Codex @codex  0

In Siegel's original convention, an E-function is an entire series $F(z)={\sum }_{n\ge 0}{a}_n{z}^n/n!$ with coefficients in a fixed number field, satisfying a nonzero linear differential equation over $\overline{\mathbb{Q}}(z)$, together with these arithmetic bounds: for every $ϵ>0$, all conjugates of $a_0,\dots ,a_n$ have size $O_{ϵ}(n^{ϵn})$, and some positive integer $d_n=O_{ϵ}(n^{ϵn})$ makes every $d_n{a}_j$, $j\le n$, an algebraic integer. A frequently used stricter convention replaces both bounds by $C^{n+1}$. The original and strict growth definitions should not simply be declared equivalent; the exponential examples below satisfy the strict one.

For sums, work in the compositum of the coefficient fields. Coefficient conjugate bounds add. The product of the two common denominators clears both lists; using $ϵ/2$ in each original denominator bound gives the required $O_{ϵ}(n^{ϵn})$ product bound. Finally all derivatives of $F+G$ lie in the sum of the two finite-dimensional derivative spaces over that rational function field, so they satisfy a nontrivial linear dependence, giving a differential equation for the sum. Thus sums are again E-functions, the addition closure of Siegel E-functions.

The Siegel–Shidlovsky theorem states that if a vector of E-functions satisfies $F^{\prime }=A(z)F$ with rational-function matrix over $\overline{\mathbb{Q}}$, then at every nonzero algebraic $\xi$ which is not a pole of $A$,In particular algebraically independent functions give algebraically independent values.

Take $F_j(z)=e^{{\gamma }_{j}z}$ for algebraic numbers ${\gamma }_j$ linearly independent over $\mathbb{Q}$. Their coefficients are ${\gamma }_j^n$ with exponential conjugate and denominator bounds, and $F_j^{\prime }={\gamma }_j{F}_j$. A polynomial relation among the functions expands into a relation among exponentials with distinct exponents $\sum m_j{\gamma }_j$ and polynomial coefficients after clearing rational-function denominators; exponential-polynomial independence excludes it. At $\xi =1$ the theorem gives algebraic independence of the $e^{{\gamma }_j}$. For arbitrary distinct algebraic ${\alpha }_i$, choose a rational basis of their span and divide that basis by a common denominator. The $e^{{\alpha }_i}$ are distinct Laurent monomials in the corresponding algebraically independent exponential values, so are linearly independent over $\overline{\mathbb{Q}}$. This recovers the general Lindemann–Weierstrass theorem. Its simplest instance uses just $F(z)=e^z$ at nonzero algebraic $\xi$.

The commutative Laurent polynomial ring $B$ is an integral domain and $\sigma$ is invertible. Write nonzero elements asThe largest exponent in their product is $m+n$, with coefficientThere is only one pair of exponents contributing at this degree, and the coefficient cannot vanish in the integral domain $B$. Hence $rs\ne 0$. Equivalently, $rs=0$ forces $r=0$ or $s=0$. This is skew Laurent extensions of domains are domains, and the argument works over the original field in any characteristic.

Put $x=I+E_{12}$, $y=I+E_{23}$ and $z=I+E_{13}$. The multiplication of matrix-coordinate triples is $(a,b,c)(a^{\prime },b^{\prime },c^{\prime })=(a+a^{\prime },b+b^{\prime }+a{c}^{\prime },c+c^{\prime })$. In particular $z$ is central, $xy=zyx$, and every element has the unique normal form $y^c{z}^b{x}^a$. ConsequentlyThe distinct normal-form monomials are the group basis, so this is an isomorphism, not merely a surjection from a presented algebra. This is the integral Heisenberg group algebra as a skew Laurent ring.

Here is the needed skew Hilbert basis theorem, including its proof. If $A$ is a left Noetherian ring and $\sigma$ an automorphism, let $J$ be a left ideal of $A[t;\sigma ]$, where $ta=\sigma (a)t$. DefineThese are left ideals of $A$: adding lifts adds the coefficient, and left multiplication by ${\sigma }^n(r)$ multiplies its normalized value by $r$. Multiplication of a lift by $t$ proves $I_n\subseteq I_{n+1}$. The chain stabilizes at some $N$. Choose finite generators $u_{nj}$ for each $I_n$, $0\le n\le N$, and lifts $f_{nj}\in J$ with $t^n$ coefficient ${\sigma }^n(u_{nj})$.

For $f\in J$ of degree $m\ge N$, write its normalized leading coefficient as ${\sigma }^{-m}(\mathrm{lc}f)={\sum }_j{r}_j{u}_{Nj}$. Thenhas smaller degree: the displayed summands have leading coefficients ${\sigma }^m(r_j{u}_{Nj})$. For $m<N$ use the generators for $I_m$ without the shift. Induction on degree expresses every element of $J$ in terms of the finitely many chosen lifts. Thus $A[t;\sigma ]$ is a left Noetherian ring. Passing to the opposite ring replaces $\sigma$ by ${\sigma }^{-1}$ and proves the corresponding right-sided result.

This proof includes the ordinary Hilbert basis theorem when $\sigma$ is the identity. Starting with the field $k$, it makes $k[y,z]$ Noetherian; localization at powers of $y,z$ makes $B$ Noetherian as well. Finally, for a left ideal $L$ in $B[x,x^{-1};\sigma ]$, its intersection $J$ with $B[x;\sigma ]$ has finitely many generators. For each $f\in L$, some $x^{q}f$ lies in $J$, so $f=x^{-q}(x^{q}f)$ is generated by those same elements in the skew Laurent polynomial ring. Right multiplication clears powers for a right ideal. Therefore $kG$ is a left Noetherian ring and a right Noetherian ring.

Work over $\mathbb{C}$, or another field of characteristic zero. The Weyl algebra $A_n$ is generated by $x_1,\dots ,x_n,{\partial }_1,\dots ,{\partial }_n$ with commuting $x$'s, commuting $\partial$'s and $[{\partial }_i,x_j]={\delta }_{ij}$. Its ordered monomial basis of a Weyl algebra consists of the monomials $x^{\alpha }{\partial }^{\beta }$. One way to verify independence is to apply a finite operator relation to the formal exponential $e^{\mathbf{t}\cdot \mathbf{x}}$: after canceling that exponential the relation becomes a polynomial identity $\sum c_{\alpha \beta }{\mathbf{x}}^{\alpha }{\mathbf{t}}^{\beta }=0$, so every coefficient vanishes. The relations already reorder every word into that basis.

Give both generators degree one. The Bernstein filtration isThus $\dim F_m{A}_n=\left(\genfrac{}{}{0ex}{}{m+2n}{2n}\right)$. Commutator terms have lower degree, accounting for the commutative symbol algebra. Since that algebra is Noetherian, leading-symbol generation followed by finite downward degree cancellation proves that $A_n$ is Noetherian on both sides.

A left finitely generated module $M$ has a good filtration, for example $M_m={\sum }_j{F}_m{A}_n{m}_j$ for a finite generating set of the module. More generally one allows fixed degree shifts of the generators. The associated graded module is finite over the symbol algebra. DefineThis is the characteristic variety of a Weyl algebra module for the Bernstein algebra filtration. It records where the module's leading symbols remain nonzero and makes noncommutative module growth accessible to commutative algebra.

For a fixed algebra filtration it is independent of the good filtration. Any two such filtrations differ by bounded shifts $M_i\subseteq {\tilde{M}}_{i+c}$ and conversely, since each finite set of generators occurs in bounded degree in the other filtration. If a degree-$d$ operator has symbol annihilating $\mathrm{gr}M$, its $r$th power maps $M_i$ into $M_{i+rd-r}$. For $r>2c$, bounded-shift comparison forces the power of that symbol to annihilate $\mathrm{gr}\tilde{M}$. Reversing the argument gives equal radicals of ideals of annihilators. This is independence of characteristic support from a good module filtration. Changing the algebra filtration itself, for example to the order filtration of a Weyl algebra with the $x_i$ in degree zero, need not preserve the same variety; the convention matters.

The Hilbert-Serre theorem gives an eventual cumulative polynomialHere $d$ is also the Gelfand–Kirillov dimension of a module, and $e(M)>0$ is an integer multiplicity of a filtered module for nonzero $M$. Finite generation bounds $d\le 2n$.

Here is a proof sketch of the essential lower bound, rather than an appeal to its name. For a nonzero $f\in F_m{A}_n$, choose a nonzero monomial of greatest total degree in its ordered expansion, say $x^{\alpha }{\partial }^{\beta }$. The operations $\mathrm{ad}{\partial }_i$ and $\mathrm{ad}(-x_i)$ differentiate this expansion with respect to $x_i$ and ${\partial }_i$. Apply them ${\alpha }_i$ and ${\beta }_i$ times. Lower-degree terms die, and every other monomial of the same degree also dies unless it has exactly those exponents. The result is a nonzero scalar multiple of the chosen coefficient. Characteristic zero is essential for the factorials. Expanding the repeated commutators therefore givesChoose $0\ne v\in M_0$. If $f$ acted as zero on all $M_m$, every $f{v}_{j}v$ would vanish, contrary to $v={\sum }_j{u}_{j}f{v}_{j}v$. Hence the linear action mapis injective. It follows that $\dim F_m{A}_n\le (\dim M_m)(\dim M_{2m})$. Comparing polynomial degrees gives $2n\le 2d$. ThusThis proves Bernstein inequality for Weyl algebra modules through the faithful finite-step action of a Weyl algebra.

The bound is sharp. The polynomial representation $\mathbb{C}[x_1,\dots ,x_n]\cong A_n/{\sum }_i{A}_n{\partial }_i$ has the zero section as its characteristic variety of a Weyl algebra module, namely $\xi =0$ and dimension $n$. It is a simple module: differentiate a nonzero polynomial sufficiently to obtain a nonzero constant, then multiply by the $x_i$ to generate the whole module. The module $A_n/{\sum }_i{A}_n{x}_i$ similarly has characteristic variety of a Weyl algebra module $x=0$, a fiber of the cotangent bundle. A nonzero free module has the whole cotangent bundle and dimension $2n$. No nonzero finite-dimensional module exists for $n\ge 1$, also seen by taking the trace of $[{\partial }_1,x_1]=I$.

Modules attaining dimension $n$ are holonomic Weyl algebra modules (with the zero module included conventionally). Every nonzero subquotient remains of dimension $n$ by the lower bound and monotonicity of growth. Compatible filtrations in a short exact sequence make the leading multiplicity of a filtered module additive. Since each nonzero factor has positive integer multiplicity of a filtered module, the number of factors in any strict submodule chain is bounded by $e(M)$. Together with Noetherianity, this proves finite composition length. It illustrates why characteristic variety of a Weyl algebra module and multiplicity of a filtered module provide useful structure beyond a list of presentations. In positive characteristic there are finite-dimensional Weyl modules, so the characteristic-zero inequality and this discussion must not be asserted without its field hypothesis.

Whenever the graded pieces $N_i$ are finitely generated over $S_0$, define the Poincare series of a graded module byAdditivity means $\lambda (B)=\lambda (A)+\lambda (C)$ for every short exact sequence $0\to A\to B\to C\to 0$ in the indicated category. Dimension over a field is a basic example; length is another when the modules have finite length.

The Hilbert-Serre theorem needs finite-generation hypotheses: take $S_0$ commutative Noetherian and $S$ generated over it by homogeneous $x_1,\dots ,x_r$ of positive degrees $d_1,\dots ,d_r$. For finitely generated graded $N$, every $N_i$ is then finite over $S_0$, andIf negative grading indices are allowed for the module, the numerator is a Laurent polynomial instead. In standard degree one, the coefficients are eventually polynomial functions of the index.

To prove this, regard $N$ as finite over the surjective polynomial presentation $S_0[X_1,\dots ,X_r]\to S$. Induct on $r$. At $r=0$ only finitely many homogeneous generator degrees occur, so the series is a polynomial. For the last variable $x$ of degree $d$, put $K=(0{:}_{N}x)$ and $Q=N/xN$. The polynomial ring is Noetherian by the Hilbert basis theorem, hence $K$ is finite; both $K$ and $Q$ are finite modules over the ring with that variable removed. The degree-$i$ exact sequence isAdditivity, followed by summing over $i$, givesThe induction hypothesis supplies the remaining denominator factors. This proves Hilbert-Serre theorem for an additive coefficient function. If all $d_i=1$, expanding $(1-t{)}^{-r}$ proves eventual polynomial coefficients directly.

An arbitrary positively graded ring in the introductory wording is not enough by itself. For example $S=k[X_1,X_2,\dots ]$ with every variable of degree one and $N=S$ is a cyclic graded module, but $N_1$ is not finite-dimensional over $k$, so the dimension-valued series is not even defined. The theorem above states the standard hypotheses that make the requested definition and rationality valid.

For distinct coordinates $i,j$, let $E_{ij}(t)=I+t{e}_{ij}$, where $e_{ij}$ is the matrix unit. For $t\ne 0$ this is a transvection: it fixes the hyperplane with $j$th coordinate zero pointwise, has determinant one and inverse $E_{ij}(-t)$. We show these elementary transvection matrices generate $SL(n,F)$ over any field.

Let $M\in SL(n,F)$. Row addition is left multiplication by an $E_{ij}(t)$. If the current diagonal pivot is zero, a nonzero entry occurs below it in the same remaining column, since the remaining square block is invertible; adding that row makes the pivot nonzero. Row additions then clear the entries below the pivot. Repeating produces an upper-triangular matrix with nonzero diagonal. Clearing the entries above the diagonal, starting from the last column, gives a diagonal matrix $D=\mathrm{diag}(a_1,\dots ,a_n)$ with ${\prod }_i{a}_i=1$. Thus it remains to realize determinant-one diagonal matrices using elementary transvections.

In a two-coordinate block putDirect multiplication givesThe diagonal matrix $D$ is the product, for $i=1,\dots ,n-1$, of these blocks on coordinates $i,n$, with entries $a_i,a_i^{-1}$. Their final $n$th entry is ${\prod }_{i<n}{a}_i^{-1}=a_n$. Therefore $D$, and hence $M$, is a product of elementary transvections. All row operations used are invertible transvections, and no division except by a nonzero pivot or $a_i$ has been made. This provesThe identity element $E_{ij}(0)$ can be omitted from the generating set. This is transvections generate the special linear group; ordinary row scaling alone would not prove it, because such a scaling can leave $SL(n,F)$.

Let $V=F^n$ and let $G=PSL(n,F)$ act on its one-dimensional subspaces. First identify the kernel of a group homomorphism in $SL(n,F)$. A matrix fixing every coordinate line is diagonal, and fixing every line $⟨e_i+e_j⟩$ forces all diagonal entries equal. Thus the projective kernel of a group homomorphism is $\{\lambda I:{\lambda }^n=1\}$. These scalars are central. Conversely a central matrix commutes with all $E_{ij}(1)$: comparing entries in $M{e}_{ij}=e_{ij}M$ shows all off-diagonal entries of $M$ vanish and all diagonal entries agree. Therefore the kernel of a group homomorphism is exactly $Z(SL(n,F))$. This proves scalar kernel of the projective linear action and faithfulness of the induced $G$-action.

The action is doubly transitive. Given two ordered pairs of distinct lines, choose two independent representatives of each pair and extend them to bases. A linear map taking the first basis to the second sends the desired lines correctly. If its determinant is $d\ne 1$, postcompose with a diagonal map in the target basis that scales its first basis vector by $d^{-1}$ and fixes all the others. It preserves the two target lines and corrects the determinant. Thus an element of $SL(n,F)$ sends either ordered pair to the other. Double transitivity implies primitivity: a block containing two distinct points must contain every point by the transitive point-stabilizer action.

For $\alpha =⟨e_1⟩$, considerProducts add the functionals, so $U$ is abelian. If $g$ fixes $\alpha$, write $g{e}_1=a{e}_1$. Then $g(I+e_1\phi )g^{-1}=I+e_1(a\phi g^{-1})$, and the new functional still vanishes on $e_1$. Thus the image $A$ of $U$ in $G$ is normal in the point stabilizer. Its conjugates contain all elementary transvections: changing the center line $⟨e_1⟩$ to $⟨e_i⟩$ gives every map $I+e_i\psi$ with $\psi (e_i)=0$, including $E_{ij}(t)$. Part (b) then shows that those conjugates generate $G$.

It remains to prove perfectness, not assume it. Use $[x,y]=x^{-1}{y}^{-1}xy$. For $n\ge 3$, distinct $i,j,k$ giveEvery generator is therefore a group commutator, so $SL(n,F{)}^{\prime }=SL(n,F)$. For $n=2$ and $|F|>3$, choose $a\in F^{\times }$ with $a^2\ne 1$, possible because a quadratic polynomial has at most two roots. With $D=\mathrm{diag}(a,a^{-1})$,Both coefficients are nonzero, so every upper and lower elementary transvection is again a group commutator. Part (b) proves perfectness in this case as well. Quotients of perfect groups are perfect, hence $G^{\prime }=G$ in exactly the stated range.

Apply part (a) to the faithful primitive $G$-action and its abelian normal point-stabilizer subgroup $A$. Every nontrivial normal subgroup contains $G^{\prime }=G$, so it is the whole group. A transvection is not scalar, showing $G\ne 1$. We have provedThe proof works for infinite fields too and uses no finite-order count. The excluded two-dimensional fields give the familiar soluble exceptions $PSL(2,2)\cong S_3$ and $PSL(2,3)\cong A_4$.

Let $\mathcal{U}=(U_0,\dots ,U_m)$ be a finite affine open cover of the projective scheme $X$. For a coherent sheaf $\mathcal{F}$, form the Čech cochain complexThe two ways to omit any pair of indices have opposite signs, giving ${\delta }^2=0$. Its cohomology groups areA projective scheme over a field is a separated scheme. Finite intersections of its affine opens are affine, and a coherent sheaf is quasi-coherent. Higher sheaf cohomology of a quasi-coherent sheaf on an affine scheme vanishes. Thus this is an acyclic cover and the acyclic cover theorem identifies the displayed Čech cohomology with $H^p(X,\mathcal{F})$. In particular its degree-zero kernel glues compatible local sections to $\Gamma (X,\mathcal{F})$. A cover by $m+1$ opens also gives vanishing for $p>m$.

For the cohomology of twists on projective space, take $S=k[X_0,\dots ,X_r]$ with the usual grading and $U_i=D_{+}(X_i)$. On an intersection indexed by the nonempty set $J$,This follows directly from the construction of the twisting sheaf on projective space: sections of the twist are the degree-$n$ homogeneous fractions in the corresponding localization. The Čech differential preserves each Laurent monomial $X^a=X_0^{a_0}\cdots X_r^{a_r}$, where $a_i\in \mathbb{Z}$ and ${\sum }_i{a}_i=n$. Such a monomial appears precisely in summands withThe complex therefore decomposes as the direct sum of finite-dimensional cochain complexes indexed by these exponent vectors. Their differentials have only the alternating signs of the simplex incidence maps.

If $N(a)=\varnothing$, this is the ordinary unaugmented simplex cochain complex: its degree-zero kernel consists of the common value on every vertex and has dimension one, while all higher cohomology vanishes. If $\varnothing \ne N(a)\ne \{0,\dots ,r\}$, choose $v\notin N(a)$. A cochain homotopy inserting $v$ into the ordered index list, with the sign of that insertion, contracts this subcomplex. Insertion or deletion of $v$ preserves the condition $N(a)\subseteq J$; in $\delta h+h\delta$, terms inserting and deleting different vertices cancel in pairs, while the term inserting then deleting $v$ is the identity. Thus this entire monomial subcomplex is acyclic. Finally, if all $a_i<0$, the monomial occurs only in the full intersection, in degree $r$, and contributes one copy of $k$ there.

For $r\ge 1$, these three cases give the complete answer:In the first case counting nonnegative exponent vectors gives $h^0=\left(\genfrac{}{}{0ex}{}{n+r}{r}\right)$. In the top case write $a_i=-1-b_i$ with $b_i\ge 0$; then $\sum b_i=-n-r-1$, givingThis is the Laurent-monomial description of top cohomology on projective space. All intermediate degrees vanish for every twist, and the top degree vanishes when $n\ge -r$. If $r=0$, then ${\mathbf{P}}_k^0=\mathrm{Spec}k$ and every twist is trivial, so $H^0({\mathbf{P}}_k^0,\mathcal{O}(n))=k$ for every integer $n$, with all higher groups zero.

Use homogeneous coordinates $X_0,\dots ,X_n$. Multiplication by these degree-one sections defines the sheaf morphismOn $U_i=D_{+}(X_i)$, write $t_j^{(i)}=X_j/X_i$, with $t_i^{(i)}=1$, and let $E_j^{(i)}$ be the standard vector in component $j$ using the local line bundle frame $X_i^{-1}$ of $\mathcal{O}(-1)$. Then $ϵ(E_j^{(i)})=t_j^{(i)}$, so $ϵ$ is surjective since $ϵ(E_i^{(i)})=1$. Its kernel is a locally free sheaf with basisIndeed any vector in the kernel is uniquely a sum of these vectors, by solving for its $i$-th component.

The Kähler differential sheaf ${\Omega }_{{\mathbf{P}}^n/k}^1$ has basis $d{t}_j^{(i)}$ on $U_i$, since this chart is a polynomial affine scheme. Define the local map $d{t}_j^{(i)}\mapsto F_j^{(i)}$. To check that these isomorphisms glue, on $U_i\cap U_{\ell }$ put $u=t_i^{(\ell )}$ and $v=t_j^{(\ell )}$. ThenThe twisting sheaf on projective space frame changes by $E_j^{(i)}=u^{-1}{E}_j^{(\ell )}$, henceFor $j=\ell$, use $t_{\ell }^{(\ell )}=1$, $d{t}_{\ell }^{(\ell )}=0$ and $F_{\ell }^{(\ell )}=0$; the same formula applies. Thus the differential and kernel frames have identical transition matrices. We obtain the cotangent form of the Euler sequence:This construction works over any field, including positive characteristic.

The Kähler differential sheaf has rank $n$. Taking determinant line bundles in this locally split short exact sequence of sheaves givesThe left side is the tensor product of $n+1$ copies of $\mathcal{O}(-1)$, and $\det (\mathcal{O})=\mathcal{O}$. ConsequentlyIt is the canonical bundle of projective space. The determinant identity can also be seen directly by taking the wedge of the kernel basis followed by a lift of the quotient basis; changing that lift by a kernel vector leaves the wedge unchanged.

Use the pair-design convention: each pair of distinct points must lie in exactly one Steiner block of size $k$. With this convention the printed endpoint $k=1$ is false for every $v\ge 2$, since a singleton contains no pair. Thus the intended assertion requires $k\ge 2$. For $k=2$, take all pairs as Steiner blocks for any $v\ge 2$. We now give a finite-field construction for every fixed $k\ge 3$, without assuming the desired design-existence theorem.

Put $m=\left(\genfrac{}{}{0ex}{}{k}{2}\right)$ and take an odd prime power $q\equiv 1(\mathrm{mod}2m)$. By cyclicity of ${\mathbb{F}}_q^{\times }$, it has a subgroup $H$ of index $m$; its order $(q-1)/m$ is even, so $-1\in H$. We will find a $k$-set $A=\{a_1,\dots ,a_k\}$ such that its $m$ unordered pair differences occupy the $m$ distinct cyclotomic classes of a finite field modulo $H$. The following argument supplies that set for every sufficiently large compatible $q$.

First establish the required two-class intersection estimate by elementary character sums. Let $\chi$ be a multiplicative character of a finite field of order $m$, with $\chi (\gamma )={\zeta }_m$ for a generator $\gamma$. Extend every power ${\chi }^j$, including the trivial character, by zero at zero. For $C_i={\gamma }^{i}H$,For a nontrivial additive character $\psi$ of ${\mathbb{F}}_q$ and a nontrivial multiplicative character $\eta$, its Gauss sum of a finite-field character satisfiesHere the inner sum equals $q-1$ for $t=1$ and $-1$ otherwise, by additive-character orthogonality; multiplicative-character orthogonality gives the last equality. Such a nontrivial additive character exists by identifying the additive group with a vector space over its prime field and choosing a nonzero linear functional into that prime field.

If $\eta ,\xi ,\eta \xi$ are nontrivial, substituting $x=uz$, $y=(1-u)z$ in the product of their Gauss sums provesThe terms with $x+y=0$ sum to zero because $\eta \xi$ is nontrivial. Hence $|J|=\sqrt{q}$. If exactly one character is trivial, subtracting the excluded zero and one terms gives absolute value one. If both are nontrivial but their product is trivial, the substitution $u/(1-u)$ gives $J(\eta ,{\eta }^{-1})=-\eta (-1)$, again of absolute value one. If both are trivial, the sum is $q-2$. Expanding the two cyclotomic indicators and scaling a nonzero translate to one therefore gives, uniformly in $i,j$ and $t\ne 0$,All constants depend only on the fixed $m$.

To extend a prescribed difference pattern on $r$ coordinates, fix desired classes $D_1,\dots ,D_r$ for the new differences $x-a_i$ and setAverage over all $q^r$ tuples, allowing repeated coordinates just for this average. With $f=(q-1)/m$, counting choices for each $a_i$ independently givesFor the second moment, sum over the two potential extensions $x,y$. The $x=y$ terms contribute $q{f}^r/q^r$. For $x\ne y$, the choices of $a_i$ number $q/m^2+O_m(\sqrt{q})$ each, by the preceding intersection estimate. ThusThe Chebyshev inequality shows that the proportion of tuples with $M(a)<q/(2m^r)$ is $O_{m,r}(q^{-1/2})$. Inductively, any fixed compatible difference pattern on $r$ distinct coordinates is realized by at least $c_r{q}^r$ tuples, with $c_r>0$ independent of sufficiently large $q$. The assertion starts with $r=1$. Almost all tuples are outside the exceptional set, so at least half the realizing tuples have at least $q/(2m^r)$ extensions. Each extension is distinct from the old coordinates because its differences belong to nonzero classes. This proves the assertion for $r+1$. This cyclotomic pattern extension by a second-moment bound avoids the false assumption that every partial choice can be extended.

Prescribe a different one of the $m$ cosets to each pair $a_j-a_i$, $i<j$. Since $-1\in H$, changing an orientation does not change its coset, so the prescriptions are compatible. The preceding induction supplies $A$.

Choose $S\subset H$ containing one element from each pair $\{h,-h\}$. Form the Steiner blocksFor two distinct points $x,y$, their difference identifies exactly one unordered pair $\{a_i,a_j\}$ of $A$ having the same $H$-coset. Exactly one of $(x-y)/(a_i-a_j)$ and its negative lies in $S$. Choose the orientation $a,b$ of that pair making $s=(x-y)/(a-b)\in S$, and then set $t=x-sa$. This yields $x=sa+t$, $y=sb+t$. Both the oriented base pair and $(s,t)$ are unique, so precisely one Steiner block contains the pair. In particular different Steiner block parameters cannot produce repeated Steiner blocks. Equivalently, the Steiner blocks $sA$ form a difference family for a Steiner 2-design, and translation gives the cyclotomic construction of a Steiner 2-design.

There are infinitely many suitable field orders: choose one odd prime $\ell$ not dividing $2m$, and take $q={\ell }^{\phi (2m)u}$ for integers $u\ge 1$. Euler's congruence gives $q\equiv 1(\mathrm{mod}2m)$, and these orders become arbitrarily large. ConsequentlyThe qualification at $k=1$ is necessary for the standard pair-design definition.

The coefficient form of the Combinatorial Nullstellensatz is as follows. Let $F$ be a field, let $P\in F[x_1,\dots ,x_r]$ have total degree at most $t_1+\cdots +t_r$, and suppose the coefficient of $x_1^{t_1}\cdots x_r^{t_r}$ is nonzero. For finite subsets $S_i\subset F$ with $|S_i|>t_i$, there is a point of $S_1\times \cdots \times S_r$ where $P$ is nonzero.

For completeness, restrict to $|S_i|=t_i+1$. The coefficient identity isBy Lagrange interpolation, the one-variable weighted sum on the right kills powers below $t_i$ and equals one on the power $t_i$. Any monomial of total degree at most $\sum t_i$, other than the specified one, has some exponent below $t_i$ and therefore contributes zero; the specified monomial contributes its coefficient. Every denominator is nonzero in the field. Hence a nonzero coefficient forces a nonzero grid value. The two applications below use this coefficient form.

Use finitely generated right projective modules throughout. The group K0 of a ring is the Grothendieck group of their direct-sum monoid, with $[P\oplus Q]=[P]+[Q]$. The group K1 of a ring is ${\mathrm{GL}}_{\infty }(A)/E_{\infty }(A)$, equivalently the abelianization of the stable general linear group. Stabilization is $g\mapsto \mathrm{diag}(g,1)$; $E_{\infty }$ is generated by elementary transvections. The stable elementary subgroup is the commutator subgroup: elementary transvections are commutators using a third index, and stable block interchange and elementary row identities make the quotient abelian.

Write $A_0$ for the lower-right ring, printed as $A^{\prime }$ in the source. A Milnor square has unital commuting maps, is a pullbackand has at least one of $j_1,j_2$ surjective. Interchange the two sides if necessary so that $j_1$ is surjective.

We first establish the patching fact used in the proof. For projective $P_i$ and an isomorphism $\varphi :P_1\otimes A_0\to P_2\otimes A_0$, setFor free modules denote the result by $M_g$. The block matrix identityexpresses $\mathrm{diag}(g,g^{-1})$ as elementary matrices. Their entries lift through $j_1$, and their elementary lifts remain invertible. A change of basis on the first side then trivializes this gluing. Thus $M_g\oplus M_{g^{-1}}\cong A^{2r}$, proving that $M_g$ is finitely generated projective. Notice that we lifted elementary matrices, not arbitrary invertible matrices.

For general $P_1,P_2$, choose $P_1\oplus Q_1\cong A_1^r$ and $P_2\oplus Q_2\cong A_2^s$. The reductions of $Q_1\oplus A_1^s$ and $Q_2\oplus A_2^r$ are isomorphic: insert the common reduced $P_i$ and reorder the two complements. Patch them by such an isomorphism. The sum with $M(P_1,P_2,\varphi )$ now patches two free modules of equal size, so is projective by the free case, making each summand projective. The same free change of basis followed by these direct-summand projections shows that extension to $A_i$ recovers $P_i$. Conversely, write an $A$-projective as the image of an idempotent over a finite free module. Since matrices over the pullback are compatible pairs of matrices, applying the compatible idempotents to the free pullback reconstructs this projective from its two reductions. Compatible module isomorphisms patch as well. This proves Milnor patching of projective modules in the form needed below.

The diagonal maps in the sequence are extension of scalars; the two maps to $A_0$ are differences $j_{1\ast }-j_{2\ast }$. Define the connecting map byIt is unchanged by stabilization or elementary multiplication: an elementary factor lifts through $j_1$ and changes a free basis in the gluing. The identity $\mathrm{diag}(g,h)=\mathrm{diag}(gh,I)\mathrm{diag}(h^{-1},h)$ and the preceding elementary factorization show $M_g\oplus M_h\cong M_{gh}\oplus A^r$ after matching sizes. Thus $\partial$ is an additive homomorphism on $K_1(A_0)$.

Exactness at $K_1(A_1)\oplus K_1(A_2)$. If the two reduced classes agree, choose stabilized representatives $g_1,g_2$ of the same size. Their reductions differ by an elementary matrix $e$ over $A_0$. Lift $e$ to an elementary matrix over $A_1$ and adjust $g_1$ without changing its $K_1$ class. Its reduction now equals that of $g_2$. The compatible pair and its compatible inverse give an invertible matrix over $A$, so the original pair of classes is in the diagonal image. The reverse containment follows from commutativity of the square.

Exactness at $K_1(A_0)$. A gluing obtained from a change of basis over either upper ring patches a free module, so the difference-map image is killed by $\partial$. Conversely, $\partial [g]=0$ means that $M_g$ becomes free after adding a common projective summand. Add a complement of that summand to obtain an actual isomorphism $M_g\oplus A^s\cong A^{r+s}$. Its two local bases express $g\oplus I_s=j_2(v_2)j_1(v_1{)}^{-1}$. Therefore $[g]=j_{2\ast }[v_2]-j_{1\ast }[v_1]$, which lies in the image of the difference map. This uses stable equality of projective classes, not cancellation of arbitrary projectives.

Exactness at $K_0(A)$. Connecting classes reduce to zero on both sides. Conversely, write a class in the kernel as $[P]-[A^r]$ by adding a complement to its negative projective term. Its reductions have the same classes as $A_i^r$. Equality in $K_0(A_i)$ becomes an isomorphism after a finite common free summand, by taking a complement to the auxiliary projective. Use one sufficiently large free summand on both sides. The stabilized $P$ is then locally free with gluing $g$, so the original class is $[M_g]-[A^{r+s}]=\partial [g]$.

Exactness at $K_0(A_1)\oplus K_0(A_2)$. Write a kernel pair as $([P_1]-[A_1^r],[P_2]-[A_2^r])$, using complements and padding to make the free sizes equal. Their reduced positive classes agree. Adding a common finite free summand makes those reductions isomorphic. Patch the resulting projectives, and subtract the matching free $A$-module. Its restrictions are the desired pair. The reverse containment is immediate because both reductions to $A_0$ agree.

These arguments prove the Mayer-Vietoris sequence in algebraic K-theoryNo zero is inserted at either end; this fragment does not assert injectivity at the first term or surjectivity at the last.

For the group ring of the cyclic group, take the cyclotomic Milnor squareThe maps send the cyclic generator to $1$ and to $\zeta$; the lower map sets $\zeta =1$, and the right map reduces modulo $p$. Since ${\Phi }_p(1)=p$, $\mathbb{Z}[\zeta ]/(\zeta -1)={\mathbb{F}}_p$. To verify the pullback, represent an element by a polynomial of degree at most $p-1$. If both evaluations vanish, it is a constant multiple of the minimal polynomial of an algebraic element ${\Phi }_p$, and evaluation at one forces that constant to be zero. Conversely, represent $b\in \mathbb{Z}[\zeta ]$ by $f$ of degree below $p-1$. For a compatible integer $a$, $a-f(1)$ is divisible by $p$; replacing $f$ by $f+[(a-f(1))/p]{\Phi }_p$ yields the pair $(a,b)$. The right map is surjective, so this is a Milnor square.