Subobject Lattices Generators and the Grothendieck Axioms — Examples

For each divisor $d$ of $12$, the cyclic group $C_{12}$ has a unique subgroup of order $d$, namely $⟨12\mathrm{/}d⟩$. So the subgroup lattice is the divisor lattice of $12$ with elements of orders $1,2,3,4,6,12$. The meet is intersection, hence gcd of orders, and the join is subgroup sum, hence lcm of orders.

The divisor lattice of a single integer is distributive, so this example is more rigid than the modular-only situation of [L2]. It therefore illustrates that modularity does not force every concrete subobject lattice to look like the $M_3$ example.

Over ${\mathbf{F}}_2$, the nonzero vectors of $V$ are $(1,0)$, $(0,1)$, and $(1,1)$, and each spans a distinct one-dimensional subspace. Any two distinct lines meet only in $0$, and because they are not equal each pair spans all of $V$. So the subspace lattice has exactly the five elements $0$, the three lines, and $V$.

This is the same five-element diamond described in [L2]. Hence the subobject lattice of a very familiar module can already be modular without being distributive.

For the subgroup $2\mathbb{Z}\le \mathbb{Z}$, the direct image is $f_{\ast }(2\mathbb{Z})=\{0,2,4\}\le \mathbb{Z}\mathrm{/}6$. For the subgroup $C=\{0,2,4\}\le \mathbb{Z}\mathrm{/}6$, the inverse image is $f^{\ast }C=2\mathbb{Z}$. Also [L3] identifies $f^{\ast }(0)=6\mathbb{Z}=\ker (f)$ and $f_{\ast }(\mathbb{Z})=\mathbb{Z}\mathrm{/}6=\mathrm{im}(f)$.

Thus $f_{\ast }(2\mathbb{Z})\le C$ and $2\mathbb{Z}\le f^{\ast }C$ are the same true statement in this example, exactly as [L1] predicts. Likewise [L2] is visible here: the meet of $\{0,2,4\}$ with $0$ pulls back to $2\mathbb{Z}\cap 6\mathbb{Z}=6\mathbb{Z}$, and the join of $2\mathbb{Z}$ with $6\mathbb{Z}$ pushes forward to $\{0,2,4\}$.

The subgroup $⟨6⟩$ has order $2$ and $⟨2⟩\mathrm{/}⟨6⟩$ has order $3$, while $(\mathbb{Z}\mathrm{/}12)\mathrm{/}⟨2⟩$ has order $2$. Likewise $⟨4⟩$ has order $3$, $⟨2⟩\mathrm{/}⟨4⟩$ has order $2$, and the top quotient is again order $2$. So both displayed chains are composition series.

The two series already display the same three simple factors up to permutation, which is the conclusion predicted abstractly by [L2] and [L3]. This is the concrete module calculation that the categorical refinement theorems package.

Every module homomorphism $u:R\to M$ is determined by $u(1)$, and every element $m\in M$ defines a homomorphism $u_m(r)=rm$. Therefore the canonical map ${\coprod }_{m\in M}R\to M$ that sends the $m$-indexed basis vector to $m$ is surjective.

By [L1], this surjectivity is exactly the generator property for $R$, and [L2] records the same conclusion abstractly at the category level.

The category $\mathbf{F}\mathbf{i}\mathbf{n}\mathbf{A}\mathbf{b}$ is abelian: kernels, cokernels, and finite biproducts of homomorphisms of finite abelian groups are again finite abelian groups. Let $P$ be a nonzero finite abelian group, and fix a prime $p$ for which $P$ has a nonzero $p$-primary quotient. Among all cyclic quotients of $P$ of the form $\mathbb{Z}\mathrm{/}{p}^m$, choose one with maximal $m$, say $u:P\twoheadrightarrow \mathbb{Z}\mathrm{/}{p}^m$.

Let $q:\mathbb{Z}\mathrm{/}{p}^{m+1}\twoheadrightarrow \mathbb{Z}\mathrm{/}{p}^m$ be the canonical quotient map. If $P$ were projective, [L1] would lift $u$ to $g:P\to \mathbb{Z}\mathrm{/}{p}^{m+1}$ with $qg=u$. Since $u$ is surjective, so is $g$. Thus $\mathbb{Z}\mathrm{/}{p}^{m+1}$ would be a quotient of $P$, contradicting maximality of $m$. Therefore no nonzero object of $\mathbf{F}\mathbf{i}\mathbf{n}\mathbf{A}\mathbf{b}$ is projective. So $\mathbf{F}\mathbf{i}\mathbf{n}\mathbf{A}\mathbf{b}$ is an abelian category with no nonzero projective object.

The subgroups $B_n$ are directed by inclusion and their union is all of $A$, since every element of $A$ has finite support. Also $B_n\cap C=⟨e_1+e_2,e_3,\dots ,e_n⟩$ for $n\ge 2$, because those are exactly the generators of $C$ lying in the first $n$ coordinates.

Taking the union of the intersections from step 1.1 recovers all of $C$, so ${\bigvee }_n(B_n\wedge C)=C=({\bigvee }_n{B}_n)\wedge C\mathrm{.}$ This is precisely the AB5 identity [L1], exactly as the abstract theorem [L2] predicts for module categories.

The chain $0<⟨6⟩<⟨2⟩<\mathbb{Z}\mathrm{/}12$ is a composition series of $A$, so $A$ has finite length and $\mathrm{\ell }(A)=3$ by [L1] and [L2]. The subgroup $B=⟨2⟩\cong \mathbb{Z}\mathrm{/}6$ has composition series $0<⟨6⟩<⟨2⟩$, so $\mathrm{\ell }(B)=2$, while $A\mathrm{/}B\cong \mathbb{Z}\mathrm{/}2$ has length $1$.

The numerical identity $\mathrm{\ell }(A)=\mathrm{\ell }(B)+\mathrm{\ell }(A\mathrm{/}B)$ becomes $3=2+1$ in this case, exactly as [L3] asserts.