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On an affine chart $\mathrm{Spec}A\subseteq V$, the module of Kähler differentials ${\Omega }_{A/k}$ is generated by symbols $da$ subject to $k$-linearity and $d(ab)=adb+bda$. It represents $k$-derivations. These modules commute with localization, so their associated quasi-coherent sheaves glue to the Kähler differential sheaf ${\Omega }_{V/k}^1$. It is coherent: if $A=k[x_1,\dots ,x_N]/I$ and $I=(f_1,\dots ,f_s)$, then it has the finite presentation of a moduleFor a closed point $P$, put $R={\mathcal{O}}_{V,P}$ with maximal ideal $\mathfrak{m}$ and residue field $k$. The Zariski tangent space is ${\mathrm{Hom}}_k(\mathfrak{m}/{\mathfrak{m}}^2,k)$, equivalently ${\mathrm{Der}}_k(R,k)$. A derivation kills constants and ${\mathfrak{m}}^2$, so it factors through $a\mapsto a-a(P)$; conversely every linear functional on $\mathfrak{m}/{\mathfrak{m}}^2$ defines such a derivation by the product rule. The universal property of Kähler differentials therefore identifies this space withIn particular, ${\Omega }_{V,P}^1/{\mathfrak{m}}_P{\Omega }_{V,P}^1\cong {\mathfrak{m}}_P/{\mathfrak{m}}_P^2$, the algebraic cotangent space.

A point is a smooth point of a variety when its local ring is a regular local ring; over this algebraically closed field this says ${\dim }_k{T}_{V,P}=n=\dim V$. Tensor the finite presentation of a module above with $k(P)$. The tangent space is the kernel of the Jacobian matrix $J(P)$, hence has dimension $N-\mathrm{rank}J(P)$. This proves the Jacobian criterionFor a smooth irreducible variety $V$, choose at each point an invertible $(N-n)$-minor and shrink the affine chart so that it remains invertible. Its relations eliminate $N-n$ of the generators of ${\Omega }_{A/k}$, yielding a surjection $A^n\to {\Omega }_{A/k}$. At the generic point, its target has dimension $n$: over a perfect ground field, a separating transcendence basis of the function field has differentials forming a basis. Equivalently this follows from the assumed density of the smooth locus. The kernel therefore becomes zero over the fraction field of the integral domain $A$. As a submodule of $A^n$, it is a torsion-free module, so it is already zero. Thus these maps give local isomorphisms with ${\mathcal{O}}_V^n$, proving local freeness of differentials on a smooth variety with rank $n$.

For an affine chart $\mathrm{Spec}A\subseteq V$, write $W\cap \mathrm{Spec}A=\mathrm{Spec}(A/I)$ and $\mathcal{M}=\tilde{M}$. The restriction of a module sheaf to a closed subvariety is $\widetilde{M{\otimes }_{A}A/I}=\widetilde{M/IM}$. The Conormal exact sequence for Kähler differentials isIt follows from the generators and relations: passing to $A/I$ imposes precisely the additional relations $df=0$ for $f\in I$. The first map is well defined because $d(I^2)$ lies in $I{\Omega }_{A/k}$. Glue these exact module sequences, using exactness of localization, to obtainIn the final assertion, interpret locally principal subvariety as a proper local hypersurface. Its ideal on each chart is $(f)$ with $f\ne 0$. Since $V$ is an irreducible variety and reduced, $f$ is a non-zero-divisor, and $A/(f)\to (f)/(f^2)$, $\bar{a}\mapsto af$, is an isomorphism. These local rank-one descriptions make the conormal sheaf ${\mathcal{I}}_W/{\mathcal{I}}_W^2$ invertible.

Because $W$ is not contained in the singular locus of $V$, there is a dense open subset of $W$ where both $V$ and $W$ are smooth varieties. At a closed point there, $\dim W=\dim V-1$ by the Krull principal ideal theorem. The tangent description $T_{W,P}=\ker (df:T_{V,P}\to k)$ then forces $df\ne 0$. Hence the conormal map is injective at the generic point of $W$. Its kernel is a subsheaf of a line bundle on the integral variety $W$, so it is a torsion-free sheaf; a torsion-free sheaf with zero generic fibre is zero. This proves conormal injectivity for a generically smooth Cartier divisor. If zero equations were allowed in the phrase locally principal, $W=V$ would be a counterexample to invertibility; the proper-hypersurface convention is essential.

Regard the bifunctor as $D^{(2)}:K\to [J,\mathcal{C}]$. Its pointwise categorical limit $E={\lim }_k{D}^{(2)}k$ satisfieswith its action on morphisms obtained by the construction in part (a). Applying the limit-preservation result from part (d) givesThe comparison isomorphism is canonical relative to the chosen categorical limit cones: it is the unique morphism identifying all the projections to $D(j,k)$. This proves commutation of iterated categorical limits, including an empty $J$ or $K$. Equivalently, both iterated constructions have the universal property of the categorical limit of $D$ on $J\times K$.

It is enough to prove the solution-set condition and then apply the general adjoint functor theorem. Fix $D\in \mathcal{D}$ and a morphism $d:D\to UC$. We will factor it through one object in a set depending only on $D$.

First construct a minimal supported subobject. Among the subobjects $m:M\hookrightarrow C$ for which $d=U(m)d_M$ for some $d_M:D\to UM$, include $1_C$ and take their intersection $m_0:C_0\hookrightarrow C$. This is a small intersection, by well-poweredness. It exists by completeness as the categorical limit of the diagram consisting of these monomorphisms into $C$. Its map to $C$ is a monomorphism: two maps with the same composite to $C$ have equal projections to every $M$, since each $m$ is monic, and are then equal by the categorical limit property. One can equivalently construct these intersections by pullbacks in a category and small products in a category, using stability of monomorphisms under pullback in a category.

Choose the factorizations $d_M$. They form a compatible categorical cone into the image diagram under $U$, all with common composite $d$ to $UC$. Preservation of small categorical limits yields $d_0:D\to U{C}_0$ with $U(m_0)d_0=d$. If $n:N\hookrightarrow C_0$ is another subobject through which $d_0$ factors after applying $U$, then $m_{0}n$ is among the original supported subobjects. The intersection property gives $r:C_0\to N$ with $m_{0}nr=m_0$. Since $m_0$ is monic, $nr=1_{C_0}$; since $n$ is monic, also $rn=1_N$. Thus every supported subobject of $C_0$ is invertible.

For each member $Q_i$ of the small cogenerating family, considerThis map is injective. If $U(q)d_0=U(q^{\prime })d_0$, preservation of the equalizer of $q,q^{\prime }$ makes $d_0$ factor through the image of that equalizer. Minimality makes its inclusion an isomorphism, so $q=q^{\prime }$.

Write $R_i=\mathcal{D}(D,U{Q}_i)$; it is a set by local smallness of $\mathcal{D}$. Let $S_i\subseteq R_i$ be the image of ${\theta }_i$. For each $s\in S_i$, there is exactly one corresponding $q_{i,s}:C_0\to Q_i$. These maps define the evaluation embedding into cogenerator productsIt is a monomorphism: if $eg=eh$ and $g\ne h$, cogeneration supplies some $q:C_0\to Q_i$ distinguishing $g,h$; that $q$ is one of the projections of $e$, a contradiction. The product is small. It is important to use the subfamilies $S_i$, since some missing coordinate in $R_i$ need not correspond to a morphism out of $C_0$.

There are only a set of possible families $(S_i\subseteq R_i{)}_{i\in I}$. For each such family form $P_S$, choose a set of representatives $N\hookrightarrow P_S$ of its subobjects, and take all pairs $(N,d_N:D\to UN)$. Local smallness of $\mathcal{D}$ and well-poweredness make their union a set. In the case just constructed, $e$ identifies $C_0$ with one chosen representative $N$. Transporting $d_0$ to $UN$ and composing the inverse identification with $m_0$ gives a factorization of the original $d$ through that pair. Therefore these pairs form a weakly initial set in $(D\downarrow U)$.

The general adjoint functor theorem now supplies a left adjoint to $U$. Conversely, a right adjoint preserves small categorical limits, either by the adjunction hom-set bijections and the hom-set detection of categorical limits, or directly from their universal properties. Hence the stated special theorem is proved in both directions.

Write $K:\mathcal{B}\to \mathcal{C}$, $J:\mathcal{C}\to \mathcal{B}$, with adjunction unit $\rho$ and adjunction counit $\epsilon$. The adjunction givesFor $f:C\to C^{\prime }$, precomposition with ${\epsilon }_C$ followed by this bijection givesby the triangle identities for an adjunction. Thus $J$ is fully faithful exactly when precomposition with ${\epsilon }_C$ is bijective for every $C,C^{\prime }$.

A morphism $e:P\to Q$ with this property is an isomorphism. Surjectivity for target $P$ supplies $u:Q\to P$ with $ue=1_P$. Since $(eu)e=e=1_{Q}e$, injectivity for target $Q$ gives $eu=1_Q$. Conversely, precomposition with an isomorphism is always bijective. Applied to every ${\epsilon }_C$, this proves the fully faithful right-adjoint criterion:When the adjunction counit is invertible, the inverse to $f\mapsto Jf$ is explicitly $h\mapsto {\epsilon }_{C^{\prime }}K(h){\epsilon }_C^{-1}$.

If ${\rho }_B:B\to KB$ is an isomorphism, then $B$ is isomorphic in $\mathcal{B}$ to an object of $\mathcal{C}$. Since $\mathcal{C}$ is a replete subcategory, it contains $B$. Thus invertibility of the unit implies membership, completing the cycle of implications:

If $B$ belongs to the full subcategory $\mathcal{C}$, the adjunction and fullness of its inclusion give, for every $B^{\prime }\in \mathcal{B}$,Therefore membership implies the hom-set condition. In addition, $J$ is fully faithful, so the adjunction counit is invertible by the preceding criterion. The triangular equation $J{\epsilon }_B{\rho }_B=1_B$ then shows that ${\rho }_B$ is an isomorphism. As in the printed inclusion notation, $J$ is suppressed when writing $KB$ as an object of $\mathcal{B}$.

Suppose every algebra action is an isomorphism. Its inverse is ${\eta }_A$, by the unit law for a monad algebra. For any underlying morphism $f:A\to B$ between two algebras for a monad, naturality of $\eta$ givesTherefore every underlying morphism is automatically a monad algebra morphism. The forgetful functor is always faithful, because a monad algebra morphism has no data beyond its underlying morphism, and it is now full as well. Hence invertible algebra actions imply

If $G^{\mathbb{T}}$ is fully faithful, the criterion proved for a right adjoint says that the adjunction counit of $F^{\mathbb{T}}⊣G^{\mathbb{T}}$ is a natural isomorphism. Its component at $(A,a)$ is precisely the algebra action $a$. In particular, its component at the free algebra for a monad $(TX,{\mu }_X)$ has underlying morphism ${\mu }_X:T^{2}X\to TX$. Thus every ${\mu }_X$ is an isomorphism, and the natural transformation $\mu$ is invertible. This closes the cycle and proves all four conditions equivalent:

The Weak Hilbert Nullstellensatz states that if $k$ is an algebraically closed field, every maximal ideal of $k[Y_1,\dots ,Y_s]$ isEquivalently, every proper ideal has a common zero. We prove the more general field statement, the Zariski lemma: a field finitely generated as a $k$-algebra is a finite field extension of $k$.

Write such a field as $L=k[b_1,\dots ,b_s]$. Choose a maximal set of algebraically independent elements $t_1,\dots ,t_r$ among these generators and denote the remaining generators by $c_1,\dots ,c_h$. Each $c_j$ is an algebraic element over $K=k(t_1,\dots ,t_r)$, so $L/K$ is a finite field extension. Clearing the denominators in their monic equations, choose a nonzero $d\in k[t_1,\dots ,t_r]$ such that each $c_j$ is an integral element overBecause $L$ is a field and $d^{-1}\in L$, we have $L=D[c_1,\dots ,c_h]$. Thus $L$ is an integral extension of $D$.

A subring over which a field is integral must itself be a field. Indeed, for $0\ne a\in D$, an integral equation for $a^{-1}$ multiplied by $a^{m-1}$ expresses $a^{-1}$ as an element of $D$. But if $r>0$, the ring $D$ is not a field: choose an irreducible polynomial $q(t_1)\in k[t_1]$ which does not divide $d$. There are infinitely many monic irreducible polynomials in $k[t_1]$, by the usual product-plus-one argument, whereas only finitely many can divide the nonzero $d$. The prime ideal $(q)$ stays proper after inverting $d$, so $q$ has no inverse in $D$. Therefore $r=0$, proving that $L/k$ is a finite field extension, and in particular an algebraic field extension.

Apply the Zariski lemma to the residue field $k[Y_1,\dots ,Y_s]/\mathfrak{m}$ of a maximal ideal. When $k$ is an algebraically closed field, this field is $k$ itself. The images $a_i$ of $Y_i$ then give $\mathfrak{m}=(Y_1-a_1,\dots ,Y_s-a_s)$, proving the Weak Hilbert Nullstellensatz.

For the real Laurent polynomial ring, the same argument applies because it is the finitely generated algebraEvery residue field at a maximal ideal is consequently a finite field extension of $\mathbb{R}$, and so is $\mathbb{R}$ or $\mathbb{C}$. Here we use the fundamental theorem of algebra, which makes $\mathbb{C}$ the algebraic closure of $\mathbb{R}$ of degree two. The images of $X_1,X_2$ must be nonzero since these elements are units. It follows that the answer iswith $(\alpha ,\beta )$ and $(\overline{\alpha },\overline{\beta })$ defining the same maximal ideal. If both coordinates are real, evaluation has image $\mathbb{R}$; otherwise its image is $\mathbb{C}$. Thus every displayed kernel is maximal, and the Zariski lemma shows that this list is exhaustive.

For explicit generating sets of an ideal, there are three cases. For real nonzero $a,b$,If $\alpha =a\in {\mathbb{R}}^{\times }$ and $\beta =s+it$ with $t\ne 0$,If $\alpha =u+iv$ with $v\ne 0$ and $\beta =s+it$, thenThe last two quotients are $\mathbb{C}$; the nonzero coordinate condition ensures that the Laurent inverses exist in them. There are no further coincidences in the classification: equal kernels induce an $\mathbb{R}$-isomorphism between the evaluation image fields. The only $\mathbb{R}$-automorphisms of $\mathbb{C}$ are the identity and complex conjugation, so the two pairs must agree or be conjugate.

To prove the intersection assertion, fix a prime ideal $P$ and an element $f\notin P$. The integral domain $B=R/P$ has nonzero image $\bar{f}$. Its localization of a ring $B[{\bar{f}}^{-1}]$ is nonzero and still a finitely generated algebra over $\mathbb{R}$, so it has a maximal ideal $\mathfrak{n}$. By the Zariski lemma, the field $L=B[{\bar{f}}^{-1}]/\mathfrak{n}$ is finite over $\mathbb{R}$.

The image of $B$ in $L$ is a subring containing $\mathbb{R}$ and is a finite-dimensional $\mathbb{R}$-vector space. It is an integral domain, hence a field: multiplication by any nonzero element is an injective linear map of a finite-dimensional vector space and therefore surjective. The kernel of $B\to L$ is thus a maximal ideal which avoids $\bar{f}$, since $\bar{f}$ is invertible in $L$. Its inverse image in $R$ is a maximal ideal containing $P$ but avoiding $f$. As this works for every $f\notin P$,This is the Jacobson ring property for this Laurent polynomial ring.

For a nonzero finitely generated module $M$, its set of associated primes isHere ${\mathrm{Ann}}_R(m)=\{a\in R:am=0\}$ is the annihilator of the element. The Laurent polynomial ring $R$ is a Noetherian ring, so the nonempty collection of annihilators of nonzero elements of $M$ has a maximal member, say $P={\mathrm{Ann}}_R(m)$. It is proper because $m\ne 0$.

If $ab\in P$ and $b\notin P$, then $bm\ne 0$ andMaximality forces ${\mathrm{Ann}}_R(bm)=P$, hence $a\in P$. Thus $P$ is a prime ideal, proving $\boxed{{\mathrm{Ass}}_R(M)\ne \varnothing }$. This argument in fact only needs that $R$ is Noetherian, not the finite generation of $M$.

A proper ideal $Q$ is $P$-primary when its radical of an ideal is $P$ andEquivalently, $ab\in Q$ and $b\notin P$ imply $a\in Q$; equivalently, every zero divisor of $R/Q$ is nilpotent. These formulations distinguish the actual primary ideal property from the weaker assertion that there is only one minimal prime ideal.

For the final example, set $u=X_1-1$ and $v=X_2-1$. ThenSince $R/P\cong \mathbb{R}[X_2,X_2^{-1}]$ is an integral domain, $P$ is a prime ideal. Moreover $I\subseteq P$ and $u^2\in I$, so $\sqrt{I}=P$. Consequently $P$ is the unique minimal prime ideal over $I$.

Nevertheless $uv\in I$, while $u\notin I$ and no power of $v$ belongs to $I$. For the first assertion, reduce modulo $v$: the image of $I$ is $(u^2)$ in $\mathbb{R}[X_1,X_1^{-1}]$, and $u$ is not a multiple of $u^2$ because $u$ is a nonunit in this integral domain. For the second, $v\notin P$, so $v^j\notin P\supseteq I$ for every $j\ge 1$. Thus $I$ is not a $P$-primary ideal: a unique minimal prime does not imply primary. In $R/I$, the nonzero class of $u$ is killed by $v$, a zero divisor which is not nilpotent.

For a finitely generated module $V={⨁}_d{V}_d$ over the graded ring $S=k[X_1,\dots ,X_n]$, with each $X_i$ of degree one, its Poincare series of a graded module isThis is also its Hilbert series. One usually takes $d\ge 0$; allowing an integer grading also permits finitely many negative degrees. Homogeneous generators of degrees $a_1,\dots ,a_h$ give a surjection ${⨁}_{j}S(-a_j)\to V$, where the graded shift is defined by $V(-a{)}_d=V_{d-a}$. Consequently every $V_d$ is a finite-dimensional vector space and $V_d=0$ for all sufficiently negative $d$.

The Hilbert-Serre theorem in the standard grading saysFor a nonnegatively graded $V$, the numerator is in $\mathbb{Z}[t]$. Equivalently, the only possible pole of this rational expression is at $t=1$, with order at most $n$.

We prove the Hilbert-Serre theorem by induction on the number of variables. For $n=0$, $S=k$ and $V$ is a finite-dimensional graded module; thus $P_V(t)$ is a Laurent polynomial. Suppose $n>0$ and putThe Hilbert basis theorem makes $S$ a Noetherian ring, so $K$ is a finitely generated module, as is $C$. Both are graded modules killed by $X_n$, hence finitely generated over $S^{\prime }=k[X_1,\dots ,X_{n-1}]$.

The graded exact sequenceidentifies the kernel and cokernel of multiplication by $X_n$. Taking the alternating sum of the finite-dimensional degree components yieldssoBy induction the right side has denominator $(1-t{)}^{n-1}$, which proves the required denominator $(1-t{)}^n$. Since $(1-t{)}^n{P}_V(t)$ has no negative powers when $V$ is nonnegatively graded, its Laurent polynomial numerator is an ordinary polynomial in that case.

For $n\ge 1$, this also implies the usual Hilbert polynomial consequence. Writing $Q(t)={\sum }_j{q}_j{t}^j$ and usinggives, for all sufficiently large $d$,a polynomial in $d$ of degree at most $n-1$. For $n=0$, the graded pieces are eventually zero.

For the final request, a free resolution of $V$ is an exact sequencewith each $F_i$ a free module. In the graded setting we take finite direct sums of shifts $S(-a)$ and maps preserving degree. The resolution has length at most $n$ if $F_i=0$ for every $i>n$. The syzygy modules are the successive kernels which record the relations among generators, then relations among those relations, and so on.

Put $\mathfrak{m}=(X_1,\dots ,X_n)$, the homogeneous maximal ideal with $S/\mathfrak{m}=k$. Choose a homogeneous $k$-basis of $V/\mathfrak{m}V$ and lift it to $V$. These lifts generate $V$ by the graded Nakayama lemma: a bounded-below graded module $W$ satisfying $W=\mathfrak{m}W$ must be zero, since a nonzero homogeneous element of least degree could not be a sum of variables times elements of lower degrees. Apply this to the quotient by the submodule generated by the chosen lifts. This gives a surjection $F_0\to V$ inducing an isomorphism modulo $\mathfrak{m}$.

Its kernel is finitely generated because $S$ is a Noetherian ring. Repeat the construction for that kernel, then for each subsequent kernel. We obtain a minimal graded free resolution, meaningIndeed, at each stage the free cover induces an isomorphism modulo $\mathfrak{m}$, so its kernel lies in $\mathfrak{m}$ times its source. The sequence is exact by construction, though it may at first appear infinite.

The Koszul complex on $X_1,\dots ,X_n$ iswith basis vectors $e_j$ given degree one, and differentialThe hat means omission of that factor. Every pair of terms in ${\partial }^2$ cancels with opposite signs, so this is a chain complex. The sequence $X_1,\dots ,X_n$ is a regular sequence: after quotienting by the first $j-1$ variables, the next variable is a non-zero-divisor in the remaining polynomial ring.

The Koszul complex is consequently exact in positive degrees and has $H_0=k$, so it is a Koszul resolution of $k$ of length $n$. For completeness, one proves this by induction: for one variable it is $0\to S\stackrel{X_1}{\to }S\to S/(X_1)\to 0$. Appending $X_j$ constructs the mapping cone of multiplication by $X_j$ on the previous chain complex. Its homology is zero in positive degrees because $X_j$ acts injectively on $S/(X_1,\dots ,X_{j-1})$; its degree-zero homology is the further quotient. Here a mapping cone combines a chain complex with a shifted copy and adds the given multiplication map to the differential, thereby measuring its kernel and cokernel on homology.

The Tor functor ${\mathrm{Tor}}_i^S(V,k)$ is defined as the degree-$i$ homology of $F_{\bullet }{\otimes }_{S}k$ for a free resolution $F_{\bullet }$ of $V$. It can equally be computed as the homology of $V{\otimes }_S{K}_{\bullet }$, using the Koszul resolution of $k$. To see the equality, form the double complex $F_{\bullet }{\otimes }_S{K}_{\bullet }$. Taking homology first along $F_{\bullet }$ leaves $V{\otimes }_S{K}_{\bullet }$, since each $K_i$ is a free module; taking homology first along $K_{\bullet }$ leaves $F_{\bullet }{\otimes }_{S}k$, since each $F_i$ is free. The finite sums along each total degree identify both with the homology of the total complex. This also explains the symmetry used in computing the Tor functor.

Since $K_i=0$ for $i>n$, we have ${\mathrm{Tor}}_i^S(V,k)=0$ for $i>n$. On the other hand, all differentials of the minimal graded free resolution become zero after tensoring with $k=S/\mathfrak{m}$, soThus $F_i/\mathfrak{m}{F}_i=0$ for $i>n$, and the graded Nakayama lemma implies $F_i=0$ for those $i$. We have proved the required Hilbert syzygy theorem:where some initial $F_i$ may be zero. The argument supplies finitely generated graded free modules, which is stronger than merely giving an ungraded free resolution.

For $M>a\ge 0$ and $N>b\ge 0$, defineThe normal form theorem for a free product giveswhere $X,Y,T$ map to $x^M,y^N,t(a,b)$. To see injectivity, expand an alternating word in $⟨x^M,y^N⟩$ and powers of $t(a,b)$. The conjugating $x,y$ factors commute with the lattice syllables, so each intervening nonidentity lattice syllable stays nonidentity in the expanded free-product normal form.

Its intersection with $T_K$ is the free group with basisThis follows by moving all $x^M,y^N$ factors to the right; the zero-exponent lattice remainder characterizes membership in the kernel.

The required isomorphisms are specified on the three generators byandEach source and target has the same presentation $⟨X,Y,T\mid [X,Y]=1⟩$, and each displayed assignment identifies its three abstract generators. Hence these are genuine isomorphisms, not merely maps of generating sets.

On the kernel basis they give exactly the machine transitions:For example, rewrite the first input as $x^{-mu}{y}^{-mv}t(a_i,b_i)x^{mu}{y}^{mv}$ and apply ${\varphi }_i$ to each factor. These identities also hold for negative $u,v$, since they are identities in the groups.

Form the HNN extension which centralizes $L$:This is the extension associated to the identity isomorphism $L\to L$; relations on the displayed finite generating set imply $k^{-1}hk=h$ for all $h\in L$. Since $K_{\mathcal{M}}$ is finitely presented and $I,J$ are finite, this is a finite group presentation.

For any $g\in K_{\mathcal{M}}$,The forward implication follows from Britton's lemma: if $g\notin L$, the word $k^{-1}gk{g}^{-1}$ is reduced with two stable letters and cannot be the identity. The reverse implication is the defining centralization relation. Consequently the computable wordsatisfiesAn algorithm for the word problem for a group $G_{\mathcal{M}}$ would therefore decide the nonrecursive set $H_0(\mathcal{M})$, a contradiction. Hence $\boxed{G_{\mathcal{M}}\text{ has unsolvable word problem}}$.

Suppose $w\ne 1$ in $P$. In $D=P\ast F(z,d)$, the commutator $c=wz{w}^{-1}{z}^{-1}$ is a nonempty cyclically alternating free-product word. Every positive power remains reduced, so $c$ has infinite order. Each $g_j$ also has infinite order: this is immediate for $z,d$; if $x_i=1$, then $x_{i}z=z$, and otherwise $x_{i}z$ is a cyclically alternating word of length two.

Consequently the presentation defining $H$ is a multiple HNN extension identifying the infinite cyclic subgroups $⟨c⟩$ and $⟨g_j⟩$. By the normal form theorem for an HNN extension, $D$ embeds in $H$.

The subgroupis a free group on exactly these $n+3$ elements. For a proof, view $D$ as $(P\ast ⟨z⟩)\ast ⟨d⟩$. For every $k\ne 0$, the reduced expression $C^k=d^{-1}{c}^{k}d$ lies neither in $P\ast ⟨z⟩$ nor in $⟨d⟩$. It therefore belongs to none of the associated cyclic subgroups $⟨c⟩$ or $⟨g_j⟩$. Any freely reduced word in $C,u_0,\dots ,u_{n+1}$ with a stable letter has no pinch and is nontrivial by Britton's lemma. A word with no stable letters is a nonzero power of $C$, also nontrivial. This proves the claimed free basis of a group.

Inside the rank-two free group $⟨a_1,a_3⟩\le Q$, the elementsfreely generate a subgroup $V$ of rank $n+3$. This is the same free-product normal-form calculation as for the conjugate basis in Question 5: between conjugates with distinct indices, a nonzero power of $a_3$ remains. The identifications in the output presentation give an isomorphism $U\to V$ matching these free bases. HenceThe normal form theorem for an amalgamated free product makes both factors embed. Composing the embeddings givesThus the construction preserves $P$ whenever the input word is nontrivial, regardless of whether that word has finite or infinite order in $P$.

In a manifold chart, write a smooth differential form as $\eta ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$, with $I$ increasing. Define its exterior derivative byTo check that this definition is global, change coordinates to $y$. The chain rule expresses $d{x}^i$ as ${\sum }_j({\partial }_j{x}^i)d{y}^j$. Differentiating it gives zero because the second partial derivatives of $x^i$ are symmetric and $d{y}^k\wedge d{y}^j$ is antisymmetric. The coordinate definition satisfies the graded Leibniz rule, so differentiating the transformed expression for $\eta$ leaves exactly ${\sum }_{I}d{a}_I\wedge d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. This agrees with the original formula and proves independence of the manifold chart. The same cancellation of mixed derivatives gives $d^2=0$.

The de Rham cohomology is thereforethe vector space of closed differential forms modulo exact differential forms.

Here is a Green-theorem proof of exactness on a disc. Identify the disc with a round disc centred at zero, and let $\eta =Pdx+Qdy$ be a closed differential form. Set $f(x)={\int }_{[0,x]}\eta$. For $x$ and $x+h$ in the disc, Green theorem on the oriented triangle with vertices $0,x,x+h$ gives zero boundary integral, since ${\partial }_{x}Q-{\partial }_{y}P=0$. HenceTaking $h=\epsilon v$ and dividing by $\epsilon$ yields $d{f}_x(v)={\eta }_x(v)$. The radial integral defining $f$ is smooth, so $\eta =df$ andDegenerate triangles follow by continuity. The same proof works on any star-shaped plane domain, including the whole plane.

Integration requires an orientation: take $M$ to be an oriented compact $n$-dimensional smooth manifold without boundary. The Generalized Stokes theorem says ${\int }_{M}d\beta =0$ for every smooth $(n-1)$-form $\beta$. Every $n$-form is closed by dimension, and changing it by an exact differential form leaves its integral unchanged. Thusis a well-defined linear map $H_{\mathrm{dR}}^n(M;\mathbb{R})\to \mathbb{R}$. Compactness alone does not supply the orientation omitted from the printed integration statement.

For an $n$-dimensional Lie group, choose a nonzero $\lambda \in {\Lambda }^n{T}_e^{\ast }G$ and define the invariant volume form on a Lie group byEach $d{L}_{g^{-1}}$ is an isomorphism, so $\omega$ is nowhere zero and defines an orientation. The composition identity $L_{(hg{)}^{-1}}\circ L_h=L_{g^{-1}}$ gives $L_h^{\ast }\omega =\omega$. Moreover, every left-invariant differential form of degree $n$ is determined by its value at $e$, so the space of such forms is one-dimensional.

Under the supplied invariant-representative assumption, all top-degree de Rham cohomology classes are multiples of $[\omega ]$. For compact $G$, orient it so $\omega$ is positive; then ${\int }_G\omega >0$, whereas the Generalized Stokes theorem makes every exact top form have integral zero. Therefore $[\omega ]\ne 0$ andunder that assumption. For actual compact Lie groups, this argument applies when $G$ is connected. The supplied assumption fails for disconnected groups: on $S^1\times \mathbb{Z}/2$, the two components have independent degree-one classes, while globally left-invariant top forms form only a one-dimensional space. In general the top-degree de Rham cohomology of a compact Lie group is ${\mathbb{R}}^c$, where $c$ is its number of connected components.

For $S^2$, use $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with the plane. The disc argument gives $\eta {|}_U=d{f}_U$ and $\eta {|}_V=d{f}_V$ for any closed differential one-form $\eta$. The overlap $U\cap V$ is connected, so $d(f_U-f_V)=0$ makes the difference a constant. Adjust one potential by that constant; they glue to a global smooth function $f$ with $df=\eta$. ThusFinally, $S^3$ is a compact connected Lie group: the identificationidentifies it smoothly with the special unitary group $SU(2)$. Matrix multiplication and inversion supply its smooth group operations. The preceding invariant-volume argument therefore gives

A monomorphism $m:A\to B$ is left cancellable: $mu=mv$ implies $u=v$ for all parallel arrows into $A$. A strong monomorphism is a monomorphism with the following lifting property: if $e:X\to Y$ is an epimorphism and $u:X\to A$, $v:Y\to B$ satisfy $mu=ve$, there is a diagonal $d:Y\to A$ with $de=u$ and $md=v$. It is unique by monicity. An extremal monomorphism is a monomorphism $m$ for which every factorization $m=he$ with $e$ an epimorphism forces $e$ to be an isomorphism. We use the strong lifting formulation for the paper's terminology; below we also justify its equivalence with the extremal formulation under the complete, well-powered hypotheses. A regular monomorphism is an equalizer of two parallel arrows.

First, a strong monomorphism is extremal. In a factorization $m=he$, apply its lifting property with upper arrow $1_A$ and lower arrow $h$. This gives $d$ with $de=1_A$. Since $e$ is an epimorphism, $ede=e$ implies $ed=1$, so $e$ is an isomorphism.

If $m:A\to B$ is an equalizer of $r,s:B\rightrightarrows C$, consider a lifting square as above. The equation $rve=smu=sve$ and the epimorphism property give $rv=sv$. The equalizer property therefore gives a unique $d$ with $md=v$. Monicity then gives $de=u$. Hence every regular monomorphism is a strong monomorphism.