Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 7 / 2

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 $N\to \mathbb{Z}$.

For any injection $u:U\hookrightarrow V$, the direct sum decomposition $F=P\oplus Q$ gives natural module isomorphismsUnder these, $u\otimes 1_F$ is the block-diagonal map $(u\otimes 1_P)\oplus (u\otimes 1_Q)$. It is injective because $F$ is a flat module. If an element lies in the kernel of $u\otimes 1_P$, insert zero in the other component; injectivity of the whole map forces that element to be zero. Hence $u\otimes 1_P$ is injective for every $u$, which proves $\boxed{P\text{ is flat}}$. This establishes that direct summands of flat modules are flat, without a finite-generation assumption.

Here the Hom groups must be taken over $A$; the printed subscript $B$ in this part has no defined coefficient ring. For an arbitrary module $P$, form the free moduleand define a module homomorphism $\pi :F\to P$ by $e_p\mapsto p$. Each element of $F$ has finite support, so this prescription extends uniquely and is well-defined. It is surjective because every $p\in P$ is the image of $e_p$. Thus every module is a quotient of a free module.

Let $K=\ker \pi$. The short exact sequence $0\to K\to F\stackrel{\pi }{\to }P\to 0$ and the defining exactness condition for a projective module makesurjective. Lift $1_P$ to a map $s:P\to F$. Then $\pi s=1_P$, so $s$ is injective. Every $f\in F$ has the decompositionwhose first term lies in $K$ and second lies in $s(P)$. If $s(p)\in K$, applying $\pi$ gives $p=0$, so the intersection is zero. ConsequentlyThis is a split short exact sequence and proves that projective modules are direct summands of free modules, including modules that are not finitely generated.

First show that free modules are flat. If $F={⨁}_{j\in J}A{e}_j$, distributivity of the tensor product of modules givesExplicitly, $u\otimes {\sum }_j{a}_j{e}_j$ maps to $(a_{j}u{)}_j$, and the inverse sends a finitely supported tuple $(u_j{)}_j$ to ${\sum }_j{u}_j\otimes e_j$. For an injection $u:U\hookrightarrow V$, the tensor map is therefore the coordinatewise injection ${⨁}_{J}U\hookrightarrow {⨁}_{J}V$. This proves flatness for any free rank.

By part (c), a projective module $P$ is a direct summand of a free module $F$. The latter is flat by the preceding argument, and part (b) proves that its summand $P$ is flat. Therefore $\boxed{\text{every projective }A\text{-module is flat}}$. This proves projective modules are flat directly from the exactness and splitting definitions.