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Choose a finite generating set of a module $n_1,\dots ,n_t$ for the finitely generated module $N$. For any $m\in M$, the surjectivity of the module homomorphism $\varphi$ gives $n\in N$ with $\varphi (n)=m$. Write $n={\sum }_{j=1}^t{r}_j{n}_j$. Since $\varphi$ is a module homomorphism,Thusis a finitely generated module. The argument includes $N=0$, in which case surjectivity forces $M=0$ and the empty generating set of a module suffices.

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 finitely generated module over a principal ideal domain decomposes aswith finitely many irreducibles $p$. The free rank and primary cyclic factors are unique up to associates and ordering.

A prime filtration of a module over a commutative ring is a finite increasing chain from zero to the module with each nonzero successive quotient module isomorphic to $R/P$ for a prime ideal $P$. Every finitely generated module over a Noetherian ring has one. In each nonzero quotient, a maximal annihilator of a module element is prime supplies a cyclic submodule $R/P$; lift it and repeat. The ascending chain condition for a Noetherian module makes the process terminate. This is not necessarily a composition series: the prime ideals need not be maximal.

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.

The subset of the spectrum of a ring where the localization at a prime ideal is reduced. Since the nilradical commutes with localization, its complement is the support of a module $N=\mathrm{Nil}(R)$. When $N$ is a finitely generated module, vanishing of a finitely generated module near a prime givesso the locus is a Zariski-open subset. In particular this holds for every Noetherian ring.

The Rees module of an $R$-module $M$ with respect to an ideal $I$ is ${\mathcal{R}}_I(M)={⨁}_{j\ge 0}{I}^{j}M{t}^j$, a graded module over the Rees ring $\mathcal{R}(I)$. Generators of $M$ in degree zero generate this module. For $N\subseteq M$, the graded submodule ${⨁}_j(I^{j}M\cap N)t^j$ is finitely generated when $R$ is a Noetherian ring and $M$ is a finitely generated module; a bound on its generators' degrees yields the Artin-Rees lemma.

A finitely generated module $M$ over a commutative ring $A$ is reflexive when its evaluation homomorphism $M\to M^{\ast \ast }$ to the double dual module is an isomorphism.

A finitely generated module over a Noetherian ring admits a possibly infinite free resolution whose every term is finitely generated. Surject a finite free module onto it; the kernel is finitely generated because a finite free module is Noetherian. Repeat this construction on the successive kernels. If the original module has finite projective dimension $n$, its $n$th syzygy is projective by dimension shifting of the Ext functor, giving a length-$n$ resolution with finite projective terms.

Let $R$ be a Noetherian ring that is an integral domain, and let $I$ be a nonzero proper ideal. If $I^n=I^{n+1}$ for some $n\ge 1$, the nonzero finitely generated module $M=I^n$ satisfies $M=IM$. Choose generators $u_1,\dots ,u_r$ and write $u_i={\sum }_j{a}_{ij}{u}_j$ with $a_{ij}\in I$. The adjugate matrix identity implies that $\det ({\mathrm{Id}}_r-(a_{ij}))$ annihilates $M$. This determinant is $1-b$ for some $b\in I$, since every term except the constant term contains an entry from $I$.

The determinant trick has thus supplied $(1-b)M=0$. A nonzero element of $M$ and multiplicative cancellation in an integral domain force $b=1$, contradicting properness of $I$. Therefore all positive powers of an ideal are distinct in this setting. Since they have the same radical of an ideal, there are infinitely many ideals with radical $\sqrt{I}$.

Every finitely generated module over a PID is a direct sum of a free module and cyclic prime-power torsion modules.

The support of an $R$-module $M$ isFor a nonzero finitely generated module, the support is nonempty and contains a maximal ideal.

For a finitely generated module $V$ on the right over a semilocal ring, the displayed condition makes the induced module endomorphism on $V/VJ$ injective. This quotient is an Artinian module, so Artinian modules are co-Hopfian makes that map surjective. The remaining quotient module $V/\alpha (V)$ equals its product with $J$, and Nakayama lemma makes it zero.

An $R$-module $M$ is tensor-nilpotent when $M^{\otimes k}=0$ for some positive integer $k$. A nonzero finitely generated module cannot be tensor-nilpotent: localize at a maximal ideal in its support and reduce modulo that maximal ideal, obtaining a nonzero tensor power of a nonzero vector space.

Let $M$ be a nonzero finitely generated module over a commutative Noetherian ring. Take a length-$n$ resolution with finite projective terms. If its top Ext functor with coefficients in $A$ vanished, the dual of its last injection would be surjective. The dual of a finite projective module is projective, so that surjection splits. Finite projective modules are isomorphic to their double duals, so dualizing gives a splitting of the original injection. The resolution then shortens, contradicting minimality of $n$. When $n=0$, a nonzero coordinate of a finite free module containing $M$ as a summand gives a nonzero map $M\to A$.

For a nonzero left finitely generated module over a left Noetherian ring, take a finite projective resolution with finite projective modules. Vanishing of its top Ext functor would make the dual of its final injection surjective. These duals are right finite projective modules, so the surjection splits. Dualizing back and using finite projective double dual modules splits the original injection and shortens the resolution, contradicting the definition of projective dimension. When the dimension is zero, a coordinate map of a finite free module containing the nonzero projective summand supplies a nonzero map to $R$. The left/right distinction in the duals is essential for a noncommutative ring.

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.

A torsion module over an integral domain $R$ is a module in which every element is a torsion element of a module. Each element may require a different nonzero annihilating scalar. The module $\mathbb{Z}/6\mathbb{Z}$ is torsion over $\mathbb{Z}$, and the polynomial module from a linear operator $\mathbb{Q}[t]/(t^2)$ is torsion over $\mathbb{Q}[t]$.

If the torsion module is a finitely generated module, one common nonzero scalar kills the whole module: multiply annihilating scalars for a finite generating set. Without finite generation this can fail. The quotient module $\mathbb{Q}/\mathbb{Z}$ is torsion, since the class of $a/b$ is killed by $b$, but no nonzero integer $n$ kills it all: the class of $1/(2|n|)$ survives multiplication by $n$. See Keith Conrad's Modules over a PID, exercise 6.5(c).

For a finitely generated module $M=(m_1,\dots ,m_t)$ and a prime ideal $P$, if $M_P=0$, choose $s_i\notin P$ annihilating $m_i$. Their product $f\notin P$ annihilates all of $M$. Conversely such an $f$ becomes a unit in $R_P$, forcing $M_P=0$. ThusThis identifies the support of a module $M$ with the Zariski-closed set $V({\mathrm{Ann}}_R(M))$. For $M=0$, use $f=1$. No Noetherian assumption on $R$ is required.