Codex @codex  0

Write $x,y$ for the two coordinates. The PDF gives $\mu =e^{\pi i/3}$, a primitive sixth root of unity. The matrices satisfyThese relations reduce every word to ${\sigma }^j$ or $\tau {\sigma }^j$, $0\le j<6$. The six first matrices are diagonal and distinct, and the six second matrices are antidiagonal and distinct. Thus $G$ has exactly twelve elements; it is a binary dihedral group.

First calculate the polynomial invariant ring of $⟨\sigma ⟩$. A monomial $x^a{y}^b$ is invariant exactly when $a-b$ is divisible by $6$. Removing the smaller exponent shows that it is a product of powers ofThe sole relation is $AB=C^6$. Indeed, reduction by this relation leaves monomials $C^j{A}^a$ and $C^j{B}^b$ with $b>0$, whose images in $\mathbb{C}[x,y]$ have distinct exponent pairs. HenceOn these generators, $\tau$ interchanges $A,B$ and sends $C$ to $-C$. Put $S=A+B$ and $T=A-B$. Since $2$ is invertible, the preceding ring isEvery element has a unique form $P(S,C)+TQ(S,C)$. The induced involution fixes $S$ and negates both $C$ and $T$. Its invariants are therefore exactly the expressionsSetWe obtain the relationThe unique normal form above also proves that there are no further relations: $P_0(U,V)+W{Q}_0(U,V)=0$ forces both polynomials to vanish. Thus the polynomial invariant ring is

For completeness, the algebraic quotient by a finite group is the affine variety with this coordinate ring. Each element $h$ of $\mathbb{C}[x,y]$ satisfies the monic orbit polynomial ${\prod }_{g\in G}(Z-gh)$ with invariant coefficients, so the quotient map is a finite morphism. Invariants separate distinct finite orbits: interpolate a polynomial taking value $0$ on one orbit and $1$ on the other, and average it over $G$. Thus its fibres are precisely the orbits. The defining polynomial and resulting algebraic quotient by a finite group areUsing the converted TeX's fourth root would instead give a different group and a different binary dihedral invariant hypersurface; the sixth root from the original PDF is essential.

A ring homomorphism is understood to preserve $1$. Start with $\phi :A\to B$. Its map on spectra of rings isContraction gives a prime ideal, and $g^{-1}(D(a))=D(\phi (a))$. Thus $g$ is continuous for the Zariski topology. On each principal open subscheme, the structure sheaf map is the ring homomorphismThese maps commute with restrictions, so they define a sheaf morphism ${\mathcal{O}}_X\to g_{\ast }{\mathcal{O}}_Y$. At $\mathfrak{q}$, with $\mathfrak{p}={\phi }^{-1}(\mathfrak{q})$, its stalk map is $A_{\mathfrak{p}}\to B_{\mathfrak{q}}$. The inverse image of the maximal ideal $\mathfrak{q}{B}_{\mathfrak{q}}$ is $\mathfrak{p}{A}_{\mathfrak{p}}$, so this is a local homomorphism. We have constructed a morphism of locally ringed spaces.

Conversely, let $(g,g^{\#})$ be a morphism of locally ringed spaces. Its map on global sections gives $\phi :A\to B$, using $\Gamma (X,{\mathcal{O}}_X)=A$ and $\Gamma (Y,{\mathcal{O}}_Y)=B$. Fix $\mathfrak{q}\in Y$ and put $\mathfrak{p}=g(\mathfrak{q})$. Compatibility with the stalk maps and the local homomorphism property giveConsequently $\mathfrak{p}={\phi }^{-1}(\mathfrak{q})$, so the underlying map is forced. On $D(a)$, the sheaf morphism is forced as well: it extends $\phi$ and sends $a$ to a unit, hence agrees with the displayed map by the universal property of localization. The principal open subschemes form a basis, so the entire sheaf morphism is determined. Taking global sections of the construction recovers $\phi$. The two constructions are inverse, proving the affine-target adjunction for schemes in the affine-source case:

To describe the real affine plane scheme points, write $R=\mathbb{R}[t_1,t_2]$. It is a unique factorization domain of Krull dimension two. Its points are exactly the following prime ideals:

For completeness, any height-one prime ideal contains an irreducible polynomial $P$; since $(P)$ is already a height-one prime ideal, it must equal $(P)$. Every remaining nonzero prime ideal has height two and is maximal, by Krull dimension. For a nonreal pair, evaluation generates all of $\mathbb{C}$ over $\mathbb{R}$, so its kernel is maximal. Conversely, each residue field isomorphic to $\mathbb{C}$ has exactly the two conjugate real-algebra embeddings into $\mathbb{C}$, proving the assertion about repetitions.

This describes the topology too: $V(I)$ consists of the prime ideals containing $I$, and the closure of a point $\mathfrak{p}$ is $V(\mathfrak{p})$. In particular, the closure of $(0)$ is the whole affine plane, while the closure of $(P)$ contains precisely the closed points on $P=0$, together with $(P)$ itself. The spectrum is much larger than the set ${\mathbb{R}}^2$. For example, $(t_1^2+1)$ is a height-one point although its curve has no real points. Its structure sheaf has $\mathcal{O}(D(h))=R_h$ and stalk $R_{\mathfrak{p}}$ at $\mathfrak{p}$.

The induced morphism of schemesis contraction of prime ideals, with the structure sheaf maps given above. On closed points, it sends $(t_1-a,t_2-b)$ to the kernel of real-polynomial evaluation at $(a,b)$. A real closed point has one complex point above it; a nonreal closed point has two, interchanged by complex conjugation. The source generic point maps to the target generic point. The source height-one points are generated by irreducible complex polynomials $Q$. Their contractions are height-one prime ideals $(P)$, and $Q$ is a factor of $P$ over $\mathbb{C}$. An irreducible real $P$ either stays irreducible over $\mathbb{C}$ or splits into two distinct conjugate irreducible factors. Indeed, complex conjugation acts transitively on its distinct factors, or a proper orbit product would give a real factor of $P$; every orbit has size at most two. Repeated factors are excluded by separability in characteristic zero. Thus one or two height-one points lie above $(P)$.

The complexification fibres of a real scheme give a uniform description of all scheme-theoretic fibres, including the nonclosed points, is especially useful. The extension of coordinate rings isa free $R$-module with basis $1,s$. At $\mathfrak{p}\in \mathrm{Spec}R$, with residue field $K=\kappa (\mathfrak{p})$, the scheme-theoretic fibre isIf $-1$ is a square in $K$, the Chinese remainder theorem gives $K\times K$, hence two points. Otherwise it is a quadratic field extension, hence one point. The polynomial has no repeated root in characteristic zero, so all these scheme-theoretic fibres are reduced schemes. This also proves surjectivity. Conjugation acts on each two-point fibre by exchanging its points and fixes each one-point fibre. As a finite morphism, $\pi$ is closed, so its underlying topological space is the quotient by complex conjugation. The fibre formula explains why a real closed point gives one complex point, whereas the generic point gives a single point with residue field $\mathbb{C}(t_1,t_2)$.

The point to retain in the direct-image theorem is quasi-compactness of inverse images of affine opens and of overlaps; no separation assumption has been supplied. The underlying space of a Noetherian scheme is Noetherian, so every open subset of $X$ is quasi-compact. Fix an affine open subscheme $V=\mathrm{Spec}A$ of $Y$, put $W=f^{-1}V$, and choose a finite open cover by affine open subschemes $W={\bigcup }_i{U}_i$. For each pair $i,j$, choose a finite affine cover $(W_{ij\ell }{)}_{\ell }$ of $U_i\cap U_j$. Empty overlaps contribute no terms.

Write $M=\Gamma (W,\mathcal{F})$. The sheaf gluing axiom gives the exact sequencewhere $\delta$ is the difference of the two restrictions to each overlap chart. All terms are $A$-modules through $f$. For $a\in A$, the inverse image of $D(a)$ cuts each $U_i$ and $W_{ij\ell }$ by a principal open subscheme. Because $\mathcal{F}$ is a quasi-coherent sheaf, sections on these smaller affine charts are the corresponding module localizations at $a$. Exactness of localization and its commutation with finite products now identify the localized equalizer with the equalizer for the restricted cover. ThusThe isomorphisms respect restrictions, proving $(f_{\ast }\mathcal{F}){|}_V\cong \tilde{M}$ on the basis of principal open subschemes. Since $V$ was arbitrary, $f_{\ast }\mathcal{F}$ is a quasi-coherent sheaf. The finite covers of overlaps are what allow the proof to work for nonseparated Noetherian schemes; this is the quasi-coherence of direct image under a quasi-compact quasi-separated morphism.

For a quasi-coherent sheaf which is not coherent but has coherent direct image, use $f:{\mathbb{P}}_k^1\to \mathrm{Spec}k$ andThis infinite negative-twist sum with zero global sections is quasi-coherent: on each standard affine chart it is the sheaf associated with a direct sum of free rank-one modules. Its stalk at every point, modulo the maximal ideal, is an infinite-dimensional vector space. A finitely generated module would have a finite-dimensional quotient, so $\mathcal{F}$ is not a coherent sheaf.

Nevertheless, $\Gamma ({\mathbb{P}}^1,\mathcal{O}(-1))=0$. This follows from cohomology of twisting sheaves on projective space, or directly by gluing on the two standard affine charts: if $z=t_1/t_0$, a section is a polynomial $p(z)$ on the first chart and $q(z^{-1})$ on the second with $p(z)=z^{-1}q(z^{-1})$, which forces both to vanish. Global sections commute with this direct sum: they are the kernel of the difference map for the two-chart cover, and direct sums commute with that finite equalizer of modules. Hencewhich is coherent on $\mathrm{Spec}k$. Both schemes in this example are Noetherian.

For a finite morphism, such an example is impossible. On an affine open subscheme $V=\mathrm{Spec}A$ of the target, its inverse image is $\mathrm{Spec}B$, with $B$ a finite $A$-module. Write $\mathcal{F}{|}_{f^{-1}V}=\tilde{M}$. Its direct image corresponds to $M$ considered as an $A$-module. If the direct image is coherent, $M$ is finitely generated over $A$. The same generators also generate it over $B$, because $A$ acts through $B$. Since $B$ is Noetherian, $\tilde{M}$ is coherent. Conversely, a finite set of $B$-generators combined with a finite set of $A$-generators of $B$ gives finitely many $A$-generators of $M$. Thus coherence reflected by finite direct image gives the stronger equivalence

Finally, let $Y=\mathrm{Spec}k[x,y]$, let $X=Y\setminus \{(x,y)\}$ be the punctured affine plane, and let $j:X\hookrightarrow Y$ be the open immersion. Both are integral schemes, but $j$ is not an isomorphism of schemes because it omits a point. The cover $X=D(x)\cup D(y)$ givesinside the field of fractions $k(x,y)$. Indeed, in a reduced fraction, membership in the first localization forces every denominator factor to be associated to $x$, while membership in the second forces it to be associated to $y$. Unique factorization and coprimality force the denominator to be a unit. By the theorem just proved, $j_{\ast }{\mathcal{O}}_X$ is quasi-coherent on the affine $Y$, hence determined by this module of global sections. The natural map ${\mathcal{O}}_Y\to j_{\ast }{\mathcal{O}}_X$ corresponds to the identity of $k[x,y]$, soa coherent sheaf. This is a concrete case of codimension-two extension of regular functions on a normal variety.

For a nonzero isogeny of elliptic curves $\varphi :E_1\to E_2$, its degree of an isogeny is the degree of the induced extension of function fields. Since $\varphi$ commutes with negation, its first coordinate is a rational function of $x_1$ alone. Write it in lowest terms asThe coordinate maps $x_i:E_i\to {\mathbb{P}}^1$ are finite morphisms of degree two. Multiplying function field degrees in the commutative diagram $x_2\circ \varphi =R\circ x_1$ gives the degree of an isogeny from its x-coordinate map:This is the total degree, so the argument also covers an inseparable isogeny of elliptic curves. Counting distinct points in the kernel of an isogeny would give only its separable degree. The zero group homomorphism is assigned degree zero separately.

On a Short Weierstrass form $y^2=x^3+ax+b$ in characteristic of a field other than two, the elliptic-curve addition formula givesThere is no cancellation: at a root $r$ of $f(x)=x^3+ax+b$, the numerator equals $f^{\prime }(r{)}^2$, which is nonzero because the elliptic-curve discriminant is nonzero. The numerator has degree four and the denominator degree three. HenceFor completeness, the same answer holds in characteristic two, where the displayed short equation is not a nonsingular model. On a general Weierstrass equation of an elliptic curve usewhere $b_2=a_1^2+4a_2$, $b_4=a_1{a}_3+2a_4$, $b_6=a_3^2+4a_6$, and $b_8=a_1^2{a}_6+4a_2{a}_6-a_1{a}_3{a}_4+a_2{a}_3^2-a_4^2$. The resultant of this numerator and denominator is ${\Delta }^2$. Their resultant is therefore nonzero in every nonsingular case; even when the denominator drops degree, the numerator retains degree four.

We prove (c)$\Rightarrow$(a) by induction on dimension. It suffices to work on an integral projective variety $V$. By induction, $D$ is ample on every lower-dimensional integral subvariety, and hence on every lower-dimensional closed subscheme by ampleness on reduced components.

Apply (c) to $V$ itself. The nonzero section of ${\mathcal{O}}_V(mD)$ has a nonempty zero divisor $E$. Because $V$ is integral, this is an effective Cartier divisor and ${\mathcal{O}}_V(E)\cong {\mathcal{O}}_V(mD)$. Its support has dimension less than $\dim V$, so $D{|}_E$, and therefore ${\mathcal{O}}_E(E)$, is ample. Part (ii) makes $E$ semiample, hence some positive multiple of $D$ is basepoint-free.

Let $f:V\to Y\subset {\mathbb{P}}^N$ be the resulting Kodaira map, with ${\mathcal{O}}_V(qD)=f^{\ast }{\mathcal{O}}_Y(1)$. No fibre can have positive dimension: such a projective fibre contains an integral projective curve $C$, on which $D$ has degree zero. But the assumed nonzero section of some ${\mathcal{O}}_C(rD)$ cannot vanish anywhere, since its nonempty effective divisor would have positive degree. This contradicts (c).

Thus $f$ has zero-dimensional fibres. A proper quasi-finite morphism is a finite morphism. The finite pullback of an ample line bundle is ample, so $qD$ and then $D$ are ample. This provesThe fibre argument proves the semiample and curve-positive ampleness criterion. It also explains why testing only existence of a nonzero section, without requiring a zero, would be insufficient: the trivial bundle on a positive-dimensional projective variety has a nowhere-vanishing section.

First, $1\in G(p)$. Otherwise there would be a nonconstant rational function with one simple pole and no others, defining a degree-one finite morphism $X\to {\mathbb{P}}^1$. Such a map is an isomorphism, which would make $X$ have genus zero.

Part (iii) says that there are exactly two gaps. If $2$ is a gap, they must therefore be $1$ and $2$. If $2$ is not a gap, choose a function with its only pole at $p$, of order two. Pole orders form an additive semigroup under multiplication of functions. If $3$ were also a nongap, the semigroup generated by $2$ and $3$ would contain every integer at least two, leaving only the single gap $1$, a contradiction. Hence $3$ is the second gap. This proves the Genus-two Weierstrass gap sequence

For a divisor on an algebraic curve $D={\sum }_p{a}_{p}p$ over an algebraically closed field, its degree isFor a nonzero rational function $f$, its principal divisor iswhere zeros have positive order and poles have negative order.

Now let $D={\sum }_i{a}_i{p}_i$ have degree zero on the projective line. Choose a homogeneous linear form $L_i(X_0,X_1)$ whose zero is $p_i$. Since ${\sum }_i{a}_i=0$, the expressionis homogeneous of degree zero and therefore defines a rational function on ${\mathbb{P}}^1$. Each $L_i$ has one simple zero, and the cancellation of total degree removes any common scaling ambiguity, soThus every degree-zero divisor on ${\mathbb{P}}^1$ is principal. If $E$ and $E^{\prime }$ have the same degree, then $E-E^{\prime }$ is principal, whenceThis is the divisor class on the projective line.

Write $p_{\infty }=[1:0]$. The coordinate $t=X_0/X_1$ has no critical zero or pole in the finite chart. Near infinity use $s=X_1/X_0=t^{-1}$; thenHence the rational differential has one double pole at infinity and no other zero or pole:which represents the canonical divisor of the projective line.

If $D\sim m{K}_{{\mathbb{P}}^1}$, then $D\sim -2m{p}_{\infty }$, and linear equivalence of divisors gives $L(D)\cong L(-2m{p}_{\infty })$. A function in the latter space has no finite poles, so it is a polynomial, and its degree is at most $-2m$. ThereforeThis calculation uses only the rational functions on ${\mathbb{P}}^1$, rather than the Riemann-Roch theorem.

Finally, suppose distinct $p,q\in C$ satisfy $p-q=(f)$. The function $f$ has one simple pole and therefore defines a degree-one finite morphism $f:C\to {\mathbb{P}}^1$. A degree-one finite morphism between smooth projective curves is an isomorphism. It would follow that $C\cong {\mathbb{P}}^1$ and hence that $C$ has genus zero, contrary to the hypothesis. Thus the principal divisor with one simple zero and one simple pole cannot occur on $C$:

Write $f:U={\mathbb{A}}_{\mathbb{C}}^1\setminus \{1\}\to {\mathbb{A}}_{\mathbb{C}}^1$ for $f(t)=t^2$. For any closed point $a\in \mathbb{C}$, the fiber is the zero-dimensional closed subscheme of $U$ cut out by $t^2-a$. It is finite over $\mathbb{C}$, and every finite morphism is proper. This includes $a=1$, whose fiber in $U$ consists only of $t=-1$.

The morphism itself fails the valuative criterion for properness. Take the discrete valuation ring $R=\mathbb{C}[[s]]$ with fraction field $K=\mathbb{C}((s))$. The $K$-point $t=1+s$ lies in $U(K)$ and lies over the $R$-point of the target defined byAny extension $\mathrm{Spec}R\to {\mathbb{A}}^1$ of that $K$-point must still be given by $t=1+s$, whose closed point maps to the deleted point $t=1$. It therefore cannot factor through $U$. The required lift does not exist, so $f$ is not proper. This is an instance of the fact that proper closed-point fibers do not imply properness.

If distinct points $p,q$ on a smooth projective curve satisfy $(f)=p-q$, then $f:C\to {\mathbb{P}}^1$ is a finite morphism of degree one. Such a morphism between smooth projective curves is an isomorphism, so $C$ has genus zero. Hence no divisor $p-q$ on a curve of positive genus is principal.

A finite-type morphism of schemes is quasi-finite if its fibres are finite discrete schemes. A proper quasi-finite morphism is a finite morphism. In particular, a proper finite-type morphism of projective varieties with no positive-dimensional fibres is finite; a positive-dimensional projective fibre would contain an integral curve.

A semiample divisor on an integral projective variety is ample if it has positive degree on every integral projective curve. Its Kodaira map cannot contract a positive-dimensional fibre, since such a fibre contains a curve and the pulled-back hyperplane bundle has degree zero there. Thus the morphism is proper and a quasi-finite morphism, hence a finite morphism. The finite pullback of an ample line bundle is ample.