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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 odd $p$, the Coleman norm operator satisfies the displayed inclusion for every $k\ge 1$. Write $M_h$ for multiplication by $h$ on the finite free module $R$ over $\varphi (R)$. Its trace is $p\varphi (S(h))$, where $S$ is the Coleman trace operator. In $\det (I+p^k{M}_h)$ the linear term is therefore divisible by $p^{k+1}$, and all higher terms are divisible by $p^{2k}$. Congruence reflection for Frobenius substitution gives the assertion. A norm-fixed element of $1+pR$ lies in $1+p^{k}R$ for every $k$ and hence equals $1$.

Apply clearing denominators relative to an independent module subset to find $0\ne a$ with $aM$ lying in a finite free module. Multiplication by $a$ is injective on a torsion-free module, giving the required embedding. A submodule need not itself be free over a general integral domain.

A free module is a module with a basis of a module $(e_i{)}_{i\in I}$: every element has exactly one expression $m={\sum }_{i\in I}{r}_i{e}_i$ with $r_i\in R$ and only finitely many $r_i$ nonzero. Equivalently, the coefficient map ${⨁}_{i\in I}R\to M$ is a module isomorphism. Existence means generation; uniqueness means linear independence in a module. For finite $I$ of size $n$, this says $M\cong R^n$, a finite free module. See Neil Strickland's Rings, Modules and Linear Algebra, section 11.

The integer lattice ${\mathbb{Z}}^2$ in the opening diagram is a free module over $\mathbb{Z}$ with basis of a module $e_1=(1,0)$, $e_2=(0,1)$. For example, $(3,2)=3e_1+2e_2$, and comparing the two coordinates makes the coefficients unique. A different basis of a module is $u=(1,0)$, $v=(1,1)$, because $(a,b)=(a-b)u+bv$; in particular $(3,2)=u+2v$.

By contrast, $w_1=(2,0)$ and $w_2=(0,1)$ are linearly independent in a module, but generate only the submodule $2\mathbb{Z}\times \mathbb{Z}$. They miss $(1,0)$ because $2a=1$ has no integer solution. They would form a basis over the rational numbers. Even generation need not let one extract a basis of a module: the elements $2,3$ generate $\mathbb{Z}$, since $1=3-2$, but neither singleton generates it, and $3\cdot 2-2\cdot 3=0$ is a nontrivial linear relation between them. These are concrete failures of the usual vector space argument that divides by any nonzero coefficient.

The cyclic module $\mathbb{Z}/6\mathbb{Z}$ has no basis of a module over $\mathbb{Z}$: every element has the nontrivial relation $6m=0$, so no nonempty subset is linearly independent in a module, while the empty subset cannot generate this nonzero module. Specifying the scalar ring is essential: over $R=\mathbb{Z}/6\mathbb{Z}$, the same additive object is a free module with the one-element basis of a module $\{\overline{1}\}$, since each coefficient is itself a unique class in $R$.

Elements of a module are linearly independent if every finite linear relation between them has all coefficients zero. An independent finite set freely generates its span, a finite free module. Over an integral domain, maximal independence within a finite generating set enables clearing denominators relative to an independent module subset.

Module theory studies abelian groups with scalar multiplication by a ring. It extends linear algebra: a module over a field is a vector space, while ring scalars let the same language describe integer arithmetic and linear operators. The new phenomena are visible before any abstract classification: some nonzero scalars cannot be divided out, some modules have torsion elements of a module, and some have no basis of a module.

The distinction is useful because an integer relation such as $6m=0$ carries information which disappears if one assumes all nonzero scalars have multiplicative inverses. A finitely generated module over a principal ideal domain can be separated into a finite free module and a torsion module; the structure theorem for finitely generated modules over a principal ideal domain unifies the classification of finitely generated abelian groups with the study of linear operators through polynomial rings. See Keith Conrad's Modules over a PID, Theorem 4.1 and the operator applications.

A historical anchor is Emmy Noether's 1921 paper ^[ref]. Its first section studies rings and ideals under the condition that every ideal is a finitely generated module over its ring, and connects this to the ascending chain condition; later sections develop primary decomposition. This is an early source for the structural viewpoint underlying Noetherian rings. ^[ref] makes those statements accessible in English.

If $B$ is a commutative $A$-algebra which is a finite free module over $A$, its norm sends $b$ to the determinant of multiplication $m_b:B\to B$. The determinant is independent of the chosen basis and is multiplicative because $m_{bc}=m_b{m}_c$. It commutes with scalar extension. On a separable field extension it becomes the product over embeddings, recovering the field norm. An invertible $b$ has invertible norm.

Put $X=1+T$ and let $\phi$ be the Frobenius substitution on cyclotomic power series, $\phi (g)(T)=g(X^p-1)$. It is injective: modulo $p$ it is the injective map $\overline{g}(T)\mapsto \overline{g}(T^p)$. If $\phi (g)=0$, this makes every coefficient of $g$ divisible by $p$; dividing by $p$ and repeating makes every coefficient zero.

We prove that $R$ is a finite free module over $S=\phi (R)$ of rank $p$, with basis $1,X,\dots ,X^{p-1}$. Modulo $p$, every series has a unique decomposition over ${\mathbb{F}}_p[[T^p]]$ in the basis $1,T,\dots ,T^{p-1}$. Changing to the powers of $X=1+T$ is an invertible triangular change of basis. Thus, modulo $p$, every $h\in R$ can be expressed as ${\sum }_{i=0}^{p-1}{X}^i\phi (h_i)$. Lift the coefficients to $R$, subtract this expression, and divide the remainder by $p$. Repeating and summing the p-adically convergent coefficient series givesUniqueness follows by reduction modulo $p$ and successive division by $p$. This proves finite freeness, not just a formal invariance of the right side in the question.

For $f\in R^{\times }$, multiplication by $f$ is an invertible $S$-linear map $m_f:R\to R$. Its determinant $b={\det }_S(m_f)$ belongs to $S^{\times }$. DefineThis is the norm for a finite free ring extension, pulled back along the injective substitution, and it is multiplicative and unit-valued.

To identify its determinant with the desired product, extend coefficients to $\mathcal{O}={\mathbb{Z}}_p[{\mu }_p]$. The substitutions $X\mapsto \xi X$ for ${\xi }^p=1$ are continuous automorphisms of $\mathcal{O}[[T]]$ fixing $\phi (\mathcal{O}[[T]])$. Evaluation of a series at $\xi X-1$ is legitimate in the $(\xi -1,T)$-adic topology, because the constant term $\xi -1$ is topologically nilpotent. On fraction fields this is a degree-$p$ extension with these $p$ distinct automorphisms: finite freeness bounds the degree by $p$, while the automorphisms force degree at least $p$. The determinant norm therefore equals the product of its conjugates. Norms commute with this scalar extension, soas required. Finally, any other unit-valued map satisfying this identity has the same image under $\phi$ for every $f$, and injectivity of $\phi$ makes it identical to $N$. The Coleman norm operator exists and is unique.

A quasi-coherent sheaf is a sheaf of modules which locally has a presentation by free module sheaves:where the index sets may be infinite. Equivalently, on affine neighbourhoods it is an affine module sheaf. A coherent sheaf is locally finitely generated, with the additional requirement that the kernel of every map from a finite free module sheaf is locally finitely generated. On a Noetherian scheme, these conditions are equivalent to being locally associated with a finitely generated module.

For an affine variety $X$ with coordinate ring $A$ and an $A$-module $M$, setRestriction on principal open subsets is the natural module localization map. On an arbitrary open subset, sections are families of these stalk values locally represented by a single fraction from a localization. This constructs the affine module sheaf; its global sections are $M$. Choose a free-module presentation $A^{(J)}\to A^{(I)}\to M\to 0$. Exactness of localization and its compatibility with direct sums yieldThus $\tilde{M}$ is a quasi-coherent sheaf, without a finite-generation assumption.

We now prove coherence of an affine module sheaf. If $M$ is finitely generated, take a surjection $A^r\to M$. Since $A$ is a Noetherian ring, its kernel is finitely generated, so $M$ has a finite presentation of a module. More generally, on every principal affine open, the kernel of a map $A_a^s\to M_a$ is a submodule of a finite free module over the Noetherian ring $A_a$, and is finitely generated. On sheafifying, this verifies both parts of the definition of a coherent sheaf.

Conversely, if $\tilde{M}$ is coherent, choose a finite principal-open cover $X={\bigcup }_{i}D(a_i)$ on which it has finitely many generators. Write those local generators as $m_{ij}/a_i^{r_{ij}}$. Their numerators generate a finite submodule $N={\sum }_{i,j}A{m}_{ij}$ of $M$, and $N_{a_i}=M_{a_i}$ for every $i$. To check that $M/N=0$, let $q\in M/N$. Each equality after localization gives an exponent $r_i$ such that $a_i^{r_i}q=0$. The ideal generated by the $a_i^{r_i}$ is the unit ideal: otherwise a maximal ideal containing all of them would give a point outside the principal-open cover. A linear combination of these powers equal to $1$ consequently annihilates $q$. Every $q$ is zero, proving

Next consider a morphism $\varphi :{\mathcal{G}}_1\to {\mathcal{G}}_2$ between quasi-coherent sheaves. On an affine open $V$ of the target variety, use the assumed affine module-sheaf equivalence to write ${\mathcal{G}}_j{|}_V=\widetilde{M_j}$, with $M_j=\Gamma (V,{\mathcal{G}}_j)$. The sheaf morphism is the sheafification of the module map $\psi =\Gamma (V,\varphi )$. Indeed, its action on a section $m/a^r$ of a principal open is forced by module-linearity and restriction to be $\psi (m)/a^r$. Therefore exactness of localization givesThese identities are compatible with restriction, so $(\ker \varphi ){|}_V=\widetilde{\ker \psi }$. Since this holds on an affine cover, the kernel sheaf is quasi-coherent on every variety. This proof of kernels of quasi-coherent sheaf morphisms does not assert that taking global sections is right exact.

For direct image along a morphism of affine varieties, write $Y$ with coordinate ring $A$, $X$ with coordinate ring $B$, and let $\alpha :A\to B$ be the induced ring homomorphism. If $\mathcal{F}=\tilde{M}$ on $X$, regard $M$ as an $A$-module by restriction of scalars. For $a\in A$,The last equality identifies localization with the scalar action of $a$. It identifies the direct image sheaf with the affine module sheaf of the restricted module, so $f_{\ast }\mathcal{F}$ is quasi-coherent.

Suppose $Y$ is affine and $X={\bigcup }_{i=1}^r{U}_i$ is a finite affine cover. Since a variety is separated, every $U_i\cap U_j$ is affine: it is identified with the inverse image of the closed diagonal morphism inside the affine product $U_i{\times }_k{U}_j$. Write $g_i=f{|}_{U_i}$ and $g_{ij}=f{|}_{U_i\cap U_j}$. Form the morphism of sheaves of modulesBoth sums are finite, and every summand is quasi-coherent by the affine calculation. For every open $V\subseteq Y$, the sheaf gluing axiom on $f^{-1}(V)$ identifies the kernel of this map on sections with $\mathcal{F}(f^{-1}(V))$. HenceThe kernel result proved above establishes quasi-coherence.

Finally, for arbitrary $Y$, take an affine neighbourhood $V\subseteq Y$ and apply the preceding argument to $f^{-1}(V)\to V$. The open subset $f^{-1}(V)$ is again a variety and has a finite affine cover, because varieties have Noetherian Zariski topology. The restriction of $f_{\ast }\mathcal{F}$ to $V$ is exactly the direct image for this restricted morphism. ThusThe finiteness and separation conditions are the relevant hypotheses, not affineness of the original source or target. More generally, quasi-coherence of direct image under a quasi-compact quasi-separated morphism is valid. If one permits nonseparated finite-type prevarieties, each overlap in the argument has a finite affine cover; replacing its summand by the sum of direct images from these charts gives the same kernel, since agreement can be tested on a cover. This also proves the assertion in that convention. It does not extend by this argument to arbitrary spaces with an unrestricted infinite cover.

Put $X=1+T$ and $Y=X^p-1$, so that $\varphi (f)=f(Y)$. The essential integrality statement for the Frobenius substitution on cyclotomic power series is the decomposition into a finite free module:To prove it, reduce modulo $p$. Since $Y\equiv T^p$, grouping the coefficients of a formal power series according to their exponents modulo $p$ gives the unique expression ${\sum }_{j=0}^{p-1}{T}^j{h}_j(T^p)$. The change from $1,T,\dots ,T^{p-1}$ to $1,X,\dots ,X^{p-1}$ is a triangular matrix with diagonal entries $1$, so this is also a basis over ${\mathbb{F}}_p[[T^p]]$. Lift these $p$ coefficient series to $R$, subtract their contribution from $f$, divide the remaining error by $p$, and repeat. The sums of the successive lifts converge in the finer of the topologies on integral formal power series, because $R$ is complete in that topology. This proves existence of the decomposition over $R$. A relation among its summands reduces to the zero relation modulo $p$, forcing each coefficient series to be divisible by $p$. Repetition forces divisibility by every power of $p$, hence every coefficient series is zero. This proves uniqueness and, in particular, injectivity of $\varphi$.

For $f={\sum }_{i=0}^{p-1}{X}^i\varphi (f_i)$, defineThis is a ${\mathbb{Z}}_p$-linear map. To calculate the average over roots of unity, work over $\mathcal{O}={\mathbb{Z}}_p[{\zeta }_p]$. The substitutions $T\mapsto \zeta X-1$ are legitimate: their constant terms are topologically nilpotent, so the defining coefficient sums converge in $\mathcal{O}$. Since $(\zeta X{)}^p=X^p$, each $\varphi (f_i)$ is unchanged by the substitution. The sums of the powers of the roots of unity are $p$ for $i=0$ and zero for $1\le i<p$. ConsequentlyThus the average is integral and belongs to $\varphi (R)$, and the constructed Coleman trace operator has the required property. Injectivity of $\varphi$ forces uniqueness of any map satisfying that property. Finally, the decomposition of $\varphi (g)$ has $f_0=g$ and all other coefficients zero, giving $S\circ \varphi ={\mathrm{id}}_R$.

We will also use the trace of multiplication by $h$ on this finite free module. After extending scalars to split the conjugates, that trace is the sum just computed. Hence

A generic linear projection over an infinite field is not enough for the stated arbitrary field $k$. Instead use Noether normalization by weighted substitutions to make the defining polynomial monic in one coordinate.

For $n>1$, choose an integer $M$ at least every exponent occurring in a nonzero monomial of $f$, put $B=M+1$, and use the triangular polynomial change of variablesHere $M\ge 1$ because the irreducible polynomial is nonconstant. This is an automorphism, with inverse $y_i=x_i-x_n^{B^i}$, $y_n=x_n$. A monomial $x_1^{{\alpha }_1}\cdots x_n^{{\alpha }_n}$ contributes a highest $y_n$ powerThese weights are all distinct: the exponents are base-$B$ digits bounded by $M$. All other terms in the expansion have smaller $y_n$ degree than that monomial's weight. Therefore the highest-degree $y_n$ term in the transformed $f$ comes from exactly one original monomial and has a nonzero coefficient in $k$, independent of the other $y_i$. No assumption about the cardinality or characteristic of $k$ enters this argument.

Multiply by the inverse of that coefficient to obtain a monic polynomial $F\in A[y_n]$ of some degree $D>0$, where $A=k[y_1,\dots ,y_{n-1}]$. The same construction for $n=1$ simply rescales $f$ to be monic over $A=k$.

The monic polynomial quotient is finite free because division by $F$ gives a unique representative of degree less than $D$. Thusas $A$-modules. Uniqueness follows because any nonzero multiple of a monic $F$ has $y_n$ degree at least $D$. In particular the map $A\to A[y_n]/(F)$ is injective.

The induced morphismis finite, since its coordinate algebra is a finite module, and flat, since tensoring with a finite free module is a finite direct sum of copies of the original module and preserves exactness. This is a finite flat Noether normalization of an affine hypersurface, with the explicit projection coordinates $y_i=x_i-x_n^{B^i}$. The construction actually applies to any nonconstant polynomial, not only an irreducible one.

The map $\phi :M\to N$, $x\mapsto rx$, is a module homomorphism, and it is injective because $r\ne 0$ and $M$ is a torsion-free module. The linear independence in a module of $T$ makes the map $R^{|T|}\to N$ sending the standard basis to $T$ an isomorphism. Thus $N$ is a finite free module, and $M\cong \phi (M)\subseteq N\cong R^{|T|}$. This proves embedding a finitely generated torsion-free module in a finite free module. No claim that the submodule itself is free over an arbitrary integral domain is needed.

An integral form of a group representation $V$ is a $G$-stable finite free $\mathcal{O}$-submodule $W\subset V$ with $K{\otimes }_{\mathcal{O}}W\cong V$. Such a form exists: take a basis lattice $L$ and replace it by ${\sum }_{g\in G}gL$, which is finite and torsion-free and therefore free over the discrete valuation ring $\mathcal{O}$.

Choose ordinary simple modules $V_i$ with forms $W_i$, and modular simple modules $S_j$ with projective covers $P_j$. With $\pi$ a uniformizer, put ${\overline{W}}_i=W_i/\pi W_i$. The decomposition matrix and Cartan matrix of a group algebra have entriesThe first numbers do not depend on the integral form: the Brauer character of its reduction is the ordinary character restricted to p-regular elements, and part 2(a) determines all composition multiplicities from this restriction.

For the assertion in (i), set $I={\mathrm{Hom}}_{\mathcal{O}G}(W,W^{\prime })$. It is a submodule of the finite free module ${\mathrm{Hom}}_{\mathcal{O}}(W,W^{\prime })$, hence is finite free over $\mathcal{O}$. There is a natural injective mapFor any homomorphism $f$ in the target, clearing the finitely many denominators of its values on an $\mathcal{O}$-basis of $W$ gives ${\pi }^{N}f(W)\subset W^{\prime }$. Thus ${\pi }^{N}f\in I$, and the map is surjective. Therefore $\boxed{I\text{ is an }\mathcal{O}\text{-form of }{\mathrm{Hom}}_{KG}(M,M^{\prime })}$, regarding this Hom space as a vector space. The original PDF supplies part (i), which is missing from the supplied TeX.

Completeness permits idempotent lifting from $kG$ to $\mathcal{O}G$. In particular, each projective cover $P_j$ lifts to a finite projective $\mathcal{O}G$-lattice ${\tilde{P}}_j$; one can lift an idempotent presenting it as a summand of a finite free module. Set $Q_j=K{\otimes }_{\mathcal{O}}{\tilde{P}}_j$.

For a simple $S_{\ell }$, every map $P_j\to S_{\ell }$ factors through its simple head $S_j$. Splitting and Schur lemma give ${\dim }_k{\mathrm{Hom}}_{kG}(P_j,S_{\ell })={\delta }_{j\ell }$. Exactness of this Hom functor along a composition series therefore givesPart (iii) identifies this with ${\dim }_K{\mathrm{Hom}}_{KG}(Q_j,V_i)$. By Maschke's theorem and splitting, $KG$ is split semisimple, soThe reductions of any two integral forms of the same ordinary module have identical Brauer characters, hence identical composition multiplicities. We may thus reduce a direct-sum form for the displayed decomposition instead of ${\tilde{P}}_j$. In the modular representation ring this givesComparing the simple basis coefficients proves $c_{\ell j}={\sum }_i{d}_{i\ell }{d}_{ij}$, or $\boxed{C=D^{T}D}$.

Suppose for a contradiction that $m>n$. Compose the given injective module homomorphism with the standard injection $B^n\hookrightarrow B^m$ that appends $m-n$ zero coordinates. This gives an injective endomorphism $T$ of the finite free module $B^m$ whose matrix has a zero final row.

Its characteristic polynomial has zero constant term, so the Cayley-Hamilton theorem givesInjectivity lets us cancel $T$. Repeating this argument eventually gives the identity endomorphism equal to zero. That would imply $B^m=0$, contrary to $B\ne 0$. Hence $m\le n$. This proves the rank inequality for an injection of finite free modules.

An $A$-module $M$ is a flat module when the tensor functor $-{\otimes }_{A}M$ preserves injections, equivalently when it is exact.

Suppose first that $M$ is flat. For any nonzero $a\in A$, tensor the injection $A\stackrel{a}{\to }A$ with $M$. The resulting map $M\stackrel{a}{\to }M$ is injective, so $am=0$ implies $m=0$. Thus $M$ is a torsion-free module.

Conversely, suppose $M$ is torsion-free over the principal ideal domain $A$. Every finitely generated submodule of $M$ is a finitely generated torsion-free module over a PID, hence a finite free module and therefore flat. The module $M$ is the filtered colimit of these submodules. Tensor products commute with filtered colimits, and filtered colimits of modules preserve exact sequences, so $M$ is flat. This proves that a torsion-free module over a principal ideal domain is flat.