Past exam of the rings and modules course of the University of Oxford / 2023 / 2 / a

A subset $E\subseteq M$ is a basis of a module if every $m\in M$ has a unique expressionUniqueness concerns the entire family of coefficients, including the zero coefficients. Equivalently, $E$ generates $M$ and is linearly independent in a module: every finite relation ${\sum }_{e\in H}{r}_{e}e=0$, with $H\subseteq E$, has all $r_e=0$.

A module is a free module if it has a basis of a module. An infinite basis of a module still uses only finite sums in each expression; arbitrary infinite sums are not part of the module operations.

Yes. An $\mathbb{F}$-module over a field is a vector space. Start with a finite generating list. If it is linearly dependent, one vector has a nonzero coefficient in a linear relation; dividing by that coefficient expresses the vector using the others, so it can be removed without changing the span. Repeating terminates with a linearly independent generating list, which is a basis and therefore a basis of a module.

Thus every finitely generated module over a field is a free module. For the zero module, the empty list is the required basis of a module.

In the direct sum of modules $M=N\oplus P$, every $m\in M$ has a unique expression $m=n+p$, with $n\in N$ and $p\in P$. DefineThe uniqueness makes this well-defined. For $m=n+p$, $m^{\prime }=n^{\prime }+p^{\prime }$ and $r\in R$,so $\pi$ is a module homomorphism. Finally, $\pi (n+p)=0$ exactly when $p=0$, equivalently $m\in N$. Therefore $\boxed{\ker \pi =N}$. Its image is $P$, and ${\pi }^2=\pi$, so it is the projection onto that direct summand.

Put $N=⟨E^{\prime }⟩$ and $P=⟨E^{\prime \prime }⟩$, the submodules generated by the two parts of the basis of a module. Every finite expression in the basis of a module splits into its $E^{\prime }$ terms and its $E^{\prime \prime }$ terms. Thus $M=N+P$.

If $m\in N\cap P$, there are finite expressionsSubtracting gives a finite relation among distinct members of $E$, because $E^{\prime }\cap E^{\prime \prime }=\varnothing$. Their linear independence in a module forces every coefficient to vanish, so $m=0$. HenceThis proves the basis partition gives a module direct sum lemma, also when one part is empty.

Choose a countable generating family $(n_k)$ for $N$. For each $k$, the expression of $n_k$ in the basis of a module $E$ has a finite set $H_k$ of nonzero coefficients, its coordinate support in a free module. LetBy the countable union of countable sets property, $E^{\prime }$ is a countable set. Every $n_k$ belongs to $⟨E^{\prime }⟩$, so all finite $R$-linear combinations of the $n_k$ do too. ThereforeHere and below, countable includes finite. The same construction applies to a finite or empty generating family. This is the countable basis support lemma.