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For a commutative ring $R$, remove the section $[1:0:0]$ from ${\mathbb{P}}_R^2$. The two remaining standard charts have coordinate rings $R[a,b]$ and $R[a/b,b^{-1}]$ inside the Laurent polynomial ring $R[a,b,b^{-1}]$. Their intersection consists of constants: their monomial exponent regions are $i\ge 0,j\ge 0$ and $i\ge 0,j\le -i$, respectively. The Čech cohomology in degree one is free over $R$ on the missing monomials $a^i{b}^{-k}$ with $1\le k<i$. Equivalently its basis consists of the classes of $x_0^{m+n}/(x_1^m{x}_2^n)$ for $m,n\ge 1$. Degree zero is $R$ and every degree at least two is zero. The charts and their intersection are affine schemes, so this computes sheaf cohomology. If $R$ is nonzero, the first sheaf cohomology is nonzero and this scheme is not an affine scheme.

The connected covering space corresponding to $\ker \alpha$ has deck transformation group $\mathbb{Z}$. Choosing a deck generator makes its integral homology a module over the Laurent polynomial ring $\mathbb{Z}[t,t^{-1}]$.

For the Integer Heisenberg group, choose $x=I+E_{12}$, $y=I+E_{23}$ and central $z=I+E_{13}$. Every element has a unique form $y^c{z}^b{x}^a$, and $xy=zyx$. The group algebra is therefore the displayed skew Laurent polynomial ring, with $\sigma (z)=z$. The base is a commutative Laurent polynomial ring and an integral domain; the skew Hilbert basis theorem gives both-sided Noetherianity and the leading coefficient of a product proves that there are no zero divisors.

A monomial allowing negative integer exponents in invertible variables. Finite sums of Laurent monomials form a Laurent polynomial ring. In a graded coordinate ring each has total degree ${\sum }_i{a}_i$.

A nonzero quotient of an $r$-variable Laurent polynomial ring is finite over an embedded Laurent polynomial ring in $m\le r$ variables. Given a nonzero Laurent relation, choose $N$ so that weights $(N,N^2,\dots ,N^{r-1},1)$ separate its finite exponent support. The displayed invertible monomial substitution makes its extreme $Y_r$ coefficients units in the remaining Laurent variables. Normalize the relation to a monic polynomial with unit constant term; both $Y_r$ and $Y_r^{-1}$ are integral over the image of the $(r-1)$-variable subring. Induction and transitivity of finite extensions prove the assertion, over finite fields as well as infinite fields. An infinite-dimensional quotient has $m\ge 1$.

The algebraic loop algebra of a Lie algebra $\mathfrak{g}$ over a field $k$ consists of finite Laurent polynomials with coefficients in $\mathfrak{g}$. Its Lie bracket is $[x\otimes t^m,y\otimes t^n]=[x,y]\otimes t^{m+n}$, extended bilinearly. Alternation and the Jacobi identity follow from those of $\mathfrak{g}$, since the Laurent polynomial ring is commutative. Appropriate smooth or formal completions give other loop-algebra variants; the affine current algebra adds a central extension to these Fourier modes. For the affine extension, see Kac and Wakimoto, Section 0.2.

Interpretation of the removed locus. Over $\mathbb{Z}$, the notation $[1:0:0]$ specifies a section $\mathrm{Spec}\mathbb{Z}\to {\mathbb{P}}_{\mathbb{Z}}^2$, with image $V_{+}(x_1,x_2)\cong \mathrm{Spec}\mathbb{Z}$. We remove this entire closed subscheme. This is the interpretation forced by the asserted open cover $X=U_1\cup U_2$. Removing one closed point in a particular fiber would not give that open cover. The native PDF uses the same notation as the transcription.

Put $a=x_0/x_1$, $b=x_2/x_1$ on $U_1$. On $U_2$ put $c=x_0/x_2$, $d=x_1/x_2$. On their intersection $c=a/b$ and $d=b^{-1}$, giving coordinate ringsThe Čech cochain complex for the structure sheaf, using increasing distinct indices, isThus ${\check{H}}^0$ is the intersection of the two subrings of $L$, while ${\check{H}}^1$ is the quotient module of $L$ by their additive sum.

The Laurent polynomial ring $L$ is a free abelian group on the monomials $a^i{b}^j$ with $i\ge 0$, $j\in \mathbb{Z}$. The first subring has the basis $j\ge 0$. The second has the basisso its allowed exponents satisfy $j\le -i$. A monomial belongs to both precisely when $i=j=0$. The monomials belonging to neither satisfy $i\ge 2$ and $-i<j<0$. Since the subgroups are spanned by subsets of the same basis, their sum introduces no other relations. ThereforeHere the square brackets denote cohomology classes, and the direct sums are as abelian groups. In particular, $[a^2{b}^{-1}]=[x_0^2/(x_1{x}_2)]$ is a nonzero cohomology class. This scheme is a separated scheme, being an open subscheme of projective space over $\mathbb{Z}$, and is a quasi-compact scheme by (d). Both charts and their intersection are affine schemes, so this Čech cohomology equals sheaf cohomology. The nonzero first sheaf cohomology of the quasi-coherent structure sheaf rules out an affine scheme:

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

Use $[g]$ for the basis element of the group algebra $A={\mathbb{F}}_p[{\mathbb{Z}}_p]$, so $[g][h]=[g+h]$ and $[0]=1$. The kernel of the augmentation map is the augmentation ideal $I$. This algebra must not initially be identified with a Laurent polynomial ring: the additive group ${\mathbb{Z}}_p$ is not abstractly generated by the integer $1$.

For $r\ge 1$, map $A$ onto ${\mathbb{F}}_p[{\mathbb{Z}}_p/p^r{\mathbb{Z}}_p]$ and call its kernel $J_r$. It is generated by $[g]-1$ for $g\in p^r{\mathbb{Z}}_p$: differences of basis elements in one coset have the form $[h]([g]-1)$. Since each such $g=p^{r}h$,by the characteristic-$p$ binomial identity. Thus $J_r\subseteq I^{p^r}$.

The quotient group is a cyclic group of order $p^r$, generated by the image of $1$. Its group algebra isThe image of $I$ is $(T)$ and its $p^r$th power vanishes, proving the reverse inclusion. HenceThis is cyclic finite quotients of the p-adic augmentation filtration.

For any $n\ge 0$, choose $p^r>n+1$. The finite quotient then identifies $I^n/I^{n+1}$ with $(T^n)/(T^{n+1})$, a one-dimensional space generated by the image of $([1]-1{)}^n$. The degree-one symbol $X$ of $[1]-1$ consequently gives a surjective graded map ${\mathbb{F}}_p[X]\to {\mathrm{gr}}_{I}A$, and no power of $X$ vanishes in its degree. ThusThe same finite quotients show that the I-adic completion is ${\mathbb{F}}_p[[T]]$, and its associated graded ring is identical. The proof therefore covers the completed group algebra interpretation too, while proving the literal ordinary-algebra statement without assuming completion. Applying part ii to a ring would require its actual completeness, not just this polynomial associated graded ring.

The multiplicative algebraic group ${\mathbb{G}}_m$ has coordinate ring $\mathbb{C}[t,t^{-1}]$. An algebraic morphism $f:{\mathbb{G}}_m\to {\mathbb{G}}_m$ therefore sends the invertible coordinate on the target to a unit of this Laurent polynomial ring.

Every unit of $\mathbb{C}[t,t^{-1}]$ has the form $c{t}^i$, with $c\in {\mathbb{C}}^{\ast }$ and $i\in \mathbb{Z}$. To prove this, suppose Laurent polynomials $P,Q$ satisfy $PQ=1$. Write their least and greatest active exponents as $a\le b$ and $c\le d$. The least and greatest exponents in their product are $a+c$ and $b+d$, since their endpoint coefficients are nonzero. Both must be zero. Consequentlyso each factor has only one active exponent. This proves the asserted unit description.

Thus $f(t)=c{t}^i$. A group homomorphism sends the identity to the identity, so $f(1)=1$ and $c=1$. Conversely, for every integer $i$, including negative integers and zero, $t\mapsto t^i$ is regular on ${\mathbb{G}}_m$ and satisfies $(st{)}^i=s^i{t}^i$. We obtainThese are exactly the characters of an algebraic torus in the one-dimensional case. The algebraic-morphism requirement is essential to the argument: the map must come from the Laurent coordinate ring.

Let $F$ be an oriented Seifert surface of genus $g$ for $K$, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is a torsion module. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed diagram, start on the outer lower arc near the lower left undercrossing and follow the arc through the lower outer loop toward the upper right undercrossing. Number successive undercrossings $0,\dots ,9$, and let $x_i$ label the incoming arc at crossing $i$. Reading the overpassing arcs and the conjugation exponents in a consistent meridian convention givesThus the Wirtinger presentation relations are $x_{i+1}=x_{o_i}^{{\epsilon }_i}{x}_i{x}_{o_i}^{-{\epsilon }_i}$, with indices modulo ten. Their Fox derivatives after abelianization give row relationsDelete the redundant last relation and the last column, setting $z_9=0$. This is a nine-generator presentation of the Alexander module. The cofactor determinant is $t^{-6}(t^4+t^3-3t^2+t+1)$; here is a smaller elimination check. Retain $z_1,z_4$. Seven relations solve the other generators asThe two remaining rows, at undercrossings two and seven, are the matrixIts determinant is $t^{-3}(t^4+t^3-3t^2+t+1)$, differing from the original cofactor only by a Laurent unit. ConsequentlyThe factor has value one at $t=1$ and is reciprocal, as required. Its Conway-normalized Alexander polynomial is $t^2+t-3+t^{-1}+t^{-2}$, and its knot determinant is three.

For the genus upper bound, the over/under Gauss code from the same traversal isNumber these visits $0,\dots ,19$. The oriented smoothing permutation of a Gauss code has cyclesThere are seven Seifert circles, so the Seifert algorithm gives ten bands and seven disks: $1-2g=7-10$, hence $g=2$. The Alexander breadth bound on Seifert genus supplies the matching lower bound from breadth four. Therefore $\boxed{g_s(K)=2}$.

Modulo two, the primitive Alexander polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root, and it is not $(t^2+t+1{)}^2=t^4+t^2+1$, the only possible product of irreducible monic quadratics over ${\mathbb{F}}_2$. It is therefore irreducible over ${\mathbb{F}}_2$, hence over $\mathbb{Q}$ and $\mathbb{Z}$ by Gauss lemma for polynomials. The full-degree irreducible Alexander polynomial implies a prime knot criterion now applies, since its breadth is exactly twice the genus. Thus the knot is prime. The full-degree condition is needed to rule out a nontrivial summand with unit Alexander polynomial.

To test reflection, calculate the Jones polynomial from the same crossing data. The diagram has nine positive crossings and one negative crossing, so its writhe of a link diagram is eight. Enumerate the $2^{10}$ bracket smoothing states, join their smoothed ends, and weight each state by $A^{a-b}(-A^2-A^{-2}{)}^{s-1}$. Correct by $(-A^3{)}^{-8}$ and put $t=A^{-4}$. This givesAs checks, $V_K(1)=1$ and $V_K(-1)=-3={\Delta }_K(-1)$. The Jones polynomial of a mirror is $V_K(t^{-1})$, which differs from this polynomial. Therefore the knot is not equivalent to its reflection. Reversing all crossing conventions reciprocates the displayed Jones polynomial and leaves the conclusion unchanged.

Let F be an oriented genus-g Seifert surface for K, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is torsion. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed knot we give a reproducible diagram calculation. Label its ten crossings A through J as follows: A and B are the two upper crossings from left to right; C,D,E are the next three crossings in descending order; F is the crossing below E on the almost vertical strand; G is the left crossing of the bottom loop; H,I,J are the remaining three bottom crossings from left to right. With one traversal direction the over/under record isThe crossings have the same sign. Choose the meridian/deck convention giving the conjugation relation below; the opposite convention replaces t by $t^{-1}$ and gives the same reciprocal polynomial up to a unit.

Number the undercrossings in the order $B,I,F,H,J,C,D,A,E,G$, starting at zero, and let $x_k$ be the incoming arc at undercrossing k. The overpassing arc numbers areThe Wirtinger presentation has $x_{k+1}=x_{o_k}{x}_k{x}_{o_k}^{-1}$, with indices modulo ten. Its lifted cellular boundaries, or equivalently Fox derivatives, give the Alexander relation matrixDelete the redundant final relation and the last column: the resulting matrix presents the Alexander module on the nine relative arc generators. To make the determinant check transparent, set $h=2t^2-3t+2$, $k=4t^2-7t+5$, and $\ell =2t^2-2t+1$. Eliminating unit pivots reduces this nine-generator presentation toFor example, writing the relative generators as $z_0,\dots ,z_8$, the eliminated relations give $z_3=t{z}_2$, $z_4=(t^2-t+1)z_2$, $z_1=(t^2-2t+2)z_2$, $z_5=h{z}_2$, $z_7=(1-t)h{z}_2+t{z}_6$, $z_0=t^{-1}(t^2-2t+2-(1-t{)}^{2}h)z_2-(1-t)z_6$, and $z_8=(1-t)k{z}_2+(2t-1)z_6$. The two remaining relations are precisely the rows of N. HenceAs a second determinant check, the unreduced nine-by-nine cofactor of M is $t^2\det N$. The polynomial has value one at $t=1$ and is reciprocal, as required.

For the genus upper bound, the oriented smoothing has seven Seifert circles. This can also be checked from the traversal record: number its twenty entries starting at one; smoothing sends an incoming visit to the entry following its other visit to the same crossing. The cycles are $(1,17,13,19,7)$, $(2,16)$, $(3,11,15)$, $(4,10)$, $(5,9)$, $(6,20,8)$, and $(12,18,14)$. The Seifert algorithm therefore gives seven disks and ten bands, so $\chi (F)=7-10=-3=1-2g(F)$ and $g(F)=2$. The Alexander breadth is four, so the Alexander breadth bound on Seifert genus gives the matching lower bound. Thus $\boxed{g_s(K)=2}$.

Finally, modulo two the primitive polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root. The only monic irreducible quadratic over ${\mathbb{F}}_2$ is $t^2+t+1$, whose square is $t^4+t^2+1$, so the quartic is irreducible. Reduction modulo two therefore proves irreducibility over the rationals and, by primitivity, over $\mathbb{Z}$ and $\Lambda$. If K were a sum of two nontrivial knots, additivity of Seifert genus would force both summands to have genus one. Their Alexander polynomials multiply, so irreducibility would make one polynomial a unit and the other equal to this breadth-four polynomial, contradicting the breadth-at-most-two bound for genus one. This uses the full-degree irreducible Alexander polynomial implies a prime knot criterion, rather than irreducibility alone. Therefore K is prime.

The dense algebraic torus has coordinate ring $k[M]$, a Laurent polynomial ring and hence a unique factorization domain. Its divisor class group is therefore zero. Restrict an arbitrary Weil divisor $D$ to the algebraic torus and choose a rational function on an algebraic variety $f$ whose divisor there is $D{|}_T$. Then $D-\mathrm{div}(f)$ is supported on the algebraic torus boundary. The prime divisors in that boundary are precisely the $F_j$ corresponding to the toric rays, by the torus orbit-cone correspondence. ConsequentlySince the variety is smooth, every Weil divisor is Cartier. This gives invariant representatives for all divisor classes, without assuming that the original divisor was invariant.

Now use such a representative. The associated vector space isThe principal divisor on a toric variety formula gives $\mathrm{div}({\chi }^m)={\sum }_j⟨m,e_j⟩F_j$. Define the lattice polytope of a toric divisorFor a maximal toric cone $\sigma$, its basic toric ray vectors are a basis of $N$, so there is a unique integral $m_{\sigma }$ with $⟨m_{\sigma },e_j⟩=-a_j$ on its toric rays. The algebraic torus character ${\chi }^{m_{\sigma }}$ is a local frame for $\mathcal{O}(D)$ on $U_{\sigma }$. Every global section restricts on the algebraic torus to a finite Laurent polynomial ${\sum }_m{c}_m{\chi }^m$. Relative to this frame, regularity on $U_{\sigma }$ means ${\sum }_m{c}_m{\chi }^{m-m_{\sigma }}\in k[{\sigma }^{\vee }\cap M]$. Since distinct algebraic torus characters are independent, this holds exactly when each exponent with nonzero coefficient satisfies the inequalities for the toric rays of $\sigma$. Taking all charts provesCompleteness makes $P_D$ bounded when nonempty: a nonzero recession vector would be nonnegative on every toric ray of a complete toric fan, and therefore on all of $N_{\mathbb{R}}$, which is impossible. The displayed space is consequently finite-dimensional.

At the torus-fixed point $p_{\sigma }$ in the smooth chart $U_{\sigma }\cong {\mathbb{A}}^n$, a section ${\chi }^m$ has local monomial ${\chi }^{m-m_{\sigma }}$. Its value is nonzero exactly when this monomial is constant, namely when $m=m_{\sigma }$. Thus if the complete linear system of a divisor has no base point, some section is nonzero at $p_{\sigma }$, forcing $m_{\sigma }\in P_D$. Conversely, if $m_{\sigma }\in P_D$ for every maximal toric cone, its algebraic torus character is a global section whose local expression on $U_{\sigma }$ is one. It therefore vanishes nowhere on that whole chart, and these charts cover $X$. This proves the toric basepoint-free criterion:The analogous toric ampleness criterion, stated without proof, replaces the inequalities off $\sigma$ by strict ones:Equivalently, the normal fan of a polytope $P_D$ is exactly $\Sigma$.

For the final fan subdivision, take $n\ge 2$ so that $\tilde{e}$ is a new toric ray, as the construction requires. The vectors $e_1,\dots ,e_n$ are a lattice basis, so their sum is primitive. Replacing any one basis vector by this sum gives another basis. Thus the star subdivision gives a smooth toric blowup at a torus-fixed point $\pi :X_{{\Sigma }^{\prime }}\to X_{\Sigma }$, with algebraic exceptional divisor $E=\tilde{F}$. On the old chart, a local equation of $D$ is ${\chi }^{-m_{\sigma }}$, whose order on the new toric ray is $-⟨m_{\sigma },\tilde{e}⟩={\sum }_{i=1}^n{a}_i=a$. All old toric ray coefficients are unchanged. ThereforeWe prove ampleness of $D_c=c{D}^{\prime }-E$ directly by inequalities, giving the ample divisor after blowing up a torus-fixed point result.

Let $u_1,\dots ,u_n$ be the basis of $M$ dual to $e_1,\dots ,e_n$. On the new maximal toric cone ${\sigma }_i$, which omits $e_i$, the required local algebraic torus character isIndeed its pairings on $e_j$, $j\ne i$, are $-c{a}_j$, while its pairing on $\tilde{e}$ is $-ca+1$, exactly the negative of the new coefficient $ca-1$. Its inequality on the omitted old toric ray has strict gap one:For any toric ray $e_j$ outside the original toric cone, put ${\delta }_j=⟨m_{\sigma },e_j⟩+a_j$. The original ampleness gives positive integers ${\delta }_j$, and the new strict gaps arefor all sufficiently large $c$.

For any unchanged maximal toric cone $\eta \ne \sigma$, the local algebraic torus character is $c{m}_{\eta }$. All strict inequalities on old toric rays follow from those for $D$. Its inequality on the new toric ray has gapEach summand is nonnegative, and at least one is positive because $\eta \ne \sigma$ cannot contain all its toric rays. Hence ${ϵ}_{\eta }$ is a positive integer, making this gap positive for $c\ge 2$. There are only finitely many remaining inequalities. For example, any integer satisfyingworks, with the inner maximum omitted when its index set is empty. The toric ampleness criterion now givesThe new-ray assumption is essential to the printed construction. In dimension one, $\tilde{e}=e_1$ and the fan subdivision is unchanged, so the two purported distinct divisors coincide. If the displayed definition of $D^{\prime }$ is applied literally on ${\mathbb{P}}^1$ with coefficients $a_1=-2$, $a_2=3$, then $D$ has degree one but $D^{\prime }$ has degree $-1$, contradicting the conclusion. The argument above applies to the intended blowup in dimension at least two.

The map is a homogeneous Laurent polynomial ring homomorphism between the two rank-one infinity towers. Choose homogeneous generators $e_i$ with absolute degrees $q_i$. Its degree isConsequently the general graded-module description isIf the grading does not allow an integer exponent, the map is zero. Negative exponents are allowed in the infinity theory. The notation $F$ here denotes the map called ${\Phi }^{\infty }$ in the question.

The standard topological statements for Heegaard Floer cobordism maps, which require no proof in this part, areIn the second case $a=\pm 1$ and the map is therefore $\pm U^r$ with the exponent determined by the displayed degree formula. Since both boundary components are rational homology spheres, the rational intersection form is nondegenerate, so $b_2^{+}=0$ means that it is negative definite, allowing also the rank-zero form.

The printed hypothesis $b_1(Y_i)=0$ does not imply $b_1(W)=0$. Thus one must not replace the second implication by an unconditional isomorphism assertion for every negative-definite cobordism. A concrete check is $W=(S^1\times S^3)\setminus (B^4\bigsqcup B^4):S^3\to S^3$. It has $b_2^{+}=0$ and $b_1(W)=1$. A one-handle followed by a three-handle gives $e\mapsto e\otimes {\theta }^{+}$ and then kills the ${\theta }^{+}$ summand; hence its infinity map is zero. Its grading shift is one, also incompatible with a nonzero map between the even-graded $S^3$ towers. With only the stated boundary assumption, the general homogeneous Laurent-module description and the two topological implications above are the valid conclusions; determining any remaining coefficient requires the actual cobordism, not just its boundary Betti numbers.

Use generators $a,b,c$ for the three one-handles in the generalized Heegaard diagram. Read a downward crossing of the cut from a left-hand hole to its right-hand partner as the corresponding generator. Traverse the attaching curve around the left-hand $A$ hole, starting at its outgoing downward strand, and then traverse the attaching curve around the left-hand $B$ hole in the same way. Their cut-crossing words areThese are the two relators imposed by the two handle attachments. The remaining boundary is a torus; no third relator should be added. The fundamental group presentation is thereforeThe exponent-sum vectors are $(1,-1,0)$ and $(0,1,-1)$. Thus the abelianization is $\mathbb{Z}$, with $a,b,c$ all mapping to its positive generator $t$.

Apply the Fox derivatives and then this abelianization. The resulting Alexander matrix isFor example, the three occurrences of $a$ in $r_1$ contribute $1,t^3,-t^2$ to the first entry. Every row sums to zero, as the Fox calculus identity predicts.

The cellular boundary in degree one for the infinite cyclic cover is $(t-1,t-1,t-1)$. Its kernel is freely generated by $e_a-e_c,e_b-e_c$. Expressing the two relation boundaries in this basis gives the two-by-two matrix obtained by deleting column $c$ of $F(t)$. Its determinant is the order of the Alexander module of a space. Equivalently, all three maximal matrix minors of $F(t)$ agree up to sign. The Alexander polynomial iswhere $\doteq$ allows multiplication by a unit $\pm t^m$ in the Laurent polynomial ring. Its value at $1$ is $1$.

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 $1,Y_r,\dots ,Y_r^{d-1}$.