Hom}_R(Re_1,X)$ with $e_1X$; its inverse takes $x\in e_1X$ to $re_1\mapsto rx$. Here $e_1P$ and $e_1Q$ are both $k$, and the induced map is the identity. Consequently $P$ fails to detect this noninvertible $f$. This distinguishes the general trace result from the additional detection argument for $\mathfrak a$.

Self-checked; no independent review has occurred. Original exposition and exercise text are dedicated to the public domain under CC0 1.0.