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The exterior derivative is an $\mathbb{R}$-linear operator $d:{\Omega }^r(M)\to {\Omega }^{r+1}(M)$ which agrees with the differential of a smooth function on degree zero, satisfies $d^2=0$, and obeys the graded Leibniz rule

For the uniqueness of the exterior derivative from its axioms, first note that these axioms force locality. If $\alpha$ vanishes near $p$, choose a smooth bump function $\chi$ equal to one near $p$ and supported where $\alpha =0$. Then $0=d(\chi \alpha )=d\chi \wedge \alpha +\chi d\alpha$ gives $d\alpha (p)=0$. Thus $d\alpha (p)$ depends only on the form near $p$.

In a coordinate chart write $\alpha ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$. Since $d(d{x}^i)=d^2{x}^i=0$, the axioms forceThis also proves existence. Define $d$ by that formula in each chart. The product rule for partial derivatives gives the graded Leibniz rule, and symmetry of mixed partial derivatives against antisymmetry of the wedge product of differential forms gives $d^2=0$. On overlaps it still agrees with the differential of every function, by the chain rule, so the preceding uniqueness argument in the second coordinate system makes the formulas agree. They therefore glue to a global exterior derivative.

For a differential one-form $\omega ={\sum }_i{a}_{i}d{x}^i$, the exterior derivative of a one-form evaluated on vector fields follows directly:In $X(\omega (Y))-Y(\omega (X))$, the additional terms are ${\sum }_i{a}_i[X(Y^i)-Y(X^i)]=\omega ([X,Y])$, where the bracket is the Lie bracket of vector fields. Hence

The de Rham cohomology iswith zero denominator in degree zero. The Poincaré lemma states that a closed differential form of positive degree on a star-shaped open subset of Euclidean space is an exact differential form; consequently every closed positive-degree form is locally exact on a smooth manifold.

For the suspension isomorphism for top de Rham cohomology of spheres, put $U=S^n\setminus \{\text{south pole}\}$ and $V=S^n\setminus \{\text{north pole}\}$. Both are diffeomorphic to ${\mathbb{R}}^n$, while $W=U\cap V$ deformation retracts onto the equatorial $S^{n-1}$. The Mayer--Vietoris sequence for de Rham cohomology comes from the short exact sequence of cochain complexesThe final map is surjective: a partition of unity ${\chi }_U+{\chi }_V=1$ lets a form $\eta$ on $W$ be lifted by extending ${\chi }_V\eta$ to $U$ and $-{\chi }_U\eta$ to $V$ by zero. The Poincaré lemma makes $H^{n-1}(U),H^{n-1}(V),H^n(U),H^n(V)$ vanish when $n>1$. Exactness givesIf $j:S^{n-1}\hookrightarrow W$ is the equatorial inclusion, homotopy invariance of de Rham cohomology makes $j^{\ast }$ an isomorphism. Thereforeis injective, indeed an isomorphism. Concretely, choose local primitives $d{\alpha }_U=\omega {|}_U$, $d{\alpha }_V=\omega {|}_V$, and send $[\omega ]$ to $[j^{\ast }({\alpha }_U-{\alpha }_V)]$. Their difference is closed; changing either primitive changes it by an exact form, so this agrees with the cohomological construction.

A $d$-dimensional embedded submanifold $F\subseteq M^m$ has the subspace topology and, near each of its points, a manifold chart identifying it with a coordinate slice ${\mathbb{R}}^d\times \{0\}$. These slice charts for an embedded submanifold give its smooth structure, and its inclusion is a smooth embedding.

A regular value $q\in N^n$ of a smooth map $f:M^m\to N^n$ is one for which $d{f}_p:T_{p}M\to T_{q}N$ is surjective at every $p\in f^{-1}(q)$. If the preimage is nonempty this forces $m\ge n$.

The inverse function theorem used here says that if $G:U\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^m$ is smooth and $D{G}_p$ is an invertible matrix, then its restriction to a suitable neighbourhood of $p$ is a diffeomorphism onto an open neighbourhood of $G(p)$, with smooth inverse.

To prove the regular level set theorem, fix $p\in f^{-1}(q)$ and choose coordinates centred at $p,q$. The coordinate representative $\tilde{f}$ has a rank-$n$ Jacobian matrix. After permuting domain coordinates, its first $n$ columns form an invertible matrix. DefineIts Jacobian matrix is block triangular with invertible diagonal blocks, so the inverse function theorem makes $G$ a local coordinate change. In these coordinates, $f^{-1}(q)$ is precisely the slice with first $n$ coordinates zero. Thus it is an embedded submanifold of dimension $m-n$, withAn empty preimage is harmless; the local assertion is then vacuous.

Three equivalent formulations of orientability of a smooth manifold are: an oriented atlas whose transition Jacobian determinants are positive; a smoothly varying orientation of a vector space on each tangent space; and a nowhere-vanishing smooth top-degree differential form. An oriented atlas gives compatible oriented coordinate frames. A partition of unity glues their positive local top forms, and a nonzero top form declares a frame positive precisely when it evaluates positively on that frame. These constructions prove the equivalence.

For the orientability of a regular fibre, write $F=f^{-1}(q)$. The derivative gives a short exact sequence of vector bundlesThis is the reason the normal bundle of a regular fibre is trivial. Choose any fixed positive basis $b_1,\dots ,b_n$ of $T_{q}N$ and an ambient orientation on $M$. Declare a frame $v_1,\dots ,v_{m-n}$ of $T_{p}F$ positive whenis positive in $T_{p}M$, where $d{f}_p({\tilde{b}}_i)=b_i$. Changing a lift adds a tangent vector and leaves the determinant sign unchanged. Local smooth lifts exist because $df$ is surjective, so this rule varies smoothly and agrees on overlaps. It gives a global orientation on $F$. Only an orientation of the single vector space $T_{q}N$ was chosen; $N$ itself need not be orientable.

Sending a matrix to its last column gives the special orthogonal sphere fibrationThe action is transitive and the stabilizer of that column is $SO(n-1)$. Local orthonormal frames, obtained by Gram-Schmidt process near a chosen unit vector, give local sections and hence local bundle trivializations. This allows the long exact sequence of homotopy groups of a fibration to be applied.

At the bottom of the induction, $SO(1)$ is a point and $SO(2)$ is the rotation circle $S^1$. For $SO(3)$, identify ${\mathbb{R}}^3$ with imaginary quaternions. A unit quaternion acts by $v\mapsto qv{q}^{-1}$, preserving length and orientation. Every rotation through angle $\theta$ about a unit imaginary axis $u$ comes from $q=\cos (\theta /2)+u\sin (\theta /2)$. The kernel is precisely $\{1,-1\}$, since a quaternion commuting with all imaginary quaternions is real. Thus this gives the twofold covering space $S^3\to SO(3)$. Its simply connected total space is the universal cover, yielding ${\pi }_{1}SO(3)=\mathbb{Z}/2$. Coverings induce isomorphisms on higher homotopy groups, so ${\pi }_{2}SO(3)={\pi }_2{S}^3=0$ and ${\pi }_{3}SO(3)={\pi }_3{S}^3=\mathbb{Z}$.

For $n\ge 4$, both ${\pi }_1{S}^{n-1}$ and ${\pi }_2{S}^{n-1}$ vanish. The Serre fibration sequence therefore identifies ${\pi }_{1}SO(n-1)$ with ${\pi }_{1}SO(n)$ and propagates vanishing of ${\pi }_2$. Path-connectedness also propagates from connected fiber and base. Consequently the full low-degree answer isHere ${\pi }_0$ records the single connected component.

For the remaining third group use $SO(3)\to SO(4)\to S^3$. The relevant exact segment, using the supplied fourth homotopy group, isThe first homomorphism is zero since a torsion group cannot map nontrivially to $\mathbb{Z}$. Thus it reduces to a short exact sequence $0\to \mathbb{Z}\to {\pi }_{3}SO(4)\to \mathbb{Z}\to 0$. These groups are abelian and the final free group has a lift of its generator, so the sequence splits. ThereforeThese are the low homotopy groups of special orthogonal groups.

We give the detailed norm-index argument for a cyclic extension $L/K$ of p-adic fields; thus $K$ is a finite extension of ${\mathbb{Q}}_p$. Write $G=⟨\sigma ⟩$, $|G|=n$. For an additive $G$-module $M$, put $D=\sigma -1$ and $N=1+\sigma +\cdots +{\sigma }^{n-1}$. The two periodic Tate cohomology of a cyclic group groups and the Herbrand quotient arewhen these groups are finite. For multiplicative modules, $N$ is the product of conjugates and $D(z)=\sigma (z)/z$.

The periodic complex alternating $D$ and $N$ gives a six-term exact sequence for every short exact sequence of modules. Taking the alternating product of the finite group orders in that exact sequence provesFor a finite module, the counting fibers gives $|NM|=|M|/|\ker N|$ and $|DM|=|M|/|\ker D|$, so the two Tate groups have equal order. This proves the Herbrand quotient of a finite module is one. It follows that replacing a module by a finite-index submodule does not change its quotient, whenever the cohomology groups are finite.

Two basic calculations drive the norm computation. For the trivial module $\mathbb{Z}$, $D=0$ and $N$ is multiplication by $n$, so $h_G(\mathbb{Z})=n$. For a regular lattice $R[G]$, with $R={\mathbb{Z}}_p$, invariants are constant coefficient vectors and every such vector is a norm: put its constant coefficient in just one coordinate before applying $N$. The group kernel of $N$ consists of the coefficient vectors of sum zero. Successively taking partial sums solves the cyclic difference equation $Dx=y$ for every such vector $y$. Thus both Tate groups vanish and $h_G(R[G])=1$. This is cyclic cohomology of a regular lattice.

Let $U_L={\mathcal{O}}_L^{\times }$ and $U_L^r=1+{\mathfrak{P}}_L^r$. For sufficiently large $r$, the p-adic logarithm and p-adic exponential are inverse $G$-equivariant isomorphismsTo make the range explicit, $r>v_L(p)/(p-1)$ suffices: the higher terms in the logarithm and exponential have greater valuation than their leading terms, their series converge there, and the identities $\log (\exp x)=x$ and $\exp (\log (1+x))=1+x$ remain valid. Thus a deep multiplicative unit group becomes an additive ${\mathbb{Z}}_p$-lattice in $L$.

By the normal basis theorem, $L\cong K[G]$ as a $K[G]$-module. As a ${\mathbb{Q}}_p[G]$-module it is therefore ${\mathbb{Q}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$. Transport the regular lattice $M_0={\mathbb{Z}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$ into $L$. Both $M_0$ and ${\mathfrak{P}}_L^r$ are full $G$-stable ${\mathbb{Z}}_p$-lattices, so their intersection has finite index in each. The exact sequences with finite quotients show that the Tate groups of ${\mathfrak{P}}_L^r$ are finite and thatSince $U_L/U_L^r$ is finite, the exact-sequence formula for the Herbrand quotient gives $h_G(U_L)=1$. Finally, the valuation sequencehas trivial $G$-action on $\mathbb{Z}$. Hence the Herbrand quotient of the local multiplicative group is

For completeness, the denominator vanishes by Hilbert theorem 90. If $N_{L/K}a=1$, put $c_0=1$ and $c_j=a\sigma (a)\cdots {\sigma }^{j-1}(a)$. Independence of distinct field automorphisms lets us choose $t\in L$ with $b={\sum }_{j=0}^{n-1}{c}_j{\sigma }^j(t)\ne 0$. Since $c_n=1$, shifting the sum gives $\sigma (b)=b/a$. Thus $a=b/\sigma (b)$, a multiplicative coboundary. Therefore ${\hat{H}}^{-1}(G,L^{\times })=1$. The invariant subgroup of $L^{\times }$ is $K^{\times }$, soThis proves the cyclic local norm index without assuming local reciprocity. If the extension has ramification index $e$ and residue degree $f$, the valuation formula $v_K(Nz)=f{v}_L(z)$ shows that the valuation contribution to the index is $f$. A norm is a unit only when $z$ is a unit, so the remaining contribution isIn particular, unit norms are surjective in an unramified cyclic extension. The norm subgroup is open: on deep units its logarithm is ${\mathrm{Tr}}_{L/K}({\mathfrak{P}}_L^r)$, a nonzero full lattice in $K$, and therefore contains a sufficiently deep principal-unit group. This connects the index calculation to the open finite-index norm subgroups classified by local class field theory.

To finish, the Hasse norm theorem for a cyclic extension of number fields states that a nonzero element is a global norm if and only if it is a norm at every finite and infinite completion. Here is its Brauer-theoretic proof outline. Associate to $a\in K^{\times }$ the cyclic algebra $A=(L/K,\sigma ,a)$, with relations $z\ell =\sigma (\ell )z$ and $z^n=a$. The splitting criterion for a cyclic algebra says that it splits exactly when $a=Nc$: if $a=Nc$, multiplication by $L$ and the operator $c\sigma$ realize it as ${\mathrm{End}}_K(L)$; conversely, in a split matrix representation, the underlying $n$-dimensional space is one-dimensional over $L$, so $z$ must act as $c\sigma$ and $z^n=Nc$.

Apply this criterion at every completion, using the restricted cyclic character and any one completion of $L$ above that place. If $a$ is a local norm everywhere, every localized Brauer class is zero. Injectivity in the Albert-Brauer-Hasse-Noether theorem forces $[A]=0$ in $\mathrm{Br}(K)$; the global splitting criterion gives $a=Nc$. Conversely, a global norm is a local norm after completion, since the norm on $L{\otimes }_K{K}_v$ is the product of the component norms, whose images coincide in the Galois case. The cyclic hypothesis is essential; the stated local-global principle does not hold for arbitrary extensions.

Use the short exact sequence of coefficient groupsThe groups of singular chains are free, so applying the cochain construction gives a short exact sequence of cochain complexes. Its long exact sequence in cohomology has the required terms and reduction map $r_{\ast }$. Define the Bockstein homomorphism to be its connecting homomorphism.

Explicitly, represent $x$ by a mod-$p$ cocycle $a$ and lift it to an integral singular cochain $A$. Since $a$ is a cocycle, $\delta A$ is divisible by $p$. ThenThis is exactly the lift-and-coboundary construction for the coefficient sequence above: the mod-$p^2$ coboundary of $A$ lies in the image of $j$. The quotient cochain is a cocycle, since ${\delta }^{2}A=0$ and integral cochains have no $p$-torsion. Replacing $A$ by another lift adds $pC$ and changes the quotient by $\delta C$ modulo $p$; replacing the representative by a coboundary likewise leaves its cohomology class unchanged. Thus the definition is independent of choices.

A degree-zero cocycle is a function on points constant on each path component. Choose an integer representative for its value on each component. This produces an integral zero-cocycle lift, with zero coboundary. Hence the Bockstein vanishes on $H^0(X;{\mathbb{F}}_p)$ for every space.

For a nonzero example in every positive degree $i$, take the Moore space $X=S^i{\cup }_p{e}^{i+1}$, attaching the cell by a map of mapping degree $p$. Its positive-degree integral cellular cochain complex has the differential $\mathbb{Z}\stackrel{p}{\to }\mathbb{Z}$ from degree $i$ to $i+1$; when $i=1$, the preceding differential from degree zero is zero. Modulo $p$ the displayed differential vanishes, and $H^i(X;{\mathbb{F}}_p)=H^{i+1}(X;{\mathbb{F}}_p)={\mathbb{F}}_p$. Lift the degree-$i$ generator to the cellular cochain with value one. Its coboundary has value $p$, so dividing by $p$ gives the generator in degree $i+1$. Naturality of cellular and singular cohomology with coefficient sequences identifies this with the Bockstein on a cyclic Moore space. Therefore $\beta$ is nonzero, indeed an isomorphism, in the requested degree for every prime $p$.

Finally take integral lifts $A,B$ of cocycles representing $x\in H^i(X;{\mathbb{F}}_p)$ and $y\in H^j(X;{\mathbb{F}}_p)$. Write $\delta A=p{A}_1$ and $\delta B=p{B}_1$. The cochain cup product obeys the graded Leibniz identity, soDivide by $p$, reduce modulo $p$ and pass to cohomology. This proves the Bockstein derivation rule

For a sheaf of abelian groups $\mathcal{F}$ on a topological space $X$, putThese arbitrary families of germs form a sheaf, and restrictions are projections. Extending a family by zero proves that $C(\mathcal{F})$ is flasque. Sending a section to all its germs gives an injective sheaf map $\mathcal{F}\to C(\mathcal{F})$, since a section with every germ zero vanishes locally and hence globally.

Set $Q^{-1}=\mathcal{F}$, $I^j=C(Q^{j-1})$ and $Q^j=\mathrm{coker}(Q^{j-1}\to I^j)$, using the sheaf cokernel. Compose $I^j\to Q^j\to I^{j+1}$ to obtain the Godement resolutionIt is exact by construction, with flasque terms. Define sheaf cohomology by the cohomology of its global-section complex:The resolution principle for sheaf cohomology identifies this with the computation using any flasque resolution.

The elementary input for acyclicity is the flasque-kernel section-lifting lemma: in a short exact sequence with flasque kernel, sections of the quotient lift over every open subset. Consequently a quotient of two flasque sheaves is flasque, because a quotient section lifts and that lift extends. If $\mathcal{F}$ itself is flasque, induction through $0\to Q^{j-1}\to I^j\to Q^j\to 0$ makes every $Q^j$ flasque. The same lifting lemma makes these sequences exact on global sections. Hence the global-section complex has no positive-degree cohomology, provingThis deduction uses the lifting lemma, not the desired cohomological vanishing as an assumption.

For the rational-section construction, add two representative sections after restricting to the intersection of their dense domains. Multiply by a representative rational function in the same way. Finite intersections of dense open sets are dense, and further restricting both representatives does not change their resulting class. The module axioms hold on their common domain. Thus rational sections form a module over $\mathrm{Rat}(X)$; irreducibility is not needed for this module statement. When $X$ is irreducible, $\mathrm{Rat}(X)$ is the function field $K=k(X)$.

Suppose now that $\mathcal{F}$ is locally free and $X$ irreducible. A germ at $P$ is represented on a nonempty open neighbourhood, which is dense, so it defines a rational section. Two representatives of the germ agree on a neighbourhood of $P$ and therefore give the same class. If the class is zero, trivialize $\mathcal{F}$ on a neighbourhood of $P$. Its coefficient functions vanish on a dense open subset; regular functions on a reduced irreducible variety with this property vanish identically. The germ is therefore zero. This proves the stalk inclusion into rational sections of a locally free sheaf:Local freeness matters: a nonzero torsion germ can vanish on a dense open set.

On an irreducible algebraic curve, every proper closed subset is finite, and every open subset is quasi-compact because the Zariski space is Noetherian. Let $R=\mathrm{Rat}(\mathcal{F})$. The constant sheaf $\mathcal{R}(\mathcal{F})$ has sections $R$ on every nonempty open set: such opens are irreducible and hence connected. DefineRestriction drops the components outside the smaller open set. For a compatible family over an open cover of $U$, take a finite subcover by quasi-compactness. The union of the finite supports of these finitely many sections is finite. Their agreeing components therefore glue to a unique finite-support tuple on $U$. This verifies the sheaf gluing axiom, including arbitrary covers, rather than merely asserting that a presheaf direct sum is always a sheaf.

A rational section is regular on some dense open set, whose complement on a curve is finite. Its classes $[s{]}_P\in R/{\mathcal{F}}_P$ therefore have finite support, defining a sheaf map $\mathcal{R}(\mathcal{F})\to \mathcal{P}(\mathcal{F})$. At $P$, the first sheaf has stalk $R$. The second has stalk $R/{\mathcal{F}}_P$: any finite tuple can be restricted to a neighbourhood omitting all its other support points, and its $P$-component is unaffected by such restriction. The resulting sequence on stalks isExactness of sheaves can be tested on stalks, so this proves the rational principal-parts resolution on an algebraic curveBoth right-hand sheaves are flasque: restrictions for the constant rational-section sheaf are identities between nonempty opens, and restrictions for the rational principal-parts sheaf are projections, with extension by zero. ConsequentlySurjectivity on stalks has not been confused with surjectivity on global sections.

For an explicit failure of global surjectivity take $X={\mathbb{P}}^1$ and $\mathcal{F}=\mathcal{O}(-2)$. Use coordinate $z=X_1/X_0$ and frame $e_0=X_0^{-2}$ on the finite chart. At infinity use $w=1/z$ and $e_{\infty }=X_1^{-2}=z^{-2}{e}_0$. Prescribe the rational principal part of $z^{-1}{e}_0$ at zero and zero at every other point. If a rational section $r(z)e_0$ realized this tuple, then $r-z^{-1}$ would be regular at zero and $r$ would be regular at every other finite point. Thus $r=z^{-1}+p(z)$ with $p\in k[z]$. Regularity at infinity would requireto be regular at $w=0$. Its $w^{-1}$ term cannot cancel with any term of the polynomial contribution, whose powers are at most $-2$. This is impossible. Hence the global rational-principal-part map need not be surjective.

For an affine curve it is always surjective, and here is a direct construction. Write $X=\mathrm{Spec}A$, $K=\mathrm{Frac}A$ and $\mathcal{F}=\tilde{M}$. By the affine module sheaf construction, $R=M{\otimes }_{A}K=:M_K$ and ${\mathcal{F}}_P=M_{{\mathfrak{m}}_P}$. Choose representatives $v_P\in M_K$ for a prescribed finite tuple, and choose one nonzero $a\in A$ with $a{v}_P\in M$ for all of them. The zero set $T=V(a)$ is finite. The ring $A/(a)$ is Artinian, since it is Noetherian of dimension zero, and its Artinian decomposition into local factors givesPrescribe the class of $a{v}_P$ in the indicated component for each requested point in $T$, and zero in the other components. A requested point outside $T$ already has zero principal part because $a$ is a unit there. Lift the tuple to $w\in M$. Then $s=w/a\in M_K$ satisfies $s-v_P\in M_{{\mathfrak{m}}_P}$ at every requested point in $T$, and is regular at every other point of $T$. Outside $T$ it is regular because $a$ is invertible. It therefore realizes exactly the prescribed tuple. This proves affine principal-parts interpolation for a locally free sheaf and givesNo nonsingularity assumption was used; the construction applies to singular irreducible affine curves as well.

A filtration of a ring is an exhaustive increasing family of additive subgroups $F_{i}R$, indexed by the integers, with $1\in F_{0}R$ and $F_{i}R{F}_{j}R\subseteq F_{i+j}R$. A compatible filtration of a module on a right module satisfies $F_{i}M{F}_{j}R\subseteq F_{i+j}M$. The associated graded module has components $F_{i}M/F_{i-1}M$, and its action by the associated graded ring isChanging either representative changes the product only by an element of $F_{i+j-1}M$, so the action is well-defined.

The subspace filtration on $N$ is $F_{i}N=N\cap F_{i}M$. The quotient filtration is $F_i(M/N)=(F_{i}M+N)/N$. Inclusion induces an injective map $\mathrm{gr}N\to \mathrm{gr}M$, and the quotient induces a surjective map to $\mathrm{gr}(M/N)$. If $m\in F_{i}M$ maps to zero in its degree-$i$ quotient, write $m=n+m^{\prime }$ with $n\in N$ and $m^{\prime }\in F_{i-1}M$. Then $n\in N\cap F_{i}M$, so the kernel is exactly the image of $\mathrm{gr}N$. We have the short exact sequencewhich is exactness of associated graded modules for induced filtrations.

For the negative case, $F_{i}R=R$ for $i\ge 0$. Completeness means that $R\to {\underleftarrow{\lim }}_{q}R/F_{-q}R$ is an isomorphism; in particular the filtration of a ring is separated. If $x\in F_{-1}R$ and $r\in R$, then $(xr{)}^q\in F_{-q}R$. The geometric seriesconverges in the complete negative filtration and is a two-sided inverse of $1-xr$, by the finite geometric identity and continuity of multiplication. The unit criterion for the Jacobson radical now gives

Let $I$ be any right ideal. The associated graded module $\mathrm{gr}I$ for its subspace filtration is a homogeneous ideal on the right in $\mathrm{gr}R$. If $\mathrm{gr}R$ is a right Noetherian ring, choose homogeneous generators that lift to $x_1,\dots ,x_s\in I$, with degrees $d_1,\dots ,d_s\le 0$.

For $x\in I\cap F_{d}R$, cancel its degree-$d$ symbol by a sum ${\sum }_i{x}_i{a}_{i,0}$ with $a_{i,0}\in F_{d-d_i}R$. The remainder belongs to $I\cap F_{d-1}R$. Repeat the cancellation in each succeeding degree. After $q$ steps,For every $i$, the coefficient series converges to some $a_i\in R$ by completeness. Passing to the limit gives $x={\sum }_i{x}_i{a}_i$. Thus these finitely many $x_i$ generate the actual right ideal $I$; no assumption that $I$ is closed is needed. Since every right ideal is finitely generated, $R$ is right Noetherian, by complete negative filtered-graded transfer of Noetherianity.

The prime radical of a noncommutative ring is $N(R)={\bigcap }_{P}P$, with $P$ ranging over all prime ideals of a noncommutative ring. Here primeness means that for two-sided ideals $A,B$, $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$. A minimal prime over an ideal $I$ is a prime ideal containing $I$ with no strictly smaller prime ideal containing $I$.

The ascending chain condition on left ideals also applies to two-sided ideals. We first prove by Noetherian induction that every proper $I$ contains a finite product of prime ideals, each containing $I$. If there were counterexamples, choose a maximal one. It is not prime, so there are strictly larger two-sided ideals $A,B$ with $AB\subseteq I$. Neither is a counterexample; products of prime ideals contained in $A$ and in $B$ then concatenate to a product contained in $I$, a contradiction.

Write $P_1\cdots P_t\subseteq I\subseteq P_j$. Every prime ideal $P$ over $I$ contains at least one $P_j$, by primeness. The inclusion-minimal members of the finite list $P_j$ are therefore precisely the minimal primes over an ideal $I$: any smaller prime over $I$ would contain a member of that same list. Consequently there are finitely many. Replacing each factor $P_j$ by a minimal prime contained in it giveswith each $Q_j$ minimal over $I$. Repeated factors may be necessary; in $k[\epsilon ]/({\epsilon }^m)$ the sole minimal prime $(\epsilon )$ requires its $m$th power to reach zero.

Apply this to $I=0$. Since $N(R)\subseteq Q_j$ for every $j$,Thus the prime radical is a nilpotent ideal, as in finite minimal primes and nilpotent prime radical.

Now assume $R$ is commutative. A prime ideal $P$ is an associated prime of a module $M$ when $P={\mathrm{Ann}}_R(m)$ for some $0\ne m\in M$. Every nonzero module over $R$ has an associated prime: the ascending chain condition gives a maximal annihilator of a nonzero element, and maximal annihilator of a module element is prime.

Since the given $M$ is a Noetherian module, repeatedly choosing such a cyclic submodule in the next quotient yields a finite prime filtrationThe process terminates by the ascending chain condition. In a short exact sequence $0\to U\to M\to W\to 0$, one has $\mathrm{Ass}(M)\subseteq \mathrm{Ass}(U)\cup \mathrm{Ass}(W)$. Indeed, if $P=\mathrm{Ann}(m)$ and $Rm\cap U=0$, the image of $m$ in $W$ still has annihilator $P$. Otherwise choose $0\ne am\in U$; then $a\notin P$, and primeness gives $\mathrm{Ann}(am)=P$. As $\mathrm{Ass}(R/P_i)=\{P_i\}$, induction along the prime filtration provesIn particular there are only finitely many associated primes of a module.

Finally let $P$ be a minimal prime ideal of $R$. Its localization at a prime ideal $R_P$ has exactly one prime, its maximal ideal $P{R}_P$. The preceding nilpotence result makes this maximal ideal nilpotent. Choose $0\ne y\in R_P$ annihilated by $P{R}_P$, for example a nonzero element in its last nonzero power, or $1$ if $P{R}_P=0$. Since elements outside the maximal ideal are units, ${\mathrm{Ann}}_{R_P}(y)=P{R}_P$.

Write $y=r/s$. The element $r/1$ is nonzero and also has annihilator $P{R}_P$. Generate $P$ by $p_1,\dots ,p_{\ell }$. For each $i$, choose $u_i\notin P$ with $u_i{p}_{i}r=0$, and put $u={\prod }_i{u}_i$. Then $m=ur$ remains nonzero after localization, while $Pm=0$. If $am=0$, localization forces $a/1\in P{R}_P$, and hence $a\in P$. Therefore ${\mathrm{Ann}}_R(m)=P$. This proves that minimal primes are associated primes.

Fix a finite-dimensional nonnegative filtration of a ring giving $R$ its almost commutative algebra structure, and choose a good filtration of a module on $M$. When $\mathrm{gr}R$ is a standard graded algebra, the Hilbert-Serre theorem makes the cumulative Hilbert function $h_M(j)={\dim }_k{F}_{j}M$ eventually a polynomial. For $M\ne 0$, define the dimension of a filtered module and multiplicity of a filtered module byUse $d(0)=-\infty$ and $m(0)=0$. With positive weighted generators instead of a standard grading, the cumulative Hilbert function is eventually a quasipolynomial with a common positive leading coefficient across residue classes. Equivalently, $h_M(j)=c{j}^d+O(j^{d-1})$, and the definitions are $d(M)=d$ and $m(M)=d!c$. The common leading coefficient follows from monotonicity: interlacing consecutive residue classes forces their leading coefficients to agree. For a standard grading, multiplicity is a positive integer, since an integer-valued polynomial has an integral expansion in the binomial polynomials; weighted normalization can instead give a positive rational number.

Two good filtrations of a module for this fixed algebra filtration bound each other after fixed index shifts. The corresponding Hilbert functions therefore satisfy $h_F(j-a)\le h_G(j)\le h_F(j+b)$ for fixed $a,b$. Comparing their leading growth shows that the degree and leading coefficient agree. Thus $d(M)$ and $m(M)$ are independent of the module filtration; the fixed algebra filtration is part of the multiplicity convention.

For a submodule $N$, use its subspace filtration and the quotient filtration on $M/N$. These are good because $\mathrm{gr}R$ is Noetherian. The short exact sequence of filtered pieces givesNonzero leading coefficients are positive, soIf $d(N)=d(M/N)$, those leading coefficients add andThis is dimension and multiplicity in a filtered exact sequence.

For the Weyl algebra $A_n(k)$ in characteristic zero, Bernstein inequality for Weyl algebra modules asserts $d(M)\ge n$ for every nonzero finitely generated module $M$. Write its generators as $x_i,D_i$, with $[D_i,x_j]={\delta }_{ij}$ and all other generator commutators zero. The ordered monomials $x^{\alpha }{D}^{\beta }$ give the Bernstein filtrationTake a finite-dimensional generating subspace $V\ne 0$ of $M$ and put $M_j=A_{j}V$.

To prove faithful finite-step action of a Weyl algebra, consider $0\ne a\in A_j$ and choose a nonzero coefficient $c_{\alpha \beta }$ of a monomial of maximal total degree $d\le j$. Applying ${\prod }_i(\mathrm{ad}{D}_i{)}^{{\alpha }_i}(\mathrm{ad}{x}_i{)}^{{\beta }_i}$ to $a$ givesIndeed, these commutators differentiate the ordered coefficients: lower-degree monomials vanish, and another degree-$d$ monomial can survive only if all its exponents dominate $(\alpha ,\beta )$, which forces equality. This is scalar extraction by iterated Weyl commutators.

Expanding these iterated commutators writes $\lambda ={\sum }_{\ell }{u}_{\ell }a{v}_{\ell }$ with $u_{\ell },v_{\ell }\in A_j$. If $a$ annihilated $M_j$, every summand would annihilate $V$, contrary to $\lambda V\ne 0$. Therefore the action map is injective:Taking dimensions and comparing polynomial degrees giveswhich proves Bernstein inequality for Weyl algebra modules. The polynomial representation $k[x_1,\dots ,x_n]$ has dimension $n$, so the bound is sharp.

The associated graded ring of the Weyl algebra is a polynomial ring, so ascending filtered-graded transfer of Noetherianity makes $A_n(k)$ a left Noetherian ring. Thus every submodule of a finitely generated module is finitely generated. If $d(M)=n$, every nonzero subquotient has dimension at most $n$ by the exact-sequence formula, and at least $n$ by Bernstein inequality for Weyl algebra modules. Every such subquotient therefore has dimension $n$ and positive integral multiplicity.

In a strictly descending chain $M=M_0⊋M_1⊋\cdots$, each nonzero quotient $M_i/M_{i+1}$ consumes at least one unit of multiplicity, by dimension and multiplicity in a filtered exact sequence. Hence there are at most $m(M)$ strict steps. This proves the descending chain condition, so $M$ is Artinian. In fact this holonomic Weyl algebra module has module length at most $m(M)$, as in multiplicity bounds the length of a holonomic Weyl module.

Choose an ordered affine open cover $\mathcal{U}=(U_i)$ of the separated scheme. For a quasi-coherent sheaf $\mathcal{F}$, the Čech cochain groups and differential areTerms corresponding to empty intersections are zero. Restriction commutes with restriction, so the terms in $d^2$ cancel in pairs. DefineHere an affine cover computes the Čech groups independently of the chosen affine cover. The reason is that finite intersections of affine opens in a separated scheme are affine: the intersection of two is the inverse image of the closed diagonal inside their affine product. Repeating gives the finite-intersection assertion. Quasi-coherent sheaves have no higher cohomology on these intersections. Apply an injective resolution of the sheaf and form the double complex of its Čech cochains. In one direction the higher cohomology vanishes on every affine intersection, leaving precisely the displayed Čech complex. In the other direction the augmented Čech complex of the flasque resolution terms is exact, leaving the global-section resolution computing sheaf cohomology. Thus the two cohomologies agree, and refinement maps between affine-cover complexes induce the same groups. No such affine-intersection claim is being made for arbitrary nonseparated schemes.

For an exact sequence of quasi-coherent sheaves, sections on an affine intersection give an exact sequence of modules, including surjectivity: this follows from the equivalence between quasi-coherent sheaves and modules on an affine scheme. Taking products yields a short exact sequence of Čech cochain complexes. To describe the connecting map, lift a degree-$p$ cocycle in $C^p({\mathcal{F}}^{\prime \prime })$ to a cochain in $C^p(\mathcal{F})$. Its differential maps to zero in $C^{p+1}({\mathcal{F}}^{\prime \prime })$ and therefore is a cocycle in $C^{p+1}({\mathcal{F}}^{\prime })$. Changing the lift adds a coboundary there; changing the original representative does the same. This defines the connecting homomorphism. The same lift-and-differentiate argument shows that a class is in its kernel exactly when it lifts to a cocycle in the middle complex, and verifies exactness at the other terms. Thus the resulting sequence isThe sequence must be in quasi-coherent sheaves for this affine-sections argument; that is the intended category here.

For the punctured affine plane, use $U=D(x)$ and $V=D(y)$. Their rings are $k[x^{\pm 1},y]$ and $k[x,y^{\pm 1}]$, and the intersection ring is $k[x^{\pm 1},y^{\pm 1}]$. The two-open Čech complex isConsequentlyThe denominator contains precisely the span of Laurent monomials for which at least one exponent is nonnegative. Every Laurent polynomial has a unique monomial expansion, so the remaining monomials are linearly independent in the quotient and span it. Thereforea countably infinite-dimensional vector space. The brackets here denote cohomology classes. This calculates the Čech cohomology of the punctured affine plane, rather than merely observing that some nonzero cohomology exists.

Use the defining criterion that a flat module $E$ makes $u\otimes 1_E$ injective for every injective module homomorphism $u$. By right exactness of the tensor product, preserving injections is equivalent to preserving every short exact sequence.

First assume $E$ is flat over $A$, and fix a maximal ideal $\mathfrak{m}$. Let $u:U\hookrightarrow V$ be an injection of $A_{\mathfrak{m}}$-modules. Regard it as an injection of $A$-modules. There are natural module isomorphismsFor example, $u_0\otimes (e/s)$ maps to $s^{-1}{u}_0\otimes e$, with inverse $u_0\otimes e\mapsto u_0\otimes (e/1)$. These are well-defined because every $s\notin \mathfrak{m}$ already acts invertibly on $U$. Under these isomorphisms the tensor map is $u{\otimes }_A{1}_E$, which is injective by flatness. Thus $E_{\mathfrak{m}}$ is flat over $A_{\mathfrak{m}}$.