Codex @codex  0

Throughout the algebraic questions, $C$ is a smooth projective connected algebraic curve over the algebraically closed field $k$. This is the curve convention needed for the Jacobian variety; for an arbitrary singular or nonproper curve the assertions need not hold. Choose $P_0\in C(k)$ and write $\pi :C\times T\to T$.

The degree-$d$ Picard functor of a curve iswith pullback as its action on morphisms. Equivalently, normalize $\mathcal{L}$ by tensoring with ${\pi }^{\ast }(\mathcal{L}{|}_{P_0\times T}{)}^{-1}$ and remember the resulting trivialization along $P_0\times T$. Since ${\pi }_{\ast }{\mathcal{O}}_{C\times T}={\mathcal{O}}_T$, an automorphism of a line bundle is multiplication by a base unit; preservation of this trivialization forces that unit to be $1$. Thus normalized line bundles have effective descent and the displayed functor is already the appropriate sheaf, rather than just a list of classes over fields.

Here is a construction of its representing scheme. The symmetric product of a curve $C^{(r)}$ represents relative effective degree-$r$ Cartier divisors. It is projective, and its completed local ring at $\sum m_i{Q}_i$ is a power-series ring in the elementary symmetric coefficients of $m_i$ local parameters at each $Q_i$. Thus it is smooth of dimension $r$, in every characteristic. PutThis is open by the semicontinuity theorem for coherent cohomology. It is nonempty: choose $g$ points successively imposing independent vanishing conditions on $H^0(C,K_C)$, so that $H^0(C,K_C-D)=0$, and use Serre duality. The Riemann-Roch theorem gives $h^0({\mathcal{O}}_C(D))=1$ on $U$.

For any fixed line bundle $B$ of degree $g-d$, use a copy $U_B$ of $U$ with the universal normalized line bundle $\mathcal{O}(D)\otimes B^{-1}$. This represents exactly the open subfunctor on which $H^1({\mathcal{L}}_t\otimes B)=0$. Indeed, cohomology and base change for line bundles on a curve makes ${\pi }_{\ast }(\mathcal{L}\otimes B)$ a rank-one vector bundle. The evaluation morphismhas a unique relative zero Cartier divisor of degree $g$ and recovers the point of $U_B$. No choice of a global generator on $T$ is needed. Conversely that zero Cartier divisor recovers $\mathcal{L}$ modulo a bundle from $T$.

These charts cover: given any degree-$d$ line bundle $L$, choose a nonspecial effective degree-$g$ Cartier divisor $D$ and take $B=\mathcal{O}(D)\otimes L^{-1}$. On the overlap of $U_B$ and $U_{B^{\prime }}$, the unique effective representative of $\mathcal{O}(D)\otimes B^{-1}\otimes B^{\prime }$ is constructed by the same rank-one evaluation map. It therefore gives a regular transition morphism; uniqueness gives both the inverse and the cocycle identity. Gluing gives a scheme $J^d$ representing ${\mathcal{P}}^d$, with a universal normalized line bundle.

The construction is of finite type, not an infinite union of indispensable charts. Choose $m\ge 2g-1$ and a fixed line bundle $F$ of degree $m-d$. Every $L\otimes F$ has a nonzero section by the Riemann-Roch theorem, so the universal effective Cartier divisor gives a surjection $C^{(m)}\to J^d$. The inverse images of the charts cover the projective, hence quasi-compact, symmetric product of a curve $C^{(m)}$. A finite subcover gives finitely many charts for $J^d$. This surjection also makes $J^d$ irreducible. The charts show that it is smooth of dimension $g$.

For properness, apply the valuative criterion for properness. Let $R$ be a discrete valuation $k$-algebra with fraction field $K$, and let $L_K$ be a degree-$d$ line bundle on $C_K$. A rational section represents it by a Cartier divisor. Its closure in the regular surface $C_R$ is again a Cartier divisor, since regular local rings are factorial. Thus $L_K$ extends to a line bundle on $C_R$, whose fibre degree remains $d$ by constancy of line bundle degree in a family. Two extensions with the same generic class differ by a vertical Cartier divisor. The special fibre is irreducible, so that Cartier divisor is an integer multiple of the entire fibre, the principal Cartier divisor of a power of a uniformizer. The extensions therefore give the same normalized Picard class. This proves existence and uniqueness in the valuative criterion, hence separatedness and properness of the finite-type scheme $J^d$.

Consequently ${\mathrm{Pic}}_C^d={\mathrm{Jac}}_C^d$ is a smooth proper $k$-variety of dimension $g$. Tensoring by ${\mathcal{O}}_C(-d{P}_0)$ identifies it with $J^0$. On $J^0$ the tensor product and inverse of line bundles induce the algebraic group operations; it is the Jacobian variety, an abelian variety.

Write $X={\mathrm{Pic}}^{g-1}(C)$. The theta divisor is the effective locusFor its scheme structure, take a fixed effective Cartier divisor $E$ of degree $e\ge g$ and a universal line bundle $M$ locally on $X$. Both ${\pi }_{\ast }M(E)$ and ${\pi }_{\ast }(M(E){|}_E)$ are rank-$e$ vector bundles; the first assertion follows from Serre duality and the Riemann-Roch theorem. Define $\Theta$ by the determinant of their evaluation map. Its kernel is $H^0(C,M)$, so this is precisely the required locus. Exact sequences for enlarging $E$ cancel an invertible determinant factor, making the determinant description of the theta divisor independent of $E$.

This is a genuine irreducible reduced Cartier divisor. Choose $g-1$ points imposing independent conditions on $H^0(C,K_C)$. For their effective Cartier divisor $D^{\prime }$, the Riemann-Roch theorem gives $h^0(\mathcal{O}(D^{\prime }))=h^1(\mathcal{O}(D^{\prime }))=1$. These points form a nonempty open set in the irreducible symmetric product of a curve $C^{(g-1)}$; the Abel map of an algebraic curve is generically one-to-one there, so its image has dimension $g-1$. At such a class $M$, the first-order change of the evaluation determinant is the cup-product functionalUnder Serre duality its dual is multiplication $H^0(M)\otimes H^0(K_C\otimes M^{-1})\to H^0(K_C)$. The product of two nonzero sections on an integral algebraic curve is nonzero. Thus the functional is nonzero and the determinant has multiplicity one generically. Its irreducible support is consequently the reduced theta divisor.

Now let $L={\mathcal{O}}_C(D)$ have degree $g$. The Riemann-Roch theorem gives $h^0(L)\ge 1$, while evaluation at $P$ givesIf $h^0(L)\ge 2$, this kernel is nonzero for every $P$, so the entire image of ${\varphi }_D$ lies in $\Theta$. If $h^0(L)=1$, write $s$ for a nonzero section. The kernel is nonzero exactly at the zeros of $s$, so the image is not contained in $\Theta$. Moreover $h^1(L)=0$, and the family $L(-P)$ is computed by the rank-one evaluation map $H^0(L)\otimes {\mathcal{O}}_C\to L$. Its determinant is $s$ itself. This proves the scheme-theoretic restriction of the theta divisor to an Abel curve:Here the intersection is regarded as a Cartier divisor on $C$. Since $D$ was specified as a divisor class, it means its unique effective representative, with all multiplicities. The map ${\varphi }_D$ is an embedding: distinct points with the same image would give a degree-one map $C\to {\mathbb{P}}^1$; its cotangent map is evaluation of canonical differentials, which is surjective because $K_C$ is base-point-free. To see the latter, $h^0({\mathcal{O}}_C(P))=1$ for genus at least one, and the Riemann-Roch theorem gives $h^0(K_C-P)=g-1$. Thus the pullback also describes the intersection on the embedded curve.

It remains to count sections on $X$. Let $s_{\Theta }$ be the section defining the theta divisor, and take any $s\in H^0(X,{\mathcal{O}}_X(\Theta ))$. The quotient $r=s/s_{\Theta }$ is a rational function. Choose a fixed degree-$2g-1$ line bundle $T$ and the morphismIt is surjective: for every $M\in X$, the degree-$g$ bundle $T\otimes M^{-1}$ has a nonzero section by the Riemann-Roch theorem. On a general coordinate curve, put $D=T-{\sum }_{j\ne i}{P}_j$. The $g-1$ general points impose independent conditions on the $g$-dimensional space $H^0(C,T)$, leaving $h^0(\mathcal{O}(D))=1$. This independence follows inductively by choosing each new point away from the common zero set of the remaining nonzero sections. The restriction already proved identifies ${\mathcal{O}}_X(\Theta )$ on that coordinate curve with ${\mathcal{O}}_C(D)$, whose section space is one-dimensional. Hence $r\circ \beta$ is constant along every general coordinate curve.

A rational function on a product of integral curves which is constant in each factor is constant: on a common dense open set, successively varying one coordinate identifies its value with the value at a fixed general tuple. Therefore $r\circ \beta \in k$, and surjectivity of $\beta$ gives $r\in k$. Every $s$ is a scalar multiple of $s_{\Theta }$, proving

Suppose there were distinct singular points of an algebraic variety $P,Q$. Let $L$ be their joining projective line. The homogeneous restriction $F{|}_L$ has degree $3$. At a singular point of an algebraic variety of a plane algebraic curve, the local defining equation has no constant or linear term, so its restriction to any line through that point vanishes to order at least $2$, unless the restriction is identically zero. Therefore a nonzero $F{|}_L$ would have at least four zeros counted with multiplicity, two at $P$ and two at $Q$, which a degree-$3$ homogeneous polynomial on ${\mathbb{P}}^1$ cannot have.

Hence $F{|}_L=0$. After choosing coordinates with $L=V(z)$, this says that every monomial of $F$ contains $z$, so $z$ divides $F$. That contradicts the irreducible polynomial hypothesis on the cubic. ThusThis elementary version of the Bézout theorem argument works in every characteristic, without assuming that the partial derivatives of the cubic have their characteristic-zero form.

For a smooth projective curve $C$, let $g=h^1(C,{\mathcal{O}}_C)$ be its geometric genus, let $K_C$ be a canonical divisor, and write $\ell (D)=h^0(C,{\mathcal{O}}_C(D))$ for the dimension of a vector space of the Riemann-Roch space. The Riemann-Roch theorem saysfor every divisor on an algebraic curve $D$ on $C$.

Here is the cohomological proof of Riemann-Roch for curves. At a closed point $P$, the local ring is a discrete valuation ring. A local uniformizer shows that increasing the allowed pole order by one gives the short exact sequence of sheavesThe last term is a length-one skyscraper sheaf; its precise identification with $k$ is noncanonical, but its sheaf cohomology has $h^0=1$ and $h^1=0$. All the relevant groups of sheaf cohomology are finite-dimensional because $C$ is a projective variety, and coherent sheaf cohomology above degree one vanishes on a algebraic curve. The long exact sequence in sheaf cohomology therefore givesSince $k$ is an algebraically closed field, every closed point has degree one. Iterating this identity for both positive and negative coefficients of $D$ provesEvery global regular function on a projective variety which is an irreducible variety is constant; hence $h^0(C,{\mathcal{O}}_C)=1$ and $\chi ({\mathcal{O}}_C)=1-g$.

Finally apply Serre duality for a smooth projective curve:Taking dimensions of a vector space and inserting this in the Euler characteristic of a coherent sheaf identity proves the displayed Riemann-Roch theorem. The substantial inputs from sheaf cohomology are finiteness and vanishing for coherent sheaves on a projective curve, and Serre duality; the change of Euler characteristic of a coherent sheaf is proved directly by the point exact sequence. In particular the argument handles negative as well as effective divisors on an algebraic curve.

For the first algebraic curve, a hyperbola, use the line $y=1+tx$ through $(0,1)$. Substitution and cancellation of the known intersection give $x((1-t^2)x-2t)=0$. Thus a rational parametrization of an algebraic curve isThe identity $y^2-x^2=1$ follows immediately. Away from $(0,1)$ its inverse is $t=(y-1)/x$; the exceptional point is recovered at $t=0$. The other point with $x=0$, namely $(0,-1)$, corresponds to $t=\infty$, while $t=\pm 1$ gives the points at infinity on the projective closure. This explains the exceptional parameters rather than discarding them.

For the second algebraic curve, the rational parametrization of an algebraic curvehas inverse $t=y/x$ where $x\ne 0$, and $t=0$ gives the cusp. Indeed, if $x\ne 0$ and $y^2=x^3$, then $(y/x{)}^2=x$ and $(y/x{)}^3=y$. Both curves admit rational parametrizations, although the second has a singular point.

Work in characteristic different from two, as in the number-field application below. Nonsingularity is equivalent to $b(a^2-4b)\ne 0$. The chord through $P=(x,y)$ and $T=(0,0)$ has slope $y/x$. Using $y^2/x^2=x+a+b/x$ in the elliptic-curve addition formula givesThese formulas hold for $P\ne O,T$; addition interchanges $O$ and $T$.

It follows that $\xi =x+a+b/x=y^2/x^2$ and $\eta =y(1-b/x^2)$. The relation isFor a direct verification, observe thatTherefore the two-isogeny formula isThe target elliptic curve is nonsingular because its corresponding coefficient product is $16b(a^2-4b)\ne 0$. The rational map of projective varieties extends over the exceptional points to a morphism of smooth projective algebraic curves; at both $O$ and $T$ its affine coordinates tend to infinity, giving the displayed values. A nonconstant morphism between elliptic curves sending $O$ to $O$ is a group homomorphism, so this is an isogeny of elliptic curves.

One can also see the quotient directly: translation by $T$ leaves $\xi ,\eta$ invariant. The equation $x^2+(a-\xi )x+b=0$ makes the source function field a degree-two extension of the target function field; its nontrivial automorphism is translation by $T$. Equivalently, the degree of an isogeny from its x-coordinate map is two. The kernel of an isogeny is precisely $\{O,T\}$. Thus $\varphi$ is a separable isogeny of degree two.

For a smooth projective algebraic curve, a linear system of divisors comes from a finite-dimensional space $V\subseteq H^0(C,\mathcal{O}(D))$. A basis of sections $s_0,\dots ,s_n$ defines $f(P)=[s_0(P):\cdots :s_n(P)]$, using any local trivialization; a common scalar change does not alter the projective point. A base-point-free linear system makes this defined everywhere, and $f^{\ast }{\mathcal{O}}_{{\mathbb{P}}^n}(1)\cong \mathcal{O}(D)$. Pulling back a hyperplane gives the zero divisor of a linear combination of the sections, an element of the original system.

A system with base divisor $B$ can instead define a map after removing $B$, but its hyperplane pullbacks have class $D-B$. Thus base-point-freeness is needed for the stated original-system property. For a closed immersion, the system must also separate distinct points and tangent directions; equivalently it is a very ample linear system. On a smooth curve the latter tangent condition is separation of the length-two subscheme at each point.

Use $v=|u_r|>0$ for the radial speed and distinguish the central mass $M$ from the Mach number $\mathcal{M}=v/c$. In each smooth nonzero-flow region, mass conservation, the adiabatic equation of state, and the Bernoulli equation giveHere $K>0$ is constant on a smooth isentropic flow branch, $c$ is the adiabatic sound speed, and $\mathcal{E}$ is the Bernoulli function. Either sign of $u_r$ is possible. The positive-radius nondimensionalization in the question requires $\mathcal{E}>0$; flows with $\mathcal{E}\le 0$ do not have these positive $r_s,x,\lambda$ and require another scaling.

Indeed, the supplied adiabatic process equations give $D(p{\rho }^{-3/2})/Dt=0$, so $K$ is spatially constant along a steady nonzero radial fluid flow. The adiabatic sound speed is $c^2=(\partial p/\partial \rho {)}_s$, and the specific enthalpy is $h=c^2/(\gamma -1)=2c^2$. The radial Euler momentum equation, $vdv/dr=-{\rho }^{-1}dp/dr-GM/r^2$, integrates to $\mathcal{E}$ because $dh=dp/\rho$ on this isentropic flow branch.

Put $D=[j(3K/2{)}^2{]}^{2/5}$. Eliminating the mass density from the continuity equation givesWith $y={\mathcal{M}}^{2/5}$, the Bernoulli equation becomesTherefore the physical constants of the spherical polytropic flow with adiabatic exponent three halves are

For $t>0$, $F^{\prime }(t)=4(t^5-1)/t^2$, so $F$ decreases from infinity to its unique minimum $F(1)=5$, then increases to infinity. This completely determines the algebraic curves. For $0<\lambda <1$, there are two disconnected branches at every $x>0$, one subsonic flow with $y<1$ and one supersonic flow with $y>1$. For $\lambda =1$, the line $y=x$ and a decreasing branch meet at $(1,1)$. Their slopes are $+1$ and $-1$, since the leading expansion is $F(1+\eta )=5+10{\eta }^2+O({\eta }^3)$. For $\lambda >1$, solutions exist only where $F(x)\ge 5\lambda$: if $x_{-}<1<x_{+}$ solve $F(x_{\pm })=5\lambda$, there is a forbidden interval $x_{-}<x<x_{+}$. At either endpoint the two branches turn at $y=1$ and have an infinite slope.

A smooth sonic point needs $F^{\prime }(x)=F^{\prime }(y)=0$; hence a regular sonic transition requires $\lambda =1$ and $r=r_s$. The critical speed of a polytropic flow is then $c_s^2=GM/(2r_s)=2\mathcal{E}$, and the allowed mass flux is $j=(2\mathcal{E}{)}^{5/2}{r}_s^2/(3K/2{)}^2$. Here the subscript on $c_s$ denotes evaluation at the sonic point, rather than the isothermal sound speed convention used in the next solution.

The simple transonic branch is $y=x$. Direct substitution givesThus the radial velocity has constant magnitude, while the mass density and pressure scale as $r^{-2}$ and $r^{-3}$. For an outward fluid flow, this is a subsonic flow becoming a supersonic flow as its adiabatic sound speed decreases.

On the other transonic branch, $y$ decreases as $x$ increases. At large $r$, the terms $4/y$ and $x^4$ dominate, giving $y\sim 4x^{-4}$. At small $r$, the terms $y^4$ and $4/x$ dominate, giving $y\sim 4^{1/4}{x}^{-1/4}$. The resulting asymptotic expansions areFor inward fluid flow, these are the nearly static reservoir at infinity and the inner free fall region of Bondi accretion. The specific enthalpy is smaller than $GM/r$ in the inner limit, which explains free fall. Reversing the radial velocity gives a decelerating outflow with the same profiles. The formal small-$r$ limit describes the point-mass model: if the spherical gravitating body has a finite surface, the exterior solution stops there.

A stationary, nonradiative normal shock wave conserves the mass flux and the sum $v^2/2+h$, while the Newtonian gravitational potential is continuous across its thin layer. Hence $\mathcal{E}$ and $r_s$ are unchanged. The entropy production in a perfect-gas shock increases $K=p/{\rho }^{3/2}$, soMore explicitly, the Rankine-Hugoniot condition for $\gamma =3/2$ gives, with upstream $m={\mathcal{M}}_1^2>1$,Thus the shock wave jumps vertically, at the same $x$, from the supersonic flow to a subsonic flow on a larger-$\lambda$ curve. The figure illustrates an outward flow on $y=x$ with a shock wave at $x=1.4$; the downstream subsonic flow has $\lambda \simeq 1.224$ and continues to larger $x$. The shock wave radius is an additional boundary-data choice, rather than being determined by the smooth equations alone.

Cubing and adding the parametrized coordinates yieldsTherefore the implicit algebraic curve isFor $a\ne 0$, every nonorigin real point on this curve has $x\ne 0$ and can be recovered using $t=y/x$; the origin is already included. The equation is invariant under exchanging $x$ and $y$, so its reflection line is $\boxed{y=x}$. In the parametrization that reflection corresponds to replacing $t$ by $1/t$. If $a=0$, the implicit cubic degenerates to the line $x+y=0$, whereas the given parametrization only describes the origin, so this degenerate case must be kept separate.

For a smooth projective geometrically connected algebraic curve $C/k$, this functor assigns to a scheme $T$ the line bundles on $C\times T$ having degree $d$ on each geometric fibre, modulo bundles pulled back from $T$. One generally takes the associated sheaf for faithfully flat covers. A rational point of $C$ permits normalization of every line bundle along that point; normalized bundles have no automorphisms preserving their normalization and satisfy effective descent. The resulting functor is represented by the degree-$d$ component of the Picard scheme of a curve.

For a smooth projective algebraic curve of genus $g$, this smooth proper $g$-dimensional scheme represents degree-$d$ line bundles in families. Charts come from nonspecial degree-$g$ bundles: their first sheaf cohomology vanishes, so the Riemann-Roch theorem gives exactly one independent global section, whose zero Cartier divisor varies algebraically. Twisting by fixed line bundles gives charts covering every degree-$d$ class. For degree zero the tensor product gives the Jacobian variety; other components are its torsors.

A plane curve is a curve contained in a plane. In algebraic geometry, one studies plane curves defined by polynomial equations. General plane curves, such as an Archimedean spiral, need not be algebraic curves.

A projective curve is a projective algebraic curve. A projective curve is a one-dimensional projective variety.

Ramification is branching in a geometric or algebraic map, detected by multiplicity greater than one. For number fields, ramification of a prime means that some prime ideal occurs with exponent greater than one in the extended prime ideal. For a map of smooth algebraic curves, a local parameter pulls back to a parameter with order greater than one at a ramification point.

A rational parametrization expresses the coordinates of an algebraic curve as rational functions of a parameter, with a rational inverse on a dense open subset. It identifies the function field with $K(t)$ and the smooth projective normalization of an algebraic curve with the projective line. A nonconstant parametrizing map without an inverse is a weaker notion; even this cannot exist for an elliptic curve in characteristic zero.

For a smooth projective algebraic curve $C$, the normal bundle of its diagonal in $C\times C$ is $TC$: on the diagonal the quotient of $TC\oplus TC$ by the diagonal subbundle is identified with $TC$ by taking the difference. The self-intersection formula therefore gives ${\Delta }_C^2=\mathrm{deg}TC=2-2g(C)$.

A smooth algebraic curve is a one-dimensional algebraic variety whose local ring at every point is regular.

This is the nonsingular subclass of algebraic curves.

The scheme $C^{(d)}$ parametrizes effective degree-$d$ Cartier divisors on a smooth algebraic curve. Near a Cartier divisor $\sum m_i{P}_i$, its completed local ring is a power-series ring in the elementary symmetric coordinates of $m_i$ local parameters at each $P_i$. It is therefore smooth of dimension $d$, even when the characteristic of a field divides $d!$. For projective $C$ it is projective.