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A ring is Artinian when its ideals satisfy the descending chain condition: every chain $I_1\supseteq I_2\supseteq \cdots$ eventually becomes constant. It is Noetherian when its ideals satisfy the ascending chain condition, equivalently when every ideal is finitely generated. We first prove finite length of a commutative Artinian ring; this gives the stronger structural reason for its Noetherian property.

If $\mathfrak{p}$ is a prime ideal of an Artinian ring, the quotient $R/\mathfrak{p}$ is an Artinian integral domain. For a nonzero element $x$ of that domain, the chain $(x)\supseteq (x^2)\supseteq \cdots$ stabilizes. Thus $x^n=b{x}^{n+1}$ for some $b$, and cancellation gives $bx=1$. The quotient is a field, so every prime ideal is a maximal ideal.

There are only finitely many maximal ideals. Otherwise, choose distinct ones ${\mathfrak{m}}_1,{\mathfrak{m}}_2,\dots$. The intersections ${\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_n$ form a strictly descending chain. To see strictness, for each $i\le n$ choose $a_i\in {\mathfrak{m}}_i\setminus {\mathfrak{m}}_{n+1}$; their product belongs to the first $n$ ideals but not to the next one, since ${\mathfrak{m}}_{n+1}$ is prime. This contradicts the descending chain condition.

Put $J={\bigcap }_i{\mathfrak{m}}_i$. This Jacobson radical is also the nilradical, since all primes are maximal. We need the stronger conclusion that $J$ is nilpotent, without assuming Noetherianity. Its powers stabilize, say $K=J^n=J^{n+1}=JK$. Suppose $K\ne 0$. By the descending chain condition, choose an ideal $L$ minimal subject to $KL\ne 0$. Some $x\in L$ has $Kx\ne 0$, so minimality gives $L=Rx$. Moreover $K(JL)=KL\ne 0$, and minimality gives $JL=L$. Hence $x=jx$ for some $j\in J$. But $1-j$ is a unit: it cannot lie in any maximal ideal, because $j$ lies in all of them. Thus $x=0$, a contradiction. Therefore $J^n=0$.

The Chinese remainder theorem gives $R/J\cong {\prod }_{i}R/{\mathfrak{m}}_i$, a finite product of fields. Each quotient $J^i/J^{i+1}$ is an Artinian module over this product, and each field component must be a finite-dimensional vector space; an infinite-dimensional vector space admits a strictly descending chain of subspaces. Consequently every layer has finite composition length. The finite filtrationshows that $R$ itself has finite composition length. A strict inclusion of submodules strictly increases length, so an ascending chain cannot continue indefinitely. Every commutative Artinian ring is therefore Noetherian. This is the Artinian rings are Noetherian result.

For the formal power series ring, let $B=R[[X]]$ with $R$ Noetherian, and let $H$ be any ideal of $B$. For $n\ge 0$ define a coefficient idealThese coefficient ideals of a formal power series ideal satisfy $A_n\subseteq A_{n+1}$, by multiplication by $X$. The ascending chain condition gives $A_n=A_N$ for all $n\ge N$. For each $0\le n\le N$, choose finitely many series $f_{n,j}\in H\cap X^{n}B$ whose coefficients at $X^n$ generate $A_n$.

We claim that these finitely many series generate $H$ as an ordinary ideal. Given $f\in H$, cancel its coefficient at $X^k$ successively. After coefficients below $k$ have vanished, its coefficient at $X^k$ lies in $A_k$. If $k\le N$, use an $R$-linear combination of the $f_{k,j}$. If $k>N$, use a combination of $X^{k-N}{f}_{N,j}$, since $A_k=A_N$. The remainder then belongs to $H\cap X^{k+1}B$.

Collect all the cancellations against each fixed generator. For $n<N$ its multiplier is a polynomial, while the multipliers of the $f_{N,j}$ are well-defined formal power series: at any fixed degree, only finitely many cancellation steps contribute. The remainder has every coefficient zero. ThusThis is a finite sum of ideal generators, rather than merely a topological closure assertion. Since $H$ was arbitrary,This coefficient-cancellation argument proves Noetherianity of a formal power series ring.

The corresponding Artinian assertion is false. For any nonzero ring $R$, the idealsin $R[[X]]$ are strictly decreasing, since $X^n$ has a nonzero coefficient in degree $n$ and no multiple of $X^{n+1}$ does. In particular, a field $k$ is Artinian, but $k[[X]]$ is not. The zero ring is the harmless exception.

Use the following finite-length version of the Hilbert-Serre theorem. Let $S={⨁}_{n\ge 0}{S}_n$ be a graded algebra generated over an Artinian ring $A=S_0$ by finitely many homogeneous elements $x_1,\dots ,x_r$ of positive degrees $e_1,\dots ,e_r$. For a finitely generated nonnegatively graded $S$-module $M$, define its Poincare series of a graded module byEvery component has finite length, and the theorem saysFor a module whose grading is merely bounded below, the same statement holds with a Laurent-polynomial numerator. When the generators all have degree one, the denominator is $(1-t{)}^r$.

Here is an induction proof. An Artinian ring is Noetherian by question 1, so the Hilbert basis theorem makes $S$ Noetherian. If there are no positive-degree generators, $S=A$ and a finite homogeneous generating set for $M$ occupies only finitely many degrees. Each component is a finite module over the Artinian ring $A$, hence has finite length of a module, and $P_M$ is a polynomial.

For $r>0$, put $x=x_r$, $e=e_r$, $N=\{m\in M:xm=0\}$ and $C=M/xM$. Both are finitely generated graded modules annihilated by $x$, hence modules over $S/(x)$, which is generated by the first $r-1$ homogeneous elements. Their multiplication exact sequence isHere $M(-e{)}_n=M_{n-e}$. Additivity of length degree by degree givesThis is the Hilbert series multiplication exact sequence. By induction, the right-hand side has denominator ${\prod }_{i<r}(1-t^{e_i})$. Division by $1-t^e$ completes the proof of the Hilbert-Serre theorem. The finite-component and finite-generation claims also follow from the finite set of positive-degree algebra and module generators; no analytic convergence of a series is involved.

For the local invariant, use the usual Noetherian local ring hypothesis of Hilbert–Samuel growth dimension. Locality alone does not guarantee finite lengths or polynomial growth; the printed question leaves this finiteness assumption implicit. For instance, in the localization at a prime ideal of the polynomial ring $k[x_1,x_2,\dots ]$ at $(x_1,x_2,\dots )$, the vector space $\mathfrak{m}/{\mathfrak{m}}^2$ has the infinitely many independent classes of the variables, so its length is not finite. Write $\mathfrak{m}=P$ and $\kappa =R/\mathfrak{m}$. Its associated graded ringis generated over $\kappa$ in degree one, because $\mathfrak{m}$ is finitely generated. Thus its Hilbert series is rational with a denominator that is a power of $1-t$. After cancelling factors, writeDefine $d(R)=d$, the pole order at $t=1$, with $d=0$ when $P_G$ is a polynomial.

Equivalently, the Hilbert–Samuel functionhas generating series $P_G(t)/(1-t)$ and eventually agrees with a polynomial of degree $d$. Indeed, coefficients of $(1-t{)}^{-d-1}$ are $\left(\genfrac{}{}{0ex}{}{n+d}{d}\right)$, and multiplication by $q(t)$ leaves leading term $q(1)n^d/d!$. Consequently $d(R)$ is the degree of the cumulative Hilbert-Samuel polynomial, rather than the degree of the individual graded-component function; the latter has degree $d-1$ when $d>0$. This pole/growth invariant also equals Krull dimension by the local dimension theorem, although that theorem is not required to define it here.

For an example, take $R=k[x,y{]}_{(x,y)}$. Its associated graded ring is $k[x,y]$ with the ordinary degree grading: degree $n$ has the $n+1$ monomials $x^i{y}^{n-i}$ as a basis. ThereforeThe length formula also follows directly by counting monomials of total degree at most $n$ in $R/(x,y{)}^{n+1}$.

A Lie group is a finite-dimensional smooth manifold with a group structure for which multiplication and inversion are smooth. The group defined here is the compact symplectic group $Sp(n)=U(2n)\cap Sp(2n,\mathbb{C})$, rather than the full complex symplectic group. Also, the displayed matrix expression requires $B$ to be $2n\times 2n$; the printed $n\times n$ size is incompatible with $J$.

Since $J^2=-I$, we have $J^{-1}=-J$. The matrix exponential commutes with conjugation, as follows term by term from its absolutely convergent power series. ConsequentlyTo construct logarithm charts for the compact symplectic group, consider the real vector spaceDifferentiating the defining identities at $I$ gives precisely these conditions. Conversely if $X\in \mathfrak{k}$, $e^X$ is unitary andso $e^X\in Sp(n)$. These are the infinitesimal conditions of the compact symplectic Lie algebra.

Near $I$, the convergent matrix logarithm series $\log (I+C)={\sum }_{m\ge 1}(-1{)}^{m+1}{C}^m/m$ is smooth and inverse to the matrix exponential near zero. These local inverses respect transpose, conjugate transpose, and conjugation; also $\log (A^{-1})=-\log A$ when both matrices are sufficiently close to $I$. Shrink their neighborhoods accordingly. If $A\in Sp(n)$ there, unitarity givesThe symplectic identity is equivalent to $A^t=J^{-1}{A}^{-1}J$, so, putting $X=\log A$, it gives $X^t=-J^{-1}XJ=JXJ$, equivalently $XJ+J{X}^t=0$. Thus this local logarithm restricts to a bijection between a neighborhood of $I$ in $Sp(n)$ and an open neighborhood of zero in $\mathfrak{k}$. Its inverse is the restricted matrix exponential. These restrictions are manifold charts; left multiplication translates them to every $A_0\in Sp(n)$ via $A\mapsto \log (A_0^{-1}A)$. The chart overlaps are smooth compositions of multiplication, exponential and logarithm. The subspace topology is Hausdorff and second countable, inherited from the finite-dimensional matrix space, so these charts give a smooth manifold.

Closure under products and inverses follows from $AJ{A}^t=J$ and unitarity. Matrix multiplication is polynomial in real and imaginary entries, and inversion on the unitary group is $A\mapsto A^{\ast }$, a real linear operation. Their restrictions are smooth in the charts just constructed. Hence $Sp(n)$ is a Lie group without needing a closed-subgroup theorem.

Write $X$ in $n\times n$ blocks. The two infinitesimal conditions giveThe skew-Hermitian matrix $P$ has $n^2$ real parameters; the complex symmetric matrix $Q$ has $n(n+1)/2$ complex parameters, hence $n(n+1)$ real parameters. ThereforeFor $n=1$, any two-by-two matrix satisfies $AJ{A}^t=(\det A)J$, so the group is $SU(2)$. Explicitly,The rows are orthonormal and the determinant is one; conversely unitarity and determinant one force this form. This is the SU(2) as the three-sphere parametrization. The map and its inverse, extraction of the first row, are smooth. Thus $Sp(1)$ is diffeomorphic to $S^3$.

The idele group is the multiplicative restricted productwith respect to ${\mathcal{O}}_v^{\times }$ at the finite places of a number field; $K_v$ is the completion of a valued field at the place $v$. Thus each tuple has nonzero components, and all but finitely many finite components are units. Its restricted product topology on the idele group has basic open sets ${\prod }_{v\in S}{W}_v\times {\prod }_{v\notin S}{\mathcal{O}}_v^{\times }$, where $S$ is finite and contains the infinite places, and each $W_v$ is open in $K_v^{\times }$. In particular $U_K$ is an open subgroup.

Embed $K^{\times }$ diagonally. Take a neighbourhood of one whose finite components all lie in ${\mathcal{O}}_v^{\times }$ and whose infinite components satisfy $|x_v-1|<1/2$ in the usual real or complex modulus. A diagonal element there is an algebraic unit $a$. If $a\ne 1$, then $a-1$ is a nonzero algebraic integer, so its field norm is a nonzero integer. Buta contradiction. Thus $K^{\times }$ is discrete. It is also closed: in a topological group, a subgroup with an isolated identity cannot have an external accumulation point, since quotients of two nearby subgroup elements would approach the identity.

Send an idele to its associated fractional ideal byOnly finitely many exponents are nonzero. This homomorphism is onto, by choosing powers of local uniformizers, and its kernel is $U_K$. Diagonal elements map to principal fractional ideals. The resulting quotient givesIt is a topological isomorphism when the ideal class group is given the discrete topology, since $U_K$ is open.

Use normalized local moduli: real modulus, squared complex modulus, and $|{\pi }_v{|}_v=(N{\mathfrak{p}}_v{)}^{-1}$ at a finite place. The idelic modulus $\Vert x\Vert ={\prod }_v|x_v{|}_v$ defines $J_K^1=\ker \Vert \cdot \Vert$, the norm-one idele group. The product formula puts $K^{\times }$ inside this kernel. Every ideal class has a representative in $J_K^1$, because an infinite component can be rescaled to correct the modulus without altering its fractional ideal. The compact space $J_K^1/K^{\times }$ therefore maps continuously onto the discrete ideal class group. Its image must be finite, proving $\mathrm{Cl}(K)$ is finite.

For the Dirichlet unit theorem, put $U^1=U_K\cap J_K^1$ and $E=K^{\times }\cap U^1={\mathcal{O}}_K^{\times }$. Let $r_1$ count real embeddings and $r_2$ count conjugate complex pairs. Infinite logarithms define a continuous surjectionusing $\log |x_v|$ at real places and $2\log |x_v|$ at complex places. Its kernel is compact: it consists of real signs, complex unit circles, and the product of compact finite-place unit groups. More generally the inverse image of a bounded closed subset of $H$ is compact. Since $E$ is closed and discrete, its intersection with each such inverse image is finite. Hence $\Lambda =\ell (E)$ is discrete in $H$, and the kernel $E\cap \ker \ell$ is a finite group. It is exactly the roots of unity $\mu (K)$, since every element of a finite multiplicative group has finite order and every root of unity has all local moduli one.

The image of $U^1$ in $J_K^1/K^{\times }$ is an open subgroup, hence also closed, and is homeomorphic to $U^1/E$. The assumed compactness therefore makes $U^1/E$ compact, and its continuous quotient $H/\Lambda$ is compact. A discrete cocompact subgroup of a real vector space is a full Euclidean lattice, of rank $\dim H=r_1+r_2-1$. Thus $E/\mu (K)\cong {\mathbb{Z}}^{r_1+r_2-1}$, and lifting a lattice basis splits off the free factor:This derives both finiteness and the unit rank from the stated compactness assumption, rather than assuming either conclusion to prove compactness.

Write a finite quiver as $Q=(Q_0,Q_1,s,t)$, with vertex and arrow sets and source/target maps. A representation of a quiver assigns a vector space $X_i$ to each vertex and a linear map $f_{\rho }:X_{s(\rho )}\to X_{t(\rho )}$ to each arrow. A quiver representation morphism $h:X\to Y$ is a family satisfying $h_{t(\rho )}{f}_{\rho }=g_{\rho }{h}_{s(\rho )}$.

The path algebra $A=kQ$ has every directed path, including each length-zero path $e_i$, as a basis. Multiplication is composition when endpoints match, and zero otherwise; in $pq$, the path $q$ is traversed first. The orthogonal idempotents satisfy $1={\sum }_i{e}_i$.

The path-algebra module equivalence is explicit. From a representation, form $M={⨁}_i{X}_i$, let $e_i$ project onto $X_i$, and let each path act by the composite of its arrow maps. Conversely, an $A$-module gives $X_i=e_{i}M$ and $f_{\rho }(x)=\rho x$. An $A$-module homomorphism restricts to the required vertex maps, and a compatible family extends by direct sum. These constructions are mutually inverse up to their evident natural identifications.

$kQ$ is finite-dimensional exactly when $Q$ is finite and has no oriented cycle. For a finite acyclic quiver, paths have length at most $|Q_0|-1$. An oriented cycle has arbitrarily many distinct powers, giving infinitely many basis paths. If arbitrary infinite quivers are allowed, finiteness of both vertices and arrows is also necessary; the unital module correspondence above uses finite $Q_0$.

Choose only the orientation $1\to 2\to 3$. The interval representations of an equioriented three-vertex quiver $I[a,b]$ have $k$ at vertices $a,\dots ,b$, zero elsewhere, and identity arrows within that interval. The complete list isHere is an elementary proof, without the Gabriel theorem. For $X_1\stackrel{f}{\to }{X}_2\stackrel{g}{\to }{X}_3$, set $K=\mathrm{im}f\cap \ker g$. Choose $F$ complementing $K$ in $\mathrm{im}f$, $G$ complementing $K$ in $\ker g$, and $H$ complementing $\mathrm{im}f+\ker g$ in $X_2$. Then $X_2=K\oplus F\oplus G\oplus H$, and $g$ is injective on $F\oplus H$. Lift bases of $K,F$ to a complement of $\ker f$ in $X_1$, and extend the bases of $gF,gH$ to $X_3$. These bases split $X$ into precisely the six kinds of interval block. Every block has endomorphism ring $k$, hence is indecomposable, and their different supports make them pairwise nonisomorphic. The same basis argument handles arbitrary vertex dimensions; each indecomposable block itself is finite-dimensional.

For a Lie bracket over $\mathbb{R}$ or $\mathbb{C}$, antisymmetry of a Lie bracket means $[x,y]=-[y,x]$. In particular $[x,x]=0$ in these characteristic-zero fields. The Jacobi identity isIt expresses compatibility of the bracket with its own adjoint action. Bilinearity must hold over the chosen base field, and the bracket must take its values in the same vector space.

For the matrix commutator, bilinearity and antisymmetry follow directly from distributivity. Associativity of matrix multiplication givesadding the two cyclic permutations cancels every monomial. Thus the Jacobi identity holds in the entire matrix algebra. For a specified linear subspace, the only additional bracket condition is closure: $AB-BA$ must belong to the subspace whenever $A,B$ do. It is unnecessary to require closure under the separate products $AB$ and $BA$.

To determine the special unitary Lie algebra, let $g(t)$ be a differentiable curve in the special unitary group with $g(0)=I$ and $A=g^{\prime }(0)$. Differentiating $g(t{)}^{†}g(t)=I$ gives $A^{†}+A=0$. Differentiating $\det g(t)=1$ at the identity gives $\mathrm{tr}A=0$. Conversely a traceless skew-Hermitian matrix has $e^{tA}$ unitary and $\det e^{tA}=e^{t\mathrm{tr}A}=1$, so it really is a tangent vector. ThereforeThis is a real Lie algebra of complex matrices: multiplication by $i$ generally leaves this real subspace. Its complexification is ${\mathfrak{sl}}_n(\mathbb{C})$, not the compact algebra itself. For $A,B\in \mathfrak{su}(n)$,Thus closure holds, and the already verified commutator identities establish all the Lie algebra axioms.

The cross-product Lie algebra on ${\mathbb{R}}^3$ is bilinear, antisymmetric and closed because the cross product has those properties. Its Jacobi identity follows from the vector triple-product identity:In the cyclic sum, the coefficients cancel by symmetry of the scalar product.

For the explicit relation to the SU(2) Lie algebra, take the three Pauli matrices and defineThese matrices are traceless and Skew-Hermitian, and the three images of the standard basis form a real basis of $\mathfrak{su}(2)$. Using the Pauli matrix commutator identity,Hence $F$ is a real Lie algebra isomorphism. The factor and sign $-i/2$ are essential for preserving the unscaled cross-product bracket. At the group level there is an Adjoint double cover from SU(2) to SO(3); the isomorphism of their tangent algebras does not identify the two global groups.

For a group presentation $\mathcal{P}=⟨X\mid R⟩$ with $X$ finite, let $F(X)$ be its free group. For a nontrivial relator defineThe p-deficiency in the unshifted convention used here isIf the weighted sum diverges the value is $-\infty$; identity relators may be omitted or assigned weight zero. Roots are taken in the free group, not in the presented quotient. Some authors subtract one from this definition; here the requested threshold is $1$.

Two elementary bounds explain why p-deficiency detects infinitude. The p-rank of a group isHere $G^p$ denotes the subgroup generated by all $p$th powers. Each relator that is not a $p$th power imposes at most one linear relation in this vector space, and a $p$th-power relator imposes none. If there are $s$ relators of the first type, then

The second bound is the index-p rewriting bound for p-deficiency. Suppose $H⊴G$ has index $p$, and its preimage in $F(X)$ is $V$. The Nielsen–Schreier formula gives $V$ rank $1+p(|X|-1)$. For a relator $r=w^{p^k}$, there are two cases in the Reidemeister–Schreier theorem. If $w\in V$, its $p$ coset-conjugates are all $p^k$th powers in $V$, with total weight at most $p{p}^{-k}$. If $w\notin V$, then $k\ge 1$, since $r\in V$. Its cosets generate $F(X)/V$, so representatives $1,w,\dots ,w^{p-1}$ show that the $p$ rewritten conjugates of $r$ are redundant up to conjugation in $V$. One relator suffices, and $r=(w^p{)}^{p^{k-1}}$ has weight at most $p^{-(k-1)}$. In both cases the total weight is at most $p$ times the old weight. Thus the induced group presentation $\mathcal{Q}$ of $H$ satisfiesThe argument applies termwise to infinitely many relators whenever the weighted sum converges.

If ${\mathrm{def}}_p(\mathcal{P})\ge 1$, the p-rank of a group bound gives a surjection to $C_p$, hence a normal subgroup of index $p$. The rewriting bound gives that subgroup another presentation of p-deficiency at least one. Iterating produces subgroups of index $p^j$ for every $j$. p-deficiency at least one implies infinitude.

Now enumerate the nonidentity elements $w_1,w_2,\dots$ of $F(x,y)$ and choose the presentationIts p-deficiency obeysThe infinitude criterion shows that $T$ is infinite. It is generated by two elements, and every element is represented by some $w_i$ or is the identity; the imposed relation makes its order a power of $p$. Thus $\boxed{T\text{ is an infinite finitely generated torsion group}}$. This is a torsion group construction by p-power relators; the presentation intentionally has infinitely many relators.

The Frankl-Wilson theorem. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}\subseteq \mathcal{P}([n])$. Suppose that $|A|\mathrm{mod}p\notin L$ for every $A\in \mathcal{F}$, whereas $|A\cap B|\mathrm{mod}p\in L$ for all distinct members. Thenwith terms beyond $n$ understood as zero.

Work over the finite field ${\mathbb{F}}_p$. For each member define its intersection polynomialReplace every positive power $x_i^d$ by $x_i$ to obtain a multilinear polynomial ${\tilde{f}}_A$ of polynomial degree at most $s$. This Boolean multilinearization preserves its values at every characteristic vector of a set. At the characteristic vector of a set $B$,It is zero if $A\ne B$ and nonzero if $A=B$. Evaluating a proposed linear relation at each $1_B$ therefore establishes linear independence of the ${\tilde{f}}_A$. The vector space of multilinear polynomials of polynomial degree at most $s$ has the basis of monomials ${\prod }_{i\in T}{x}_i$, $|T|\le s$, of size ${\sum }_{j=0}^s\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. This proves the Frankl-Wilson theorem by the polynomial method in combinatorics. The empty product for $s=0$ is $1$, and the argument still applies.

The constant-intersection family bound, directly. Write $\mathcal{F}=\{F_1,\dots ,F_q\}$ and assume $n\ge 1$. The cases $q\le 1$ are immediate. If $\lambda =0$, the nonempty members are pairwise disjoint and there can also be the empty set, givingFor $\lambda >0$, every member has size at least $\lambda$. If every member has size strictly greater than $\lambda$, form the real matrix $X$ whose rows are their characteristic vectors of sets. Its Gram matrix satisfiesFor any nonzero real vector $z$,Thus the rows are linearly independent and $q\le n$. If a member $F_1$ has size $\lambda$, it lies inside every other member. The sets $F_i\setminus F_1$, $i>1$, are nonempty and pairwise disjoint, so $q-1\le n-\lambda$. This also givesThese bounds are sharp: the singletons together with $\varnothing$ attain $n+1$ at $\lambda =0$, while all complements of singletons attain $n$ at $\lambda =n-2>0$ when $n\ge 3$.

The constant t-wise intersection dichotomy. In its substantive form the last argument requires $m\ge t$. The printed statement does not explicitly impose this. Without it the $t$-fold condition can be vacuous and the claimed dichotomy is false: take $t=3$, $m=2$, $n=1$, $A_1=\varnothing$, $A_2=\{1\}$ and $\lambda =1$. There is no $3$-tuple to test, no common $1$-set, and $k=0$, so $m=2>k+t-2=1$. We therefore prove the intended assertion for $m\ge t$ and record this necessary qualification.

If $\lambda =0$, the first alternative always holds with $S=\varnothing$. Suppose $\lambda >0$. Choose $t-2$ indices attaining the minimum intersection size $k$, and writeFor each of the $q=m-t+2\ge 2$ remaining indices put $B_j=K\cap A_j$. The hypothesis gives $|B_j\cap B_{\ell }|=\lambda$ for distinct remaining indices. This reduces the constant t-wise intersection dichotomy to a constant-intersection family bound on the $k$-element set $K$. The two possibilities are treated below.

Apply the polynomial method in combinatorics to the finite-field Kakeya set $A$. We will actually obtain the stronger estimateLet $\mathcal{P}$ be the vector space over the prime field ${\mathbb{F}}_p$ of multivariate polynomials in $n$ variables of total degree of a polynomial at most $p-1$. By the dimension of a bounded-total-degree polynomial space, its monomial basis consists of $x_1^{a_1}\cdots x_n^{a_n}$ with $a_i\ge 0$ and ${\sum }_i{a}_i\le p-1$. Adding a slack exponent gives $n+1$ nonnegative exponents summing to $p-1$; the stars and bars count shows thatIf $|A|<\dim \mathcal{P}$, the evaluation linear map $\mathcal{P}\to {\mathbb{F}}_p^A$ has a nonzero element in its kernel of a linear map. Thus there is a nonzero polynomial $P$ of total degree of a polynomial $d\le p-1$ vanishing at every point of $A$. Its degree cannot be zero, since $A$ contains a line and is nonempty. Write $P_d$ for its nonzero top homogeneous polynomial part.

For every nonzero $v\in {\mathbb{F}}_p^n$, the directional hypothesis supplies an affine line in a vector spaceDirections are one-dimensional vector subspaces; rescaling a representative does not change this line. The univariate polynomial $P(a_v+tv)$ has degree at most $d<p$ and vanishes for all $p$ values of $t$. The root bound for a polynomial makes it the zero polynomial. Its coefficient of $t^d$ is exactly $P_d(v)$, so $P_d(v)=0$ for every nonzero $v$. Because $d>0$, $P_d(0)=0$ too.

To finish, we prove the relevant polynomial nonvanishing below the field size. A polynomial over ${\mathbb{F}}_p$ of degree at most $p-1$ in each variable cannot vanish on all of ${\mathbb{F}}_p^n$ unless it is zero. Induct on $n$. The one-variable case is the root bound for a polynomial. For more variables, writeFixing the first $n-1$ coordinates gives a univariate polynomial with $p$ roots of a polynomial, so all its coefficients are zero. Hence every $Q_j$ vanishes on ${\mathbb{F}}_p^{n-1}$, and the induction hypothesis makes every $Q_j$ zero. Applying this to $P_d$, whose total degree of a polynomial is less than $p$, contradicts its choice as nonzero.

Therefore $|A|\ge \dim \mathcal{P}$. Finally,which proves the requested lower bound. The crucial observation in the finite-field Kakeya polynomial bound is that complete lines force the highest homogeneous polynomial part to vanish in every direction.

A prime ideal $P$ is a proper ideal such that $ab\in P$ implies $a\in P$ or $b\in P$. Equivalently $R/P$ is an integral domain.

For each $i$, and each $j\ne i$, choose $a_{ji}\in P_j\setminus P_i$, using incomparability. PutThen $b_i$ belongs to the other two prime ideals and does not belong to $P_i$, by primality. In particular each $b_i$ lies in the union. But $b_1+b_2+b_3\notin P_i$ for every $i$: modulo $P_i$, the sum equals $b_i\ne 0$. The union is not closed under addition and is not an ideal. This is a concrete instance of prime avoidance.

The union of three incomparable nonprime ideals example can retain pairwise incomparability. In the ringtakeThese distinct two-element ideals are pairwise incomparable, yetwhich is an ideal. None is prime: its quotient ring still contains a nonzero element of square zero. The three one-dimensional vector subspaces cover the two-dimensional vector space over ${\mathbb{F}}_2$.

The key is a Laurent normalization by a monomial change of variables. We prove the slightly stronger assertion that every nonzero quotient of a Laurent polynomial ring in $r$ variables is finite as a module over an embedded Laurent polynomial ring in some number $m\le r$ of variables.

Induct on $r$. If the defining ideal is zero, the ring itself works. Otherwise choose a nonzero Laurent polynomialin that ideal, with nonzero coefficients. For a sufficiently large positive integer $N$, the weightsare distinct for all $\mathbf{a}\in F$. Indeed, each difference gives a nonzero polynomial in $N$, which has only finitely many integer roots. Make the invertible monomial substitutionIts inverse is $Y_j=X_j{X}_r^{-N^j}$, so it is an automorphism of the Laurent polynomial ring. When $r=1$ there is no substitution and distinct exponents already give distinct weights.

As a Laurent polynomial in $Y_r$, the transformed relation has unique lowest and highest exponents. Their coefficients are nonzero scalar multiples of monomials in $Y_1,\dots ,Y_{r-1}$, hence units inMultiplying by a power of $Y_r$ and the inverse leading coefficient givesHere $d>0$: a single-term Laurent polynomial would be a unit and the quotient would be zero. Reversing the equation and dividing by $b_0{Y}_r^d$ gives a monic equation for $Y_r^{-1}$. Thus both $Y_r$ and $Y_r^{-1}$ are integral over the image $B$ of $B_0$ in $T$. In fact $b_0$ being a unit also expresses $Y_r^{-1}$ as a polynomial in $Y_r$, so $T$ is generated as a $B$-module by $1,Y_r,\dots ,Y_r^{d-1}$.

The ring $B$ is itself a nonzero quotient of an $(r-1)$-variable Laurent polynomial ring. By induction, it is finite over an embedded Laurent polynomial ring in $m\le r-1$ variables. Finiteness of modules is transitive, so the same is true of $T$; a finite ring extension is an integral extension, by the determinant trick applied to multiplication by an element.

Applying this to $r=3$ yields an embedded subringThe base case $m=0$ means the subring $k$; finiteness over it would make $T$ a finite-dimensional vector space. The hypothesis excludes this, so $1\le m\le 3$, as required. No infinitude assumption on $k$ is used.

For $a\in R$, write $m_a$ for multiplication by $a$ in ${\mathrm{End}}_k(R)$. The algebraic differential operator ring is defined by the filtrationThe order of an algebraic differential operator is the least $j$ for which $\theta \in D_k^j(R)$; the zero operator may be assigned order $-\infty$. In degree zero, commuting with all multiplication maps means $\theta (a)=a\theta (1)$, so $D_k^0(R)=\{m_a:a\in R\}\cong R$. The commutator identity $[UV,m_a]=U[V,m_a]+[U,m_a]V$ proves by induction that $D_k^r{D}_k^s\subseteq D_k^{r+s}$, so the union is a ring under addition and composition.

The derivations of an algebra formEvery such derivation has $\delta (1)=0$ and $[\delta ,m_a]=m_{\delta (a)}$, so has differential order at most one. Conversely, if $\theta \in D_k^1(R)$, set $c=\theta (1)$ and $\delta =\theta -m_c$. Then $\delta (1)=0$ and $[\delta ,m_a]$ is a multiplication map. Evaluating at $1$ shows it is $m_{\delta (a)}$; evaluating at $b$ gives the Leibniz rule. Hence the first-order algebraic differential operator decomposition isIt is a direct sum as $k$-vector spaces: a multiplication map that is a derivation vanishes at $1$, and is zero.

The complex Weyl algebra has generators $x_1,\dots ,x_n,{\partial }_1,\dots ,{\partial }_n$ and relationsTake $n\ge 1$. If a nonzero unital module $M$ had finite dimension of a vector space over $\mathbb{C}$ $m$, its representing matrices would satisfy $[{\partial }_1,x_1]=I_m$. Taking the matrix trace gives $0=m$, impossible in characteristic zero. The Weyl algebra has no nonzero finite-dimensional modules. For $n=0$ the algebra is $\mathbb{C}$ and this assertion has the obvious exception.

For a nonzero finitely generated Weyl algebra module, use the Bernstein filtrationThe ordered monomials form a vector space basis, soTo justify the ordered basis, the relations move all ${\partial }_i$ to the right and give spanning. Independence follows in the action on $\mathbb{C}[x_1,\dots ,x_n]$: apply a putative zero operator ${\sum }_{\mathbf{b}}{p}_{\mathbf{b}}(x){\partial }^{\mathbf{b}}$ coefficientwise to the formal exponential $e^{t\cdot x}$, obtaining ${\sum }_{\mathbf{b}}{p}_{\mathbf{b}}(x)t^{\mathbf{b}}=0$. Thus every coefficient vanishes.

Choose a finite-dimensional nonzero generating subspace $M_0\subseteq M$ and set $M_j=F_j{A}_n{M}_0$. The associated graded module is finitely generated over the polynomial associated graded ring, so the Hilbert-Serre theorem implies ${\dim }_{\mathbb{C}}{M}_j$ is eventually a polynomial. DefineThis Bernstein growth dimension of a Weyl algebra module equals the Gelfand–Kirillov dimension of a module, not its ordinary vector-space dimension. If a different finite generating subspace is used, each lies in some fixed filtration step of the other. The corresponding $M_j$ are sandwiched between shifted filtrations, so the degree is unchanged.