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Assume $H=\ker \psi$ is a finitely generated module. The restriction $q:H\to L$ is a surjective module homomorphism, so part (a)(i) makes $L$ a finitely generated module.

The free module $R^m$ has its finite standard basis of a module. Its quotient module $R^m/\mathrm{Im}q$ is therefore a finitely generated module by part (a)(i). Part (c) transfers this property to the isomorphic quotient module $\ker \varphi /L$. Finally apply part (a)(ii) to the submodule $L\subseteq \ker \varphi$. It follows thatMore concretely, take the images under $q$ of finitely many generators of $H$, and adjoin lifts in $\ker \varphi$ of finitely many generators of $\ker \varphi /L$. This finite list generates $\ker \varphi$. The argument shows why finite presentation of a module is independent of the chosen finite generating set of a module.

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.

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.

For $j\in {\mathbb{N}}_0$, let $e_j\in B$ have $e_j(j)=1$ and $e_j(k)=0$ for $k\ne j$. If $f,g\in F$ and $z\in \mathbb{Z}$, then $f+g$ can be nonzero only where $f$ or $g$ is nonzero, and $zf$ can be nonzero only where $f$ is nonzero. These are finite sets. Since $0\in F$, this proves that $F$ is a submodule.

Every $f\in F$ has the finite expressionEvaluating at coordinate $k$ recovers the coefficient of $e_k$, so the expression is unique. Thus $\{e_j:j\in {\mathbb{N}}_0\}$ is a basis of a module for $F$, andis a free module. By contrast, the ambient module of integer sequences $B={\prod }_{j\in {\mathbb{N}}_0}\mathbb{Z}$ allows arbitrary integer sequences.

A linear operator $T:V\to V$ on a vector space over a field $k$ makes $V$ a module over the polynomial ring $k[t]$: let $t$ act as $T$, let constants act as the original scalars, and define $p(t)\cdot v=p(T)v$. The identity $(pq)(T)=p(T)q(T)$ for polynomials $p,q$ verifies the multiplication axiom. The submodules are precisely the invariant subspaces of $T$. This construction is described in Keith Conrad's Introductory Notes on Modules, section 6.

For a computed example, take $k=\mathbb{Q}$, $V=k^2$, andThen $T{e}_2=e_1$, $T{e}_1=0$, and $T^2=0$. ConsequentlyEvery $(a,b)$ equals $(b+at)\cdot e_2$. The module homomorphism $k[t]\to V$, $p\mapsto p(T)e_2$, is therefore surjective, and its kernel is the ideal $(t^2)$. It induces a module isomorphism $V\cong k[t]/(t^2)$. This is a cyclic module and a torsion module over $k[t]$, since $t^{2}V=0$. Over $k$ it is a free module with basis $(e_1,e_2)$ and has no nonzero torsion elements of a module. Over $\mathbb{Q}$ its nonzero elements also have infinite additive order of a group element. Changing scalars changes the questions one asks about the same addition law.

Every module $P$ is a quotient of a free module $F={⨁}_{p\in P}A{e}_p$, via $e_p\mapsto p$. If $P$ is a projective module, lift $1_P$ through this surjection to obtain a section $s:P\to F$. The resulting split short exact sequence identifies $F$ with $\ker \pi \oplus P$. Conversely, if $P$ is a summand of a free module, extend a prescribed map out of $P$ to the free module using the summand projection, lift each basis image through the given surjection, and restrict the lift to $P$. This proves the equivalence with the lifting definition of a projective module, for arbitrary ranks.

A projective module is a direct summand of a free module. Free modules are flat because tensoring with a free module produces a direct sum of copies of the original module and map. If $F=P\oplus Q$ is flat, tensoring an injection with $F$ is the direct sum of its tensor maps with $P$ and $Q$, so each component is injective. Applying this to a free module containing $P$ proves that $P$ is flat.

For a module $M$ over an integral domain $R$ with field of fractions $K$, its rank is the dimension of $K{\otimes }_{R}M$. The tensor product of modules here extends scalars to a field, turning the question into a dimension calculation. For a free module $R^r$, the result is $r$.

For a finitely generated module over a principal ideal domain, the structure theorem for finitely generated modules over a principal ideal domain gives $M\cong R^r\oplus {⨁}_{i}R/(d_i)$, with $d_i$ nonzero nonunits. Each torsion element of a module vanishes after extension of scalars, because its annihilating nonzero scalar becomes invertible in $K$. Thus $K{\otimes }_{R}M\cong K^r$, proving that the rank is the number of free summands. The rank does not record the torsion submodule: for example, $R\oplus R/(d)$ and $R$ have the same rank.

Let $A$ be a Noetherian ring that is an integral domain. Ifis exact, $N$ is a finitely generated free module, and $P$ is a torsion-free module, then $M$ is a reflexive module. In particular, the dual of every finitely generated $A$-module is reflexive.

A module $M$ over an integral domain $R$ is torsion-free when $rm=0$ with $0\ne r\in R$ implies $m=0$, equivalently when its torsion submodule is zero. Every free module over an integral domain is torsion-free, as one can check in each coefficient. Every vector space is torsion-free over its scalar field.

The converse requires hypotheses. A finitely generated module over a principal ideal domain is free if it is torsion-free, but the module $\mathbb{Q}$ over $\mathbb{Z}$ is torsion-free and not free. Any two nonzero rational numbers have a nontrivial integer relation, so a possible basis of a module could have at most one element; no single $q$ generates all rational numbers, since $q/2\notin \mathbb{Z}q$ for $q\ne 0$. The finite-generation hypothesis cannot be dropped. See Keith Conrad's Modules over a PID, Corollary 2.6.

Order the linearly independent subsets of a vector space by inclusion. A chain's union is independent, since every finite relation is contained in one chain member. Zorn's lemma gives a maximal independent subset. A vector outside its span could be adjoined while preserving independence, so it spans. Sending finite-support coefficient families to their linear combinations gives an isomorphism from the free module on that subset to $V$. This supplies an arbitrary, possibly infinite-dimensional basis using the axiom of choice; no pre-existing basis theorem is invoked.