Free Modules and Exact Sequences: Examples and Counterexamples

The displayed formula is additive and $R$-linear by the module axioms, and it sends $e_i$ to $m_i$.

If $S:R^3\to M$ is another homomorphism with $S(e_i)=m_i$, then [F1] gives $(r_1,r_2,r_3)=r_1{e}_1+r_2{e}_2+r_3{e}_3$, so linearity gives $S(r_1,r_2,r_3)=r_1{m}_1+r_2{m}_2+r_3{m}_3=T(r_1,r_2,r_3)$.

Therefore $T$ is the unique universal extension asserted by [L1].

Every class $k+n\mathbb{Z}$ equals $k(1+n\mathbb{Z})$, so $M$ is generated by one element according to [F2].

Every element of $M$ is killed by the nonzero integer $n$.

If $M$ were free on $X$, it would be nonzero because $n>1$, so [F1] would force $X\mathrm{\ne }\mathrm{\varnothing }$; choose $x\in X$. The standard vector $e_x$ is not killed by $n$, since its $x$-coordinate is the nonzero integer $n$, contradicting step 1.2.

Thus $M$ is generated but not free. At the excluded boundary $n=1$, $\mathbb{Z}\mathrm{/}\mathbb{Z}=0$, which is free on the empty basis by [F1].

Define ${\alpha }_0(e_k)=e_{2k}$ and ${\alpha }_1(e_k)=e_{2k+1}$. Define ${\beta }_0(e_{2k})=e_k$, ${\beta }_0(e_{2k+1})=0$, ${\beta }_1(e_{2k})=0$, and ${\beta }_1(e_{2k+1})=e_k$. Extending the formulas across the unique finite basis expressions from [F5] gives four endomorphisms.

On every basis vector, ${\beta }_i{\alpha }_j$ is the identity when $i=j$ and zero otherwise, while ${\alpha }_0{\beta }_0+{\alpha }_1{\beta }_1={\mathrm{id}}_V$. Unique finite basis expressions from [F5] make these identities hold on all of $V$.

The operations from [F2] and [F3] satisfy the ring axioms [F4], so $R={\mathrm{End}}_F(V)$ is a unital ring. Define left $R$-linear maps $T:R\to R^2$ and $S:R^2\to R$ by $T(f)=(f{\alpha }_0,f{\alpha }_1),S(g,h)=g{\beta }_0+h{\beta }_1\mathrm{.}$ Associativity and distributivity give left linearity.

The identities in step 2.1 give $T(S(g,h))=(g,h)$ and $S(T(f))=f$, so $T$ and $S$ are inverse isomorphisms.

Therefore ${}_{R}R\cong {}_R{R}^2$, refuting finite rank invariance in this ring as [F1] records.

The element $3$ is idempotent modulo $6$, and $1-3=4$ is also idempotent, with $3\cdot 4=0$ and $3+4=1$. Hence $R=3R\oplus 4R$.

The module $P$ has two elements. A free module on the empty set has one element; a free module on a nonempty finite set of size $k$ has $6^k$ elements; and a free module on an infinite set has infinitely many distinct standard basis vectors. None has two elements.

Thus $P=3R$ is a direct summand of the rank-one free module $R$, which is projective without AC by [L2]. Precomposing a map from $P$ with the projection $R\to P$, lifting from $R$, and restricting the lift to $P$ proves directly that $P$ is projective. The map $\mathbb{Z}\mathrm{/}2\mathbb{Z}\to P$ sending $1$ to $3$ is an $R$-module isomorphism.

Hence $P\cong \mathbb{Z}\mathrm{/}2\mathbb{Z}$ is projective and nonfree over $\mathbb{Z}\mathrm{/}6\mathbb{Z}$.

Multiplication by two is injective, $q$ is surjective, and $\ker q=2\mathbb{Z}$, the image of multiplication by two; hence the sequence is short exact.

If $s:\mathbb{Z}\mathrm{/}2\mathbb{Z}\to \mathbb{Z}$ were a section, put $x=s(1+2\mathbb{Z})$. Then $q(x)=1+2\mathbb{Z}$, so $x$ is odd, while $2x=s(0)=0$, forcing $x=0$, a contradiction.

Therefore no section exists, and [F1] and [L1] show that the short exact sequence does not split. Equivalently, a retraction $r$ would satisfy $2r(1)=r(2)=1$.

The map $i$ is injective, $p$ is surjective, and $\ker p=\{(a,0):a\in A\}=\mathrm{im}i$, so the sequence is short exact.

The formulas give $p(s(c))=c$ and $r(i(a))=a$, so $s$ is a section and $r$ is a retraction by [F2].

By [L1] the sequence splits, and the resulting map $A\oplus C\to A\oplus C$, $(a,c)\mapsto i(a)+s(c)$, is the identity.

Every homomorphism $\mathbb{Z}\mathrm{/}2\mathbb{Z}\to \mathbb{Z}$ is zero, because the image of its generator would be killed by two; and evaluation at $1$ identifies ${\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z})$ with $\mathbb{Z}$.

Under this identification, precomposition with multiplication by two is the map $\mathbb{Z}\stackrel{\times 2}{\to }\mathbb{Z}$, because $(f\circ (\times 2))(1)=2f(1)$.

Applying the functor therefore produces $0\to 0\to \mathbb{Z}\stackrel{\times 2}{\to }\mathbb{Z},$ which is exact as [L1] predicts but whose final map is not surjective. Thus the functor is not right exact.

Multiplication by two is injective, $q$ is surjective, and its kernel is $2\mathbb{Z}$, so the sequence is short exact.

If a section $s$ existed, $x=s(1+2\mathbb{Z})$ would be odd because $q(x)=1+2\mathbb{Z}$, but $2x=s(0)=0$, so $x=0$, a contradiction.

Hence this short exact sequence does not split, refuting the statement. The hypotheses in [L1] and [L2] are sufficient conditions, not properties of every endpoint.

Modulo six, $3$ and $4$ are orthogonal idempotents with sum one, so $R=3R\oplus 4R$. The rank-one free module $R$ is projective without AC by [L2], and the direct lifting argument for a summand makes $P$ projective; this is the finite instance of [L1].

The module $P$ has two elements. By [F1], a free module on no basis vectors has one element, one on a nonempty finite set of size $k$ has $6^k$ elements, and one on an infinite set is infinite. Therefore $P$ is not free.

The projective nonfree module $P$ refutes the statement.

The group $I$ is divisible: for $q+\mathbb{Z}\in I$ and nonzero integer $n$, the class $q\mathrm{/}n+\mathbb{Z}$ is an $n$-th preimage. Thus $I$ is injective by [L1].

The class $1\mathrm{/}2+\mathbb{Z}$ is nonzero and killed by two, so $I$ has nonzero torsion.

Suppose $I$ were projective. By [L2], it would be isomorphic to a direct summand, hence a subgroup, of a free abelian group.

A free abelian group is torsion-free by uniqueness of the finite coordinate expression in [F1], and every subgroup of a torsion-free group is torsion-free. This contradicts step 1.2.

Therefore the injective module $\mathbb{Q}\mathrm{/}\mathbb{Z}$ is not projective, refuting the statement.