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For a local ring $(R,\mathfrak{m})$ with residue field $\kappa =R/\mathfrak{m}$, its cotangent space is the $\kappa$-vector space $\mathfrak{m}/{\mathfrak{m}}^2$. In a Noetherian local ring, its dimension is the minimal number of generators of $\mathfrak{m}$, by the Nakayama lemma. Equality of this dimension with the Krull dimension defines a regular local ring.

Let $(R,\mathfrak{m})$ be a Noetherian local ring, let $M$ be nonzero and finitely generated, and let every $f_i$ lie in $\mathfrak{m}$. Then positive Koszul homology vanishes exactly when $f$ is a regular sequence on a module $M$. The mapping cone exact sequence proves the forward implication by induction; surjectivity of the last generator on earlier homology and the Nakayama lemma prove the converse.

In a Noetherian ring, every prime ideal minimal over a proper principal ideal has height at most one.

For a proof, suppose instead that ${\mathfrak{p}}_0⊊\mathfrak{q}⊊\mathfrak{p}$ and that $\mathfrak{p}$ is minimal over $(a)$. Quotient by ${\mathfrak{p}}_0$ and localize at $\mathfrak{p}$; one may assume that $R$ is a Noetherian local domain, $0⊊\mathfrak{q}⊊\mathfrak{p}$ with $\mathfrak{p}$ maximal, and $\mathfrak{p}$ is the only prime containing $(a)$. For the symbolic powerseach ${\mathfrak{q}}^{(n)}$ is $\mathfrak{q}$-primary. The quotient $R/(a)$ is a zero-dimensional Noetherian ring and hence Artinian, so the descending chain $({\mathfrak{q}}^{(n)}+(a))/(a)$ stabilizes. For large $n$, writing an element of ${\mathfrak{q}}^{(n)}$ modulo ${\mathfrak{q}}^{(n+1)}$ and using $a\notin \mathfrak{q}$ givesThe Nakayama lemma applied to ${\mathfrak{q}}^{(n)}/{\mathfrak{q}}^{(n+1)}$ gives ${\mathfrak{q}}^{(n)}={\mathfrak{q}}^{(n+1)}$. Localizing at $\mathfrak{q}$ would therefore make the powers of the nonzero maximal ideal $\mathfrak{q}{R}_{\mathfrak{q}}$ stabilize at a nonzero ideal, contradicting the Krull intersection theorem for the Noetherian local domain $R_{\mathfrak{q}}$.

In a right Artinian ring, the powers of its Jacobson radical stabilize at an ideal $I$ satisfying $I^2=I$. If $I\ne 0$, choose a minimal right ideal $K$ with $KI\ne 0$. Since $(KI)I=KI$, minimality gives $KI=K$. Choosing $k\in K$ with $kI\ne 0$ gives $K=kR$ by the same minimality. Consequently $k=kj$ for some $j\in J(R)$. The unit criterion for the Jacobson radical makes $1-j$ invertible, forcing $k=0$, a contradiction. This proof avoids incorrectly applying the finitely generated form of Nakayama lemma to an arbitrary Artinian module.

The descending powers $J\supseteq J^2\supseteq \cdots$ stabilize because they are right ideals. Choose $m\ge 1$ with $J^m=J^{m+1}$ and put $I=J^m$. Then $I=J^{2m}=I^2$.

Suppose $I\ne 0$. Among the right ideals $K$ such that $KI\ne 0$, choose a minimal one; the set is nonempty since $RI=I\ne 0$. The right ideal $KI$ lies in $K$, and $(KI)I=K{I}^2=KI\ne 0$, so minimality forces $KI=K$. Choose $k\in K$ with $kI\ne 0$. Then $(kR)I=kI\ne 0$, and $kR\subseteq K$, so the same minimality gives $K=kR$.

Since $K=KI$ and $I\subseteq J$, its generator satisfies $k={\sum }_{\nu }k{r}_{\nu }{i}_{\nu }=kj$ for an element $j={\sum }_{\nu }{r}_{\nu }{i}_{\nu }\in J$. Thus $k(1-j)=0$. The unit criterion for the Jacobson radical makes $1-j$ invertible, implying $k=0$, contrary to $kI\ne 0$. Therefore $I=0$, andThe minimal-right-ideal proof of radical nilpotence is important here: one cannot apply the finite-generator form of Nakayama lemma directly to $J^m$ before proving it finitely generated. We made a suitable auxiliary right ideal cyclic before using the unit argument.

Work in the classical setting of varieties over an algebraically closed field $k$. A prevariety is separated when its diagonal morphism ${\Delta }_X:X\to X{\times }_{k}X$ is a closed immersion; for prevarieties the diagonal is already an immersion, so this is equivalent to its image being closed. A complete variety has the following closed-projection property: for every variety $Y$, projection $X{\times }_{k}Y\to Y$ maps closed subsets to closed subsets. This is universal closedness; since a variety is separated and of finite type, it is equivalent to the structural map being proper.

First consider projective space. In two sets of homogeneous coordinates its diagonal is defined byThese equations hold exactly when the two nonzero coordinate vectors are proportional. They are homogeneous in each coordinate set, so they define a closed subset of ${\mathbb{P}}^n\times {\mathbb{P}}^n$. On a common affine chart, the diagonal map is the usual closed immersion defined by equality of the affine coordinates. Thus projective space is separated. If $X\hookrightarrow {\mathbb{P}}^n$ is a projective embedding, its diagonal is the intersection of that closed diagonal with $X\times X$. Consequently every projective variety is separated.

Here is an algebraic proof of completeness, rather than an appeal to projective compactness in an analytic topology. It suffices to prove closedness of projection from projective space over an affine chart $Y$, with coordinate ring $A=k[Y]$. A closed subset $Z\subseteq {\mathbb{P}}^n\times Y$ is cut out by a homogeneous ideal $I\subseteq A[T_0,\dots ,T_n]$. Put $B=A[T_0,\dots ,T_n]/I$ and write $B_d$ for its degree-$d$ part. Each $B_d$ is a finitely generated $A$-module, because it is a quotient of the free module on the degree-$d$ monomials.

Suppose the fibre over $y\in Y$ is empty. The specialized homogeneous equations then have no nonzero common zero. By the Hilbert Nullstellensatz, some power of each $T_i$ belongs to the specialized ideal. If those powers are $T_i^{r_i}$, then every monomial of degree $d=1+{\sum }_i(r_i-1)$ belongs to that ideal. HenceThe Nakayama lemma gives $(B_d{)}_{{\mathfrak{m}}_y}=0$. Since $B_d$ is finitely generated, there is $a\in A\setminus {\mathfrak{m}}_y$ with $(B_d{)}_a=0$: choose generators and multiply finitely many elements annihilating their localizations. On the principal open subset $D(a)$, all graded pieces of degree at least $d$ vanish too, since the quotient is generated in degree one. Every fibre over $D(a)$ is therefore empty: a projective point would have a nonzero coordinate whose degree-$d$ power could not vanish. Thus the complement of the projection of $Z$ is open, proving the required closed-projection property.

An affine open cover of an arbitrary $Y$ establishes the same result there. Finally, a closed subset of $X\times Y$ is closed in ${\mathbb{P}}^n\times Y$ when $X$ is a projective variety, so its image is closed by the preceding argument. ThereforeOver a non-algebraically-closed field, the corresponding statement concerns schemes or geometric points. Projection of rational points alone need not be closed, so it must not replace the completeness definition.

The Jacobson radical is $J(R)={\bigcap }_{L}L$, where $L$ runs over the maximal right ideals. Every proper right ideal $I$ lies in such an $L$: the union of a chain of proper right ideals containing $I$ is a right ideal and remains proper because it omits $1$. Thus Zorn lemma applies.

The right-module form of Nakayama lemma says that a finitely generated module $M$ with $MJ=M$ is zero. More generally, if $N$ is a submodule and $M=N+MJ$, then $M=N$. To prove the first statement, suppose $M\ne 0$ and take a generating set $m_1,\dots ,m_s$ of minimal size. Since $MJ=M$, there are $a_i\in J$ withThe unit criterion for the Jacobson radical makes $1-a_s$ a unit. Hence $m_s$ is generated by the other $m_i$, a contradiction. Apply this result to the quotient module $M/N$ to obtain the general form.

For an injective module endomorphism $f$ of an Artinian module $M$, the descending chain condition givesfor some $n$. Given $x\in M$, write $f^n(x)=f^{n+1}(y)$; injectivity of $f^n$ gives $x=f(y)$. Thus $f$ is surjective, as asserted by Artinian modules are co-Hopfian.

Now $R$ is a semilocal ring and $\overline{V}=V/VJ$ is a finitely generated module over the right Artinian ring $\overline{R}=R/J$. A finite direct sum of copies of $\overline{R}$, and hence its quotient module $\overline{V}$, is an Artinian module. The module endomorphism $\alpha$ induces $\overline{\alpha }$ on $\overline{V}$, and the given inverse-image condition says precisely that $\ker \overline{\alpha }=0$. The preceding argument makes $\overline{\alpha }$ surjective, soThe quotient module $W=V/\alpha (V)$ is finitely generated and satisfies $WJ=W$. Applying Nakayama lemma yields $W=0$. Therefore $\alpha$ is surjective, by surjectivity from injectivity modulo the Jacobson radical.

Here $k$ is algebraically closed and $C$ is a connected smooth projective curve. For a scheme $T$ over $k$, let $\pi :C_T=C{\times }_{k}T\to T$. The degree-$d$ Picard functor of a curve assignsOne can equivalently take the associated faithfully flat sheaf. It is essential to divide out by line bundles from the base: a parameter space should remember the family on $C$, not an independent base line bundle.

Choose $p_0\in C(k)$. A class can be normalized as $L\otimes {\pi }^{\ast }(L{|}_{p_0\times T}{)}^{-1}$, with its canonical trivialization along $p_0\times T$. Since ${\pi }_{\ast }{\mathcal{O}}_{C_T}={\mathcal{O}}_T$, an automorphism of a line bundle is multiplication by a base unit, and preserving this normalization forces that unit to be $1$. Descent of line bundles is effective, so normalized bundles satisfy the faithfully flat sheaf condition with no automorphism ambiguity. Thus this normalized description works on families, including nonreduced bases, rather than only on $k$-points.

We construct a representing scheme by Picard charts from nonspecial divisors. Fix $N\ge 2g$ and put $r=N-g$. The symmetric product of a curve $C^{(g)}$ represents relative effective Cartier divisors on a curve of degree $g$ and has a universal divisor $\Delta$. This uses the scheme symmetric product, not the naive set of unordered $T$-points. Locally on a smooth curve, a divisor of length $m$ is described by a monic polynomial in a smooth coordinate; its coefficients are elementary symmetric functions, so this construction includes coincident points in any characteristic. These local descriptions give the usual effective-divisor interpretation of the symmetric product.

Let $W\subset C^{(g)}$ be the open set where $H^1(C,\mathcal{O}(E))=0$. It is open by semicontinuity theorem for coherent cohomology. The Riemann-Roch theorem gives $h^0(\mathcal{O}(E))=1$ there, so $E$ is the unique effective divisor in its line-bundle class. For every fixed effective divisor $B$ of degree $r$, take a copy $W_B$ of $W$, with familyof degree $N$. Normalize it along $p_0$ as above.

This chart represents the open subfunctor defined byIndeed, for a family satisfying that condition, cohomology and base change for line bundles on a curve makes $V={\pi }_{\ast }L(-B)$ a line bundle on $T$, with formation commuting with base change. The rank is one by Riemann-Roch theorem. The evaluation mapis a nonzero section on each fibre, up to the base twist. Its zero scheme is a relative effective Cartier divisor on a curve $E$ of degree $g$. To check the family assertion, source and target are flat over $T$ and the fibre maps are injections of line bundles on an integral curve, so their cokernel is flat over $T$. Their zeros are Cartier, and the zero scheme is proper with finite fibres of length $g$, hence finite flat of degree $g$. It therefore defines a unique map $T\to W_B$. Conversely this construction recovers the family $\mathcal{O}(B+E)$ modulo a base line bundle.

The base-change step controls infinitesimal families as well. Locally, a finite free complex computes the two cohomology groups. Vanishing of its degree-one cokernel on fibres makes the last differential surjective by the Nakayama lemma. Splitting that differential leaves a locally free degree-zero kernel of rank one and commutes with base change. This is the standard finite-complex form of the permitted cohomology semicontinuity and base-change results.

These chart subfunctors cover all degree-$N$ families. For a geometric-fibre line bundle $L$ of degree $N$, the preceding solution gives $H^1(L)=0$ and $h^0(L)=N+1-g=r+1$. Choose $r$ points successively so that each imposes one independent condition on the current section space. At each stage that space is nonzero; a nonzero section has finitely many zeros, so a point outside its zero set can be chosen, also outside the finitely many previously chosen points. For their sum $B$,and Riemann-Roch theorem forces $h^1(L(-B))=0$. The points may be chosen in $C(k)$ even when the fibre field extends $k$, since $C(k)$ is infinite and a nonzero section on the base-changed curve has only finitely many zeros. Thus fixed divisors $B$ defined over $k$ suffice. Upper semicontinuity gives an open neighbourhood on the base for each successful $B$.

It remains to glue the charts explicitly. On $W_B$, the locusis open. The line bundle inside the braces has degree $g$; its unique section up to scale has an effective zero divisor $E^{\prime }$. Base change and the evaluation construction make $E\mapsto E^{\prime }$ a morphism $W_{B{B}^{\prime }}\to W_{B^{\prime }}$. It satisfies $\mathcal{O}(B+E)\cong \mathcal{O}(B^{\prime }+E^{\prime })$. Reversing $B,B^{\prime }$ gives its inverse. On triple overlaps the transition maps compose correctly because the residual effective divisor is unique. Hence the copies $W_B$ glue along these open isomorphisms to a scheme $P^N$.

The normalized bundles $L_B$ glue as well: on overlaps their normalized isomorphisms exist and are unique, so they satisfy the cocycle condition. Let ${\mathcal{P}}_N$ denote the resulting universal normalized line bundle on $C\times P^N$. For any $T$, the open cover where the conditions for some $B$ hold gives compatible maps $T\to W_B$, and thus a unique map $T\to P^N$. Pulling back ${\mathcal{P}}_N$ reverses this procedure. We have established a natural bijectionThis proves representability of the full functor, not just a parametrization of individual line bundles. The nonspecial-divisor construction is also recorded in Stacks Project, Picard scheme of a curve.

Finally tensoring by ${\mathcal{O}}_C((N-d)p_0)$ is a natural isomorphism ${\mathrm{Pic}}_C^d\to {\mathrm{Pic}}_C^N$ for every integer $d$. Transporting the universal bundle back constructs its degree-$d$ representative $P^d$, usually denoted ${\mathrm{Pic}}_{C/k}^d$. Thus every degree component is represented. These identifications depend on the chosen point and twist; intrinsically the degree-zero component is a group scheme and each degree component is its torsor.

The root-of-unity specialization requires $q\ne 1$: we take $q$ to be a primitive cube root of unity. The displayed presentation contains $(q-q^{-1}{)}^{-1}$, so substituting $q=1$ literally is undefined. Its classical $U({\mathfrak{sl}}_2)$ limit instead has simple finite-dimensional modules of arbitrarily large dimension. This is why the primitive-root qualification matters.

First consider simple modules occurring in the quantum plane. Here $[3]=0$, $K^3=1$, andwith the analogous formula $F^3=0$. Thus the action factors through the finite-dimensional quotient with $E^3=F^3=0$, $K^3=1$, spanned by $F^i{K}^j{E}^l$ with $0\le i,j,l<3$. Any simple module over this quotient is finite-dimensional, since it is generated by one vector. In a nonzero simple submodule choose a $K$ eigenvector $v$ in $\ker E$. This is possible because $E$ is nilpotent and its kernel is $K$-invariant. Then the highest-weight string for quantum sl2 shows thatis invariant under $E,K$, and $F^{3}v=0$ makes it $F$-invariant. Simplicity gives $W=V$, proving $\dim V\le 3$.

The same bound holds even for arbitrary finite-dimensional simple modules over the unrestricted quantum enveloping algebra of sl2. Indeed $E^3,F^3,K^3$ are central: the string commutation identity has a factor $[3]=0$, and the $K$ commutation factors satisfy $q^6=1$. By Schur lemma, they act as scalars $\alpha ,\beta ,\kappa$, with $\kappa \ne 0$. The elementis also central. For example $[E,FE]=(K-K^{-1})E/(q-q^{-1})$, while $[E,\varphi (K)]=[\varphi (q^{-2}K)-\varphi (K)]E=-(K-K^{-1})E/(q-q^{-1})$; the $F$ commutator cancels in the same way. Let $\Omega$ act as $\gamma$.

If $\alpha =0$, choose the highest vector $v$ as above. Its three-vector $F$ string is invariant, now using $F^{3}v=\beta v$, so again its dimension is at most three. If $\alpha \ne 0$, choose a $K$ eigenvector $v$. The space $W=⟨v,Ev,E^{2}v⟩$ is stable under $E,K$ because $E^3=\alpha$, and it is stable under $F$ becauseIt therefore equals the simple module and has dimension at most three.

Over $\mathbb{C}$ this also excludes infinite-dimensional simple modules of the unrestricted algebra. It is a finite module over the central finitely generated commutative algebra over a field $Z_0=\mathbb{C}[E^3,F^3,K^3,K^{-3}]$, using the same 27 ordered spanning elements. A simple module $M$, generated by one vector, is consequently finite over $Z_0$. Choose a maximal ideal $\mathfrak{m}$ in its support. Nakayama lemma makes $M/\mathfrak{m}M\ne 0$; since $\mathfrak{m}M$ is a submodule, simplicity gives $\mathfrak{m}M=0$. The Hilbert Nullstellensatz gives $Z_0/\mathfrak{m}=\mathbb{C}$, so $M$ is finite-dimensional and the preceding argument applies. Thus the usual complex, primitive-root interpretation yields

Put $R={\mathcal{O}}_{X,x}$, $\mathfrak{m}={\mathfrak{m}}_x$ and $d=\dim R$. Smoothness makes $R$ a regular local ring, with residue field $k$. A list of local parameters $t_1,\dots ,t_d$ is a regular system of parameters: its images form a basis of the cotangent space of a local ring $\mathfrak{m}/{\mathfrak{m}}^2$. Set $J=(t_1,\dots ,t_d)$. The basis condition says $\mathfrak{m}=J+{\mathfrak{m}}^2$, so the finitely generated module $\mathfrak{m}/J$ equals $\mathfrak{m}(\mathfrak{m}/J)$. The Nakayama lemma gives

We now construct the requested formal power series, rather than assert that a local function has a convergent expansion. The associated graded ring of a regular local ring isIndeed these initial forms generate the graded ring, since the $t_i$ generate $\mathfrak{m}$. The Dimension theorem for Noetherian local rings gives $\dim {\mathrm{gr}}_{\mathfrak{m}}R=d$. A nonzero kernel in the surjection from the polynomial ring would make its quotient have dimension at most $d-1$, so that kernel is zero.

For $f\in R$, choose $P_0\in k$ to be its residue at $x$. Inductively suppose homogeneous polynomials $P_0,\dots ,P_{n-1}$ have been chosen such thatThe graded-ring isomorphism gives a unique homogeneous polynomial $P_n(T_1,\dots ,T_d)$ of degree $n$ representing the class of $r_n$ in ${\mathfrak{m}}^n/{\mathfrak{m}}^{n+1}$. Subtracting $P_n(t)$ makes the next remainder lie in ${\mathfrak{m}}^{n+1}$. HenceThis equality is interpreted in the adic completion of a module $\hat{R}=\underleftarrow{\lim }R/{\mathfrak{m}}^n$. The same induction applies to every compatible sequence of residues in this inverse limit, so evaluation $T_i\mapsto t_i$ defines an isomorphism $k[[T_1,\dots ,T_d]]\cong \hat{R}$. It is injective because the lowest nonzero homogeneous term of a series has nonzero initial form. The Krull intersection theorem gives ${\bigcap }_n{\mathfrak{m}}^n=0$, so $R$ embeds in its completion and the series determines $f$. This is the formal expansion at a smooth point; it depends on the chosen regular system of parameters, and does not claim that the infinite sum converges inside $R$. If $d=0$, the local ring is just $k$, and only the constant term is needed.

Use the usual Krull–Schmidt theorem setting, in which indecomposable summands have local endomorphism rings; the paper's rings $\mathcal{O},k$ satisfy it for the subgroups involved. The relative projective module condition for $H$ means that $M$ is a direct summand of a module induced from $H$, equivalently, $M$ is a summand of ${\mathrm{Ind}}_H^G{\mathrm{Res}}_H^{G}M$. A vertex of an indecomposable module is an inclusion-minimal such subgroup. A source of an indecomposable module at its module vertex $D$ is an indecomposable $RD$-summand $S$ of $M{\downarrow }_D$ such that $M\mid S{\uparrow }^G$.

Existence of a module source follows by decomposing $M{\downarrow }_D$ and using Krull–Schmidt theorem on the induced decomposition: the indecomposable summand $M$ must occur in the induction of one indecomposable summand $S$. Such $S$ has module vertex $D$ as a $D$-module, since relative projectivity of $S$ for a proper subgroup would make $M$ relatively projective for that subgroup.

In the paper's modular coefficient rings, an indecomposable module is a trivial source module when its module source is the rank-one trivial module $R$ of its module vertex. It is then a summand of $R[G/D]$, a permutation module. Conversely suppose $M$ is an indecomposable summand of a permutation module. Restriction to a module vertex $D$ is a sum of modules $R[D/E]$, one for each $D$-orbit of the permutation basis. For a $p$-group $D$, each transitive permutation module is indecomposable: over $k$ it has simple head of a module over the local ring $kD$, and over $\mathcal{O}$ a nontrivial splitting would give a nontrivial splitting after reduction. A module source $S$ must be one of these summands by Krull–Schmidt theorem. Its module vertex is $D$, whereas $R[D/E]$ is relatively $E$-projective. Thus $E=D$, and $S\cong R$. Therefore trivial module source is equivalent to being an indecomposable permutation-module summand.

The rank-one formulation uses the paper's coefficient rings. If the earlier general coefficient-ring wording were applied to a disconnected ring, it would need qualification: for $R=k\times k$ and $G=1$, $(k,0)$ is an indecomposable summand of the permutation module $R$, but is not the rank-one trivial module $R$. One can instead allow indecomposable summands of the trivial coefficient module as module sources in that generality.

For the homomorphism request, take $M_i=\mathcal{O}[G/H_i]$. Frobenius reciprocity and Mackey decomposition giveThe summand for $g$ is generated by the sum of the cosets in the $H_1$-orbit of $g{H}_2$. The identical orbit-sum construction over $k$ gives one copy of $k$ per double coset. These are free bases in both characteristics, since invariance means coefficients are constant on orbits, with no division by orbit size. HenceThe same argument applies to direct sums of coset permutation modules.

If $M$ is a trivial module source $kG$-module, choose a permutation lattice $X$ with $M\mid \bar{X}$. Its projection is an idempotent $f\in {\mathrm{End}}_{kG}(\bar{X})$. The endomorphism reduction is surjective by the preceding calculation. Its kernel is $\mathfrak{p}{\mathrm{End}}_{\mathcal{O}G}(X)$, and the endomorphism algebra is complete over $\mathcal{O}$. By idempotent refinement theorem, $f$ lifts to an idempotent $\hat{f}$. The lattice $\hat{f}X$ is a permutation-module summand, hence a trivial module source lattice, and reduces to $M$.

To prove uniqueness within the class of trivial module source lattices, let $L,L^{\prime }$ be two such lifts. They are summands of permutation lattices. Lift a modular isomorphism and its inverse between them by the same homomorphism-surjectivity argument, inserting the summand projections on either side. The lifted composites are identity modulo $\mathfrak{p}$ and therefore invertible over $\mathcal{O}$, by Nakayama lemma or the determinant criterion. The lifted maps are consequently inverse up to an invertible composite and give $L\cong L^{\prime }$. Thus each trivial module source module has a unique trivial module source lift up to isomorphism. This is permutation homomorphisms lift through modular reduction; uniqueness is not claimed among arbitrary lattice lifts.

A finite-dimensional group algebra has finite representation type of a group algebra when there are only finitely many isomorphism classes of finite-dimensional indecomposable modules. The definition concerns all dimensions, not merely finitely many modules in each fixed dimension.

Let $e_B$ be the block idempotent and choose a maximal Brauer pair $(D,b_D)$: $b_D$ is a block idempotent of $k{C}_G(D)$ with $b_D{\mathrm{Br}}_D(e_B)\ne 0$. The inertial index of a block iswhere $N_G(D,b_D)$ is the stabilizer of the chosen pair. Changing the pair by conjugation does not alter this number. For cyclic $D$, it is a prime-to-$p$ divisor of $p-1$. Here $n\ge 1$, as required for the order-$p$ subgroup $Q$ to exist.

Put $H=N_G(Q)$. Since $Q$ is characteristic in $D$, $N_G(D)\le H$, so Green correspondence applies. In fact the nonprojective modules in $B$ have a particularly simple correspondence. If $g\notin H$, a nontrivial intersection $D\cap {}^{g}D$ would contain the unique order-$p$ subgroup of both cyclic groups, giving $Q={}^{g}Q$ and hence $g\in H$, a contradiction. Thus every such intersection is trivial. The error modules in induction are projective. In restriction, an intersection $H\cap {}^{g}D$ cannot contain a conjugate of a nontrivial subgroup of $D$ by an element of $H$: such a conjugate contains $Q$, and that would again force ${}^{g}Q=Q$. The general Green correspondence therefore pairs all indecomposable modules with nontrivial module vertex contained in a conjugate of $D$. In particular it pairs the nonprojective modules of $B$ with the nonprojective modules of its Brauer correspondent $b$ in $H$; the block-compatible form of the correspondence selects $b$. Explicitly,where $P$ is projective and the restriction error family is $\mathcal{Y}=\{H\cap {}^{g}D:g\notin H\}$. One should not replace this restriction error term by projectives without an additional argument about $H$.

For the final structure calculation we use the general cyclic-defect structure theorem, which is permitted here as a general fact: the basic algebra of a split block with cyclic defect group is a Brauer tree algebra. Its tree has $e(B)$ edges, one simple module per edge, and exceptional multiplicityFor a one-edge tree the basic algebra presentation is $k[t]/(t^{m+1})$. This is the local presentation of the tree algebra, not an assertion that an arbitrary block with one simple module is automatically uniserial.

With $e(B)=1$ the tree has one edge and $m=p^n-1$. Thus the basic algebra of $B$ isIt has exactly one simple module, $k=A_0/(t)$. Its ideals are precisely $(t^i)$: for any nonzero element, factor out its lowest power of $t$, after which the remaining factor is a unit. The regular projective therefore has the unique submodule chainwith each consecutive factor one-dimensional and simple. A Morita equivalence from $A_0$ to $B$ preserves simple modules, projective covers, submodule lattices and composition length. HenceIn the split setting one may also write $B\cong M_{{\dim }_{k}S}(A_0)$. The projective cover has $p^n$ composition factors all isomorphic to $S$, so its vector-space dimension is $p^n{\dim }_{k}S$, not generally $p^n$.

There is also an elementary reason why the uniserial conclusion follows once the one-simple-module conclusion and finite type are known. If the basic local algebra had $\dim J/J^2\ge 2$, quotienting its radical square and all but two generators would give $k[x,y]/(x,y{)}^2$. The two-dimensional parameter family from Question 3(b) would then give infinitely many indecomposables. Thus $J/J^2$ is one-dimensional. Nakayama lemma makes $J$ generated by one element $t$, so its powers yield a chain and the basic algebra is $k[t]/(t^r)$. The cyclic-defect calculation above identifies $r=p^n$. This explains why finite representation type rules out branching in this projective cover.

Choose $a_1,\dots ,a_d\in \mathfrak{m}$ whose classes form a basis of the cotangent space of a local ring $\mathfrak{m}/{\mathfrak{m}}^2$, and put $J=(a_1,\dots ,a_d)$. Spanning gives $\mathfrak{m}=J+{\mathfrak{m}}^2$. Thus the quotient module $N=\mathfrak{m}/J$ satisfies $N=\mathfrak{m}N$. It is a finitely generated module, since $A$ is a Noetherian ring.

Here is the needed form of the Nakayama lemma, with proof. Choose generators $n_1,\dots ,n_r$ of $N$ and write $n_i={\sum }_j{c}_{ij}{n}_j$ with $c_{ij}\in \mathfrak{m}$. For the column vector $n$, $(I-C)n=0$. Multiply by the adjugate to obtain $\det (I-C)n=0$. Since $\det (I-C)\equiv 1(\mathrm{mod}\mathfrak{m})$, this determinant is a unit in the local ring $A$, so every $n_i=0$. Therefore $N=0$, proving $\mathfrak{m}=J$. Conversely, any generators of $\mathfrak{m}$ must span its cotangent space, so at least $d$ are needed. HenceHere $\mu (M)$ denotes the minimal number of generators of a module $M$.

For the additional minimum, the maximal ideal $\mathfrak{m}$ is itself $\mathfrak{m}$-primary, so it supplies an eligible ideal generated by $d$ elements. ConsequentlyThe minimum here is over primary ideals. A bound $\mu (Q)\le d$ for every $\mathfrak{m}$-primary ideal would be false. Indeed, for $A=k[[u,v]]$ and $\mathfrak{m}=(u,v)$, $d=2$ but $Q={\mathfrak{m}}^2=(u^2,uv,v^2)$ is primary and needs three generators. To verify primaryness, if $ab\in Q$ and $a\notin Q$, then $b$ cannot be a unit, so $b\in \mathfrak{m}$ and $b^2\in Q$. The three displayed monomials form a basis of $Q/\mathfrak{m}Q={\mathfrak{m}}^2/{\mathfrak{m}}^3$, proving the lower bound of three, while they already generate $Q$. This illustrates that primary ideals can need more generators than the maximal ideal.

For a finite-dimensional algebra over an algebraically closed field, basic algebra means $A/J(A)\cong k^n$, so its simple modules are one-dimensional and occur once in the semisimple quotient of the regular module. A hereditary ring has every submodule of a projective module projective; equivalently its global dimension is at most one.

Choose representatives $S_1,\dots ,S_n$ of the simple modules. Our Ext quiver convention gives one vertex $i$ for $S_i$ andWith primitive orthogonal idempotents $e_i$, put $P_i=A{e}_i$ and $J=J(A)$. Applying ${\mathrm{Hom}}_A(-,S_j)$ to $0\to J{P}_i\to P_i\to S_i\to 0$ givesThe restriction map from $\mathrm{Hom}(P_i,S_j)$ is zero since every map to a simple kills $J{P}_i$, and $P_i$ is projective. Thus the arrow multiplicity is $\dim e_j(J/J^2)e_i$. This specifies the orientation rather than silently replacing the quiver by its opposite.

Because $A$ is hereditary, $J{P}_i$ is projective. Its semisimple top has $a_{ij}$ copies of $S_j$, where $a_{ij}$ is the arrow multiplicity. Its projective cover therefore supplies an epimorphismThis splits by projectivity of $J{P}_i$. Its kernel is a summand contained in the radical of the source; projection onto that summand and Nakayama lemma force the kernel to be zero. HenceEvery arrow $i\to j$ forces $\dim P_i>\dim P_j$, so $Q$ has no directed cycles.

Choose lifts in $e_{j}J{e}_i$ of bases of the arrow spaces. Send the vertices and arrows of the path algebra to these idempotents and lifts, multiplying paths in the order $\beta \alpha$ for consecutive arrows $\alpha :i\to j$, $\beta :j\to l$. This defines an algebra homomorphism $kQ\to A$. It is surjective: vertices span $A/J$, arrows span $J/J^2$, and products show that paths of length $r$ span $J^r/J^{r+1}$. Nilpotence of $J$ then gives all of $A$.

Let $N_i$ be the number of directed paths starting at $i$, including the length-zero path. Acyclicity gives the same recursion $N_i=1+{\sum }_j{a}_{ij}{N}_j$, with value one at a sink. Induction from sinks gives $N_i=\dim P_i$. ThereforeThe surjective homomorphism is an isomorphism. This proves the basic hereditary algebra path-algebra theorem:

For each connected component $C$ of the underlying unoriented graph, $e_C={\sum }_{i\in C}{e}_i$ is a central idempotent. Paths never connect distinct components, so $kQ={⨁}_C{e}_{C}kQ$ as a direct sum of two-sided ideals. To prove each is a block, let $z$ be any central idempotent. Modulo the arrow ideal it has coordinates $c_i\in \{0,1\}$. Commuting $z$ with an arrow $i\to j$ and reducing modulo the square of that ideal gives $c_i=c_j$. Thus these coordinates are constant on each component. If they are zero on a component, the restriction of $z$ there is a nilpotent idempotent, hence zero; if they are one, apply that argument to $1-z$. Consequently

Write $a=t_1$, $b=t_2$, $c=t_3$, and $A=\mathbb{C}[a,b,c]/(ac-b^2)$. It is the quadratic cone invariant ringwhere the involution negates both $u$ and $v$. The invariant monomials have even total degree and are generated by $u^2,uv,v^2$; reduction by $ac=b^2$ proves the asserted presentation.

To verify the normal variety property, suppose $q\in \mathrm{Frac}(A)$ is integral over $A$. The same monic equation makes $q$ integral over $\mathbb{C}[u,v]$. This polynomial ring is an integrally closed domain, so $q\in \mathbb{C}[u,v]$. Since $q$ belongs to the fraction field of $A$, it is invariant under the involution, and hence belongs to $A$. Thus $X$ is normal. The gradient of its equation is $(c,-2b,a)$, so the Jacobian criterion shows that the origin $o$ is its only singular point of an algebraic variety. In particular $X$ is not smooth.

Take the prime Weil divisor $D=V_X(a,b)$ and $x=o$. We show that $D$ is not a Cartier divisor at $o$. In the local ringits prime ideal is $I=(a,b)R$. The quotient $I/\mathfrak{m}I$ has dimension of a vector space $2$ over $\mathbb{C}$: $a,b$ generate it, and they are independent since $\mathfrak{m}I\subset {\mathfrak{m}}^2$ while the relation $ac-b^2$ has no linear term. By Nakayama lemma, $I$ cannot be generated by one element.

On a normal variety, the ideal of an effective prime Weil divisor is the divisorial ideal of functions with order at least one along that divisor and order at least zero along every other prime Weil divisor. If $D$ were a Cartier divisor at $o$, a local defining rational function $h$ would identify this ideal with $hR$, because an integrally closed domain is the intersection of its height-one localizations at a prime ideal inside its fraction field. That would make $I$ principal, contradicting the calculation. Therefore $D$ has a nonzero class in the local divisor class group.

Now let $D^{\prime }=D+\mathrm{div}(q)$ be any linearly equivalent Weil divisor. If $o$ were outside its support of a Weil divisor, there would be an open subset containing $o$ on which $D^{\prime }$ is zero. On that neighbourhood $D=-\mathrm{div}(q)$ would be a principal Weil divisor, hence a Cartier divisor, which is impossible. Therefore this point is unavoidable in every representative:The argument does not assume that $D^{\prime }$ is effective. For comparison, $\mathrm{div}(a)=2D$ because on the chart $c\ne 0$, $a=b^2/c$; the obstruction has order two, as in the Divisor class group of an A-type surface singularity.

Let $R$ be a commutative ring and $f=(f_1,\dots ,f_r)$. The Koszul complex packages these elements and their relations in a finite chain complex of free modules. With $E=R^r$ and basis $e_1,\dots ,e_r$, set $K_i(f;R)={\bigwedge }^{i}E$ for $0\le i\le r$, and defineDeleting two distinct basis elements in the two possible orders gives opposite signs and the same coefficient, so ${\partial }^2=0$. Equivalently $K(f;R)$ is the tensor product of chain complexes of $r$ two-term complexes $[R\stackrel{f_i}{\to }R]$ in degrees one and zero. The tensor differential is $\partial (u\otimes v)=\partial u\otimes v+(-1{)}^{\mathrm{deg}u}u\otimes \partial v$. This fixes the sign convention. On the exterior algebra, the differential is a graded derivation determined by $\partial e_i=f_i$.

For an $R$-module $M$, define the Koszul complex with module coefficients by $K(f;M)=K(f;R){\otimes }_{R}M$. Its degree-zero Koszul homology isIts higher Koszul homology measures the failure of these equations to form a regular sequence on a module. Each $f_j$ annihilates every homology group. Indeed the Koszul homotopy for multiplication by a generator is $h_j(u)=e_j\wedge u$, and the graded product rule givesFor coefficients in $M$ the same identity holds after tensoring. Thus multiplication by $f_j$ is zero on homology, and the homology groups are naturally modules over $R/(f)$. If $(f)=R$, choose ${\sum }_j{a}_j{f}_j=1$; then ${\sum }_j{a}_j{h}_j$ is a contracting homotopy, so the entire complex is contractible. These identities are also useful after localization of a ring: wherever one generator is a unit, the complex has zero homology.

The construction is functorial under a ring homomorphism, and base change gives $K(f;R){\otimes }_{R}S=K(\bar{f};S)$, since its terms are free with the displayed basis and differential. An invertible change of generators $f_i={\sum }_j{u}_{ij}{g}_j$ gives a chain isomorphism by sending $e_i$ to ${\sum }_j{u}_{ij}{e}_j$ and extending to exterior powers. In particular, permuting the generators changes only the exterior signs, not the isomorphism class of the complex.

The essential exactness theorem is that a regular sequence on a module gives zero positive Koszul homology. Recall that $f_i$ must act injectively on $M/(f_1,\dots ,f_{i-1})M$, with final quotient nonzero. Put $f^{\prime }=(f_1,\dots ,f_{r-1})$. Adding the final two-term complex identifies $K(f;M)$ with the mapping cone of multiplication by $f_r$ on $K(f^{\prime };M)$, with the chosen tensor signs. The long exact sequence in homology consequently has segmentsFor $r=1$, positive homology is exactly $\ker (f_1:M\to M)$. By induction the preceding homology vanishes above zero. The final injectivity assumption kills $H_1(f;M)$, while the exact sequence kills every $H_i$ for $i>1$. Thereforefor a regular sequence on a module. In particular $K(f;R)$ is a finite free resolution of $R/(f)$ when $f$ is a regular sequence, with ranks $\left(\genfrac{}{}{0ex}{}{r}{i}\right)$ and length $r$.

There is a precise converse under local finiteness assumptions. Suppose $(R,\mathfrak{m})$ is a Noetherian local ring, $0\ne M$ is finitely generated, and all $f_i\in \mathfrak{m}$. If positive Koszul homology vanishes, the same exact sequence makes multiplication by $f_r$ surjective on every $H_i(f^{\prime };M)$ for $i>0$. These modules are finitely generated because the ring is Noetherian and the terms of the complex are finite modules. The Nakayama lemma gives $H_i(f^{\prime };M)=0$. Induction makes $f^{\prime }$ regular on $M$, and the degree-one part of the exact sequence makes $f_r$ injective on $H_0(f^{\prime };M)$. The final quotient is nonzero, again by the Nakayama lemma. This proves the Koszul acyclicity criterion in a Noetherian local ring:The hypotheses matter: a unit ideal makes the complex contractible but cannot be a proper regular sequence.

Examples make both sides visible. For $R=k[x,y]$ and $f=(x,y)$, the Koszul resolution isThe signs agree with $\partial (e_1\wedge e_2)=x{e}_2-y{e}_1$. For $R=k[x]$ and $f=(x,x)$, the degree-one cycles are $R(1,-1)$ and the boundaries are $xR(1,-1)$; hence $H_1\cong R/(x)$, while $H_2=0$. The repeated element has exposed a relation that the first element already kills.

The Koszul complex also computes derived functors. If $f$ is a regular sequence on $R$, its free resolution givesHere no regularity of $f$ on $M$ is assumed. For $S=k[x_1,\dots ,x_r]$, tensor the Koszul resolution on the variables with $k$. Every differential becomes zero, soThe top group is nonzero, showing that a free resolution of $k$ cannot be shorter than $r$.

Finally, wedging complementary exterior degrees gives a perfect pairing $K_i{\otimes }_R{K}_{r-i}\to {\bigwedge }^r{R}^r\cong R$. It identifies the dual cochain complex with the degree-reversed Koszul complex, after the appropriate signs. This self-duality of the Koszul complex shows, for a regular sequence, thatThus the same explicit complex simultaneously records quotient equations, regularity, relations, Tor functor computations and Ext functor computations. Its finiteness and exterior structure are what make it especially effective in commutative algebra.