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If $R$ is Noetherian, $I$ is an ideal, and $N\subseteq M$ are finitely generated modules, then some $k$ satisfiesfor every $n\ge k$.

Let $A\subseteq B\subseteq C$ be commutative rings. If $A$ is Noetherian, $C$ is a finitely generated algebra over $A$, and $C$ is a finitely generated module over $B$, then $B$ is a finitely generated algebra over $A$.^[ref]

Choose algebra generators $x_1,\dots ,x_r$ for $C$ over $A$ and module generators $c_1=1,c_2,\dots ,c_m$ for $C$ over $B$. Express every $x_i{c}_j$ as a $B$-linear combination of the $c_k$. Let $B_0$ be the $A$-subalgebra generated by the finitely many coefficients in these expressions. The $B_0$-module $M={\sum }_k{B}_0{c}_k$ contains one and is stable under multiplication by every $x_i$, so $M=C$. Thus $C$ is finite as a $B_0$-module.

The Hilbert basis theorem makes $B_0$ Noetherian. Hence its submodule $B\subseteq C$ is finitely generated over $B_0$. Combining those module generators with the finite algebra generators of $B_0$ gives finitely many algebra generators for $B$ over $A$.

A good Bernstein filtration on a finitely generated module produces a finite graded module over $\mathrm{gr}{A}_n\cong \mathbb{C}[x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n]$. Its support is the characteristic variety for this filtration. Its algebraic dimension is the Bernstein growth dimension of a Weyl algebra module, which is independent of the choice of good filtration.

Let a finitely generated module over an integral domain have a finite generating set $S$ and a maximal independent subset $T$. For each $s\notin T$, dependence of $T\cup \{s\}$ gives a nonzero coefficient $a_s$ with $a_{s}s\in ⟨T⟩$. The product of those coefficients is nonzero and carries the whole module into $⟨T⟩$.

For a Noetherian ring $A$, a finitely generated module $M$ has a finite presentation of a module, and kernels of maps from finite free modules to $M$ are finitely generated. Its affine module sheaf is therefore a coherent sheaf. Conversely, if $\tilde{M}$ is coherent, choose a finite principal open subset cover on which its section modules are finitely generated. Lift their finitely many generators to numerators in $M$. The submodule generated by these numerators becomes all of $M$ on every chart, and hence is all of $M$: its quotient has zero localization at every point. This proves finite generation without assuming that global sections preserve arbitrary sheaf surjections.

For a nonzero finitely generated module over a complex Weyl algebra, choose a good filtration for the order filtration and take the Krull dimension of its support of a module over $\mathrm{gr}R=\mathbb{C}[x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n]$. This is the dimension of its order characteristic variety of a Weyl algebra module. One may set $d(0)=-\infty$.

For independence from the good module filtration, any two such filtrations of a module are bounded shifts of one another, by their finite shifted-generating descriptions. If a homogeneous symbol $\sigma (a)$ of degree $d$ annihilates one associated graded module, then $a{F}_{j}M\subseteq F_{j+d-1}M$ and its $q$-th power lowers the expected degree by $q$. For shift bound $c$ and $q>2c$, comparison shows that $\sigma (a{)}^q$ annihilates the other associated graded module. Reverse the roles to obtain equality of the radicals of ideals of their annihilators, hence equality of their support of a module and Krull dimension. This argument holds the algebra filtration fixed.

If a submodule $N\subseteq M$ and the quotient module $M/N$ are finitely generated modules, then $M$ is a finitely generated module. Choose finitely many generators of $N$ and lift finitely many generators of $M/N$. For any $m\in M$, subtract an $R$-linear combination of the lifts to obtain an element of $N$, then express this using the generators of $N$. The union of the two finite lists generates $M$.

Suppose $A\subseteq B$ are unital rings and $B$ is a finitely generated module over $A$ on the right. If $A$ is a right Noetherian ring, then $B_A$ is a Noetherian module. Every right ideal of $B$ is an $A$-submodule, hence has finitely many generators over $A$; these also generate it over $B$. Thus $B$ is a right Noetherian ring. The same argument works on the left under the corresponding hypotheses. Finite-index group ring extensions give an important example, using coset representatives as a finite module basis.

A module is finitely presented when it admits the displayed exact sequence for finite integers $r,s$. Thus it has finitely many generators and finitely many relations. On a Noetherian ring, every finitely generated module has a finite presentation.

A finite projective module is a finitely generated module that is a projective module. Equivalently it is a direct summand of a finite free module: a finite free surjection splits by projectivity. Its dual module is finite projective and its evaluation map to its double dual module is an isomorphism, since both properties hold for finite free modules and pass to direct summands. Finite generation is needed for this elementary duality argument.

For a nonzero right module $M$ that is a finitely generated module, choose a finite-dimensional generating vector subspace $W$ and define ${\mathrm{GKdim}}_{A}M={\mathrm{limsup}}_{n\to \infty }\log ({\dim }_{k}W{V}^n)/\log n$, where $V$ generates the algebra and contains $1$. For a left module use $V^{n}W$. It is independent of these choices and measures polynomial-order growth of a generating filtration of a module.

A subset $S$ of a module $M$ generates $M$ when every element of $M$ is a finite sum ${\sum }_{s\in S}{r}_{s}s$, with coefficients in the scalar ring. The existence of a finite such subset defines a finitely generated module.

For a filtered algebra $A$ with finite-dimensional filtration pieces, a good filtration of a module on a finitely generated module is compatible with multiplication and has finitely generated associated graded module over the associated graded ring. Taking bounded-degree translates of a finite set of module generators supplies such a filtration of a module when the associated graded ring is Noetherian.

A graded module over a graded ring $R={⨁}_i{R}_i$ is an $R$-module with a direct sum decomposition $M={⨁}_d{M}_d$ satisfying $R_i{M}_d\subseteq M_{i+d}$. A graded homomorphism preserves these degrees. Over a polynomial ring with positive variable degrees, a finitely generated module of this kind is bounded below and has finite-dimensional graded components over the coefficient field.

For an ideal of definition $I$ of a Noetherian local ring and a finitely generated module $M$, the Hilbert–Samuel function isFor sufficiently large $n$, it equals the Hilbert-Samuel polynomial.

Every finitely generated module with a grading over the polynomial ring $S=k[X_1,\dots ,X_n]$ has a finite graded free resolution of length at most $n$. The Koszul resolution of $k$ has length $n$, so ${\mathrm{Tor}}_i^S(M,k)=0$ for $i>n$. In a minimal graded free resolution, this Tor functor is $F_i/(X_1,\dots ,X_n)F_i$; the graded Nakayama lemma forces $F_i=0$ for $i>n$.

For a finitely generated module over a principal ideal domain, its invariant factors are nonzero nonunits $d_i$ in a decomposition $M\cong R^r\oplus {⨁}_{i}R/(d_i)$, ordered by $d_i\mid d_{i+1}$. They are unique up to multiplication by units and describe the torsion submodule. Over $F[t]$, making $t$ act as a linear operator recovers the invariant factors of a linear operator used in its rational canonical form.

An invertible module over a commutative ring $A$ is a finitely generated module that is a projective module and whose localization at a prime ideal is free of rank one at every prime. Its dual module $L^{\ast }={\mathrm{Hom}}_A(L,A)$ is also finite projective. Evaluation $L{\otimes }_A{L}^{\ast }\to A$ is an isomorphism after every prime localization, hence is an isomorphism globally by localization detects zero elements. The Picard group of a ring consists of the isomorphism classes of these modules under the tensor product of modules.

Let $R$ be a Noetherian local ring, $I$ an ideal contained in its maximal ideal, and $M$ a finitely generated module. The Krull intersection theorem statesIn particular ${\bigcap }_n{I}^n=0$ by taking $M=R$. It says that the I-adic topology is separated on every finite module.

For a proof, let $N={\bigcap }_n{I}^{n}M$. It is finitely generated because $M$ is Noetherian. The Artin-Rees lemma gives, for some $k$, $I^{k+1}M\cap N=I(I^{k}M\cap N)$. Both intersections equal $N$, so $N=IN$. The Nakayama lemma then gives $N=0$. Finite generation is essential to this argument; it cannot be omitted merely because the scalar ring is Noetherian.

For a commutative ring $R$ that is a Noetherian ring and a multiplicatively closed set $S$, a finitely generated module $N$ has a free resolution with every term finitely generated. Hom from a finite free module commutes with localization, and exactness of localization commutes with cohomology. Thus the displayed isomorphism holds for every $i\ge 0$ and arbitrary $M$. The finiteness condition is on the first argument; the second need not be finitely generated.